Solution Manual For 21st Century Astronomy Seventh Edition Stacy Palen George Blumenthal

Contents Preface

vii

Part 1

Instructor’s Manual

Chapter 1

Thinking Like an Astronomer

3

Chapter 2 ​Patterns in the Sky—­Motions of Earth and the Moon 10 Chapter 3 ​Motion of Astronomical Bodies Chapter 4 ​Gravity and Orbits Chapter 5 ​Light

19

28

36

Chapter 6 ​The Tools of the Astronomer

43

Chapter 7 ​The Formation of Planetary Systems 51 Chapter 8 ​The Terrestrial Planets and Earth’s Moon 60 Chapter 9 ​Atmospheres of the Terrestrial Planets 67 Chapter 10

​Worlds of Gas and Liquid—­The G ­ iant Planets 75

Chapter 11

​Planetary Moons and Rings 83

Chapter 12

D ​ warf Planets and Small Solar System Bodies 90

Chapter 13

​Taking the Mea­sure of Stars

Chapter 14

O ​ ur Star—­The Sun

Chapter 15

​The Interstellar Medium and Star Formation 113

Chapter 16

E​ volution of Low-­Mass Stars 120

Chapter 17

​Evolution of High-­Mass Stars 128

Chapter 18

R ​ elativity and Black Holes 136

Chapter 19

G ​ alaxies 143

Chapter 20

​The Milky Way—­A Normal Spiral Galaxy 151

98

105

v

vi ◆ Contents

Chapter 21

​The Expanding Universe 158

Chapter 22

​Cosmology

Chapter 23

L​ arge-­Scale Structure in the Universe

Chapter 24

​Life 178

Part 2

Answers to Starry Night Workbook Exercises

Exercise 1

The Celestial Sphere

Exercise 2

Earth’s Rotation Period 188

Exercise 3

Motion of the Sun along the Ecliptic

Exercise 4

Motion of the Moon

Exercise 5

Earth and Moon Phases

Exercise 6

Sunrise on Mars 192

Exercise 7

Precession 194

Exercise 8

Kepler’s Laws 194

Exercise 9

Flying to Mars

Exercise 10

The Moons of Jupiter 195

Exercise 11

The Rings of Saturn 196

Exercise 12

Pluto and Kuiper ­Belt Objects

Exercise 13

Asteroids

Exercise 14

The Magnitude Scale and Distances

Exercise 15

Stars and the H-­R Diagram 199

Exercise 16

Nebulae: The Birth and Death of Stars 200

Exercise 17

Pulsars and Supernova Remnants

Exercise 18

Galaxy Classification 201

Exercise 19

Quasars and Active Galaxies 201

Exercise 20

Views of the Milky

202

Exercise 21

Globular Clusters

203

Exercise 22

The Neighborhood of the Sun

Exercise 23

Beyond the Milky Way

165 171

188 189

189 190

195

197

197

Credits 205

204

197

200

203

Preface For each chapter of the textbook, you w ­ ill find a corresponding chapter in the Instructor’s Manual that contains all or most of the following sections:

INSTRUCTOR NOTES • This section provides a brief overview of the chapter and a list of major topics discussed. It often includes common misconceptions to address and recommendations for additional resources.

INTERACTIVE SIMULATIONS • Textbook author Stacy Palen has created seven Interactive Simulations that pair with selected Exploration activities. This section briefly describes each Interactive Simulation associated with the chapter and notes the corresponding section of the textbook.

TEACHING READING ASTRONOMY NEWS

DISCUSSION POINTS

• This section contains the worked solutions and discussion of the Reading Astronomy News questions.

• This section suggests impor­tant discussion topics and activities. The chapter Learning Goal associated with each item is noted.

CHECK YOUR UNDERSTANDING SOLUTIONS

PRO­CESS OF SCIENCE

• This section provides answers and supporting information for all of the in-­chapter Check Your Understanding questions.

• This section provides a brief overview of the chapter’s Pro­cess of Science figure. We also have questions in Smartwork5 that relate to the figure.

ASTROTOUR ANIMATIONS • The AstroTour animations are narrated, conceptual overviews with a consistent structure of Introduction— ­E xplanation—­Conclusion. This section of the Instructor’s Manual briefly describes each AstroTour animation associated with the chapter and notes the corresponding section of the textbook.

ASTRONOMY IN ACTION VIDEOS • The Astronomy in Action Videos are a series of mini-­ lectures and demos done by textbook author Stacy Palen. This section of the Instructor’s Manual briefly describes each Astronomy in Action Video associated with the chapter and notes the corresponding section of the textbook as well as the length of the video.

END-­OF-­CHAPTER SOLUTIONS • This section provides worked solutions to all of the end-­of-­chapter questions and prob­lems (Test Your Understanding, Thinking about the Concepts, and Applying the Concepts).

EXPLORATION • This section briefly describes the Exploration activity and provides worked solutions to each question.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES • This section introduces activities from the Learning Astronomy by ­Doing Astronomy workbook that are relevant to the chapter. The textbook reference of the associated topic is noted. vii

viii ◆ Preface For adopters of The Norton Starry Night Workbook, the answers to the exercises are included at the end of the manual. ­ ill find the information in this manual We hope that you w useful. We welcome your comments, questions, and suggestions (contact your local Norton representative: http://­books​.­w wnorton​.­com​/­books​/­fi nd​-­your​-­rep​/­).

Fi­nally, we would like to thank Ricardo Covarrubias of Milikin University and Jay Dunn of Georgia State University, Perimeter College, whose careful review improved the accuracy and usefulness of this manual.

Additional resources: Norton Interactive Instructor’s Guide (IIG) iig​.­w wnorton​.­com​/­astro6 This new and searchable online resource is designed to help instructors prepare for lecture in real time. All of the content in this Instructor’s Manual, and more, is located on the IIG. In addition to this manual’s content, you w ­ ill find: the Test Bank, AstroTour animations, Astronomy in

Action videos, Interactive Simulations, Lecture Power-­ Point slides, all of the textbook’s art, photos, and ­t ables, and Learning Management System Coursepacks (available in Blackboard, Canvas, Desire2Learn, and Moodle formats).

Smartwork5 Online Activities and Assessment digital​.­w wnorton​.­com​/­astro6 More than 2,00 questions support 21st ­Century Astronomy, Sixth Edition—­a ll with answer-­specific feedback, hints, and ebook links. Norton offers pre-­made assignments for each chapter of the text to make it easy to get started, but Smartwork5 is also fully customizable.

Questions include ranking, labeling, and sorting exercises based on book and NASA art, selected end-­of-­chapter questions, versions of the Explorations (based on AstroTours and new Simulations), and questions that accompany the Reading Astronomy News feature in each textbook chapter. Astronomy in Action video questions focus on overcoming common misconceptions, while Pro­cess of Science guided inquiry assignments take students through the steps of a discovery and ask them to participate in the decision-­making pro­cess that led to the discovery.

PA RT I :

Instructor’s Manual

CHAPTER 1

Thinking Like an Astronomer

INSTRUCTOR NOTES Chapter 1 is an introduction to the mea­sures and methods of astronomy. Major topics include: • our cosmic address; that is, the hierarchy of structures from the Solar System to the Laniakea Supercluster.

• relevant and relative distance scales, including the light-­ year, and relating t­ hose scales to an equivalent time.

• the scientific method and relevant vocabulary, with an emphasis on distinguishing a theory from an idea.

• mathe­matics as a way to recognize patterns, how to read both linear and logarithmic graphs, and unit conversion and scientific notation.

Students enroll for introductory astronomy courses for many reasons, but the most common one is to fulfill a general education requirement. We have found that in a large lecture hall (say 200 students or more) of a major university, the spread of educational backgrounds can be large.

We teach students at ­every level, from incoming freshmen with no major declared to graduating se­niors who have majored in a science, technology, engineering, and mathe­ matics (STEM) field. For an open-­enrollment, two-­or four-­year college, we have students who are earning college credit while still in high school, students who are also working full time, and some who are raising a f­ amily along with attending school. Returning students may not have had any exposure to basic math for a de­cade or more.

The goal of this first chapter is to ease all t­ hese students into their study of astronomy. It captures their interest by covering our cosmic address and translating huge distances into terrestrial time examples for a better grasp of large numbers. We may be quite comfortable discussing wavelengths in nanometers, particle densities in atoms per cubic centimeter, masses in 1030 kilograms, and distances in megaparsecs, but students are not. We find it useful to conduct exercises with Figures 1.1 and 1.3 or show a version of the “Powers of Ten” montages available online to provide students with some visual context for the ranges of size, mass, speed, and time that are discussed in this course.

Too often we hear the phrase “it’s just a theory” as a way of dismissing facts that are personally unacceptable. ­Here

the concepts of questioning, predicting, testing, disproving, and more about the scientific method are presented in a way that is relatable to everyday occurrences. Students are given practice with scientific notation and the graphing of data. The mathe­matics is presented as the language of, as well as a valuable tool for, science. Much of the quantitative prob­lem solving in this course can be achieved through proportional reasoning, so in addition to asking questions about scientific notation and unit conversion, we can also introduce some basic ideas of how area or volume changes with size.

For many of our students, this is the last formal science course they ­w ill ever take. We have included learning outcomes that help them learn the pro­cess of science, gain scientific literacy, and understand the difference between science and pseudoscience. The seeds of ­these outcomes are sown in this first chapter, not only through discussion of the scientific method, but also through the vari­ous logical fallacies presented in the Exploration. Although science is ideally in­de­pen­dent of culture or creed, it has often collided with religious or other strongly held beliefs. Therefore, ­because science is a ­human activity carried out by individuals who may hold nonscientific beliefs, we emphasize that we must construct safeguards within our work to counteract any personal bias that might taint the results.

The emphasis is on how science is all about searching for objective truths that lead to conclusions that cannot be falsified, at least at this time.

DISCUSSION POINTS • Have students look at the sketches shown in Figures 1.1 and 1.3. Ask them if they are familiar with any of the shapes and structures shown. Where have they encountered them before? (LG 1) • Have students think about the times given in Figure 1.3.

Discuss the distances and times between our planet and nearby stars, and relate that to the likelihood that we w ­ ill communicate with extraterrestrials in our lifetime (remind students that we have only been broadcasting and listening for less than 100 years).

(LG 1) 3

4 ◆ CHAPTER 1 Thinking Like an Astronomer • Astronomers need to keep collecting data from the objects in the universe to find unexpected trends and to test hypotheses. Discuss how this pro­cess has analogies in students’ own experiences. Have they ever had to collect data to learn something or to explore the unknown?

Share in pairs and then with the class. (LG 2) • Students could formulate a testable hypothesis about how the number of hours spent studying in a course ­w ill affect their grade in that course. Would the outcome depend on w ­ hether the course was in their chosen major? What other variables might affect the outcome of their efforts in a course? (LG 2) • Ask students if they are familiar with any scientific equations. Discuss differences and similarities between a well-­k nown scientific equation (for example, F = ma or E = mc2) and a world-­renowned work of art. What pro­cesses went into creating each of t­ hese?

How is each example used? (LG 3) • How can we describe our astronomical origins? Start with the hydrogen atom minutes ­after the Big Bang and have students build a story of the “life” of that hydrogen atom from its beginning to its presence in their DNA. Set up a think-­pair activity, and then randomly select a pair or pairs to share their stories. (LG 4) • The chapter-­opening image is the one taken of Earth by astronaut William (Bill) Anders in 1968 as part of the Apollo 8 mission, as the orbiter came around the Moon and Earth appeared to “rise.” What do students already know about the image? Do they think it was taken from a lander or an astronaut on the surface of the Moon? The question asked is, “Why is this considered one of the most famous photos ever taken?” The answer is that it was the first image taken of Earth from space by ­humans on a mission to another world. NASA put together a short 45th anniversary commemorative video that combined a­ ctual flight recordings, photo mosaics, and elevation data from the Lunar Reconnaissance Orbiter. The video is narrated by Andrew Chaikin, and sketches of the orbiter containing Anders, Jim Lovell, and Frank Borman are included for a more complete perspective of how the Earthrise picture was obtained. The video can be found at https://­youtu​.­be​ /­d E​-­v OscpiNc.

PROCESS OF SCIENCE The Pro­cess of Science figure for this chapter introduces the basics of the scientific method. The emphasis is on how the pro­cess w ­ ill repeat in­def­initely as scientists continue to test the theory. Subsequent chapters in this text w ­ ill emphasize a­ ctual examples of how the scientific method “works.”

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. ­Because light has a finite speed, it takes time for the light to get to us from distant galaxies. We assume that galaxies in the universe all formed at roughly the same time.

For t­hose galaxies that are the farthest away, light left them when they w ­ ere very young, just forming. Th ­ ese galaxies w ­ ill look the youn­gest to us. Galaxies that are closer ­will look older. Telescopes are like time machines.

2. If we are looking extremely far away, such as when the galaxies w ­ ere only a few billion years old (a look-­back time of around 12 billion years), and we see many more small galaxies than large galaxies, then that suggests that galaxies started small.

3. The observable universe is that part where light has had time to reach us. We cannot see the entire universe ­because ­there ­hasn’t been enough time for light beyond ­there to reach us.

4. Since the question asks for learner statements, answers ­w ill vary.

5. Students are asked to do a search on galaxies. Answers will vary.

CHECK YOUR UNDERSTANDING SOLUTIONS 1.1 (d) radius of Earth (a) light-­m inute (e) distance from Earth to the Sun (c) light-­hour (f) radius of the Solar System (b) light-­year. Use Figure 1.3.

1.2 (b) Theories must be testable, and a theory is valid up ­until a test fails.

1.3 (c) Patterns and order are indicative of a physical pro­ cess at play.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (f, e, c, b, a, h, g, d) is the correct order from smallest to largest size.

CHAPTER 1 Thinking Like an Astronomer

2. (a) See Figure 1.3.

3. (b) We can connect facts through an under­lying idea.

Note that (a) one must accumulate facts to consider how they are related, and (c, d) that science makes predictions based on ­t hese relationships, but “understanding” is the development of t­ hese relationships.

4. (a) The universe is understood to be homogeneous and isotropic on its largest scales.

5. (d) The Sun is the center of our Solar System, just one of the billions of stars in our galaxy, and one of the billions of galaxies in the universe.

6. (a) It is the distance that light travels in one year.

7. (c) Occam’s razor suggests that nature relies on the simplest (or most straightforward) pro­cesses.

8. (d) Distance units in terms of light speed are very con­ve­nient but sometimes odd to think about at first ­because we seem to be using time to refer to distance.

This prob­lem shows us two ways of considering the meaning of light distance.

9. (d) A reading of Figure 1.7b indicates that the answer cannot be obtained from a linear plot. A reading of the log-­l inear plot in Figure 1.7c, however, shows that at time step 4, the number of viruses is more than 10 times what it was at the start.

10. (d) As the answer indicates, science relates only to the natu­ral world.

11. (c) Our understanding must be tested, and at any time, a test could show that it is wrong. Note that this is not an issue of being worthless or incomplete, but merely reflects the fact that we are constantly testing our theories and hypotheses.

12. (c) Light travels a light-­year in one year, so a star that is 10 light-­years away emitted its light 10 years ago for us to see it ­today.

13. (d) Carbon is made inside stars.

14. (b) Except for hydrogen and helium (and a tiny bit of lithium), all of the ele­ments found on Earth ­were produced in stars. Note that the beryllium produced in the Big Bang was unstable and decayed long before Earth formed.

15. (b, d, a, c, e) The material for the Sun had to come before the Sun could be formed. Gas came first (b), the gas formed stars to make heavier ele­ments (d), the stars blew up to spread t­ hose ele­ments around (a), and then the gas had to collect (c) before it could form the Sun and the planets (e).

Thinking about the Concepts 16. Tau Ceti e, Tau Ceti, Milky Way Galaxy, Local Group, Virgo Supercluster, Laniakea Supercluster, universe 17. The cosmological princi­ple essentially states that observers in the universe should find that the natu­ral

◆ 5

laws governing their local region are representative of the natu­ral laws governing the universe as a w ­ hole.

Consequently, they should derive the same natu­ral laws that an observer on Earth derives.

18. 8.3 minutes (see Figure 1.3).

19. Andromeda is about 2.5 million light-­years away (see Figure 1.3), so it would take 2.5 million years for the light from the exploding star to reach Earth.

20. Answers w ­ ill vary. An example is general relativity superseding Newtonian mechanics, which began at the first step of the Pro­cess of Science for this chapter when Einstein reinvisioned gravity and spacetime.

Another example occurred in the 1920s when it was predicted that the Andromeda galaxy was just in a distant part of the Milky Way, perhaps 300,000 light years distant. Edwin Hubble employed the “observe” step and found that the result did not support the hypothesis: the Andromeda galaxy was, in fact, millions of light years away.

21. This does not qualify as a scientific theory b­ ecause it is not falsifiable. Although it is pos­si­ble that we may someday stumble upon irrefutable evidence that aliens visited Earth in the remote past, the absence of evidence ­today cannot be used to refute the hypothesis. ­ ill simply argue In fact, proponents of the theory w that we just have not found any evidence yet. Evidence that could support the theory would include finding advanced technology in ancient archaeological sites or buried in old geological layers. The only tests we can think of to refute the hypothesis are ­either to demonstrate that ­every piece of technology and archaeological monument could have been reasonably constructed with ­human knowledge of the time or e­ lse invent a time machine and return to the most likely times for aliens to have visited Earth. ­Because option 2 is utterly implausible and option 1 does not preclude alien visitations, it is impossible to falsify the hypothesis.

22. Falsifiable means that something can be tested and shown to be false or incorrect through an experiment or observation. Some examples of ideas that cannot be falsified are religious beliefs, po­liti­cal views, emotional statements, and opinions. Students may have a wide variety of ­these and other ideas, but all sacred cows are usually considered to be not falsifiable by the ­people holding t­ hose beliefs. Falsifiable ideas include cause and effect and logic.

23. A theory is generally understood to mean an idea, regardless of any proof, evidence, or way to test it. A scientific theory is an explanation for an occurrence in nature that must be based on observations and data and make testable predictions.

6 ◆ CHAPTER 1 Thinking Like an Astronomer 24. A hypothesis is an idea that might explain some physical occurrence. A scientific theory is a hypothesis that has been rigorously tested.

25. (a) Yes, this is falsifiable. (b) Find a sample of a few hundred ­children born during dif­fer­ent moon phases, who come from similar backgrounds and go to similar schools and follow their pro­gress for a number of years.

26. In 1945, our distance-­measuring methods w ­ ere not correctly calibrated and, as a result, our calculation of the distance to Andromeda was wrong. As we improved that calibration, we found dif­fer­ent and more reliable mea­sure­ments of its distance. In science, statements of “fact” reflect our current best understanding of the natu­ral universe. A scientific “fact” does not imply that science has determined absolute truth; rather, it is simply a statement that this is the best understanding of nature that our current knowledge and technology supports. Over time, all scientific “facts” evolve as our knowledge base and technology grow.

27. Answers w ­ ill vary. Depending on the generality of the horoscopes, students may provide a wide array of answers for this question. For general statements, students might find that several, if not all, of the horoscopes on a given day could describe their experience. For a very specific horoscope, we expect that it should match approximately 1/12th of the students, regardless of their astrological sign. In any event, if astrology accurately reflected some natu­ral truth, we would expect nearly every­one to find one and only one horoscope each day that describes his or her experience, that the horoscope would match the person’s astrological sign, and that the daily horoscope would be accurate for each person for the entire week of rec­ord keeping. Students should perform this experiment and be honest with themselves about the results. Students need to be aware of any bias by t­ hose who have prior beliefs in horoscopes and who ­will likely not provide objective data.

28. Taken at face value, this is a ridicu­lous statement, but ­there are several items to consider critically before we apply a label of “not reputable.” First, was this statement a sound bite taken out of context? Did the scientist simply misspeak when he or she might have been trying to say that we have not yet found extraterrestrial life? If, in fact, the statement can be taken at face value, then the credibility of the scientist might be called into question ­because he or she has forgotten that absolute truth is not falsifiable (and therefore not scientific) by definition.

29. Some scientific fields rely heavi­ly on math, but not all.

Following the scientific method is the impor­tant part, not use of mathe­matics.

30. Only hydrogen and helium (with perhaps a trace amount of lithium) ­were created in the Big Bang.

Heavier ele­ments such as carbon, oxygen, nitrogen, sodium, silicon, iron, and more are manufactured in the interiors of massive stars. At least one generation (and more likely several generations) of stars must die in massive supernova explosions to make heavy ele­ments available to construct planets and serve as the building blocks for life. Therefore, ­because all the heavy ele­ments in our bodies w ­ ere originally manufactured in stars, it is fair to claim that we are truly made of stardust.

Applying the Concepts 31. Setup: To convert to scientific notation, remember that when the number is greater than one count all digits to the right of the first non-­zero digit, and when the number is less than one count the number of digits between the decimal point and first non-­zero digit.

Solve: (a) 7 × 109 (b) 3.46 × 10 −3 (c) 1.238 × 103 Review: A good way to check is to use a scientific calculator, where “times 10 to the” is usually the “EE” key.

32. Setup: To convert scientific to standard notation, move the decimal point the number of digits indicated in the exponent, to the right if the number is positive, and left if negative.

Solve: (a) 534,000,000 (b) 4,100 (c) 0.0000624 Review: Again, you can test this by using your calculator.

33. Setup: Distance is given in terms of speed and time by d = vt, where v is speed and t is time. If speed is in km/h, then use time in hours, for which we may have to convert. Remember that ­there are 60 minutes in an hour. km × 1h = 35 km Solve: (a) d = vt = 35 h km 1 (b) d = vt = 35 × h = 17.5 km h 2 km 1 (c) d = vt = 35 × h = 0.58 km h 60 Review: A good sanity check is to make sure the distance traveled becomes smaller if the time traveled decreases.

34. Setup: We are given the relationship that surface area, A, is proportional to radius, r, squared, or A ∝ r 2 .

To work a proportional-­reasoning prob­lem, insert the ­factor by which one variable changes into the formula to see how the result changes.

Solve: (a) If r doubles, then r → 2r, or r changes by a f actor of 2. Putting this in our formula shows ­ A ∝ 22 = 4, or that the area changes by a f­ actor of 4. (b) If r ­triples, then it changes by a ­factor of 3, or A ∝ 32 = 9.

CHAPTER 1 Thinking Like an Astronomer

1 (c) If r is halved, then r → r , or r changes by a ­factor 2 2 1 1 1 of ; therefore, A ∝ ⎛ ⎞ = . (d) If r is divided by ⎝ ⎠ 2 4 2 2 1 1 1 3, then r → r and A ∝ ⎛ ⎞ = . ⎝ 3⎠ 3 9 Review: Note first that the change in area is much larger than the change in radius, which reflects the dependence on radius squared. Note also how easy it is to do proportional reasoning rather than using the full surface-­area formula (A = 4πr 2), when all we need to know is how much the result changes.

35. Setup: In this prob­lem, we convert among distance, rate, and time with d = vt, or solving for time, t = d/v.

The prob­lem is straightforward ­because the units of distance are already the same. d 384,000 km There are Solve: t = = = 480 h. ­ v 800 km/h 24 hours in a day, so this would take 1day 480 h × = 20 days, or about two-­thirds of a 24 h month (assuming the typical month is 30 days).

Review: A typical flight from New York to London takes about 7 hours and covers a distance of about 5,600 km. The moon is about 69 times farther away, so it would take about 69 × 7 h = 483 hours to reach the Moon using ­these estimates. This is basically the same amount of time as we found by exactly solving the prob­lem.

36. Setup: In this prob­lem, we ­w ill convert between distance, rate, and time with d = vt or, solving for speed, v = d/t. The prob­lem is straightforward ­because the units of distance are already the same. d 384,000 km 1day Solve: v = = × = 5,333 km/h. t 3 days 24 h 5,333 km/h This is about = 6.7 times faster than a 800 km/h jet plane.

Review: Using the result from prob­lem 35, we have to 20 days travel = 6.7 times faster than a jet plane, 3 days which agrees with our solution.

37. Setup: We are given the prob­lem in relative units, so we d­ on’t need to use our speed equation or use the ­actual speed of light. Instead, we ­w ill use ratios.

Solve: (a) If light takes 8.3 minutes to reach Earth, then it takes 8.3 min × 2 = 16.6 minutes to go twice as far. Neptune is 30 times farther from than the Sun than Earth is. When Neptune and the Earth are on the same side of the Sun (Neptune is in opposition), Neptune is 29 AUs from Earth. This means that light

◆ 7

­ ill take 8.3 min/AU × 29 AU = 240.7 minutes, or w 240.7/60 = 4.0 hours. (b) This means that sharing just one sentence ­w ill take about 8 hours, so it could take a few days just to say hello, talk about the weather, and think of something ­else to say. This is the shortest pos­si­ble time happening when Neptune is at its closest approach to Earth.

Review: If you watch 2001: A Space Odyssey, you ­w ill note that the televised interview between Earth and David Bowman had to be conducted over many hours and then edited for time delays. This was factually correct. ­Because Neptune is about six times farther away than Jupiter, it stands to reason that it would take light and communication a lot longer still.

38. Setup: Light travels at approximately 3 × 105 km/s. d To find the travel time, use d = vt or t = . v d 5.6 × 10 7 km Solve: t = = = 187 s v 3 × 105 km/s 4 × 108 km d = 1330 s.

Likewise, using t = = v 3 × 105 km/s Review: Light takes about 8.3 min to travel from the Sun to Earth, or about 498 s. If, as an estimate, Mars’s distance from Earth ranges from 0.52 AU to 2.52 AU, then the light travel time ­w ill range from 260 s to 1255 s. Given the eccentricity of the orbit of Mars, our numbers are consistent with t­ hese approximations.

39. Setup: We are given the relationship surface area A ∝ r2 . To work a proportional-­reasoning prob­lem, insert the f­actor by which one variable changes into the formula to see how the result changes.

Solve: The Moon’s radius is about one-­fourth that of the surface of Earth; therefore, its surface area is 2 1 1 A ∝⎛ ⎞ = the area of Earth’s surface. ⎝ 4 ⎠ 16 Review: We saw this same be­hav­ior in prob­lem 34.

40. Setup: We w ­ ill use the equation d = vt, where the distance is 3.6 × 10 4 km, one way, and light travels at v = c = 3 × 105 km/s. d 2 × (3.6 × 10 4 km) Solve: t = = = 0.24 s, or about 3 × 105 km/s c 1 of a second 4 Review: If we are receiving information by an Internet satellite on a regular basis, we almost never notice a lag so the time has to be short, on the order of what we found (much less than 1 second).

41. Setup: Let the horizontal axis be time and vertical be population. If we choose to plot a graph in linear space, then a constant population w ­ ill be a horizon-

8 ◆ CHAPTER 1 Thinking Like an Astronomer tal line followed by a sharp, curved increase in the population, resembling Figure 1.7b. An exponential growth ­w ill look similar to the log-­l inear graph of Figure 1.7c. ­A fter a 50-­year time step, an almost vertical line down w ­ ill represent a crash, back down to the original population number. Two of ­these would be represented by a ramp up to a 50-­year time followed by a vertical line down, w ­ hether a linear or long-­l inear graph is used. ­ ill vary. The question gives an examSolve: Answers w ple in which the baseline population is 1 unit.

Review: Note the growth starts out very slowly, jumps up very rapidly, and takes a nosedive down.

42. Setup: We need our assumptions of speed. A car goes 60 miles per hour, on average, if we include filling up with gas, eating, and rest­room breaks. We also need to relate distance, rate, and time with the formula distance equals rate times time, or d = vt.

Solve: Solving for time, t = d/v, so by car, t = 2,444 miles/60 mph = 41 “car” hours. B ­ ecause ­there are 24 hours in a day, the car takes 41/24 = 1.7 “car” days. Note ­t hese assume you travel around the clock, which we do not usually do ­u nless working with a team.

Review: If you drive “almost” nonstop, you can go from NY to LA in 3 days. This is consistent with our value, ­because that assumed no stops at all.

43. Setup: For ­water to freeze, it has to cool down to 0°C; then the liquid has to become solid.

Solve: (a) This theory makes no sense to us ­because hot ­water ­w ill have to cool down to the starting temperature of the cold ­water before it starts to freeze. That takes time that was not needed by the w ­ ater that started out cold. (b) Yes, this is easily testable. Simply try it in a freezer. (c) We tried it, and it took about five times longer for the hot ­water to freeze, confirming our hypothesis.

Review: ­Going back to our original physical reasoning, we see that this theory could be easily refuted without experimentation. Sometimes refuting a theory is not as straightforward, however, and experiments must be performed.

44. Setup: On the surface, it seems that the two pizzas cost the same number of dollars per inch; but remember that each pizza is a thin cylinder, so we eat the volume, not the dia­meter.

Solve: If both pizzas have the same thickness, then we only need to worry about area, A = πr2. Since the area goes as radius squared, the area of the 18-­inch pizza is four times larger than the area of the 9-­inch one. Since the 18-­inch pizza costs only twice as much, it is more eco­nom­ical to buy the larger pizza.

Review: Often, larger items cost less per unit than smaller ones ­because almost the same amount of ­labor went into making them, and ­labor is generally the highest part of the cost. This is why you should always check the unit price when buying t­ hings.

45. Setup: For part (a), use the formula given, = 2πr. For part (b), we need to relate distance, speed, and time by d = vt, where we ­w ill solve for time. We use the formula again for part (c), where we must remember ­there are 24 hours in one day.

Solve: (a) C = 2πr = 2π × (1.5 × 108 km) = 9.4 × 108 km (b)

v=

d 9.4 × 108 km = ≈1.075 × 105 km/h, t 8,766 h

or

about 107,500 km/h (c) ­Because d = vt, and b­ ecause ­there are 24 hours in one day, Earth moves about 2,580,000 km per day.

Review: It is amazing that we are hurtling around the Sun at more than 100,000 km/h and do not even realize it. Why? It’s ­because every­thing e­ lse (planets, the Sun, other stars) is so far away that we have no reference point to enable us to observe this breakneck speed.

Using the Web 46. Europa, Eris, Pluto, and Triton are about the size of the United States. Venus is about the size of Earth. Lots of stars are larger than an astronomical unit, but Rigel, Gacrux, and Alnitak are about that size. Voyager 1 is 17 billion km, 0.002 light-­years, or 0.73 light-­days away.

The Stingray and Cat’s Eye Nebulae and Gomez’s Hamburger are about the same size as the distance from the Sun to nearest star. The Milky Way is 120,000 ly / 2 ly = 60,000 times larger than the Solar System, which includes the Oort Cloud. The Local Group is 10 million ly across while the Milky Way is 120,000 ly in dia­meter, a ratio of ~80. The observable universe is 93 billion ly across; ratio of the universe to the Local Group is 9,300.

47.(a) Radio broadcast programs have traveled 70 ly. The Sun is not particularly luminous once we get beyond the nearby red dwarf stars. Answers w ­ ill vary about the video’s effectiveness. (b) Answers w ­ ill vary concerning the “Powers of 10” circles in explaining the size and scale of the universe.

48. Students go to the Astronomy Picture of the Day and choose an archived picture or video to describe.

Answers w ­ ill vary.

49. Involves investigating news articles about astronomy or space. Answers w ­ ill vary.

50. Involves ­going to a blog about astronomy or space.

Answers w ­ ill vary.

CHAPTER 1 Thinking Like an Astronomer

EXPLORATION This exploration asks students to consider the logic ­behind a variety of statements, many of which they had not considered before. Besides the examples given in the exploration, ask students to come up with one of their own examples that can be classified as one of the logical fallacies included ­here.

EXPLORATION SOLUTIONS 1. This is an example of post hoc ergo propter hoc, in which we assume that the chain mail caused the car accident.

2. This is a slippery slope, ­because we are assuming that our per­for­mance on the first event must influence the next.

3. This is a biased sample, or the use of small-­number statistics, ­because we assume that our small circle of friends represents every­one.

4. This is an appeal to belief, where we argue that ­because most ­people believe it, it must be true.

5. By attacking the professor rather than the theory, we are committing an ad hominem fallacy.

6. This is an example of begging the question (a bit of a syllogism, too) in which the proof of the assertion comes from another person’s assertions.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES The Learning Astronomy by D ­ oing Astronomy workbook activities that are relevant to Chapter 1 are introduced here. For more information, please see the Learning ­ Astronomy by ­Doing Astronomy workbook, the Instructor’s Manual for the workbook, and the Power­Point clicker question slides associated with the workbook. Our goal is to have complete coverage across all topics in an introductory astronomy course.

ACTIVITY 1: MATHEMATICAL AND SCIENTIFIC METHODS This activity reviews the mathe­matics that students may encounter in this course. This activity helps with tools

◆ 9

such as working with logarithms, the small-­angle formula, scientific notation, or scaling exercises, like ­those used to find the scale of a map; laboratory techniques concerning mea­sure­ments; the uncertainties in ­those mea­sure­ments; and statistical analy­sis. In the first six sections, students review specific mathematical topics and laboratory techniques. ­These sections include explanations and practice prob­lems. In the last section, students pull multiple concepts together to analyze images of three galaxies. Activity 1 covers the majority of math concepts presented in the workbook. Each individual activity has a set of preactivity questions that tutor students on the math that is included. Specifically, students ­w ill: • demonstrate knowledge of the essentials of mathe­ matics through practice and review of: • scientific notation and powers of 10; • algebra; • logarithms; • the small-­angle formula; • the use of scale ­factors and scaling; • statistics and uncertainties in mea­sure­ments.

• describe the pro­cess of science and the scientific approach as personally experienced.

ACTIVITY 2: ASTRONOMICAL MEA­SURE­MENTS: EXAMPLES FROM ASTRONOMICAL RESEARCH In this activity students explore the relationship among apparent brightness, luminosity, and distance and learn to manipulate more advanced equations used in astronomy.

Specifically, students ­w ill: • apply the small-­angle formula.

• distinguish between apparent magnitude and absolute magnitude and relate them correctly to the concepts of apparent brightness and luminosity.

• relate the ratio of distances to the brightness ratio for stars of equal luminosity.

• solve for the distance to a star using the parallax a­ ngle.

• find absolute magnitude from apparent magnitude and distance.

• demonstrate proficiency in manipulating more advanced equations used in astronomical research.

CHAPTER 2

Patterns in the Sky—­Motions of Earth and the Moon INSTRUCTOR NOTES Chapter 2 covers the c­ auses and effects of the apparent motions of the stars, the Sun, and the Moon in our sky.

Major topics include: • the celestial sphere.

• the daily and yearly paths and motions of stars in our sky.

• Earth’s axial tilt as the cause of the seasons.

• Moon phases.

• solar and lunar eclipses.

Many students come into our classes believing that Earth’s shadow ­causes Moon phases and that summer happens ­because Earth is closer to the Sun. ­There is a video from the late 1980s called A Private Universe that shows schoolchildren trying to learn the real c­ auses of the seasons and the phases of the Moon. ­A fter a full lesson, most of the students have incorporated ­these original misconceptions into the ­actual reasons to make new, but still incorrect, explanations. The video then takes us to Harvard University, where it shows graduating se­niors and faculty being asked the same questions, and only a physics professor answers both correctly. The point is not to humiliate any of the students, but rather to highlight that: (1) it is very hard to unravel a misconception and replace it with a correct model; and (2) a very large number of ­people believe in incorrect explanations of the regular rhythms of our world. ­There is, of course, much more in this chapter than just the ­causes of the Moon phases and the seasons. During ­these lessons, many of our students discover that they never consciously noticed that the days are longer or that the Sun is higher in the sky during summer than winter; that stars move across our sky on angled paths rather than ­going from due east to the zenith to due west; or that the Moon can be vis­i­ble during the day. Thus, we find it advisable to devote extra time to the topics presented in this chapter, as needed or as the curriculum allows. Out of every­thing that students w ­ ill learn in the w ­ hole course that can be of practical use for the rest of their lives, we find the contents of this chapter are most relevant.

We encourage our students to take full advantage of the AstroTours, Astronomy in Action videos, and simulations 10

included in this chapter, and emphasize that they should develop the habit of d­ oing so whenever they come across a link in the text. Th ­ ese supplementary materials, plus t­ hose given in the Learning Astronomy by ­Doing Astronomy workbook, are excellent for helping students master ­these confusing but fundamental concepts outside class.

DISCUSSION POINTS • Apply the dependence of the perspective on the celestial motions shown in Figure 2.8 to the location of the classroom. Engage students coming from other countries or states to discuss the perspective on the sky from their birthplaces. The goal is to have them realize that ­these perspectives are only similar for places with similar latitudes but do not change with longitude, even though they may be very far apart. (LG 1) • Figure 2.7 shows the brightest stars around the north celestial pole during timed exposures from dif­fer­ent locations. The current north celestial pole (NCP) is marked, and it lies very close to the “North Star,” Polaris. To familiarize the students with the night sky, discuss how to identify Polaris from your location, based on Figure 2.9. (LG 1) • What are the consequences of similar perspectives of the sky on the perception of the seasons in dif­fer­ent places?

What are the differences in regions with dif­fer­ent latitudes? Have students make individual lists and then share with their neighbors and then the class. Discuss how it all depends on the altitude of the Sun. (LG 2) • During winter in the Northern Hemi­sphere, Earth is closer to the Sun than at any other time of the year.

Discuss how this observation disproves the conception that the changing Earth–Sun distance is the reason for Earth’s seasons. (LG 2) • Use Figure 2.20 to show why Earth’s shadow does not cause the phases of the Moon. Start by illustrating what a typical shadow looks like, and then have the students try to create a shadow that “hits” a first quarter Moon. (LG 3) • Use the same figure to explain why we can sometimes see the Moon during the day. Discuss how this figure

CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon ◆ 11

can incorrectly imply that if p­ eople living on one side of Earth have a full Moon on the meridian during their local midnight, ­people living on the opposite side of Earth have a new Moon in their daytime sky. (LG 3) • Figures 2.24 through 2.31 provide spectacular pictures and pedagogical graphics of eclipses. To test that students have mastered the understanding of eclipses, discuss limiting cases: What would happen if Earth’s rotation rate and the Moon’s revolution rate ­were perfectly synchronized? What if the Moon’s orbit ­were not tilted with re­spect to the ecliptic? What about if the apparent size of the Moon w ­ ere double that of the Sun? What about if the apparent size of the Moon w ­ ere half that of the Sun? (LG 4) • The opening figure for this chapter shows the Sun during a total solar eclipse. The question asked is, “What ­causes a solar eclipse?” ­There was the total eclipse of the Sun viewed across the United States on August 21, 2017, and a f­ uture one is coming on April 8, 2024, with the path of totality r­ unning from Mazatlan, Mexico, to the Maritime Provinces of Canada. Find out how many students witnessed e­ ither a partial or total solar eclipse in 2017. Have them share what they experienced. Have students share any experiences with lunar eclipses and discuss why total eclipses of the Moon occur more often than total eclipses of the Sun.

PROCESS OF SCIENCE The Pro­cess of Science for this chapter covers “Theories Must Fit All the Known Facts.” The three hypotheses about the reason for the seasons are the (1) Earth is closer to the Sun in the summer and farther away in winter, (2) the tilt of Earth’s axis c­ auses one hemi­sphere to be closer to the Sun, and (3) the tilt of Earth’s axis changes the distribution of energy between the hemi­spheres. Reasons are given for the first two hypotheses being falsified.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 2 and are available from the f­ree Student Site. ­These animations are also integrated into assignable Smartwork5 online homework exercises.

THE CELESTIAL SPHERE AND THE ECLIPTIC This animation starts with the ancient model of Earth and the celestial sphere, and why it is a useful concept.

This is followed by side-­by-­side perspectives of the view of the revolving night sky from backyards in both the

Northern and Southern Hemi­spheres and an “outside” view of Earth embedded in a concentric celestial sphere.

The observations of planets are next. The content of the rest of the animation focuses on the ecliptic, showing the motion of the Sun, the Moon, and constellations relative to one another.

Text reference: Sections 2.1 and 2.2 THE VIEW FROM THE POLES This animation shows how Earth’s rotation corresponds to the movement of the stars in the sky and the rotation of the stars around Polaris, which is very nearly at the north celestial pole. It also explores the precession of Earth’s rotation, including how the movement of the stars ­w ill look when Polaris is no longer the North Star.

Text reference: Sections 2.1 and 2.2 EARTH SPINS AND REVOLVES This animation shows Earth as it is positioned with re­spect to the Sun, including motion along its orbit; the direction of its spin as viewed from the North Pole; the apparent daily motion of the celestial sphere; spin axis tilt; a discussion of the ­causes for the seasonal variation in climate in terms of latitude and the a­ ngle of incident sunlight; and comparisons of the pro­cesses to pro­cesses on other planets.

Text reference: Sections 2.1 and 2.2 THE MOON’S ORBIT: ECLIPSES AND PHASES This interactive animation explores the Earth-­Moon-­Sun system, building on the ele­ments of two previous animations (“The Celestial Sphere and the Ecliptic” and “Earth Spins and Revolves”). It shows a changing point of view from the size scale of Earth’s orbit down to the size scale of the Moon’s orbit, followed by emphasis on the Moon’s orbit to distinguish the concept of an eclipse versus a phase, and shows the relative configurations of Earth, the Moon, and the Sun. Text reference: Sections 2.3 and 2.5

ASTRONOMY IN ACTION VIDEOS ­ ese videos mix live demos with mini-­lectures develTh oped by Stacy Palen. They provide students visual preparation for class or a review of what they have learned; they are listed ­here in the order presented in the text. All videos are available on the ­free Student Site. Assignable assessment questions can be found in Smartwork5 and the Coursepack.

12 ◆ CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon VOCABULARY OF THE CELESTIAL SPHERE

INTERACTIVE SIMULATIONS

Dr. Palen divides the basic vocabulary used when we refer to locations on the celestial sphere into two categories: the vocabulary that is specific to each individual and the vocabulary that is used in general ­because it applies to every­one everywhere. The meaning of each word or phrase is actively demonstrated. Specific words: zenith, nadir, horizon, and meridian. General terms: North and South Pole, equator; north and south celestial poles, celestial equator; ecliptic, tilt of Earth’s axis, rotation of Earth. (Length: 6:49) Text reference: Section 2.1

Interactive simulations are available ­free at the Digital Resources Page, on the Student Site.

PHASES OF THE MOON The interactive simulation Phases of the Moon is included as part of the Exploration for this chapter. Students are able to locate the Sun, Earth, and the Moon at vari­ous positions and times and see the geometry of the orbits as well as the phase of the Moon as seen from Earth.

Text reference: Section 2.3 and Ch2 Exploration

THE CAUSE OF EARTH’S SEASONS The cause of Earth’s seasons is twofold: the varying length of the days throughout a year and how light from the Sun affects the surface of Earth. An unusual but effective demonstration involves a large amount of uncooked spaghetti wrapped so that it stays in a column. The individual noodles represent rays from the Sun. Dr. Palen draws a circle around the column when the rays are striking perpendicular to the surface of a sheet of paper and compares this to the larger-­area, elongated oval when the “rays” hit at an ­angle. (Length: 2:23) Text reference: Section 2.2 THE EARTH-­MOON-­SUN SYSTEM Three students work together with Dr. Palen in an open field to demonstrate the rotation of Earth, the orbit of the Moon while Earth also rotates, and fi­nally the rotation of the Sun while the (still-­rotating) Earth orbits it and the Moon orbits Earth (keeping one side always facing Earth).

The part showing the motion from high overhead is especially effective. This demonstration must have taken a lot of practice. (Length: 3:21) Text reference: Section 2.2 THE PHASES OF THE MOON This video ends with a request that the students go and do this demonstration “right now.” It ­w ill enhance their understanding of why the Moon goes through phases prob­ably more than any other demonstration. A bright light source, Dr. Palen’s head representing Earth, and an orange held at arm’s length are used to show what the Moon phase ­w ill be based on the Sun-­ Earth-­ Moon relative positions. The names of the phases (new, waxing crescent, first quarter, waxing gibbous, full, waning gibbous, third quarter, and waning crescent) are all given. (Length: 3:42) Text reference: Section 2.3

TEACHING READING ASTRONOMY NEWS 1. The shadow of the Moon w ­ ill travel across Earth’s surface—­whether on land or over ­water—­because the Moon is orbiting Earth from west to east, “dragging its shadow” with it.

2. Scientists had not had extended time coverage of the corona of the Sun, part of the Sun’s atmosphere that is best seen during a total solar eclipse.

3. The corona is so diffuse that the brilliance of the photosphere totally blocks our view of it.

4. A total eclipse of the Sun occurs during a new Moon phase when the Moon is on one of the nodes of its orbit.

A full Moon phase happens around 14 or 15 days ­later, so the answer may depend on the interpretation of “near the date” of the solar eclipse. A quick check shows that occasionally a partial eclipse of the Sun may be followed by a total eclipse of the Moon about 2 weeks ­later, but definitely not during e­ very solar eclipse.

5. During the period while this instructor’s manual was being developed, no results had been published, although ­there ­were many articles on the science goals.

Good science requires time to analyze the data and publish results.

CHECK YOUR UNDERSTANDING SOLUTIONS 2.1A. (b) Earth rotates about the North and South Poles, and the celestial poles are just an extension of Earth’s rotation axis to the sky.

2.1B. (d) If we are at 90° latitude at the North Pole, Polaris is directly overhead. If we are at 0° latitude at the equator, it is at 0° altitude above the northern horizon. It follows that its altitude in our sky and our latitude are the same number of degrees.

2.2. (d) The greater the tilt, the more severe the seasons.

CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon ◆ 13

2.3. A full Moon is rising as the Sun sets. At noon, a third-­quarter Moon ­w ill be setting. See Figure 2.20 for both answers.

2.4. If the synodic period of the Moon phases w ­ ere precisely 30 days, and Earth took precisely 12 months to orbit the Sun, the current “wandering” holidays would always fall on the same days from year to year.

2.5. (b) If the Moon w ­ ere twice as far away, its angular size would be smaller (half the size it is now), and it could no longer cover the Sun to make a total eclipse, but could create annular eclipses. However, Earth’s shadow would still cover the Moon, so lunar eclipses would happen. They would not last as long since Earth’s shadow would be smaller at the new distance.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (c) Constellations are groupings of stars that appear close together on the sky and generally form a familiar pattern. However, in real­ity the stars are quite far from one another.

2. (c) Use Figure 2.8.

3. (b) The rotation of Earth on its axis brings the Sun into and out of our sky, causing day and night.

4. (b) Currently Polaris is located very close to the north celestial pole, meaning that to the naked eye, it does not appear to move as the sky rotates.

5. (a) Refer to Figure 2.16, which shows how the Sun’s path compares to our orbital plane and the celestial equator.

6. (e) Seasons happen both ­because (b) days are longer in the summer and ­because (c) light is more direct in the summer.

7. (d) On an equinox, the ecliptic crosses the celestial equator, the Sun rises due east, and one has exactly 12 hours of daylight u­ nless at the North or South Pole.

8. (b) The Moon is in a “tidal lock” with Earth, so it spins at the same rate as it orbits.

9. (d) Using Figure 2.20, the Moon must be in the third quarter phase.

10. (a) A lunar eclipse happens when Earth’s shadow falls on the Moon.

11. (c) Dif­fer­ent civilizations merged their cultural beliefs into the daily, monthly, and yearly noninteger lengths of time. Some let holidays wander, some held end-­of-­ year festivals, and some adjusted their calendar periodically to solve the prob­lem.

12. (d) The stars that we see at night depend on all of the options listed.

13. (c) The Tropic of Cancer is the northern tropic, and so in summer, the Sun is above the Tropic of Cancer and ­there is 24 hours of daylight above the Arctic (northern) Circle.

14. (b) If you read carefully, all other answers are incorrect.

15. (a) “On the meridian” means the highest the Moon ­w ill be in the sky. If the Moon is in the first quarter phase and at this position, then by using Figure 2.20, we find it is sunset and the Sun is on the western horizon.

Thinking about the Concepts 16. Magellan could not use the North Star (Polaris) for navigation ­because he was in the Southern Hemi­ sphere; thus, Polaris was never above the horizon.

17. ­Because the north celestial pole is an extension of the North Pole on Earth, if you are standing on the North ­ ill see the north celestial pole at your zenith.

Pole, you w 18. If Gemini is high in the night sky in winter, it is high in the daytime sky in summer, and the Sun’s light blots out all light from the stars.

19. If I am f lying in a jetliner, (a) I can tell that I am moving by watching stationary objects go past me below; (b) backward, ­toward the way the jetliner just came.

20. Our first question as an expert witness is ­whether the full Moon casts very pronounced shadows or illuminates t­ hings quite brightly and, in this reader’s opinion, that only happens if one is in an area that is other ­w ise extremely dark, which does not ­really happen in cities. That being said, the next question is ­whether the full Moon can cast a long shadow at midnight. To cast a long shadow, the object (Sun or Moon) must be very low in the sky, but at midnight the Moon ­w ill be at the meridian. For most observers, this is relatively high in the sky, which negates the defendant’s claim. However, if one ­were living around the Arctic Circle, then this argument might have some credibility b­ ecause the Moon would never rise to be very high in the sky.

21. The observation is “seasons happen.” In Take 1, the hypothesis is that “seasons are due to distance from the Sun.” The prediction is that “both hemi­spheres ­w ill be hot at the same time.” The test is that “the seasons are opposite for the two hemi­spheres.” B ­ ecause this is a failure, we return to a new hypothesis in Take 2. ­Here, the hypothesis is that “one hemi­sphere is closer to the Sun than the other b­ ecause of our tilted axis.” The prediction is that “the closer hemi­sphere ­w ill be hotter.” The test is that “the difference in distance is too small to account for such temperature differences.” Our hypothesis failed, so on to a new

14 ◆ CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon hypothesis in Take 3. ­Here, we propose that “the tilt of our axis changes the distribution of energy on the surface.” The prediction is that “more energy ­w ill hit Earth in summer,” which we test and confirm. With a confirmed test, we make a new prediction: “summer days are longer.” This is true, and we make another prediction, and so on.

22. The average temperatures on Earth lag a bit b­ ehind the formal change in seasons ­because it takes time for Earth to heat up or cool down, the temperatures being modulated by the vast oceans. Thus, although the winter starts officially in December, it takes 1 to 2 months for Earth to cool down, making the coldest months January and February.

23. (a) The Earth takes 24 hours to complete one rotation (about its axis) with re­spect to the Sun—23 h 56 m with re­spect to the stars. (b) The Earth takes 26,000 years to complete one “wobble.”

24. The full Moon crosses the meridian around midnight, and the first quarter Moon rises (i.e., on the eastern horizon) around noon. To answer t­ hese, use a figure such as Figure 2.20.

25. (a) Over the course of one orbit, Earth ­w ill stay in a fixed position in the observer’s sky, ­because the same side of the Moon always f­aces Earth. (It would also rotate almost 30 times over the course of the orbit.) (b) The phases of Earth as viewed from the Moon ­w ill be the opposite of t­hose of the Moon as viewed on Earth; that is, if on Earth we see a full Moon, then on the Moon we would see a new Earth.

26. We would see a solar eclipse from the Moon. See Figure 2.29.

27. A total eclipse of the Sun casts a very small shadow on Earth and thus can be seen from only very narrow strips of Earth, whereas the partial shadow covers a much larger area and can thus be seen by many more observers.

28. To see an eclipse at each full or new Moon requires that the Moon’s orbit be in the same plane as Earth’s orbit around the Sun. B ­ ecause this is not the case, we see eclipses only on ­those occasions when the two planes line up, about twice a year, which are referred to as nodes.

29. Let’s start at the North Pole. If Earth’s tilt axis ­were 90°, the Sun would be at the zenith at the North Pole and the South Pole would be completely in the dark and extremely cold. Over the course of 3 months, the Sun would be at the zenith at decreasing northern latitudes as it moved slowly down to being over the equator. It would then be at the zenith over increasing southern latitudes ­until it reached the South Pole at 90° south.

The South Pole would be very hot, and the North Pole would be in a deep freeze. The Sun would then be at the zenith for decreasing southern latitudes, over the equator in 3 months, and over the North Pole again in another 3 months. The Sun would swing 180° in the sky compared to the 47° it swings now.

30. A cyclic change in the tilt would vary the height of the Sun in the sky, thereby changing the seasonal temperatures from their current pattern and ­either making the temperatures more extreme in their range or less than what they are now.

Applying the Concepts 31. Setup: We know that it takes 24 hours for Earth to make one revolution, so using d = vt, we can find the circumference of Earth. To find the dia­meter, we ­w ill use C = πd, where d is the dia­meter.

Solve: If the speed v = 1,674 km/h at the equator, then the total distance traveled in 24 hours is km 1,674 × 24 h = 40,176 km. Earth’s dia­meter is h then C 40,196 km d= = = 12,795 km. π 3.1416 Review: We can think of two ways to check this answer. First, look in the Appendix 4 of the textbook, which lists 6,378 km as Earth’s equatorial radius. Second, it is about 3,000 miles from New York City to Los Angeles, and ­there are 3 hours of time change between them, which means t­ here are about 1,000 miles, or 1,600 km per hour of change. Th ­ ere are 24 time zones total, so the circumference of Earth is about 24,000 miles, or 38,000 km. This is close enough to the 40,176 km we calculated, so it is a reasonable check.

32. Setup: For this prob­lem, use Figure 2.6, which gives the a­ ngle between the horizon and the north celestial pole (NCP) for an observer at 60° north latitude and Figure 2.8, which shows the celestial sphere as viewed by observers at four latitudes. The NCP and the motion of the stars around it clearly show the NCP altitude decreasing above the northern horizon.

Solve: (a) Answers ­w ill vary with latitude but they should resemble Figure 2.6. Students should start by mea­sur­ing rhe center-­equator-­latitude ­angle that equals their latitude. They then draw a line from Earth’s center, through their location, and out long enough to identify the zenith. They can then draw a tangent line on Earth’s surface at their latitude to represent their horizon. The direction arrow to the NCP w ­ ill always be

CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon ◆ 15

parallel to the N direction and the direction arrow to the celestial equator ­w ill always be parallel to the plane of the equator. They would now have all of the a­ ngles defined and measureable. (b) Combining Figures 2.6 and 2.8, one finds the maximum and minimum altitude of the Sun at noon on the solstices w ­ ill be 23.5° above or below the celestial equator, and the celestial equator appears L degrees below the zenith, or 90° − L degrees above the horizon, where L is your latitude. Thus, the Sun reaches (90° − L) ± 23.5° on the solstices.

Review: Activity 5: Altitudes of Objects on Your Meridian in Learning Astronomy by D ­ oing Astronomy: Collaborative Lecture Activities, 1st ed., has students do this very ­thing. If a student has access to a plastic celestial sphere with a movable Sun inside, using this tool is the best way to review the motion of the Sun in our sky and the relative positions of the zenith, celestial equator, and north celestial pole. It is worth investing in at least one of ­these for ­every introductory class.

33. Setup: For this prob­lem, use Figures 2.7 and 2.8, which show in graphics and time-­lapse photography how stars move around the north celestial pole. This prob­lem may be easier for students to solve if they have just completed the previous prob­lem.

Solve: If Polaris is D degrees from the zenith, then your latitude is L = 90° − D. In this prob­lem, Polaris is 40° from the zenith, therefore we are at latitude 50° N, which is in the southernmost part of Canada, and where Polaris is 50° above the northern horizon.

Review: Activity 5: Altitudes of Objects on Your Meridian in Learning Astronomy by D ­ oing Astronomy: Collaborative Lecture Activities, 1st ed., has students do this very ­thing. If a student has access to a plastic celestial sphere, using this tool is the best way to review the answer. It is worth investing in at least one of ­these for ­every introductory class.

34. Setup: For this prob­lem, use Figure 2.8, which shows how the stars move across our celestial sphere for four dif­fer­ent locations.

Solve: If you are living in the United States, then, as shown in Figure 2.8, you could see a star in the southern part of the celestial sphere if it is more than L degrees from the southern celestial pole. Therefore, if you want to see a star 65° from the celestial equator, that means it is 90° − 65° = 25° of the south celestial pole. To see it in the Northern Hemi­sphere, you would need to be at this latitude or below. The only states that reach this low latitude are Florida and Hawaii.

Review: If a student has access to a plastic celestial sphere, using this tool is the best way to review the

answer. It is worth investing in at least one of ­these for ­every introductory class.

35. Setup: For this prob­lem, use Figure 2.14 and note that panel (b) corresponds to the summer solstice in the Southern Hemi­sphere. Have students also locate the Antarctic Circle in this figure.

Solve: As Earth rotates, an observer on the South Pole w ­ ill not move with re­spect to the Sun. In other words, the Sun w ­ ill stay at the same height in the sky all day long on the solstice. Using the same argument as in prob­lem 32, the maximum height the Sun ­w ill reach in your sky is (90° − L) + 23.5°, or 23.5° above the horizon. This is the height of the Sun (a) at noon and (b) at midnight on the solstice.

Review: One can also visualize this by viewing Figure 2.17, which shows the midnight Sun for latitudes greater than 66.5° north or south.

36. Setup: For this prob­lem, use Figure 2.14, which shows how the Sun illuminates Earth at dif­fer­ent times of year.

Solve: ­Because the tropics are D degrees above (north latitude) and below (south latitude) our equator, and the (Ant)Arctic circles are D degrees below (decreasing latitude) the poles, we see that for this prob­lem, the tropics would be at latitudes 10° north and 10° south of the equator. The Arctic Circle would be at north 80° and the Antarctic Circle at south 80°.

Review: Imagine Earth had no tilt; then where would the tropics and circles be? (­There would be no tropics or circles ­because ­there would be no seasons.) Now tilt Earth by a tiny amount and then answer the question. One can easily derive the logic used in the solution in this way to verify that it is correct.

37. Setup: For this prob­lem, use Figure 2.16, which shows how the Sun’s path compares to our orbital plane and the celestial equator.

Solve: On the equinox, the highest the Sun in your sky w ­ ill be L degrees below the zenith or 90° − L degrees above the horizon, where L is your latitude.

Therefore, the highest the Sun ­w ill be on the summer solstice w ­ ill be (90° − L) + 23.5°, and if the Moon can be up to 5° above that, then in Philadelphia, the Moon can be as high as (90° − 40°) + 23.5° + 5° = 78.5° above the horizon.

Review: The Figure 2.16 shows this situation as described.

38. Setup: For this prob­lem, use Figure 2.14a that shows how the Sun illuminates Earth when it is summer in North Amer­i­ca.

Solve: According to this figure, (a) we w ­ ill have to travel to latitudes of 66.5° or higher and (b) make this

16 ◆ CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon trip as close to the summer solstice as pos­si­ble, then the Sun w ­ ill never set.

Review: ­There are a number of small towns north of the Arctic Circle that you could visit in Rus­sia, ­Finland, Sweden, Norway, Greenland, Canada, and Alaska.

39. Setup: Follow the worked example in Working It Out 2.1 but replace 185 meters with 157.5 meters. Alternatively, in the final answer given in the example, divide by 185 and multiply by 157.5 to remove one unit and replace the other.

Solve: Following the first method, we use: 1 km 1 m × C = 5,000 stadia × 157.5 × 50 stadion 1,000 m = 5 × 157.5 km = 787.5 km C = 787.5 km × 50 = 39,375 km The modern value is 40,075 km; this is only 1.7% off.

Review: Check using the second method. Using 185 meters per stadion, the example found a circumfer157.5 = 39,375 km. ence of 46,250 km: 46,250 km × 185 40. Setup: (a) We need to know how much time it takes for the vernal equinox to move from one constellation to another. Let us assume all constellations are equally distant from the next, then how much time is spent in each constellation for a total cir­cuit of 26,000 years (the period of Earth’s wobble)? (b) Use Figure 2.18b.

Solve: (a) ­There are 12 constellations, and if they are distributed roughly uniformly around the zodiac (as in Figure 2.13, although the astronomical zodiac constellations are not evenly divided), it takes roughly 26,000 yr ≈ 2,200 yr for the equinox to move from 12 constellations

Solve: 13,500 yr − 2,000 yr = 11,500 yr.

Review: We could also estimate this from Figure 2.18a as an a­ ngle. It looks like the ­angle between Polaris and Vega (as mea­sured from the center of the image) is almost 180°. That means it ­w ill take about 1 × 26,000 = 13,000 half of the total period, or 2 years, which is close to what we found.

42. Setup: To solve this prob­lem, remember that a circle mea­s ures 360°, and it takes about 27.3 days for the Moon to orbit once with re­spect to the fixed stars on the sky. We must find out how long it takes to move 1°.

Solve: If the Moon moves 360° in 27.3 days, then 27.3 days it takes = 0.076 days to move 1°, or 360° 24 h 0.076 day × ≈ 1.8 hours to move 1°. We want 1 day the Moon to move half a degree, it angular size, which ­w ill take half as long, or about 55 minutes.

Review: On average, the Moon rises about 50 minutes ­later each day. This can be explained by Earth having to rotate just a bit more for a given location to “catch up” to it.

43. Setup: The prob­ lem gives us the formula that diameter apparent size ∝ . We need to compare the distance ratio of the values for the Sun and the Moon. Since dia­meter is stated for the formula, but the values are for radii, we w ­ ill solve this as many students might by converting radii to dia­meters. We are talking about the apparent sizes of the Sun and the Moon.

Solve: For the Sun, the ratio is 6 2 × 696,000 km 1.39 × 10 km = = 9.48 × 10 −3 , and 1.469 × 108 km 1.469 × 108 km 3,474 km 2 × 1,737 km for the Moon, = = 9.19 × 10 −3.

5 3.78 × 10 km 3.78 × 105 km

one constellation to the next. (b) If the vernal equinox has moved from Aries to Pisces over about 2,200 years, then 4,000 years ago it would have been in Taurus. 2 × 1,737 km 3,474 km = 9.19 × 10 −3. The two are indeed just about the same Review: (a) Note that the question 5asked=how long 5 3.78 × 10 km 3.78 × 10 km size. the equinox spends in a constellation, not how much Review: If the two w ­ ere not roughly the same size, time it takes to move between the two. If we assume then the Moon would not seem to almost perfectly that they are all about the same size and equally cover the face of the Sun during a solar eclipse. The spaced apart, then the two questions are ­really the fact that the orbit of the Moon is not a perfect circle same. (b) Given that we determined it takes about and the distance to the Moon varies means that the 2,000 years to move one constellation, we need to apparent size of the Moon varies by about 12 ­percent, move only two constellations back from Pisces. and we view annular eclipses.

41. Setup: Looking at Figure 2.18b, Vega is around 44. Setup: Using the form of the small-­angle formula 13,000 to 14,000 CE. The question is, how long ­w ill it from the previous prob­ lem, we know that the take to move from Polaris (­today) to Vega? We have diameter all the information we need, as long as we remember apparent size ∝ . In the case of our prob­lem, that ­today is 2,000 CE. distance

CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon ◆ 17

the Earth–­Moon distance is the same, and so the ratio between the apparent size of the Moon as seen on Earth, and the apparent size of Earth as seen on the Moon, w ­ ill be the ratio Apparent size (Earth) Diameter (Earth) = .

Apparent size (Moon) Diameter (Moon)

pelling ­those misconceptions. Combine with the other methods presented, this exploration gives students another approach to learning about where the Moon is located in the sky at the new, first quarter, full, and third quarter phases, as well as when it rises and sets at certain phases.

EXPLORATION SOLUTIONS Solve: Again, as many students might insist on using 1. At startup, the time is noon, ­because the Sun is ­going the dia­meters, and we might want the chance to to be at the meridian (i.e., the highest in our sky that it show how some numbers or variables cancel: Given ­w illkm reach).

Apparent size (Earth) 2 × 6,371 the values in the text, = 3.67. = 2. The Moon is crossing the meridian ­because it is located Apparent size (Moon) 2 × 1,737 km at the same position in the sky as the Sun. arent size (Earth) 2 × 6,371 km = = 3.67. Earth appears 3.67 times 3. The Moon is new, as shown on the top right panel. arent size (Moon) 2 × 1,737 km Also, none of the illuminated face is ­toward us. larger to the observer on the Moon.

4. I would see a “full” Earth, that is, fully illuminated.

Review: Try setting out a tennis and basketball some 5. The Moon orbits counterclockwise. distance apart, and then observe each object from the 6. The right side of the Moon is illuminated first. other’s location. It ­w ill become clear that the change 7. If the horns of the crescent Moon point right, it must in angular size varies with the size of each object. be a waning Moon.

45. Setup: Refer to Figure 2.31, which shows why eclipses 8. At midnight, the first quarter Moon is setting on the happen only at certain times of the year. western horizon.

Solve: If the inclination of the Moon’s orbit drops, then 9. The full Moon crosses the meridian at midnight. ­there is a longer period of time during which the Moon 10. At noon, the third quarter Moon is on the western can pass through Earth’s shadow. Hence, the lunar horizon. eclipse seasons would become longer. The solar eclipse 11. At 6 p.m., ­there is a full Moon rising. still requires a very accurate alignment of the Sun-­Earth-­ Moon system, so this season would prob­ably not change.

LEARNING ASTRONOMY BY ­DOING Review: You can try this for yourself by ­going into a ASTRONOMY: COLLABORATIVE dark room with a single lightbulb and holding a tenLECTURE ACTIVITIES nis ball at arm’s length. The bulb represents the Sun, The Learning Astronomy by D ­ oing Astronomy workbook your head is Earth, and the ball is the Moon. By turnactivities that are relevant to Chapter 2 are introduced ­here. ing around, you can cause solar and lunar eclipses.

For more information, please see the Learning Astronomy by ­Doing Astronomy workbook, the Instructor’s Manual for Using the Web the workbook, and the Power­Point clicker question slides 46. Involves looking up times for sunrise and sunset and associated with the workbook. changing lengths of the days. Answers ­w ill vary.

47.Involves viewing Earth from dif­fer­ent locations, dates, ACTIVITY 3: WHERE ON EARTH ARE YOU? and times. Answers ­w ill vary.

48. Involves phases of the Moon on current date and Students learn about the coordinate systems that are used changes with time. Answers w ­ ill vary. on Earth and how our location on Earth and Earth’s orbit 49. Involves observing the Moon over a period of time, around the Sun are related to the seasons. They should sketching, and analyzing. Answers ­w ill vary. also be able to: 50. Involves determining when the next lunar and solar • recognize that the Sun and stars appear differently at eclipses ­w ill occur at students’ locations, and the dif­fer­ent locations on Earth. types of eclipses to be observed. Answers ­w ill vary.

• summarize how ­these differences lead to seasons on Earth.

• state where the seasons are most and least extreme on EXPLORATION Earth and how this difference follows from the locaMany students seem to have difficulty letting go of their tion of the Sun in the sky. misconceptions about the phases of the Moon, and this exploration continues the emphasis this chapter has on disText reference: Section 2.2

18 ◆ CHAPTER 2 Patterns in the Sky—­Motions of Earth and the Moon ACTIVITY 4: STUDYING THE PHASES OF THE MOON FROM A PRIVILEGED VIEW

ACTIVITY 5: ALTITUDE OF OBJECTS ON THE MERIDIAN AT YOUR LOCATION

This activity requires visualizing the Earth-­Moon-­Sun system in three dimensions. In this activity, students develop this ability by learning how to:

Students learn to connect the altitudes of celestial objects as ­those objects cross the celestial meridian to their latitudes on Earth. In addition to helping them develop abstract reasoning visualization skills, this activity w ­ ill also help them:

• successfully replicate the motions of Earth and the Moon, as well as their positions with re­spect to the Sun at each lunar phase.

• explain the continuity of the Moon phases worldwide.

• use an Earth-­Moon figure to disprove a common misconception that Moon phases are caused by Earth’s shadow.

• correctly order the phases of the Moon.

Text reference: Section 2.3

• reproduce the altitudes of the north celestial pole, celestial equator, a star, and the Sun (at solstices and equinoxes) on the meridian for their locations.

• explain why t­ here are dif­fer­ent seasons throughout the year.

• explain how the stars appear to move through the sky and how the motion of the stars differs when viewed from dif­fer­ent latitudes on Earth.

Text reference: Sections 2.1 and 2.2

CHAPTER 3

Motion of Astronomical Bodies

INSTRUCTOR NOTES Chapter 3 covers force and Kepler’s (empirical) laws.

Major topics include: • the history and details of the geocentric and heliocentric models.

• the importance of Tycho Brahe’s observations to Kepler’s formulation of his laws of planetary motion.

• the significance of Galileo’s observations to the advancement of astronomy and physics.

• an introduction to the crucial contributions Newton made, starting with his laws of motion.

Part of our motivation for teaching an introductory astronomy course is to teach the pro­cess of science. This chapter is replete with valuable lessons, such as the dangers of dogma or belief in science; scientific revolutions; evidence-­based modeling; and testing theories.

The material in this chapter can be confusing to students, some of whom find applications of the physics to be counterintuitive (e.g., why does the mass of a planet not affect the period of its orbit?). Also, the nomenclature can be somewhat confusing b­ ecause Kepler’s laws w ­ ere first empirical discoveries l­ imited to the Solar System, but as a result of Newton’s subsequent work, ­those laws are now considered universal laws of physics.

Students may need extra time to work through ­these concepts. Taking some extra time on them ­w ill help, of course, and so does bringing up ­these fundamental laws when we deal with the applications in other chapters. ­There are supplemental videos, explorations, and simulations available that ­w ill enhance learning of ­these topics, as well as activities found in the companion Learning Astronomy by ­Doing Astronomy workbook.

DISCUSSION POINTS • Based on what they learned from Chapter 2, students should now be able to describe the motion of the objects on the celestial sphere, both on a daily and on a yearly basis. Suggest a 1-­or 2-­m inute writing exercise where they describe the motion of the stars, the Sun,

and the Moon. Discuss if they are describing a model with Earth at the center or the Sun. (LG 1) • Have students use the time stamps to finger-­trace the motion of Mars in Figure 3.2. Show a video of the ­retrograde motion of Mars; for example, the one on YouTube at https://­youtu​.­be​/­1E3vlf​-­E sqk. ­These ­simulations, from programs like Starry Night Planetarium, can also show the motion of the Moon as it progresses from west to east through its phases. Have the students come up with their guesses as to why Mars appears to stop in the night sky and change direction. Can they invent a mechanism to replicate this motion? (LG 1) • Kepler’s second law seems to be the hardest one for students to grasp. What does “sweeping out” mean?

Students should put this law into their own words. Use a think-­pair-­share activity. (LG 2) • Figure 3.16 illustrates Kepler’s third law of planetary motion, his harmonic law. Working It Out 3.2 and ­Table 3.2 show the logic for the calculations and the results for eight planets and three dwarf planets of the Solar System. While we w ­ ill cover exoplanets in much greater detail in Chapter 5, let’s just test to see if Kepler’s third law is universal. Data ­were obtained from the exoplanet website http://­exoplanet​.­eu​/­catalog​/­ for two systems: HD 10180 and TRAPPIST-1. Ask, “Does Kepler’s third law also work for the extrasolar planetary systems that have multiple planets orbiting each star?” Students should examine the charts shown in Figure IM-3.1 and answer ­these initial questions: (LG 2) • Compare Figure IM 3.1A for the HD 10180 system ­ able 3.2 and IM 3.1C for the TRAPPIST-1 system to T that lists the planets in the Solar System. How do the orbital periods and Semimajor Axes A (AU) compare?

Comment on what it might be like to live on a planet in ­either of ­those systems.

• Take a close look at Figures IM 3.1B and IM 3.1D, where we show Kepler’s third law, P 2 = A 3, relationship.

What can you state about ­these two systems and Kepler’s third law? (Note for instructors: The mass of HD 10180 is 1.062 ± 0.017 MSun. and the mass of Trappist-1 is 0.089 ± 0.006 MSun. The slopes of ­these data 19

20 ◆ CHAPTER 3 Motion of Astronomical Bodies are representative of each star’s mass. Both systems follow Kepler’s third law.)

• ­These two systems seem to be very dif­fer­ent from the Solar System. Take a ­couple of minutes to come up with a hypothesis of what ­there must be in common to give t­ hese results. Start your hypothesis with, “If t­ hese two planetary systems follow Kepler’s third law, then . . . .” (Note for instructors: Each star must be much more massive than any of its planets and it is not part of a binary.)

• What made Galileo so sure that the Copernican model of the Solar System was correct and the geocentric model was wrong? Did he have enough evidence to convince o­ thers? (LG 3)

FIGURE IM 3.1

Data source: Exoplanet.eu

• In addition to Jupiter having orbiting moons and Venus ­going through phases with the phases corresponding to its change in apparent size, Galileo also noted that the Moon was imperfect: it had craters of all shapes and sizes—­blemishes. Galileo used his telescope to proj­ect the Sun onto paper, and he sketched sunspots over many cycles of the Sun’s rotation. If the Sun rotated, why not Earth? In your opinion, which one of ­these four observations was the most convincing for claiming that the Copernican model was correct? Be prepared to support your answer. (LG 3) • Galileo may have been the first to use experimentation, testing, and observations. Newton took what Galileo started and formalized it through prediction,

CHAPTER 3 Motion of Astronomical Bodies ◆ 21

testing, and mathematical analy­sis. List Newton’s three laws of motion and give personal examples of how you have experienced each one of them. (LG 4) • The opening image for this chapter asks the question, “Who discovered Jupiter’s moons?” Students may know only a ­l ittle about Jupiter and even less about its moons. Start with a­ ctual images taken by the spacecraft Juno as it was on its last approach to Jupiter, in summer 2016 (suggested links follow). Have students describe what they observe and hypothesize about what is happening.

• https://­youtu​.­be​/­wJiYoNN9lKk (21 sec) • https://­youtu​.­be​/­XpsQimYhNkA (2 min, 58 sec)

PROCESS OF SCIENCE Theories are falsifiable. The Pro­cess of Science for this chapter addresses one of the key, and often misunderstood, component of the scientific method: in order for a theory to be scientific, it must be falsifiable. The emphasis is on Kepler using Tycho’s data and falsifying the Copernicus hypothesis that the orbits of the planets must be perfect circles. It took about a ­century for that to happen.

ASTROTOUR ANIMATIONS KEPLER’S LAWS This animation of Kepler’s three laws contrasts Earth’s orbit with that of other bodies in the Solar System. Using slightly eccentric (Mars) and extremely eccentric (Comet Halley) orbits for comparison, the three laws are illustrated showing: (1) the difference between an elliptical and near-­circular orbit; (2) changes in speed as an object moves along its orbit; and (3) how period changes with larger semimajor axes. The initial statement about Kepler rejecting his original hypothesis that the planets orbit the Sun in perfect circles is particularly relevant in our coverage of the scientific method. Kepler did not manipulate the data but rather let his analy­sis of the data lead him to his three laws.

Text reference: Section 3.2 VELOCITY, ACCELERATION, AND INERTIA This animation uses examples that ­w ill be familiar to most students—­the movement of an automobile in both a straight line and a circle—to define and address the ideas of velocity and acceleration. It also defines and explores the concept of inertia, first with just the car and second

with the addition of a trailer to the automobile to increase its mass as it slams into traffic barrels.

Text reference: Sections 3.3 and 3.4

ASTRONOMY IN ACTION VIDEOS ­ ese videos mix live demos with mini-­lectures develTh oped by Stacy Palen. They provide students visual preparation for class or a review of what they have learned; they are listed h­ ere in the order presented in the text. All videos are available on the ­free Student Site. Assignable assessment questions can be found in Smartwork5 and the Coursepack.

VELOCITY, FORCE, AND ACCELERATION Dr. Palen starts the demonstrations by clarifying the difference between velocity and acceleration: “Velocity is telling me what the object is d­ oing and acceleration is all about what the object is about to do.” Velocity, acceleration, and force are then demonstrated using a ball tossed up it the air and a car on a track. (Length: 4:43) Text reference: Section 3.4

INTERACTIVE SIMULATIONS PLANETARY ORBITS The interactive simulation for this chapter is included as ­ ill be able to animate part of the Exploration. Students w the orbits of the planets and control basically all of the orbital par­a meters. The emphasis is on Kepler’s laws and Mercury’s orbit.

Text reference: Section 3.2, Ch3 Exploration, and Ch4 Exploration

TEACHING READING ASTRONOMY NEWS 1. Conjunctions occur e­ very 26 months. This is the synodic period of Mars.

2. ­Because Mars is ­behind the Sun during a solar conjunction, transmissions could be made only during the day. The Sun would disrupt or dwarf signals to and from Mars. The blackout lasts from a c­ ouple of weeks to a month or so, leaving the rovers and orbiters to function on their own. Mars is the farthest it can be from Earth at this point in its orbit.

3. Earth w ­ ill complete two full revolutions and then move an additional 2 months, or 2/12 of a revolution.

This means that Earth and Mars ­w ill move the angular equivalent of about 2 hours on a clock or about 60

22 ◆ CHAPTER 3 Motion of Astronomical Bodies 3.3. (c, d) The geocentric model fails for moons orbiting Jupiter ­because the model proposes that all objects orbit Earth. The changing of the apparent size and the order of the phases of Venus indisputably disproved the geocentric model. Note that (a) sunspots and (b) craters on the Moon—­observations that Galileo did make and noted—­are not covered in the text at this point in the book.

3.4. (b) Acceleration occurs with a change in the magnitude or direction of that velocity. The AstroTour Animation, “Velocity, Acceleration, Inertia,” covers this using familiar objects.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS

FIGURE IM 3.2

degrees. Figure IM 3.2, seen ­here, is similar to the text’s Figure 3.25, but with the new positions noted over parts of the 26 months.

4. Answers w ­ ill vary. We think it does explain the concepts well and w ­ ill also help the students understand the issues involved. Perhaps they ­w ill be better able to explain the videos to ­others.

5. Answers ­w ill vary. NASA could put satellites into the same orbit as Mars, and spaced at the LaGrangian points L3, L4, and L5, at 60° ahead, 60° b­ ehind, and opposite the Sun, such that astronauts on Mars could bounce their signals off ­these satellites and back to Earth.

CHECK YOUR UNDERSTANDING SOLUTIONS 3.1A. Ancient civilizations knew planets ­were dif­fer­ent from stars ­because planets wandered through the sky—­hence the name planet, from the Greek planetes, meaning “wanderer.”

3.1B. (c) Retrograde motion requires one planet to overtake the other. ­Because Uranus is farther from the Sun, Earth moves faster and w ­ ill always overtake Uranus.

3.2. (c, b, d, a) Largest to smallest semimajor axis ranks with longest to shortest orbital period.

Test Your Understanding 1. (d) Empirical means based only observation and not on a theoretical foundation.

2. (a) This is a description of retrograde motion that makes the other planet appear to move backward for a while and then continue forward again on the sky.

3. The correct answer is (a). Choice (b) is incorrect ­because Copernicus insisted on circular orbits. Copernicus ended up using adjustments similar to epicycles, making (c) incorrect. For (d), Copernicus’s model did not fit the telescopic data any better than did Ptolemy’s.

It was, however, a less complex model to use.

4. (c) See text Figure 3.7.

5. (a) Synodic positions refer to t­ hose made with re­spect to Earth and the Sun in conjunction.

6. (a) Kepler’s second law, a planet sweeps out equal areas in equal time, tells us that when the planet is close to the star, it moves faster, but when it is far away it moves slower.

7. (d) See Working It Out 3.1.

8. (d) All the answers listed are examples of an un­balanced force that leads to acceleration.

9. (a) Kepler’s third law tells us that P2 = A 3 for all planets orbiting the Sun. 10. (b) The phases of Venus are pos­si­ble only if Venus changes its ­angle with re­spect to Earth and the Sun by 360°, as the Moon does.

11. (d) Kepler’s second law, a planet sweeps out equal areas in equal time, says that planets move fastest when they are closest to the Sun. 12. (c) This claim violates Kepler’s second law ­because the planet is said to be moving slowest when it is closest to the Sun.

CHAPTER 3 Motion of Astronomical Bodies ◆

13. (d) The other three answers have no logical relationship to the fact that moons orbit Jupiter.

14. (b) See Working It Out 3.3. How long the change takes to happen equals how much the velocity changes divided by the acceleration, 60 mph − 0 mph t= = 6 s.

10 mph/s 15. (c) We assume that life w ­ ill develop in relatively stable environments. Environments are unlikely to remain stable if the planet spends considerable amounts of time at radically dif­fer­ent distances from its central star.

Thinking about the Concepts 16. Figure 3.2 has a view ­toward the south, with west on the right side and east on the left. During normal motion, Mars moves ­toward the east. Retrograde motion for Mars is motion ­toward the west. In this image, Mars was in retrograde motion for about 2.5 months, or about 75 days. This can be seen by following the dates of each observation. Mars observers would have seen Earth in retrograde motion as well.

17. Empirical means basing conclusions on what one has found in observations and data, rather than basing them on calculations and mathematical theory.

18. Saturn takes 29.5 years to orbit the Sun, so ­after 1 Earth orbit it has not moved far in its orbit. Earth catches up to and passes Saturn in just over one Earth year.

19. Figures 3.7 and 3.8 show a heliocentric model and Figure 3.19 shows four of the phases for that model.

The geocentric model and pos­si­ble corresponding phases of Venus are shown ­here in Figure IM 3.3 In the geocentric model, Venus w ­ ill always appear to be about the same apparent size and never in a full phase.

In the heliocentric model, Venus w ­ ill go through all the phases from “new” to “full,” and ­w ill appear its largest when new and smallest when full. B ­ ecause the size and phase of Venus are observable only through a telescope (the changes are not vis­i­ble to the unaided eye), ­these models cannot be distinguished without a telescope.

20. Two solid objects of similar shapes w ­ ill always hit the ground at the same time even if they have very dif­fer­ ent masses. They experience the same acceleration. If the dropped objects have dif­fer­ent masses, then the more massive object ­w ill hit the ground with more force. Sheets of paper and tissue ­w ill not ­because they are much more sensitive to air re­sis­tance.

21. By Kepler’s second law, a planet is always moving the fastest when it is closest to the Sun. By the same law, a

23

planet moves the slowest when it is farthest from the Sun. 22. The eccentricity of Earth’s orbit is currently about 0.017, which is much less than the 0.055 mean eccentricity of the Moon. The Moon’s eccentric orbit brings it closer or farther from Earth over time and this changes its apparent size on the sky. This leads to total versus annular solar eclipses.

23. Inertia is a description of how much an object wants to oppose a change in motion. An object with a lot of inertia is not very easy to get moving if it is at rest, and it is not very easy to change its motion once it is moving.

24. Kepler would have deduced exactly the same three laws of planetary motion. The only difference is that the ­actual value of an AU would be dif­fer­ent. But ­because a Martian year is dif­fer­ent than a year on Earth, the relationship P2 ∝ A 3 would still be true.

25. Velocity is the speed of an object plus its direction.

Acceleration is how quickly the velocity of an object is changing. However, an object can travel at constant

FIGURE IM 3.3

24 ◆ CHAPTER 3 Motion of Astronomical Bodies speed but change direction (i.e., turn), and thus, the velocity is changing, which means the object is accelerating. The AstroTour titled “Velocity, Acceleration, Inertia” does an excellent job of showing examples of each.

26. Newton’s first law tells us the object in motion stays in motion ­unless a net force acts on it to change its motion. The inertia of the person not wearing a seat ­belt ­causes them to continue forward ­until they feel the force of the windshield.

27. We wear seat b­ elts to protect us from acceleration.

We travel at the same speed as the vehicle, but ­because we have inertia of our own, we ­w ill keep moving forward (Newton’s first law) u­ ntil something stops us (that is, the seat ­belt) if the vehicle suddenly decelerates.

28. In the space station, all objects w ­ ill be participating in the same free-­fall motion and thus have zero weight so they can all be maneuvered; however, a more massive object has more inertia, which means it is more difficult to change that object’s state of motion. Thus, a “heavy” (more massive) piece of equipment accelerates more slowly than a “light” (less massive) one ­because even though the astronaut is applying the same force, F, the masses are dif­fer­ent, leading to dif­ fer­ent accelerations (Newton’s second law: a = F/m).

29. Answers ­w ill vary. Scientific theories must be testable, and tests are not only confirmations of hypotheses but also refutations of challenges. The pro­cess of meeting challenges and testing to disprove them w ­ ill lead to a greater understanding of the science and strengthen the theory.

30. This is a more complicated question to answer than it seems. If Earth’s eccentricity w ­ ere 0.17, our planet would be substantially closer or farther from the Sun at the points of closest and farthest approach. Earth would be moving faster at perihelion and slower at aphelion. The seasons in the Northern versus Southern Hemi­spheres would drastically change, considering Earth is at its closest approach to the Sun during the northern winter (southern summer) and farthest from the Sun during the northern summer (southern winter). The temperature range would be much more extreme, and the length of the summer and winter seasons would also change, with Northern Hemi­sphere summers being longer and its winters shorter. Life would have a hard time adapting to such extremes.

single counting numbers (1, 2, 3 . . .), whereas logarithmic graphs increase by powers of 10 (10, 100, 1,000 . . .). To read off any graph, draw a line from the point in question to each axis and find the number (or estimate the closest number) on that axis.

Solve: This is a log-­log graph (both axes are logarithmic). The authors have already drawn in lines to the axes. But note that the axes are P2 and A 3. We estimate that the P 2 is halfway between 102 and 10 4. This gives us P 2 ≈ 103 or P ≈ 1000 ≈ 32. We estimate the A 3 is halfway between 102 and 10 4 as well, giving A 3 ≈ 103 or A ≈ 3 1000 = 10 AU. Review: We can compare t­ hese with the Appendix; also, note that ­because P2 = A 3, we can test the numbers. Indeed, 322 ≈ 103.

32. Setup: The reader can use a ruler to mea­sure the vertical size of Venus at the two phases. An object’s apparent (or angular) size scales with distance, so one-­half the apparent size means twice as far away.

Solve: Using a ruler for mea­sure­ments on Figure 3.19, we find that the right-­most crescent is about 3 times larger than the left-­most image. This implies Venus is about 3 times farther when the image at the far left (gibbous) was taken than it was when the far right (thinnest crescent) image was taken.

Review: This is a very hard prob­lem to check mathematically ­because we do not know at what point in the orbit each image was taken. Venus ­w ill, however, be much farther from Earth during its full phase, and closer during its new phase. The orbit of Venus takes it from approximately 0.3 AU to 1.7 AU from Earth.

Therefore, it is pos­si­ble for its distance from Earth to change by a f­ actor of 3.

33. Setup: To relate period to semimajor axis for a planet orbiting the Sun, use Kepler’s third law: P 2 = A 3 or P = A 3/2 .

Solve: P = 503/2 = 354 years.

Review: We can check this by comparing the period of a known object of roughly the same distance. Pluto’s average distance is 40 AU with period 248 years, whereas Eris’s distance is 68 AU and period 557 years.

Our hy­po­thet­i­cal planet is between ­these, as expected.

34. Setup: For the semimajor axis, use Kepler’s third law P 2 = A 3, solving for A. Oppositions happen on the synodic period, so use Working It Out 3.1 to find S given P is the period of Uranus and E is 1 year.

Solve: For the semimajor axis, A = P 2/3 = 842/3 = ​ 19.2 AU. To solve for the synodic period, we use

Applying the Concepts 31. Setup: The difference between linear and logarithmic graphs is that linear graphs have axes that increase by

−1 1 1 1 1 = − . This leads to S = ⎛ 1 − ⎞ = 1.012 ⎝ S E P 84 ⎠ years, or ~366 days.

CHAPTER 3 Motion of Astronomical Bodies ◆ 25

Review: We can check the semimajor axis in the textbook’s appendices, and b­ ecause Uranus hardly moves in its orbit during one year, we expected a number just over one year.

1 1 1 35. Setup: From Working It Out 3.1, we use = − , S E P whereas Kepler’s Law is P 2 = A 3. In both cases, we are solving for P. Solve: (a) Using a synodic period of 1.278 years, we 1 ⎞ −1 = 4.597 years. find P = ⎛ 1 − ⎝ 1.278 ⎠

(b) By Kepler’s third law, P = A3/2 = 2.773/2 = 4.61 years. (c) The two answers are very close, less than 0.3 ­percent from each other.

Review: We expect the periods to match b­ ecause the two methods must give the same results to within rounding errors from the values of each variable used.

36. Setup: Kepler’s third law tells us that P2 = A 3. To test if this assertion is correct, we figure out what distance corresponds to a period of 3 years.

Solve: P2 = A 3 implies that A 3 = 32 or A = 3 9 = 2.08 AU. That is not 2 AU, although it is very close.

Review: What period corresponds to a planet at a distance of 2 AU? Using Kepler’s third law, we find P 2 = 23 or P = 8 = 2.83 years. Although close, that period is not 3 years as claimed, making the discovery suspect.

37. Setup: ­Here are the data we used: Io has P = 1.77 d, A = 0.421 Gm (giga-­meters). Europa has P = 3.55 d, A = 0.671 Gm. Ganymede has P = 7.14 d, A = 1.07 Gm. Callisto has P = 16.7 d, A = 1.88 Gm. As long as we use the same units, we can compute the ratio P2 = A 3 with the numbers given previously.

1.77 2 = 42.0. Europa has Solve: We find: Io has 0.4213 7.152 3.552 Ganymede has = 41.7. Callisto = 41.7.

1.07 3 0.6713 16.7 2 has = 42.0.

1.883 Review: ­These are all very close, considering ­there is gravitational interaction among the moons. We expected to find nearly the same ratio ­because all four moons, and indeed all orbiting satellites, w ­ ill obey Kepler’s third law.

38. Setup: To find the speed of the planet, we could divide distance by time, but be sure to convert distances to meters and months to seconds. Or we could notice that we are talking about distances traveled during the same 3-­months time period; thus, a larger distance means a correspondingly higher speed. For the second part,

recall Kepler’s second law, which tells us that planets move faster when they are closer to the Sun. Solve: ­Because the planet travels 65,000 km in 3 months rather than 30,000 km, we know the average 65,000 km = 2.17 times higher than speed must be 30,000 km the first time period. Thus, the planet travels 2.17 × 3.8 m/s = 8.2 m/s. The planet is closer to the Sun during this second period, as indicated by Kepler’s second law.

Review: To check this prob­lem directly compute the speed needed to travel the given distance in the given time. Given the distances and time (and using 30-­day months), we get a velocity of 3.9 m/s for the slowest period and 8.4 m/s for the fastest. That’s close enough.

39. Setup: No calculations needed for this prob­ lem. ­Because we are asked to compare what Mars sees of Earth to what Earth sees of Mars, we know the same ­things happen at the same time.

Solve: Mars undergoes retrograde motion to an Earth-­based observer ­every 25.6 months. So Earth also enters into retrograde motion to a Mars-­based observer ­every 25.6 months.

Review: Recall that two observers moving with re­spect to each other must agree on what they see each other ­doing.

40. Setup: Remember the semimajor axis is half the major axis. The time to go around the Sun is the period, which we find using Kepler’s third law P2 = A 3.

The eccentricity can be computed from using the distance from the center of the ellipse to the Sun. Solve: A sample image of this situation is shown in Figure IM 3.4. (a) The comet is moving fastest when it is closest to the Sun (at A, perihelion) and slowest when farthest from the Sun (at B, aphelion). (b) The semimajor axis is half of the full axis, or (1.24AU + 5.68AU)/2 = 3.46AU. This gives a period of P = A 2/3 = 3.463/2 = 6.44 year. (c) The distance from the Sun to the center is the semimajor axis less 1.24AU, or 3.46AU − 1.24AU ​ = 2.22AU. The eccentricity e × A = 2.22AU or 2.22AU 2.22AU e= = = 0.64.

A 3.46AU Review: Compare to Figure 3.13, where the orbital eccentricity of that example is about 0.77, determined by a rough count of the dashes of the semimajor axis and the distance between the center and the focus.

41. Setup: This is a question of relativity and reference frames that are not covered in this chapter. We sug-

26 ◆ CHAPTER 3 Motion of Astronomical Bodies

FIGURE IM 3.4 gest that material be reviewed as well as Newton’s first law.

Solve: (a) We see the other car approaching us at 90 km/h + 110 km/h = 200 km/h, while (b) the other driver sees us approaching at 110 km/h + 90 km/h = 200 km/h. In other words, both (a) and (b) see the same speed of 200 km/h.

Review: Remember that the total speed of two approaching objects is equal to the sum of the individual speeds, and ­because a + b = b + a, both ­drivers have to see the same total speed.

42. Setup: To compute the relative orbital periods of ­Vulcan and Mercury, we use Kepler’s third law, nor2

3

⎛ P ⎞ ⎛ A ⎞ malizing to Mercury’s orbit: ⎜ Volcan ⎟ = ⎜ Volcan ⎟ . ⎝ PMercury ⎠ ⎝ A Mercury ⎠ ­Doing some algebra, we solve for Vulcan’s orbital period and find 3

⎛ A ⎞ PVolcan = PMercury × ⎜ Volcan ⎟ . ⎝ A Mercury ⎠ Solve: ­Because Vulcan’s orbit is one-­fourth the radius of Mercury’s orbit, we can write the ratio between the 3

1 ⎛ 1⎞ two orbits and solve PVolcan = ⎝ ⎠ = PMercury or 4 8 11 days.

Review: Closer to the star means a shorter period.

Vulcan, if it ­were real, would orbit the Sun eight times for ­every one orbit of Mercury.

43. Setup: This prob­lem asks for acceleration and gives us mass and force. Use Newton’s second law: F = ma. For part (b), remember that speed equals acceleration times time or v = at.

F 100 N Solve: (a) a = = = 2 m/s2 m 50 kg

(b) Let’s find the time u­ ntil the refrigerator travels at v 10 m/s 10 m/s: t = = = 5s a 2 m/s2 Review: If the mass and force given are reasonable, then we can see it taking 4 to 5 seconds u­ ntil we could no longer keep up with the refrigerator.

44. Setup: To compute the acceleration of the astronaut, use Newton’s second (F = ma) and third laws. We could calculate the force on the satellite and then the acceleration of the astronaut directly, or notice that ­because they both experience the same force, (ma)satellite = (ma)astronaut, the astronaut’s acceleration ­w ill be greater by the ratio of the masses.

Solve: The astronaut’s acceleration is msatellite 5000 aastronaut = a = × 0.1 m/s2 = 5 m/s2 . mastronaut satellite 100 Review: To confirm, calculate the acceleration directly.

The force on the satellite is F = ma = (5000 kg) × (0.1 m​ /s2) = 500 N. The astronaut also feels 500 N, so her F 500 N acceleration is a = = = 5 m/s2 . m 100 kg 45. Setup: Remember that the semimajor axis is half the major axis. The time to go around the Sun is the period, which we find using Kepler’s third law P2 = A 3.

The eccentricity can be computed from using the distance from the center of the ellipse to the Sun. Solve: The figure ­w ill look similar to Figure 3.25, where both orbits are shown. (a) The major axis is 1.38 AU + 1.67 AU = 3.05 AU, and the semimajor axis is 3.05 AU/2 = 1.53 AU. (b) The distance from the center of the ellipse to the Sun is A less the 1.38 AU, or 1.53 AU − 1.38 AU = 0.15 AU. 0.15AU 0.15AU = = 0.098.

The eccentricity is e = A 1.53AU Earth’s orbit has an eccentricity of 0.017 so Mars’ orbit is more eccentric.

Review: The Origins section notes that Mars has an eccentricity of 9 ­percent, which we confirmed in this prob­lem.

Using the Web 46. Involves seeing which planets are observable in the sky during the current week as well as their current rising and setting times. Answers w ­ ill vary.

47.Involves looking up dates of opposition for the outer planets and observations of any planets in retrograde motion. Answers w ­ ill vary.

48. Relates to the findings from question 47. Answers ­w ill vary.

CHAPTER 3 Motion of Astronomical Bodies ◆ 27

49. Galileo’s telescope looks like long bamboo stick with a lens at each end. He also used a thermoscope, geometric and military compasses, a water-­lifting machine, and a microscope (to name just a few). He used a micrometer for mea­ sur­ ing the distances between Jupiter and its moons. He combined the distances of the moons from Jupiter and their orbital periods to show the moons obeyed Kepler’s laws. He worked from demonstrations, experiments, mea­sure­ments, and data, and he used the scientific method. His ­m iddle fin­ger is on display ­because “This item exemplifies the cele­bration of Galileo as a hero and martyr of science” https://­catalogue​.­museogalileo​.­it​/­object​ /­M iddleFingerGalileosRightHand​.­html.

50. a. ­There are over a dozen planets with masses between 0.8 and 1.2 Earth masses that have semimajor axes listed. Answers w ­ ill vary. b. Involves sorting planets by name and selecting a star with multiple planets to test Kepler’s laws. For example (to name just a few): Kepler-150 has 5 planets; Kepler-167 has 4; Kepler-20 has 6; Kepler-444 has 5; and, of course, Trappist-1 has 6. Answers ­w ill vary.

EXPLORATION This exploration ­w ill give students experience with Newton’s laws of motion by interacting with the simulation of Mercury’s orbit. B ­ ecause Mercury is the planet with the largest orbital eccentricity, the shift from Kepler’s laws to Newton’s becomes more apparent. It also introduces the concepts of vectors and the magnitudes of ­t hose vectors.

EXPLORATION SOLUTIONS 1. Mercury’s orbit is definitely an ellipse, with the Sun noticeably far from the center of the orbit.

2. Using the grid provided, we estimated the semimajor axis is about 7.8 units and the semiminor axis is about 7.6 AU 7.6, for a ratio of = 0.974.

7.8 AU

3. Using the eccentricity formula given in the Exploration, e = 1 − r 2 = 1 − 0.974 2 = 0.23.

4. The shade region from when we swept the first time (i.e., closest to the Sun) is yellow. The shade region from when we swept the second time (i.e., farthest from the Sun) is magenta. In the yellow region, we counted about 22.5 squares, and in the magenta region, we also counted about 22.5 squares. When we hit fractional squares, we tried to find another one that was shaded with the amount needed to make a full square. We found about the same total area per region, as we should ­because Kepler’s second law tells us that we should find equal area in equal time.

5. As we changed the eccentricity, the orbit becomes very elongated if e is close to 1 and very circular when e = 0, as expected.

6. The position of the planet on the graph of axis versus period does not change, which confirms that ­these orbital characteristics do not depend on the eccentricity (or shape) of the orbit, just on the semimajor axis.

7. As we made the semimajor axis smaller, the period decreased.

8. As we made the semimajor axis larger, the period increased.

9. This confirms that the larger the semimajor axis, the longer is the period of a planet.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 6: WORKING WITH KEPLER’S LAWS In this activity, students learn about Kepler’s geometric model of planetary orbits and learn to: • determine the properties of an ellipse.

• apply ­these properties to planetary orbits.

Text reference: Section 3.2

CHAPTER 4

Gravity and Orbits

INSTRUCTOR NOTES Chapter 4 covers the force of gravity and the physics of orbits. Major topics include: • Newton’s universal law of gravitation.

• orbital motion; gravity and orbits, shape of orbits.

• Kepler’s third law extended to a law of physics.

• why astronauts in orbit are not in “zero gravity,” but rather are in a continuous f­ ree fall.

• the cause of tidal forces and the effect on solid bodies.

The material in this chapter can pose considerable confusion to students, who find many applications of the physics to be quite counterintuitive. Why does the mass of the planet not affect the period of its orbit? How can two objects of unequal mass (that is, Earth and the Sun) exert the same gravitational force on each other? How is a bound orbit (that is, g­ oing around something) a perpetual state of ­free fall (that is, g­ oing ­toward something)? Also, the nomenclature can be somewhat confusing, as Kepler’s laws ­were first empirical discoveries ­limited to our Solar System, but thanks to Newton’s work, they are considered universal laws of physics.

As with the previous chapter, we have found that students need extra time to work through t­ hese concepts. You can easily devote an entire class to Section 4.1 on gravity, a full class period on the physics of orbits in Section 4.2, and another full class on the physical (as opposed to empirical) form of Kepler’s third law. Although one can pre­sent Newton’s universal law of gravitation and do a sample prob­lem in about 5 minutes, ­there are a large number of nuances that we might take for granted but which students need time and guidance to master, such as how surface gravity changes with planet mass and size; how to rank the strength of gravity between bodies of dif­fer­ent size and mass; how the net strength of gravity on one object changes with distance as you move it from one body to another (e.g., a space probe as it travels from Earth to the Moon). When it comes to orbits, most students believe orbits must close on themselves and have a hard time visualizing a crash or a parabolic trajectory as an orbit. ­There are many supplemental explorations and simulations that are helpful for ­these topics, as well as exercises 28

to be found in the Learning Astronomy by D ­ oing Astronomy workbook.

DISCUSSION POINTS • Discuss how the laws of physics presented in this chapter could be used to learn about the masses and/or sizes of objects other than the Sun, such as moons of planets or other stars. (LG 1) ­ ere to remain exactly • Think, pair, share: If Earth w the same size and mass, but was placed at a dif­fer­ent place in the Solar System, how would the gravitational force between it and the Sun compare to what it is now? Earth’s distance now is 1 AU. Figure out how the Sun’s gravitational force that it experiences would change at 2 AUs, 8 AUs, and 40 AUs. How does the inverse square law of gravity affect the gravitational environment of the Solar System? (LG 1) • How could astronomers use their knowledge of orbits to navigate the Solar System? Discuss this in the context of bound and unbound orbits, as well as an acceleration given by the gravitational force. (LG 2) • Discuss the differences between tidal effects leading to ocean bulges and t­ hose that stretch the solid parts of a planet. What happens to a planet ­because the solid parts are not as f­ ree to move as the liquid parts? What does this mean for an object that experiences a large tidal force but not one large enough to break it apart?

(LG 3, LG 4) • The opening image for this chapter shows the International Space Station (ISS) with a very blue Earth and white clouds b­ ehind it. The question is, “What keeps a space station in orbit?” Ask that question to the class and give students a few minutes to write down their answers, join with o­ thers in small teams, and put together a team statement. Another suggestion would be to have them write a 1-­m inute essay answering that question, collect the answers (or set up an online quiz), and check what they learned at the end of the term.

CHAPTER 4 Gravity and Orbits

PROCESS OF SCIENCE

ASTRONOMY IN ACTION VIDEOS

Universality: The laws of physics are the same everywhere and at all times. Galileo claimed that for all objects on Earth; Newton extended his laws to all of the Solar System. We know that a rock and a feather ­will not fall at the same rate on Earth due to the greater air re­sis­tance felt by the feather.

A hammer and a feather ­were dropped on the Moon as part of the Apollo 15 mission and hit the surface at exactly the ­ nder universame time. Galileo was right (Newton too)! U sality, we can make predictions about very distant worlds.

CENTER OF MASS

ASTROTOUR ANIMATIONS NEWTON’S LAWS AND UNIVERSAL GRAVITATION This interactive animation uses a classic thought experiment to show how force (an explosion of gunpowder) is used to move an object (a cannonball). Students can control the amount of gunpowder the gunner puts into the cannon to see what happens as increasing amounts of gunpowder are used. It briefly explores the connection between this thought experiment and the real-­world actions of satellites and space probes, as well as what would change if one fired a cannon on the Moon or on Jupiter. This classic thought experiment is extremely effective in helping students see how orbits “work.”

Text reference: Section 4.2 ELLIPTICAL ORBITS This animation shows how increasing the speed of a planet in a circular orbit results in an elliptical orbit. Higher speeds result in more elongated elliptical orbits.

Text reference: Section 4.2 TIDES AND THE MOON This AstroTour starts with the question, “But how are tides formed?” The answer lies with gravity and the interaction between Earth and the Moon. Section 2 explains self-­g ravity, why we are pulled ­toward the center of Earth, and why t­ here is no gravitational force at the center. The explanation of tidal forces on a solid body and the definition of tidal stress are clearly presented; that is, objects of real physical size experience tidal forces due to the inverse-­square dependence of gravitational force with distance. The interplay of the gravity of the Sun and that of the Moon is given as the explanation for Earth experiencing both high and low tides.

Text reference: Section 4.3

◆ 29

Dr. Palen shows how we each use our own center of mass to balance, and how two ­people can spin around a common center of mass by linking hands. (Length: 2:19) Text reference: Section 4.2 TIDES Stacy Palen uses a loose jacket, her nose, and a ball representing the Moon to demonstrate why most locations on Earth usually experience two high and two low tides during a day. (Length: 3:26) Text reference: Section 4.3

INTERACTIVE SIMULATIONS PLANETARY ORBITS This interactive simulation is included as part of the Exploration for this chapter. It expands on the use of the simulator to explore Kepler’s laws for Mercury by adding in the Newtonian features of Mercury’s orbit.

Text reference: Section 3.2, Ch3 Exploration, and Ch4 Exploration TIDAL BULGES This simulation shows how the effects of the Moon and the Sun combine to create stronger and weaker tides at different times of the lunar cycle. The simulation also shows that Earth’s rotation is faster than the speed of the Moon in its orbit, so the tidal bulge is dragged slightly ahead of the Moon’s position. Students can add or subtract the effect of the Sun’s gravity and the effects of Earth’s rotation. This simulation shows how the effects of the Moon and the Sun combine to create stronger and weaker tides at different times of the lunar cycle. The simulation also shows that Earth’s rotation is faster than the speed of the Moon in its orbit, so the tidal bulge is dragged slightly ahead of the Moon’s position. Options are provided so that students can start, pause, or reset the animation during their analysis of tides.

Text reference: Section 4.3

TEACHING READING ASTRONOMY NEWS 1. Light travels at a finite speed, so the light we receive ­today from a distant supernova was emitted a long

30 ◆ CHAPTER 4 Gravity and Orbits time ago. If the supernova happened in a galaxy 5 billion light years away, then we are seeing it as it was 5 billion years ago, not as it is t­ oday.

2. The scientific method requires testing, even of universal constants. Additionally, we must test w ­ hether the constant is actually “universal” as Newton proposed.

3. If G changes over time, then all forces of gravity would as well, namely, orbital periods and distances, tidal forces, and sizes of self-­gravitating objects (from asteroids to superclusters).

4. In­de­pen­dent tests are needed to verify a theory; other­w ise one can only state that the theory worked ­under narrow circumstances.

5. If the Sun ­were closer, tidal forces would be stronger.

Tidal force is proportional to M 3 . To calculate how d much of an effect a changing Sun-­Earth distance has compared with a changing Moon–­Earth distance, consider the ratio of t­ hese. For the Earth-­Moon system: M Moon 7.35 × 1022 kg = = 1.3 × 106 kg/km 3 .

3 (3.84 × 105 km)3 dMoon For the Earth-­Sun system: M Sun 1.99 × 1030 kg = = 6.0 × 105 kg/km 3 .

3 (1.49 × 108 km)3 dSun Thus, we see the Moon is currently more impor­tant to our tides by over a f­actor of 2 times. A slight reduction in the distance to the Sun would make the tides slightly stronger. If the distance to the Moon ­were one-­third of what it is now (its estimated distance right a­ fter it formed), we would have: M Moon 7.35 × 1022 kg = = 3.5 × 10 7 kg/km 3 , or over 3 (1.5 × 105 km)3 dMoon 25 times stronger tides than we have now.

CHECK YOUR UNDERSTANDING SOLUTIONS 4.1A. (d) The force of gravity scales as d-2 (inverse-­square law).

4.1B. (c) The force of gravity between the centers of two objects does not depend on the size of the object. (a) If Earth’s radius ­were smaller, the acceleration due to gravity would increase, making every­thing weigh more.

4.2 (c, a, b) is the ranking from smallest orbital velocity to largest. The orbital velocities are proportional to M/r, where M is the mass of Earth, and r the distance to the object. The object farthest away ­w ill have the smallest circular velocity; the closest one, the greatest.

4.3. (d, b, a, c) Weakest to strongest. Earth is closest to the Sun in January and farthest away in July. The Sun’s strongest tidal force—­the spring tides—­will thus be strongest in January and weakest in July. The full Moon in January (c) w ­ ill be the strongest, followed by the new Moon in July (a). The neap tide in July (b) w ­ ill be slightly stronger than in January (d) b­ ecause the solar tidal force ­will be smaller. It is hard to quantify this answer b­ ecause so many other f­actors enter into the “strength” of any given tide.

4.4. (a) Tidal locking occurred as a result of tidal forces between Earth and the Moon.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (a) Newton’s law of gravity states that the force is proportional to the product of the two masses and inversely proportional to the square of the distance.

2. (d) This corresponds to Kepler’s third law as well as the circular velocity at a given distance.

3. (c) This is the definition of bound and unbound orbits.

4. (d) Mass is in­de­pen­dent of the gravitational field. It is an intrinsic property of an object.

5. (c) Plugging the mass and radius of Mars into Newton’s universal law of gravitation shows that the gravitational force on Mars is smaller than that on Earth.

M 6. (d) The scaling of gravity depends on Fg ∝ 2 . If r Venus has a mass 0.8 times that of Earth and a radius 0.95 times that of Earth, your weight ­ w ill be 0.8 = 0.89 times that on Earth, or about 10 ­percent 0.952 less.

7. (c) Gravity depends on mass; thus, probing the effects of gravity from an object (i.e., orbits) allows one to calculate the mass of an object.

8. (a) Tidal force depends on the mass of the object ­c reating the tide. This ­w ill make high tides higher, removing more ­water from the low tide areas.

9. (c) Tidal force depends on the size of the object feeling the tide, so if Earth became smaller, high tides would be lower, meaning low tides would be higher.

10. (b) The strength of gravity depends on 1 2 . If the d Moon w ­ ere 1/2 the distance away, the force would be 1 = 2 2 = 4 times stronger.

1 2 2

( )

CHAPTER 4 Gravity and Orbits

11. (c) The strength of tidal forces depends on 1 3 . If d the Moon w ­ ere 1/2 the distance away, the tidal force 1 would be = 2 3 = 8 times stronger.

1 3 2 12. (d) If both objects are tidally locked, then they orbit and rotate about their axes with the same period, so (a) and (b) are both correct.

13. (b) Spring tides are the strongest and require the Sun and the Moon to be aligned.

14. (d) The Roche limit is the point at which an object that is held together by its self-­g ravity can no longer support itself against the tidal pull of what­ever it is orbiting.

15. (c) Self-­g ravity is the total gravitational force felt by an object ­because of all of its constituent parts.

( )

Thinking about the Concepts 16. The fundamental difference is that Kepler’s laws are empirical laws. They ­were deduced purely from observation of the motions in the Solar System. Newton’s laws are theoretical laws that describe the general be­hav­ior of all motions in the universe. A large mea­sure of our confidence in Newton’s general laws is their ability to reduce to the more specific cases of Kepler’s laws.

17. Circular velocity is the speed needed to orbit a body in a circular path. Escape velocity is the speed needed to escape the gravitational force of a body.

Escape velocity ­w ill always be larger ­because more energy is needed to escape the gravitational force of an object.

18. Mass is a mea­sure­ment of an object’s inertia. Weight is how strongly gravity acts on the object. Mass is an internal property of an object and never changes.

Weight depends on the strength of the gravitational field, and in some cases it can be zero.

19. The constant of proportionality is the universal gravitational constant, G. The gravitational accelerations, g, are dif­fer­ent ­because the Moon’s size and mass are dif­fer­ent from Earth’s size and mass. Earth’s gravitational acceleration at its surface is 9.81 m/s2 , whereas the Moon’s gravitational acceleration at its surface is 1.67 m/s2 .

20. The comet in an elliptical orbit ­w ill eventually return to the immediate vicinity of the Sun, depending on the size of its orbit. The comet in an unbound orbit ­w ill escape the Sun entirely and head out into interstellar space, never to return.

21. Earth rotates the fastest at the equator; thus, launching satellites from ­here gives them the highest initial speed. Launches are always to the east in order to

◆ 31

u­ tilize the inherent inertia due to Earth’s rotation.

Launching ­toward the west would work against this natu­ral slingshot effect.

22. The period and distance of an orbiting object yield, via Newton’s formulation of Kepler’s third law, the sum of the masses of the object being orbited and the object that is orbiting. In the case in which the central object is much more massive than the orbiting object (i.e., a planet around a star), the smaller object can be ignored; thus, the orbital information yields the mass of the central object.

23. If an object is approaching the Sun in an unbound orbit, then e­ ither (a) the object came from outside the Solar System or (b) the object experienced some very intense gravitational acceleration from very close encounters with a massive object such as Jupiter.

24. As you move out from the center of Earth (where the force of gravity equals zero), t­ here is more mass interior to your radius. ­Because this amount of mass is what c­ auses gravity, the strength of gravity increases.

25. The ocean reaches its maximum recession from the beach during the low spring tides that occur at the new and full Moon phases. High spring tides occur near noon and midnight when the beach lies close to the Sun-­Earth-­Moon line (see Figure 4.16). Therefore, low spring tides occur approximately 6 hours ­earlier or l­ ater, near sunrise and sunset. The best time to go digging for clams is near sunrise or sunset during the new and full Moon phases.

26. Due to the rotation of Earth dragging the tidal bulge around, ­there is a lag between when the Moon is on the meridian and when a high tide actually occurs.

See Figure 14.12.

27. The Moon attracts mass on the near side of Earth more strongly than it attracts mass at the center of Earth ­because of the inverse-­squared relationship for gravity. For the same reason, the Moon also gravitationally attracts mass at the center of Earth more strongly than it attracts mass on the far side. As the near side of Earth is pulled away from the center, we intuitively see the near side bulge. However, b­ ecause the center of Earth is also being pulled away from the far side, we see a less intuitive second bulge along the Earth–­Moon line as well. Given that the tidal bulges are farther from Earth’s center than average, they represent low spots in Earth’s gravitational field. Therefore, ­water preferentially flows ­toward the centers of both bulges, raising sea levels on both the near and far sides of the planet si­mul­ta­neously.

28. Solid bodies can be tidally distorted, and tidal distortion occurs on Earth. That distortion of the planet’s landmasses is dissipated in frictional heating of rock.

32 ◆ CHAPTER 4 Gravity and Orbits 29. Oceans cover 70 ­percent of Earth’s surface area, so a 1-­meter increase in sea level in the ocean represents a tremendous volume of ­water. If that large volume of ­water is squeezed into a smaller area along a continental shoreline, the subsequent rise in the local sea level ­w ill be substantially higher than 1 meter. Combine that situation with receding depth as ­water approaches shorelines and resonant effects like t­ hose seen in the Bay of Fundy, and some places on Earth can experience 5-­to 10-­meter increases in sea level at high tide.

30. The Roche limit is the distance from a planet at which objects with ­little to no mechanical strength are ripped apart by tidal stresses. Small objects like communications satellites have a high intrinsic mechanical strength b­ ecause of the chemical bonds that hold them together. However, larger objects like planetesimals tend to be held together only by self-­g ravity.

Furthermore, the tidal stresses inside a small communications satellite are minimal ­because the near and far sides of the satellite are practically the same distance from the center of Earth. Meanwhile, the tidal stresses inside planetesimals can be enormous ­because tens or even hundreds of kilo­meters may separate the near and far sides of a planetesimal.

Applying the Concepts 31. Setup: Newton’s law of gravity tells us that the acceleration of gravity (g) scales with the mass (M) of what makes the gravity and how far (R) you are from it; that is, g ∝ M/R 2 . We can compute the ratio of acceleration on Mars and Earth as

g Mars R2 M = Mars Earth .

2 g Earth M Earth R Mars

Note that weight is your mass times gravitational acceleration.

Solve: Mars has 0.1 times the mass of Earth and half Earth’s radius; that is, Earth is twice as large.

Using ­t hese values in our equation above, the ratio of acceleration is (0.1) × (22) = 0.4, that is, gravitational acceleration on Mars is about 40 ­percent that on Earth. This means we would weigh 40 ­percent less than on Earth, but note that our masses would not change ­because mass is an intrinsic property.

If Earthlings invading Mars weigh less but have the same strength, they would be able to jump twice as high and lift objects that w ­ ere twice as massive. Hollywood appears to miss the mark in their movies.

Review: It is the combination of its smaller mass and radius that makes the gravity on Mars so much less than that on Earth.

32. Setup: As hinted in the prob­lem, distance equals rate times time or d = vt, where the distance is the circumference 2πr of the orbit.

Solve: Notice that the prob­lem gives Earth’s speed in kilo­meters per second (km/s), so we r­eally should convert years to seconds: v=

1yr 4.5 × 10 9 km d = 2π × = 5.4 km/sec.

3.15 × 10 7 s 164.8 yr t

Review: We look this up in the Appendix and find the reported orbital speed is 5.4 km/s.

33. Setup: We can write the circular velocity for any object in orbit as vcirc =

GM primary r

=

v2 GM Sun r or M Sun = Venus circ . rVenus G

Solve: Inserting values from the text a­ fter changing km to m and km/s to m/s, and then solving for the mass of the Sun: km 1.082 × 108 km = 1.082 × 1011 m and 35.03 s 10 4 m = 3.503 × . s (1.08 × 1011 m) × (3.503 × 10 4 m/s)2 M Sun = 6.67 × 10 −11 Nm 2/kg 2 = 1.99 × 1030 kg.

Review: This is exactly the mass of the Sun listed in this text.

34. Setup: The escape velocity from any planetary sur2GM planet face is expressed as vesc = . ­Because we rplanet are looking to compare the asteroid’s escape velocity v to that of Earth’s, we w ­ ill take a ratio esc , asteroid . This vesc , Earth w ill also have the nice result of cancelling ­ ­common ­f actors: Simplifying yields vesc , asteroid vesc , Earth

=

M asteroid rEarth .

M Earth rasteroid

Solve: Substituting numbers into the equation (remembering that they must be in Earth units), vesc , asteroid 10 −12 1 = = 10 −8 = 10−4 .

1 10−4 we obtain. vesc , Earth The escape velocity from the asteroid is only 10 −4 times Earth’s escape velocity, that is, (10 −4) × (11.2 km/s) × (1,000 m/km) = 1.12 m/s.

Review: This converts to about 2.5 mph. Even a good jump ­w ill launch you off the asteroid. This stands to

CHAPTER 4 Gravity and Orbits

◆ 33

reason considering how so much smaller in mass and another way, F = ma, where a is the acceleration due to radius this asteroid is.

M gravity, a = G 2 . Do not forget to use the correct 35. Setup: Although this question involves an orbit, we R can solve it using our familiar formula that distance units, that is, mass in kilograms and distance in meters. equals rate times time, or d = vt. We are looking for Solve: Setting the two forces equal, we find time, we know the speed, and we just need the dism 3 ⎞ 5.97 × 1024 M ⎛ Mm tance, which is the circumference of Earth: F = G 2 = ma or a = G 2 = ⎜ 6.67 × 10 −11 kg⋅s2 ⎟⎠ (6.37 × 106 R R ⎝ C = 2πR = 2 × π × 6378 km = 40131 km.

24 3 d 40131 km a = G M = ⎛ 6.67 × 10 −11 m ⎞ 5.97 × 10 kg = 9.81m/s2 , as expected. = 5080 s, 2 or ⎜ about 85 Solve: t = = kg⋅s2 ⎟⎠ (6.37 × 106 m)2 R ⎝ 7.9 km/s v minutes.

Review: Notice that the acceleration of gravity Review: It takes the International Space Station depends on the mass and radius of the planet. Indeed, about 90 (92.49) minutes to orbit Earth at an average a planet with a small radius and large mass could have altitude of 400 km, and an orbital speed of 7.67 km/s, its acceleration due to gravity much higher than what consistent with our calculations. we feel on Earth.

36. Setup: The circular velocity of an orbit is 38. Setup: We could compute the ­actual force of gravity acting on the rocket at both locations or just recogGM vcirc = and the escape velocity is vesc = 2vcirc .

M r nize that gravity scales with distance as Fg ∝ 2 , so d We must travel above the escape speed to leave Earth’s w ­ ill the ratio of force on the ground to force in orbit orbit and travel to Mars.

2 Solve: (a) If the spacecraft is orbiting at its circular Fground ⎛ dorbit ⎞ =⎜ velocity, then it is orbiting Earth in a circular orbit. ⎟ . be Forbit ⎝ dground ⎠ (b) The spacecraft must be sped up by the difference between the escape velocity needed and the circular Solve: Using numbers given in the text, velocity it already has. We show one set of logical 2 Fground ⎛ 6370 km + 350 km ⎞ steps that students might take. =⎜ ⎟⎠ = 1.11. Gravity on 6370 km Forbit ⎝ 2GM GM Start with: vesc − vcirc = − . the ground is about 11 ­percent greater than when the r r rocket is in orbit.

Divide both sides by vesc and carry through the Review: We know that gravity drops with altitude, calculations: and 350 is about 5 ­percent of 6,370, so we expect the change in gravity to be 1.052 = 1.10, as we found.

GM 2GM − GM vesc − vcirc r =1− 1 , r and 39. Setup: Circular velocity vcirc = = then mulr 2GM vesc 2 M = 5.97 × 10 24 kg, r = 6,370 km + 350 km = ​ r 6.72 × 10 6 m. tiply both sides by vesc. We find that Solve: Inserting t­ hese values into our formula: 1 ⎞ ⎛ vesc − vcirc = ⎜ 1 − ⎟ v = 0.29vesc . ⎛ m 3 ⎞ 5.97 × 1024 kg m ⎝ 2 ⎠ esc vcirc = ⎜ 6.67 × 10 −11 = 7698 ≈ 7.7 km 2⎟ 6 kg⋅s ⎠ 6.72 × 10 m s ⎝ Review: In many cases, we launch space probes 24 3 ⎞ in ⎛ 5.97 × 10 kg m m −11 6.67 × 10fuel = 7698 ≈ 7.7 km / s. multiple steps, so thatvcirc we =can⎜conserve and use kg⋅s2 ⎟⎠ 6.72 × 106 m s ⎝ gravity to help the ship travel faster. In this case, we could put the probe into a lower-­earth orbit, and give it a smaller push for it to escape Earth’s gravity.

37. Setup: To solve this prob­lem, set the gravitational force from Newton’s universal law of gravitation, Mm F = G 2 , equal to the force you would feel standR ing on Earth with mass M and radius R. Stating it

Review: Wikipedia gives an average height of the ISS as ~404 km, and an orbital speed of 7.67 km/s, consistent with the lower altitude number we used ­here.

40. Setup: Start with the last equation in Working It Out M 4.1: = G 2 .

R

34 ◆ CHAPTER 4 Gravity and Orbits Solve: This is exactly the same prob­lem as prob­lem 37:

m3 ⎞ M ⎛ a = G 2 = ⎜ 6.67 × 10 −11 × 3 ⎞ 5.97 × 1024 kg ⎛ m M kg⋅s2 ⎟⎠ R 2 ⎝ g = G 2 = ⎜ 6.67 × 10 −11 = 9.81 m/s . kg⋅s2 ⎟⎠ (6.37 × 106 m)2 R ⎝ 5.97 × 1024 kg = 9.78 m/s2 ⎞ 5.97 × 1024 kg 6 2 (6.371 × 10 m + 3048 m) 2 ⎟⎠ (6.37 × 106 m)2 = 9.81 m/s .

Review: The value of g varies slightly across Earth’s surRemember that the acceleration of gravity acts on all face, ranging from 9.78 m/s2 at the equator to 9.83 m/ objects the same way, so we feel the gravitational s2 at the poles; 9.81 m/s2 is an average of this range. acceleration, g, of 9.78 m/s2 .

41. Setup: Tidal force is directly proportional to mass (b) We ­ w ill feel exactly the same acceleration g and inversely proportional to the cube of the distance ­whether one instructor is strapped to us or not. between the objects. Writing this mathematically, we (c) Using Newton’s second law, the force of gravity on M me w ­ ill be F = mg = 70 kg × 9.78 m/s2 = 684 N. have Ftidal ∝ 3 . d (d) If we ­were strapped to instructors, the total Solve: Substituting numbers into this equation and mass ­w ill be doubled so the total force ­w ill be as taking the ratio of the tidal forces, we find well. Our acceleration, however, ­ w ill still be 2 3 . g = 9.78 m/s FJupiter ⎛ 1.9 × 1027 kg ⎞ ⎛ ⎞ 1m Review: This prob­lem highlights the very impor­tant =⎜ = 5 × 10−11.

80 kg ⎟⎠ ⎜⎝ 7.8 × 1011 m ⎟⎠ Fdoctor ⎝ fact that the acceleration of gravity on dif­fer­ent objects does not change, but the force that they feel does ­because Take the inverse of this and get it depends on their mass.

Fdoctor 44.

Setup: This question is asking about orbits, so = 20 billion.

FJupiter chances are we ­w ill need Kepler’s third law. However, it mentions a star other than the Sun, in terms of the Review: The tidal force from Jupiter is incredibly tiny Sun’s quantities, so we need it in its fully expanded compared with the tidal force from the doctor. In fact, form P2 = A 3/M, where M is the mass of the star as inverting this ratio shows that the tidal force from the mea­sured in solar masses; its planet has an insignifidoctor at your birth is approximately 20 billion times cant mass in comparison, so we ignore it, along with stronger than that from Jupiter!

4π 2 42. Setup: We need to use Kepler’s third law in its full ­because we are taking ratios and this quanthe G expression, but ignoring the very small mass of tity equals one when we are using solar masses and 4π 2 A 3 . ­Dactyl: P 2 = AUs as the units.

G M Solve: (a) Using Kepler’s law, where the distance A is Solve: The orbital separation of Ida and Dactyl is still 1 AU and M = 2 solar masses, we find the period is A = 90 km = 9 × 104 m. Adjusting units and plugging into A3 13 1 P= = = = 0.71.

4 3 2 (9 × 10 m) 4π M 2 2 5 = 1.01 × 10 s, our formula: P = m 3 4.2 × 1016 kg (b) For this planet to complete an orbit with the same 6.67 × 10 −11 kg⋅s2 circumference as Earth but in 0.71 times the time, it (9 × 104 m)3 must be traveling that much faster than Earth. 0.71 is = 1.01 × 105 s, or about 28 hours.

16 a ­factor of 1/0.71 = 1.4 times smaller than Earth’s 4.2 × 10 kg Review: We expect a very short period considering 2 period, so the planet travels 1.4 times faster than the low mass and close orbit.

Earth. Earth travels at 29.8 km/s, so this planet trav43. Setup: This prob­lem is asking about gravitational els 1.4 times that, or 41.7 km/s. ­ ill need Newton’s second law: acceleration, so we w Review: Notice that the previous solution, using proF = ma. Skydivers usually jump at altitudes of 10,000 portional reasoning, saved us from a ­g reat deal of cal–­12,000 feet (3048 – 3658 meters), which adds very culation. Do the numbers make sense? Vega is more ­l ittle to the radius of the Earth. massive, so the planet must orbit faster ­because the Solve: (a) We ­w ill calculate the gravitational accelerstar’s gravity is higher, and thus, the circular velocity ation for a drop height of 3048 meters. Using the must be higher as well.

1 mean radius of Earth, this gives us an acceleration 45. Setup: Tidal force scales as Ftidal ∝ 3 . when leaving the plane of d

CHAPTER 4 Gravity and Orbits

Solve: If the Moon w ­ ere 0.8 its current distance then 1 = 1.95 the strength of the tidal force would be 0.83 times stronger. With a stronger gravitational pull, the spring tide would be significantly higher than it is ­today, and the neap tide would be a l­ ittle higher.

Review: One could always use the equation of tidal force in Working It Out 4.4 to calculate the strength of tidal force with the Moon at its current distance and again at 0.8 times that value; however, noting that none of the other par­ameters change except d, it seems like more work than needed.

Using the Web 46. The hammer and feather drop on the Moon showed that Galileo was right! Objects fall at the same acceleration. ­Unless it ­were carried out in a vacuum chamber, the experiment would not work on Earth ­because Earth has an atmosphere that interferes with the feather’s fall. A hammer and a big nail would fall with the same acceleration, depending on their surface areas and the orientation when dropped. Objects fall on the Moon with 1/6 of Earth’s acceleration.

47. Our weights would be higher on Jupiter, Saturn, the Sun, a white dwarf, and a neutron star. The difficulties to list for h­ uman bodies u­ nder a higher-­g ravity environment ­w ill vary depending on how high the higher-­ gravity environment is. Pos­ si­ ble short-­ term and long-­term effects of lower gravity on the h­ uman body might include less stress on the skeleton, ankles, knees; ­humans might grow taller overall. Answers ­w ill vary as to what types of life might adapt well to a dif­fer­ent gravity. How dif­fer­ent ­w ill the force of ­g ravity be?

48. The tennis ball appears to float when dropped at the top of the roller coaster ­because it is in ­free fall along with the riders. Newton extended the physics of Earth to the Moon in its orbit; it is a short jump to the cannonball example. The cannonball needs to fall just exactly as much as Earth’s surface curves away. The technical difficulty in launching a satellite was getting enough thrust to overcome Earth’s gravity.

49. Involves analy­ sis of tides at specific locations.

Answers w ­ ill vary.

50. The tidal tails result from the mutual gravitational pull of the two galaxies on the dust and gas in each galaxy. While the stars might pass right through each galaxy, the gas and dust in each galaxy ­w ill be affected.

◆ 35

EXPLORATION This exploration ­w ill give students experience with Newton’s laws of motion by interacting with the simulation of ­ ecause Mercury is the planet with the Mercury’s orbit. B largest orbital eccentricity, the shift from Kepler’s laws to Newton’s becomes more apparent. It also introduces the concepts of vectors and the magnitudes of t­ hose vectors.

EXPLORATION SOLUTIONS 1. Mercury is farthest from the Sun, at aphelion, when its acceleration is smallest.

2. Mercury is closest to the Sun, at perihelion, when its acceleration is largest.

3. The largest and smallest accelerations are 0.0628 and 0.0272 m/s2 , respectively.

4. The acceleration vector always points ­toward the Sun. 5. The force on the planet is always ­toward the Sun ­because Newton’s second law says that F = ma (i.e., acceleration and force always point in the same direction).

6. Mercury moves the slowest when it is farthest from the Sun. 7. Mercury moves the fastest when it is closest to the Sun. 8. The largest and smallest speeds are 59.0 and 38.9 m/s.

9. The ­angle between the acceleration and velocity vectors is not always 90°.

10. If the orbit becomes a perfect circle, then that ­angle becomes 90°.

11. When A is changed to 0.8 AU but the eccentricity is left alone, the orbit becomes larger and the period increases, but the shape of the orbital path remains the same. The planet’s highest speed is about 41 km/s.

12. When A changes to 0.1 AU, the opposite happens from question 11; namely, the planet is zipping around the Sun now at a maximum speed of 116 km/s.

13. Answers w ­ ill vary. As we played with the value of A, we confirmed that when planets are closer, they move faster, and when they are farther, they move slower.

When the planet is 0.1 AU from the Sun on a circular orbit, its speed is 94.2 km/s, at 0.2 AU it is 66.6 km/s, at 0.4 AU it is 47.1 km/s, and at 0.8 AU it is 33.3 km/s.

Using a scientific calculator, we find the best fit is A ⎞ −1/2 v = 29.8 km/s × ⎛ , which shows the de­­ ⎝ 1 AU ⎠ pendence on A as expected.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES None for this chapter.

CHAPTER 5

Light

INSTRUCTOR NOTES Chapter 5 covers the nature of light and ­matter. Major topics include: • the nature of light as a bringer of information and energy.

• the particle and wave nature of light.

• the electromagnetic spectrum.

• the quantum-­mechanical model of atoms (energy levels, emission and absorption).

• the Doppler effect and Doppler shifts.

• the role of temperature in determining luminosity and color.

• Stefan-­Boltzmann and Wien’s laws.

• the inverse-­square law and the luminosity-­d istance-­ brightness relationship.

Some of this material is quite intuitive, such as the inverse square law of brightness, as every­one knows that lights appear dimmer when they are farther away. The relationships between color and temperature or among luminosity, temperature, and size, on the other hand, can be much more problematic. Students often assume that “more” always requires “bigger,” so it can be quite counterintuitive that a hotter object is bluer or that a cooler object can be more luminous than a hotter one if the sizes are correct.

A few other topics have consistently been challenging for students. First, the notion that all wavelengths of light travel at the same speed in a vacuum is quite counterintuitive, especially when we add the fact that bluer light is more energetic. It seems that students are trying to equate higher energy with faster speed. Second, color and temperature can be very challenging: Our w ­ ater faucets teach us that red is hotter than blue, and we generally talk about metal as being “red hot” or “white hot” rather than “blue hot.” The differences between luminosity, flux, and brightness are often difficult for students to understand as the terms apply to light. Working It Out 5.3 helps students differentiate between flux and luminosity, and Section 5.5 introduces brightness. Most spectra described in books show only hydrogen, which can lead students to think that all atomic spectra look like hydrogen. It is useful to point 36

out that each atom has its own complicated series of energy levels, or spectral fingerprints.

DISCUSSION POINTS • Light behaves as an electromagnetic wave that has variable amplitude and frequency, but, in a vacuum, it always travels at a constant speed, c, which is a universal constant. Discuss how this fact constrains our ability to obtain information from distant astronomical objects and how it limits the possibilities of interstellar travel. (LG 1) • ­After covering Section 5.1, have students list instances where light behaves as a particle and where light behaves as a wave. What determines the be­hav­ior of light? Students compare lists and come up with a “master” list. (LG 1) • Using the relation between the frequency of the electromagnetic wave and the energy carried by each photon (Section 5.1). Discuss what types of natu­ral pro­cesses could emit t­ hose dif­fer­ent types of energies.

(LG 1) • Have students compare the energy levels of an atom with the bookshelf analogy. The energy levels of atoms are unique to each ele­ment. Are the levels of all bookshelves unique? (LG 2) • Have students compare the structure of an atom with that of the solar system. Discuss the differences and similarities. (LG 2) • Think of dif­fer­ent situations in which the Doppler shift of sound occurs and what we have taught ourselves to learn from them (for example, we know ­whether a motorcycle or emergency vehicle is approaching or receding based on its pitch). Discuss ways in which the Doppler shift of sound can help us understand the Doppler shift of light. (LG 3) • Divide the class into a c­ ouple of groups. Have one group prepare a short summary that compares the physical situations that lead to blackbody radiation versus emission lines. Have the other group prepare a short summary of the physical situations that lead to emission versus absorption lines. Groups then share

CHAPTER 5

their summaries. Having students make large posters (18″ × 24″ is usually large enough) and display them ­w ill serve as a valuable reminder of t­ hese critical concepts. (LG 4) • Let students choose between explaining what is meant by luminosity or what is meant by brightness, and assign a short (1–2 minute) writing exercise for them to rec­ord their explanations. Then, have them literally illustrate how distance affects their choices. This could be via sketches or demonstrations. (LG 5) • The chapter-­opening image question for this chapter is, “What is light?” How many dif­fer­ent experiences have students had where dif­fer­ent wavelengths of light are involved? What definitions of light arise from each experience?

PROCESS OF SCIENCE Astronomy and physics have dif­fer­ent fields of research as well as dif­fer­ent emphases in similar fields. The value of the fundamental constant of the speed of light in a vacuum was first estimated from astronomical observations. Light as an electromagnetic wave and the fact that its speed is the same for all observers came from theoretical physics. Dif­fer­ent prob­lems in dif­fer­ent fields came up with identical results: light has a speed that can be mea­sured.

ASTROTOUR ANIMATIONS LIGHT AS A WAVE, LIGHT AS A PARTICLE Is light a wave, a disturbance that travels from one point to another? Or is light a particle, a discrete package of energy?

This interactive animation defines the basic ele­ments of an electromagnetic wave and a photon, linking the concepts of wavelength and energy through the use of a wave whose wavelength and frequency can be adjusted. Students should be able to answer the initial questions ­after viewing this video.

Text reference: Section 5.1

Light ◆ 37

ATOMIC ENERGY LEVELS AND LIGHT EMISSION AND ABSORPTION This animation explores the atom in connection with emission and absorption of light, linking the frequency, energy, and wavelength of light to the difference in the two atomic energy levels. The module assumes that the students have been introduced to some of the concepts in previous modules. Th ­ ere is a shortcoming to the animation that may, in fact, confuse students more than help them. One of the sections has the user make the electron go up and down energy levels. The wavelengths that are generated do not correspond to differences in the energy levels, and the lower energy levels represent the longest wavelengths versus the shortest wavelengths. For example, from E2 to E1 a wavelength of 706 nanometers (nm), red, is generated. The gap between E3 and E2 generates a wavelength of 587 nm, whereas the smaller gap (about one-­half as large) between E4 and E3 generates a wavelength of 492 nm. In another example, the animation shows a cloud that has transitions from just E2 to E1, but instead of the wavelength generated being red, as represented ­earlier, it is now green. Smaller energy differences mean longer wavelengths; larger energy differences mean shorter wavelengths. This animation misses this impor­tant concept.

Text reference: Section 5.2 DOPPLER SHIFT This animation uses the analogy of stationary and moving police cars with sirens blaring to illustrate the Doppler shift for sound. It then clarifies how this relates to the redshift and blueshift of light from stars, emphasizing that although the sources and wave characteristics are dif­fer­ent, the princi­ple is the same. It finishes with our use of the Doppler shift in the discovery of an expanding universe.

Text reference: Section 5.3

ASTRONOMY IN ACTION VIDEOS

ATOMIC ENERGY LEVELS AND THE BOHR MODEL

EMISSION AND ABSORPTION

This animation explores the Bohr model of the atom in connection with emission and absorption of light, linking the energy and wavelength of light to the difference in the two atomic energy levels. The animation starts with exactly what ­matter is, including the parts of an atom, the charges, and locations of each part. ­A fter introducing the Bohr model, the animation then describes the electron clouds or electron waves, and how they can assume only specific forms. The bookcase analogy of energy levels is introduced.

Text reference: Section 5.2

The demonstration of energy levels involves a large set of outdoor stairs. Stacy Palen starts at the bottom, emphasizes that she needs precisely the right amount of energy to move up a level, and emits that energy to step back down. Just like an electron in an atom, she cannot go up half a step. A blue “photon” sends her up two steps. She also demonstrates the energy of an electron in the lowest level and then at higher energy levels using dance steps.

(Length: 2:44) Text reference: Section 5.2

38 ◆ CHAPTER 5 Light DOPPLER SHIFT

INTERACTIVE SIMULATIONS

An a­ ctual classroom demonstration of the Doppler shift of light is, of course, impossible to carry out. In this video, Dr. Palen uses a buzz­er located at the end of a long string to demonstrate the Doppler shift of sound while swinging the buzz­er in a circular motion above her head. Emphasis is placed on the pitch of the sound as the buzz­er comes ­toward the viewer and away. (Length: 2:14) Text reference: Section 5.3

None for this chapter.

CHANGING EQUILIBRIUM ­ ater, Using an innovative stack of three containers of dyed w Stacy Palen demonstrates energy coming from a star to a planet, and then from the planet back into space. Equilibrium is reached initially, and the analogy to the temperature of a planet is covered: the planet is returning as much energy as it is receiving. Dr. Palen then restricts the energy release from the planet, and we watch as the planet “heats up” ­because equilibrium has been lost. The relationship to our climate change is brought in at the end. (Length: 2:18) Text reference: Section 5.4 WIEN’S LAW In this video, Dr. Palen uses an incandescent bulb and dimmer to demonstrate that for light cooler means looking red and hotter means looking white to blue. It also clearly shows that hotter means a lot more luminosity overall than cooler, bringing in Stefan-­Boltzmann law. The difference between Wien’s law and our use of red to mean hot and blue to mean cold is addressed. (Length: 4:05) With our current use of cool LED light bulbs, the connection between temperature, luminosity, and color of light bulbs is rapidly being lost. We suggest that this video be followed by a classroom demonstration where students can witness the increase in luminosity and the increase in temperature with the change in the dominant colors in the spectrum.

Text reference: Section 5.4 INVERSE SQUARE LAW In this demonstration, Stacy Palen uses a balloon that she partially inflates to talk about the area of it at this point. She then blows it up to about twice its size and then relates the increased area that occurs at the larger size. Increase the radius by 2, and the area increases by 4. The application of the inverse-­square law is stated for gravity, light, and electrical charges. (Length: 2:12) Text reference: Section 5.5

TEACHING READING ASTRONOMY NEWS 1. The light took roughly 7,300 years to reach Earth.

2. The young stars are blue ­because they are very hot.

3. The stars w ­ ill give absorption spectra, whereas the gas ­w ill give emission lines.

4. Red is the lowest energy transition of hydrogen at vis­ i­ble wavelengths (E3 –­E2 , Figure 5.14), and is also the most likely to happen; thus, excited gas clouds of hydrogen generally appear red.

5. If users of this text are in the continental United States, the Gum nebula would not be vis­i­ble, as it lies in the constellation of Centaurus.

CHECK YOUR UNDERSTANDING 5.1A. (e, c, b, d, a) See Figure 5.6.

5.1B. (c) Energy and frequency are inversely proportional to wavelength.

5.2. The spectral features correspond to the composition of ele­ments pre­sent in the stars’ atmospheres. Each ele­ment—­even its ions and isotopes—­has a unique set of lines.

5.3. (d) A blueshift is a negative shift, and the larger the value, the greater the speed.

5.4. (d, b, a, c, e) Ranking of stars from coolest to hottest based on color.

5.5. (c) Brightness drops with distance squared.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (b) Earth is about 8.3 light minutes away from the sun.

2. (b) A given ele­ment is defined by its number of protons.

3. (a, b, c) b­ ecause we can have 10-7, 7-5, and 10-5.

4. (d) The Doppler effect explains why waves are compressed in the direction of motion and stretched out in the opposite direction.

5. (b) This is a statement of the Stefan-­Boltzmann law and Wien’s law.

6. (d) As surprising as it seems, size drops out of the planet’s temperature (see Working It Out 5.4).

7. (e) Star A is farther, so it must appear fainter. B ­ ecause 2 2 , it w ­ ill appear 3 = 9 times brightness scales as 1/d fainter, or 1/9 as bright.

CHAPTER 5

8. (a) If less energy is radiated away, the energy ­w ill be trapped, making the temperature increase to a new equilibrium level.

9. (c) The fastest speed of light is that in a vacuum. In any medium, light slows down.

10. (b) A downward transition gives off energy, meaning a photon is emitted.

11. (c) Red light has the longest wavelength and is the least energetic of the transitions to E2 .

12. (a) Brightness scales as L/d2 so if star A is twice as far away, it must have 22 = 4 times more luminosity to appear as bright as star B. 2.9 × 106 nmK = 10,000 K 13. (d) Using Wien’s law: T = 290 nm 14. (c) Equilibrium means that “stuff out” equals “stuff in.”

If that were not the case, t­ here would be an imbalance.

15. (b) Temperature is a mea­sure of the energy in an object, and ­because the object is made of atoms, it is equivalently a mea­sure of the speed of the atoms.

Thinking about the Concepts 16. Light can travel at a speed slower than 3 × 105 km/sec when it is traveling in a medium. Light, like sound, travels at a certain speed in each kind of medium.

17. Light is both a wave and a particle, depending on the scale of the objects on which it is operating. When light interacts with atomic and subatomic particles, at the smallest scale, it acts like a particle. When light interacts with ­things on a larger scale, larger than atoms and molecules, it acts like a wave. Light acts as a wave as it passes through the vacuum of space.

18. According to the scientific method, if one test of a theory fails, then the theory must be reconsidered and revised and tested again. Other­w ise, it has been disproved.

19. A beam of red light can have as much energy as a blue beam if the red beam has many more photons.

20. According to the cosmological princi­ple, observers should rec­ord the same spectrum for a given ele­ment or compound w ­ hether they are making their observations from Earth or from anywhere e­ lse in the cosmos. Consider the example of hydrogen. The same set of spectral lines for hydrogen that is mea­sured in laboratories on Earth has been observed in the atmosphere of the Sun, in the emissions of interstellar nebulae, and in the spectra of distant galaxies. Given that we see the same spectrum for hydrogen (or any other ele­ment or compound) no m ­ atter where we look in the universe, we gain confidence in the cosmological princi­ple.

Light ◆ 39

21. Atomic be­hav­ior is based on probabilities, so all we can do is say what the most likely average lifetime of the state ­w ill be through theoretical calculations.

22. Based on the atoms responsible for the spectral lines, we find the composition, temperature, and density of the object, as well as its velocity, based on the line strength and any line broadening. Emission lines come from electrons cascading to lower energy levels.

23. Redshifted objects are moving away from us; blueshifted objects are moving t­ oward us. Th ­ ese shifts are small and are detected in the changes of the wavelengths from their rest values. The shifts do not indicate that an object looks red or blue.

24. The frequency is higher, so the object is moving ­toward you.

25. This is an example of dynamic equilibrium ­because the individual visitors to the museum are constantly changing. The museum remains in equilibrium ­because the flow of visitors into the museum equals the flow of visitors out of the museum.

26. The blue star is hotter than the yellow one.

27. If we assume that all stars have the same luminosity, then this would tell us that the stars are at a variety of distances. Knowing instead that they vary in luminosity as well, it tells us that the stars cover an even greater range of distances.

28. ­There are at least two answers to this question. The first is that u­ nder Wien’s law, the peak wavelength for the Sun with a surface temperature of about 5800 K is 500 nm, about in the ­m iddle of the vis­i­ble part of the spectrum. The second is anthropological.

Life on Earth evolved with the Sun as the source of light, so creatures—at least ­those that are active during daylight hours—­adapted to see in the Sun’s dominant wavelength range, or what we ­humans call “vis­i­ble.”

29. Mercury has the greatest range in temperatures basically ­because it is not in equilibrium. We expect it to be much hotter than the other planets due to its being the closest planet to the Sun. We check the albedo of Mercury and find it is low, about 0.11, so a side would absorb heat when facing the Sun. For the side facing away from the Sun, t­ here is no atmosphere to hold any heat, which rapidly escapes.

30. Assuming all t­ hese stars are about the same size as the Sun, a red star would put out less energy per second than the Sun, so a planet would have to be fairly close to a red star to have roughly the same temperature as Earth. A planet around a yellow star, like the Sun, needs to be at roughly the same distance as Earth, and a planet around a blue star would have to be much farther away ­because of the star’s higher temperature.

40 ◆ CHAPTER 5 Light Applying the Concepts 31. Setup: Frequency, wavelength, and speed are related by c = λ × f. Energy is related to frequency by E = hf, where h = 6.63 × 10 −34 Js. Solve: 3 × 108 m/s c = 380 m for 790 AM, and (a) λ = = f 7.90 × 105 s −1 3 × 108 m/s c = = 3 m at 98.3 FM. f 9.83 × 10 7 s −1 (b) FM radio broadcasts at higher frequencies (frequency modulation). (c) E = hf = (6.63 × 10−34 Js)(7.90 × 105s−1) = 5.2 × 10−28 J E = hf = (6.63 × 10 −34 Js)(9.83 × 107 s−1) = 6.5 × 10 −26 J Review: Since radio waves are around 1 meter or longer in length, we have confirmed that t­ hese are in the radio spectrum. Note that lower frequency corresponds to longer wavelength and lower photon energy, as expected.

32. Setup: Frequency, wavelength, and speed are related by c = λ × f.

Solve: f = 2.45 × 109 Hz, so following the solution to question 31, λ = 0.12 m = 12 cm.

Review: Is it odd that a “micro­wave” emits radiation with wavelengths of centimeters rather than microns?

Micro­wave does not actually mean “microns,” but rather is the region between infrared and radio, so it includes centimeter wavelengths. Δf v 33. Setup: The Doppler formula for sound is = , f0 c s

λ=

where Δf is the change in observed frequency, f 0 is the rest frequency, and cs is the speed of sound (about 340 m/s) for this par­tic­u ­lar atmospheric pressure and temperature. v Solve: Solving for the frequency shift: Δf = f0 cs 25 m/s 440 Hz = 32.4 Hz. ­Because leading to Δf = 340 m/s the object is approaching, the pitch goes up. The new frequency w ­ ill be 440 Hz + 32.4 Hz = 472.4 Hz. Review: Note that the amount of Doppler shift depends on the speed of the wave, that is, we can travel at slow speeds and still produce audible Doppler shifts, whereas very high speeds are required to produce noticeable shifts in light. The frequency of 440 Hz is the musical note of A above ­m iddle C. The Doppler shift ­here takes it between an A# (466 Hz) and a B (494 Hz). Students can hear this frequency shift at http://­w ww​.­szynalski​.­com​/­tone​-­generator​/­. Δλ v 34. Setup: The Doppler formula is = , where Δ λ is λ0 c the change in observed wavelength, λ0 is the rest

wavelength, c is the speed of light, and v is the velocity that we need. Δλ c, we find Solve: Solving for the velocity, = λ0 ⎛ 502.3 nm − 486.1 nm ⎞ 8 7 v=⎜ ⎟⎠ 3 × 10 m/s = 10 m/s 486.1 nm ⎝ = 10,000 km/s.

Review: In the optical, a good rule of thumb is that 1 nm of Doppler shift corresponds to a speed of approximately 600 km/s. ­Here, we have about 16 nm of shift, or about 16 × 600 = 9,600 km/s, as found.

35. Setup: This is a half-­life prob­lem in which during each half-­l ife, 1/2 of the sample decays. In 2 half-­l ives, 1/2 of what is remaining decays, that is, ­there is 1/2 × 1/2 = 1/4, and in general, ­a fter n half-­l ives, 1/2 n remains.

Solve: ­Here, the half-­life is 30 minutes, so ­a fter 2 hours, 4 half-­lives have passed, so only 1/2 4 = 1/16 remains. B ­ ecause brightness depends on the number of atoms, the toy ­w ill be 1/16 times as bright.

Review: Note that the half-­life decay pattern means the number of particles remaining decreases exponentially.

36. Setup: Remember that brightness scales as 1/d2 , where d is the distance.

Solve: Neptune is 30 times farther from the Sun than 1 1 Earth, so the Sun ­w ill appear to be 2 = , or 30 900 0.1 ­percent as bright. To Voyager 1, the Sun ­w ill appear 1 1 or 0.005 ­percent of its brightness as = 2 140 19,600 seen from Earth.

Review: Although the Sun appears as a huge, blazing bright disk in our sky, it becomes just another star as seen from the most distant parts of the Solar System.

L 37. Setup: Brightness scales as 2 , so for two objects of d L L firefly equal brightness, 2bulb = 2 . d bulb dfirefly Solve: Solving for the distance of the bulb, Lbulb 2 2 d bulb = dfirefly Lfirefly d bulb = dfirefly

Lbulb = 5m 106 = 5,000 m Lfirefly

Review: This may seem very far, but remember that your eye is very sensitive to light at night. If this w ­ ere a 60-­watt bulb, imagine how far a star must be to have the same brightness as that firefly.

CHAPTER 5

38. Setup: Brightness scales as

L so for two objects of d2

L1 L2 = . d12 d22 Solve: Solving for luminosity,

equal brightness, 2

Light ◆ 41

Review: The h­uman body is putting out a fair amount of energy in the near infrared. Examining images of the body taken at t­hese wavelengths shows the heat loss varies by the location on the 2 ⎛ d2 ⎞ 2 L2 = L1 ⎜ ⎟ = 3 L1 =body.

9L1 43. Setup: In general, the blackbody temperature of a ⎝ d1 ⎠

1 ⎛d ⎞ 4 ⎡ LSun (1 − a) ⎤ 1 L2 = L1 ⎜ 2 ⎟ = 32 L1 = 9L1 or the more distant star must be 9 times planet is expressed as T = . Given ⎝ d1 ⎠ ⎢ 16πσ ⎥ d ⎣ ⎦ more luminous. that in this prob­lem, all the f­actors are equal except Review: Another way to think about this is just in the albedo, we can use the formula from Working terms of distance. Three times farther means It Out 5.4: T = 279K × (1 − a)¼ (also expressed as 1/32 = 1/9 times as bright, so the star must be 9 times T = 279 K 4 1− a ). more luminous to have the same brightness.

Solve: (a) If a = 0.1, then T = 272 K. (b) If a = 0.9, 39. Setup: The luminosity is equal to the flux times the then T = 157 K. area, where flux = σT4, where σ is the Stefan-­Boltzmann Review: Note that as albedo increases, less light constant (Working It Out 5.3). We are given an area reaches the surface, so the planet is not heated as of 1 m2 . much.

Solve: For a temperature of 500 K, this gives a lumi44.

Setup: Start with the planet-­temperature equation nosity of L = (5.67 × 10 −8W/m 2K4)(500 K)4(1m 2) ​ 1 4 ⎡ LSun (1 − a) ⎤ 1 = 3540 W. T=⎢ , which simplifies to ⎥ Review: That is a lot of energy radiated into its d ⎣ 16πσ ⎦ 1 surroundings.

4 ⎛1 − a⎞ 40. Setup: The luminosity from a blackbody of size R and T = 279K × ⎜ 2 ⎟ temperature T is L = 4πR 2 σT4. ⎝ d AU ⎠ 8 ⎛ 1 − 0.8 Solve: Using the values given R = 6.96 × 10 m and Solve: With a = 0.8 and d = 97.7 AU, T = 279K × ⎜ 8 2 ⎝ 97.7 2 A T = 5780K, we find, L = 4π(6.96 × 10 m) (5.67 × 1 4 1 − 0.8 −8 2 4 4 26 ⎞ = 19 K 10 W/m K )(5780K) = 3.85 × 10 W T = 279K × ⎛⎜ ⎟⎠ 2 26 ⎝ 97.7 AU Review: The quoted solar luminosity is 3.84 × 10 W; a blackbody approximation is quite good.

Review: Solar heating is almost non­ex­is­tent at ­these 2.9 × 106 nmK distances. .

41. Setup: Wien’s law tells us that λ peak = 45.

Setup: According to our text, the blackbody temperaT 1 5 4 Solve: (a) If T = 10 K, λ peak = 29 nm, which lies in the ⎡ LSun (1 − a) ⎤ 1 . From ture of a planet is T = ⎢ extreme UV part of the spectrum. ⎥ d ⎣ 16πσ ⎦ Review: At 100,000 K, t­ hese stars essentially emit all 1 their energy in the extreme UV range. this we see that T ∝ L4Sun , which is all we need for 42. Setup: The peak radiation for a h­ uman being can be this prob­lem. determined from Wien’s law, but first we need to con1 1 vert 98.6°F to Kelvin: (98.6°F − 32)/1.8 + 273.15 ​ Solve: T ∝ L4Sun = 10 4 = 1.8, meaning Earth would = 310.15 K. The ­ human body temperature is be about 1.8 times hotter. ap­­ ­ proximately 310 K. Then use Wien’s law, Review: It stands to reason that the planet would 2.9 × 103 µmK warm, b­ ecause the sun would put out more energy to 2.9 × 106 nmK or λ peak = . λ peak = be received.

T T For part (b), the luminosity of a blackbody is Using the Web L = σAT 4, where A represents the surface area of the 46. a. The UV light is showing hotter temperatures, blackbody. revealing higher-­energy objects and events. DifferSolve: (a) (98.6°F − 32)/1.8 + 273.15 = 310.15 K. 3 ences between the two images include UV light high2.9 × 10 µmK λ peak = = 9.35 µm. This wavelength is lights galaxy arms and center; vis­i­ble wavelengths 310K show a smoother galaxy and more stars. b. Searching in the near-­infrared part of the spectrum. for “false colors” turned up over 200 objects. Answers (b) L = σAT4 = (5.67 × 10 −8w/m2K4)(0.25m2)(310K)4 ​ ­w ill vary. = 131 watts.

42 ◆ CHAPTER 5 Light 47. Specially tuned UV light wavelengths of light are used to search for blood and saliva. A tunable UV light source that diminishes background glow and fluo­ rescent powder are used for fingerprints. Having access to dif­fer­ent kinds of light useful for an investigator is needed b­ ecause crime scenes ­w ill have multiple types of evidence within multiple types of environments.

Searching for “forensic spectroscopy” and selecting a recent report w ­ ill lead to a variety of answers.

48. Night vision goggles and thermal-­ imaging devices work differently from regular binoculars or cameras ­because ­these devices can intensify the images through a photomultiplier tube. They combine that technology with devices that are sensitive to the IR range of 700– 1,000 nm. Thermal-­imaging cameras detect thermal radiation and allow one to compare temperature differences. Night vision or thermal-­imaging tools are useful when they are commonly used by the military, when looking for bodies of ­humans or animals in areas of dense smoke, and on vehicles to enhance d­ rivers’ ability to see in darkness or poor weather (to name just a few of a number of uses). wave 49. The TSA scanning devices use millimeter-­ machines that create a three-­dimensional image. Backscatter machines send narrow, low-­density X-­ray beams over the subject. Passengers are concerned that the devices can see inside the body, can see details about their “private parts,” and use harmful radiation, among other concerns.

50. Checking the albedo of parts of Earth is time sensitive for this question, and thus answers ­w ill vary.

EXPLORATION This exploration has students work with the AstroTour, Light as a Wave, Light as a Photon. It should help students understand the relationship among the ways we can and do describe light. For ­those who may not be as ­adept as ­others in interpreting equations, this exploration w ­ ill visually decipher the relationships for them.

EXPLORATION SOLUTIONS 1. As the wavelength increases, frequency gets lower.

2. As the wavelength decreases, frequency gets higher.

3. Wavelength and frequency are inversely related, that is, if wavelength increases, frequency decreases, and vice versa.

4. If we increase the frequency, the wavelength should shorten.

5. Yes, the wavelength decreased (shortened) as expected.

6. Nothing. Frequency and wavelength are in­de­pen­dent of the amplitude.

7. Nothing, for the same reason as question 6.

8. Amplitude has no relationship to frequency or wavelength.

9. This is a wave of light in a vacuum, meaning its speed never changes.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 8: LIGHT AND SPECTRA In this activity students explore the properties of continu­ fter working through this activous and line emission. A ity, they ­w ill be able to: • explain how the temperature of an incandescent lightbulb affects the intensity and colors of the observed spectrum.

• compare the observed continuous spectrum with a series of Planck (blackbody) curves.

• examine emission spectra of five ele­ments, noting the patterns and intensities of the lines.

• identify three “unknown” ele­ments.

• state the significance of each ele­ment having a unique spectral signature of emission lines.

Text references: Sections 5.2 and 5.4

CHAPTER 6

The Tools of the Astronomer

INSTRUCTOR NOTES Chapter 6 covers the vari­ous means by which astrono­ mers collect data. Major topics include: • basic optics, such as refraction, reflection, and optical and atmospheric limitations.

• the design and function of ground-­based optical telescopes.

• how light is digitally recorded by cameras and spectrometers.

• the design and function of telescopes beyond the vis­i­ble region of the spectrum.

• planetary spacecraft, including flybys, orbiters, landers, rovers, and probes.

• other tools not always associated with astronomy, like particle accelerators, neutrino and gravity waves detec­ tors, and computers.

This chapter contains material that may often be cov­ ered in one session. D ­ oing so can be advantageous b­ ecause by this point in the class, many students are hungry for something that feels astronomical and not so much like physics. We suggest that a thorough exposition of this chapter is impor­tant if you want to make the distinction between saying, “­Here are a lot of facts,” and saying, “­Here is how we discovered t­ hose facts,” as you teach the l­ater chapters; ­a fter all, ­were it not for telescopes, detectors, and spacecraft, we would not have the rest of this book.

DISCUSSION POINTS • Students with eyeglasses can quickly find the focal point of their lenses by holding them up to a wall oppo­ site a win­dow or other light source and moving the lens back and forth from the wall ­until an image of the win­ dow appears. Ask students with differing eyesight to try this with their lenses to show differing focal lengths. (LG 1) • Have students list the many devices that are used in modern astronomy t­ oday. The charge-­coupled device (CCD) that many of them possess, e­ ither in a smart phone or a digital camera, is an integral tool. If they have a smart phone, they may also have an app dedi­

cated to astronomy. Have students discuss about what it means to have such astronomical tools at their fin­ gertips. (LG 2, LG 5) • The basics of CCDs are much easier to teach ­today than, say, a few de­cades ago, b­ ecause just about every­ body carries a small CCD in a smart phone. Some stu­ dents may even be familiar with image editing.

Zooming in on an image on a smart phone can usually be done in “edit” mode. With an image editor, one can zoom in ­until the image is pixelated. Discuss how t­ hese basics relate to astronomical cameras capable of recording very faint object. (LG 2) • Select some of the less-­technical descriptions of spec­ trographs included in the instrument lists available at the National Optical Astronomy Observatory (NOAO) website (http://­ast​.­noao​.­edu​/­observing​ /­current​-­telescopes​-­instruments). What part of the electromagnetic spectrum is covered by each of ­those spectrographs? Discuss the connection among spec­ trographs, detectors, and wavelength coverage. (LG 2) • ­Table 6.1 lists four telescopes that are e­ ither ­under construction or ­w ill be soon. Assign ­these to groups for team researching details about their assigned tele­ scope. Each group should then report to the other three. Ask students, “What are the qualities of t­ hese telescopes? What ­w ill they be able to do that the cur­ rent telescopes cannot?” (LG 3) • Have students collect information about telescopes around the world or in space that have primary aper­ tures larger than 3.5 meters but less than 8 m (similar to ­Table 6.1). Are t­ hese telescopes refractive or reflective?

Where are they located? What kind of light do they observe? Discuss the trends the students find. (LG 3) • Discuss the merits and drawbacks of putting telescopes in space versus on Earth when t­ here is an option. Is ­there any reason to place a radio dish in space? B ­ ecause ­there are win­dows of atmospheric transparency at infrared wavelengths, why would we put an infrared telescope in space? This could tie into the remarkable discoveries by the Spitzer Space Telescope. (LG 3) • The recent discoveries made by the vari­ous spacecraft on and orbiting Mars and the flybys and orbiters to 43

44 ◆ CHAPTER 6 The Tools of the Astronomer Jupiter, Saturn, and Pluto have been richly covered by the media. Students may have their favorites that they can share with the class. What information is available on social media? Are any of the students following astronomy-­interest groups? (LG 5) • The question in the opening image for this chapter is, “Why are most telescopes on remote mountaintops?”

The Internet has listings of the highest permanent observatories. The highest one is at 18,500 feet above sea level and is located in the Atacama Desert, Chile; while the lowest one (a cut-­off of about 10,000 ft) at 9,961 feet, is on Haleakala, Hawai’i. Students should be able to answer this opening question by reviewing the wavelength regions covered by the instruments in ­these observatories.

PROCESS OF SCIENCE One takeaway from this chapter’s Pro­cess of Science is that it often takes a very long time—­a ­century in this case—­for technology to catch up with the science. The prediction that a changing gravitational field should create gravitational waves and the ­actual detection of them by the L ­ aser Inter­ ferometer Gravitational-­Wave Observatory (LIGO) is an impor­tant example of the symbiosis of technology and science.

ASTROTOUR ANIMATIONS GEOMETRIC OPTICS AND LENSES This interactive animation starts with Galileo Galilei as the first person to explore the night sky using a telescope. It then explores the concepts of focal point and length by showing how lenses are used to bend light and magnify images using light rays, defining refraction in the pro­cess.

In the interactive portion of the animation, students can adjust the distance of the object and the eye from the lens.

The coverage on reflecting telescopes brings in discussions of the focal plane and aperture size, emphasizing the impor­ tance of telescopes as light-­gathering “buckets” to see dim objects.

Text references: Section 6.1 and Exploration

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Astronomers want to build at Cerro Armazones ­because this location is extremely high, away from bright cities, and dry and cloudless almost year round.

2. Larger telescopes collect more light, meaning fainter objects can be observed. Also, larger telescopes pro­ vide higher resolution, meaning more detail can be observed.

3. Adaptive optics removes the blurring effects of the atmosphere.

4. Most other wavelengths (X-­ray, ultraviolet, mid-­to-­far infrared) are not observable from Earth, no ­matter how high the observatory is. This means that this telescope can observe only in the optical and near-­infrared.

5. Answers w ­ ill vary. At pre­sent, the telescope is fully funded and has a first-­light date between 2024 and 2026. All of the Eu­ ro­ pean Southern Observatory (ESO) member states have approved this proj­ect, including Austria, Belgium, Brazil, the Czech Repub­ lic, Denmark, Finland, France, Germany, Italy, Neth­ erlands, Poland, Portugal, Spain, Sweden, Switzerland, and the United Kingdom; Chile is also a partner.

Although ­there has been considerable controversy over other mountaintop observatories, t­here did not seem to be significant opposition to the building of the Extremely Large Telescope (ELT) at the time of writ­ ing this instructor’s manual.

CHECK YOUR UNDERSTANDING SOLUTIONS 6.1A. (a, b, c) Without curved glass, a reflecting telescope cannot focus light. which makes choice (d) wrong.

6.1B. (a) In ground-­based telescopes, the atmospheric blurring is almost always larger than the resolution limit of the telescope.

6.2. (c) A higher quantum efficiency means that it takes fewer photons to excite the electrons that lead to a signal.

6.3. (d) More expensive to launch into orbit and no way to repair if something breaks.

6.4. (c) A spacecraft can mea­sure many more t­ hings about a planet than just its images. For example, even during a flyby, a satellite can still mea­sure gravity, magnetic fields, and charged particles near the planet. None of ­these can be done as effectively from Earth.

6.5. (a, b, d) Generally speaking, pointing a telescope may not “require” a high-­performance telescope. How­ ever, with adaptive optics and offset pointing to distant galaxies, it becomes a necessity. Therefore, all of the choices could also be a correct answer.

CHAPTER 6 The Tools of the Astronomer ◆ 45

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS

Thinking about the Concepts 16. ­Simple lenses suffer from chromatic aberration; that is, they bend dif­fer­ent colors by dif­fer­ent amounts.

Test Your Understanding The resulting image is not nearly as crisp as in tele­ 1. (c) The primary function of a telescope is light-­ scopes with multiple lenses. gathering power, so a larger dia­meter is better.

17. Refracting telescopes require a straight-­line light path 2. (a, e) Our atmosphere blocks most UV light and all from lens to focus. Building refractors with long tubes gamma ray and X-­ray light. to support increasing focal lengths becomes impracti­ 3. (a) 3, (b) 4, (c) 2, (d) 6, (e) 7, (f) 1, (g) 5 cal. As the tubes grow longer, telescopes are increas­ 4. (b) The resolution of an interferometer is given by ingly front-­loaded with weight and become unwieldy. the ratio of the observing wavelength and separation Another prob­lem that occurs with large refractors is 85) = 0.005 of the telescopes: θ = 2.06 × 105 (λ D) = 2.06 × 105 (2 × 10−6aperture size. If you increase the aperture of a tele­ 5 λ ( D) = 2.06 × 105 (2 × 10 −6 85) = 0.005 arcseconds. See Working It scope, you increase its light-­gathering power. Unfortu­ Out 6.2. nately, increasing the aperture size also increases the 5. (d) Interferometry allows a much higher angular res­ size and weight of the primary lens. Glass is heavy, and olution than with a single telescope. it flows like a liquid ­under its own weight within rela­ 6. (c) Depending on the trajectory of the light beam, it tively short timescales. Large pieces of glass ­w ill also can bend as it moves from one medium to another have imperfections like air b­ ubbles that w ­ ill affect the due to its changing speeds among the media. image. If astronomers go to the expense of building 7. (a) The light-­gathering power of a telescope depends such massive telescopes, they want the instruments to on the area of the aperture, and b­ ecause area scales as last for de­cades or centuries without having to replace size squared, two times bigger size is 22 = 4 times deformed optics e­ very few years. more light-­gathering power.

18. The telephoto lens w ­ ill produce a larger image in the 8. (d) Having a telescope setup that has high resolution focal plane. The longer focal length leads to greater allows astronomers to examine small-­scale features, magnification of the subject (Working It Out 6.1). which also helps them to distinguish very distant The trade-­off for magnification is image brightness. objects or separate many objects that are close The same amount of light is received from the subject, together. but the light is spread over a larger area in the focal 9. (a) Ret­ina. See Figure 6.1. plane, so the image appears fainter.

10. (b) Turbulence in the atmosphere distorts light as it 19. Dif­fer­ent types of astronomical objects are at dif­fer­ passes from space to the telescope; adaptive optics ent temperatures and thus emit light at dif­fer­ent peak attempts to remove ­these effects. wavelengths. Many astronomical sources that look 11. (a) Interferometry is used to increase the resolution normal at vis­i­ble wavelengths may be much brighter of a telescopic observation. or more complex in other regions of the spectrum 12. (d) The blurring effects of our atmosphere generally 20. The sketch w ­ ill look similar to Figure 6.9. limit how well we can resolve objects in ground-­based 21. This question can be answered in a few dif­fer­ent ways. telescope images.

The first way to look at this is to argue that brighter 13. (d) Spectra come from the diffraction of light as it stars appear to be bigger in images, and that the larger passes through a slit at the focal point of a telescope dia­meter telescope ­w ill collect more light and make and is reflected to a diffraction grating that spreads the star appear larger. Second, a star is a point source, out the light. so it cannot be made to look larger or smaller; thus, 14. (b) ­Water vapor is very effective at blocking infrared another answer is that a star ­w ill appear to be the same light, so to observe in ­these wavelengths, one must size through all telescopes. A third argument is that find a very dry location, such as a high mountaintop magnification depends on the ratio of the focal lengths or an aircraft such as the Boeing 747 used in the of the objective and eyepiece, so if both telescopes Stratospheric Observatory for Infrared Astronomy have the same focal lengths, then they ­w ill have the (SOFIA). same magnifications, making both stars appear the 15. (b) The quantum efficiency of a detector—­including same size. our eyes—is how well it turns light (photons) into a 22. Answers w ­ ill vary. It started with Einstein’s theory of signal. gravity and his prediction of gravitational waves. About a ­century ­later, it is at the stage where gravitational

46 ◆ CHAPTER 6 The Tools of the Astronomer waves have been detected due to the advanced technology—­a nd the high degree of sensitivity—­ needed. That scientific discovery is leading to an improvement of the technology with the L ­ aser Interfer­ ometer Space Antenna (LISA), with a planned launch date of 2034, or another ­couple of de­cades a­ fter the first detection.

23. An adaptive optic system ­w ill observe e­ ither a real or fake (laser-­ generated) star and use sophisticated computer programs to control a deformable mirror in an attempt to keep the star’s image steady and as small as pos­si­ble. B ­ ecause all the other stars or objects one is looking at have had their light pass through the same air as our reference star, our deformable mirror corrects for much of the atmospheric seeing and pro­ vides us with higher resolution images.

24. Our eye takes an “image” about 14 times per second.

Thus, we are ­limited by how much light we can col­ lect—in other words, how faint an object we can see. A ­ hether using film or a CCD chip, can collect camera, w light for much longer—­the amount of time it collects light is called the integration time—­and taking a much longer exposure collects much more light, which allows us to distinguish faint objects from nearby bright ones or the background. Quantum efficiency contributes to the detection of faint objects by getting the most signal out of the photons entering the detector.

25. The stars are so far away that a telescope in space is only infinitesimally closer to the stars than a telescope on Earth. Telescopes in space avoid atmospheric effects and can observe at wavelengths or energies that Earth’s atmosphere blocks.

26. Flyby spacecraft can view a large number of objects during their mission b­ ecause they travel over large distances in the solar system. However, ­there is a limit to how much they can learn about any single object ­because they spend very ­l ittle time at one place (being flybys) and generally cannot fly too close to anything.

An orbiter can do a few flybys on the way to its target and can then spend its entire mission studying one object at very high resolution. While it can study the magnetic fields and mass distributions of a planet and thus get some insight as to its internal structure, it is ­limited in how much it can teach us about the a­ ctual surface properties of the object. For this, we need a lander, which can yield an enormous amount of infor­ mation but only about one very small region of the planet, a moon, or the Moon. The next step up is a rover that can cover a greater area of the surface in more detail, but is still ­l imited.

27. The meteorites give us a very ­l imited view of the state of Mars. We only have a few rocks, we do not know

where they came from on Mars, and they have under­ gone change and contamination as a result of being ejected from Mars and landing on Earth. To r­eally study the geology of Mars (or any planet), we need to do controlled experiments, such as collecting mineral samples from known and diverse locations.

28. The Moon is tidally locked to Earth, keeping the same face ­toward Earth at all times. It is not pos­si­ble for an Earth-­bound observer to see the far side of the Moon directly (save for the glimpses over the limb provided by libration). Our first look at the far side did not come u­ ntil 1959, when the Soviets launched a space­ craft that looped around the Moon.

29. Neutrino detectors are located deep inside mines, for example, inside mountains, so that the rocks protect the detectors from other kinds of particles that come from space. Neutrinos are neutral and interact very weakly with ­matter, and the only way to detect one is to watch the result of it impacting another atom. ­Because this is very rare, it makes detection quite tricky.

30. The leftover radiation from the Big Bang is in the micro­wave region of the spectrum; ­there are numer­ ous sources of micro­wave signals on Earth. Examples include signals that come from the atmosphere and cell phone use. Thus, it is better to go into space, above and out of the way of all of t­hese other sources of noise. Even then, Doppler shifts and the noise coming from the Galaxy must be dealt with.

Applying the Concepts 31. Setup: The light-­ gathering power of a telescope depends on the area of the aperture, A = πr 2 , where r is the radius of the telescope. Since we are comparing the sizes of both telescopes, we can use the stated val­ ues for the dia­meters (the ratio of the radii is the same as the ratio of the dia­meters). Note also that we must use consistent units of size. In the example, 30 meters equals 30,000 mm, and 10 meters equals 10,000 mm.

Solve: ­Because light gathering depends on the square of the telescope’s aperture, the larger astronomical tele­ scope has more collecting power. The Thirty Meter Tele­ (3 × 10 4 mm)2 = 1.4 × 10 7 scope, therefore, collects (8 mm)2 times more light than the ­human eye. A comparison with the 10-­ meter Keck telescope gives 4 2 (3 × 10 mm) = 9 times more light gathering.

(104 mm)2 Review: ­The 30-meter telescope gathers 14 million time more light than our eyes and 9 times more light than one of the Keck 10-meter telescopes.

CHAPTER 6 The Tools of the Astronomer ◆

47

32. Setup: ­There is a vertical line pre­sent at the entrance Solve: In the infrared, and exit of the l­aser. Mea­sure the a­ ngle with re­spect to 1,000 nm 1m D = (2.06 × 105 arcsec) × × 9 = 0.21arcsec. this.

1m 10 nm Solve: As the light enters, we mea­sured an a­ ngle of In the optical violet region, this is 37°. As it exits, we also mea­sured 37°. ­These are the 400 nm 1m same ­because light rays are bent (or refracted) by the θ = (2.06 × 105 arcsec) × × 9 = 0.08 arcsec.

1m 10 nm same amount on entrance and exit. This also confirms that the block is a rectangle and its sides are parallel.

The optical observations ­w ill be higher resolution.

Review: Note that if the sides ­were not parallel, the Review: Our formula tells us that the diffraction light beams would come out at dif­fer­ ent ­ angles. limit scales directly with wavelength, so we can con­ Prisms use the index of refraction to spread out light firm that infrared ­w ill have a larger diffraction limit, into its component colors. thereby reducing our maximum resolution. Note that 33. Setup: The light-­ gathering power of a telescope from the ground, seeing generally limits us to 0.2 to depends on the area of the aperture, A = πr 2 , where r is 0.5 arcseconds anyway, so both of t­ hese wavelengths the radius of the telescope. could be seen without a prob­lem.

Solve: ­Because light gathering depends on the square 36. Setup: The diffraction limit of an optical system is of the telescope’s aperture, a 16-­inch telescope has a λ arcsec. Be sure given by θ = (2.06 × 105 arcsec) size that is 4 times larger than the 4-­inch one; thus, D the larger telescope collects 42 = 16 times more light. to use the same units for wavelength and dia­meter.

The larger telescope should be able to detect stars 1/4 Solve: (a) For the h­ uman eye, as bright, or stars that are 1–2 magnitudes dimmer.

5.5 × 10 −4 mm Review: Note that we did not need to use the radius θ = (2.06 × 105 arcsec) = 14 arcsec.

8 mm (or dia­meter), b­ ecause we compared the sizes of both (b) This limit is 5 to 10 times smaller than the stated telescopes (the ratio of the radii is the same as the limit of 60–120 arcsec. ratio of the dia­meters). ­Because light-­gathering scales (c) The diffraction limit of an optical system simply as the square of size, we can see that even a modest reports the theoretical limit and does not include real increase in aperture size yields a large gain in the light effects like seeing, imperfections in optics, and the collecting of a telescope. resolution limit of the detector. In the case of our eye, 34. Setup: The diffraction limit of an optical system is seeing is not much of an issue, but our lens is not per­ λ given by arcseconds, D = (2.06 × 105 arcsec) . Be fect, and our ret­ina does not have rods and cones θ placed to provide a resolution any higher than 1–2 sure to use angular mea­sure­ments and lengths in the arcmin. same units.

Review: If we found a diffraction limit in (a) that was Solve: Solve for the aperture of the radio telescope, larger than 1–2 arcmin, we would know we made a 21cm 1arcmin D = (2.06 × 105 arcsec) × = 4.8 × 10 4 cm,­m istake ­because the resolution limit of an optical sys­ 1.5 arcmin 60 arcsec tem cannot be better than the diffraction limit. This is 1arcmin 1cm 4 = 4.8 × 10 cm, or 480 meters. Assuming that we × a good check for a numeric calculation. rcmin 60 arcsec could detect radio waves with our eyes, we see that our 37. Setup: Image size is directly proportional to the focal pupil’s tiny size would mean we would have no angular length of the telescope. We can write this as a ­simple resolution at all, so we would not be able to detect any proportion with the subscript “big” representing the shapes or structure, even on the largest scales. big telescope and the subscript “amateur” referring to Review: Note the huge size of the radio dish that the amateur telescope. If we let x represent image size would be needed just to resolve a few arcminutes. x big f big and let f represent focal length, then = .

This is why we have built huge interferometers and x amateur famateur the dish at Arecibo.

Solve: Solving for the image size of the Moon in the 35. Setup: Not including atmospheric seeing or manu­ facturing defects in the optics, the resolution of a ⎛ f big ⎞ ⎛ 250 m ⎞ x amateur = ⎜ big telescope, we find x big = ⎜ telescope is given by the diffraction limit ⎟ ⎝ 1.5 m ⎟⎠ ⎝ famateur ⎠ λ θ = (2.06 × 105 arcsec) arcsec. Be sure to⎛ usef the⎞ ⎛ 250 m ⎞ big D x big = ⎜ (13.8 mm) = 2,300 mm or 2.3 meters. x amateur = ⎜ ⎟ same units for wavelength and dia­meter. ⎝ 1.5 m ⎟⎠ ⎝ famateur ⎠

48 ◆ CHAPTER 6 The Tools of the Astronomer Review: We can directly see why it is advantageous using standard observing methods is far harder than to have very large telescopes, not only for light collec­ the submilliarcsecond resolution that is regularly tion but for resolution as well. achieved in the radio using the VLBA. However, the 38. Setup: The number of bits produced by a detector is question states that the two large telescopes are using given by the product of the number of pixels and the interferometry, a method that adjusts for atmospheric levels of brightness per pixel. seeing.

Solve: The old detector produced a total of 41. Setup: The diffraction limit of an optical system is 160,000 × 256 = 4.1 × 107 bits of data per image. λ arcsec. Be sure to given by θ = (2.06 × 105 arcsec) ­Today’s detector produces 109 × 32,768 = 3.3 × 1013 D bits, or almost a million (800,000) times more data use angular mea­sure­ments and lengths in the same per image. units.

Review: Does this make sense? Think about the first 17mm 1 km digital cameras, which could use flashcards with only θ = (2.06 × 105 arcsec) 5 × 6 = 3.5 × 10 −5 ar Solve: 10 10 km mm 16 megabytes of memory, whereas ­today’s17mm cameras 1 km × = (2.06 × 105 arcsec) = 3.5 × 10 −5 arcsec, 0.035 milliarcseconds or 35 require cards with θmany gigabytes of memory.

105 kmIt is106 mm microarcseconds. the same effect.

Review: The VLBA currently gives us a resolution of 39. Setup: The higher the efficiency, the shorter the expo­ about 0.35 milliarcseconds, so this would be an sure time needed to make the same observation, so improvement of a ­factor of 10. the ratio of efficiency ­w ill equal the inverse ratio of 42.

Setup: The small a­ ngle formula tells us that the physi­ exposure times. dθ Solve: The formula for the setup statement could read , where d is the cal size s = 5 2.06 × 10 arcsec Q 1 t2 distance.

Q1t1 = Q2t 2 . leading to = . The CCD camera Q 2 t1 Solve: The size of an object with an angular size 280 km × 0.2 arcsec has an improvement in efficiency that is 80 times that = 2.7 ×10 −4 k of 0.2 arcsec w ­ill be s = 5 of the photographic plate, so we ­w ill need an exposure 2.06 × 10 arcsec 280 km × 0.2 arcsec time that is 1/80 times shorter. Assuming s = a 60-­m inute5 = 2.7 ×10 −4 km, or 27cm. In practice, we need two to 2.06 × 10 arcsec exposure with the photographic plate, we get three resolution ele­ments to r­ eally detect an object, so 60 min 60 sec the orbiter can detect objects about two to three times × = 45 sec. We save 59 minutes, 15 80 1 min 27 cm, or slightly less than 1 meter. seconds for each image.

Review: As an analogy, the orbiter would be able to Review: This is 1 hour versus 45 seconds. With some see where every­one on a football field was standing telescopes, we can now take an image of almost the when a certain play was being made, and it would entire sky ­every night, whereas with photographic prob­ably know where the ball was if in midair. plates this was a highly ambitious feat that could take 43. Setup: To convert AU to km, remember that ­t here many years. are 1.5 × 10 8 km in one AU. To find travel times of 40. Setup: The diffraction limit of an optical system is light signals, use d = vt, where h­ ere, v is the speed of λ light. arcseconds. Be given by θ = (2.06 × 105 arcsec) Solve: (a) Voyager 1 is 140 AU = (140 AU)(1.50 × 108 D sure to use angular mea­sure­ments and lengths in the km/AU) = 2.1 × 1010 km from the Sun. same units. (b) At the speed of light, the one way time for a radio 1.35 cm 1 km signal to travel between Earth and Voyager 1 is Solve: (a) θ = (2.06 × 105 arcsec) × = 3.5 × 10 −4 arcsec, 1 hr d 2.1 × 1010 km 8,000 km 105 cm t = = × = 19 hr.

1 km 5 cm 5 −4 v 3 × 10 km/s 3,600 s × 5 = 3.5 × 10 arcsec, or 0.35 milliarcseconds.

0 km 10 cm (c) Assume that Voyager 1 was on a direct line to 550 nm 1m (b) θ = (2.06 × 105 arcsec) × 9 = 1.1 × 10 −3Proxima arcsec, Centauri. Proxima Centauri is currently 100 m 10 nm 1m 4.24 ly ≈ 268,000 AU from the Sun. That implies × = 1.1 × 10 −3 arcsec, or 1.1 milliarcsecond. Although m 10 9 nm ­these are fairly close, we see that if we r­eally want that Voyager 1 would have completed 140 AU ultrahigh resolution, we ­w ill use VLBA imaging. × 100 percent = 0.05 percent of its jour­ Review: Note that b­ ecause of atmospheric seeing, 268,000 AU achieving 1 milliarcsecond resolution in the optical ney to the nearest star, if it ­were heading ­toward it.

CHAPTER 6 The Tools of the Astronomer ◆ 49

That means it would have to travel 2,000 times as long as it has so far.

Review: Voyager I is heading ­toward the star AC +79 3888, which lies 17.6 light-­years from Earth, or over 4 times farther than the nearest star. This prob­lem helps us to see how far away even “nearby” objects are in space.

44. Setup: All waves follow = λf, so solve h­ ere for wave­ length λ = c f .

3 × 108 m/s Solve: λ = = 3 × 107 m or 30,000 km.

10 Hz Review: Just to make this more readily understand­ able, the average distance to the Moon is 384,400 km.

The wavelength of this gravitational wave is about 8 ­percent of the average distance to the Moon. Geo­ synchronous satellites orbit about 36,000 km from Earth.

45. Setup: Wien’s law gives the peak of a blackbody 2.9 × 106 nm K .

T Solve: For a blackbody at 2.73 K, the peak wavelength

λ peak =

2.9 × 10 nm K = 1.06 × 106 nm = 1.06 mm.

2.73 K The wavelength of this light puts it in the millimeter or micro­wave regime.

Review: The cosmic micro­wave background is at 2.73 K and is so named ­because the spectrum peaks in the micro­wave region of the spectrum. is λ peak =

6

Using the Web 46. Involves picking 2 out of 80 telescopes that can be accessed as part of the “Around the World” event by ESO. Answers ­w ill vary.

47. Involves picking a specific telescope and observatory. ­ ill vary.

Answers w 48. Answers w ­ ill vary a­ fter viewing the film. Ecological arguments against light pollution includes that it disrupts plant growth, contributes to the death of land and sea animals, migrating birds lose their way, newly born turtles go inland to bright lights rather than the glow of the sea, and the pollution of the land, air, and ­w ater by coal and other sources of energy.

49. Answers ­w ill vary for the current status of the James Webb Space Telescope. The JWST w ­ ill observe just at IR wavelengths, with a mirror that is about 3 times that of the HST. Instruments include an NIR camera, NIR spectrograph, mid-­ IR instrument, and com­ bined guidance sensor, image, and slitless spectro­

graph. The goal is to understand the entire history and evolution of the universe. The estimated cost as of 2006 was close to $10 billion.

50. Involves picking a mission from Appendix 5. Answers ­w ill vary.

EXPLORATION This exploration delves into the essentials of optics. It explores the concepts of focal point and length by show­ ing how lenses are used to bend light and magnify images using light rays, defining refraction in the pro­cess. In the interactive portion of the animation, students w ­ ill adjust the distance of the object and the eye from the lens. The coverage on reflecting telescopes brings in discussions of the focal plane and aperture size, emphasizing the impor­ tance of telescopes as light-­gathering “buckets” used to see dim objects.

EXPLORATION SOLUTIONS 1. The eraser was at the bottom of the a­ ctual pencil.

2. The eye sees the pencil upside down.

3. If we need an object to move from the top to the cen­ ter of our field of view, we need to move the telescope down.

4. We first saw a “crisp” image at an eye position of 3.2.

5. The pencil was upside down.

6. The eyepiece of a telescope is small and makes a small image at a certain distance that the eye is away from the lens. If we move away from it, seeing the object in the telescope becomes very challenging ­because the eye must be exactly aligned.

7. If we rotated one way and the image became blurrier, we simply rotated the knob the other way.

8. Each person has their own par­tic­u ­lar focal length to see correctly. The focus of the telescope might allow one person without glasses or with a certain prescrip­ tion lens to see clearly, but someone with a dif­fer­ent eye focal length or dif­fer­ent prescription lens ­w ill need to adjust the focus of the telescope.

9. The bottom telescope has the longer focal length.

10. The longer focal length has larger separation of the red and blue stars.

11. A long focal length is a much larger telescope, which means that it ­w ill be heavier and take up more physi­ cal space; this can be a prob­lem with its own structure or fitting it into the space available. Also, long focal lengths do not view as much of the sky so a small focal length is needed if one wants a very wide view. Images ­w ill be dimmer as well.

50 ◆ CHAPTER 6 The Tools of the Astronomer

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 9: WHERE TO PUT THE TELESCOPE? In this activity, students review and work with the impact of the atmosphere on astronomical observations and observatories. A ­ fter completing this activity, they should be able to: • summarize how Earth’s atmosphere affects decisions about which telescopes are built and where they are located.

• distinguish among telescopes that can be used on the ground versus t­ hose that must be located in space in order to observe celestial objects at specific wavelengths.

• explain the advantages and disadvantages of ground-­ based telescopes and telescopes in orbit.

Text references: Sections 6.1 and 6.3

CHAPTER 7

The Formation of Planetary Systems

INSTRUCTOR NOTES

DISCUSSION POINTS

Chapter 7 covers the formation of stars and planets. Major topics include:

• For a think-­pair-­share activity, have students describe what they think planetary scientist do, as opposed to stellar scientists. As pairs form, ensure that ideas go further into detail other than “study planets” versus “study stars.” Discuss why astronomy has been split into separate, distinct disciplines. (LG 1) ­ ill use • Divide students into seven teams. Each team w the equation for spin angular momentum to calculate the values for the Sun, Jupiter, Saturn, Neptune, Earth, Pluto, or a typical molecular cloud (once ­every million years). Discuss the results in the context of conservation of angular momentum. (LG 2) • Assign the task of finding the “freezing” temperatures of iron, silicon (silicates), carbon (organic molecules), ­water, ammonia, methane, and nitrogen to students.

Convert temperatures to Kelvin and create a chart that runs from 2,000 to 20 K for the y-­a xis and 0.1 to 50 AU for the x-­a xis. Students slot each option into where it would first condense and at distances farther than that.

Discuss how this relates to the division between terrestrial planets and g­ iant planets. (LG 3) • For a 1–5-­m inute writing exercise, have students list as many of the pro­cesses as they can recall that are involved in the formation of the planets and other objects in the Solar System. Which ones ­were easy to recall, and which ones ­were hard? Discuss the interconnectedness of t­ hese pro­cesses and rank them according to when they occur during the formation stages. (LG 4) • If students had to choose just one technique among the leading choices for finding extrasolar planets that might be as habitable as Earth, which one they would pick? What makes that method advantageous over the ­others? (LG 5) • Ask students to search the Internet to find out how many extrasolar planets have been found in their star’s habitable zone. What cultural impact does that have?

What questions does it provoke? (LG 5) • This chapter’s opening image question is, “How did our Solar System form?” Student should return to Chapter 2 and list ­things they find that need to be

• the historical development of the nebular hypothesis for the Solar System formation.

• the collapse and conservation of angular momentum in a nebula.

• the formation and characteristics of a protostellar accretion disk.

• the hierarchical formation of planetesimals, planet migration, and formation of atmospheres.

• the formation of terrestrial and ­g iant planets.

• explanations as to why terrestrial and g­ iant planets segregate into dif­fer­ent regions.

• overviews of the vari­ous methods for the detection of extrasolar planets.

• types of extrasolar planets.

We have found that one of the most challenging aspects of this chapter for students is the cause for the segregation of planets into terrestrial and ­g iant bodies.

As students work to incorporate the physics they learned in the first chapters into this discussion, many of them mistakenly conclude that the accretion disk spins most of the gas to its outer region, so ­t here is very ­little material to form terrestrials; that heavy ele­ments are attracted to the center of the accretion disk b­ ecause they are more massive and feel more gravity, which is why terrestrial planets are rocky; or that ­g iant planets form purely out of gas and never have any rocky material in their cores. In addition to the discussion in this text that cooler temperatures lead to ele­ments in solid phase, we have found it useful to frame the prob­lem in terms of escape speeds of hot and cold atoms. In the inner regions of the accretion disk, atoms had very high temperatures and w ­ ere moving faster than the escape speed of the forming planetesimals, whereas in the outer accretion disk, ­these gases ­were cold and moving slowly, such that ­t hose planetesimals could gravitationally capture abundant gas and grow to very large sizes.

51

52 ◆ CHAPTER 7 The Formation of Planetary Systems explained or could be explained by the nebular hypothesis. Relate ­things on the lists to stages or pro­ cesses of a planetary system formation.

PROCESS OF SCIENCE Two ­things that students prob­ably listed as part of their discussion of the chapter-­opening image question was that all of the planets orbit the Sun in the same direction, and all of the major planets orbit close to the same plane.

The Pro­cess of Science for this chapter shows the power­ ful results when lines of inquiry converge from theorists, astronomers, and planetary scientists, in the case of the formation of the Solar System.

ASTROTOUR ANIMATIONS SOLAR SYSTEM FORMATION This animation begins with a rotating disk of gas that evolves to form a protostar and protoplanetary disk, and then fi­nally forms the other objects in the Solar System.

The animation emphasizes the timescales involved and the significant physical features that define bound­a ries between dif­fer­ent eras. Additional concepts covered are: the growth of grains to planetesimals, the effect the temperature differential in the disk had on planet formation, the final inventory of planets, and observational evidence for this scenario coming from observations of protoplanetary disks surrounding distant stars.

Text reference: Section 7.1 TRAFFIC CIRCLE This animation, which is also included as part of the AstroTour, Solar System Formation, shows an increasing number of cars entering a traffic circle. Although they come in from dif­fer­ent directions and at dif­fer­ent entry ­angles, all the cars end up spinning in a uniform disk, analogous to how material in a collapsing cloud forms an accretion disk. This analogy has a c­ ouple of shortcomings, however. The road is not a three-­d imensional disk and the cars are not spinning around the center due to the conservation of angular momentum.

Text reference: Section 7.2

ASTRONOMY IN ACTION VIDEOS ANGULAR MOMENTUM This video uses a “rotating platform of doom,” on which Dr. Palen stands. She is given a small shove to her out-

stretched arms, and then brings t­ hose arms in close to the body to demonstrate the conservation of angular momentum. The demonstration is repeated with barbell weights now held out away from the body. A slow spin is started, and then the barbells are brought in closer to the body to demonstrate an even greater increase in the spin rate. The dependence of the amount of angular momentum on the mass, the spin rate, and the distribution of the mass is emphasized. (Length: 2:58) Text reference: Section 7.2 DOPPLER SHIFT An ­actual classroom demonstration of the Doppler shift of light is, of course, impossible to carry out. In this video, Dr. Palen uses a buzz­er located at the end of a long string to demonstrate the Doppler shift of sound while swinging the buzz­er in a circular motion above her head. Emphasis is made on the pitch of the sound as the buzz­er comes ­toward the viewer and away. (Length: 2:14) Text reference: Section 7.5

INTERACTIVE SIMULATIONS RADIAL VELOCITY This simulation provides the means for students to visualize multiple views of a star and an orbiting planet. Another panel shows the observed radial velocity against phase. Students can adjust inclination, longitude, masses of the star and planet, semimajor axis, and eccentricity of the orbit.

They can even add noise to the data and control the number of observations to simulate ­actual mea­sure­ments.

Text reference: Section 7.5 and Ch7

TEACHING READING ASTRONOMY NEWS 1. Planets orbiting stars are gravitationally tugging on them, just as the stars are tugging on the planets.

Planets and stars are orbiting their common centers of mass. This means the more massive a given planet is, the greater the star ­w ill wobble.

2. All stars have some level of surface activity that is detectable using spectroscopy. Some types of surface activity that is periodical (such as ­those connected to the rotation of the star) can, in fact, be mistaken for an orbiting planet.

3. The astronomers ­were able to untangle the noise due to surface activity from the ­actual signal of an orbiting planet by observing a given star’s activity at many dif­fer­ent wavelengths. They also improved the sensitivity of their instruments.

CHAPTER 7 The Formation of Planetary Systems

4. A debris disk is basically all leftover material in the accretion disk that did not get put into a planet.

5. A planet that is tidally locked to its star would always have one side facing ­toward and the other facing away from the star. One side of the planet would always be heated, while the other side would not. It is not a disadvantage to the planet itself, per se, but rather to the development of life on a planet that lies in its star’s habitable zone.

CHECK YOUR UNDERSTANDING SOLUTIONS 7.1. (a, b, d) The number and size of planetary moons are not part of the nebular hypothesis.

7.2. (a) The orbital angular momentum of planets is much higher than the spin angular momentum of the central stars, especially for planets that are very massive and far from the star.

7.3. (b) Temperatures in the inner part of the forming planetary system ­were too high for volatile materials to condense. Refractory materials such as rocks (rocky silicates, metals) do condense at the high temperatures pre­sent within around 2 AU from the new Sun. 7.4. (a) A terrestrial planet (meaning one large enough to clear out its orbit) is unlikely to have formed far out beyond Neptune ­because material moved too slowly and was too sparse in the formation period of our Solar System to accrete into a regular planet. Rather, the planet prob­ably migrated out.

7.5. (c) Transit methods generally are the most sensitive to Earth-­l ike planets orbiting close to their stars.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (a, b, h, c, e, d, f, g) Sections 7.2 and 7.3.

2. (c) ­Because angular momentum L = mvr, we see that if the distance r is halved, the speed v must be doubled.

3. (b) ­Because the inner material was hotter than the outer Solar System, the inner planets w ­ ere not able to capture the large amounts of gas that are pre­sent in the outer planets.

4. (b) In the inner Solar System, the temperatures ­were too hot for the larger planetesimals to attract a significant amount of the gas that was pre­sent in the solar nebula. Any gas that was captured ­later escaped.

However, the outer planets ­were able to hold onto their primary atmospheres of hydrogen and helium.

◆ 53

5. (a) Large planets close to the star tug the stars by a large amount with short periods, making them the easiest to detect.

6. (c) Current theories of protoplanetary disks suggest that they dissipate in about half the time it would have taken Jupiter to form by accretion alone.

7. (a) When ice skaters through their arms out are increasing the radius. Since the mass d­ oesn’t change, the velocity must decrease for angular momentum to be conserved.

8. (b) In the early Solar System, collisions between particles and clumps cause them to stick together and grow in size.

9. (a) Observing a transit means detecting a drop in the light reaching us from a star, which ­favors a large planet that is close to the star, thereby blocking a significant amount of the star’s light. Plus, ­these close planets ­w ill also have short-­period orbits and do not require as much observation time to confirm their existence.

10. (a) As shown in Working It Out 7.1, spin angular momentum scales as L ∝ R 2 / P, so if the radius is cut in half, the period must drop by a ­factor of 22 = 4.

11. (d) Angular momentum of a s­ pherical object, w ­ hether spin or orbital, does not depend on its temperature but on mvr.

12. (c) Volatiles are easily evaporated or broken up with heat, and it was too hot in the inner Solar System for ­these compounds to be pre­sent once the accretion disk formed.

13. (d) Asteroids and comets are made almost exclusively of the pristine material out of which the Solar System formed.

14. (c) Conservation of angular momentum means that the direction of initial rotation dictates the direction of final rotation.

15. (d) The distance and size of the habitable zone for a star depend on the star’s type, but in general, the habitable zone is considered to be the region in which liquid ­water could exist on a planet’s surface.

Thinking about the Concepts 16. The hydrogen and most helium atoms that make up the Sun and the rest of the Solar System came from the Big Bang. Every­thing else—­a ll of the other ele­ ments and the molecules formed from them—­was formed in previous generations of stars and in the physical and chemical pro­cesses that took place, and are currently taking place, in the Solar System.

17. Stellar astronomers looking at young stellar objects have noticed that many of them are located within

54 ◆ CHAPTER 7 The Formation of Planetary Systems dark, dusty disks, as shown in Figure 7.2. Planetary scientists looking at our own Solar System noticed that the planets all lie in a disk orbiting in the same direction, and by studying meteorites they found evidence of larger objects being built up from smaller ones. This suggests that our Solar System formed from a disk of gas and dust. The two findings suggest a common origin for planetary systems.

18. A protoplanetary disk is the spinning disk of gas and dust out of which planets form around the central star.

The inner part ­w ill be hotter ­because it is closer to the central star and b­ecause the inner material gained energy when it moved in from the outer regions.

19. “Conservation” of a quantity means that the sum total of that quantity within a system must keep a fixed level.

Another way to think about this is that conserved quantities are like transactions at a bank: if you borrow from the bank, you have to pay it back. Consider ice-­ skaters, who have a fixed amount of energy in their bodies. If they skate very quickly (i.e., expending a lot of energy of motion), they must get that energy from somewhere; in this case, they convert some of their own biochemical energy. An object can always gain or lose mass, energy, momentum, or angular momentum, but it has to come from or go to another object.

20. Scientists ­w ill have to propose a formation mechanism consistent with all other evidence for planet formation. Perhaps ­these planets ­were ejected from one planetary system and ­were captured by this star, or underwent numerous close gravitational encounters, for example. Scientists w ­ ill consider all the relevant physics, propose the most likely scenarios, and then propose tests of ­these scenarios and search to validate or invalidate them.

21. Spin is a conserved quantity, but it is mea­sured by the product of how fast you rotate and how far away parts of you are from your axis of rotation. Thus, when an ice-­skater spins slowly with her arms extended, she has a fixed amount of spin. As she brings her arms in ­toward her body, that distance gets smaller, so the rate of rotation has to increase to keep the product of the two constant, and thus the skater spins much faster.

22. An accretion disk is the thin, rotating disk that forms as a gas cloud collapses on itself. It is out of this disk that the central star and planets form. Broadly speaking, this disk allows material to spiral into the star by converting its orbital energy into heat. When material spins in a disk around a star, friction between the dif­ fer­ent particles heats up that material and allows it to radiate energy. As particles give off that energy, they spiral deeper down the disk t­oward the star, eventually landing on the star.

23. Small grains of dust moving about the Sun (or other stars) in similar orbits randomly collide with each other at gentle speeds. ­These small grains stick to each other electrostatically to form increasingly larger seeds. As the seeds grow, gravity becomes more impor­ tant. Rather than awaiting random collisions, the largest seeds begin to gravitationally attract nearby particles, and their growth rate increases exponentially.

Eventually, the largest seeds reach planetesimal sizes.

When planetesimals collide, the resulting growth leaves planet-­sized objects.

24. When we blow “dust bunnies” ­toward each other (provided they ­don’t fly apart), they tend to become tangled with each other and grow in size. Although on Earth they w ­ ill never grow enough in size to pull on each other in a meaningful way, this could certainly happen in outer space. Earth’s gravity, airflow, and friction with the floor keep the bunnies from forming self-­g ravitating collections ­under our beds.

25. Rocky materials are refractory, meaning they condense at very high temperatures. As such, we expect that rocky materials w ­ ill solidify in all regions of the nebula, including near the proto-­Sun. Volatiles condense only at very low temperatures. Only the regions of the solar nebula far from the proto-­Sun ­were cool enough for ­these materials to solidify. This picture is consistent with modern observations of our planetary system.

26. The ­giant planets had several advantages over the terrestrial planets to enhance the growth of their atmospheres. (1) ­Giant planet seeds formed from more abundant volatile materials, whereas terrestrial planet seeds formed from the much less abundant refractory materials. (2) ­Giant planets had a much larger area from which to accumulate raw materials, whereas terrestrial planets could draw material only from a narrow zone in the immediate vicinity of the forming proto-­ Sun. (3) Accretion in the outer nebula left g­ iant planets less vulnerable to the power­f ul solar winds emanating from the proto-­Sun; the terrestrial planets ­were scoured clean ­because of their proximity to the proto-­Sun. (4) ­Because of their ability to quickly grow very large, the Jovian planets captured and retained significant amounts of the primordial hydrogen and helium from the solar nebula. Meanwhile, the terrestrial planets ­were too small and too exposed to the young Sun to capture or retain significant atmospheres from the solar nebula.

The secondary atmospheres of the terrestrial planets came from volcanism, with contributions from comets that collided with them.

27. The four methods that astronomers currently use to search for extrasolar planets, and the limitations of each method are as follows:

CHAPTER 7 The Formation of Planetary Systems

◆ 55

(1) spectroscopic radial velocity method: we watch the 67,000 mph = 910 smaller than (or 0.1 ­percent of) periodic Doppler shifts of emission lines from the star 74 mph due to an orbiting planet dragging the star around in Earth’s orbital speed around the Sun. space. This method preferentially selects large, massive Review: As expected, a star moves much less than the planets close to their stars; detecting an Earth-­like planet orbiting it. We see ­here that astronomers are planet would be extremely difficult. This method gives indeed detecting very small radial motions in a star to us just the minimum mass of a planet. By using just this detect the planet orbiting it. method, we cannot rule out a brown dwarf as causing 32. Setup: The ­table gives us values of planetary masses the Doppler shifts. in terms of Earth’s mass, so for this question we must (2) transit method: we watch a star regularly change sum up the planet masses and then compare them to brightness due to a planet passing in front of the star Jupiter and Earth. and eclipsing it. This method works only if we are viewSolve: (a) The sum of the eight planets in our Solar ing the orbital plane of the planets nearly edge on.

System is Mplanets = (0.055 + 0.815 ​+ 1.00 + 0.107 + ​ (3) microlensing method: a planet passes in front of a 317.83 + 95.16 + 14.54 + 17.15) M Earth = 446.66 M Earth. distant star and its gravity bends some of the star light (b) To find Jupiter’s percentage of that mass, divide ­toward Earth, making the star brighten for a short its mass by the total planetary mass: period of time. The microlensing effect is brief, the MJ 317.83 M Earth change in brightness minute for a planet, and ­there are = = 0.7116 or 71.16 percent.

M planets 446.66 M Earth millions of stars to monitor.

(4) direct imaging: we take an image that resolves (c) The same logic for finding Earth’s percentage the planet from its central star. Method involves of the total planetary mass yields detecting a dim object in the glare of an extremely M Earth 1 M Earth = = 0.0022 or 0.22 percent. bright star, and eliminating the possibility that what M planets 446.66 M Earth is detected is just a much more distant star. This Review: It is no surprise that Jupiter dominates the method works if the planet is far from its host star Solar System’s planetary mass, given how large and and large enough to reflect a significant amount of massive it is compared to every­thing e­ lse. We can the star’s light. simply see from inspection that almost all the mass is 28. Stars are very bright and far away, and the very faint in Jupiter and Saturn. light from the exoplanets would appear extremely Setup: For this prob­lem, we w ­ ill use the equation of 33. close to their stars. Thus, it is difficult to mask out the 4π mR 2 light of the star and still see close enough to it to see and orbital spin angular momentum Lspin = the reflected light of a planet.

5 P 29. Many of the initial discoveries of exoplanets ­were angular momentum L orbital = mvr. Remember to use made using the radial velocity method, which at the consistent units (e.g., time in seconds, distance in time could do no better than resolve massive planets in meters). orbits close to their stars with periods of about a week.

Solve: For Earth’s spin angular momentum, P = 24 h ​ 30. The Kepler satellite used the planetary transit method = 86,400 s, so and the monitoring of hundreds of millions of stars in 4π (5.97 × 1024 kg) (6.378 × 106 m)2 Lspin = = 7.0 × 1033 kg m 2 / the constellation of Cygnus to search for exoplanets.

5 86,400s By “Earth-­like” we mean planets that are terrestrial 4π (5.97 × 1024 kg) (6.378 × 106 m)2 = 7.0 × 1033 kg m 2 /s. Calculating the orbital angular and have roughlyLspin the =same mass as the Earth, have 5 86,400s momentum by comparison, using 1 AU = 1.5 × 1011 m, thick protective atmospheres, and orbit in the habitLorbital = (5.97 × 1024 kg)(2.98 × 104 m/s)(1.5 × 1011 m) ​ able zones of their stars. = 2.7 × 10 40 kg m2/s.

This is about 3.8 million times more than the spin Applying the Concepts angular momentum.

31. Setup: The conversion from m/s to mph is Review: Recall that angular momentum depends on 1 km 1 mi 3600 s m how fast an object is spinning and how far that object 1 = × × = 2.3 mph.

1 hr s 1000 m 1.6 km is away from its spin axis. Thus, it stands to reason Solve: From Figure 7.17, half the full range of radial that all the angular momentum of a planet is in its velocities is about 32 m/s. This gives us 32 m/s × 2.3 orbit ­because the planet’s distance from the Sun is so mph per m/s or about 74 mph. This is a ­factor of much greater than its size.

56 ◆ CHAPTER 7 The Formation of Planetary Systems Solve: (a) The density of Vesta is 2.7 × 1020 kg ρ = 4π = 3,464 kg/m 3 .

5 3 (2.65 × 10 m) 3 (b) Vesta’s density is higher than that of w ­ ater 3 (1,000 kg/m ) or the average for rocks (range from ~2,500 to ~2,800 kg/m3). This implies that Vesta is mostly made up of rock but with some metals as well in its differentiated interior.

2 Review: While we have meteorites that are thought to LVenus ⎛ m Venus ⎞ ⎛ R Venus ⎞ ⎛ PEarth ⎞ 1 2 Solve: =⎜ 0.815 × 0.949come × from = a0.003 have large impact that Vesta experienced a = ⎜ ⎜ ⎟ ⎟ ⎟ LEarth ⎝ mEarth ⎠ ⎝ R Earth ⎠ ⎝ PVenus ⎠ 243ago or so, t­ hese meteorites are all stony, and billion years 2 ⎞ ⎛ PEarth ⎞ 1 prob­ably came from Vesta’s mantle. The density we cals 2 ⎟ ⎜P ⎟ = 0.815 × 0.949 × 243 = 0.003 culated ­here confirms that Vesta is large enough that it ⎠ ⎝ Venus ⎠ must have under­gone differentiation as it was forming.

Review: Although Venus is practically Earth’s twin 37.

Setup: To compute the densities of Venus, Jupiter, in terms of mass and size, Venus rotates very slowly and Neptune, we ­w ill use the equation for the volume about its axis, which is why it has so l­ ittle spin angular 4π 3 momentum. of a sphere, V = R , and divide the mass of each 3 35. Setup: For this prob­lem, we w ­ ill use the equation of planet by its volume. orbital angular momentum L orbital = mvr. However, Solve: The text does not give any values for densities, instead of computing L for each planet, let’s directly so we have assumed an average for rocks and the dencompare the values of both by constructing a ratio. sity of hydrogen gas at standard temperature and This has two added advantages: (1) it gets rid of the pressure. See Table IM 7.1, below. constants in our formula, and (2) we can use any units Review: The comparisons imply that Venus has a we want as long as we are consistent, that is, masses in substantial percentage of metals in its interior, Jupiter terms of Earth mass, speeds in km/s, and distance in must be made up of more than just ­water and gas astronomical units (AU). (although it is massive enough to have compressed L Jupiter ⎛ m Jupiter ⎞ ⎛ v Jupiter ⎞ ⎛ rJupiter ⎞ 13.1 km/s its=interior), and Neptune may have some Solve: 727 = = 318 × hydrogen×in5.2 29.8 km/s LEarth ⎜⎝ mEarth ⎟⎠ ⎜⎝ v Earth ⎟⎠ ⎜⎝ rEarth ⎟⎠ rock along with ­water and gas in its composition.

38. Setup: Kepler’s Laws tell us that closer planets orbit r 13.1 km/s ter ⎞ ⎛ Jupiter ⎞ × × 5.2 = 727 = 318 faster. ⎟ ⎟ ⎜ r 29.8 km/s th ⎠ ⎝ Earth ⎠ Solve: (a) The red planet is the closest so it is the fastest, Review: Given Jupiter’s huge mass and greater disfollowed by the blue planet and then the purple one. tance from the Sun, it is no won­der that it has so much (b) The purple planet has the longest duration in its tranmore angular momentum around the Sun than Earth. sit ­because it moves the slowest, and thus it takes the lon36. Setup: To compute the density of Vesta, we ­w ill gest amount of time to cross the disk of its parent star. assume that it is a sphere, and use the equation, Review: On the figure, orbital speed is represented 4π 3 by how often the transit occurs. The red planet has the V= R , where R is the radiius. To find the den3 most transits so it travels the fastest, followed by blue sity of a sphere, we divide its mass by its volume. and then purple.

34. Setup: For this prob­lem, we w ­ ill use the equation of 4π mR 2 . However, spin angular momentum, Lspin = 5 P instead of computing the spin for each planet, let’s directly compare the spins of both by constructing a ratio. This has two added advantages: (1) it gets rid of the constants in our formula, and (2) we can use what­ ever units we want as long as we are consistent, that is, masses in terms of Earth mass and periods in days.

Table IM 7.1 Comparisons average for rocks: ~2,500 to ~2,800 kg/m3 average for H gas at STP: 0.09 kg/m3

Planet

Mass (kg)

Radius (m)

Density (kg/m3)

Venus

4.87 × 1024

6.052 × 106

5,245

Much greater than rock, ­water, and gas

Jupiter

1.9 × 1027

7.1492 × 107

1,241

Less than rock, but greater than ­water and gas

Neptune

1.02 × 1026

2.4764 × 107

1,603

Less than rock, but greater than ­water and gas

CHAPTER 7 The Formation of Planetary Systems

◆ 57

39. Setup: Doppler shifting of light tells us that the relative shift of a line away from its rest wavelength equals the relative speed of the moving object compared to the speed of light. Thus, the formula can be written as

Solve: (a) We require the period in years, so convert1yr = 0.55yr. The average orbital ing 200 days × 365 days distance is therefore A = P2/3 = 0.552/3 = 0.67 AU. Δλ v (b) This planet orbits farther from its parent star than = .

Mercury orbits our Sun. In fact, it is close to the λ0 c orbital distance of Venus. The planet can be expected Solve: Solving the Doppler shift equation for light to Earth b­ ecause it is 33 ­percent 0.3 m/sto be warmer than v −7 find the wavelength shift, Δλ = λ0 = 575 nm × = 5.75 × 10 nm. closer to its parent star.

3 × 108 m/s c Review: Venus has an orbital period of 225 days and a 0.3 m/s v −7 λ0 = 575 nm × = 5.75 × 10 nm. distance of 0.72 AU. ­Because this hypothesized planet 3 × 108 m/s c has a period of 200 days, we know that its distance Review: This shows that to observe Doppler shifts in must be just ­under that of Venus, as found. vis­i­ble light, large velocities of the objects, high resolu43. Setup: Following Working It Out 7.3, we can estition spectra, and precise mea­sure­ments are required. mate the percentage reduction of light during a transit 40. Setup: Doppler shifting of light tells us that the rela2 ⎛ R Planet ⎞ tive shift of a line away from its rest wavelength equals as ⎜ ⎟ . In this prob­lem, we are given units in the relative speed of the moving object compared to ⎝ R Star ⎠ the speed of light. Thus, the formula can be written as terms of a solar radius (7 × 108 m) and Earth’s radius Δλ v (6.4 × 106 m). = . and in this prob­lem we are looking for the Solve: Given the values in the prob­lem, the per­ λ0 c centage decrease in light during the transit is shift, Δλ, away from a central wavelength, λ = 500 nm.

0

2

Solve: Solving our Doppler shift equation for ⎡ (4.5)(6.4 × 106 m) ⎤ ⎢ ⎥ = 0.0014 or 0.14 p­ ercent. light to find the wavelength shift, 8 ⎣ (1.1)(7 × 10 m) ⎦ 0.09 m/s v Review: Transits of planets should produce very Δλ = λ0 = 500 nm × = 1.5 × 10−7 nm.

3 × 108 m/s c small drops in brightness given how tiny the planet is Converting to meters, Δλ = 1.5 × 10 −16 m compared to the star, as we have found h­ ere. Note that Review: This is smaller than the size of subatomic in Working It Out 7.3, we see that Kepler-11C proparticles (~10 -15 m), and thus shows how very difficult duces a drop of only 0.00077, or ~0.08 ­percent, which it would be to detect such a gravitational signal using confirms that this planet for this prob­lem is larger the radial velocity method. than Kepler-11C. 4π 3 41. Setup: The Sun’s brightness ­w ill drop in proportion R , where R, is 44. Setup: The equation to use is V = to how much area of the Sun is blocked.

3 Solve: The Sun’s radius is 7 × 108 m, and Jupiter’s the radius. Density is mass divided by volume. How­ ecause area is proportional to radius is 7.2 × 107 m. B ever, let’s directly compare the values of Earth and 7 2 this planet by constructing a ratio. This has two added (7.2 × 10 m) ≅ 0.01 radius squared, Jupiter’s area is advantages: (1) it gets rid of the constants in our for8 2 (7 × 10 m) mula, and (2) we can use what­ever units we want as as much as the Sun’s. This means the Sun’s light long as we are consistent, that is, relative units as would drop by this amount, or about 1 p­ ercent. given in this prob­lem.

Review: Aliens should easily be able to detect this Solve: (a) This planet has a volume drop with even crude instruments. However, this 3 Vplanet ⎛ Dplanet ⎞ drop happens only once ­every 11.8 years, so they ­w ill = = 1.7 3 = 4.9 times that of Earth. have to watch the Sun very carefully and for a long VEarth ⎜⎝ DEarth ⎟⎠ time to detect it. (b) If both planets have the same density, then 42. Setup: To determine the orbital radius of a body, we M planet Vplanet M planet M Earth use Kepler’s third law. ­Here, the star has 1 solar mass, so or, rearranging, = = 4.9; = 2 3 we can use P = A . Mercury’s semimajor axis and M Earth VEarth VPlanet VEarth orbital period are 0.39 AU and 88 days; Venus’s are V M planet planet = 4.9; that is, the planet is about five times as massive 0.72 AU and 225 days. We predict that this exoplanet= M Earth VEarth as Earth. values ­w ill lie in between t­ hese quantities from its star.

58 ◆ CHAPTER 7 The Formation of Planetary Systems Review: A planet of this mass may be considered more of a gas ­g iant than a terrestrial planet, so its average density may, in fact, be lower. It w ­ ill be in­ter­est­ing to see if we can learn w ­ hether this planet is terrestrial or g­ iant in nature.

45. Setup: We are comparing to Jupiter, so note that MJupiter = 1.9 × 1027 kg and RJupiter = 7.15 × 107 m. For volume: V =

4π 3 R . For density: ρ = M .

3 V

Solve: (a) M = 2.33 MJupiter = (2.33)(1.9 × 1027 kg) ​ = 4.43 × 1027 kg. (b) R = 1.43RJupiter = (1.43)(7.15 × 107 m) = 1.02 × 108 m. (c) V =

4π 3 4π R = (1.02 × 108 m)3 = 4.44 × 1024 m 3 .

3 3

(d) ρ =

M 4.43 × 1027 kg = = 998 kg/m 3 .

4.44 × 1024 m 3 V

The value for its density is just slightly less than the density of w ­ ater (1,000 kg/m3). This planet must be gaseous.

Review: We expect that planets having this large of a dia­meter and mass ­w ill all be gas ­g iants, and the density we found confirms this expectation.

Using the Web 46. Exoplanet cata­logs: a. Involves sorting exoplanets by name and selecting a star with multiple planets to graph distances from the star, masses and sizes, and comparing to the Solar System. For example (to name just a few): Kepler-150 has 5 planets; Kepler-167 has 4; Kepler-20 has 6; Kepler-444 has 5; and, of w ill vary. b. course, Trappist-1 has 6. Answers ­ Involves investigating the “­Table” option for exoplanets, and recently discovered planets. Answers ­w ill vary.

47. Space missions: a. Kepler Mission: 2,332 confirmed planets and 2,424 candidate planets. The Kepler mission was ended in October, 2018. Most systems as seen in the latest version of the “Kepler Orrery,” an animation that shows multi-­planet systems discovered by Kepler are definitely not like ours. Most planets are very close to their star. b. TESS mission: First image was taken on August 7, 2018; released on September 17, 2018. In­ter­est­ing orbit; answers ­w ill vary with time.

48. Citizen science proj­ects I: a. Prompts students to sign up as citizen scientists. Answers ­w ill vary. b. Prompts students to go to “Planet Hunters” and classify a few images. Answers w ­ ill vary.

49. Citizen science proj­ects II: Prompts students to go to “Disk Detective” and classify a few images. Answers ­w ill vary.

50. Students are to go to the “Super Planet Crash” web page and build systems and test stability. Answers ­w ill vary. (This site involving the crashing of galaxies is informative, educational, and enjoyable for students.)

EXPLORATION This simulation provides the means for students to visualize multiple views of a star and an orbiting planet. Another panel shows the observed radial velocity against phase.

Students can adjust inclination, longitude, masses of the star and planet, semimajor axis, and eccentricity of the orbit. They can even add noise to the data and conactual trol the number of observations to simulate ­ mea­sure­ments.

EXPLORATION SOLUTIONS 1. Earth’s view is most like the side view.

2. This orbit is circular.

3. The maximum radial velocity of the star is about 30 m/s.

4. Between the minimum and maximum radial velocity, the phase changes by 0.5, meaning half of the full period of 1 year, or 182.5 days.

5. If the planet moves away from Earth, the radial velocity of the planet would be positive (away from Earth) and that of the star would be negative (­toward Earth).

6. The orbit is now eccentric.

7. The planet moves fastest when it is closest to the star (this is Kepler’s second law).

8. When the star moves fastest, the planet is closest to the star (this is Newton’s third law).

9. The shape of the radial velocity curve is very dif­fer­ent for a circular and elongated (or eccentric) orbit. An astronomer would model the path of the orbit based on the shape of the radial velocity curve.

10. This planet is a good candidate for a transit observation b­ ecause the orbital plane has the planet eclipse the star.

11. With inclination set to 0.0, the Earth view is now like the orbit view.

12. We now do not detect any radial velocity.

CHAPTER 7 The Formation of Planetary Systems

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 10: 51 PEGASI: THE DISCOVERY OF A NEW PLANET Students determine a planet’s orbital period and the radial velocity amplitude of the parent star and then use ­these data with equations derived from Kepler’s and Newton’s laws in order to find the radius of the orbit and a lower limit on the mass of the planet. By comparing this planetary system to the Solar System, students also gain an appreciation for why astronomers ­were so surprised and skeptical when this discovery was first announced in 1995. On completion of the activity, students should be able to: • apply Kepler’s and Newton’s laws to find orbital and physical characteristics of an exoplanet.

• compare an extrasolar planet to the more familiar planets of our Solar System.

Text reference: Section 7.5

◆ 59

ACTIVITY 11: RANKING TASK FOR FORMATION OF PLANETARY SYSTEMS In this activity students rank the events that take place during star and planet formation. The goal is for them to visualize the birth of a star and planets in steps that begin from a ­g iant, cool, rotating molecular cloud and end with fully formed planets orbiting a genuine star. The students ­w ill accomplish this by: • correctly ordering the stages of the formation of a planetary system.

• identifying the stages where overlapping of events occurs.

• recounting the modern theory of planetary system formation.

Text references: Sections 7.2–7.4

CHAPTER 8

The Terrestrial Planets and Earth’s Moon INSTRUCTOR NOTES Chapter 8 covers the terrestrial worlds of the inner solar system—­Mercury, Venus, Earth, and Mars. The Moon is included ­because it has many characteristics similar to ­these planets. The chapter uses comparative planetology to introduce similarities, differences, and pos­si­ble c­ auses and effects to help students understand the wide variety of characteristics and properties for t­hese inner worlds.

Major topics include: • the four methods of resurfacing (impacts and craters, tectonism, volcanism, erosion).

• radioactive dating as the means to establish ­actual ages.

• means by which we probe the interior pro­cesses of the terrestrial worlds.

• an overview of the magnetic fields (or lack thereof) for the terrestrial worlds.

• tectonism and volcanism, especially for Earth.

• the geological evidence and search for ­water.

DISCUSSION POINTS • Show images of vari­ous parts of the lunar highlands and Maria to students. From t­ hese images, have ­students deduce the cratering rate on the Moon, characterize a few dif­fer­ent types of craters, and hypothesize why the near side of the Moon has so many large impact basins while the far side has very few. (An in­ter­est­i ng article on this topic can be found at: https://­w ww​.­scientificamerican​.­com​/­a rticle​ /­g ravity​-­maps​-­reveal​-­why​-­dark​-­side​-­moon​-­covered​ -­i n​-­craters​/­) (LG 1) • Compare and contrast images of cratering features on ­these worlds and deduce ­whether the pro­cesses found on Earth w ­ ere ever pre­sent on t­ hese other bodies. For example, compare Figures 8.1, 8.4, and 8.6: What can we deduce about the dif­fer­ent characteristics of each crater? (LG 1) • Carbon-14 is a radioisotope with a half-­l ife of 5,700 years. Discuss why dating of fossils using carbon-14 is dif­fer­ent from the dating of minerals using 60

the radioisotopes mentioned in Working It Out 8.1.

(LG 2) • Look at the sizes and masses of the bodies in ­Table 8.1.

Find the current state of geologic activity of each object and relate the information to the calculations in Working It Out 8.2. How recently did the dead planets become dead? How soon are the active ones likely to become inactive? (LG 3) • As a class, share knowledge and experiences students have with tectonism and volcanism on Earth. Are ­there any recent media reports about e­ ither pro­cess?

Assign Mercury, Venus, the Moon, and Mars to groups and have each report back on their summaries of similar pro­cesses on t­ hese other worlds. (LG 4) • ­Water ice is abundant on the surfaces Earth and Mars, and it is detectable on the Moon and Mercury. Th ­ ere is ­l ittle evidence that Venus has ­water pre­sent in its thick atmosphere. We discussed the importance of our efforts to detect w ­ ater on extrasolar planets. Do we have the same reasons for wanting to verify ­water (liquid or ice) on the inner planets, or are we interested in dif­fer­ent information? (LG 5) • The chapter-­opening image shows a close-up image of a steep cliff on Mars and what appears to be erosion of the cliffsides by liquid w ­ ater. This image and many more of Mars’s surface are giving us tantalizing evidence for the presence of past liquid w ­ ater on our nearby planet. What would be the pos­si­ble outcomes if it ­were ascertained that Mars did, in fact, have liquid ­water on its surface at one time?

PROCESS OF SCIENCE It was not very long ago (although it was before plate tectonics was well understood) that a popu­lar hypothesis had the Moon being “thrown out” from the volume of the Pacific Ocean due to Earth’s rapid rotation at the time. While the theory that the Moon was formed when a Mars-­sized object impacted Earth when it was still in a slightly molten state is currently the favored scenario, the other three theories cannot yet be falsified and discarded.

Chapter 8 The Terrestrial Planets and Earth’s Moon ◆ 61

ASTROTOUR ANIMATIONS PRO­CESSES THAT SHAPE THE PLANETS This interactive animation illustrates three f­ actors affecting the shape and resurfacing of a planet or moon, including tectonics, impact cratering, and weather erosion. The tectonics option covers continental drift, convection, and types of faults. The impact-­cratering option covers the kinetic energy generated during the impact, crater features, and why Earth has so few craters. The erosion option covers weathering of rocks, chemical reactions, and freeze-­thaw cycles. For all options, comparisons are made between Earth and other terrestrial planets.

Text reference: Section 8.1 CONTINENTAL DRIFT This animation shows the original “Pangaea” supercontinent drifting over time to become the continents we know ­today.

Text reference: Section 8.4 HOT SPOT CREATING A CHAIN OF ISLANDS This interactive animation starts with an animation of a mantle plume. It then shows how a stationary source of magma coupled with a moving continental plate on top of it results in the formation of volcanic islands. It ends with the erosion of the older islands, reducing them to seamounts.

Text reference: Section 8.5

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. The Moon’s radius is about one-­fourth that of Earth’s, and b­ ecause of its small size—­not its mass—it would have retained its internal heat for only a short time, cooling about 4 times as fast as Earth did.

2. The other cause of the dynamo weakening for the Moon was its gradual moving away from Earth and Earth’s tidal forces generating internal friction and heating.

3. The presence of a magnetic field around a planet or moon would mean that its surface would be shielded from the large flux of charged particles from the Sun and the rest of space.

4. The Moon is still moving away from Earth due to the Moon lagging ­behind in its orbit when compared to Earth’s rotation. Billions of years ago, the Moon was a lot closer to Earth and Earth rotated much faster. That extremely strong tidal “tug” from Earth caused the Moon to move to a higher orbit.

CHECK YOUR UNDERSTANDING SOLUTIONS 8.1. (b) Younger craters always lie on top of older ones based on what’s called the law of superposition.

8.2. (d) 40 seconds is two half-­lives, so (1/2)2 = 1/4 ­w ill remain.

8.3A. (d) Differentiation takes place based on an object’s mass and volume, which means its density.

8.3B. (c) A dynamo requires fast rotation and a liquid core that responds to it.

8.4. (c) Only Earth has evidence of plate tectonic activity now.

8.5. (b) The distance from the Sun has no bearing on the geologic activity of Mars.

8.6. (a, c, d, e) Venus may have small traces of ­water vapor in its atmosphere, but ­t here is no evidence for liquid on the surface, which is why we ­can’t include it.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (b) Impacts, tectonism, and volcanism build up structures on the terrestrial planets, whereas erosion tears them down.

2. (b) Newer material is always laid down on top of older material; therefore, the top layers ­w ill be younger (or the same age) as material that is lower down.

3. (e) All the answers that are given are ways to learn about the interiors of planets.

4. (c) Tidal effects and radioactive decay are the best answers. We eliminated the other options by reasoning. Angular momentum ­w ill keep the core of Earth spinning but w ­ ill not heat it. Tidal effects frictionally heat an object. Radioactive decay injects energy into the core at a continual rate, which keeps the core warm. Gravitational contraction is impor­ tant in gas ­g iants, but not for the solid/liquid core of Earth.

62 ◆ Chapter 8 The Terrestrial Planets and Earth’s Moon 5. (d) During each half-­life, one-­half of the population decays, so in 4 half-­l ives, (1/2)4 = 1/16 remains.

6. (c) Earth’s surface is constantly moving and changing, so its surface is renewed.

7. (c) We can study the orientation of structures in rocks to determine where the field lines pointed; by dating ­those rocks, we can trace the changes of Earth’s magnetic field.

8. (a) S-­waves cannot travel through liquid. Answer (e) is also a possibility, but since the scientists are on “the other side of the planet,” a liquid interior is more plausible than that of a fully liquid mantle.

9. (d) Radiometric dating can be done on any sample that is collected and analyzed by the appropriate equipment. Cratering rates are relatively well known, so the ages of other surfaces may be determined, although with less certainty.

10. (c) Cratering rates ­today are low. They ­were much higher during the early days of the Solar System.

11. (d) Earth experiences constant erosion and resurfacing through ­human activity, wind, ­water, and volcanism. This resurfaces the planet faster than on Venus.

12. (a) The Moon has gone through periods of partial resurfacing, so the fact that it is highly cratered means that t­hese happened long ago, due to a heavy bombardment.

13. (b) In a liquid, denser material always sinks to the bottom. If Earth had not been molten at formation, one must ask how its layers achieved the same density structure as one would get from a molten planet.

14. (c) Although tidal effects do contribute a small amount, the predominant source of internal heating is radioactive decay.

15. (a) From Working It Out 8.2, heating scales roughly with volume (V ∝ R 3), so a planet half as large as Earth ­w ill heat (1/2)3 = 1/8 or 0.125 times as fast.

Thinking about the Concepts 16. Our Moon is similar to the other terrestrial planets (especially given our theory of how it formed) and gives us a good basis for understanding Mercury and Mars, as well as the moons of the ­g iant planets.

17. Igneous rocks can be dated radiometrically ­because all parts of the rock formed and cooled in the same place at the same time. Therefore, any parts of an igneous rock w ­ ill be the same age as any other part of the rock. Sedimentary rocks are formed as an amalgamation of deposited materials that formed in dif­fer­ent places and at dif­fer­ent times. As a result, dif­fer­ent parts of the same sedimentary rock may have vastly dif­fer­ent ages. Metamorphic rocks experience physical changes that may include one or more heating

events that can falsely reset their chronometers. Complex physical histories make it difficult to determine reliable ages for most metamorphic rocks.

18. We can select igneous rocks from vari­ous levels of the ­Grand Canyon and mea­sure their ages using radiometric dating. We then compare ages of rocks at the rim to ­those along the river and note that rocks increase with age as we pro­g ress deeper into the canyon. In geology, this is called the law of superposition, and it generally holds true in most geologic strata ­because younger sediments are deposited on top of older sediments.

19. Earth’s internal heat comes primarily from its early accretion, radioactive decay, and friction from motion in its interior. Magma originates deep in the crust and in the upper mantle where ­these sources of thermal energy combine.

20. The Moon is smaller than Earth, so it contained less thermal energy in its core to begin with, and a smaller core cools faster than a larger one. Thus, the Moon’s core is cool and solid, whereas the larger Earth’s core has still retained enough heat to remain liquid.

21. A transverse wave is one in which the direction of motion of the disturbance is perpendicular to the direction of motion, whereas a longitudinal wave has both in the same direction. Shaking a taut string makes a transverse wave, whereas stretching a spring and then giving it a flick makes a longitudinal wave.

22. Earthquakes generate two types of seismic waves that pass through Earth’s interior: primary and secondary waves. Primary waves pass through both solid and liquid materials, but secondary waves pass through only solid materials. Seismometers located near an earthquake source detect both primary and secondary waves. However, seismometers located 104° or more from an earthquake source detect only primary waves. This shadow zone for secondary waves is a clear indication of a liquid layer in Earth’s interior.

23. The impactor theory of the Moon is supported by the compositional similarities of Earth and Moon, the lack of volatiles on the Moon, and the inclination of Earth’s spin axis. However, evidence of the impacting object’s own composition has yet to be observed. It is hypothesized that part of the object lies deep within Earth.

24. Venus, Earth, and Mercury all show evidence of tectonic pro­cesses. ­There is no active tectonism on Mercury t­ oday, but the planet displays several large cliffs that appear to be the result of buckling as the planet shrank early in its history. All the tectonic features on

Chapter 8 The Terrestrial Planets and Earth’s Moon ◆

Mercury are compressive. Venus displays large circular tectonic fractures that some planetary scientists have interpreted as the upwelling of mantle hot spots.

Over geologic timescales, large chunks of the lithosphere centered over t­ hese hot spots may melt, causing Venus’s entire surface to overturn. Tectonic activity on Earth is dominated by plate motions. The plates are moved around the surface by convection in the upper mantle. It is not clear why Venus and Earth exhibit such dif­fer­ent tectonic be­hav­ior.

25. Plate tectonics is the fractured pieces of the planet’s upper crust floating on the denser mantle. Convection in the mantle moves t­ hese pieces of the planet’s surface around, generating new seafloor through volcanic activity in some places and subducting old seafloor back into the mantle at ocean trenches. Plate tectonics have been observed only on Earth.

26. Mars has the highest volcanoes in the inner Solar System: Olympus Mons, 21 km; Ascraeus Mons, 18 km; Arsia Mons, 17.8 km; Pavonis Mons, 14 km; and Elysium Mons, 13.9 km. Maxwell Montes on Venus is next at a height of 11 km. When mea­sured from the ocean floor, Earth’s Mauna Kea has a height of 10.2 km, making it number 7 on this list.

27. Small, ­simple, bowl-­shaped craters with no internal structure are likely to be impact craters. A crater formed by a volcanic eruption should be easily associated with nearby volcanic action and a buildup of lava and basalt. Large, complex impact craters may also display a central peak surrounded by a series of concentric rims. A pattern of smaller, secondary impact craters is usually found near young impact craters. If the resolution is sufficiently high, the rim should display an inverted geologic stratigraphy.

28. Venus, Earth, and Mars all have atmospheres, so ­there are many pro­cesses of gradation/erosion that can change each planet’s surface. ­These three planets show ample evidence of resurfacing through tectonics and volcanism. One can do a crater-­density analy­ sis and note that t­hese three planets have far fewer craters than the Moon and Mercury. The Moon and Mercury are geologically dead and have only impacts to change their surfaces.

29. Mars shows dry riverbeds, canyons, and teardrop-­ shaped erosions around craters, indicating that ­water was flowing and eroding the surface.

30. The Chicxulub crater is dated to the same time as the K-­T boundary, and both contain anomalous amounts of iridium and other radioactive ele­ments; both are suggestive of a major impact. ­There are no dinosaur fossils found above the K-­T boundary, and abundant ones below it, suggestive that this boundary marks a

63

cataclysmic event. The K-­T boundary itself is a thin layer found all over Earth, suggestive of ash, dust, or debris on a global scale.

Applying the Concepts 31. Setup: Figure 8.8 is presented in linear scale on the x-­a xis, and the y-­a xis is arbitrary, where higher equals more. We see that the cratering rate drops enormously up through about 3.3 billion years ago and then decreases very slowly thereafter.

Solve: (a) The cratering rate has changed quite abruptly around 3.3 billion years ago. (b) At pre­sent, the rate is negligible compared to 4 billion years ago. (c) If t­ here ­were many small masses flying around the early solar system, we would expect t­ here to be many collisions early on. As time goes by, t­ hose collisions clean up the debris, meaning t­ here are fewer objects for a planet or moon to hit, and the cratering rate must drop and slow at the same time.

Review: Note that very young surfaces should have very l­ittle cratering, according to the figure, which agrees with what we observe for young surfaces on Earth.

32. Setup: When looking at axes, remember that linear ones change by integer values, whereas logarithmic ones change by powers of some number, such as 2 or 10.

Solve: Figure 8.7 is shown with linear spacing. ­A fter one half-­l ife, the number of parents and ­daughters are the same. This must be a general feature ­because when we start the experiment, we have all parent and no ­daughter isotopes. ­A fter one half-­l ife, half the parents have become ­daughters, giving us the same number of both.

Review: The y-­a xis of this graph might appear logarithmic only ­because the numbers labeled each increase in spacing. But in a log plot, the spacing would not change, as t­ hese numbers increased by a ­factor of 2.

33. Setup: Exponential growth or decay means that the rate changes as a function of time, that is, it doubles or halves each time step.

Solve: The ­daughter growth is exponential ­because the number of d­ aughters doubles with each step.

Review: Note that b­ ecause parents become d­ aughters, if the change of parents is exponential decay, then the change of d­ aughters must be exponential growth.

34. Setup: When comparing ages of a surface, look for signs of erosion. Few large craters, or filled in bottoms, indicate a younger surface. The more craters seen, the older the surface.

Solve: Figure 8.18 shows a relatively smooth surface with lots of small craters. Figure 8.23 also shows a

64 ◆ Chapter 8 The Terrestrial Planets and Earth’s Moon very smooth surface with fewer small craters, suggesting this region is younger.

Review: Without radiometric dating, it is very hard to tell the ages of regions with roughly similar cratering, so the answer given to this question could be wrong. We also do not know the surface area of e­ ither image and thus cannot estimate the crater density nor the sizes of the craters.

35. Setup: Remember that the surface area of a sphere is expressed as 4πR 2 , where R represents the radius of the sphere.

Solve: (a) Surface area of Earth: 2 A Earth = 4π R Earth = 4π (6,378 km)2 = 5.11 × 108 km 2 .

(b) Similarly, surface area of Mars: 2 A Mars = 4π R Mars = 4π (3,390 km)2 = 1.44 × 108 km 2 .

(c) If 72 ­percent of Earth’s surface is covered with ­water, then the remaining 28 ­percent is land. The total land area of Earth’s surface is, then, A land = (0.28) (5.11 × 108 km2) = 1.43 × 108 km2 . In other words, the surface area of Mars is approximately the same as the total land area of Earth.

Review: Mars’s size is about one-­half that of Earth, so Mars has about one-­fourth the surface area. Given that about three-­quarters of Earth’s surface area is ­water, that means only one-­fourth of our surface area is land. An approximate calculation confirms our finding.

1 36. Setup: Kinetic energy: K = mv 2 .

2 Solve: The kinetic energy of the ice is KEice =

1 (0.001 kg)(5.0 × 10 4 m/s)2 = 1.25 × 106 J. 2

The kinetic energy of the SUV is

KESUV =

1 (2,000 kg)(90 km/hr × 1,000 m/1 km 2 × 1 hr /3,600 s)2 = 6.25 × 105 J.

Review: The SUV speeding at 90 km/hr has only half the kinetic energy of the 1-­g ram chunk of ice hitting Earth’s atmosphere at 50 km/s. This dramatic example illustrates how speed contributes much more to kinetic energy than mass does.

37. Setup: Density, ρ, is equal to the mass, M, divided by the volume, V, where we assume the impactor was a M 4 sphere: ρ = , and where V = π R 3 , R being the V 3 radius of the sphere.

Solve: The prob­lem gives the mass and radius, so 3.0 × 108 kg ρ = 4π = 4,580 kg/m 3 .

3 (25 m) 3 Review: ­Water has a density of 1,000 kg/m3, rock a few times this, and metals a few times more. Our conclusion is that this meteor was largely composed of metals.

38. Setup: Working It Out 8.2 explains that the rate of cooling scales as ratio of surface area to volume, or 1/R. Solve: Using data in ­Table 8.1, the Moon’s size is 0.272 that of Earth; therefore, Earth ­w ill cool off 0.272 times more slowly than the Moon, or alternatively, the Moon cools 1/0.272 ≈ 3.7 times faster.

Review: The larger an object, the less surface area it has for its volume; thus, the slower it cools. This is why Earth is cooling more slowly than the Moon did.

39. Setup: Density, ρ, is equal to the mass, M, divided by M the volume, V: ρ = . Since Earth is a sphere, V 4 V = π R 3, R being the radius of Earth.

3 Solve: (a) The prob­lem gives the mass and radius, 6 × 1024 kg making ρ = = 5,520 kg/m 3 .

4π 6 3 (6.378 × 10 m) 3 (b) ­Because crustal rock has a density of 2,600 kg/m3, Earth must have a sizable chunk of metal in its interior to bring the average density up to 5,500 kg/m3.

Average metal densities are 5,000 kg/m3 and higher. (c) Similar calculations for the Moon (mass = 0.012 M Earth, radius = 0.272 R Earth; from ­Table 8.1), giving a density of 3,292 kg/m3. (d) The Moon’s density is much lower than that of Earth, meaning it contains more rocks and fewer metals. The density of the Moon is similar to that of Earth’s crust.

Review: We know from this chapter that Earth has a metallic core, which this prob­lem confirms numerically.

40. Setup: ­Every half-­l ife reduces the radioactive sample size by half. In one half-­l ife, half of the radium would have decayed to radon. In a second half-­life, half of the remaining radium would have decayed to radon, meaning one-­quarter of the radium would remain, and so on.

Solve: In this prob­lem, we are told that ­there are three atoms of radon for each radium atom, which means that of four original radium atoms, three-­quarters have decayed. Based on the logic in the setup, this pottery has been around for two half-­ l ives, or

Chapter 8 The Terrestrial Planets and Earth’s Moon ◆

65

43. Setup: This prob­lem asks us for the relationship 2 × 1,620 yr = 3,240 yr. Therefore, the pottery is about among distance, speed, and time, which we know is 3,240 years old. d = vt. We also need to adjust the units for distance: Review: Radioactive dating is not exact, and uncer1,000 m 100 cm tainties in mea­sure­ments are usually stated. This 4,500 km × × = 4.5 × 108 cm. shard could be 3,000 to 3,500 years old and still give 1 km 1m 4.5 × 108 cm the same result. Since we know that pottery was being = 1.2 × 108 yr.

Solve: Solving for time yields t = made that long ago, our answer is reasonable.

3.75 cm/yr 4.5 × 108 cm 41. Setup: Working It Out 8.1 tells us ­every = 1.2 × 108 yr. South Amer­i­ca and Africa would have t = half-­life 3.75 cm/yr reduces the radioactive sample size by half. In the first begun separating about 120 million years ago if we half-­l ife, half of the carbon-14 would have decayed. In can assume that the separation rate has been constant a second half-­life, half of the remaining carbon-14 through geologic time. would have decayed, meaning a quarter of the carReview: Our first real­ity check is that this timescale bon-14 would remain, and so on. We can write this as is much shorter than the age of Earth, which is impor­ a formula where the fraction remaining equals (1/2)n, tant. Our second check is that animal and plant fossil where n is the number of half-­l ives. remnants on dif­f er­ent continents match, which means Solve: (a) Using the previous formula, 1/64 = (1/2)6, that the supercontinent could not have broken up bilso this sample is 6 half-­l ives old. lions of years ago; therefore, this number has about (b) ­Because each half-­l ife is 5,700 years, the sample the right value. is 6 × 5,700 yr = 34,200 years old.

44. Setup: The volume, V, of a cone is equal to the area of (c) If ­daughter nuclei ­were pre­sent in the sample when its base, b, multiplied by one-­third of its height, h: it formed, then we would mea­ sure a parent-­ to-­ π 1 V = bh = r 2h. We are making a comparison and daughter ratio that is smaller than actually occurred, 3 3 π meaning we would calculate the sample to have gone ­w ill use ratios, canceling out the in the pro­cess.

3 through more half-­lives and thus overestimate its Solve: The ratio of Olympus Mons (OM) to Mauna age.

Loa (ML) is Review: Part (c) shows us an impor­tant issue with 2 VOM ⎛ rOM ⎞ hOM ⎛ 275 ⎞ 2 27 radioactive dating, which is why we usually try to date = = × = 63. objects using many dif­fer­ent radioactive isotopes.

VML ⎜⎝ rML ⎟⎠ hML ⎝ 60 ⎠ 9 42. Setup: Working It Out 8.1 tells us that ­every half-­l ife Review: Olympus Mons has about 63 times more reduces the radioactive sample size by half. volume than Mauna Loa. Solve: (a) The age of the Solar System is about 4.5 bil45. Setup: Using Newton’s law of universal gravitation, 4.5 × 10 9 yr 5 the strength of gravitational force on the surface of a ≈ 8 × 10 lion years old, which is about 5,700 yr planet scales as M/R 2 . We could construct a ratio of ⎛R half-­l ives of carbon-14. In that time, all the carbon-14 F M this quotient for Mars and Earth, namely, Mars = Mars ⎜ pre­sent in an initial sample w ­ ill have decayed. This 2 FEarth MEarth ⎝ R MMars ⎛ R Earth ⎞ means radioactive isotopes with a short half-­life are FMars = .­Table 8.1 gives values for Mars in useless when estimating ages of objects that are many FEarth MEarth ⎜⎝ R Mars ⎟⎠ M R times older. relation to Earth: Mars = 0.107, Mars = 0.533; how­ MEarth R Earth (b) The number of half-­lives of rubidium-87 that have passed in the last 14 billion years are ever, it also gives the values of the surface gravity for 14 billion yr each planet, which we w ­ ill use ­here. = 0.29. In this many half-­l ives, (1/2)0.29 = ​ Solve: Using values from ­ Table 8.1, 49 billion yr 2 Surface − gravity 3.71 m/s Mars 0.82 of the rubidium pre­sent in a sample ­w ill remain. = = 0.38. This low Surface − gravity Earth 9.781 m/s2 In other words, some rubidium (100 ­ percent -­ 82 ­percent, or about 18 ­percent) has indeed decayed. surface gravity, which is 38 ­percent that of Earth’s Review: It is impor­tant to always date an object with surface gravity, would have meant lava would have an isotope whose half-­life is of the same order as the been less compressed than on Earth, perhaps explainage of the object. For human-­made objects and other ing why it was easier for volcanic mountains on Mars ­ atter, carbon-14 is very useful, whereas the organic m to be much higher than shield volcanoes on Earth. other isotopes listed would be used for objects that Review: The completeness of values in ­Table 8.1 date back millions or billions of years. expedited the solution to this prob­lem.

66 ◆ Chapter 8 The Terrestrial Planets and Earth’s Moon Using the Web 46. Involves the U.S. Geological Survey’s “Earthquake” website and checking for recent earthquakes. Answers ­w ill vary.

47. a. Involves the citizen science site and examining images of Mars. Answers ­w ill vary. b. Prompts students to create an account for the “Cosmoquest” proj­ ects and classify some images. Answers w ­ ill vary. c.

Prompts students to click on “Moon Mappers” on the “Cosmoquest” web site and classify a few craters.

Answers w ­ ill vary.

48. Space missions: Mars—­I nvolves three dif­fer­ent Mars missions and finding the latest science results.

Answers w ­ ill vary.

49. Space missions: The Moon—­I nvolves India’s Chandrayaan-­2 mission and the joint mission to the Moon planned by NASA, SpaceX, and ­w hether any missions are happening soon. Answers w ­ ill vary.

50. a. Students watch documentaries about the Apollo missions and comment. Answers w ­ ill vary. b. Students watch A Trip to the Moon (George Méliès, 1902, restored). The Selenians live beneath the surface of the Moon. The Moon’s surface is coarse, with high, sharp, jagged peaks. To answer h­ ere what the ­humans do to the Selenians would be a spoiler.

EXPLORATION The goal for this exploration is to give students a hands-on experiment meant to help them understand what is meant by an exponential function and what it means for radioactive decay. The experiment involves emptying the M&M’s from a 1.69-­ounce bag into their hands, shaking them, and pouring them onto a t­able. Counts are made for M-­side-up (­daughters), which are set aside. This continues for 10 t­ rials, a­ fter which sums are calculated and graphed: sum versus time step.

EXPLORATION SOLUTIONS 1. Early on, the number of M&Ms landing “M” side up decreases quickly. As time goes on, the rate of decrease slows.

2. If we treat radioactive decay the same way, then the rate of decay starts out very quickly and slows over time.

3. As time goes by, the number of M&Ms that remain gets smaller, as the number in the pile set aside grows.

4. If we treat radioactive decay the same way, then the number of parents decreases with time as the number of d­ aughters grows.

5. The plot given shows the sum, not the average. We did 10 t­ rials, so we have to divide the y-­a xis by 10 to find the average number remaining over time. Thus, to find the time at which 10 remain, we must look for 100 on the y-­a xis. This happens around 2.5 time steps.

6. If we know how many parents the sample started with, or the ratio of parents to d­ aughters, then we can compare the number or ratio of isotopes found to the decay rate and determine how many half-­l ives have passed.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 12: AGE DATING THE LUNAR SURFACE USING CRATERING STATISTICS Students find the relative ages of regions on the Moon by comparing the distribution of crater sizes in dif­fer­ent regions. This is a method that planetary scientists still use ­today. Whereas the scientists have a computer program designed to do this, students w ­ ill use two much smaller contrasting regions of the Moon to work with manually.

Specifically, students ­w ill be able to: • set criteria for mea­sur­ing the sizes of craters on the Moon.

• determine the distribution of sizes for both heavi­ly and lightly cratered regions.

• state the relative ages of the heavi­ly and lightly cratered regions of the Moon and the a­ ctual range of ages derived.

• list advantages and disadvantages between finding relative versus a­ ctual ages of surfaces.

• evaluate the effectiveness of this method for dating cratering events in the Solar System.

Text reference: Section 8.1

CHAPTER 9

Atmospheres of the Terrestrial Planets INSTRUCTOR NOTES Chapter 9 covers atmospheres, with par­tic­u ­lar applica­ tion to the terrestrial planets. Major topics include: • formation and loss of primary and secondary atmospheres.

• evolution of secondary atmospheres; comparative planetology for the green­house effect.

• the detailed structure and composition of Earth’s atmosphere.

• Earth’s magnetosphere and auroras.

• the interplay of air circulation and weather.

• comparative studies of the atmospheres of Earth, Venus, and Mars.

• the green­house effect and anthropogenic global cli­ mate change.

We find that some students can easily lose themselves in the vocabulary of dif­fer­ent atmospheric layers or the temperature/pressure changes of atmospheres with alti­ tude. It may help to focus less on memorizing the par­tic­u­ lar terms from the text and more on understanding why ­there are dif­fer­ent atmospheric layers and why the terres­ trial planets have dif­fer­ent atmospheric structure. To refer back to the scaffolding of learning, a comparative plane­ tology approach teaches and strengthens the higher-­order pro­cesses of analy­sis and synthesis and w ­ ill allow the stu­ dent to interpret the reported discovery of a new atmo­ spheric world (e.g., a moon or an exoplanet).

DISCUSSION POINTS • Let’s return to Section 7.3, The Compositions of Plan­ ets, as a bridge to this chapter. As a review, have dif­fer­ ent groups research the answers to the following questions and share their reviews with the rest of the class: What volatile materials ­were pre­sent in the pro­ toplanetary disk of the Solar System? How did the temperature of the solar accretion disk determine the composition of the primary atmospheres? Which plan­ ets captured gas from the protoplanetary disk? Which ones retained their primary atmospheres, and why?

Which planets lost their primary atmospheres, and why? What environments or events formed secondary atmospheres, and on which planets? (LG 1) • Does Mars offer conditions that make it a target for a ­f uture visit by h­ umans? Discuss the pros and cons of actually sending ­humans to visit the red planet. (LG 2) • Earth’s atmosphere is far from primordial. Discuss how the current main components of our atmosphere (nitrogen and oxygen) came to be dominant and why our atmospheric composition differs from ­those of Venus and Mars. (LG 3) • The atmospheric pressure on the surface of Venus is 92 times greater than that on the surface of Earth, and the average global temperature is 2.56 times greater. ­Going up through the atmosphere, the pressure and temperature drop. At an altitude of 50 km, Venus’s atmosphere has an average pressure and temperature similar to Earth’s pressure and temperature at sea level. Could h­ umans survive if stationed high in Venus’s atmosphere? What would they need to do to create habitable conditions? What advantages and dis­ advantages would t­ here be to living on Venus as opposed to Earth? (LG 3, 4) • By changing the composition of Earth’s atmosphere, ­humans are carry­ing out a dangerous experiment ­because their survival is linked to the outcome of the atmospheric changes. Discuss what could be done to keep this experiment ­under control so that h­ umans do not create an inhospitable environment. (LG 5) • The question asked by this chapter’s opening figure is, “Why can you breathe only on Earth?” Set this up as a think-­pair-­share or a 2-­m inute writing exercise by reversing the question: “Why ­can’t we breathe on Mer­ cury, Venus, the Moon, or Mars?” Students should include specific conditions on ­those worlds or take an evolutionary approach, or both.

PROCESS OF SCIENCE The Pro­cess of Science feature addresses directly the complexity of the scientific issue of climate change. The content guides students to ask themselves the questions 67

68 ◆ Chapter 9 Atmospheres of the Terrestrial Planets that should be asked of all scientific issues: What is the quality of the evidence? Do the basic physics princi­ples make sense? Is the claim falsifiable? If someone’s mind cannot be changed, then they are not making decisions based on evidence.

ASTROTOUR ANIMATIONS ATMOSPHERES: FORMATION AND ESCAPE This animation explores the formation of planetary atmo­ spheres, first from a planet’s original formation and then from its volcanic pro­cesses. It shows how a planet can lose or retain gas molecules depending on temperature and surface gravity. It also discusses the differences in chemi­ cal composition of the planetary atmospheres of Earth and Jupiter.

Text reference: Section 9.1 GREEN­HOUSE EFFECT This animation shows the mechanism and importance of the green­house effect for Earth, as well as its effects on Venus and Mars. It also includes a clear explanation of what is meant by equilibrium temperature.

Text reference: Section 9.2

ASTRONOMY IN ACTION VIDEOS CHANGING EQUILIBRIUM Using an innovative stack of three containers of dyed ­water, Stacy Palen demonstrates energy coming from a star to a planet, and then from the planet back into space.

Equilibrium is reached initially, and the analogy to the temperature of a planet is covered: the planet is returning as much energy as it is receiving. Dr. Palen then restricts the energy release from the planet, and we watch as the planet “heats up” ­because equilibrium has been lost. The relationship to our climate change is brought in at the end. (2:18) Text reference: Section 9.2 CHARGED PARTICLES AND MAGNETIC FORCES The opening remarks are about the importance of charged particles moving in magnetic fields to Earth’s having auroras and understanding synchrotron radiation. Stacy Palen uses a C-­magnet and points out the direction of the field. A wire is placed inside the magnet. When she closes the switch and a current is generated (charges are moving),

the wire jumps out of the magnetic field. This result dra­ matically demonstrates the “right-­hand rule.” Using a copper hoop and a pool ball, Dr. Palen demonstrates tan­ gential motion and a perpendicular centripetal force.

(2:44) Text reference: Section 9.3

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Scientists chose argon to study ­because as one of the noble gases, it basically never reacts chemically with other ele­ments. It would have escaped from Mars’s atmosphere only by a pro­cess called “sputtering” and not by being chemically bound to rocks.

2. The primary reason why Mars lost its atmosphere was ­because the Sun was emitting more ultraviolet radia­ tion and high-­energy particles that caused sputtering at that time than it does now.

3. From the context in which it is used, volatile ultravio­ let radiation describes the wavelengths at which the molecules in Mars atmosphere ­were most susceptible to interacting with the radiation. It does not have the same meaning as “volatile gases.” (Note that volatile usually applies to liquids or the sublimation of some solid substances.)

4. ­Because it w ­ ill be centuries before t­ here is any possi­ bility of transforming the atmosphere of Mars, the first astronauts on Mars ­w ill be living in the current state of its atmosphere, which is much too thin for ­humans.

5. News searches for “Mars atmosphere” are time depen­ dent; answers w ­ ill vary.

CHECK YOUR UNDERSTANDING SOLUTIONS 9.1. (a, b, c, d) The presence or lack of a moon does not enter into ­ whether a planet has l­ittle gas in its atmosphere.

9.2. (c) Note that methane and ammonia are green­house gases but are not as abundant as carbon dioxide and ­water.

9.3A. (c, a, e, b, d) Layers of the atmosphere from nearest to farthest from the surface. See Figure 9.6.

9.3B. (a) See Figure 9.6. The troposphere contains about 90 ­percent of our atmosphere and is both the air we breathe and the location of weather on Earth.

Chapter 9 Atmospheres of the Terrestrial Planets ◆

69

9.4. (a, d, c, b) Greatest to smallest seasonal variations.

Mercury has an eccentric orbit as well as a 3:2 orbital resonance. Mars’s orbit is more eccentric than Earth’s.

Venus’s green­house effect stabilizes its surface tem­ perature around the entire planet.

9.5. (a, b, c) See Figure 9.20.

14. (d) It stands to reason that Earth’s average tempera­ ture tracks the amount of carbon dioxide in the atmo­ sphere b­ ecause carbon dioxide is a major green­house gas on Earth.

15. (c) The troposphere contains about 90 ­percent of our atmosphere and is both the air we breathe and the location of weather on Earth.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS

Thinking about the Concepts 16. Hydrogen (~75 ­percent) and helium (~25 ­percent) ­were by far the dominant ele­ments in the solar neb­ ula, and both existed as gases in the region of space where the terrestrial planets formed. Combined, all other volatile compounds represented only a tiny fraction (less than 1 ­percent) of a terrestrial planet’s gas content. Therefore, the primary atmospheres of the terrestrial planets w ­ ere approximately three-­ quarters hydrogen and one-­ quarter helium, by mass.

17. ­A fter the primary atmospheres ­were stripped from the terrestrial planets, the planets outgassed volatile material from their interiors. The gas expelled from planetary interiors combined with ­those materials delivered by comet impacts to produce secondary atmospheres.

18. Although ­f ree nitrogen was poorly represented in the protostellar disk, ammonia was abundant. Most of the ammonia froze as ice in the outer Solar System, becoming a significant component of comets. The ammonia was prob­ably delivered to Earth through comet impacts during the heavy bombardment period early in the history of the Solar System. ­Under the conditions on early Earth, ammonia was dissociated into its constituent nitrogen and hydrogen atoms. The hydrogen was light enough to escape Earth, but the much heavier nitrogen stayed ­behind.

19. Impacts from icy comets and outgassing from volcanoes.

20. Green­house gases trap heat inside the planet, like a blanket, helping the planet to become warmer than it could if it ­were just heated by sunlight alone. On Earth, this raised the average temperature above the freezing point of w ­ ater, making the planet hospitable for terrestrial life.

21. Plants remove carbon dioxide from the atmosphere through photosynthesis and produce oxygen as a waste product.

22. The troposphere is the lowest layer of the atmosphere; this is where ozone is a pollutant since it has a nega­ tive effect on health. In the higher layers (i.e., strato­ sphere), ozone absorbs UV photons, protecting the planet from this harmful radiation.

Test Your Understanding 1. (d, b, c, f, a, e) Chronological order for formation and evolution of Earth’s atmosphere.

2. (a) The green­house effect is strongest on Venus, where it has heated the planet to temperatures of around 870° F, much higher than we use to bake bread.

3. (c) Oxygen was created by plant life out of the CO2-­ rich secondary atmosphere.

4. (c) Although Mars, Earth, and Venus are all at dif­ fer­ent distances from the Sun, their temperatures depend much more strongly on the strength of the green­house effect, which itself is dominated by the numbers of molecules of each of the green­house gases that are pre­sent. The radical differences between ­t hese three planets are dominated by the amount of green­house gas, that is, the thickness (difference in weight between pressure levels) of the atmosphere.

5. (d) As noted in Sections 9.3 and 9.5, weather is small scale and short term, and climate describes the aver­ age state of an atmosphere.

6. (b) If particles have the same kinetic energy, then the less massive ones are moving much faster, which helps them escape the gravity of the planet.

7. (b) Venus’s exceptionally thick (in the sense of den­ sity or pressure) atmosphere translates into a very strong green­house effect.

8. (a, b, c, d) All of t­ hese are excellent reasons to study climate in a varied range of environments.

9. (d) Dust storms cause an atmosphere to take on a reddish hue, and b­ ecause dust storms on Mars are quite common, the atmosphere often takes on this appearance.

10. (a) Auroras occur when charged particles trapped in Earth’s atmosphere and coming from the Sun interact with Earth’s magnetic field.

11. (c) Ozone is a thin layer of O3 that protects Earth from harmful UV rays from the Sun. 12. (d) The rotation of Earth breaks global circulation into zones.

13. (c) The more molecules of green­house gas pre­sent, the stronger the green­house effect.

70 ◆ Chapter 9 Atmospheres of the Terrestrial Planets 23. Uneven solar heating produces temperature differ­ ences at the surfaces of the terrestrial planets. ­Temperature differences produce pressure gradients in the atmosphere, creating regions of high and low pressure. According to the second law of thermody­ namics, air flows from high pressure to low pressure, and we detect this flow of air as wind. Steeper pres­ sure gradients lead to higher wind speeds and stron­ ger wind.

24. (a) The Arctic ice is already floating in the ocean, so it does not add any additional mass to the oceans when it melts. (b) The glaciers that cover Greenland and Antarctica currently are located on land. When ­those ices melt and flow to the oceans, they add to the mass of w ­ ater that is already in the oceans, causing the aver­ age sea level to rise.

25. Venus has a perpetual opaque cloud cover due to its runaway green­house atmosphere.

26. The three terrestrial planets with atmospheres all have surface temperatures that are much higher than can be explained through the balance of energy received by the Sun and given off by the heated sur­ face. This means t­ here must be some effect at play that retains heat, that is, the green­house effect.

27. First of all, the runaway green­house effect on Venus means that the planet holds on to its heat efficiently, ­ hether the Sun is shining or not, the planet stays so w warm through its very effective green­house “blan­ ket.” Venus’s slow rotation means that t­here are no strong wind currents across the planet’s longitude, only in latitude, so hot gas at the equator is constantly mixed with cooler gas at the poles to make a uniform temperature.

28. Venus must have been cooler and had lower pres­ sure in the past, before the runaway green­house effect led to its current conditions. The evidence for past flowing ­w ater on the surface of Mars implies Mars must have been warmer and had higher atmo­ spheric pressure than it has now. Thus, both planets had periods when they could have been more hospitable.

29. Answers ­w ill vary.

30. Answers w ­ ill vary. Is it easier to evacuate Venus of its excess carbon dioxide and acidic atmospheric com­ pounds or enrich the atmosphere of Mars?

Applying the Concepts 31. Setup: Remember that linear axes change by con­ stant additive increments, and log axes by multiplica­ tive ones.

Solve: (a) The axes are linear.

(b) The variations of CO2 and temperature are very similar; that is, the low periods and high peaks in tem­ perature correlate in time with the variations of CO2 . (c) The highest CO2 level was about 300 ppm, but it is 410 ppm ­today. This is an increase of about 37 ­percent.

Review: ­Because the amount of green­house gases in the atmosphere dictates the strength of the green­ house effect, we expect the temperature and gas levels to be correlated, as indeed they are.

32. Setup: This question relies primarily on panel (a) of Figure 9.22.

Solve: (a) The amount of green­house gases began rising exponentially about 150–200 years ago (~1800–1850). (b) Using panel (a), it seems the be­hav­ior of the three gases generally varied in unrelated ways u­ ntil the last 100 years or so. (c) The rise of all three gases in modern times is most likely due to industrialization and the exhausting green­house gases into the atmosphere.

Review: This confirms the general consensus among scientists about anthropogenic (human-­caused) cli­ mate change.

33. Setup: Figure 9.6b shows that commercial jetliners fly about 12 to 15 km above the ground.

Solve: It seems advantageous for jetliners to fly above the bulk of atmospheric convection, for a smoother ­r ide, and as close to the stratospheric winds as pos­si­ ble, so the winds help carry the planes along. This requires that they fly above 10 km or so (where ­there is still enough atmosphere to provide lift), which is what they do (Figure 9.6b).

Review: This height is also the location of the jet streams that have become extremely impor­tant in aviation. Flying at that height takes advantage of the stream to decrease flight time as well as conserve fuel.

34. Setup: This question asks about the effects of pres­ sure in dif­fer­ent environments. B ­ ecause ­there is a closed bag of chips, we can assume the pressure inside the bag remains constant. What happens as the pres­ sure outside the bag changes?

Solve: At higher altitudes, the air pressure in the cabin drops, so the pressure inside the bag increases. (a) Inside the plane, the pressure is lower than when the plane is on the ground. (b) Airplane cabins are pressurized, which means they are at higher pressure than the out­ side air when they are flying. A bag outside the plane would expand as much as it could, about five times its size, based on Figure 9.6. (In real­ity, it would prob­ably burst ­because the outside pressure is so low.)

Review: This explains why b­ ottles of w ­ ater some­ times leak if you open them before takeoff and lay

Chapter 9 Atmospheres of the Terrestrial Planets ◆ 71

them on their sides while the plane climbs. Similarly, at cruising altitude the pressure is lower than on the ground, so a b­ ottle that you drank while flying w ­ ill crush itself as you descend b­ ecause the pressure inside is lower than the pressure on the ground.

35. Setup: Force F = pA, where A is the area on which the pressure, p, acts, or F = ma, where m is the mass being accelerated at a.

Solve: (a) A head has a typical circumference, C, of 50 cm, meaning a radius of r = C 2π ≈ 8 cm, yielding an area A = πr 2 = π (0.08 m)2 = 0.02 m2 . Then the force of air on a typical person’s head is F = pA = 105 N/m2 × 0.02 m2 = 2,000 N. (b) If this force ­ were due to a kangaroo, the k angaroo would have to have a mass of ­ m = F/a = 2,000 N/9.8 m​/s2 ≈ 200 kg. (c) Given that a typical kangaroo has a mass around 60 kg, we would need 200/60 ≈ 3 kangaroos on a per­ son’s head. (d) We are not crushed by this force ­because the bones and muscles in our body hold us up, and the pressure inside our bodies is equal to the pressure outside of our bodies. That is, we have the equivalent of 3 kangaroos pushing up to offset the 3 pushing down.

Review: First of all, why would we ever put kanga­ roos on our heads? That being said, we evolved to sup­ port the atmospheric weight on top of us. This shows, however, why we could not simply “jaunt” off to another planet such as Venus ­because the atmosphere would crush us.

36. Setup: Venus has an atmospheric pressure about 92 times that on Earth, or 9.2 × 106 N/m2 , and an accel­ eration of gravity about 0.9 times that on Earth, or 8.9 m/s2 .

Solve: (a) Using the area of a person’s head as 0.02 m 2 (see prob­lem 35), then the force of air on our head is about F = pA = 9.2 × 106 N/m2 × 0.02 m2 = ​1.08 × 105 N. (b) If this force ­were due to a kangaroo, the kangaroo would have to have a mass of m = F/a = 1.8 × 105 N/8.9 m/s2 ≈ 20,700 kg. (c) Given that a typical kangaroo has a mass of around 60 kg, we would need 20,700/60 ≈ 350 kangaroos on a person’s head. (d) We would certainly be crushed by this force ­because the bones and muscles in our body could not hold us up and our internal pressure would be a lot less.

Review: When we think of traveling to other planets, we often forget about the effects of pressure and think only about temperature and w ­ hether we could breathe

t­ here. The composition of the atmosphere seems to be more impor­tant to us.

37. Setup: Mars has an atmospheric pressure about 0.006 times that on Earth, or about 600 N/m2 , and an accel­ eration of gravity about 0.4 times that on Earth, around 3.8 m/s2 .

Solve: (a) Using the area of a person’s head as 0.02 m2 , then the force of air on our head is about F = pA = 600 N/m2 × 0.02 m2 N = 12 N. (b) If this force ­ were due to a kangaroo, the k angaroo would have to have a mass of ­ m = F/a = 12 N / 3.8 m​/s2 ​≈ 3.2 N. (c) Given that a typical kangaroo has a mass around 60 kg, we would need 3.2/60 ≈ 0.05 kangaroos, or a tiny joey, on a person’s head. (d) We would certainly not be crushed by this force ­because the bones and muscles in our body already hold us up so well on Earth. But, since our internal body pressure is much higher than the ­s urface pressure of Mars, we would need pressure suits.

Review: Given that the atmospheric pressure on Mars is so low, but the pressure in our body is about that of the atmosphere on Earth, we would prob­ ably plump up like a balloon, and our blood would boil.

38. Setup: Looking at Figure 9.6, we see temperature ver­ sus altitude in panel (a) and pressure versus altitude in panel (b). To relate temperature to pressure we must consider both panels at once.

Solve: Figure 9.6 shows that in the lowest atmo­ spheric level—­the troposphere—­temperature and pressure increase as altitude decreases, which means that pressure and temperature are related. Thus, if ­either one rises, the other must rise as well.

Review: This is the essence of how refrigerators work.

Gas is pressurized and then cooled down with a fan, and the gas is released into a lower pressure chamber.

Lower pressure means lower temperature, so the gas ends up cooling down significantly.

39. Setup: Remember that p ­percent = 0.01 × p and per­ cent increase is equal to the added mass divided by the current mass times 100.

Solve: (a) The mass of carbon dioxide in the atmo­ sphere is 0.06 ­percent of the total mass of the atmo­ 0.06 ⎞ × (5 × 1018 kg) = 3 × 1015 kg. sphere, or ⎛ ⎝ 100 ⎠ (b) If we add 3 × 1013 kg per year, then we are increas­ ing the total amount by (3 × 1013 kg) / (3 × 1015 kg) ​ = ​0.01, or 1 ­percent per year. (c) The number of molecules is (3 × 1013 kg) / (7.3 × ​ 10 −26 kg/molecule) = 3.9 × 1038 molecules per year.

72 ◆ Chapter 9 Atmospheres of the Terrestrial Planets (b) The pressure in the Antarctic night: (d) E ­ very molecule of carbon dioxide absorbs infra­ red energy from the surface and helps heat it. Thus, T 203 K even a 1 ­percent increase in the total amount (which PAntarctic = Antarctic Pstd = × 105 N/m 2 T 273 K std itself is only 0.06 ­percent of the atmosphere) means = 7.43 × 10 4 N/m2 that much more energy is being stored.

Review: Although the annual increase might be (c) If the container walls w ­ ere not rigid, they would small, the cumulative increase is quite large. That is, if bend outward in a convex shape when placed in the we double the amount of green­house gas in the atmo­ desert shade. In this case, the internal pressure sphere, we also double its effect, which ­w ill warm the would greatly exceed the surrounding atmospheric planet. (Note that this prob­lem does not state if we pressure. When placed in an Antarctic night, the are considering a net increase of CO2 or just the container walls would bend inward in a concave annual h­ uman output. A more thorough consider­ shape. In this case, the surrounding atmospheric ation would include the amount of CO2 rained out, pressure would greatly exceed the pressure inside the absorbed by the oceans, and scrubbed by planet life container. each year.)

Review: The same can be observed in your own 40. Setup: This prob­ lem involves calculating kinetic room by putting a balloon on top of a large jar and 1 2 watching it grow or shrink as the temperature energy, KE = mv .

2 changes.

Solve: (a) Kinetic energy of parcels of air on Earth and 42. Setup: To calculate speed, we w ­ ill assume the kinetic 1 2 1 KE = (1.23 kg) × (10 m/s) = 61.5 J Venus are Earth energy of all atoms in air is the same, and KE = mv 2 ; 2 2 1 then we can solve that the average speed of a particle and KEVenus = (64.8 kg) × (1 m/s)2 = 32.4 J. Tak­ 2 1 v∝ . ing a ratio of the two numbers shows that a parcel of m air on Earth has about 1.90 times the kinetic energy of Solve: (a) If oxygen molecules are 16 times as massive a parcel of air on Venus. as hydrogen, and carbon dioxide molecules are 22 1 2 times as massive, then carbon dioxide is 22/16 = 1.38 (b) KEMars = (0.015 kg) × (50 m/s) = 18.75 J. 2 times as massive as oxygen. This means they ­w ill Taking a ratio of the two numbers shows that a parcel 1 = 0.85 times as fast. travel v = of air on Earth has about 3.28 times the kinetic 1.38 energy of a parcel of air on Mars. (c) Wind erosion is (b) Note that as temperature increases, both mole­ only one of the gradation pro­cesses that operate on cules gain the same kinetic energy. This fixes the ratio Earth. We must consider w ­ ater erosion, volcanism, as long as the gas is in equilibrium. landslides, and plate tectonics as well.

Review: Working It Out 9.1 gives the velocity of H 2 Review: In many places on Earth, the other pro­cesses at 288 K as 1.88 km/s and the velocity of O2 at that are more dominant than wind erosion. Wind erosion same temperature as 0.47 km/s. Using the same logic, 288 is best observed in deserts where abrasion of build­ the velocity of CO2 at 288 K would be v = 0.157 = 0.4 k ings and scouring of the ground are common sights.

44 288 41. Setup: The prob­lem states that we seal a rigid con­ v = 0.157 = 0.4 km/s. This gives 0.4 0.47 = 0.85 as we 44 ­ ill assume that the volume of tainer full of air, so we w found. the container does not change. Then according to the 43. Setup: The average speed of a molecule at tempera­ ideal gas law, the pressure in the container is directly 3kT T proportional to the Kelvin temperature of the gas in = 0.157 k ture T is given in Working It Out 9.1 as v = Pnew Tnew m A = , or, equivalently, 3kT the container; that is, T v= = 0.157 km/s, where A is the molecule’s atomic Pstd Tstd T A m Pnew = new Pstd . mass. Using T ­ able 9.1, the average gas temperature on Tstd Earth is 288 K, and on Mars it is 210 K. CO2 has a Solve: (a) The pressure in the desert: molecular weight of 12 + 2 × 16 = 44.

T 323 K Pdesert = desert Pstd = × 105 N/m 2 273 K Tstd = 1.18 × 105 N/m2

Chapter 9 Atmospheres of the Terrestrial Planets ◆

Solve: On Earth, the average speed of a CO2 molecule 288 288 km/s 0.157 km/s == 0.402 0.402 km/s, km/s,and on Mars it is vv == 0.157 44 44 210 210 km/s km/s==0.343 0.343km/s. km/s. ­Table 8.1 lists 44 44 the escape velocities of Earth and Mars as 11.2 and 5.0 km/s. Earth, therefore, holds onto its gas much better than Mars holds onto its gas.

Review: Working It Out 9.1 suggests that if the mole­ cular speed is less than one-­sixth that of the escape velocity, the planet w ­ ill hold onto its atmosphere over its lifetime. For Earth and Mars, ­t hese are 1.9 and 0.83 km/s. ­These are both higher than the average molecular speed of carbon dioxide on each planet, explaining why each planet has held on to some of its gas. But ­because the difference is much less for Mars than for Earth, we see why Mars has lost most of its carbon dioxide whereas Earth has lost much less.

Plus, volcanism on Earth keeps replenishing CO2 ­here, whereas the volcanoes on Mars are extinct.

44. Setup: We translate the average surface pressure on Mars of 6.4 millibars to 0.0064 bars.

Solve: Using Figure 9.6(b), we see that we could have to climb higher than 40 km above sea level to reach such low pressures.

Review: This altitude puts us close to the top of the stratosphere, which shows just how thin the atmo­ sphere is on Mars.

45. Setup: The atmospheric pressure on Venus is 92 bars.

Solve: At 10 m per bar, you would have to dive 920 m below the surface of the ocean to achieve the same pressure you would feel at Venus’s surface.

Review: 920 m is about 0.6 of a mile. This is about 25 ­percent deeper than any person has dived (700 m) using specially designed, self-­contained hard suits called atmospheric diving suits.

0.157 is vv==0.157

Using the Web 46. Students are asked to look up the data on the current ­ ill vary. year’s ozone hole. Answers w 47. Mars: a. Prompts students to create a “Zooniverse” account and classify images. Answers ­w ill vary. b. The scientific goals of the MAVEN mission are to study “the upper atmosphere of the Red Planet and how it interacts with the Sun and the solar wind.” The mis­ sion involves an orbiter that contains a spectrometer, spectrograph, magnetometer, electron analyzer, ther­ mal ion composition, probe and waves antenna, solar wind ion analyzer. “Scientists w ­ ill use MAVEN data to determine the role that loss of volatiles from the Mars atmosphere to space has played through time,

73

giving insight into the history of Mars’ atmosphere and climate, liquid ­water, and planetary habitability.”

Answers w ­ ill vary concerning the results of the mission.

48. Earth: a. Involves students ­going to “National Snow and Ice Data Center” and getting updates on the Arctic sea ice and Antarctic ice. Answers w ­ ill vary. b. Students investigate the “Global Annual Mean ­Surface Air Temperature Change” at Goddard Institute for Space Studies. Answers ­w ill vary. c. Students go to NOAA’S “Trend in Atmospheric Carbon Dioxide” web page on CO2 levels at the Mauna Loa observatory.

Answers will vary.

49. Climate change: a. Th ­ ere ­were a number of dates where investigations hinted at the cause of climate change, but it was in 1971 that scientists reported the danger caused by ­ humans. Other anthropogenic green­house gases w ­ ere identified in 1975 and 1976 as CFCs, CO2 , methane, and ozone. Scientific opinions about global warming started to converge in 1985, although ­there are certainly definite hints of that a de­ cade or so ­ earlier. Observations of Mars and Venus contributed to our knowledge b­ ecause Mars was undergoing a huge planet-­w ide dust storm rein­ forced by a temporary warming. The Soviet Venera 4 showed Venus’s atmosphere to be mostly CO2 and its surface to be extremely hot and dry. “The ­human chemical emissions, aerosols, tended to turn into sulfate particles whose sizes fell exactly within the range most effective for scattering sunlight.” b. Involves stu­ dents reviewing the “Pre­sen­ta­t ion” that is included in the Fifth Assessment report from the Intergovern­ mental Panel on Climate Change. Students w ­ ill select some ­causes, effects, mea­sure­ments, and conclusions. ­ ill vary. c. As an advanced activity, students Answers w are urged to use NASA GISS modeling software and fun a functional three-­d imensional global climate model on their computers. Answers w ­ ill vary.

50. Mars movies: a. Involves watching a science fiction ­ ill vary. b. film about p­ eople g­ oing to Mars. Answers w Involves students watching the end of Total Recall (1990) and evaluating the last few scenes where Mars suddenly becomes terraformed and habitable.

Answers ­w ill vary. (Note that the original movie (1990) was rated R for strong sci-fi vio­lence and gore, language, sexuality, and some nudity.)

EXPLORATION This exploration involves a series of experiments. Students ­w ill use permanent markers, translucent plastic cups, ­water, and ice cubes to test the implications of the melting of the polar ice caps as the planet warms.

74 ◆ Chapter 9 Atmospheres of the Terrestrial Planets EXPLORATION SOLUTIONS 1. This question involves a prediction. We naïvely expected the w ­ ater level to rise.

2. The ­water level did not rise.

3. As the Arctic sheet melts, the ocean levels should not rise.

4. We expected the w ­ ater level to rise.

5. The ­water level did indeed rise.

6. When surface ice melts, the ocean levels w ­ ill rise.

7. Most population centers are near the coasts, so sea rising ­w ill dramatically affect a large amount of the global population via flooding.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES None for this chapter.

CHAPTER 10

Worlds of Gas and Liquid—­The ­Giant Planets INSTRUCTOR NOTES Chapter 10 covers the g­ iant planets. Major topics include: • overview of common and dif­fer­ent characteristics of the Jovian planets.

• comparisons of atmospheric circulation, turbulence, and weather.

• atmospheric structure down to a few hundred kilo­ meters below the surface layers.

• comparisons of the interior compositions and struc­ tures, down to the cores.

• sizes and shapes of the magnetic fields, radiation ­belts, and auroras.

• overview of why our planetary system may be aty­pi­cal: exoplanets and planetary migration.

Chapter 10 offers another opportunity to use “compar­ ative planetology” to teach the ­g iant (or “Jovian”) planets. ­Under this comparative approach, memorizing the names and chemical compositions of dif­fer­ent atmospheric lay­ ers and the inclinations of the magnetic fields (as exam­ tant than knowing why the ples) become less impor­ atmospheres are differentiated and how magnetic fields influence the planets. ­There are two good opportunities to discuss planet migration. First, one can motivate it by comparing the high methane content of Uranus and Neptune with the primor­ dial content of Jupiter and Saturn, and also by noting that at Neptune’s distance, planet formation timescales are longer than the age of the Solar System. The Origins section of this text gives an alternate entry point. You might also choose to wait ­u ntil Chapter 12 ­because planet migration comes from a planet clearing out the space around it, which is the distinguishing feature between planets and dwarf planets.

We recommend the excellent videos of planet migration at http://­jila​.­colorado​.­edu​/­~pja​/­planet​_­migration​.­html.

DISCUSSION POINTS • Being made up of primarily hydrogen and helium, the atmospheres of Jupiter and Saturn are considered pri­ mordial. The atmospheres of Mars, Earth, and Venus,

on the other hand, have no trapped hydrogen or helium. Rather, ­these terrestrial planets have second­ ary atmospheres (Earth, a tertiary atmosphere due to life) of carbon dioxide, nitrogen and other molecules.

Discuss the connection between the mass of a planet and the composition of its atmosphere. Within the framework of the formation of the Solar System, how is this relationship understood? (LG 1) • Working in groups, have students make 18 × 24-­inch posters (sheets of newsprint are ­g reat) of the atmo­ spheres of the ­g iant planets. Groups then post their work for a short poster section where they describe the structure, composition, temperature variance, and pressure with depth. Students can ask questions a­ fter each pre­sen­ta­tion. (LG 2) • The temperatures and pressures of the interiors of the ­g iant planets create in­ter­est­ing phenomena. For exam­ ple, the metallic nature of the liquefied hydrogen is not something easily produced in our laboratories. Discuss the implications of the existence of ­these phenomena and the scientific creativity needed to create theories for ­these and many other not-­before-­seen phenomena in the universe. (LG 3) • Discuss the basic physics leading to the creation of a magnetic field. Start by having students list what they know about magnets. A changing magnetic field cre­ ates electricity. How do we use this fact to our benefit?

Huge cranes are capable of lifting heavy containers and then dropping them without hooks or cables. How do ­these work? (LG 4) • For a quick writing exercise, have students compare the magnetospheres of the ­g iant planets to that of Earth. (LG 4) • Based on our four ­g iant planets, what characteristics do we expect g­ iant extrasolar planets to have? Turbu­ lent and banded atmospheres? Storms? How might we observe ­these characteristics? (LG 5) • The opening image for this chapter shows a color-­ enhanced image of the chaotic clouds of Jupiter as observed by the spacecraft Juno during its close flybys of this ­g iant planet. Have students compare this image to one in natu­ral colors (https://­w ww​.­nasa​.­gov​/­sites​ 75

76 ◆ Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets /­default​/­fi les​/­thumbnails​/­image​/­pia22427​-­16​.­jpg), which is what we would see if we ­were at Juno’s location.

Compare t­ hese two images. Which one gives more information about the atmosphere of Jupiter? Why does NASA spend the time ­doing the enhancements?

Do ­these enhanced images give false impressions?

PROCESS OF SCIENCE How does the saying from (or attributed most frequently to) Isaac Asimov go? “The most exciting phrase to hear in science, the one that heralds new discoveries, is not ‘Eureka!’ (I found it!) but ‘That’s funny.’ ” The Pro­cess of Science for this chapter covers the point that scientific laws make testable predictions. It was “odd” that Uranus’s orbit did not match predictions. Working with observa­ tions of the planet and Newton’s laws, the then-­unknown planet Neptune was discovered.

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS

detectors, a radio/plasma wave experiment, an ultra­ violet imager/spectrometer, an infrared imager/spec­ trometer, and a color camera called JunoCam that gives images that are broadcast nearly instanta­ neously to the public.

CHECK YOUR UNDERSTANDING 10.1. (a) Uranus and Neptune are ice ­g iants ­because they have higher percentages of ices in their interiors.

10.2. (b) The color of an atmosphere depends on how the par­tic­u ­lar ele­ments that make it up absorb and reflect light.

10.3. (b) If a planet gives off more energy than it receives, it must have a source of its own.

10.4. (c) The radiation b­ elts are created from charged particles being dragged around the planet due to its rotation.

10.5. (a, d, b, c) Sorted in order of increasing dia­meter.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS

Test Your Understanding 1. (a, c, f, e, b, d) See Figure 9.7, which shows the convec­ None for this chapter. tion cycle on Earth.

2. (b) The state of m ­ atter (solid, liquid, gas) depends not INTERACTIVE SIMULATIONS only on the temperature but on the pressure as well. ­ ater at room tempera­ For example, at low pressures, w None for this chapter. ture can boil and at very high pressures, even ultra-­ hot materials can stay liquid or solid.

TEACHING READING ASTRONOMY NEWS 3. (b) For calculating the radius of a Solar System 60 sec 1. The ­Great Red Spot has been seen only since the × 30 km/s = 46,800 km. that is occulting a star: 26 min × invention of the telescope, which was a60l­ittle over 1 min sec 30 km/s = 46,800 km. Half of this is about 23,000 km.

26 min bands × 400 years ago. It and the equatorial are×not 1 min 4. (a) Uranus’s motion did not follow the path predicted viewable with just the naked eye.

by Newtonian physics, suggesting that another large 2. Jupiter’s cloud tops are opaque, making it hard to see mass was tugging on it. below them.

5. (c) Uranus’s seasons are the most extreme ­because the 3. The answers may vary depending on how many years inclination of its rotation axis is the highest. have passed since this edition was published. Noticeable 6. (b) The g­ iant planets’ magnetic fields are very strong changes in the size of the ­Great Red Spot over a period and extend far into outer space. However, at the cloud of a few years, if so reported, would be significant. tops, they are as weak as the field on Earth, and as 4. Juno is using “citizen scientists” to pro­cess the images shown in Figure 10.14, the fields do not necessarily as a means to get multiple ave­nues of input to their sci­ pass through the planet centers nor are they necessar­ ence results. The wealth of the information contained ily aligned with the rotation axes. in the images means it is beneficial to have “more eyes” 7. (b) Note that passing in front of the Sun is still an view the data. The public inputs are used in educa­ occultation ­because the Sun is a star. However, we tional outreach and as part of published articles. usually refer to this as a transit.

5. Juno’s scientific payload includes a gravity/radio sci­ 8. (a) The timing of an occultation gives us the size of ence system; a six-­wavelength micro­wave radiometer, the planet. a vector magnetometer, plasma and energetic particle

Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets

9. (d) Jupiter and Saturn are composed almost entirely of hydrogen and helium, similar to the Sun. 10. (d) Temperature changes with altitude in cloud layers.

11. (b) The ­Great Red Spot was first observed by Hooke and Cassini in 1664 to 1665 and has been raging con­ tinuously since.

12. (a) This is why they are called ice ­g iants.

13. (d) Resonant orbits can cause planetary migration.

14. (b) ­Giant planets rotate in about half the number of hours as an Earth day, and they are much larger, mean­ ing they have much higher rotational velocities that yield much faster zonal winds.

15. (d) “Hot Jupiters” are ­g iant planets found very close to their central stars.

Thinking about the Concepts 16. Several differences between terrestrial and ­g iant plan­ ets are summarized in Table IM 10.1.

17. The composition of the solar nebula was approxi­ mately 98 ­percent hydrogen and helium, with only 2 ­percent heavy ele­ments. Jupiter’s protoplanetary seed grew very quickly (less than 10 million years) beyond the “snowline,” allowing it to capture and hold the hydrogen, helium, and other volatile materi­ als that prevailed in the solar nebula. Earth, however, was too small to retain hydrogen and helium and too hot to absorb many volatiles. As such, Earth contains mostly the refractory ele­ments that ­were pre­sent in the solar nebula inside the snowline.

18. The first answer is the size of a planet. We know how fast the planet moves and its distance from us, so the duration of an occultation gives us its physical size.

However, we can also watch the variation of starlight as the occultation takes place, and study the presence and properties of the planet’s atmosphere and rings.

19. Rapid rotation drives the zonal winds and also pro­ duces strong Coriolis forces.

◆ 77

20. Uranus is discovered. ® Hypothesis: We can predict its orbit. ® Prediction: Based on its distance from the Sun, we predict Uranus’s orbit. ® Test: Uranus’s orbit does not match predictions. ® FAIL ® New hypothesis: ­There might be something pulling on Uranus. ® New prediction: ­There might be another planet. ® New test: Find the planet. ® Success: We found Neptune.

21. ­Giant planets rotate very rapidly, and rapidly rotating spheres tend to expand along the equator and contract along the poles, making the planet oblate, or “squished.”

22. Variations in color are ­because we are seeing dif­fer­ent depths into the atmosphere. For example, the white clouds are high in the atmosphere and highly reflec­ tive; thus, we are basically seeing reflected sunlight.

Colorful regions are ­those for which we see deeper into the atmosphere and effectively see “smog,” that is, chemical pollutants (e.g., sulfur, organic compounds) that have caused the gases and ices to take on reddish colors. Methane gas, which is abundant in the atmo­ spheres of Uranus and Neptune, strongly absorbs the longer red wavelengths of vis­i­ble light. Therefore, we see only the shorter blue and green wavelengths of light reflected by ­these planets. The absence of clouds in the atmospheres also prevents the strong color from being washed out by scattering effects.

23. All the ­giant planets have rotation axes that are ­ ill expe­ inclined, and a planet with an inclined axis w rience seasons. Of ­these, Jupiter’s inclination is very small, so its seasonal effect w ­ ill be very low. Saturn’s and Neptune’s axes are similar to t­ hose of Earth, so the seasons for t­ hese planets w ­ ill be similar. On the other hand, Uranus’s inclination places its axis in its orbital plane, so its seasons are very severe.

24. The extraordinarily high pressure in Jupiter’s interior prevents individual atoms and molecules from mov­ ing around randomly in such a confined space.

Instead, atoms and molecules are forced into such close proximity that they remain loosely bound to

Table IM 10.1 Terrestrial

­Giant

Small worlds with solid surfaces

Large worlds with no solid surfaces

Tightly spaced within 2 AU of the Sun

Spread out beyond 5 AU from the Sun

Dense rock and metal composition

Lightweight gaseous composition

Secondary or tertiary atmospheres

Thick primary atmospheres

Weak or no magnetic fields

Power­f ul magnetic fields

Few moons and no ring systems

Numerous moons and rings

78 ◆ Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets each other even at high temperatures. Random parti­ cle motions in a system are characteristic of a gaseous state. Loose bonding between particles in a system is characteristic of a liquid state.

25. We mea­sure the positions of clouds as they move across the surface of the planet, and compare ­those speeds with each other (e.g., in latitude) and also with our assumed rotation speed of the interior of the planet, based on its magnetic fields.

26. The ­Great Red Spot is a massive storm that has been raging on Jupiter for at least 300 years. In general, its motions and changes are unpredictable. It has been shrinking in size, and nobody ­really knows why.

27. The extra energy comes from ongoing gravitational contraction and differentiation in the interiors of Jupiter, Saturn, and Neptune.

28. The bright radio source emanating from Jupiter is synchrotron radiation. The energetic motion of elec­ trons accelerating within Jupiter’s power­f ul magnetic field c­ auses emission at radio wavelengths.

29. Auroras on the ­g iant planets are caused by the same mechanism as on Earth. Charged particles from the solar wind and ­g iant planets’ atmospheres become trapped in the planet’s magnetic fields and spiral around the magnetic field lines and move back and forth between the poles. Th ­ ose energetic particles col­ lide with atoms and molecules in the planet’s atmo­ sphere, knocking them into excited energy states that then decay and emit light.

30. Migration of the outer ­g iant planets could leave inner ones alone completely as they eject objects that are orbiting nearby, or they could completely disrupt their orbits and send them flying out into space.

Applying the Concepts 31. Setup: Read the figure caption to learn the difference between the images.

Solve: The top panel shows the comparative planets’ ­actual sizes scaled to Jupiter’s size, mea­sured in kilo­ meters. The bottom panel shows how they would appear to an observer on Earth mea­sured in angular arcseconds. (It is likely that this is an average mea­ sured over many oppositions.)

Review: Note that Uranus and Neptune are so far away that from Earth they only appear like small blue dots, even though you can see ­g reat detail on Jupiter and Saturn. This is why we had to send the Voyager 2 flyby spacecraft to take close up images of ­these ice ­g iants.

32. Setup: Remember that Neptune is very distant and hard to observe from Earth and has been visited only one time by the Voyager 2 satellite as it flew past.

Solve: ­Because the Northern Hemi­sphere is cur­ rently tilted away from Earth and not viewable, it is most likely that we have ­little information about the winds in that region and a lack of line indicates that we do not know the be­hav­ior. If t­ here ­were zero veloc­ ity, the line would be vertical in that region.

Review: It is impor­tant to recognize that even though we know much about our Solar System, ­there is still much to be discovered.

33. Setup: When a gas is compressed to very high densities and pressures, the electrons are stripped from the nuclei, and the resulting plasma behaves like a liquid metal.

Solve: The interiors of Jupiter and Saturn reach very high pressures and temperatures over 10,000 K, vio­ lent and hot enough to strip the electrons off the hydrogen atoms.

Review: Note that Uranus and Neptune do not have enough internal gravity to reach ­these high temperatures.

34. Setup: Using a ruler, we find that the distance from the edge of Jupiter to the torus is about 2.5 times the dia­meter of Jupiter. Accounting for the half of Jupiter that is its own radius, this is about 3 Jupiter dia­meters or 6 Jupiter radii.

Solve: Jupiter’s radius is about 7 × 10 4 km. At six times Jupiter’s radius, the torus is about 6 × (7 × 104 km) ​ = 4.2 × 105 km.

Review: Online, we find that the torus is about the same distance from Jupiter as the moon Io itself, which is 4.2 × 105 km.

35. Setup: From Working It Out 5.4, the relationship between distance and luminosity indicates that the observed brightness of light from the Sun, b, decreases L with the square of the distance, d; that is, b = Sun2 .

4π d ­Because we ­w ill be using proportional reasoning, we 1 1 set up the equation as b ∝ 2 or d ∝ . d b Solve: The question asks us how far we would have to be from the Sun for it to appear as bright as the full Moon, that is, 400,000 times fainter than it does now, which means brightness changes by a ­factor of 1 b= . Inserting this into our distance-­ 400,000 1 brightness equation, d ∝ = 632 AU, 1 400,000 which means that the distance is 632 times farther than where we are currently. B ­ ecause we are 1 AU from the Sun, we would have to move to 632 AU to see the Sun as bright as the full Moon.

Review: Note that this distance is about 21 times far­ ther than Neptune’s orbit of 30 AU.

Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets

◆ 79

M 318 36. Setup: To find the size of Uranus, we must compute 1,320 times larger than Earth’s, then ρ = = = 0.2 distance based on speed and time, that is, d = vt. How­ V 1,320 M 318 ever, note that we are given time in minutes, not sec­ ρ= = = 0.24 times that of Earth. In other words, even 60 s V 1,320 = 2,220 s. The total occultation onds. 37 min × though Jupiter is more massive and bigger than Earth, 1 min its density is about one-­quarter that of Earth. time of 37 minutes, 2 seconds is 2,220 + 2 = 2,222 s.

Review: (a) Taking the ratio of the average radii, we Solve: (a) d = vt = 23 km / s × 2,222 s = 51,100 km 69,900 km (b) If Uranus had not passed directly in front of the find that Jupiter’s size is = 10.97 times star, the occultation time would be shorter than 6,370 km expected and would have only given us a lower limit larger than Earth. Using our volume-­radius relation­ to the planet’s a­ ctual size. ship, the volume ­w ill be V ∝ R 3 = (10.97)3 = 1,320 Review: We are told throughout this text that Uranus times larger. is a ­g iant planet. The size we computed, 51,100 kilo­ (b) We compare the densities using the information meters, is 4 times larger than Earth’s dia­meter of in ­Tables 8.1 and 10.1. Average density of Jupiter is 12,756 kilo­meters, and is close to Uranus’s mean dia­ 1.33, and the average density for Earth is 5.52. This meter of 50,724 ± 7 km. gives 1.33 / 5.52 = 0.24, or approximately 1/4.

37. Setup: Oblateness is defined as O = (R eq—­R pole)/R eq.

40. Setup: Referring back to Chapter 2, recall that the maxi­ (This is not shown in this chapter.) mum height of the Sun at a pole is equal to the planet’s Solve: Given the oblateness (0.065) and equatorial inclination; for example, in the summer at the North radius (71,500 km) of Jupiter, we solve R pole = ​ Pole, the Sun is 23.5 degrees above the horizon.

R eq(1—­O) = 66,852 km.

Solve: Using this logic, the Sun on Uranus would Review: NASA’s Jupiter fact sheet lists a polar radius be 98 degrees above the horizon, or 8 degrees from of 66,854 km (and an equatorial radius of 71,492 km), the zenith on its summer solstice. consistent with our calculations.

Review: Eight degrees from the zenith is very high in 38. Setup: Nitrogen has an atomic mass of 14, sulfur is the sky at the instance of the summer solstice at one of 32, and hydrogen is 1. the poles. Uranus has an orbital period of 84 years, so Solve: ­There is 1 N, 1 S, and 5 H, so the total molecu­ each “season” is about 21 years long. ­There is a time when lar mass is 14 + 32 + 5 × 1 = 51 atomic mass units. one pole is pointing ­toward the Sun, 21 years ­later, the Review: Ammonium hydrosulfide, as a molecule, has Sun would be above the equator, 21 years ­a fter that about the same mass as vanadium and just a ­l ittle less the Sun is above the opposite pole. Extreme seasons, than an iron atom. indeed, as one or the other pole is dark for many years.

39. Setup: ­Because volume scales with radius cubed, R 3, 41. Setup: The cloud has moved from 122° to 118°, or a we can compare the volumes of the two planets and total of 4° in 10 hours. To compute the a­ ctual speed, assume that this is the number of Earths that can fit we need to know what distance corresponds to 1°. We inside. If we imagine stacking golf balls in a box, t­ here can compute this distance as a ratio of the circumfer­ ­w ill be space between each ball, so in real­ity the num­ ence, C = 2πR. To compute speed, remember that dis­ ber we get w ­ ill be larger than the a­ ctual number of tance equals rate multiplied by time, or d = vt. When Earths that could fit inside. For part (b), the formula solving for speed, we take v = d/t.

M Solve: ­Because the cloud has moved from 122° W to for density is ρ = , where ρ is the density, M is the V 4π 3 118° W, it has traveled eastward. Winds blowing mass, and V is the volume: V = R. from the west to the east are westerly winds. The 3 Solve: (a) Since we know that the ratio of the volumes circumference of Jupiter at its equator is C = 2πR = ​ is directly proportional to the ratio of the average 2π × 71,500 km = 449,248 km. Th ­ ere are 360 VJ rJ3 (69,900degrees km)3 in a circle, so 4° is 4/360 = 1/90 of a circle. = = = 1,320, radii cubed, we can set up the equation 449,248 km VE rE3 (6,370 km)3 3 3 VJ rJ (69,900 km) Then the distance covered is 1/90th of C, = 4,992 k 90 = 3= = 1,320, where the subscripts are J449,248 and km 3 VE rE (6,370 km) = 4,992 km. It took 10 hours to travel this distance, E for Jupiter and Earth. This suggests that, at most, 90 giving a speed of 4,992 km / 10 h = 499.2 km/h, or 1,320 Earths could fit inside Jupiter. approximately 500 km/h. (b) To compare the relative density of Jupiter to Review: Figure 10.10 lists wind speeds of about Earth, use the density equation, in which Jupiter’s 550 km/h at Jupiter’s equator, in agreement with what mass is 318 times larger than Earth, and its volume is we find h­ ere.

80 ◆ Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets 10.9(b), the pressure changes from 0.1 to 10 atmo­ 42. Setup: The power (or flux) given off by a blackbody is spheres as one travels from 0 to 240 kilo­meters, for a related to temperature by the Stefan-­Boltzmann law, 4 (10 − 0.1) atm ; so we can compute the ratio of energy F∝T rate of change of = 0.04 atm/km; that absorbed to energy radiated by comparing the ratio of 240 km is, pressure increases much more rapidly on Jupiter T4 F temperatures, that is, in = in4 . than on Saturn. Repeating the same exercise for Uranus, Fout Tout Figure 10.9(c), the pressure increases to 200 atmo­ 4 Tin4 95 Fin Solve: Using the previous ratio, we find that = 4 = spheres = 1.8.at a depth of about 150 kilo­meters, or a rate of 4 Fout Tout 82 change of 1.33 atmospheres per kilo­meter, whereas Fin Tin4 954 = 4 = 4 = 1.8. Saturn gives off 1.8 times (80 ­percent) for Neptune, Figure 10.9(d), the pressure increases by Fout Tout 82 about 500 atmospheres at a depth of about 125 kilo­ more energy than it receives. meters, or a rate of change of about 4 atmospheres per Review: How is this pos­si­ble? Gravitational contrac­ kilo­meter. Thus, the pressure change is greatest for tion and differentiation (helium rain) of the planet Neptune.

Based on ­these values, Saturn has the low­ must be continuing in order to explain that the planet’s est rate of change, and Neptune has the highest. energy output is greater than that provided by solar Review: Are ­these results surprising? The ice ­g iants warming alone. It must include the planet’s own inter­ are much denser than the gas ­g iants and thus, should nal energy. have a larger increase of pressure with depth, as we 43. Setup: The power (or flux) given off by a blackbody is found.

Using this same logic, Saturn is less dense related to temperature by the Stefan-­Boltzmann law, (overall) than Jupiter and shows the same trend.

F ∝ T4, so we can compute the ratio of energy 45.

Setup: Look at Figure 10.9 and read off the tempera­ absorbed to energy radiated by comparing the ratio ture from the red trace at an altitude of 100 kilo­ Tin4 Fin meters. It may help to use a ruler to draw lines from = 4 . Rearranging of temperatures, that is, Fout Tout 100 kilo­meters to the trace and then from the trace to to solve for the ­actual temperature gives us, the temperature scale at the bottom.

1/4 Solve: Jupiter: about 210 K; Saturn: about 100 K; ⎛F ⎞ Tout = Tin ⎜ out ⎟ .

Uranus: about 90 K; Neptune: about 100 K. ⎝ Fin ⎠ Review: Are t­hese results surprising? One might expect that b­ ecause the ice g­ iants are so far from the Solve: ­Because Neptune gives off 2.6 times more Sun, their temperatures would be much colder than in energy than it receives, we find that the closer gas g­ iants. However, remember that for gas, 1/4 ⎛ Fout ⎞ temperature and pressure are related by the ideal gas 4 Tout = Tin ⎜ ⎟ = 47 K × (2.6) = 60 K. law, and the ice g­ iants’ atmospheres are u­ nder signifi­ F ⎝ in ⎠ cantly higher pressure.

Review: For the planet to give off more energy than it receives, it must be warmer than its predicted temper­ Using the Web ature based just on solar heating, as we found. From 46. a. The spacecraft Juno looped around the Sun and passed NASA’s fact sheet for Neptune, we find T = 72 K at 1 Earth again in 2013 before heading to Jupiter ­because it bar, and T = 55 K at 0.1 bar, for an average of 64 K. needed a flyby gravity assist to gain enough speed to get 44. Setup: Looking at the panels in Figure 10.9, note that to Jupiter. Galileo discovered that Jupiter had moons the rate of increase of pressure with depth is the orbiting it. The moons Io, Europa, Ganymede, and Cal­ change in pressure compared with the change in listo are known as the Galilean moons. The plaque is depth, that is, the ratio of the two. However, ­because to honor his discoveries. b. The goals of the mission each panel gives a dif­fer­ent range of depths, we need ­ ater in Jupiter’s are: determining the amount of w to compute and compare the a­ ctual ratios rather than atmosphere; mea­ sure composition, temperature, simply inspect them visually. cloud motions and other properties of deep in its Solve: Starting with Figure 10.9(a), note that for Jupi­ atmosphere; map magnetic and gravity fields; study ter the pressure changes from 0.1 to 10 atmospheres its magnetosphere. It has had many recent discover­ as one travels from 0 to ~110 kilo­meters below the ies, so answers ­w ill vary. The end of the mission is surface. Thus, the rate of change for Jupiter is about expected in July 2021.

(10 − 0.1) atm 47. a. In autumn, 2018, Voyager I was 143 AUs from the = 0.09 atm/km. For Saturn, Figure 110 km Sun (interstellar space); Voyager 2 was 119 AUs from

Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets

◆ 81

4. Using a similar method as described in question 2, we the Sun. Much was learned about Uranus and Nep­ find the distance the storm has moved from the first tune from the flybys; answers w ­ ill vary. b. Students to second image is about 11 mm. So the total distance are asked to design a new version of the Golden Rec­ it traveled around the planet is about 138 + 11 = ​ ­ ill vary. ord. Answers w 149 mm.

48. NASA crashed Cassini into Saturn to protect the 5. As the question suggests, we take a ratio to find the moons of Saturn that could have conditions suitable for life. Main discoveries from that mission include Titan T C = . Solving for the rotation rotation period: is one of the most Earth-­like worlds w ­ e’ve encountered. t D ⎛ 138 mm ⎞ Each of Saturn’s moons is unique; some may harbor C = 16.4 h. period, we getT = t = 17.6 h ⎜ ­water. Enceladus has cryovolcanism. The rings of Sat­ D ⎝ 149 mm ⎟⎠ urn have complex structure. Many discoveries ­were This is less than 17.6 hours, as suggested in the hint, made; Space​.­com lists 13 of its greatest discoveries. A ­because in one period the storm would have moved search in the NASA ADS (Astrophysics Data System) one time around the planet. That it moved more than shows many newly published papers. one rotation means that the elapsed time of the image 49. a. Involves students searching for super-­Jupiter plan­ was longer than the a­ ctual period of rotation. ets with masses much greater than that of Jupiter at 6. Repeating the previous exercise, we find that the the Extrasolar Planets encyclopedia. Answers w ­ ill limb-­to-­center radius at the latitude of the GRS is vary. b. Involves students searching for a super-­ about 26 mm, and the storm was 5 mm short of Earth on that same website, and listing the full char­ returning to its original location. The circumference acteristics of the exoplanet they find. Answers w ­ ill of this circle is 163 mm and, in this case, the storm vary. moved 163 − 5 = 158 mm around the planet. Using 50. Students are asked to find a star that has multiple ⎛ 163 mm ⎞ C planets on the exoplanet archive site, and graph dis­ = 18.2 h. our period equation, T = t = 17.6 h ⎜ D ⎝ 158 mm ⎟⎠ tances, masses, and sizes, and compare that system to ⎛ 163 mm ⎞ C = 18.2 h. This is larger than the elapsed time ­because = t = 17.6 h ⎜ the Solar System. Answers w ­ illTvary.

D ⎝ 158 mm ⎟⎠ the ­Great Dark Spot did not complete one full rota­ tion; it needed a bit more time to do so (about one-­ EXPLORATION half hour).

7. ­These two periods differ by about 1.8 hours (16.4 ver­ The exploration for this chapter investigates the rotation sus 18.2 hours) or about 10 ­percent. So they are close of Neptune based on mea­sure­ments of the movement of but certainly not the same. the G ­ reat Dark Spot and a smaller spot nearer the planet’s 8. We can use this information to find the velocity at south pole over a period of time. each latitude on Neptune. For the small storm: d 138 mm mm d 158 mm = = = 7.8 ; for the GDS: v = = EXPLORATION SOLUTIONS 17.6 h 17.6 h t h t 1. Neptune is an ice ­g iant and significantly larger than v = d = 158 mm = 9.0 mm/h. ­ These findings tell us that 17.6 h t any of the terrestrial planets. For it to have completed ­either (a) the equatorial region of the planet rotates a single rotation in 17.6 hours (about three-­quarters faster than the polar regions or (b) this method is sub­ of an Earth day) means that it is rotating very quickly. ject to large enough errors that we cannot ­really trust ­Because the ­Great Dark Spot (GDS) and this small the results. Indeed, when reviewing the strengths of storm rotated dif­fer­ent amounts within the same time the winds on Neptune, we may need to take into period, it means that the cloud tops experience “dif­ account how t­ hose velocities affect what we mea­sure ferential rotation”; that is, the speed of clouds change as a rotation period. with latitude.

9. We can use periodic radio bursts caused by rotation of 2. To mea­sure the approximate radius at which the small the planet’s magnetic fields to determine the rotation storm travels, we used a copy of Figure 10.23 in an rate. We could also improve the method done above image editor with a metric ruler overlaid. We found by fitting circles to the orbits of the storms rather than the segment from the limb to the line through the straight lines similar to the ones shown in Fig­ center of the planet is about 22 mm. ure 10.24 (Neptune is a three-­d imensional object, 3. The circumference at this latitude is C = 2πR = 2 × 3.14 ​ ­a fter all). × 22 mm = 138 mm. v=

82 ◆ Chapter 10 Worlds of Gas and Liquid—­The G ­ iant Planets

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 13: COMPARISON OF STORMS ON JUPITER AND EARTH Through the use of Voyager I spacecraft observations of the ­Great Red Spot on Jupiter, students learn about the wind conditions in Jupiter’s upper atmosphere. They also make mea­sure­ments in order to:

• find the circumference of the ­Great Red Spot.

• calculate the rotation period and speed of the ­Great Red Spot.

• compare the speed of the ­Great Red Spot and its life­ time to one of the strongest storms on Earth.

• compare the results with published values and con­ sider sources of errors.

Text reference: Section 10.2

CHAPTER 11

Planetary Moons and Rings

INSTRUCTOR NOTES Chapter 11 covers rings and moons of planets. Major top­ ics include: • an overview of moons of the Solar System, including distribution and orbits.

• formation and be­hav­ior of regular and irregular moons.

• comparative investigation of the geologic activity, pres­ ence of w ­ ater, and history of moons.

• the discovery, orbits, characteristics, composition, evo­ lution, and structure of planetary rings.

This is a very exciting time in the field of habitable moons. Obviously Europa has been a target of interest for quite some time, and currently, an extensive flyby mis­ sion, named the Europa Multiple Flyby Mission (formerly Europa Clipper), has received extensive funding to move into development stages. Astronomers have concluded that Saturn’s moon Enceladus also has a subsurface ocean of liquid ­water. An in­ter­est­ing question for students is ­whether we should try to land on ­either moon and enter its ocean. On the one hand, what could be more exciting than turning on the lights and seeing a pos­si­ble extrater­ restrial ecosystem? On the other hand, bacteria are such tenacious survivors that we can be almost guaranteed that we would contaminate that environment as soon as our probe entered it.

DISCUSSION POINTS • Figure 11.1 shows the major moons of the Solar ­System, to scale. Are t­ hese moons “regular” or “irregular”? What are the orbital characteristics of regular moons? How do the orbits and characteris­ tics of irregular moons compare to regular moons?

(LG 1) • A hundred years ago, Mars was suspected to be the most likely place to host extraterrestrial life in the Solar System. In the last few de­c ades, it was Europa, and in the last few years, Enceladus took that distinction. Why does the object of our focus

keep changing? Is this a strength or weakness of sci­ ence? (LG 2) • This chapter introduces the exciting possibility that other worlds in the Solar System may harbor life, which would most likely be primitive (extremophiles).

Students in groups pick one of the following worlds and research current status to report back to the class.

(LG 2) • Enceladus • Europa • Titan • Mars • One of the g­ reat unsolved mysteries in planetary sci­ ence is the diversity of satellites in each of the ­g iant planetary systems. Why did some moons in the same system develop active geology whereas o­ thers did not?

Why did some moons maintain active geology whereas o­ thers became inactive early in their histo­ ries? (LG 2) • Each of the four Galilean moons of Jupiter display dif­ fer­ent levels of geologic activity. Discuss why each one has the level it does in context of its distance from Jupi­ ter. (LG 2) • All four g­ iant planets sport ring systems, but none of them are alike. Assign each ring system to a team. Have them make 18″ × 24″ detailed posters of the ring systems of their g­ iant planet and pre­sent it to the rest of the class. Students discuss the simi­ larities and the differences among the ring systems.

(LG 3) • Saturn has what have been named “shepherd moons.”

Why is this name appropriate for ­these small moons embedded within its rings? What role do gravity and the orbits of the ring particles and small moons play in the structure of Saturn’s rings? (LG 4) • The chapter-­opening image is a spectacular one of Saturn and its rings. The question is, “Why do some planets have rings?” We might want to start with, “Why ­don’t the inner planets have rings?” Stu­ dents have prob­ably seen many pictures of Saturn and its rings. What ­were their impressions? What do they know about what makes up ­t hose rings?

83

84 ◆ Chapter 11 Planetary Moons and Rings

PROCESS OF SCIENCE Before the flyby mission of Voyager 1, it was thought that the orbits of the ring particles w ­ ere orderly and obeyed Kepler’s laws. ­Those laws would not account for substruc­ ture of kink, knots, braiding within a ring. Was our phys­ ics of orbits wrong in some way? Further calculations, investigations, and detailed observations discovered the culprits: small moons.

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Cassini came from Earth, and even though strong mea­sures are taken to shield satellites from bio­ logical contamination—­bacteria, viruses, and other microorganisms—­before they are launched, we ­can’t guarantee a perfectly sterile craft. Some microorgan­ isms can withstand the environment of space, and thus could contaminate any moon if they hitchhiked a ­r ide on the lander.

2​.­ In addition to searching on YouTube to see the end of Cassini, students may wish to get the information direct from the source by visiting the NASA web site: https://saturn​.­j pl​.­n asa​.­gov​/­m ission​/­g rand​-­finale​ /­overview​/­. A major find is that the rings are relatively young compared to their planet, perhaps forming at the same time dinosaurs roamed Earth.

3. Besides the pos­si­ble missions mentioned in the arti­ cle, at the time of this writing, NASA has a ­couple of missions in concept development, to the moon Titan and one to enter Saturn’s atmosphere as was done for Jupiter.

4. It appears that e­ very planet, moon, minor body is affected by that policy, including Earth for returning spacecraft. The protection mea­sures for the Mars landers and rovers are/­were extreme, and t­here is a long list of them at: https://­planetaryprotection​.­nasa​ .­gov​/­solarsystem​/­mars​/­

CHECK YOUR UNDERSTANDING SOLUTIONS 11.1. (a, b, c) Note that many moons are about the size of Mercury. Answer (d) would be right if it said most known planets, but it said all.

11.2A. (a) Io formed too close to Jupiter to hold on to any volatile material.

11.2B. (a, d, b, c) Ranked from the highest density of impact craters to least.

11.3. (a) Ring systems are not stable over long periods of time.

11.4. (b) Rings are brightest when looking back ­toward the Sun, in the shadow of the planet, something known as “backlighting.”

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (a) This is the basis of comparative planetology.

2. (b) The proximity of Io and Europa to Jupiter means ­these moons experience greater tidal stresses that have continually heated the interiors enough for both moons to stay geologically active.

3. (d) Note that by life, we mean “life as we know it.”

4. (d) Tidal stresses in Io’s orbit around Jupiter deform the moon’s radius by about 100 meters per orbit.

5. (a) Of the choices given, the narrow gaps and oscillat­ ing edges seen in rings are due to the gravitational effects of small moons.

6. (a, d) ­These are the best choices. We do not know how recently the rings formed.

7. (c) Physics was able to provide an explanation of this ring even though it was not obvious at first.

8. (a) Moons are expected to form out of the same mate­ rial as the host planet; thus, a retrograde moon is likely to have been captured.

9. (b) Rings are best observed when lit from ­behind; thus, being ­behind the planet looking ­toward the Sun. 10. (b) ­There was more material in the outer Solar System from which the planets and their moons could form.

11. (a) According to the IAU definition, a moon is the sat­ ellite of another body.

12. (c) Although (a) and (b) are ways to date objects, one has to land on a moon to perform them, which only leaves us the study of surface features from imaging.

13. (d) More precisely, it is the tidal stretching of Io by Jupiter’s gravity and Europa’s pull keeping its orbit elliptical that keeps Io’s interior molten hot.

14. (c) If a moon was formerly active, it would have cov­ ered some of its older craters. This does not preclude

Chapter 11 Planetary Moons and Rings

­those regions receiving a few new craters, but b­ ecause most of the cratering happened early in the Solar Sys­ tem’s history, one expects few new craters in regions that ­were resurfaced.

15. (c) The volcanic ejections inject new particles into the ring system.

Thinking about the Concepts 16. Tidal stresses from Jupiter continuously flex Io by many tens of meters per orbit and keep the interior mantle molten.

17. Like terrestrial volcanism, cryovolcanism involves the ejection of liquefied materials at high pressures. Cry­ ovolcanism is volcanic activity that occurs with mate­ rials that are liquid at very low temperatures. For example, Enceladus exhibits cryovolcanism with liq­ uid w ­ ater substituting for magma. Triton exhibits cryovolcanism with liquid nitrogen substituting for magma. Io exhibits it through constant volcanic erup­ tions of magma made of sulfur compounds.

18. Europa’s surface is spectroscopically dominated by ­water ice. This implies that the source of this ice is liq­ uid ­water just below the surface. Observations sup­ port a subsurface ocean as we note the surface ice of Europa is broken into pieces that fit together like pieces of a jigsaw puzzle, just like the broken pieces of the North Pole ice and tectonic plates that float on Earth’s molten mantle. The absence of impact craters also indicates that the surface must be relatively young, and the material used for resurfacing the planet could come only from its interior.

19. Methane is not expected in Titan’s atmosphere ­because UV photons from the Sun photodissociate the molecule.

20. Titan resembles Earth with the nitrogen composition of its atmosphere and surface features that are similar to rivers and lakes. On Titan, methane appears to play the same role that w ­ ater plays in the hydrologic cycle on Earth.

21. Ganymede shows grooves and ridges on its bright ter­ rain, clearly indicating that it had tectonic activity at some point in its past. Other moons, such as Tethys, Ariel, and Miranda, also show evidence of past cryo­ volcanic or tectonic activity. Tethys has a huge can­ yon system that spans most of its circumference.

Ariel’s surface exhibits smooth plains and complex rifts, indicating that cryovolcanism and tectonic activity operated over portions of its surface. Fi­nally, Miranda displays enormous fault scarps, suggesting ­either that its history was marred by a catastrophic breakup and accretion or that differentiation s­ topped prematurely when the moon cooled.

◆ 85

22. Inner planets lacked the debris that outer planets had during formation, and therefore, ­there is not a current source of material to form rings t­oday. Additionally, the geologically active moons of the outer planets continuously replenish rings with outgassed material.

The moons of the inner planets are all dead. Occulta­ tions of stars by planets are the best ground-­based method used to look for rings. If the star’s brightness varies just before or ­a fter the planet occults the star, then the chances are that a ring of material has caused that dimming.

23. Ring arcs, such as ­those seen at Neptune, are denser regions of a ring, so that when viewed, one sees only the denser material and could interpret the ring as being incomplete—­that is, an arc of material rather than a complete ring.

24. Planetary ring material is likely produced by the breakup of moons, asteroids, or comets that fall inside a planet’s Roche limit. It is also produced as nearby moons shed material through volcanic pro­cesses or impact events.

25. Gaps can be created by small moons, which clear out a lane of material as they orbit within the ring. Larger gaps are created by orbital resonances of shepherd moons.

26. Planetary rings are fairly dense so that particles con­ stantly collide with each other and maintain a fairly tight, thin ring. This comes about when particles crossing through a ring from opposite directions col­ lide, lose energy, and ­settle in with the other particles.

Diffuse rings are sufficiently low density so that colli­ sions are rare, allowing the particles to have noncircu­ lar orbits. As a result, diffuse rings are often thick (puffed up).

27. The discovery of kinks and knots in Saturn’s F ring meant that ­either our understanding of rings was completely wrong or that we had an incomplete the­ ory. B ­ ecause scientific understanding is provisional, our understanding of rings was correct ­until we learned more about them and required a theory that incorporated more physics. Thus, our understanding grew with more observations to become a more com­ plete explanation.

28. Particles in rings are constantly undergoing colli­ sions, which sometimes impart enough energy to the particle that it escapes from the ring. Thus, all rings are doomed to dissipate ­unless t­ here is a source feed­ ing ring material to make up for the particle losses.

However, shepherd moons on e­ ither side of a ring herd the ring particles back into position. This shep­ herding pro­cess keeps the ring confined to a narrow region of space.

86 ◆ Chapter 11 Planetary Moons and Rings 29. The E ring of Saturn is being fed by eruptions from Review: The reported mass of Jupiter is 1.898 × 1027 Enceladus and may survive in­def­initely, or at least as kg, confirming our result. long as Enceladus is active.

33. Setup: Using any moon of Saturn, calculate the ratio 30. Answers ­w ill vary considerably. Europa is one of the 4π 2 A3 A3 M= × 2 = [5.9 × 1020 (kg s2)/km 3] × 2 , most commonly cited moons b­ ecause it may harbor G P P life as we know it in the liquid w ­ ater ­under its surface. and calculate the mass shown in Working It Out 11.1 In such a case, a probe would have to land on the and prob­lem 32. moon, bury itself into the ice ­until it reaches the sur­ Solve: Titan orbits at a distance of 1.22 × 106 km face ocean, and return both imaging and microscopic 86,400 s and has a period of P = 15.95 days × = 1.38 × 106 s. analyses of the w ­ ater.

1 day 86,400 s = 1.38 × 106 s. For the mass of Saturn, this gives us P = 15.95 days × Applying the Concepts 1 day (1.22 × 106 km)3 31. Setup: To solve this prob­lem we ­w ill use the escape M = [5.9 × 1020 (kg s2)/km 3] × M (1.38 × 106 s)2 velocity equation, vesc = 2G , as given. Note that = 5.63 × 1026 kg.

R the radius of the moon is given in kilo­meters but must Review: The reported mass of Saturn is 5.68 × 1026 kg. be converted to meters.

34. Setup: Linear axes increase in integer increments, Solve: (a) logarithmic in multiples of 10.

M Solve: Both diagrams have linear axes. Panel (b) has vesc = 2G axes with ranges that are 15 times larger than in panel R 22 (a). The amount of space covered in ­these images is 8.9 × 10 kg = (2)(6.67 × 10 −11 m 3 kg −1s −2 ) 4 × 1012 km2 for panel (a) and 9 × 1014 km2 for panel 6 1.82 ×10 m (b). ­There is 152 = 225 times as much space repre­ ≈ 2,500 m/s, sented in panel (b).

Review: Note how small the orbits from panel (a) are or 2.5 kilo­meters per second (km/s) (b) The escape in (b), which confirms how much more space is being 2.5 km/s shown.

4 velocity is = 2.5 times faster than the typi­ 35. Setup: Surface area is 4πR 2 and volume is V = π R3 .

1 km/s 3 cal speed of gas escaping from Io’s vents. This means The relationship between the two is that (volume) = ​ that most of the vented material ­w ill fall back on the (surface area) × (thickness). The time is takes for moon, rather than escaping into space. something to happen is, (time) = (amount) / (rate).

Review: The book explains that Io has essentially Solve: (a) Using Io’s radius, we find a surface area of turned itself inside out a few times, meaning that A = 4πR 2 = 4π (1.82 × 106 m)2 = 4.16 × 1013 m2 and a the moon has expelled almost its full mass of mate­ 4 4 rial from its volcanic eruptions. This is only pos­si­ble volume of V = π R3 = π (1.82 × 106 m)3 = 2.53 × 1019 m 3.

3 3 4 to pull 4 all that material if its gravity is sufficient V = π R3 = π (1.82 × 106 m)3 = 2.53 × 1019 m 3. back onto the moon rather than 3 3 letting it escape (b) The volume of material deposited on the surface into space. Thus, our finding is consistent with this of Io each year is Vdeposit = surface area × thick­ necessity. ness = (4.16 × 1013 m2) × (0.003 m) = 1.25 × 1011 m3.

32. Setup: The first equation of Working It Out 11.1 gives (c) The time, t, that it takes to deposit the equivalent 4π 2 A3 A3 20 2 3 of Io’s entire volume onto the surface is determined × 2 = [5.9 × 10 (kg s )/km ] × 2 us M = G P P by dividing Io’s volume by the deposition rate, 2 4π . when using P in sec­ by inserting the value for 2.53 × 1019 m3 Volume G 8 onds and A in km. t = = 86,400 s Deposit Rate5 1.25 × 1011 m3/yr = 2.03 × 10 yr. = 3.07 × 10 s Solve: Europa has an orbital period of P = 3.55 days × 1 day Io ­ w ill essentially turn inside out ­every 200 million 86,400 s P = 3.55 days × = 3.07 × 105 s and a semimajor axis A = ​ years.

1 day (d) The Solar System is about 4.5 billion years old, so 6.71 × 105 km. Inserting t­ hese into our formula yields Io has turned “inside out” about 22 times.

5 3 (6.71 × 10 km) 20 2 3 27 M = [5.9 × 10 (kg s )/km ] × = 1.88 × 10 kg.

Review: This prob­lem shows us just how “active” Io .71 × 105 km)3 (3.07 × 105 s)2 27 ­ r eally is. = 1.88 × 10 kg.

3.07 × 105 s)2

Chapter 11 Planetary Moons and Rings

36. Setup: According to Working It Out 11.2, Ftotal ∝ Mplanet Mmoon R moon Ftotal ∝ . Use this as the guide to scale the d3 quantities to see how changes in some ­factors affect the tidal force.

Solve: (a) If the radius of the moon R moon increases but its mass stays constant, the tidal force ­will increase. (b) If the orbital radius of the moon decreases, the tidal force ­will increase. (c) The formula shows us that if the mass of the central planet increases, the tidal force w ­ ill increase.

Review: The strength of the tidal force scales with the force of gravity, with the extra term of the radius of the moon divided by its distance from the planet.

Thus, we can see that the masses and sizes of the planet and moon directly affect the tidal force.

37. Setup: Use the full equation for tidal force and calcu­ late it for Jupiter-­ Io and for Earth-­ Moon: Mplanet Mmoon R moon Ftotal = 2G . d3 Solve: Substituting the values from the book for masses, radius, and distance, we find the tidal forces for each system:

Mplanet Mmoon R moon Review: This . total force is so small that the astronaut d3 would never feel it. As a comparison, the mass of a paperclip is about 0.001 kg (1 gram). Its weight on the surface of Earth is about 1 × 10 −2 N. 39. Setup: If all e­ lse is held equal, then tidal force scales as 1 d 3 and gravity scales as 1 d 2 . The radius of the planet’s moon m ­ atters for the tidal force. We have set it to 1 m for this example to highlight just the effect of changing the distance for the two forces. We have set the masses of the planet and moon to be 1 × 108 kg and 1 kg. This sets the value of the forces to be in Newtons. ­ ill decrease much faster than that Solve: The tidal force w of gravity as the distance between a planet and its moon increases. We show a graph of tidal force (bottom curve) versus gravitational force (top curve) in Figure IM 11.1 for the mass of a planet (1 × 108 kg) and its moon (1 kg), with the radius of the moon set to 1 m. An increase in the radius of the moon would shift the tidal force curve up.

FIGURE IM 11.1 1.E+08

FJupiter/Io = (2)(6.67 × 10 −11 m 3 kg −1s −2 ) (1.9 × 1027 kg)(8.93 × 1022 kg)(1.822 × 106 m) × (4.23 × 108 m)3 20 = 5.45 × 10 N 3

−1 −2

5.45 × 1020 N = 303. The tidal ­Doing the comparison: 1.8 × 1018 N force between Jupiter and Io is about 300 times that between Earth and the Moon.

Review: The calculations w ­ ere done using the aver­ age distance of Io from Jupiter. With the orbital eccen­ tricity forced on it by Europa and Ganymede, the tidal force differs by about 2 ­percent.

38. Setup: Use the tidal force formula, Mplanet Mmoon R moon Ftotal = 2G , where “moon” is now d3 replaced by the astronaut’s values for mass and size.

Solve: We ­w ill use a mass of 60 kg and size of 1.7 m ­because that’s the average size of a ­woman in the United States. The tidal force of Earth on the astronaut is FEarth/astronaut = (2)(6.67 × 10 −11 ) (5.97 × 1024 kg)(60 kg)(1.7 m) × [(3.80 × 105 ) + (6.378 × 106 )]3 = 2.6 × 10 −4 N.

Tidal Force versus Gravita onal Forece for Arbitrary Values F tidal α 1 d3

1.E+07

F gravity α 1 d2

Tidal Fore (N) Gravita onal Fore (N)

1.E+06 1.E+05 Force (N)

FEarth/Moon = (2)(6.67 × 10 m kg s ) (5.97 × 1024 kg)(7.35 × 1022 kg)(1.74 × 106 m) × (3.844 × 108 m)3 18 = 1.8 × 10 N −11

◆ 87

1.E+04 1.E+03 1.E+02 1.E+01 1.E+00

0

100 200 300 400 500 600 700 800 900 1,000 Distance “Moon” is from the “Planet” (m)

Review: Note that the tidal force w ­ ill approach that of gravity, but it happens when a moon is very close to its planet.

40. Setup: A 7:6 resonance implies that the ring particles orbit Saturn seven times for e­ very six orbits of Janus.

This means that their orbital periods are 6/7 as long as Janus’s orbital period.

Solve: Pring =

6 6 (16 h 41m ) = (16.6833 hr) = 14.3 hours 7 7

Review: The orbital period of the ring particles is 14.3 hours or 14h18m.

41. Setup: Escape velocity vesc = 2G M .

R Solve: Janus has a mass of 1.9 × 1018 kg and a radius ­ ill be of 8.95 × 10 4 m. The escape velocity w

88 ◆ Chapter 11 Planetary Moons and Rings vesc = 2G

M R

= (2)(6.67 × 10 −11 m 3 kg −1s −2 ) = 53 m/s.

1.9 × 1018 kg 8.95 × 10 4 m

Review: This is about 120 miles per hour; that is, an airplane could escape from Janus.

42. Setup: To compute the total mass of material in the ring, we need to compute the volume of the ring mate­ rial. We can approximate the ring as a disk with a smaller, disk-­shaped hole in the center. The volume of a disk with radius r and thickness h is V = πr 2h. The volume of the ring is the volume of the ­whole disk minus the volume of the inner material, that is, 2 V = π h(rout − rin2 ). Fi­nally, to compute the mass of material in the ring, we substitute the volume of the ring into the well-­k nown relationship between mass, M or M = ρV. Solving for density, and volume, ρ = V 2 mass gives M = πρh(rout − rin2 ).

Solve: Inserting values given in the text, and using 2 M = πρh(rout − rin2 ), we find M = π(150 kg / m3) (10 m)

[(2.35 × 108)2 − (1.84 × 108)2] = 1.01 × 1020 kg.

Review: We know from the text that the total mass in the rings is small. In this case, the total mass is 0.0017 ­percent that of Earth, and 0.14 ­percent that of the Moon.

43. Setup: From the previous prob­lem, the mass of the mate­ rial in B ring is 1020 kg, whereas Mimas is 3.8 × 1019 kg.

1.01 × 1020 kg Solve: The ring mass is = 2.7 times 3.8 × 1019 kg larger than that of Mimas.

Review: This is significant ­because it suggests that the ring was plausibly formed by the tidal disruption of a medium-­sized moon rather than from the slow escape of particles from nearby moons.

44. Setup: To relate period and distance, use Kepler’s law, that is, P 2 = A 3. ⎛ 2.35 × 105 km ⎞ P Solve: outer = ⎜ Pinner ⎝ 1.84 ×105 km ⎟⎠

3/2

= 1.44. The outer

ring particles take 1.44 times longer to complete an orbit than the inner ring particles. The particles are orbiting individually according to Kepler’s laws.

Review: We usually think of rings as being solid bodies, but this shows us that the ring is clearly made up of material that is ­doing dif­fer­ent ­things at dif­fer­ent places.

M from the 45. Setup: Escape velocity scales as vesc ∝ R formula in Working It Out 11.3.

Solve: The escape velocity formula tells us that a (a) more massive planet ­w ill have a higher escape velocity. (b) larger planet ­w ill have a lower escape velocity ­because making R larger makes 1/R smaller.

M . (c) The escape velocity of a planet scales as vesc ∝ R Even if we know the escape velocity of a planet, we still need to know the radius to find the planet’s mass.

Review: Why is the escape velocity smaller if the planet’s ­ ecause gravity is weaker the farther you size is larger? B are from the center of the mass for solid, ­spherical worlds.

Using the Web 46. Students go to Sky & Telescope’s “Jupiter Moons” web page and update the positions of the Galilean moons and follow their orbits. Answers w ­ ill vary.

47. Students investigate updated lists of g­ iant planet moons for recently discovered moons and information about their orbits and characteristics. Answers ­w ill vary.

48. Enceladus c­auses Saturn’s aurora. This is an accurate link to the video: https://­solarsystem​.­nasa​.­gov​/­resources​ /­10732​/­hiss​-­from​-­aurora​-­caused​-­by​-­enceladus​/­ 49. Missions to the moons: a. The scientific plan for the Clipper mission is to perform a detailed investigation of Europa and its potentially habitable environment. It ­w ill orbit Jupiter instead of Europa b­ ecause Europa is bathed in radiation from Jupiter. The orbiter must get out of that radiation. Planned launch date is ~ 2023. It ­w ill get to Europa 5–7 years ­a fter launch. It ­w ill not include a lander, but it ­w ill investigate pos­si­ble landing sites. b. The JUICE mission ­w ill study Ganymede, Cal­ listo, and Europa. The mission goals are the interrela­ tionship of Jupiter and its moons, Ganymede as a planetary body and potential habitat, comparative study of the moons, and more. The mission looks like it is still a “go.” JUICE ­w ill take longer to get to Jupiter than the Clipper mission b­ ecause it ­w ill undergo five gravity assists. c. Students do a search to see ­whether a new mission to one of the moons of Saturn has been approved. Answers ­w ill vary.

50. Involves students searching to see w ­ hether moons or rings have been confirmed on exoplanets. Answers ­w ill vary. Astronomers might consider it in­ter­est­ing to see how common rings are to help them under­ stand how they are formed.

EXPLORATION This exploration has students find out the scale of an image by relating it to finding the scale on a map. We have found with the quick availability of digital maps and GPS devices

Chapter 11 Planetary Moons and Rings

that most students d­ on’t understand what scaling means.

We have had students simply convert millimeters to kilo­ meters without any mea­sure­ments. We recommend giving a number of demonstrations of how it is done, even beyond what is given in the text.

EXPLORATION SOLUTIONS 1. In the copy we have, the image is 50 mm in height.

2. This is prob­ably accurate to 2 mm, which means the mea­sure­ment could range from 48 mm to 52 mm.

3. The radius is half the dia­meter, or 25 mm.

4. Io has a radius of 1,821 km.

5. The scale is 1,821 km / 25 mm = 72.8 km/mm.

6. km per mm 7. The black ring is about 6.5 mm across.

8. 6.5 mm × 72.8 km/mm = 473 km. Answers ­will vary for an example. The area of the geyser is about 175,000 km2, greater than about 50 p­ ercent of our states.

9. The plume is about 4 mm, or 4 mm × 72.8 km/ mm = 290 km tall. Comparing to objects that are or ­were in orbit: ISS, 330-420 km; Hubble Space Tele­ scope, 540 km; Gravity Field and Steady-­State Ocean Circulation Explorer (GOCE, launched by Eu­ro­pean Space Agency), 255 km.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 14: THE CLEARING AND HERDING OF SATURN’S RING PARTICLES In this activity, students use Kepler’s and Newton’s laws to analyze the be­h av­ior of ring patterns in the rings of

◆ 89

Saturn. A ­ fter completing this activity, they ­w ill be able to: • apply Kepler’s third law to sections of the ring system of Saturn.

• summarize the pro­cess by which a single moon ­w ill clear a gap in a ring system like Saturn’s.

• summarize how the two shepherd moons manage to keep the particles that orbit between their orbits tightly confined within a narrow ring.

Text reference: Sections 11.3 and 11.4 ACTIVITY 15: MOONS OF THE ­ GIANT PLANETS Students compare major moons of the g­ iant planets, both against one another and against Earth’s Moon. This activ­ ity continues the method of “comparative planetology,” which is useful for finding commonalities among the bod­ ies of the Solar System. As students work on this activity, they: • compare and contrast the surface features of Jupiter’s moon Ganymede with t­ hose of the Moon.

• discuss evidence of resurfacing and cratering on four moons of Saturn.

• compare and contrast surface features of Saturn’s moon Titan to similar ones on Earth.

• compare the evidence for the repro­cessing of part of Neptune’s moon Triton.

Text references: Sections 11.1 and 11.2

CHAPTER 12

Dwarf Planets and Small Solar System Bodies INSTRUCTOR NOTES Chapter 12 rounds out the Solar System with every­thing that is not a planet or moon. Major topics include: • dwarf planets as found in the asteroid ­belt and the Kui­ per ­Belt.

• Pluto’s “demotion.”

• the distribution, composition, and classification of asteroids.

• the homes, orbits, and anatomy of comets.

• meteors, meteoroids, and meteorites.

• impacts with Jupiter and Earth.

DISCUSSION POINTS • The categories of the small bodies of the Solar Sys­ tem are dwarf planets (Kuiper ­Belt and Ceres), ­a steroids (the main asteroid ­belt and the asteroids orbiting outside it), comets (Oort Cloud and Kui­ ­ elt), and meteorites (vari­ous classifications). per B Have students create 18″ × 24″ posters of their assigned topic. Posters should include characteristics, classifications, compositions, orbits, and connection to the formation of the Solar System. Students pre­ sent their posters to the class in a poster session. (LG 1, 2, 3, 4) • ­Earlier in ­human history, the sudden appearance of a comet in the sky was often regarded as an omen, and often as a fear-­inducing event. Without the knowledge that we have t­ oday, why do you think that would be the case? (LG 4) • Asteroids and comets often collide with the major planets. Discuss what we have learned about the his­ tory of the Solar System and what astronomers are ­doing now to help assure us that t­ here ­w ill not be an impact with Earth without advance notice. (LG 5) • ­After working through the Applying the Concepts prob­lems number 35 and 37, discuss the implications of an asteroid of that mass hitting one of the largest cities on Earth. Chances are, the asteroid would hit an ocean. What would be the consequences if that hap­ pened? (LG 5) 90

• The chapter-­opening image asks, “Why land a spacecraft on a comet?” Have students do a 3-­m inute writing ­exercise before they have studied the section on comets, answering the question, and then ­later on, ­a fter they have worked through the section. Discuss how their views e­ ither changed or did not change. Does the knowl­ edge we gain through t­ hese missions offset the costs?

PROCESS OF SCIENCE The resolution made by the International Astronomical Union in 2006 reclassifying Pluto and other solar bodies as dwarf planets raised a ­great deal of controversy at the time, some of which remains t­oday. You may wish to discuss ­whether such a controversy is based on scientific grounds or ­whether t­here are also some historical and cultural aspects before covering the Pro­cess of Science for this chap­ ter. By following a bit of the history of the new classification, the conclusion ­here is that scientific decision making must follow the weight of the evidence, not historical pre­ce­dent.

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Spectral lines and isotope ratios are both atomic pro­ cesses, which follow fixed laws of physics, such that observations of ­either spectral lines or isotope ratios tell us impor­tant information about the medium we are investigating. Isotopes have unique spectral fingerprints just like the individual ele­ments we study.

Chapter 12

2. ­Water in Comet 67P has a dif­fer­ent ratio of hydrogen to deuterium than ­water on Earth, whereas the ­water on Comet Hartley 2 matches that of our ­water. It is, therefore, unlikely that Earth’s w ­ ater came from a comet like 67P, but it could have come from a comet like Hartley 2. Perhaps ­these two comets had dif­fer­ ent origins and histories.

3. At pre­sent, the w ­ ater on Comet 67P is a closer match to that found on other comets and not like that found on Earth or Hartley 2.

4. Other sources of Earth’s ­water include that it was on the planet at the time of its formation, that it was “beaten out of other minerals” as the planet formed, or that “wet planetesimals” from near Jupiter collided with the early Earth.

5. Dawn arrived at Ceres on March 6, 2015. At the time of the writing of this manual, Dawn had taken pictures of the floor of Ceres’s Juling Crater. The crater floor showed evidence of the flow of ice and rock. The area of ice appeared to increase between two dif­fer­ent images. The mission is nearing its end, and perhaps more information w ­ ill be published soon.

CHECK YOUR UNDERSTANDING SOLUTIONS 12.1. According to the IAU definitions, ­these objects are dwarf planets and not moons b­ ecause they do not orbit another body other than the Sun. 12.2. (c) Volcanic activity implies the body was, at one time, geologically active.

12.3A. (c) Oort cloud comets must have a highly elliptical orbit in order to enter the inner Solar System.

12.3B. (c)The text describes comet nuclei as similar to deep-­f ried ice cream “with a soft and porous interior.”

We know that the interior of a comet is made of ices (­water and carbon dioxide) mixed with dust and organic material.

12.4. (a, b, c, e) Meteorites can offer no information about cratering rates, eliminating choice (d).

12.5. The origins of a meteorite come from its precollision orbit as mea­sured by multiple observations, its struc­ ture, and its chemical makeup.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (c) Most planetesimals w ­ ere swept up by larger bodies (like planets).

Dwarf Planets and Small Solar System Bodies ◆ 91

2. (b) Stony-­iron meteorites come from the transition zone between the iron core and rocky mantle. Com­ pared with the size of a planetesimal, this is a very small region.

3. (d) A comet’s ion tail always points away from the Sun. 4. (a) Near Earth objects may collide with Earth; hence, our need to try to detect and prepare for such events.

5. (c) Meteorites are not debris from comets.

6. (a) Perihelion is the point in an orbit closest to the Sun; aphelion is the point in an orbit farthest from the Sun. 7. (b) Comet nuclei form comae and tails only when they are heated by the Sun’s radiation as they approach perihelion.

8. (a) Asteroids are primitive planetesimals orbiting the Sun. Most are found between Mars and Jupiter, but they are not restricted only to that location.

9. (b) Short-­period comets have had many more passes by the Sun and thus have lost more of their ice compared to long-­period comets that visit the Sun infrequently.

10. (d) Averages mean that one must consider a large sample.

In this case, a long period of time ­w ill yield roughly the average number of comets found.

11. (b) This is the “asteroid ­belt.”

12. (b) Comets contain a large amount of carbon dioxide ­ ater ice; carbonaceous chondrite meteorites con­ and w tain w ­ ater bound in molecules. The impacts could have delivered this w ­ ater to the planet.

13. (b) A large clump of iron is likely from an object which differentiated such that the heavier metals sank into a central region together, whereas lighter rocky material moved to the upper regions of the object.

14. (c) Comets leave ­behind particles of dust and ice from their tails as they pass the Sun; Earth passes through ­these producing meteor showers at roughly the same time each year.

15. (d) Dwarf planets do not have enough mass to have cleared out the space around them. This means they have lower mass than planets.

Thinking about the Concepts 16. Pluto differs from the other planets in a number of ways. It does not have a nearly circular orbit and does not orbit in the plane of the other planets, which sug­ gests that its formation was dif­fer­ent than the eight “classical” planets. Pluto also is not massive enough to have gravitationally swept out the space around it, whereas all the other planets are, which has allowed them to remain in the nearly circular and planar orbits.

17. Pluto is no longer classified as a planet ­because it did not clear out the orbital space around it. It was origi­ nally classified as a planet b­ ecause we did not know

92 ◆ Chapter 12 Dwarf Planets and Small Solar System Bodies anything about it other than it orbited the Sun; as we learned more—­such as the fact that t­ here ­were more Pluto-­like bodies just beyond Pluto’s orbit—we dis­ covered we had to change our formal understanding and categorization.

18. Asteroids are loose agglomerations of refractory rock and metal held together only by self-­g ravity. Comet nuclei are “dirty snowballs” that are composed primar­ ily of ices and organic materials held together by a loose rocky matrix.

19. Meteoroids are particles of space debris smaller than about 100 meters in dia­meter. Meteors are ionized trails left by the passage of meteoroids and interplanetary dust through the upper atmosphere. Meteorites are chunks of meteoroids that survive their atmospheric passage to land on the surface.

20. Comet nuclei are kilo­meters across compared with the typical millimeter sizes of the meteoroids that produce meteors. Comet tails can be hundreds of millions of kilo­meters long compared with meteor streaks that are only a few kilo­meters long. Even at closest approach, most comets still remain tens of millions of kilo­meters away from Earth. Meteor streaks occur less than 100 kilo­meters above Earth’s surface. Comets remain bright for weeks as they pass through the inner Solar System. Meteors remain vis­i­ble for at most a few seconds.

21. The age of a meteorite is determined by the time at which it cooled from its original molten state. ­Because C-­t ype meteorites are older than S-­t ype and M-­t ype meteorites, we can conclude that C-­type asteroids cooled quickly and prob­ably remained undifferenti­ ated. On the other hand, S-­t ype and M-­t ype asteroids prob­ably came from larger, differentiated planetesi­ mals that took longer to cool. We can also conclude that ­because of their older ages, C-­t ype asteroids ­were likely never incorporated into larger planetesimals.

22. The vast majority of asteroids lie on stable orbits in the asteroid ­belt, posing no threat to Earth. However, if an asteroid happens to cross Earth’s orbit at the same time Earth is passing through that region of space, the like­ lihood of a catastrophic collision increases. Earth-­ crossing asteroids are also much easier to reach with spacecraft b­ ecause the Sun’s gravity brings the aster­ oids to us. In the f­ uture, near-­Earth asteroids might be mined for their resources long before their main b­ elt cousins.

23. Position yourselves several kilo­meters apart and watch the same section of sky, maintaining contact with your cell phones. When you see a meteor, you should each describe the path of the meteor against the background stars. ­Because the meteor is an atmospheric phenom­

enon, you and your friend should notice a very large shift of the meteor against the background stars.

24. With an origin of “only” 1 billion years, the meteorite did not come from a primitive Solar System object like an asteroid or a comet, nor did it come from a world like Mercury or the Moon that was geologically inactive 1 billion years ago. Instead, the meteorite prob­ably had to come from Mars, from a surface that had its “clock” reset by melting.

25. The Kuiper ­Belt is a wide disk of cometary nuclei orbit­ ing the Sun beyond Neptune’s and Pluto’s orbits out to about 50 AU, whereas the Oort cloud is s­ pherical and may extend out to a ­couple of light years from the Sun. Also, the Kuiper ­Belt is very close to the Sun com­ pared with the Oort cloud. As a result, Kuiper B ­ elt comets w ­ ill orbit roughly in the disk of the Solar Sys­ tem and w ­ ill have short periods (hundreds of years), whereas Oort cloud comets w ­ ill orbit randomly and ­w ill have very long periods (many thousands of years).

The ultimate fate of the Kuiper ­Belt comets may be to eventually “burn out” a­ fter too many close encounters with the Sun. The ultimate fate of Oort cloud comets that fall into the inner Solar System may be to end up having an unbound orbit.

26. Comets contain a nucleus (head), a coma, an ion tail, and a dust tail. The nucleus is the a­ ctual “dirty snow­ ball” source of the material that forms the coma and the tails. It is the smallest part of the comet, usually only a few dozen kilo­meters in dia­meter. It also contains almost all of the mass of the comet. The coma is the bright, fuzzy ball of gas that surrounds the nucleus, typi­ cally reaching a million kilo­meters in dia­meter. The tails are the largest and most impressive parts of a comet.

27. ­There is very ­little mass in the tail of a comet. What ­little cyanide Earth absorbed as it passed through the comet’s tail was utterly dwarfed by the amount of cya­ nide that already exists at Earth’s surface.

28. Comets have a dust and an ion tail. Ion tails are pushed directly away from the comet by the charged particles of the solar wind, whereas the dust tails are pushed away both by solar wind and solar radiation. The two tails need not point in the same direction b­ ecause the shape of the dust tail depends on the particle makeup of the dust and how it is accelerated away from the comet by the Sun’s light. The dust ends up orbiting the Sun just as the comet did.

29. Zodiacal light is sunlight that is scattering from a disk of dust that is a mixture of cometary debris and groundup asteroid material that permeates the inner Solar System. The source has long been debated. One thought is that the particles are being generated by col­ lisions among the asteroids.

Chapter 12

Dwarf Planets and Small Solar System Bodies ◆

93

Review: Although space is mostly empty and the dis­ tances between the stars are vast, we see that weakly bound objects, such as comet nuclei, could still fill much of the interstellar space.

34. Setup: Pluto’s semimajor axis is 39.5 AU and its eccen­ tricity is 0.249. For Eris, t­hese mea­s ure­ments are 67.7 AU and 0.441, respectively. The perihelion and Applying the Concepts aphelion are given by (1 ± e) A, where e is the eccen­ M tricity and A is the semimajor axis. where 31. Setup: Density is calculated as ρ = V Solve: For Pluto, the perihelion and aphelion are 29.7 4 3 AU and 49.3 AU, respectively. For Eris, they are 37.8 V = π R . Escape velocity is calculated as 3 AU and 97.6 AU, respectively.

M Review: Note that the more eccentric the orbit, the vesc = 2G .

R larger the change in distances between nearest and closest approach. Pluto’s perihelion distance is actu­ Solve: (a) Using the numbers given in the text ally smaller than the semimajor axis of Neptune’s orbit for the comet, a mass of 1013 kg and a radius of (30 AU).

2,000 m, we find for the density of the comet, 1 2 KE = mv , 35.

Setup: The kinetic energy of an object is, 13 10 kg 2 ρ= 4 = 298 kg/m3 . The density of expressed in Joules. A 1-­megaton hydrogen bomb has π (2000 m)3 a yield of 4.184 × 1015 J. 3 3 ­water is 1,000 kg/m , so the comet is about 1/3 the Solve: For a mass of 4.6 × 1011 kg and speed of density of w ­ ater.

1 2 1 11 (1013 kg) M 40,000 kinetic −11 the (b) The escape velocity is, vesc = 2G = (2)(6.67 × 10m/s, m 3 kg −1 s −2 ) energy is KE = 2 mv = 2 (4.6 × 10 (2000 m) R1 1 13 KE13=kg)mv 2 = (4.6 × 1011 kg)(40,000 m)2 = 3.7 × 1020 J. This is the (10 M (10 kg) M −11 3 −1 −2 −11 3 ×−110−2 m kg s = 2G =× 10 (2)(6.67 ) 0.82 m/s.

2 2 vesc = 2G vesc == (2)(6.67 m kg s ) equivalent of about 88,000 hydrogen bombs. Note R (2000 m) (2000 m) R that this does not depend on what is hit, just on what is = 0.82 m/s. = 0.82 m/s. ­doing the hitting.

Review: We expect the comet to be much less Review: We clearly see how much energy is contained dense than w ­ ater b­ ecause it is mostly made of ices within a fast-­moving object, and why the impacts of and dust grains. We also expect the escape velocity even smallish objects with Earth are of high concern. to be very low b­ ecause the comet has such a low 36. Setup: Use the small-­angle formula, s = dθ, where s is density. the physical size of the object, d is its distance, and θ is 32. Setup: The rotation speed is (circumference)/(period) its angular size in radians. d Solve: Using the angular size given by the small-­angle = π , where, d is the dia­meter. formula, θ = s/d, we find that Charon appears P θ = 1,186/19,600 = 0.06 radians, or 3.5°.

Solve: Rotational speed is equal to the circumfer­ (a) Charon appears about 7 times larger in Pluto’s sky ence of the object divided by its period, or than our Moon. π d 3.14 × 975 km vrot = = = 340 km/hr. (b) Charon and Pluto share a locked orbit, so Charon P 9 hr ­w ill always appear in the same place to a Pluto observer Review: This is about 210 mi/hr or the speed of a small who lives on the side facing Charon. Charon would aircraft. never be vis­i­ble to someone living on the side of Pluto 33. Setup: Figure 12.11 shows a box 200,000 AU on a side, that ­faces away from its moon. with the Oort cloud filling up most of it. The figure Review: Note that Charon’s proximity means that it ­will notes that Proxima Centauri is about 270,000 AU have a much stronger tidal effect on Pluto than the Moon away. does on Earth. This explains the tidal lock in part (b).

Solve: If we assume the Oort cloud extends all the way 37. Setup: The probability of being struck by a meteorite to the side of the box, then it is about 100,000 AU in in your lifetime is 100,000 AU radius. This is = 0.37, or about 37 ­percent ⎛ A you ⎞ 270,000 AU Phit = ⎜ ⎟ × Rate × Time.

A ⎝ ⎠ Earth of the way to the nearest star.

30. Comets and asteroids rich in complex organic material might have brought the chemical seeds of life to Earth in the form of amino acids, or the impacts might have created enough energy to cause chemical reactions that formed self-­replicating proteins in our “primordial soup.”

94 ◆ Chapter 12 Dwarf Planets and Small Solar System Bodies Solve: Substituting appropriate numbers into the

period of an object at this distance is one third that of Jupiter. We can use Kepler’s third law to find the dis­ ⎛ 0.25 m 2 ⎞ ⎛ 365.25 ⎞ P2 = A 3. dayswith tance × (800/day) × equation yields, Phit = ⎜ −8 14 2⎟ ⎜⎝ ⎟⎠ × (100 yr) = 1.5 × 10 . ⎝ 5×10 m ⎠ 1 yr 11.86 yr = 3.95 yr; A = P 2/3 = 2.5 AU Solve: PKirk = 3 ⎞ ⎛ ⎞ × (800/day) × 365.25 days × (100 yr) = 1.5 × 10 −8. This is a Review: A quick Internet search confirms that the gap 2⎟ ⎜ ⎟⎠ ⎠ 1 yr ⎝ is indeed at 2.5 AU. probability of 0.0000015 ­percent.

40. Setup: ­Because the Sun is the primary mass for all Review: This is approximately the same odds as win­ three comets, we can use the s­ imple version of Kepler’s ning the lottery. Calculations are based on ­people third law to compute their semimajor axes P 2 = A 3, being equally spaced across the globe. Have students where P is in years and A is in AU. If we assume that think about ­whether the largest cities in the world that the perihelion distances for all three comets approach have populations exceeding 24 million might be zero, then the aphelion distances ­w ill be approximately increasing their probability of being hit. Review logi­ twice the semimajor axes. Table IM 12.1 summarizes cal fallacies in Chapter 1. the results.

38. Setup: (a) Newton’s form of Kepler’s third law allows Solve: (a) Using Kepler’s third law for Comet Encke, us to compute the mass of Electra. For part (b), we the period is 3.3 years, yielding a semimajor axis of 2.2 need to compute volume and density. The dia­meter is AU. For Comet Halley, the period is 76 years yielding, given as 182 km for a radius of 91 km or 91,000 m. a semimajor axis of 17.9 AU. For Comet Hale-­Bopp, the Solve: (a) Using the equation for Newton’s form of period is 2,530 years yielding, a semimajor axis of Kepler’s third law to solve for the mass, 185.7 AU. (b) Using Working It Out 12.1 for the perihelion and 4π 2 A3 MElectra = aphelion distances, they are 0.34 AU and 4.1 AU for G P2 Comet Encke, 0.59 AU and 35.3 AU for Comet Halley, 4π 2 and 0.93 AU and 370.4 AU for Comet Hale-­Bopp. = (6.67 × 10 −11 m 3 kg −1s −1 ) (c) Encke prob­ably came from the Kuiper B ­ elt but suf­ (1.35 × 106 m)3 fered an orbital change. As a short-­period comet, × Halley prob­ably came from the Kuiper B ­ elt, while (3.92 days × 86,400 s/day)2 B opp came from the inner part of the Oort Cloud.

Hale-­ = 1.27 × 1019 kg. (d) Comet Hale-­Bopp should be the most pristine ­because its long orbital period means it has made (b) The volume of Electra ­w ill be fewer passes around the Sun and lost less of its volatile 4 4 V = π R 3 = π × (9.1 × 10 4 m)3 = 3.16 × 1015 m3. inventory. Comet Encke should be the least pristine 3 3 ­because its frequent trips around the Sun rapidly To compute the d­ ensity, ρ, we substitute the results deplete its volatile inventory. into the standard relationship for mass and density Review: To check our distances, consider Figure 12.12, M 1.27 × 1019 kg which shows many comets relative to the Solar System.

3 ρ= = = 4,020 kg/m .

V 3.16 × 1015 m3 Encke is about at two-­thirds the distance of Jupiter Review: The density is about double what a typical rock (about 4 AU), Halley is past Neptune (about 30 AU), would have. We can conclude that Electra contains a sig­ and Hale-­Bopp is far beyond Pluto. nificant amount of metal as part of its composition.

41. Setup: From 240 BCE to its most recent appearance 39. Setup: The Kirkwood Gap is in a 3:1 resonance with in 1986 is a total of about 2,230 years.

Jupiter, meaning asteroids in this gap complete three Solve: (a) 2,230 years divided by the period of orbits during one orbit of Jupiter. In other words, the 76.4 years yields 29 orbits.

Table IM 12.1 Comet Encke Halley Hale-­Bopp

Period (yr) 3.3 76 2,530

A (AU) 2.2 17.9 185.7

Data source: Larry Marschall and Christy Zuidema, Gettysburg College

e 0.847 0.967 0.995

perihelion (AU) 0.34 0.59 0.93

aphelion (AU) 4.1 35.3 370.4

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(b) 29 orbits times 3 × 1011 kg lost per orbit is about 9 × 1012 kg lost.

9 × 1012 kg = 0.04, or (c) This represents about 2.2 × 1014 kg 4 ­percent of its total mass.

Review: If Comet Halley lost a huge amount of its mass at each passage, then we should expect to have seen con­ siderable changes in the comet’s brightness over the past few millennia. Furthermore, the comet would be extremely short-­lived, which would raise the question “Why are we able to see it at all?” The fact that it has been losing only a small fraction of its mass during this time frame is consistent with our observations.

42. Setup: We solve this prob­lem by using our equation for M density: ρ = .

V Solve: (a) Using the numbers given in the text and not­ ing that the radius of the comet is 5 km or 5,000 m, we 2.2 × 1014 kg M ρ = = = 420 kg/m3. The den­ find 4 V 3 π (5000 m) 3 sity of w ­ ater is 1,000 kg/m 3, so the comet is about 0.42 the density of w ­ ater.

Review: We expect the comet to be much less dense than ­water ­because it is mostly made of ices and dust.

43. Setup: Just like mass density, number density N n = , where N is the total number of particles pre­ V sent, and n is the number of particles per unit volume.

Solve: At 200 molecules per cubic centimeter, the comet tail would need a volume of 1019 particles N V= = = 5 × 1016 cm 3 to con­ n 200 particles/cm 3 tain 1019 molecules.

Review: We expect this volume to be huge given how few molecules ­t here are per volume ele­ment in the comet’s tail. Indeed, this volume is equivalent to a cube that is 3.7 × 105 centimeters (or 3.7 kilo­ meters) on a side. A quick Internet search gives the volume of Mount Everest (base camp to summit) of 2,640,000 cubic yards or 2 × 1012 cm 3. The volume of the comet’s tail would hold 25,000 mountains like Mount Everest.

44. Setup: The energy from an impact is kinetic, for which 1 we use KE = mv 2 . We convert the velocity to 2 meters, which gives us 5,000 m/s.

Solve: Using the mass and speeds given, 1 1 KE = mv 2 = (4.6 × 1011 kg)(5,000 m/s)2 2 2 = 5.8 × 1018 J.

Dwarf Planets and Small Solar System Bodies ◆ 95

Review: For comparison, this is about 1,374 H-­bombs.

45. Setup: The removal rate is (total mass removed)/(time to remove it). If the total amount of dust remains approximately constant, this represents equilibrium between the dust removed from the system and the dust being produced by comets and asteroid collisions.

Solve: (a) The removal rate would be 1 yr 1016 kg × = 10,550 kg/s, and if the 30,000 yr 3.16 × 10 7 s zodiacal dust remains constant, this would have to be the average production rate as well. (b) In dynamic equilibrium, the rate of replacement equals that of removal.

Review: Dust is removed and replaced at a rate of approximately 10,550 kg e­ very second, or about 12 tons a second. This is about the same weight as an anchor of a cruise ship, two adult elephants, or two tyrannosaurus rex dinosaurs.

Using the Web 46. Dwarf planets: a. As of June 21, 2018, ­t here ­were 10 objects that are nearly certainly dwarf planets, 26 objects that are highly likely to be dwarf planets, 64 objects that are likely to be dwarf planets, 127 objects that are prob­ably dwarf planets, and 661 objects that are possibly dwarf planets. It is difficult officially to certify an object as a dwarf planet b­ ecause it must be round. It is hard to figure out the size of an object to determine if it has enough mass to force it into a sphere. b. What has been learned from the New Horizons mission? So much was learned about Pluto that the answers vary. The press kit has a very good sum­ mary in bullet points: http://­pluto​.­jhuapl​.­edu​/­News​ -­Center​/­R esources​/­P ress​-­K its​/­N HPlutoFly​by​ PressKitJuly2015​.­pdf 47. Citizen science proj­ects: Directs students to “Back­ yard Worlds: Planet 9” where they w ­ ill read the sections and classify some of the images. Answers w ­ ill vary.

48. Involves ­going to NASA’s “Asteroid Watch” website and report on what is new. Answers ­w ill vary.

49. Students visit the “Space Weather” website and report on any comets that are viewable, any close encounters coming up, and information on any NEOs. Answers ­w ill vary.

50. “Do a search for the Japa­nese Hayabusa2 and the NASA OSIRIS-­R Ex missions, each of which is sched­ uled to arrive at an asteroid in 2018 and bring back a sample to Earth a­ fter making observations.” The land­ ing of the Hayabusa2 spacecraft on the asteroid 162173 Ryugu was successful, and as of Septem­ ber 2018, the four small rovers ­were in good health

96 ◆ Chapter 12 Dwarf Planets and Small Solar System Bodies and commencing the survey of the asteroid. Discover­ ies and the return of any samples are yet to come. On October 1, 2018, NASA’s OSIRIS-­R Ex spacecraft exe­ cuted its first Asteroid Approach Maneuver (AAM1), which put it on course for its arrival at the asteroid 101955 Bennu in December, 2018. As of early 2019, the spacecraft is in orbit around Bennu. Return of samples is expected in September, 2023.

EXPLORATION This exploration has students try to identify the location of a passing asteroid by comparing 4 images. This could be FIGURE IM 12.1

done by making a flip book, as suggested in the text, or by subtracting one image from another. One could also copy the Figure 12.26 on a transparency, separate the frames, and stack them.

EXPLORATION SOLUTIONS 1. The following image shows the four images with the location of the asteroid marked with a circle in each frame.

2. Early h­ umans eating on the savannah had very l­ittle to defend themselves from prey other than their wits. Even now, if we detect subtle movements, then

Chapter 12

we are more likely to see approaching danger and escape before it is too late.

3. As noted in the text, a more sophisticated method is to subtract two images, which gets rid of every­thing that is constant and only shows what has changed. Let’s say we chose the first image as our reference. In t­ hese nega­ tive images, we would expect a white background with dark spots showing where the asteroid was in the first and second, then third, and fi­nally fourth images.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 16: CLASSIFYING METEORITES (Note: Th ­ ere are color images of the meteorites available from the online resources for instructors at the W. W. Nor­

Dwarf Planets and Small Solar System Bodies ◆ 97

ton website). Students learn about dif­fer­ent types of mete­ orites and their origins. As they work through the activity, they: • examine images of samples of meteorites and note dominant features and pos­si­ble classification.

• distinguish among iron, stony-­iron, and stony meteorites.

• classify “unknown” meteorites using supportive explanations.

• compare spectra of an asteroid and a meteorite and evaluate the evidence for the origin of the meteorite.

Text reference: Section 12.4

CHAPTER 13

Taking the Mea­sure of Stars

INSTRUCTOR NOTES Chapter 13 describes how we mea­sure most stellar properties. Major topics include: • parallax and spectroscopic parallax.

• working with degrees, arcminutes, and arcseconds, and the introduction of the parsec.

• the relationship between luminosity, brightness, distance, and magnitudes.

• determining the temperature of a star using Wien’s law and stellar spectral types.

• the composition of stars.

• using the Stefan-­Boltzmann law to find luminosity, radius, or temperature.

• binary stars and how they are used to determine stellar masses and size.

• the Hertzsprung-­Russell (H-­R) diagram.

Some of this material is quite intuitive, such as the inverse square law of brightness, as every­one knows that lights appear dimmer when they are farther away. On the other hand, the relationships of color and temperature or among luminosity, temperature, and size can be much more problematic. Students often associate that “more” always requires “bigger,” so it can be quite counterintuitive that a hotter object is bluer (meaning a shorter wavelength) or that a cooler object can be more luminous than a hotter one if the sizes are correct.

DISCUSSION POINTS • Students w ­ ill learn in Chapter 21 that the Milky Way is about 100,000 light years, or 31,000 parsecs, in dia­ meter. Given that we can reliably mea­sure parallax to 0.001 arcsec (out to 1,000 parsecs), have students discuss what this means for mea­sur­ing distances to most objects in the Milky Way. What does this say about mea­sur­ing distances in the rest of the universe? (LG 1) • Combine Figure 13.5 (distribution of the luminosities of stars) and ­Table 13.2 (Properties of Main-­Sequence Stars) for a discussion about how the Sun fits in in terms of its size, temperature, mass, and luminosity. (LG 2) 98

• Follow up on the previous discussion by having students work in teams to prepare to teach the rest of the class about how astronomers obtain the temperatures, sizes, composition, or masses of stars. ­These reports could then be used again when they reach Chapter 14, on the Sun, for comparisons of the methods and techniques. (LG 2, LG 3) • Discuss the history of the stellar classification system, including its original incarnation and its rearrangement once temperature was realized to be the impor­ tant quantity. How is this an example of the scientific method at work? (LG 4) • Discuss the utility and ambiguity of the spectral classification of stars. Have students look for information about the new spectral classes defined in the twenty-­ first c­ entury, such as the L, T, and Y dwarfs. Discuss their spectral properties. (LG 4) • Figure 13.18 graphically shows the relationships between the mass of a star and its luminosity, radius, and temperature. Give students a 3-­m inute writing exercise that has them explain why t­ hese relationships exist. (LG 5) • This chapter’s opening figure asks the question, “Why do stars have dif­fer­ent colors?” How many students recognize what constellation or set of stars (asterism) is being shown in this image? Only the star Betelgeuse is noticeably dif­fer­ent in color, being definitely orangish versus blue-­white. The red color that is shown is not a star at all. Rigel is partially hidden by the question inlay. Why does it appear so much brighter than the other stars in the image?

PROCESS OF SCIENCE The Pro­cess of Science emphasizes that a gain in understanding what stellar data mean physically may take a ­century or more. It covers the classification work of Annie Jump Cannon for over 400,000 photo­g raphs of stellar spectra, through development of the H-­R diagram, and fi­nally the work of Cecilia Payne-­Gaposchkin and Maghnad Saha, which led to our quantum understanding of what the spectra mean and the basis of the classifications.

Chapter 13

ASTROTOUR ANIMATIONS STELLAR SPECTRUM This interactive animation starts with what characteristics astronomers can determine from the light from a star: temperature and size. The colors of stars corresponding to their temperatures is related to Wien’s law and the peak wavelength of the light emitted. This is followed by the use of blue and visual (yellow) filters and the ratio of the brightness through ­these two filters to compare stars. The Stefan-­ Boltzmann law for determining radii is explained. Spectra of stars containing e­ ither absorption lines or emission lines are shown with the Planck curve for the star. This animation also explores how the spectra emitted by dif­fer­ent atoms can be used to determine the chemical composition of a star, and students find out that a spectrum indicates that the example star contains hydrogen, helium, and calcium.

Text reference: Section 13.2 H-­R DIAGRAM In this animation, students move a position marker across an interactive H-­R diagram. An offset win­dow shows an example of a typical star that would be found in that place, with basic information about luminosity, surface temperature (and color), and the life stage of the star. The animation references modules that students needed to view before viewing this one, such as the one that covers what can be determined from a stellar spectrum. The animation covers the in­de­pen­dent work of Hertzsprung and Russell that led to the development of this diagram.

Text reference: Section 13.4

ASTRONOMY IN ACTION VIDEOS PARALLAX Stacy Palen starts the explanation of parallax with the example of her right thumb held up close to her face and then at arm’s length and the differences in the apparent “jump” of her thumb. She then moves to the example where the baseline goes from her two eyes to two cameras set up in the lecture hall. The emphasis at the end is the use of 1 astronomical unit (AU) to get the best pos­si­ble baseline for mea­sure­ments of the parallaxes of stars. (Length: 2:02) Text reference: Section 13.1

INTERACTIVE SIMULATIONS H-­R EXPLORER This simulation sets up the H-­R diagram with luminosity expressed in terms of the luminosity of the Sun and loga-

Taking the Mea­sure of Stars

◆ 99

rithmic temperature values from 40,000 to 2,500 Kelvin.

Students ­w ill be able to manipulate the temperature, luminosity, and radius of a star using sliders and see the effects of t­ hose changes in quantities on the location of the star on the H-­R diagram and its characteristics when compared to the Sun. Text reference: Exploration

TEACHING READING ASTRONOMY NEWS 1. One parsec (pc) is 3.26 ly, so 7,500 ly is about 2,300 pc, and 10,000 ly is about 3,070 pc.

2. Since d = 1/p, a star 7,500 ly away has a parallax of 1/2,300 pc = 0.0004 arcsec. A star at 10,000 ly has a parallax of 0.0003 arcsec.

3. Exosures are taken 6 months apart to provide the largest baseline; see Figure 13.1.

4. From space, images of stars are not subject to the atmospheric turbulence on Earth; thus, their positions can be recorded much more accurately.

5. Mea­sure­ments of farther distances are always calibrated using nearer methods; the best calibration is trigonometric parallax, which provides the strongest foundation on which all other methods are based. However, spectroscopic parallax is fairly uncertain ­because all stars of a given spectral type are treated as identical, which is not necessarily true. Thus, the improvements h­ ere w ­ ill be ­limited for the par­tic­u­lar case of spectroscopic parallax.

CHECK YOUR UNDERSTANDING SOLUTIONS 13.1. (b) Brightness follows an inverse square law. For a star to be twice as far away but appear equally bright, the star has to be 22 = 4 times more luminous.

13.2. (d) Using the Working It Out 13.3, L ∝ R 2T4 so if L is the same but T is doubled then R must be quartered.

13.3. (a, b, d) Period and distance give the sum of the two masses, whereas velocities allow the individual masses to be determined.

13.4. (b, d) Stars at the lower right of the H-­R diagram are cool, low luminosity, small, and if they are main sequence stars, they have low mass as well.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (b) ­Because the two stars are at the same distance, the only reason one would appear brighter is if it ­were more

100 ◆ Chapter 13 Taking the Mea­sure of Stars luminous. One star is twice as bright, so it must be twice as luminous.

2. (a) ­Because a star acts like a blackbody, we know that a bluer intrinsic color means the star’s surface is hotter.

3. (c) ­Because the stars are at the same distance and appear the same brightness, they must have the same luminosity. The fact that one is red means it is cooler; therefore, it must be larger to have the same luminosity as the blue star.

4. (a) For a given line to appear, the atoms must be pre­ sent and in the right energetic configuration, which is highly dependent on temperature.

5. (b) Star A is moving three times faster than star B; therefore, it must be less massive.

6. (b) If stars are at the same distance, then brightness indicates luminosity. ­Because star A is brighter than B but at the same temperature, it must be larger; that is, A is a g­ iant, and B is a dwarf or main sequence star.

7. (d) If stars are at the same distance, then brightness indicates luminosity. ­Because star A is brighter than B but the same color (same temperature), then it must be larger. Star A cannot be a main-­sequence star for that to be true, ruling out (b). We can rule out ­either star being a red ­g iant ­because of the blue color. The star B being a white dwarf and A being a blue g­ iant works. The “many times brighter” statement, however, should be “many million times brighter.”

8. (a) See Figure 13.20. From the text: “Main-­sequence stars more massive than the Sun are hotter and have larger habitable zones.”

9. (a, c, e) ­Because both stars are main sequence, more massive implies hotter, larger, and more luminous.

10. (d) ­Because parallax depends on the distance of the planet from the Sun, a larger orbital distance would make the parsec longer and would allow us to see parallax motion of more distant stars.

11. (a) More distant stars have smaller parallax ­angles.

12. (a) Brightness b changes as L/d2 , so the fact that star A appears twice as bright but is twice as far away means that L ∝ b × d2 = 2 × 22 = 8, using proportional reasoning.

13. (b) Helium has four times the mass of hydrogen.

14. (b) Capella ­w ill be made of yellow (G-­t ype) and red (M-­t ype) stars, and ­because both pairs are close to ­ ill dominate. each other, the brighter G-­t ype stars w Therefore, the color ­w ill appear yellow.

15. (a) See Figure 13.13. Star A dominates the light from this system.

Thinking about the Concepts 16. As parallax ­angles become smaller, they approach the precision limits of our instruments. For example, consider an instrument that can mea­sure any parallax with

a precision of ± 1 milliarcsecond. For an object with a 50 milliarcsecond parallax, the uncertainty is only 1 milliarcsecond out of 50 milliarcseconds, or 2 ­percent.

On the other hand, an object with a 2 milliarcsecond parallax has an uncertainty of 1 milliarcsecond out of 2 milliarcseconds, or 50 ­percent. Therefore, due to the ­l imited precision of our instruments, uncertainties in distance grow as parallax ­angles get smaller.

17. The masses of stars are determined accurately only for binary stars. Finding ­these masses depends on distance ­because we need to know the size of the semimajor axis A. Finding the size of a star would not need any distance determination if it is part of an eclipsing binary. If not, the luminosity needs to be found in order to use the Stefan-­Boltzmann law, and that depends on distance. Temperature, color, spectral type, and composition are all distance-­independent quantities that can be determined from a star’s spectrum. ­Because stellar spectra show the same lines ­whether viewed from up close or from halfway across the galaxy, the mea­sured properties of a star derived from its spectrum are not influenced by the star’s distance.

18. (a) The color of a star is a direct mea­sure of its temperature. Blue stars have hot surface temperatures, and red stars have cooler surface temperatures. ­Because the golden star lies closer to the redder side of the spectrum than the sapphire blue star, we conclude that the sapphire blue star has the higher temperature of the pair. (b) The obvious answer would that blue stars emit more light per unit area than yellow stars. Therefore, if the blue star w ­ ere the same size as or larger than the golden star, it would be the brighter star in the pair. ­Because our observations demonstrate that the golden star is actually the brighter of the two, we must conclude that it has a much larger surface area than the blue star to appear more luminous. However, it has not been shown that ­these two stars are, in fact, at the same distance from us, and may be at vastly dif­fer­ent distances. Plus, the “golden” star is actually a double star, consisting of a K3 ­g iant and a B9 main sequence star.

Thus, the system is more complicated than just being of dif­fer­ent colors.

19. ­Every blackbody emits light at all wavelengths; the question is just how much light is emitted. A star with a 2500 K blackbody emits mostly in the red and infrared but still gives off a mea­sur­able amount of blue light.

20. Betelgeuse is cooler than Rigel. Nothing can be said about their luminosity or size without knowing their distance.

21. The original classification of stars was based on the strength of hydrogen absorption lines in their spectra, with A-­class stars exhibiting the strongest absorption.

Chapter 13

Hydrogen absorption lines are strongest at temperatures near 10,000 K, but stellar surface temperatures span the range from 3000 K to 30,000 K. At surface temperatures well below 10,000 K (F, G, K, and M stars), much of the hydrogen in a star’s upper atmosphere remains in an unexcited state, so fewer electron transitions are observed in the stellar spectrum.

At temperatures well above 10,000 K (O and B stars), most of the hydrogen in a star’s upper atmosphere is ionized, so no electrons remain in the hydrogen to produce spectral lines. When spectral classes are arranged as a temperature sequence, the O and B spectral classes move ahead of the A-­class stars despite their weaker hydrogen lines. As a final note, the missing letters in the OBAFGKM sequence are a result of merging the original 22 spectral classes into the seven larger spectral classes that we have ­today.

22. We can directly mea­sure stellar masses only by using Newton’s version of Kepler’s third law, which requires an observation of the orbit of a companion. To accurately mea­sure the semimajor axis of the companion’s orbit, we need to know how its orbit is inclined to our line of sight. Only eclipsing or visual binaries allow us to mea­sure both the orbital period of a companion star and its semimajor axis directly. Note that we can use Kepler’s third law with spectroscopic binaries as well; however, b­ ecause we do not know the orbital inclination of the companion star, we can determine only a minimum mass for the system, not the exact mass. For stars not in a binary system, we would need to ascertain ­whether it is a main-­sequence star. If it is, then we can use its spectral type and estimate its mass based on known masses of other stars having the same spectral type.

23. Stars are all governed by the same laws of physics, and their evolution depends almost exclusively on their mass (which controls their entire interior characteristics). We assume that if two stars have similar composition and mass, they w ­ ill act and evolve the same way ­because ­there is no physical mechanism to change this.

24. Binary stars can transit each other just like planets, so any mission that is looking for regular time variability in the light from the stars is ­going to discover binary stars (and pulsating variables as well).

25. Answers ­w ill vary. Scientific collaboration requires the open sharing of data, division of ­labor, participation of many dif­fer­ent ­people in the analy­sis and interpretation, and corroboration of findings through many dif­ fer­ent ­people performing the same work.

26. As the distance from the Sun to the planet increases, 1 arc second corresponds to a farther distance, so if we move to Mars we ­w ill be able to mea­sure the distances to even farther stars (about 1.5 times as far using the

Taking the Mea­sure of Stars

◆ 101

same methods). Jupiter would open up the sky to even more distant stars (about 5 times farther away). If we ­were on Venus, we would be able to mea­sure parallaxes only for stars closer than about 70 ­percent of the distance we currently mea­sure.

27. Hydrogen absorption is weak in O stars b­ ecause most of the star’s atmospheric hydrogen is ionized; that is, ­there are no electrons bound to the nucleus to absorb photons. In G stars, the hydrogen is neutral but also sufficiently cool that most of its electrons are in the ground state for which only UV photons can be absorbed to excite electrons to the second energy level.

A-­t ype stars have the right temperature to have many hydrogen atoms, both neutral and in the first excited state (energy level 2) ready to absorb optical photons.

28. Edge-on observations are best for eclipsing and spectroscopic binaries ­because ­there is no inclination effect to change. Face-on observations are best for visual binaries.

29. (a) Star 2 has a smaller orbit ­because it has a higher mass; therefore it moves less about the center of mass. (b) A sketch showing the case in which star 1 is a very low, planetary-­mass companion would have star 2 in an even smaller orbit (center of mass would be closer to star 2), and the companion in a larger orbit. (c) For stars having equal mass in a binary system, the orbits ­will be circular with the center of mass at the center, and each star located opposite the other as the orbit.

30. If the Sun w ­ ere a blue, main-sequence star, its luminosity would be thousands or even tens of thousands times higher than it is now. Figure 13.20 shows that Earth would be inside the habitable zone in the direction of the star (scorching). If the Sun ­were a red, main sequence star, its luminosity would be hundreds or even thousands of times lower than it is now. Figure 13.20 shows the orbit of Earth would lie within the habitable zone of cooler stars. An exact answer would depend on the mass of the main-­sequence star.

Applying the Concepts 31. Setup: Parallax ­a ngle scales inversely with distance: p = 1/d, so if a star is twice as far away as another star, its parallax a­ ngle is half of that of the other star.

Solve: If a star C w ­ ere located four times farther than star A, it would appear to move one-­fourth as much each year than star A does. ­Because it is only two times farther than star B, it would move one-­half as much than B. Review: Students can try this directly using the method shown in Figure 13.1a, by changing the farthest distance of the two pencils they are holding.

32. Setup: Parallax ­a ngle scales inversely with distance: p = 1/d, so if a star is twice as far away as another star, its parallax a­ ngle is half of that of the other star. Note

102 ◆ Chapter 13 Taking the Mea­sure of Stars that one-­third of a degree is two-­thirds as small as one-­ 36. Setup: Parallax uses the parallax formula d = 1/p, half of a degree. where d is in parsecs and p is in arcseconds. To convert Solve: An object that moves two-­thirds as much w ­ ill parsecs to light-­years, recall that 1 parsec = 3.26 ly.

Solve: The distance to Sirius is d = 1/0.379” = 2.64 pc 1 3 be 2 = 2 or 1.5 times farther away.

3.26 ly = 8.6 ly. It takes light 8.6 years to or 2.64 pc × 3 1 pc Review: This object is 1.5 times the distance. Does reach Earth. this make sense? If the parallax a­ ngle in this case w ­ ere Review: Remember that a parallax of 1 arcsecond one-­fourth a degree, it would be at twice the distance. equals a distance of 1 parsec. Sirius has a distance close One-­third is between one-­half and one-­fourth, so yes, to one-­third of an arcsecond, so its distance should be this makes sense. about 3 parsecs, as we found.

33. Setup: Quite often in astronomy, graphs need to cover 37.

Setup: Remember that luminosity L ∝ R 2 T4, and that a huge range of par­a meters, which is easier to visualif two stars are the same spectral type, then they have ize using logarithms as they spread out the w ­ hole range the same temperature. of values so that one can identify the positions of points Solve: Both stars are the same spectral type (A) and with both small and large values equally well. To find therefore have the same temperature. If the “Pup” the difference in luminosity between the least and most 1 luminous stars, read off the values on the y-­a xis. appears 6,800 times as bright and is the same disSolve: The most luminous star shown in Figure 13.15 tance as the “Dog Star” (which they are ­because they has a luminosity of almost 107 LSun, whereas the least 1 are a binary pair), then the “Pup” is 6,800 times as luminous red dwarfs have luminosities of about 10 −4 luminous. ­Because luminosity L ∝ R 2T4, we can use the LSun. The total range of luminosities is about 1011 (one above information to infer that the “Pup Star” must hundred billion). This range in values would be extremely hard to display in a linear plot. The horizonbe R ∝ L T 4 = 1 6,800 ≈ 1 83 times the size, or tal axis in Figure 13.15 showing surface temperature about 1.2 ­percent the size of the “Dog Star.” is logarithmic, just not with powers of 10.

Review: If two stars at the same distance are the same Review: It is remarkable that stars that have a small temperature and one is very faint, then it must be very range of masses and temperatures can have such a huge small. Why? All blackbodies with a temperature T give range of luminosities; we ­w ill learn why this happens off the same amount of energy per tiny patch of area. in f­ uture chapters.

This means that if stars having the same temperature 34. Setup: This prob­lem again works with Figure 13.15. are part of a binary, and one has a smaller luminosity, Remember that when reading logarithmic axes, each it must have a smaller surface area. “tick mark” is a power of 10.

38. Setup: Based on the calculations shown in working It Solve: (a) On the right, the stars are 107, or ten million Out 13.4. For binary stars, we know the ratio of the times more luminous than the Sun. masses is equal to the inverse ratio of the orbital veloci(b) On the left side of the graph, the stars have 10 −4 ties. We ­w ill use that and Newton’s version of Kepler’s solar luminosities, or are 0.0001 times less luminous third law, solving for the semimajor axis of the orbit, A. than the Sun. v M Review: This graph shows stars spanning 1011 or a Solve: (a) For binary stars, 1 = 2 . If the companM2 v1 range of one hundred billion in luminosity. ion has 2.35 times higher velocity, it has 2.35 times 35. Setup: In Figure 13.17, temperature is on the bottom axis, luminosity is on the left side, mass is indicated lower mass. Its mass is then about equal to 2 2.35 = 0.85 along the main sequence, and size is shown by diagotimes that of the Sun ­because Sirius A has a mass 2 nal lines in the figure panel. times larger than the Sun. Solve: A star 10 times the mass of the Sun is found at (b) Using a total mass of 0.85 + 2.0 = 2.85 solar masses the upper left of the main sequence. It has a luminosand a period of 50 years, we find A = 3 (M1 + M 2 )P 2 = 3 (2.85) × 502 ity of around 10,000 solar luminosities, a temperature = 3 (M + M 2 )P 2 = 3 (2.85) × 502 = 19 AU. around 23,000 K, and a radius around 8Asolar radii.

1 Review: In our Solar System, a period of 50 years Review: Note that this star, although only 10 times as would place a planet between Saturn and Uranus massive as the Sun, is almost 10 times larger, 4 times (about 14 AU). ­Because the Sirius system has more hotter, and 10,000 times more luminous. This is conmass, the period of a given object would be shorter sistent with Working It Out 13.3.

Chapter 13 Taking the Mea­sure of Stars

than if it ­were orbiting the Sun. It therefore makes sense that the companion must be more than 14 AU away to have a period of 50 years.

39. Setup: To solve this prob­lem, we begin by noting that the apparent brightness (b) of a star is proportional to its luminosity (L) and inversely proportional to its disL tance squared (d2) as b ∝ 2 , and we ­w ill solve in d L terms of the distance, d ∝ and use ratios. b Solve: Taking a ratio of the brightness of Polaris to Sirius and solving for distance yields dPolaris = dSirius

LPolaris bSirius × = LSirius bPolaris

2,500LSun × 24 = 49.

25LSun

Polaris is 49 times farther from us than Sirius is.

Knowing from prob­lem 36 that Sirius is 8.6 ly from the Sun, we compute that Polaris must be 421 ly from the Sun (49 × 8.6 ly = 421 ly).

Review: When we talk about “nearby” stars, we are ­really including t­ hose up to many hundreds of light-­ years from Earth.

40. Setup: Parallax uses the parallax formula d = 1/p, where d is in pc and p is in arcseconds.

Solve: Betelgeuse has a parallax of 0.00451 ± ​ 0.00080 arcsec. This corresponds to a distance of 221.73 parsecs. The uncertainty means the mea­sured parallax could be as small as 0.00451 − ​0.00080 = ​ 0.00371 arcsec, or as large as 0.00451 + 0.0080 ​ = 0.00531 arcsec. ­These values yield distances of 270 and 188 parsecs. The uncertainty is therefore half this difference, or about 40 parsecs.

Review: Remember that a parallax of 1 arcsecond equals a distance of 1 parsec. Its distance by the astro48 pc, which is what metric method is stated as 222 +−34 we found h­ ere.

41. Setup: Parallax uses the formula d = 1/p, where d is in pc and p is in arcsec. Luminosity, brightness, and distance are related by L ∝ bd2 .

Solve: Achernar is at a distance of d = 1/0.02339 arcsec = 42.75 pc based on its parallax, or about 140 light years. The distance to Betelgeuse is about 1.6 times that of Achernar, so Betelgeuse must be about 1.592 ≈ 2.5 times more luminous than Achernar to appear the same brightness. Achernar appears bluish and is thus hotter ­under Wien’s law.

Review: As a confirmation, use Wien’s law. Betelgeuse’s surface temperature is 3590 K; Achernar’s, ~15,000 K. Peak wavelength for Betelgeuse ­u nder 2.9 × 106 nm ⋅ K = 808 nm , in the Wien’s law: λ = 3590 K

◆ 103

near infrared. Peak wavelength for Achernar ­u nder 2.9 × 106 nm ⋅ K Wien’s law: λ = = 193 nm , in the 15,000 K ultraviolet.

42. Setup: We know that the brightness of an object decreases as the square of the distance according to L b ∝ 2 . We can form a ratio of the brightness of the d Sun to the brightness of a distant solar-­mass star, b yielding a distance dstar = dSun × Sun . bstar Solve: (a) Using the numbers in the prob­lem, we find, bSun = 1 AU × 1.6 × 1013 bstar

dstar = dSun ×

= 4 × 106 AU. (b) Converting this distance to light-­years yields, 1.496 × 108 km 1 ly × 1 AU 9.46 × 1012 km = 63 ly

4 × 106 AU ×

.

Review: The limiting apparent magnitude for naked-­ eye observations at a very dark site is about 6th magnitude. The star HD 130948 is a G1V star at 58.5 ± 0.9 ly away. Its apparent V magnitude is 5.99, making it a good candidate for supporting our calculations ­here.

43. Setup: Figure 13.9 suggests that the center of mass is moved t­ oward the heavier mass by the f­ actor by which it is greater than the lighter one.

Solve: If the masses are equal, the center of mass ­w ill be at the m ­ iddle of the beam, or directly between the two masses; that is, d1 = d2 . If one mass is twice that of the other, then d2 = 2d1.

Review: It makes sense that the more massive one object is than the other, the center of mass must move closer to the more massive one. The Sun is about 1,000 times as massive as Jupiter. The center of mass for Jupiter and the Sun is just outside the Sun’s surface.

44. Setup: Use Wien’s law.

Solve: At 10,000 K, Wien’s law gives 2.9 × 106 nm ⋅ K 2.9 × 106 nm ⋅ K λ= = = 290 nm.

T 10,000 K This is in the near-­U V. This star ­w ill appear blue-­ white to us.

Review: Although the majority of the energy comes out in the near-­U V, this star is still emitting copious amounts of radiation in the vis­i­ble spectrum, which is why we can see a blackbody having a peak wavelength outside the vis­i­ble range of light.

104 ◆ Chapter 13 Taking the Mea­sure of Stars 45. Setup: Using the relationship between brightness, b; luminosity, L; and distance, d; L = 4πd2b, so, knowing we are 1 AU, or 1.496 × 1011 m, away and b = 1,470 W/ m2 , we can solve for the luminosity of the Sun. Solve: This yields a luminosity of 4.1 × 1026 W. Review: This is very close to the a­ctual value of 3.828 × 1026 W. Using the Web 46. GAIA ­w ill mea­sure positions of all objects brighter than 15th magnitude to an accuracy of 24 microarcseconds, monitoring each of its target stars about 70 times over a five-­year period, thus giving astronomers high-­quality data to determine accurate distances to more stars. Making parallax mea­sure­ments from space is better than mea­sur­ing from the ground on Earth ­because Earth’s atmosphere distorts the images and locations of stars. A search for released data through GAIA news (for example, at http://­sci​.­esa​.­i nt​/­gaia​/­) ­w ill yield a variety of answers.

47. Students visit the “Kepler Eclipsing Binary Cata­log” and see what new observations look like. Answers w ­ ill vary.

48. Students search for a photo­g raph of their favorite constellation. Answers ­w ill vary.

49. Citizen science: Students go to the website for the “Stellar Classification Online Public Exploration” and classify stars. Answers w ­ ill vary.

50. The video about planets in binary star systems can be found at: https://svs.gsfc.nasa.gov/11106. Current theory suggests that it would be extremely hard for planets to form around binary stars, and thus they would be rare. There have been a number of circumbinary planets detected, and perhaps more will be found under research projects currently underway.

EXPLORATION This exploration allows students to gain more understanding of what the H-­R diagram is telling us about stars by letting them manipulate the temperature and luminosity of a star and see the resulting change to its radius. They are also able to compare the closest stars to the brightest stars and draw conclusions.

EXPLORATION SOLUTIONS 1. As the star moved to the left of the H-­R diagram, its radius became smaller.

2. As the star moved to the right of the H-­R diagram, its radius became larger.

3. As the star moved up the H-­R diagram, its radius became larger.

4. As the star moved down the H-­R diagram, its radius became smaller.

5. For a star to move into the white dwarf area of the H-­R diagram, its temperature must be very high and its luminosity must be very low.

6. In general, the nearest stars are cooler than the Sun. 7. In general, the nearest stars are less luminous than the Sun. 8. In general, the brightest stars are hotter than the Sun, but ­there are still many that are cooler.

9. Except for one straggler, all the stars are more luminous than the Sun. 10. The brightest stars all have high luminosities and a range of temperatures. The nearest stars almost all have lower temperatures and luminosities than our Sun. This shows us that the nearest stars are not always the brightest stars, meaning that space is filled with more low-­luminosity than high-­luminosity stars.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 17: SPECTRAL CLASSIFICATIONS OF STARS Students learn to determine fundamental properties of stars from a study of their spectra. The activity involves: • applying Wien’s law to real spectra and estimating the surface temperatures of three stars.

• classifying stars by their surface temperatures.

Text reference: Section 13.2 ACTIVITY 18: FINDING DISTANCES TO STARS USING PARALLAX MEA­SURE­MENTS Students determine a relationship between distance and apparent motion of a nearby object when viewed from two vantage points. They then apply this relationship to the mea­sur­ing of distances to stars. As part of this activity they w ­ ill also: • demonstrate parallax of a nearby object, at dif­fer­ent distances from their eyes, relative to a distant scale.

• derive the relationship between the distance of the nearby object and the sizes of its apparent shifts relative to a distant scale.

• apply this knowledge to the mea­sured parallax a­ ngles of stars.

Text reference: Section 13.1

CHAPTER 14

Our Star—­The Sun

INSTRUCTOR NOTES Chapter 14 is an extensive overview of the Sun. Major topics include: • hydrostatic equilibrium—­the balance between energy generation and energy loss, pressure and gravity.

• energy generation, explained through a detailed description of the proton-­proton chain.

• energy transport through each region of the interior of the Sun and the study of helioseismology.

• probing the solar interior with neutrinos and helioseismology.

• the atmosphere of the Sun, comprising the photosphere, chromosphere, and corona.

• all aspects of solar activity including sunspots, such activity’s relationship to the magnetic field of the Sun, and its effects on Earth.

We highly recommend incorporating the following websites into your lectures on the Sun: Solar Dynamics Observatory (SDO): http://­sdo​.­gsfc​ .­nasa​.­gov Spaceweather​.­com: News and information about the Sun-­Earth environment Solar and Heliospheric Observatory (SOHO): https://­sohowww​.­nascom​.­nasa​.­gov This chapter gives students the opportunity to observe their very own star and learn more about it. In this age of high technology, space stations, and power grids, students tend to take a keen interest in the Sun-­Earth connection.

DISCUSSION POINTS • The balance of forces inside the Sun determine its structure. Have students summarize t­ hese forces and then relate them to the balance of forces on Earth. What are the similarities? Differences? Is the force between each student and the Earth balanced? What would happen if it w ­ asn’t balanced? Are t­ here examples of unbalanced forces for the Sun? (LG 1)

• Work through the details of Working It Out 14.1.

What does this tell us about the ­f uture of life on Earth? We generate energy through chemical burning.

What do we do when the chemical source (e.g., wood, gasoline, coal) starts to run out on a small scale (as in a campfire, car, or electricity)? In the 2007 film Sunshine the Sun dies in 2057. A crew is sent to reignite it using all of the “fissile materials” on Earth. Beyond science fiction, what would it take to give the Sun more hydrogen in its core to prolong the proton-­ proton cycle? (LG 2) • Have students list as many everyday examples of energy transport by conduction, convection, and radiation as they can. Then form teams to make master lists to be shared with the class. Link each method of energy transport to a region or regions of the Sun. (LG 3) • Have students create their own models of the regions of the interior and atmosphere of the Sun based on how they interpret the relative importance of each region to the Sun. Would their models change if they ­were to draw the relative importance of each region to the Earth? (LG 3) • Discuss how the solar neutrino prob­lem exemplifies the scientific method. At which step ­were discrepancies found that needed to be investigated? What pos­ ere considered, and sibilities for the discrepancies w why was one discarded? Did the testability of a hypothesis enter into the scientists’ decision? (LG 4) • An active field of research is the investigation ­ hether variations in solar activity may into w induce changes in Earth’s weather. A pos­si­ble link between t­ hese two phenomena is the so-­c alled ­L ittle Ice Age that took place during the Maunder Minimum. Discuss ­w hether Earth’s climate change during the past ­century could be related to solar activity. What kind of evidence would be needed to ascertain a cause and effect to support your viewpoint? (LG 5) • What makes the Sun shine? The chapter-­opening question could be discussed before the chapter is covered to gain initial thoughts from the students about something they might not have considered before, or afterward as a 105

106 ◆ Chapter 14 Our Star—­The Sun review of what they have learned, or both. We think a before-­and-­after discussion would be the most in­ter­est­ing.

PROCESS OF SCIENCE This chapter investigates the “Learning from Failure” aspect of the pro­cess of science. Th ­ ere was a “neutrino prob­lem” for a very long time u­ ntil the development of technology and instruments that ­were capable of detecting that t­ here ­were three dif­fer­ent types of neutrinos.

ASTROTOUR ANIMATIONS The following AstroTour animations are referenced in Chapter 14 and are available from the ­free Student Site. ­These animations are also integrated into assignable Smartwork5 online homework exercises.

THE SOLAR CORE This animation shows the collapse of molecular clouds, their fragmentation into cloud cores, the formation of an accretion disk, and the creation of stellar winds and bipolar jets. It ends with the question of “What happens next?”

It follows with a statement of the balance of energy generation and gravity. The emphasis is on the pro­cess of fusion, which generates the radiation at the core of a star; the star’s pressure balances the self-­g ravitation of the gas of which the star is made. The animation also details the transport of energy through the regions of the Sun. Text reference: Section 14.1

ASTRONOMY IN ACTION VIDEOS ­ ese videos mix live demos with mini-­lectures developed Th

by Stacy Palen. They provide students visual preparation for class or a review of what they have learned; they are listed ­here in the order presented in the text. All videos are available on the ­free Student Site. Assignable assessment questions can be found in Smartwork5 and the Coursepack.

RANDOM WALK Stacy Palen demonstrates a random walk by having students in a lecture hall randomly toss a wad of paper between one another. The time it takes the wad of paper to reach the back of the lecture hall is compared with the time it would take a neutrino to leave the Sun directly, without any interaction.

This can be a very effective demonstration, and a wake-up moment for the lecture, if a few balloons are used instead of a wad of paper. Start with the balloons at the “core” of the Sun (the students in the front row). Have students randomly hit them in dif­fer­ent directions ­until you feel enough time (which you have noted) has been spent on this. Have a teaching assistant or a student then run the distance—­say, from the front of the room to the back—­for comparison. (Length: 1:21) Text reference: Section 14.2

INTERACTIVE SIMULATIONS PROTON-­PROTON CHAIN The proton-­proton chain animation provided in this chapter guides the students through each reaction of the fusion cycle that powers the Sun. Students identify the particles involved and t­hose carried away. Students are asked to define the vari­ous particles and gamma rays, as well as ­whether charge was conserved. Students then make a t­ able of inputs and outputs to summarize the fusion cycle.

Text reference: Exploration

TEACHING READING ASTRONOMY NEWS 1. A CME is a coronal mass ejection; that is, a large mass of solar material composed of charged particles (helium nuclei, protons, electrons) that is blasted into space from the Sun. 2. A CME contains huge numbers of moving charged particles that create strong electrical and magnetic storms. Th ­ ese storms interact with the energy generation, current through wires, and general cir­cuits of every­thing that runs on electricity. They could disable electronic devices including GPS satellites.

3. Due to the rotation of the Sun, the CME occurred ­after the active region had been oriented t­ oward Earth, and thus the material missed us.

4. The loss of electricity, computers, and telecommunications lines would bring radical change to almost ­every aspect of modern society and our way of life.

One can imagine that if a storm ­were strong enough and knocked all of ­these out permanently, we would be forced to adopt a much lower-­tech lifestyle—­ perhaps not as bad as the ­M iddle Ages, but certainly rivaling that of the late 1800s.

5. The video contains the same text as this article. It illuminates some of the pieces of the article through graphics and animation, which makes it helpful to a nonspecialist. From the video: The Carrington Event of September 1859 was a series of power­f ul CMEs that hit Earth head-on, sparking northern lights as far south

Chapter 14 Our Star—­The Sun

as Tahiti. Intense geomagnetic storms caused global telegraph lines to spark, setting fire to some telegraph offices and disabling the “Victorian Internet.” Updated link for the video: https://science.nasa.gov/science-news​ /science-at-nasa/2014/02may_superstorm/

CHECK YOUR UNDERSTANDING SOLUTIONS 14.1A (c) One clue is the name “hydro” ­because fluids always deal with pressure instead of force.

14.1B (b) Fusion works via E = mc2; mass is converted to energy.

14.2 (c) Neutrinos stream directly out of the Sun’s core, with ­l ittle if any interaction with interplanetary material, thus allowing us to corroborate our understanding of nuclear fusion.

14.3 (b) The photosphere’s thinness ­causes it to look well defined instead of fuzzy.

14.4 (d) Sunspots are still quite hot, just not as hot as the photosphere; thus, they look dark in comparison.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (a) Only neutrinos can escape from the center of the Sun unchanged, and helioseismology allows us to probe the Sun’s interior just as earthquakes on Earth allow us to probe our planet’s core.

2. (e = a = g; f; h = b; d = c) See Figure 14.4.

3. (a) Magnetic activity of the Sun c­ auses ­these phenomena, as explained in the text.

4. (b) The first statement is hydrostatic equilibrium, and the second is energy equilibrium. Both conditions are necessary for a star.

5. (b) Fusion requires protons to overcome their mutual repulsion, which requires very high temperatures and densities.

6. (b) Sometimes not finding what we are looking for indicates that we do not understand that t­hing, not that the t­ hing is missing.

7. (c) Sunspots are a magnetic phenomenon.

8. (d) Neutrinos stream out of the core of the Sun almost instantaneously.

9. (d) ­There are very few atoms emitting energy in the corona, thus it is very hot and very faint.

10. (c) As the Sun spins, Earth moves around the Sun in its own orbit.

11. (b, c, d, f, g, e, a) This is shown graphically in Figure 14.10.

◆ 107

12. (b) Charged particles are effectively trapped along magnetic fields, so when t­hose field lines break, the particles are whipped into escape velocity.

13. (c) See Figure 14.10.

14. (d) Nuclear fusion requires very high temperatures (tens of millions of degrees) and pressures that are found only at the centers of stars.

15. (c) The solar wind is a stream of neutral and charged particles streaming out of the Sun that carries with it the Sun’s magnetic field. The interaction between the Sun’s charged particles and Earth’s magnetosphere changes the magnetosphere’s shape.

Thinking about the Concepts 16. If the star starts making too much energy, hydrostatic equilibrium ­causes the star to expand, which releases that excess and slows the production of energy in the core. Having lost the excess energy, gravity contracts the star ­u ntil hydrostatic equilibrium is again achieved, and the star is back to its original size, temperature, and luminosity. As a result, the star keeps its energy generation and energy loss at a constant, self-­regulated value.

17. Nuclei all have positive charges, which repel each other.

To fuse two atoms, very high temperatures (tens of millions of degrees), pressures, and densities are required so the two nuclei can overcome this amazingly strong repulsion. The oceans, and indeed ­every natu­ral environment on and in Earth, do not satisfy ­these conditions.

18. Neutrinos are very hard to detect b­ ecause they carry no charge, travel at nearly the speed of light, and interact very weakly with ­matter. To detect a particle, we have to watch the reaction of that particle striking or scattering off another one. Neutrinos hardly interact ­ atter, and so it is very rare that we detect them. with m 19. In the proton-­proton chain, four hydrogen atoms do not come together at one time to produce a helium atom; rather, the hydrogen atoms are slowly built up to helium.

First, two hydrogen atoms come together to form a heavy hydrogen or deuteron. For this to occur, one proton must turn into a neutron, which releases a positron and a neutrino. Then the deuteron fuses with a proton to form light helium. The first two steps of the chain must happen twice; once this occurs, two of the light helium atoms fuse to form a helium nucleus, with two hydrogen nuclei left over. This is how the pathway looks: Step 1: 11 H + 11 H → 21 H + υ + e + Step 2: 21 H + 11 H → 23 He + γ Final step: 23 He + 23 He → 24 He + 2 11 H

108 ◆ Chapter 14 Our Star—­The Sun 20. Fission requires that a large number of heavy ele­ments already exist (uranium’s atomic number is 92). The Sun is made up almost exclusively of light ele­ments (mainly H and He), so ­there are very few nuclei available for fission within the Sun, or any other star.

21. If a huge amount of hydrogen suddenly fused in the Sun, the core would heat up, causing it to expand. The core would then cool down, fusing less hydrogen per second and producing less energy. The core would then contract, and hydrostatic equilibrium would return; the Sun would be stable again. Now all that energy ­ ill go into needs to get out of the Sun; some of it w making the Sun expand a ­little, which would also make the Sun become a ­l ittle brighter. Of course, we might not see this for hundreds of thousands of years.

22. Answers w ­ ill vary. If neutrino oscillations had not been discovered, a variety of other neutrino-­oscillation detection experiments would have to be conducted to conclusively rule out this idea. If the oscillations still ­were not discovered, then our understanding of nuclear physics and/or how fusion powers the Sun would need to be revised.

23. The solar neutrino prob­lem stems from a mea­sure­ ment of the solar neutrino flux that was only one-­third as large as the flux predicted by stellar structure models.

Initially, astronomers thought t­here was a prob­lem with the structure models of the Sun, but ­today we know that neutrinos carry a tiny mass that allows them to oscillate between three dif­fer­ent flavors: electron, muon, and tau. Our detectors on Earth are sensitive only to electron neutrinos, but at any given time ­there is only a one in three chance that a neutrino ­w ill be an electron neutrino. Thus, our mea­sure­ments of solar neutrinos missed the remaining two-­thirds of the flux that contained muon or tau neutrinos.

24. The Sun is made of gas and plasma only, so it has no ­actual surface. The photosphere is as deep into the Sun as we can see ­because the Sun becomes opaque at the base of this layer. Think of it a ­little like looking down into the ocean—at some depth one can no longer see any farther b­ ecause the w ­ ater, which is not a solid surface, becomes too opaque.

25. The winds from solar flares contain huge amounts of charged particles that can disrupt or destroy electronics. Thus, a flare could knock out any satellites, including telecommunication, spy, and weather satellites. ­These periods of increased solar activity also cause Earth’s upper atmosphere to expand slightly. This increases the drag on orbiting satellites, causing their orbits to decay. U ­ nless the satellites are boosted back up again, this effect can cause them to crash to the ground much sooner than expected or wanted.

26. The solar corona is the outermost part of the atmosphere of the Sun and is a plasma heated to millions of degrees. It has an extremely low density. ­Because it is faint, we can see it only if we block out the light of the rest of the Sun, ­either during a total eclipse of the Sun or with a special circular disk called a coronagraph.

27. By Einstein’s E = mc 2 , the difference in mass becomes energy emitted in the form of gamma rays, with an extremely tiny amount disappearing with each neutrino.

28. Sunspots tell us the Sun’s overall rotation period, and, more importantly, that the Sun has differential rotation, whereby the equator rotates faster than the poles.

29. Dif­fer­ent parts of the Sun give off their light at dif­fer­ ent electromagnetic wavelengths ­ because of their temperatures or the par­tic­u ­lar radiative pro­cesses that happen in ­those regions. Anything that gives off UV or X-­rays (e.g., the chromosphere and corona) must be studied from space ­because our atmosphere blocks ­those wavelengths. Our atmosphere does not block visual wavelengths but does distort sunlight; thus studying the photosphere at high resolution is also best done from space.

30. The interaction between the solar wind and the interstellar medium is an impor­tant region of study, as it probes our understanding of both environments. It is especially impor­tant to us ­because the heliosphere protects us from high-­energy cosmic rays; when the Sun is in a period of lower activity, we are more susceptible to damage from ­these particles.

Applying the Concepts 31. Setup: To determine ­whether a graph is linear or logarithmic, check ­whether major tick marks increase by a constant additive f­ actor or by powers of 10.

Solve: (a) The height axis increases by adding 1,000 kilo­meters (km) to each tick mark; thus, it is linear. (b) The density axis increases by a multiplicative ­factor of 100 for each tick mark; thus, it is logarithmic. (c) The temperature axis increases by a multiplicative ­factor of 10 between each tick mark; thus, it is logarithmic.

Review: The range of values of the Sun’s characteristics (such as temperature, density, or pressure) is very large and thus requires a log scale to show all of t­ hose values.

32. Setup: Figure 14.19 shows (b) the location and number of sunspots versus time and (c) the strength and position of magnetic fields versus time. If we are to argue that spots result from strong magnetic fields, then they should appear in roughly the same regions over time.

Chapter 14 Our Star—­The Sun

Solve: Inspection of the figures shows that the butterfly shape seen in panel (b) is reproduced in panel (c), confirming that the two effects are correlated.

Review: If sunspots w ­ ere not correlated with magnetic field positions, then we would expect t­ here to be no visual similarity between the two panels.

33. Setup: Figure 14.17(a) shows a small patch of the Sun’s surface with (b) showing a high-­resolution image of the sunspots pres­ent in that area. Figure 14.20 shows the number of sunspots pres­ent as a function of time over the last few centuries. We can answer question (a) with Figure 14.17(a) and question (b) with Figure 14.20, and (d) with Figure 14.24.

Solve: (a) We superimposed a grid over Figure 14.17(a) and counted the number of ­whole squares and estimated the number of partials; we came up with 80 squares, 4 of which ­were covered by the region of sunspots. This is 5 ­percent of this side of the Sun, or 2.5 ­percent of the entire surface. (b) The number of spots that occur at solar maximum varies between 100 and 200, making the average about 150 sunspots. (c) On average, near solar maximum, ­there are hundreds of spots, whereas in question (a), the date was near solar minimum. This would suggest that at maximum, the area of the Sun covered by spots is a few hundred times greater than at minimum—­about 2 to 3 ­percent of its surface. (d) Compared to the graph of irradiance shown in Figure 14.24, we see that ­t here is a direct correlation between the number of spots and the Sun’s brightness. This seems counterintuitive ­because the sunspots are cooler than the surrounding photosphere, and the more ­t here are, the cooler the Sun should be. It is the other phenomena, however, that accompany the sunspots that make up for the reduction in radiation from the sunspots, resulting in an overall increase in the Sun’s brightness by about 0.1 ­percent.

Review: Note that although a small percentage of the Sun’s surface may be covered with spots, the overall energy coming out of the Sun (the right-­hand panel in Figure 14.24) is essentially constant with time. Sunspots have a very small effect ­unless something like the Maunder Minimum happens.

34. Setup: For a gamma ray to emerge 2.3 seconds a­ fter being formed in the core, it must travel straight out of the Sun. Is this pos­si­ble?

Solve: We have learned that a photon travels only a short distance in the Sun’s radiative zone, traveling about 1 centimeter (cm) before being absorbed and reemitted or scattered by m ­ atter. Thus, the distance

◆ 109

required for a photon to escape the Sun is much longer than the straight path out.

Review: Note that neutrinos can stream freely out of the Sun, and so they emerge in about 2.3 seconds.

35. Setup: Density is calculated by dividing the mass of the sphere by its volume. We ­w ill use ratios for our M M logic: ρ = ∝ V R3 Solve: (a) Putting in the values for the Sun’s mass and radius as compared to Earth, we find that the Sun’s 300,000 density is ρ = = 0.3 times that of Earth, or 1003 about 1,650 kg/m3. (b) This is 0.3 times the average density of Earth ­ ater, which is and about 65 ­percent denser than w 1,000 kg/m3.

Review: We know the gas g­ iants Jupiter and Saturn have very low densities and are made mostly of hydrogen and helium, like a star, so it stands to reason that a star’s average density w ­ ill be low as well.

36. Setup: Einstein’s equation E = mc2 can be solved for mass: m = E/c2 . A joule is 1 kg m2/s2 (note that the units match Einstein’s equation). A watt is a joule/ second, so the amount of energy (in joules) given off in 1 second is numerically equal to the luminosity of the star.

Solve: Inserting the solar luminosity and speed of light, we find: E 3.85 ×1026 kg m 2/s2 m= 2 = = 4.3 ×10 9 kg (3 ×108 m/s)2 c ­ ere are 1,000 kilograms in a metric ton, so this mass Th equivalent is 4.2 million metric tons burned each second.

Review: The mass-­burning rate of the Sun is about 4 billion kilograms per second, or about 9 billion pounds per second. Although this may seem huge, remember the mass of the Sun is 2 × 1030 kg, so the amount of mass burned is tiny in comparison.

37. Setup: This prob­lem asks for the mass lost over a given amount of time. In the previous prob­lem, we found the mass-­burning rate is 4.3 × 109 kg/s, so the mass lost in a given time is just the rate times the time.

To calculate the p­ ercent mass lost, remember: M % mass = lost ×100 M total Solve: (a) If the Sun’s rate of burning has remained constant over the last 4.6 billion years, then the total mass lost is M lost = ( 4.3 ×10 9 kg/s ) × (1.4 ×1017 s) = 6 ×1026 kg

110 ◆ Chapter 14 Our Star—­The Sun ­u ntil 4–5 days ­later. The most energetic coronal (b) The fraction of the Sun’s mass that has already mass ejections, on the other hand, can reach velocibeen converted to energy is 26 ties about twice this speed, reaching Earth in a few 6 ×10 kg % mass = ×100 = 0.03%. days.

2 ×1030 kg 41. Setup: ­Because the photosphere and sunspot both act Review: Note that despite converting 100 Earth like blackbodies, we know that the brightness b ∝ T4. masses to energy in 4.6 billion years, the Sun still We can construct a ratio: retains an astounding 99.97 ­percent of its original 4 mass. ⎛ Tspot ⎞ bspot =⎜ 38. Setup: Use Working It Out 14.1. We are given that ⎟ bphotosphere ⎝ Tphotosphere ⎠ the star is an A5 main-­sequence star with 2 times the mass and 12 times the luminosity of the Sun. Solve: The sunspot is 70 ­percent as bright as the photo- 4 b spot Solve: We ­w ill give the Sun a mass of 1 and a lumisphere, which means for our ratio above Tspot = Tphotosphere × bphotosphere nosity of 1 and work with the final ratios and the information given in the question to determine the 4 bspot = Tphotosphere × = 5780K 4 0.7 = 5780K × 0.9 = 5290K approximate lifetime of this star as comparedTspot to the bphotosphere lifetime of the Sun. The lifetime of a star scales roughly as a ratio of its mass to luminosity: Review: Blackbody brightness scales strongly mass lifetime ∝ /luminosity with temperature (T4), so the spot appears fainter than the photosphere ­because its temperature is 2 1 This star w ­ ill live about = as long as the Sun. lower.

12 6 42. Setup: Solving E = mc 2 for the mass ­w ill yield the Review: A more exact result is that the lifetime of a amount of mass that is converted to energy. To figure star scales roughly as M–­2 .5. We substitute 2 for the out how much energy would come from our textbook, 1 −2.5 we w ­ ill consider how many 5-­megaton bombs are in mass and get 2 = , or about one-­sixth.

5.7 the book, based on our answer to part (a).

Solve: (a) For the 1.6 kg textbook 39. Setup: Neutrinos and photons both travel at the speed of light, and the Sun is about 8.3 light-­m inutes away.

E =1.6kg × (3 ×108 m/s) 2 =1.44 ×1017 J Solve: (a) Neutrinos hardly interact with any other If a 5-­ megaton bomb releases 2 × 1016 J, then ­matter, so they essentially stream freely out of the Sun. 1.44 ×1017 J Thus, they w ­ ill arrive at Earth 8.3 minutes a­ fter they = 7.2 equivalent 5-­megaton bombs. are created.

2 ×1016 J (b) A photon takes about 300,000 years or longer to 2 ×1016 J E travel from the Sun’s core to its exterior, so any physical = 0.22kg . If one 5-­megaton (b) m = 2 = c (3 ×108 m/s)2 ­ ill not be change that occurs in the heart of the Sun w bomb uses up 0.22 kg of mass, and the Earth’s mass is seen in vis­i­ble light ­until approximately 300,000 years 6 × 1024 kg, then the Earth lost ­after the change occurs.

Review: A good example of this is SN 1987A., Neutri0.22kg ×100 = 3.7 ×10 −24 percent. nos ­were detected before the vis­i­ble light reached us 6 ×1024 kg ­because the neutrinos streamed right out of the collapsing star, whereas the light took much longer to Review: This prob­lem shows us just how much energy escape. ­there is inside a small amount of mass and why fission 40. Setup: The Sun is 150 million (1.5 × 108) kilo­meters and fusion are such amazingly energetic pro­cesses. Even from Earth (on average). To find the time it takes for a so, just an extremely small amount of Earth’s mass was particle in the solar wind to reach us, remember d = vt. lost.

Solve: Solving for time, 43. Setup: ­There are a number of Internet resources that give statistics from which students can retrieve d 1.5 ×108 km t= = = 3.75 ×105 s. This is about 4.3 data related to this prob­lem. We use just one source v 400km/s ­ h ere. days.

Solve: The Global Energy Statistical Yearbook 2018 Review: In other words, on average, particles reachestimate for the total energy production from all ing escape velocity from the sun ­w ill not reach Earth sources and all countries was 26,000 TWh (terawatt

Chapter 14 Our Star—­The Sun

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­ ill round that to 1020 Solve: We first express the units of the gravitational hours) or 9.4 × 1019 joules. We w joules. If this rate is maintained over a half-­m illion constant in a dif­fer­ent form to make cancelling of years, then the global production ­w ill be 1020 J/ terms easier: m3/kg ⋅ s2 can also be stated as J ⋅ m/kg2.

5 25 2 yr × 5 × 10 yr = 5 × 10 J. The Sun produces about 2 2 × 1030 kg ) M Sun ( 26 −11 2 G = 6.67 × 10 J ⋅ m / kg × 4 × 10 joules each second, or about 10 times the R Sun LSun ( 7 × 108 m )(3.8 × 1026 J/s global production over 500,000 years.

Review: The Sun does indeed outproduce 2what all M Sunthan a (2 × 1030 kg )2 current power sources would produce G in more = 6.67 × 10 −11 J ⋅ m / kg 2 × = 1015 s, RtSun ( 7 × 108 m )(3.8 × 1026 J/s ) er­eLst­ ing is half-­m illion years. What might also be in­ Sun the breakdowns given by current sources of energy or about 32 million years. and the forecasts for t­ hese sources of energy over the Review: The Sun could power itself via gravitational next 50 years. contraction for around 30 million years. It did just 44. Setup: Let’s treat this prob­lem by rounding numbers that as a pre-­main-­sequence star. Theoretical calculato powers of 10. For example, the mass of the Sun is tions give an estimate of about that length of time for about 1030 kg, and it is mostly hydrogen, which has an the Sun to form and start fusion as a main-­sequence atomic mass of 10 –27 kg per atom. Thus, ­there are about star.

1030 kg =1057 atoms in the Sun. The text sug10 −27 kg/atom

gests that each atom could produce 10 –19J of energy in a chemical reaction.

Solve: The total amount of energy stored in the Sun ­under chemical reactions would be the product of [the energy per reaction] times [the number of pos­si­ ble reactions], which is the number of atoms. Thus, the total energy available is about 10 −19 J × 1057 J = 1038 J. At an energy-­output rate of 1026 W, the Sun would use up its fuel in 1012 seconds, or about 32,000 years.

This is a small fraction (~7 millionths) of the age of the Earth.

Review: It may be enlightening for students to compare our first approximate approach to a more robust set of calculations, using more precise values. We follow the same logic given in the solution and assume that the Sun is 75 ­percent hydrogen by mass, with only hydrogen taking part in the reactions, ­g iving us 1.5 × 1030 kg for the total mass available and (1.5 × 1030 kg) / (1.67 × 10 −27 kg/atom) = 9.0 × 1056 hydrogen atoms. The energy per reaction is given as 1.6 × 10 −19J. The total energy from chemical reactions would then be: 9.0 × 1056 reactions × 1.6 × 10 −19 J/reaction = 1.4 × 1038 J. The mea­sured power output for the Sun is 3.8 × 10 26 J/s. The expected ­lifetime of the Sun if chemical reactions w ­ ere its source of energy would be: (1.4 × 1038 J)/ (3.8 × 10 26 J/s) = 3.8 × 1011 s. In years this would be (3.8 × 1011 s)/​(3.16 × 10 7 s/yr) ≈ 1.2 × 10 4 yr. By ­either method, energy production by chemical reactions is disproved.

45. Setup: Use all values in MKS units and the lifetime ­w ill be in seconds. Remember ­there are 3.16 × 107 ­seconds in one year.

Using the Web 46. Students visit the a. SDO and report on sunspots, news, and latest images. Answers w ­ ill vary. b. Students go to the STEREO mission’s website. Updated link for the website is https://­w ww​.­nasa​.­gov​/­m ission​ _­pages​/­stereo​/­main​/­index​.­html. The two nearly identical observatories are located ahead of and trailing ­behind Earth in its orbit. By using that spacing—­ they are 180 degrees apart in their orbits—we can see the entire Sun. 47. a. Goals and discoveries of NASA’s Interface Region Imaging Spectrograph are: “IRIS ­w ill make use of high-­resolution images, data and advanced computer models to unravel how ­ matter, light, and energy move from the Sun’s 6000 K surface to its million K outer atmosphere or corona.” Many discoveries listed; answers ­w ill vary. b. Students visit the SOHO website and report on the “Sun Now.” Answers w ­ ill vary. c. The Sun’s magnetic field is what drives its activity levels that lead to sunspots, flares, CME’s, the solar wind, and solar variability, and is thus impor­tant. The advantage of a ground-­based telescope instead of a space-­based telescope is that it is much cheaper to build and maintain. The DKIST telescope observes at near-­i nfrared and vis­i­ble wavelengths, to start. Haleakalā is over 3,000 m (10,000 feet) high, and thus a good location for viewing at IR wavelengths. The telescope is scheduled to be completed in 2019.

48. a. Students go to the Space Weather website to report current solar flares and sunspot numbers, if any.

Answers ­w ill vary. b. Citizen Science: Students are directed to set up accounts for “Zooniverse” and work through the tutorial. Answers ­w ill vary.

112 ◆ Chapter 14 Our Star—­The Sun 49. a. In experiments to develop controlled nuclear fusion on Earth, 192 ­lasers converge on a target the size of a peppercorn filled with hydrogen atoms. This differs from the Sun ­because it happens at about 3 million degrees C (compared to 15 million Kelvin) for a very short time over a very small volume. b. Space-­faring nations (USA, Rus­sia, ESA), and India are talking about g­oing to the Moon for mining helium-3. ­There is no timeline yet ­because the mission ­hasn’t gotten beyond the interest stage. Difficulties w ­ ill include cost, distance, mining, profit margin, determining even if helium-3 ­w ill produce a safe fusion solution.

50. a. Students report on the Voyager spacecraft. As of February, 2019, Voyager I was 145 AUs from Sun (interstellar space); Voyager 2, 120 AUs. The Voyager 2 left the heliosphere for interstellar space on November 5, 2018. b. The IBEX mission has been active for 5 years with a long list of discoveries about the solar wind and the interstellar medium. Answers ­w ill vary.

EXPLORATION This exploration steps the students through the proton-­ proton chain in detail through the Interactive Simulation “Proton-­Proton Chain.”

EXPLORATION SOLUTIONS 1. The positron carried away the positive charge.

2. A neutrino is an almost massless particle that is produced in nuclear pro­cesses involving the fusion of two protons producing a positron and a neutrino.

3. Yes, the same reactions happened in both panels.

4. Deuterium (a neutron and a proton) and hydrogen (a proton) nuclei entered the reaction, and light helium (3He, two protons and a neutron) resulted.

5. Two positive charges came in and two left, so charge was conserved.

6. A gamma ray is the highest energy photon. A gamma ray was produced during the reaction.

7. Two 3He entered the reaction, and an ordinary helium, 4 He (two protons and two neutrons), was produced.

8. ­Because six hydrogen nuclei are used but two are returned, it is pos­si­ble to consider two of the hydrogen nuclei as catalysts.

9. The positron created in the first step w ­ ill meet an electron; that annihilation produces two gamma rays. The second step, in which the deuterium nucleus fuses with a proton, produces another gamma ray. ­Because ­these steps happen twice, a total of six gamma rays are produced that then travel through the Sun to its surface.

10. The two hydrogen nuclei that are released at the production of 4He could be involved in additional reactions.

Also, the positrons ­will be used almost immediately when they annihilate with ­free electrons.

11. The helium nuclei that are produced w ­ ill stay as they are for a very long time.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES The Learning Astronomy by D ­ oing Astronomy workbook activities that are relevant to Chapter 14 are introduced ­here. For more information, please see the Learning Astronomy by ­Doing Astronomy workbook, the Instructor’s Manual for the workbook, and the Power­Point clicker question slides associated with the workbook.

ACTIVITY 19: ANALYZING A CORONAL MASS EJECTION FROM THE SUN Students take a close look at a coronal mass ejection (CME) from the Sun, calculate the speed of a clump of material, and determine how much notice scientists have before the charged particles reach Earth. A ­ fter completing the activity, students ­w ill be able to: • calculate the speed of the material ejected.

• determine how much advance warning we would have.

• consider the source(s) of errors and uncertainties.

• summarize why it is impor­tant for us to predict CMEs accurately.

Text reference: Section 14.4

CHAPTER 15

The Interstellar Medium and Star Formation INSTRUCTOR NOTES Chapter 15 covers the interstellar medium and star formation. Major topics include: • the dif­fer­ent components, densities, and phases of the interstellar medium.

• the effects of interstellar dust on starlight.

• molecular clouds as sites of star formation.

• stability, collapse, and fragmentation of molecular clouds.

• the formation and evolution of a protostar.

• the role of hydrostatic equilibrium during star formation.

• a summary of our knowledge of brown dwarfs.

• details of the luminosity and temperature as a star forms.

• bipolar outflows leading to such ­things as Herbig Haro objects and T Tauri stars.

Dust not only reddens and extinguishes light, it also reflects blue light, creating eerie blue “reflection nebulae.”

This can be seen in the bottom left of Figure 15.8a and surrounding the Pleiades cluster in Figure 15.24. It is worth a quick drawing showing how the stars must be in front of the dust for us to see the reflected blue light, and furthermore, that ­every dust cloud that we see extinguishing starlight from Earth could look like a blue reflection nebula to an observer on the other side of the cloud.

DISCUSSION POINTS • Inspire a 1-­m inute writing exercise by telling the students, “Interstellar dust makes every­thing look fainter and redder than it r­ eally is. This is called interstellar reddening. Some astronomers prefer to call this ‘interstellar de-­bluing.’ Why would that be a more appropriate description of what is happening?” (LG 1) • List as many t­ hings as come quickly to mind where objects on Earth have the same composition as the Sun and the interstellar medium. Compare lists and discuss if this gives additional clues as to how Earth formed.

(Mentioned in this chapter are hydrogen, helium, soot, iron, silicon, carbon, carbon monoxide, and methanol.

Additional clues are polycyclic aromatic hydrocarbons [PAH], formaldehyde, the sugar molecule glycolaldehyde, and more.) (LG 1) • In this chapter, we learn that stars form inside molecular clouds. Why? What is special about molecular clouds, as opposed to other regions of the interstellar medium? (LG 2) • How likely is it that the Sun has siblings that formed from the same parental molecular cloud? Discuss with students w ­ hether ­those solar siblings exist and, if they do, where they might be. (LG 2) • Compare and contrast the information in this ­chapter concerning the collapse of a molecular cloud to a protostar to the discussion in Chapter 7 related to the chapter-­opening image of the formation of planetary systems. (LG 2) • Section 15.4 covers the evolution of a protostar.

Review that material, especially the information contained in Figure 15.18, and quickly make a list of what you initially find confusing. Which of the steps in the evolution of a protostar are the most difficult to understand? Compare your list with t­ hose of a few of your classmates and make a master list to give to your instructor. (LG 3) • Figure 15.20 shows the paths of protostars for masses ­ ere one ranging from 0.8 MSun up to 60 MSun. Say you w of t­ hose stars—­pick one. What would you be like at the start? At the end? How would you describe the changes in temperature, luminosity, and radius you went through as you ­were “being born”? (LG 4) • The chapter-­opening image asks the question, “How do stars form?” Extend that to asking, “Why do stars form? Where do stars form? When do stars form?”

PROCESS OF SCIENCE Astronomy covers the full range of the “powers of ten.” This Pro­cess of Science brings in the quantum physics of the “electron flip” in hydrogen atoms. This knowledge is then applied to observations of the structure of neutral hydrogen in the Milky Way, which is 1030 times the size of an atom.

113

114 ◆ Chapter 15 The Interstellar Medium and Star Formation

ASTROTOUR ANIMATIONS STAR FORMATION The star formation animation starts with the overview of the birth of stars within the interstellar medium in the spiral arms of galaxies. It then covers the cooler, g­ iant molecular clouds. The following concepts are covered: self-­gravity, thermal pressure, fragmentation, “inside-­out” collapse, protostar characteristics, formation of an accretion disk, and onset of fusion.

Text reference: Section 15.2

ASTRONOMY IN ACTION VIDEOS ANGULAR MOMENTUM This video uses a “rotating platform of doom” on which Dr. Palen stands. She is given a small shove to her outstretched arms, and then brings t­ hose arms in close to the body to demonstrate the conservation of angular momentum. The demonstration is repeated with barbell weights now held out away from the body. A slow spin is started, and then the barbells are brought in to demonstrate an even greater increase in the spin rate. The dependence of the amount of angular momentum on the mass, on the spin rate, and on the distribution of the mass is emphasized.

(Length: 2:58) Text reference: Section 15.3

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Answers w ­ ill vary. Scientists typically want to identify interstellar dust grains b­ ecause that is the material out of which the Solar System—­and we ourselves—­formed.

2. The researchers looked for chemical signs of interstellar origin and travel to distinguish interstellar dust from Solar System dust. They also looked for dust entering away from the Sun in the direction of the interstellar dust stream.

3. A complete analy­sis of ­these particles ­w ill involve destroying them, for example in scanning-­electron microscopes. Thus, they cannot be further studied.

4. Citizen scientists aided in searching for trails in the aerogel and informing the astronomers of t­ hese discoveries.

This rapidly sped up the pro­cess of analyzing the gel.

5. This prob­lem sends the user to Stardust@Home, to look for t­ hese tiny tracks of dust in the aerogel. New results ­w ill vary

CHECK YOUR UNDERSTANDING SOLUTIONS 15.1. (a, d) Interstellar dust both dims and reddens distant objects.

15.2. (c) Collapse takes place when regions become denser than their surroundings and draw other material in.

15.3. (a) Gravitational energy is converted to thermal energy (increasing the temperature) as the protostar compresses itself.

15.4. (b) Although the mechanism driving the bipolar outflows is not totally understood, current models have been able to replicate some of the structure using magnetic interactions.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (a, b, d) It is hard to make hot gas very dense b­ ecause the pressure would be very high.

2. (b) Although the effects of dust can be observed in the optical (extinction), dust can be directly observed by its thermal glow in the infrared.

3. (a) The Sun’s composition of heavy ele­ments is fairly close to the interstellar medium b­ ecause the Sun formed out of this gas and it takes a very long time to significantly change the chemical composition of a star.

4. (c) Dust grains are about the size of optical and UV wavelengths, which make them very effective at stopping t­ hose “colors” of light.

5. (c) Supernovae (and luminous, hot O and B stars) are quite effective at heating up interstellar gas to extremely high temperatures.

6. (a) Spectra tell us the composition, density, and temperature of the interstellar medium.

7. (a) “Bi” means two and “polar” means “the end” so bipolar means “related to both ends.”

8. (c) The pro­cess of star formation is too slow to see it happen in real time, but we have observed many objects at dif­fer­ent stages, allowing us to piece together a complete picture of the pro­cess.

9. (d) When the spins of the nucleus and electron “flip” with re­spect to each other, a 21-cm photon is emitted.

10. (c) A vertical line on the H-­R diagram means changing luminosity at constant temperature (or color).

11. (a) Gravity pulls inward, and pressure pushes outward.

12. (c) A bar or lane of dust in the distant galaxy is blocking its light from reaching us, making the region look dark.

Chapter 15

13. (d) A 21-cm photon is released when the spin state of an electron flips over, from being aligned with the nucleus to unaligned.

14. (a) The dark bands that we see are ­either a con­spic­u­ ous lack of material (which is unlikely) or the presence of something such as dust-­blocking light. We have confirmed the presence of dust by looking at t­ hese objects in the infrared.

15. (a) The mass of a star dictates nearly all aspects of its evolution.

Thinking about the Concepts 16. Dust grains are comparable in size to visible-­light wavelengths, so they act as a barrier to vis­i­ble light. Gas molecules are much smaller than dust grains, so they do not obstruct light at vis­i­ble wavelengths.

17. Star formation occurs only where the interstellar medium is dense, cold, and dusty. ­Because of the dust, ­t hese regions are hidden from view in vis­i­ble light.

However, infrared radiation is less affected by interstellar dust, making it easier to peer through the veil at ­t hese wavelengths. In addition, the low surface temperatures of protostars mean that they emit significantly in the infrared region of the spectrum.

18. Interstellar clouds are dense and are distributed sporadically ­because only about 2 ­percent of the volume of our galaxy is filled with them. The rest of the galaxy is filled with hot, low-­density intercloud gas.

19. Much of the material of the cloud collapses into the disk that forms the star, or stars, and planets of that system. If the cloud is large enough, then gas and dust from the cloud can remain and enshroud the star. If the star is very luminous (such as an O or B star), then it ­w ill ionize the gas and can disperse it through its power­f ul radiation. A molecular cloud cannot remain cold and dark with one or more stars inside it.

20. Temperature mea­sures the average kinetic energy of particles in a system, whereas heat mea­sures the total thermal energy of the system. Even though the w ­ ater is 100°C cooler than the oven, the number of hot-­water molecules touching your skin in boiling ­water is far greater than the number of hot-­air molecules touching your skin in the oven. This means that even though the oven is hotter, your skin experiences far more collisions each second with ­water molecules in the pan than it does with air molecules in the oven. As a result, the boiling w ­ ater transfers heat to your skin much more rapidly than the oven does. This same princi­ple applies in the interstellar medium. Even though temperatures reach millions of kelvins in the space between stars, ­there are very few particles to carry this energy. Assuming that you could float in the interstellar medium, the

The Interstellar Medium and Star Formation

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density of the surrounding material is so low that your body would radiate its internal heat much faster than it would absorb heat from the surrounding environment. This is the reason why astronauts wear heated spacesuits when they go outside of the International Space Station.

21. Scientists believe we are located in a large b­ ubble of hot gas b­ ecause much of the interstellar medium surrounding the Solar System is very hot—­glowing at X-­ray wavelengths—­and very low density, whereas outside a radius of about 325 ly, the interstellar medium shows all of its dif­fer­ent phases.

22. Tenuous gases have very few collisions with other atoms or dust; thus, the atoms have a hard time getting rid of their energy, and as a result, they stay hot. On the other hand, dense gases have frequent collisions; thus, they can cool easily.

23. We observe 21-cm radiation from all directions in the sky and across the universe, showing that the laws of quantum physics we discovered h­ ere on Earth are valid everywhere in the observable universe.

24. In interstellar clouds, no mechanism (such as gravity) exists to separate hydrogen from the carbon and oxygen, so we can safely assume that if carbon and oxygen are pre­sent in the cloud, hydrogen is pre­sent as well.

When temperatures are low, carbon and oxygen combine with each other to produce carbon monoxide (CO). U ­ nder t­ hose same low-­temperature conditions, atomic hydrogen ­w ill chemically bond with itself to produce molecular hydrogen (H 2). Thus, by observing CO emission and making the assumption that if ­there is CO, ­there is also molecular hydrogen, we can map the distribution of molecular hydrogen in g­ iant molecular clouds.

25. A ­g iant molecular cloud (GMC) is a cradle of star formation. It is a huge, cold, dense, and dark interstellar gas cloud. In ­these clouds, gas is sufficiently cold and of such high density that small perturbations can cause a cloudlet to collapse and form a star.

26. While it is true that all parts of the cloud experience the force of gravity, ­there are smaller regions (clumps) that are denser and tend to contract into individual stars, most likely within clusters. As the cloud collapses and fragments, the central regions of the protostars in ­these clumps become more massive, more compressed, and hotter. Eventually, the center of each protostar gets hot and dense enough to start fusion, and a single star is born.

27. As a protostar collapses, more mass implies higher gravity, but that higher gravity compresses the gas further. This compression raises its temperature and thus its pressure, so the star maintains hydrostatic equilibrium at all times.

116 ◆ Chapter 15 The Interstellar Medium and Star Formation 28. Brown dwarfs and giant planets are similar in the fact that both are not massive enough for fusion of hydrogen to helium to occur in their cores. They differ in the ways that they form: brown dwarfs form as binary stars do, while giant planets form from material in the circumstellar disk. In the range from about 10 to 80 times the mass of Jupiter, it is difficult to classify a brown dwarf as ­either a planet or a failed star. Some brown dwarfs in this mass range can sustain low levels of nuclear fusion of deuterium or even lithium for a brief time in their cores, but ­others cannot. Some are hot enough to produce a vis­i­ble light glow, but o­ thers are not. In the absence of a formal definition, we hesitate to classify anything larger than about 10 Jupiter masses as a planet. The term brown dwarf already describes ­these objects well.

29. At a certain range of temperatures, the H− ion is a strong source of opacity; that is, it very efficiently blocks radiation from escaping a star. Once it exists in a gas, it is able to absorb low energy photons and use that energy to split the extra electron off of the ion, rather than allowing the photon to escape. If a protostar cools to the point that a lot of H− ions can form, ­those ions make that gas opaque and thus trap radiation trying to escape. This reheats the gas, causing the star to expand. As it expands, it cools, and the H− ions can once again form. This cyclical pro­cess acts as a thermostat in controlling the surface temperature of a protostar.

30. A planet can form only out of the ele­ments available, so a molecular cloud that has very few heavy ele­ments ­w ill be unlikely to form any terrestrial planets. Higher levels of heavy ele­ments help the gas cool faster, which helps the gas collapse faster and can affect the kinds of stars and planets formed within the cloud.

Applying the Concepts 31. Setup: Recall that dust grains are comparable in size to visible-­light wavelengths, but they scatter UV and blue light much more effectively than redder wavelengths.

Solve: Dust “reddens,” which means it makes stars look redder than they r­ eally are by scattering the blue wavelengths in random directions away from us. Thus, if we gauge the temperature of a star by its color, we ­w ill always estimate it as being cooler than it ­really is.

Review: Note that dust also makes a star fainter, which makes us think it is farther away than it ­really is.

32. Setup: Review the material in Section 5.2, “The Quantum View of ­Matter,” and Section 5.4, “Temperature.”

Solve: (a) The two components are the proton and electron.

(b) ­Because the kinetic energy depends on the mass and speed of a particle, if two particles have the same energy, the more massive one ­w ill travel slower. The proton is 1,837 times more massive than the electron, so it ­w ill move much slower.

Review: In general, astronomers talk about the electron temperature or density of a gas, and this is b­ ecause the electrons move so much more rapidly and therefore collide more often, producing most of the pressure in a gas.

33. Setup: From T ­ able 15.1, we know the density of the vari­ous components of gas in the interstellar medium; thus, we can estimate that of dust by noting that the mass of dust in the interstellar medium is about 1/100 that of the gas component. The prob­lem gives the mass of a typical grain of dust as about 10 -17 kg.

Solve: We know the mass density of dust is about 0.01 times that of gas. ­Because the gas density is given around 10 −27 kg/cm3 (~1 hydrogen atom per cm3), dust has a mass density of 10 −29 kg/cm3, or dividing by the mass of a dust grain, (10 −29 kg / cm 3) / (10 −17 kg / grain) yields a number density of 10 −12 grains/cm3.

Review: Earth has a volume of about 1027 cm3 (1 billion km3), which means that in a volume the size of Earth, t­ here would be about 1015 (one million billion) dust grains. Seems like a large number, but t­ here are estimates of 1050 atoms of vari­ous kinds that make up Earth.

34. Setup: Look at Figure 15.3, which shows the spectra of a star seen with and without dust. We ­w ill use Wien’s law and solve for the temperature.

Solve: Examining Figure 15.3 closely, we get a peak of about 460 nm for the star without dust and 720 nm for the star with dust. We need to use Wien’s law to calculate the surface temperatures of t­hese stars, 2.9 × 106 nm K λmax = . Rewriting Wien’s law to solve T 2.9 × 106 nm K . ­These peak for temperatures, T = λmax wavelengths yield temperatures of about 6300 K without dust, and 4030 K with it. Obviously, the differences in temperatures are significant. If we did not include the effect that dust has on dimming and reddening light from stars, our conclusions could be radically wrong.

Review: Note that the star not only appears far too cool but its spectrum faded from a peak around 1 to 0.08; that is, the star became about 12 times fainter ­because of the dust.

35. Setup: We ­w ill use Wien’s law and solve for the peak wavelength.

Chapter 15

Solve: For a temperature of 8000 K, 2.9 × 106 nm K 2.9 × 106 nm K λ peak = = = 363 nm.

T 8000 K Review: Note that the gas w ­ ill not actually radiate a blackbody spectrum peaking around 363 nm b­ ecause this material is low density and thus produces mostly emission lines.

36. Setup: We w ­ ill use Wien’s law solved for the temperature. Th ­ ere are 1,000 nm in 1 μm.

2.9 × 106 nm K 1 µm × = 10,000 K. Solve: T = 0.29 µm 1,000 nm Review: Per ­Table 15.1, this component of the interstellar medium would be classified as having ionized or neutral hydrogen gas. 10,000K is about the perfect temperature to excite the atom up to the second energy state.

37. Setup: To ionize a hydrogen atom, the electron must be stripped away from the nucleus; that is, it must be given more than its binding energy to the nucleus. If the atom is in the ground state, this is 13.6 eV. Solve: (a) A B8 star is very hot, meaning that it produces many high-­energy UV photons that can ionize the hydrogen from its ground state, whereas a K0 ­giant is cool and produces mostly low-­energy red photons.

According to Wien’s law, its spectrum would peak between 600 and 700 nm. (b) It is clear that the temperature of the star, and not its luminosity, is most impor­tant if one wishes to ionize the interstellar medium.

Review: Most bright nebulae glow in emission lines ­because ­there are hot stars (such as O and B types) that are heating the gas so that the atoms can produce emission lines in the optical.

38. Setup: We ­w ill use the equation for calculating the kinetic energy, and work with ratios.

Solve: If a proton and an electron have the same kinetic 1 1 energy, we can write, me ve2 = mp v 2p . Rewriting the 2 2 equation and solving for the ratio of electron speed to proton speed gives

ve = vp

mp me

= 1,850 = 43.

Review: The electron travels over 40 times faster than the proton, which explains why pressure and temperature are usually mea­sured by the electron properties ­because ­these are the particles that interact the most.

39. Setup: This is a ­simple d = vt prob­lem with conversions.

Solve: In terms of seconds: 1.5 × 1012 km t= = 1 × 1012 s. Converting to years: 1.5 km/s

The Interstellar Medium and Star Formation

◆ 117

1 yr = 33,000 years.

3 × 10 7 s Review: This assumes a constant velocity. The a­ ctual time of collapse w ­ ill differ ­because the particle ­w ill be accelerating and interacting with other particles as it falls.

40. Setup: Observations indicate that t­ here is 100 MSun of carbon monoxide (CO) in a ­g iant molecular cloud. ­Because carbon represents 3/7 of the mass of a CO molecule, the mass of carbon in the molecular cloud is 3 MC = × 100 MSun ≈ 43 M Sun . Next, we consider 7 the mass of hydrogen in the Sun relative to the mass of carbon. The Sun contains 2,400 hydrogen atoms for ­every carbon atom. However, each carbon atom has an atomic weight of 12, whereas each hydrogen atom has an atomic weight of only 1. In other words, it requires 12 hydrogen atoms to equal the mass of a single carbon atom. Therefore, the Sun contains only 2,400/12 = 200 times as much hydrogen as carbon by mass.

Solve: The mass of carbon in the molecular cloud is 3 MC = × 100 MSun ≈ 43 MSun . We assume that the 7 ratio of hydrogen mass to carbon mass in the Sun is representative of that in a molecular cloud. This means the molecular cloud of interest contains, 1 × 1012 s ×

MH cloud =

MH Sun M = 200 × 43 M Sun = 8,600 MSun , MC Sun C cloud

most of it in the form of molecular hydrogen.

Review: Note that the a­ ctual fraction of carbon, by mass, is 43/8,600 = 0.5 ­percent, which reinforces the fact that most of a cloud’s mass is hydrogen. Note that current research implies that the ratio of CO to H 2 found for the Milky Way may not apply to other galaxies.

41. Setup: If it takes N seconds for something to happen on average, then the chances of it happening in 1 second are just 1/N. The rate at which photons are emitted is the probability of emission (p) times the number of objects (n) that can emit.

Solve: (a) The probability that any atom makes this transition in 1 second is, simply, 1 p= = 2.86 × 10 −15.

3.5 × 1014 (b) The rate at which photons are emitted ­w ill be p × n, where n = 6 × 1059 atoms. The rate of photon emission ­w ill be about (2.86 × 10 −15) × (6 × 1059) = 1.7 × 10 45 photons/second. (c) Comparing this rate to the rate of photon emission from a solar-­t ype star (1.8 × 1045), we find that the rates are almost identical.

118 ◆ Chapter 15 The Interstellar Medium and Star Formation they would then become a ­little over 2,500 times Review: The difference between the 21-cm emission brighter than t­ hose faintest stars. and that of the star is that the solar-­t ype star primarily Review: The 30 Doradus region is full of the most emits more energetic, visible-­l ight photons, whereas luminous O-­and B-­t ype stars; this shows just how the 21-cm neutral hydrogen radiation is found at the luminous ­those stars ­really are. less energetic radio wavelengths. Even though the 45. Setup: We solve this using the Stefan-­Boltzmann law probability of an atom dropping from a higher-­energy and ratios. The brown dwarf’s size is about that of Jupispin state to a lower one is very small, ­there are so many ter, and Jupiter’s dia­meter is about 1/10 that of the neutral hydrogen atoms in the interstellar medium that Sun. We set the Sun’s surface temperature as 5780 K, we are able to map out the Galaxy using the 21-cm line. making the brown dwarf ’s surface temperature of 42. Setup: ­There are 86,400 seconds in 1 day, so if we are 1000 K to be 1/5.78 TSun. compressing time, we must determine the relative fracSolve: Using the Stefan-­Boltzmann equation, we set tion of a certain phase out of the Sun’s w ­ hole lifetime 2 4 and then multiply that by the length of a day to find out Lbd ⎛ R bd ⎞ ⎛ Tbd ⎞ up the ratios the scaled time. = . Solving for LSun ⎜⎝ R Sun ⎟⎠ ⎜⎝ TSun ⎟⎠ Solve: (a) The collapse of the Sun takes only 0.3 ­percent 3 × 10 7 the luminosity the brown dwarf in terms of the = 0.003 of totaloftime. of the Sun’s total lifetime: fraction collapse time = 1 × 1010 luminosity of the Sun and substituting the values, we 7 3 × 10 collapse time = = 0.003 of total time. find that the brown dwarf ­w ill have a luminosity of 1 × 1010 4 1 2 1 ⎞ If we shrink the Sun’s lifetime down to a single day, the Lbd = ⎛ ⎞ ⎛ L ≈ 9 × 10 −6 LSun . It would ⎝ 10 ⎠ ⎝ 5.78 ⎠ Sun collapse time is tcollapse = (0.003)(86,400 s) = 259 s, or take around11,000 brown dwarfs to produce the 4.32 min. luminosity of a star like the Sun. (b) The Sun spends one-­t hird of its protostellar colReview: Brown dwarfs are indeed very faint, which is lapse on the Hayashi track, or 1/3 × 4.32 = 1.44 min. ­ ill be extremely hard to detect directly why they w Review: If the Sun’s lifetime is compressed into a sinaround bright stars without specialized imaging gle day with its birth occurring at midnight, the techniques. entire protostar collapse period is finished by about 12:04:20 a.m.

Using the Web 43. Setup: We ­w ill use the Stefan-­Boltzmann law and set 46. Students w ­ ill visit the APOD site to search for molecup ratios for the protostar and the Sun. ular cloud images. Answers w ­ ill vary.

Solve: Stefan-­Boltzmann law tells us L = 4πR 2σT4. We 47.

Students w ­ ill visit the Spitzer Space Telescope website use ratios and solve for the radius, R, of the protostar and find recent stories about star formation. Answers LPS 200 ­w ill vary.

LSun R 14.14 1 (PS): PS = = 38.4.

48. Updated link: https://­w ww​.­sofia​.­usra​.­edu​/­multimedia​ 2 = 2 = 3500 K TPS R Sun 0.37 /­about​-­sofia​/­sofia​-­aircraft. An infrared telescope in an 5770 K T Sun S ­w ill be high above most of Earth’s blocking airplane 200 LSun 14.14 1 = atmosphere. In addition, the airplane returns a­ fter = 38.4.

2 = 2 3500 K 0.37 ­every observing session, making repairs easier and 5770 K TSun The protostar was a ­little over 38 times larger than the overall costs of maintenance cheaper. ­There have been Sun is now. numerous discoveries as the telescope is still active.

Review: We see clearly h­ ere how large protostars w ­ ere, ­ ill vary.

Answers w and also how a cool. This protostar has some more con49. Students visit the Stardust website, create an account, traction to go through before it is the size of the Sun. and help search for stardust. Answers ­w ill vary.

44. Setup: Brightness scales as 1d 2 . 30 Doradus is 50. Students do a news search for a story about brown 160,000 ly / 1,300 ly = 123 times farther than the dwarfs. Answers ­w ill vary.

Orion Nebula.

Solve: Orion is about 1/123 times the distance to 30 EXPLORATION Doradus. This means that the stars in the 30 Doradus This Exploration was designed to show how the relative would be 1232 = 15,148 times brighter if they ­were presence or absence of the H-­ ion can have a significant located at the distance of the Orion Nebula. We are effect on the heating and cooling, and hence the size, of a given, however, that they appear about 1/6 as bright as star having just the right range of temperatures. the faintest stars vis­i­ble to the naked eye. This means

(

) (

)

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)

Chapter 15

EXPLORATION SOLUTIONS 1. ­There ­were 20 coins used for our exploration experi­ ment.

2. We removed 14 coins (or ions) to hit the 5000 K mark.

That leaves six in the star.

3. With ions removed, heat can escape and the star cools down.

4. We put about 10 coins (or ions) back, for a total of 16 to reach 5000 K. 5. Now that t­ here are more ions, the temperature of the star ­w ill rise.

6. Based on this pattern, the presence of ions heats up the star, at which point the ions are eventually destroyed, and radiation escapes. This ­causes the cooling of the star. When it is cool enough, ions reform, the opacity increases, radiation has a harder time escaping, and the star heats back up. This is a heating and cooling cycle.

7. The spreadsheet w ­ ill contain the following: star heats → ions are destroyed → star loses energy and cools → ions reform → star heats. Note: Obviously, inside the star, t­ here are not 20 coins but rather huge numbers of hydrogen ions, which the coins represent.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES ACTIVITY 11: RANKING TASK FOR FORMATION OF PLANETARY SYSTEMS In this activity students rank the events that take place during star and planet formation. It encompasses concepts

The Interstellar Medium and Star Formation

◆ 119

covered in both Chapters 7 and 15. The goal is for them to visualize the birth of a star and planets in steps that begin from a ­g iant, cool, rotating molecular cloud and end with fully formed planets orbiting a genuine star. The students ­w ill accomplish this by: • correctly ordering the stages of the formation of a planetary system.

• identifying the stages where overlapping of events occurs.

• recounting the modern theory of planetary system formation.

Text reference: Section 15.3 ACTIVITY 20: THE STUFF IN BETWEEN THE STARS In this activity students study images of vari­ous locations in the interstellar medium to learn to distinguish among the reasons an object might look red. Students accomplish this by: • distinguishing among the red emission radiation created in H II regions, interstellar reddening (de-­bluing) caused by dust, and dust clouds emitting infrared light due to thermal radiation.

• explaining how astronomers help us to “see” objects and features that give off light that is outside of the vis­i­ble part of the spectrum.

Text reference: Sections 15.1 and 15.2

CHAPTER 16

Evolution of Low-­Mass Stars

INSTRUCTOR NOTES Chapter 16 covers the stellar evolution, primarily focusing on low-­mass stars. Major topics include: • how stellar mass determines main-­sequence lifetime.

• stages of single low-­mass stellar evolution: the main sequence, subgiant, red g­ iant, horizontal, and asymptotic branches, mass loss, planetary nebula, and white dwarf.

• changes in the core and envelope, and location on the H-­R diagram, as single low-­mass stars evolve from the main sequence to become white dwarfs.

• degeneracy and its effects in stellar cores.

• evolution in close binary systems, including mass transfer, novae, and Type Ia supernovae.

A wonderful resource for stellar evolution is ­http://­rainman​.­a stro​.­i llinois​.­edu​/­ddr​/­stellar​/­beginner​ .­html, which contains a number of movies showing stars of dif­fer­ent mass evolving across the H-­R diagram. The “Advanced Page” at the bottom of the screen lets you create customized movies. You control the masses, metallicities, and duration of the movie and can add in starbursts and control the star-­formation rate. Not only do ­t hese show each star ­going through each of its evolutionary phases, but they also clearly show how H-­R diagrams reveal the age of the cluster. ­These can be interwoven between Chapters 16 and 17, to show the relative timescales of each branch of evolution and to help students visualize how the stars evolve through the H-­R diagram.

DISCUSSION POINTS • Section 16.1 uses the analogy of gas-­g uzzling trucks to describe why high-­mass stars have shorter lifetimes than low-­mass stars. Have students form small teams and discuss ­whether or not this analogy helps them understand main-­sequence lifetimes. Have students suggest equally applicable analogies. (LG 1) • Section 16.2 contains some of the most difficult concepts to understand, including electron degeneracy, degenerate pressure, shell fusion, and the changes in luminosity and temperature that the stars go through as 120

they evolve. Have students identify t­ hose concepts that are the hardest for them to grasp or have them rank the concepts from the hardest to easiest. Students should form groups and work to help each other learn. (LG 2) • Compare and contrast the evolution from main-­ sequence to RGB with that from horizontal branch to AGB. If a star could start fusing carbon, predict how the star would subsequently evolve. (LG 3) • How w ­ ill the evolution of the Sun from its beginning as a protostar to its end as a white dwarf affect Earth?

In par­tic­u ­lar, have students estimate or investigate ­whether the Sun ­w ill swallow Earth when it becomes an AGB star. (LG 3) • Planetary nebulae are formed from stellar winds, and since stars are ­spherical, one might naïvely expect the nebulae to be as well. Have students investigate the morphology of planetary nebulae to test this hypothesis. Hypothesize as to what physical mechanisms could be at work to explain the diversity of planetary nebulae seen in ­those images. (LG 4) • Compare and contrast novae and Type Ia supernovae. ­Will one always occur before the other? Are all binary stars doomed to experience one or both? (LG 5) • This chapter-­opening image shows the Ring Nebula as observed through three dif­fer­ent large telescopes. It is a composite image that combines a narrowband hydrogen image and images of vis­i­ble and infrared light emission.

The question is, “Is this the f­ uture of our Sun?” By now, students have studied the Sun and may have investigated the dif­fer­ent kinds of planetary nebulae as prompted by one of the learning goals. Have students do a short writing exercise where they answer the question.

PROCESS OF SCIENCE At conferences, ­after astronomers have given pre­sen­ta­tions of their research on observational data, we often hear, “We need more data to solve this question.” This Pro­cess of Science gives a good example of how much of science remains unfinished. Pieces of the puzzle get slowly put together through a combination of additional observational and theoretical efforts.

Chapter 16 Evolution of Low-­Mass Stars

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS H-­R Diagram Explorer This interactive simulation was first introduced in Chapter 13. The Exploration prompts students to an even greater understanding of stars. In this animation, students move a position marker across an interactive H-­R diagram. An offset win­dow shows an example of a typical star that would be found in that place, with basic information about luminosity, surface temperature (and color), and the life stage of the star. The animation references modules that students needed to view before viewing this one, such as the one that covers what can be determined from a stellar spectrum. The animation covers the in­de­pen­dent work of Hertzsprung and Russell that led to the development of this diagram.

Text reference: Exploration

TEACHING READING ASTRONOMY NEWS 1. In this context, “young” means a white dwarf that was recently formed, or more precisely, exposed ­because the outer layers of the progenitor star have been blown away.

2. The spectra from the white dwarfs show the emission or absorption patterns corresponding to vari­ous chemical ele­ments; see Chapter 5.2. Each ele­ment has a unique spectral pattern, just like fingerprints.

3. Stars are mostly made of H and He, so one would expect that to make up most of any gas left over on a white dwarf. The presence of significant amounts of metals, ­ ere pre­sent especially iron or silicon suggest they w before the star formed and may have been accreted onto the star’s core if it engulfed its own planets during its g­ iant phases.

4. Far-­U V wavelengths do not penetrate Earth’s atmosphere; a telescope must therefore orbit above the atmosphere to observe at t­ hese wavelengths.

5. In this study, white dwarfs with significant amounts of metals in their atmospheres constitute around one-­ third of all white dwarfs. This is of the same order of magnitude as the number of Sun-­l ike stars with planets (22 ­percent). If a white dwarf needs planets to become enriched with t­hese metals, then it makes sense that the two numbers should be close.

◆ 121

CHECK YOUR UNDERSTANDING SOLUTIONS 16.1. (a) More massive stars have higher core temperatures and therefore fuse their fuel faster.

16.2. (b) See Figure 16.5.

16.3. (a), (c) The energy released by the gravitational collapse of the core and the hydrogen-to-helium shell fusion both contribute to the increase in temperature of the core.

16.4. (a) The “planetary nebula” is a misnomer and has nothing to do with planets; it is the outer layers of the star that have been blown into space and excited by the UV photons from the exposed, hot white dwarf.

William Herschel called ­these ­spherical clouds planetary nebulae b­ ecause they w ­ ere thought to be the material that would form planets.

16.5. (d) To explode, the white dwarf must reach the critical mass of about 1.4 solar masses, which it does by accreting mass from its overgrown companion.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (c, e, f, b, d, a) From shortest lifetime to longest, which is also the ranking from highest to lowest mass in that list. The higher the mass, the shorter the lifetime.

2. (a, e, d, c, f, b, g) This can be confirmed using Figure 16.13, which shows the evolution of a low-­mass star like the Sun. 3. (c) A horizontal path means the star changes surface temperature at a constant luminosity.

4. (a) Degenerate ­matter does not expand when you heat it, and as you increase the mass, the density increases as well so a clump becomes smaller as it grows more massive.

5. (a) Higher mass means more gravitational compression (pressure) in the core, which yields higher temperatures that cause the fuel to be fused at a faster rate.

6. (b) This situation is shown in Figure 16.5 as the red ­g iant branch.

7. (b) The helium flash ends the climb up the red g­ iant branch.

8. (d) A ­g iant star is so large that its surface gravity is very low, allowing the radiation pressure to accelerate material away from the surface.

9. (d) The white dwarf at the center of a planetary nebula emits UV photons that excite the electrons in the surrounding gas, creating emission lines.

10. (c) As an AGB star loses its mass, it peels away the outer layers, exposing the hotter layers under­neath.

122 ◆ Chapter 16 Evolution of Low-­Mass Stars 11. (c) Helium ash from shell fusion falls onto the degenerate core.

12. (c) A Type Ia supernova results from a white dwarf that grew to 1.4 MSun in mass and can be seen billions of light years away, 13. (c) Novae and supernovae make stars grow much brighter, and in general, they appear as new stars in the sky in which the prenovae or pre-­supernovae stars ­were too faint to see.

14. (a) Shell fusion ­ causes the core temperature to increase, which in turn increases the fusion rate of the H-­f using shell, which ­causes the star to become larger and more luminous.

15. (c) The “lower left” of the H-­R diagram is for stars that are very hot and have very low luminosity.

Thinking about the Concepts 16. To fuse helium, a star must have a massive amount of it in its core and the core must be hotter than about 100 million K. With ­today’s chemical makeup of stars, the first is unlikely. More importantly, if a star did have a central temperature hot enough to fuse helium, it would still fuse all the hydrogen that was pre­sent ­because H fusion starts at much lower temperatures.

17. If a main-­sequence star started fusing H at a faster rate, the energy generated would increase. That energy increase would then cause the star to grow in size and become more luminous. The expansion would then cool the core, the fusion rate would drop, and the star would return to its original equilibrium.

18. When a star starts fusing a new ele­ment, it raises its temperature but expands to reach hydrostatic equilibrium; thus, its density drops.

19. The birth mass of the small star in a close binary pair does not determine that star’s ultimate destiny. As its higher-­mass companion evolves into a red g­ iant (or supergiant), it w ­ ill overfill its Roche lobe and transfer mass to the smaller companion. The addition of mass to the smaller companion shifts it higher on the main sequence and changes its evolutionary track.

20. If all Type Ia supernovae ­were shown to be more luminous than previously thought, then our distance estimates to them would increase. We mea­sure their brightness, and since that ­won’t change, the only way that the brightness would remain the same for a higher luminosity would be if the supernovae ­were farther away.

21. Stars do change structure while on the main sequence.

Fusion turns four H into one He, which takes up less space, so the core slowly contracts and heats over the main-­sequence lifetime. Any change in the composition of the core affects the rest of the star.

22. We consider the mass-­luminosity relationship for dim red main-­sequence stars to answer this question: L / LSun = (M / MSun)2.5. (a) Depending on where it happens to be in its orbit, Jupiter is anywhere from 4.2 to ­ ere a 6.2 AU from Earth at any given time. If Jupiter w G5 star, it would be about 60 ­percent as luminous as the Sun, and thus would appear fainter even when at its closest. During Jupiter’s opposition, when it would be at its closest to Earth, we would experience daylight for all 24 hours of the day, but Jupiter’s light would provide only a small p­ ercent of the light that the Sun provided. Thus, the half-­day when light was coming from Jupiter would be equivalent to a deep twilight. The greater concern for life on Earth would be how adding a 0.8 solar mass star at Jupiter’s distance would affect Earth’s current stable orbit. (b) As the Sun evolved into a red g­ iant, it could overfill its Roche lobe. If so, the G5 star might gain some mass through an accretion disk made up of the former outer atmosphere of the Sun. Eventually, the Sun would ­settle into a white dwarf. ­Later, as the G5 star evolved into a red g­ iant, the white dwarf Sun could steal back some hydrogen from the G5 star and form its own accretion disk. If enough hydrogen from the accretion disk landed on the white dwarf ’s hot surface, it could cause the Sun to explode as a nova.

23. Both follow the Hayashi track, but in dif­fer­ent directions. The protostar is approaching the main sequence while the red g­ iant is leaving it.

24. A horizontal branch star is less luminous than when it was a red g­ iant star b­ ecause it has reached another state of equilibrium. Core helium fusion keeps the core of the star puffed up, gravity has become stronger than the outward pressure, and the star shrinks.

25. We assume we are not talking about a very high mass star’s evolution as discussed in Chapter 17. If the core temperature of a star is high enough for the star to begin oxygen fusion ­a fter passing through the carbon-­ to-­helium fusion stage, then we might expect a path similar to that taken by a horizontal branch star.

Depending on the mass of the star, it might increase its luminosity a bit, or not at all and move horizontally to the right on the H-­R diagram. It would have a series of shell fusion events: H to He, He to C, and C to O surrounding its core.

26. A white dwarf is just a cooling blackbody of constant size. Based on the Stefan-­Boltzmann law, as it emits radiation and cools, its temperature ­w ill decrease, and hence its luminosity drops as well.

27. A degenerate gas does not expand when you heat it, so ­there is no safety valve to slow down the nuclear reactions.

Chapter 16 Evolution of Low-­Mass Stars

28. If binary stars merge, the result could form a single star with a more massive hydrogen-­fusing core. It would thus appear as a higher-­mass, main-­sequence star. This is one theory for “blue stragglers.”

29. (a) T Coronae Borealis is a binary system. We know of no other mechanism for generating novae except through mass transfer in a binary system. (b) Hydrogen from the accretion disk falls onto the hot surface of the white dwarf, causing periodic outbursts of uncontained nuclear fusion. (c) As long as ­there is hydrogen streaming from the red g­ iant companion to the accretion disk, and as long as the mass of the infalling hydrogen does not cause the white dwarf to exceed the Chandrasekhar limit and detonate, then novae can recur.

30. Low-­mass stars are also low luminosity. The lower mass means it is easier for a planet to tug the star, yielding larger Doppler shifts in the star’s spectra that would be easier to detect than for a high-­mass star. The lower luminosity means that one could more easily block the star’s light to look for orbiting planets around the star in imaging. A third, impor­tant reason is that lower luminosity stars are more in­ter­est­ing as they live significantly longer than higher mass stars, which could allow for stable environments on exoplanets to create life..

Applying the Concepts 31. Setup: On linear axes, a straight line appears straight, whereas an exponential appears curved. On logarithmic axes, however, a straight line is curved and an exponential is straight. This is ­because the logarithm undoes an exponential.

Solve: In Figure 16.1, the axes are logarithmic, so the line that is drawn is a straight line; however, it does not represent a linear fit, but rather an exponential function. It is drawn that way ­because the range in luminosities is about 1010, and the luminosities depend on the masses raised to a power.

Review: Note that our approximation predicts that some stars ­w ill be a ­little brighter or fainter than ­ ese errors may be a f­ actor of two they r­ eally are. Th or more, so although they are good for instructive purposes, one must be careful when using them in calculations that require high precision. The correct exponent to use depends on the masses of the stars.

32. Setup: Figure 16.2 shows the percentage of the mass pre­sent that is hydrogen or helium as a function of a fraction of the Sun’s radius. The amount of the Sun that is turned from H into He is the area of He in the last panel above 30 ­percent. We ­w ill approximate that by a triangle drawn within a rect-

◆ 123

angle having a base of R/R Sun = 0.3 and a height of 70 ­percent.

Solve: The area of a triangle is 1/2 base × height, or 0.15 × 70 = 10.5 ­percent.

Review: The Sun uses up its core hydrogen during its main-­sequence lifetime, so we expect the amount of hydrogen inside it to drop over time, as we found.

33. Setup: Figure 16.13 shows the evolution of a 1MSun star.

The axes are logarithmic in luminosity, with the current star having a luminosity of 1LSun. To compare the brightness of the star between two stages we need to simply compare the luminosities on the y-­a xis. Note that the question asks us to compare stages of evolution to the top of the g­ iant branch, which we assume means compare the luminosity at the top of the red g­ iant branch to when it was on the main sequence, and compare the luminosity at the top of the asymptotic g­ iant branch to what it was on the horizontal branch as well as on the main sequence.

Solve: Table IM 16.1 Stage

Luminosity

Radius

1 LSun

1 RSun

top red g ­ iant branch

1,500 LSun

100 RSun

horizontal branch

100 LSun

10 RSun

top asymptotic branch

8,000 LSun

200 RSun

main sequence

(a) On the main sequence, the star has a luminosity of 1 LSun, and at the top of the red g­ iant branch the star is about 1,500 times brighter than the main-­sequence star. (b) While on the horizontal branch, the star is around 100 LSun, and at the top of the asymptotic branch, the ­giant star is about 80 times brighter than when it was a horizontal branch star, and about 8,000 times brighter than it was when it was a main-­sequence star.

Review: Stars have a large range of luminosities and radii as they “move” through the vari­ous stages of evolution.

34. Setup: We need temperatures and timescales for each phase shown in the H-­R diagram in Figure 16.13. From the information given in the text, the main sequence lasts about 10 billion years at 6000 K. The red g­ iant branch is about 200 million years at 4000 K. The horizontal branch is about 100 million years at 5000 K. The AGB is about 1 million years at 3500 K. The planetary nebula is about 50,000 years up to 20,000 K. The white dwarf cools for about 10 million years to reach the dot on the white dwarf path around 100,000 K.

124 ◆ Chapter 16 Evolution of Low-­Mass Stars FIGURE IM 16.1

Solve: FIGURE IM 16.1 Review: The stages ­r unning from the top of the AGB through the planetary nebula phase to a white dwarf is just a “blink of an eye” in the evolution of a star.

35. Setup: The mass-­luminosity relationship can be written as L/(LSun) = M/(MSun)3.5 ­because our Sun has one solar mass and one solar luminosity.

Solve: Luminosity scales with mass as M3.5. We find (a) 0.09 LSun, (b) 530 LSun, and (c) 1.7 × 106 LSun. For comparison, ­Table 16.1 gives luminosities of (a) 0.08 LSun, (b) 480 LSun, and (c) 500,000 L sun.

Review: Referring to ­Table 16.1, we see that the luminosities we found are on the same order as a­ ctual stars.

The differences, especially for the 60-­solar-­mass star, are prob­ably due to the fact that the exponent in the mass-­luminosity relationship varies a bit with mass.

36. Setup: The stellar-­lifetime formula can be normali­ zed to the Sun’s life as lifetime as shown in Working 2.5 ⎛ MSun ⎞ 10 It Out 16.1: Lifetime MS = 10 × ⎜ ⎟ . ⎝ MMS ⎠ Solve: (a) 57 billion years (b) 110 million years (c) 360,000 years.

For comparison, T ­ able 16.1 gives approximate lifetimes around (a) 54 billion years for a 0.51 solar-­mass star, (b) 120 million years for a 5.9 solar-­mass star, and (c) 360,000 years for a 60 solar-­mass star.

Review: Our answers are close for the 0.5 and 6 solar-­ mass stars, and exactly the same for the 60 solar-­mass stars.

37. Setup: To solve, we construct a ratio of the escape velocity of the Sun as an asymptotic red g­ iant (AGB) and a main-­sequence star (MS) and solve in terms of the Sun’s current escape velocity, 617 km/s.

Solve: Using the formula for calculating the escape M velocity, vesc = 2G , and rearranging to solve for R the escape velocity of the Sun as an asymptotic branch red ­g iant, we get vAGB = vMS

MAGB

MMS

R MS

RAGB

= 0.7 1 1 200 = 0.059 vMS .

That is 0.059 of its escape velocity value as a main-­ sequence star. The escape velocity as an AGB star ­w ill be 0.059 × 617 km/s = 37 km/s. As escape velocity

Chapter 16 Evolution of Low-­Mass Stars

◆ 125

drops ­because the outer layers are extremely far from (1.33 × 10 −3 )3 A3 the center of the Sun, it is easier for mass to leave it, in size. The period is, P = = = 6.3 × 10 −5 M 0.6 meaning that the Sun’s rate of mass lossA­w3 ill increase.

3.15 × 10 (1.33 × 10 −3 )3 = = 6.3 × 10 −5 yr, in years, and P = 6.3 × 10 −5 yr × P = Review: Remember that the force of M gravity scales0.6 1 yr 3.15 × 10 7 s inversely with size squared (F ∝ 1/R 2); thus, it makes −5 P = 6.3 × 10 yr × = 1.98 × 103 s, in seconds. sense that the strength of gravity at the surface of the 1 yr C 2π R 2π (2 × 105 km) Sun ­w ill drop considerably as it becomes a red and = = 635 km/s.

The orbital speed is v = = asymptotic g­ iant, and with it the escape velocity.

1.98 × 103 s P P 5 C 2π R 2π (2 × 10 km) 38. Setup: If the rate scales as Tn, where n is some number, = 635 km/s. = v= = 1.98 × 103 dwarf s PReview: then an increase of x ­percent in temperature results in a P A white is very small in astronomical change of (1 + x)n in energy generation. We are looking terms, and therefore, we expect that ­matter in an at relative increases with this prob­lem. accretion disk w ­ ill be orbiting rapidly.

Solve: (a) If the temperature increases by 10 ­percent, the 42. Setup: Density, ρ, is mass over volume, so it is inversely rate of hydrogen fusion increases by (1.1)4 = 1.46 times. proportional to the cube of a sphere’s dia­meter, D. We (b) If the temperature increases by 10 ­percent, the solve for the dia­meter of Earth as a white dwarf by takrate of helium fusion increases by (1.1)40 = 45.3 times. ing the ratio of densities between Earth and the white Review: The fusion rate for the proton-­proton cycle is dwarf.

D ρ proportional to T4; that of triple-­a lpha fusion, T40.

Solve: We use the ratio, WD = 3 Earth , and solve DEarth ρ WD Not only higher temperatures but also higher fusion rates are needed to maintain balance in stars as they for what the size of Earth would be if it had the same evolve. density as a white dwarf: 39. Setup: This is a s­ imple d = vt prob­lem where we must ρ DWD = 3 Earth DEarth convert units.

7 3.15 × 10 s ρ WD Solve: d = vt = 20 km/s × 50,000 yr × 7 17 yr 3.15 × 10 3.15 × 10 ss 5,500 kg/m3 km/s× ×50,000 50,000 d d= =vt vt= =2020km/s =3 × 12,700 km = 224 km.

1 ly yryr× × 1 yr years. yr 10 9 kg/m3 × = 3.31 light 12 19.5ly× 10 km 1 ly Review: 224 kilo­meters is about 140 miles, or about lightyears. years. ×× = =3.33.3 light 12 12 km 9.59.5 ×× 1010 km 6 ­percent of the distance from the West Coast to the East Review: A planetary nebula may expand to be about as large as the space between stars.

40. Setup: The star starts at 0.6 MSun and must grow to 1.4 MSun, so it has to gain 0.8 MSun. We must therefore find out how long it takes to gain this much mass at a rate of 10 −9 solar masses per year.

Solve: Mass gained equals rate times time or M = rt.

Solving for the time it would take to transfer 0.8 MSun M = −9 = 8 × 108 yr, or 0.8 MSun is t = r 10 MSun/yr 800 million years. A typical low-­mass main-­sequence star generally lives on the order of 10 billion years.

Review: The fact that mass transfer happens faster than the typical lifetime of a low-­mass star is an impor­tant confirmation of our theory of how Type Ia supernovae happen.

A3 , ­w ill 41. Setup: Kepler’s law in general form, P = M give us the period. To get the velocity, we divide the circumference of the orbit by the period.

Solve: The white dwarf has a mass of 0.6 MSun and the accretion disk is 2 × 105 kilo­meters, or 1.33 × 10 −3 AU

Coast of the United States (half the width of C ­ alifornia).

That is a very small size in which to pack the ­whole Earth.

43. Setup: Density is mass divided by volume. We ­w ill be solving for the radius of the Sun, R Sun, if it ­were ­ atter. made of degenerate m Solve: The text says that degenerate material can have a density of 1,000 kg/cm3 or more. If the Sun’s mass had this density, its size would be M R Sun = 3 4 Sun πρ 3 =3

2 × 1030 kg ≈ 7.8 × 108 cm, ⎛ 4 π ⎞ × (103 kg/cm3) ⎝3 ⎠

or

about 7,800 kilo­meters.

Review: Earth’s radius is 6,378 kilo­meters, so the Sun’s size would be about 20 ­percent larger than Earth if the Sun ­were degenerate.

44. Setup: Write the luminosity law in terms of solar units L/(LSun) = R/(R Sun)2T/(TSun)4. The white dwarf’s radius is given as 107 meters, or 0.014 that of the Sun’s radius.

126 ◆ Chapter 16 Evolution of Low-­Mass Stars Solve: Use the ratios from the Stefan-­Boltzmann law: 2 4 Lwd ⎛ R wd ⎞ ⎛ Twd ⎞ = , and ratios of the white LSun ⎜⎝ R Sun ⎟⎠ ⎜⎝ TSun ⎟⎠ dwarf radius to the radius of the Sun, R Sun 107 m × = 0.014 R Sun .

7 × 108 m 4

⎛ T ⎞ Lwd = (0.014) ⎜ wd ⎟ LSun . ⎝ 5780 K ⎠ 2

(a) L

4

⎛ 108 ⎞ = (0.014)2 ⎜ = 1.8 × 1013 LSun wd ⎝ 5780 K ⎟⎠ 4

106 ⎞ (b) Lwd = (0.014)2 ⎛⎜ L = 1.8 × 105 LSun ⎝ 5780 K ⎟⎠ Sun

tors all types of variable stars through professional and amateur efforts. Answers ­w ill vary.

48. Students go to the University of Washington’s “Properties of Planetary Nebulae” web page and download and complete the lab exercise. Students w ­ ill create their own sketches, t­ ables, and graphs. Answers w ­ ill vary.

49. Students are directed to the Hubble Space Telescope’s planetary nebula gallery to find examples of dif­fer­ent types of symmetry in the nebulae. Answers ­w ill vary.

50. Students look up press releases on planetary nebulae and white dwarf stars in the Hubble telescope news archive. Answers ­w ill vary.

4

⎛ 10 4 ⎞ (c) Lwd = (0.014)2 ⎜ L = 1.8 × 10 −3 LSun ⎝ 5780 K ⎟⎠ Sun 4

⎛ 102 ⎞ (d) Lwd = (0.014)2 ⎜ L = 1.8 × 10 −11 LSun ⎝ 5780 K ⎟⎠ Sun Review: We see that a white dwarf is amazingly bright when it is first exposed but becomes so dim when it cools, that it would be very hard to observe.

45. Setup: For a star to have already evolved to the red g­ iant stage t­ oday and be as old as pos­si­ble, the star would need to have a main-­sequence lifetime just short of the age of the universe. Use T ­ able 16.1 to answer this.

Solve: Looking at the values in the ­table, the star should be between a G5 (lifetime of 12 billion years) and K0 (lifetime 18 billion years) type or 0.92 to 0.79 solar masses.

Review: Remembering that our Sun has a main-­ sequence lifetime around 10 billion years, a star whose ­ ill main-­sequence lifetime is closer to 13 billion years w not be significantly less massive.

Using the Web 46. “The Katzman Automatic Imaging Telescope (KAIT) is an entirely robotic telescope dedicated to the search for supernovae and the monitoring of celestial objects.

It is a 76 cm dia­meter reflecting telescope equipped with a CCD camera and automatic guider.” The search can be automated ­because it involves taking lots of images and comparing new with old. No one knows when or where the next supernova w ­ ill occur, so wide coverage of the night sky is done. Students are then asked to pick a current year and report. Answers ­w ill vary.

47. Students visit the American Association of Variable Star Observers (AAVSO), an organ­ization that moni-

EXPLORATION Students return to the H-­R Diagram Explorer in Chapter 16 to further investigate the changes in luminosity, temperature, and radius as a star evolves. The test star is a twin of the Sun, and the solutions are based on our working with the interactive simulation.

EXPLORATION SOLUTIONS 1. 5800 K 2. One solar luminosity 3. One solar radius 4. As a star moves up and to the right in the H-­R diagram, the test star becomes larger, more luminous, and redder.

5. The test star’s temperature changed to 4900 K. The temperature change made the star became redder (for a blackbody, redder means cooler).

6. The star is now at 3.5 solar luminosities.

7. The star is now at 2.6 solar radii.

8. When a blackbody grows cooler but larger, it is pos­­­si­ble for the change in size to outpace the change in temperature. As such, each “patch” of the star may give off a ­little less energy than before, but ­there is so much more area that the total amount of energy given off is higher, that is, the star has higher luminosity. This demonstrates the Stefan Boltzmann law.

9. The changes in the star’s size and color are even more dramatic than what happened when I played with the star for question 4 ­because the test star moved through a greater range in temperature and luminosity.

10. The size of the star is the most noticeable change to us.

11. As the star was dragged horizontally across the H-­R diagram, it shrank in size and became blue.

Chapter 16 Evolution of Low-­Mass Stars

12. By the time the star reached the far left of the diagram, it was about 3 times the size of the Sun. 13. Dragging the star down the left side took its radius from about 3 times the size of the Sun down to 0.01 times the radius of the Sun. 14. The star is now about 0.01 times the radius. Since the radius ­won’t change much as the white dwarf cools, it ­w ill slowly get dimmer and cooler. (The question

◆ 127

d­ idn’t ask that, but it was enlightening to add that to the simulation.)

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES None for this chapter.

CHAPTER 17

Evolution of High-­Mass Stars

INSTRUCTOR NOTES Chapter 17 covers the life and death of high-­mass stars.

Major topics include: • details of the CNO cycle.

• changes in the core and envelope as single high-­mass stars evolve from the main sequence to become Type II supernovae.

• pulsating variables and the instability strip.

• core-­collapse supernovae.

• the energy and chemical enrichment from supernovae.

• neutron stars and pulsars.

• the study of star clusters as a means for testing theories of stellar evolution.

A wonderful resource for stellar evolution is http://­rainman​.­a stro​.­i llinois​.­edu​/­ddr​/­stellar​/­beginner​ .­html, which contains a number of movies showing stars of dif­fer­ent mass evolving across the H-­R diagram. The “Advanced Page” at the bottom of the screen lets you create customized movies. You control the masses, metallicities, and duration of the movie and can add in starbursts and control the star-­formation rate. Not only do ­these show each star g­ oing through each of its evolutionary phases, but they also show how H-­R diagrams reveal the age of the cluster.

DISCUSSION POINTS • The interior structure of the vari­ous fusion cycles of massive stars is almost always likened to the cross section of a large onion bulb. Have students form small groups and draw the cross section of a large onion bulb. Then have them transfer the information on the structure and fusion cycles of a massive star nearing its end—­Figure 17.4—to their drawing. Is the simile helpful? Figure 16.13 has a similar figure for a one-­ solar-­mass star. Have students sketch the cross section of an asymptotic g­ iant branch star and label the structure and fusion cycles. Brainstorm ­whether ­there is a fruit or vegetable that can be similar to that in structure. (LG 1 and 2) 128

• We are taught that chemical ele­ments heavier than iron cannot be fused as a source of energy, yet supernovae produce them. Discuss how this is pos­si­ble. (LG 3) • Compare and contrast Type Ia and Type II (core collapse) supernovae. ­Under what conditions w ­ ill they occur? How common are they? What remnant (if any) is left ­behind? Are heavy ele­ments created? Which is likely to be responsible for the heavy ele­ments out of which we formed? (LG 3) • Graphs such as that shown in Figure 17.19 tend to confuse students b­ ecause they show so much in one image. Have students discuss how they interpret or what they learn from Figures 17.17, 17.18, and 17.19.

Which figure helps them understand best what we learn from our study of star clusters? Which one helps them understand that t­ hese are plots of star data? (LG 4) • Return to the concepts covered in Chapter 13. What pro­cesses do astronomers need to follow in order to get the visual luminosities and surface temperatures of the individual stars in a cluster? (LG 4) • The chapter-­opening image is a composite image of the Crab Nebula. The Chandra X-­ray image is shown in blue, the Hubble Space Telescope optical image is in red and yellow, and the Spitzer Space Telescope’s infrared image is in purple. The question asked is, “What caused this star to explode?” Where is the once massive star located in this image? Have students do a 3-­minute writing exercise where they tell the story about what caused this star to explode.

PROCESS OF SCIENCE The Pro­cess of Science feature in this chapter covers the history of the discovery of pulsars and the discarding of a much more complicated (and untestable) hypothesis that the radio signals ­were coming from extraterrestrial ­little green men. The simpler conclusion—­a rotating neutron star with strong magnetic fields—­was accepted.

ASTROTOUR ANIMATIONS None for this chapter.

Chapter 17 Evolution of High-­Mass Stars

ASTRONOMY IN ACTION VIDEOS TYPE II SUPERNOVAE Dr. Palen shows how core collapse and rebound ­causes the ejection of the star’s outer layers, with the classic demonstration of dropping two balls on top of each other, making the top ball fly off with enormous energy ­a fter the pair hits the ground. Length: 2:01 Text reference: Section 17.2 PULSAR ROTATION Dr. Palen uses a foam ball and a few flashlights embedded within it to demonstrate the light­house effect of pulsars.

Length: 2:00 Text reference: Section 17.3

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neutron star; anything “more than” becomes a black hole.

2. Astronomers are trying to figure out what caused a “kilonova” that involved the merger of two massive objects and sent out gravitational waves. Two neutron stars? Two black holes? A neutron star and a black hole?

3. Gravitational waves as predicted by Einstein.

4. The main distinction among ­those three types of explosions is the amount of energy released. Ranked from least to most power­f ul: nova, supernova, and kilonova.

Each of t­ hose explosions also involves dif­fer­ent progenitors. Nova is recurrent and involves a white dwarf with an accretion disk. ­There are two types of supernova: Type II (core collapse) and Type Ia (mass transfer onto a white dwarf). Kilonova involve neutron stars and/or black holes in some combination. That is what the researchers want to know.

INTERACTIVE SIMULATIONS

CHECK YOUR UNDERSTANDING

H-­R EXPLORER

17.1. (b, d) High-­mass stars have higher internal temperatures, so they burn fuel faster and use the CNO cycle to burn H to He. 17.2. (b) Having the highest nucleus binding energy, iron ­can’t fuse and give off energy, and that c­ auses the star to implode.

17.3. (c) Neutron stars as pulsars ­were discovered to rotate as fast as 1 millisecond. Something massive that rotated fast and the conservation of angular momentum as it collapsed would explain that observation.

17.4. (b) Stars at the top of the main sequence live for a very short amount of time, so if the main-­sequence turn off for a cluster is at this location, the cluster must be very young. (However, ­there are no “very young” globular clusters in the Milky Way.)

This interactive simulation was first introduced in Chapter 13. The Exploration prompts students to an even greater understanding of stars. In this animation, students move a position marker across an interactive H-­R diagram. An offset win­dow shows an example of a typical star that would be found in that place, with basic information about luminosity, surface temperature (and color), and the life stage of the star. The animation references modules that students needed to view before viewing this one, such as the one that covers what can be determined from a stellar spectrum. The animation covers the in­de­pen­dent work of Hertzsprung and Russell that led to the development of this diagram.

Text reference: Section 17.1 CNO CYCLE The carbon-­nitrogen-­oxygen (CNO) cycle is a more complicated fusion pro­cess than the proton-­proton chain in that more nuclei are involved. This animation prompts students for identifying which nuclei are involved, what colors represent each nucleus, what is created, and what the end result is. Students should be able to recognize how the carbon nucleus acts as a catalyst for this fusion cycle.

Text reference: Exploration

TEACHING READING ASTRONOMY NEWS 1. The astronomers have found the “upper limit” on neutron star mass. Any leftover object a­ fter a supernova that is “less than” 2.16 times the Sun’s mass becomes a

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (c) High-­mass stars burn successively heavier ele­ ments in their cores while continuing to fuse the lighter ele­ments in a shell. This creates an “onion skin” of fusing layers.

2. (e, a, f, b, c, d) See Figure 17.4.

3. (d) The iron formed as the last fusion product in the cores of massive stars is destroyed by photodissociation during the explosion. Thus, iron that exists t­ oday was created in explosive events such as supernovae.

4. (d) A pulsar is a spinning neutron star whose magnetic axis is dif­fer­ent from its spin axis; the magnetic axis acts as a light­house.

130 ◆ Chapter 17 Evolution of High-­Mass Stars 5. (a) As the star moves to the right in the H-­R diagram, it cools off and becomes redder.

6. (a) The star moves to the right at roughly constant luminosity, so it becomes red and gigantic.

7. (e) The Sun is 5 billion years old, so it must lie between panels (e) and (f), which show clusters at 1 and 10 billion years, respectively.

8. (b) High-­mass stars fuse hydrogen via the CNO cycle, low-­mass stars by the proton-­proton chain.

9. (a) The layers of high-­mass stars correspond to which ele­ments are being fused, and heavier ele­ments (that is, higher atomic number) are fused at higher temperatures, meaning closer to the core.

10. (a) Eta Carinae is a very high mass star, and as a result is driving off a huge amount of its material into space, yet another reason it is an extreme example of a star.

11. (b) To fuse two iron nuclei, energy must be absorbed ­because ele­ments heavier than iron are also more massive than their constituents.

12. (a) All core collapses produce supernovae; many ­w ill produce a neutron star.

13. (c) Chinese astronomers recorded the “new star” in 1054; the pulsar was discovered mid-20th ­century; we date the explosion from analy­sis of the velocity of the expanding nebula.

14. (b) Neutron degeneracy pressure holds up a neutron star.

15. (b) Very young stars that are turning off the main sequence ­w ill have very short lifetimes, meaning they are very high mass. ­These are located at the upper left of the H-­R diagram.

Thinking about the Concepts 16. In low-­mass stars, the proton-­proton chain involves only protons, deuterium, and helium-3 to produce a helium-4 nucleus. With a lower repulsive (that is, Coulomb) barrier to overcome, this fusion pro­cess operates most efficiently at the temperatures and pressures in the cores of low-­mass (that is, low-­core-­temperature) stars. The CNO cycle uses 12C as a catalyst to fuse hydrogen into helium. The carbon nucleus collides with a proton to become nitrogen. ­A fter several stages of fusion and radioactive decay, 4He splits from 16O, leaving a 12C nucleus to start the pro­cess again. The CNO cycle generates more energy to support a star, but it requires the high temperatures and pressures in the cores of high-­mass stars to overcome the large Coulomb barrier between carbon and hydrogen. mass stars have difficulty achieving ­ t hese Low-­ conditions.

17. When it begins to fuse helium, a low-­mass star has a degenerate helium core. As such, the core begins

fusing helium in a sudden thermonuclear runaway that ­causes a thermal flash. In the high-­mass star, the core is not degenerate, so the high-­mass star eases into helium fusion. Both fuse helium by the triple-­a lpha pro­cess; however, the way this pro­cess begins is dramatically dif­fer­ent.

18. The core of a high-­mass star is convective and well mixed, plus temperatures are higher, which all oppose degeneracy ­because the nuclei do not become forced too close together.

19. Post-­helium fusion for high-­mass stars becomes shorter for each cycle ­because in each successive fusion cycle, ­there are fewer and fewer nuclei to be fused (numerous lighter nuclei are combined to make a single heavier nucleus). Second, the fusion of heavier nuclei requires much higher energy meaning the collisions happen far more often.

20. Cepheid variables are supergiant stars that are vis­i­ble over intergalactic distances. B ­ ecause the Cepheid has a regular period, it is easy to detect even at large distances by watching the brightness variations of stars in a galaxy. The period of a Cepheid’s variability is directly related to its luminosity, so all we need to do is mea­sure the period over which its brightness fluctuates. Once the luminosity of the Cepheid is known, we can use the distance-­luminosity relationship to accurately determine the star’s distance. Over intergalactic scales, the distance to the Cepheid is essentially the same as the distance to its host galaxy.

21. Supernovae influence new stars by synthesizing nuclei that are heavier than iron and seeding the clouds in the interstellar medium with t­ hese heavy ele­ments. When new generations of stars and their planets are formed, they incorporate the heavy ele­ments into their compositions. Furthermore, the shock wave from the expanding supernova b­ ubble can compress interstellar clouds, triggering their collapse into new stellar systems.

22. The neutron star explanation is simpler ­because it was a testable hypothesis. Theoretical predictions ­were made that ­were then tested by observations and confirmed. The theory survived multiple tests. The existence of radio-­transmitting l­ittle green men would have been difficult to test.

23. As a luminous blue variable, Eta Carinae w ­ ill form a Type II supernova that w ­ ill be bright enough to observe during daylight. Maybe not as bright as a full moon, but close.

24. The neutrinos from SN 1987A escaped directly from the core, whereas the light from the shock wave from the core collapse took a few hours to break out of the envelope.

Chapter 17 Evolution of High-­Mass Stars

25. The farther material falls, the higher the kinetic energy grows. ­Because neutron stars are much smaller than white dwarfs (city-­ sized rather than Earth-­ sized), inward-­falling material has more distance to fall and gain speed before reaching the accretion disk.

The kinetic energy of the inward-­falling material varies as the square of its speed, so even a modest increase in speed translates to a tremendous increase in kinetic energy.

26. The ejecta from a supernova may start out traveling at 10 ­percent the speed of light, but that material loses energy as it blows into low-­density interstellar gas and dust, and thus the ejecta slow down.

27. Astronomers assume that all stars in a cluster formed at approximately the same time, from the same molecular cloud, and are at the same distance from us.

Based on our observations of star-­forming regions, this appears to be a reasonable assumption. We know that high-­mass stars evolve much more quickly than low-­mass stars. This means that when we plot all of the stars in a cluster on an H-­R diagram, t­ here ­w ill be some “turnoff point” on the main sequence at which stars are beginning to evolve into ­g iants. A young cluster ­w ill still have its high-­mass stars residing on the main sequence, but an old cluster w ­ ill not. In a very old cluster, even some of the lower-­mass stars ­w ill have evolved into subgiants and red g­ iants. Thus, by noting the point at which its stars are turning off the main sequence, astronomers can estimate the age of a cluster.

28. The Big Bang produced only hydrogen and helium, yet even the oldest stars that we see show traces of heavier ele­ments in their atmospheres, which must have come from a previous generation of stars.

29. Binding energy refers to the amount of energy needed to break a nucleus into its components. One can also think of it as the ability of the strong nuclear force to overcome the electrostatic repulsion between atomic nuclei. Higher binding energies require more effort to break nuclear bonds, but ­those bonds also release more energy when broken. Therefore, when lighter ele­ments are fused into heavier ele­ments (up to iron), the lost mass is converted to energy via Einstein’s equation: E = mc2 . The difference in binding energies between the beginning and end products of a reaction is the energy given off in that reaction.

30. All the ele­ments that are heavier than helium w ­ ere made inside stars. Although AGB stars create some C ­ ere made and O, most of the “metals” in the universe w in supernovae and distributed into space in their ejecta. Thus, most e­ very atom on Earth that is heavier than helium came from a stellar explosion.

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Applying the Concepts 31. Setup: Figure 17.2 shows how much of the star is composed of hydrogen or helium at dif­fer­ent radii.

The core is demarcated by an arrow out to about 33 ­percent of the total radius on the x-­a xis. The fraction of the core and the rest of the star that is helium can be read off of the y-­a xis.

Solve: At t = 0 years, the helium content of the entire star is 30 ­percent. At t = 7 million years, the core is 100 ­percent helium, while the rest of the star, about 67 ­percent, remains at 30 ­percent He. Figuring out the fraction of the star that is helium involves working with volumes. Let the volume of the star be 1. The relative volume of the core is 0.333 or 0.036 and the volume of the star outside the core is (13 − 0.333) = 0.96. We now consider the fractional helium content in each part: (0.036 × 1) + (0.96 × 0.3) = 0.32. The fractional helium content of the star increased by 2 ­percent over 7 million years.

Review: The main sequence ends when the core has run out of hydrogen to fuse; therefore, this figure shows us that the main-­sequence lifetime of a 25 MSun star is about 7 million years. This is short compared to the 10 billion years for our Sun. 32. Setup: To gauge how a star compares to the Sun, remember that the Sun has values of one solar mass or solar radius and a surface temperature of 5,780 K. The uppermost main-­sequence star is the l­ ittle blue dot at top left.

Solve: The uppermost main-­sequence star has a luminosity of about 2 × 105 times more than the luminosity of the Sun. Its radius is about 10 R Sun. Its temperature is about 37,000 K, which is 37,000 K / 5780 K = 6.4 times hotter than the Sun. Review: This star is about a ­factor of 10 larger and over 6 hotter, but 200,000 times more luminous. ­These stars burn through their fuel quickly, and die fast.

33. Setup: The radius of the core of a typical star takes up about 25 ­percent to 33 ­percent of the star’s interior.

To judge ­whether or not each star is to scale, compare the size of the core to the ­whole star, and read the figure caption.

Solve: For the first three stars shown, the core takes up about 25 ­percent to 33 ­percent of the radius of the star, which is about what we expect, and thus ­these are to scale. In the bottom star, the core takes up about half of the radius of the star. The scale below indicates 1,000 R Sun to the edge of the star, with a core radius of 0.01 R Sun. The bottom image is not to scale.

Review: Providing the interior details of an onionskin model, such as that shown in the bottom star in

132 ◆ Chapter 17 Evolution of High-­Mass Stars Figure 17.4, is hard to show to scale to the star’s ­actual M vesc = 2G NS size, which is why the core was exaggerated for the R NS highly evolved, massive star.

2.8(2 × 1030 kg) 34. Setup: Use d = vt where v = 1,500 km/s and Solve: = 2(6.67 × 10 −11 m 3/kg ⋅ s2 ) 10 4 m t = 2,019–1,054 = 965 yr.

8 = 2.73 × 10 m/s 3 × 10 7 s = 2.9 × 1010 s. This gives Solve: t = 965 yr × Review: This is 90 p­ ercent the speed of light.

1 yr 39.

Setup: This prob­lem gives us all the information we the size of the nebula as need in terms of solar masses and solar luminosities.

10 13 d = vt = 1,500 km/s × 2.9 × 10 s = 4.3 × 10 km.

We need to construct the ratios necessary to answer Review: 4.3 × 1013 km is about 4.6 light years. If the the questions. neutron star of the Crab Nebula ­were located at the Solve: (a) The total luminosity of the 50,000 average-­ Sun, the nebula would extend past our nearest stars. mass stars, 0.5 MSun, is the number of stars times the 35. Setup: Figure 17.6b shows the period-­ luminosity luminosity per star (0.08 LSun), which yields 4,000 relationship for a classical Cepheid. To determine the . A single 20 MSun star puts out 10,000 LSun, or 2.5 L Sun luminosity, simply find the point on the top line that times the luminosity of 50,000, 0.5 MSun stars. matches the period. (b) The total mass in average-­mass stars is the number Solve: For a period of 10 days, the luminosity is of stars, 50,000, times the average mass of each star, approximately 4,000 LSun. However, a distance can0.5 MSun, which yields 25,000 MSun. This is 1,250 not be determined to this star ­until we also know the more mass than a single 20 MSun star. star’s apparent brightness. We can then use the equa(c) Based on t­hese calculations, average-­mass stars L contain more stellar mass in the galaxy; however, the , where b is the brightness and d is the tion d = 4π b high-­mass stars produce most of the light. distance.

Review: We know that low-­mass stars are long-­lived Review: It is easy to mea­sure the average brightness and much more common, so it makes sense that they of a Cepheid by observing it regularly over its cycle. would make up most of the stellar mass of a galaxy.

We si­mul­ta­neously get its period, which then lets us Given that luminosity scales so strongly with mass determine its distance.

(M3.5), it also makes sense that massive stars ­w ill make 36. Setup: Linear plots increase by single units whereas up most of the light output. log plots increase by powers of 10.

40. Setup: This prob­lem is an exercise in unit conversion.

Solve: Figure 17.13 is a logarithmic plot. Oxygen is ­ ill use 2 × 1030 kg for one solar mass, and the fact We w found at about 10-3, number of atoms relative to hydrothat ­there are 5.26 × 105 min in one year. gen. The abundance of oxygen is 0.001 that of hydro0.1 MSun 2 × 1030 kg 1 yr gen, or ­there is 1 oxygen atom for ­every 1,000 hydrogen Solve: (a) × × = 3.8 × 1023 k 5 5.26 × 10 M 1 yr min Sun atoms.

0.1 MSun 2 × 1030 kg 1 yr 23 = 3.8 × 10 kg Review: We expect ­every ele­ment to have ×abundances × 5.26 × 105 min M 1 yr much less than hydrogen and helium ­because Sun they 3.8 × 1023 kg (b) = 5.2 Eta Carinae was losing 5.2 ­were made in stars a­ fter the first two ­were created.

7.35 × 1022 kg GM 37. Setup: Math Tools 17.2 tells us that gravity g = 2 .

“Moon masses” ­every minute during that outburst.

R Review: The overall mass-­loss rate of Eta Carinae is For this neutron star, M NS = 2.8MSun and the radius is truly huge compared to most stars.

10 km.

41. Setup: Mass-­loss rate is given by (mass lost)/(time).

GMNS (2.8)(2 × 1030 kg) Solve: A 25MSun star that loses 20 p­ ercent of its mass Solve: g = R 2 = 6.67 × 10−11 m 3/kg ⋅ s2 (10 4 m)2 NS loses a total of 5MSun. If it loses this over 7 million = 3.74 × 1012 m/s2 years, the mass loss rate is Review: This is almost 400 billion times stronger 5 MSun than what we have on Earth. = 7.14 × 10 −7 M Sun /yr.

7 × 106 yr 38. Setup: Math Tools 17.2 tells us that escape velocity M . For this neutron star, M NS = 2.8MSun R and the radius is 10 km. vesc = 2G

Review: This shows how extreme the mass loss rate is for eta Carinae. The fraction of its total mass lost for the Sun is around 10 -13 MSun / yr.

Chapter 17 Evolution of High-­Mass Stars

42. Setup: Using the period-­ luminosity relationship L/(LSun) = 335 P, we find the luminosity of Delta Cephei is 1,809 LSun and the more distant one is 18,090 LSun. Using our equation for parallax-­ d istance, Delta Cephei is at d = 1 p = 1 0.0033 = 303 pc. To find the distance to the more distant Cepheid, we use the L inverse-­square law for brightness, b ∝ 2 , to solve for d the distance in terms of the relative luminosities and brightness.

Solve: (a) In this case, the more distant star is 10 times more luminous and 0.001 times as bright, so solving for distance, we find that the more distant star is

d2 = d1

L2 b1 = L1 b2

10 = 100 times farther away, 0.001

or 30,300 parsecs. (b) The more distant star is more than 30 times farther away than our limit for mea­sur­ing parallaxes of stars, and thus we cannot use parallax to mea­sure its distance.

Review: In this prob­lem, we computed two rungs of the distance ladder: first we found the distance to a star via its mea­sured parallax and its luminosity via the period-­luminosity relation, which confirmed what we would have calculated just using its brightness and parallax.

Having the a­ ctual distance to a variable star, we can then use the period-­luminosity relation to find the luminosity of a more distant star—­a star well beyond our parallax limit—­and given its mea­sured brightness, determine its distance.

43. Setup: This prob­lem refers to orbits, which we usually solve using Kepler’s third law. However, note that the co-­rotating satellite must complete 30 orbits of the neutron star ­every second. In other words, it travels the circumference of its orbit 30 times each second, where circumference is C = 2πR and we know the speed is c, the speed of light.

Solve: If it is traveling at the speed of light, the circumference of the co-­rotating satellite is, 300,000 km/s C = cω = = 10,000 km. ­ Because the 30 rev/s circumference C = 2πR, we find the radius of the orbit C 10,000 km = = 1,590 km. This is almost to be r = 2π 6.2831 160 times the radius of a pulsar (10 km).

Review: This radius is impor­tant, for the following reason. The magnetic field of the pulsar rotates at the same rate as the pulsar itself, and particles travel on

◆ 133

­those field lines. Beyond about 1,600 kilo­meters, the particle would start rotating faster than light. However, this does not happen. The magnetic field would be much weaker and would prob­ably spiral much like the magnetic field of the Sun does as it extends out into the Solar System.

44. Setup: We w ­ ill use the equations for the density and volume of a sphere. Combining the equations and solving for the radius, gives R = 3 4

M

πρ

.

3 Solve: The mass of Earth is about 6 × 1024 kg, and the density of a neutron star is listed in the text as 1018 kg/m3.

Using our formula, the equivalent radius of Earth would be 3 M 3 (6 × 1024 kg) R=3 =3 = 113 m.

4π ρ 4π (1018 kg/m3) Now, 113 meters is about 124 yards. A football field includes 100 yards of playing field, and 20 yards of end zones, and that is pretty close to what the radius of Earth would be.

Review: Hard to imagine, but it emphasizes how ­really dense a neutron star is.

45. Setup: We are given the density of a neutron star is about 1018 kg/m3. Use the setup from prob­lem 44 and M = 1 MSun to find the radius of the object.

Solve: R = 3

3 M 3 (2 × 1030 kg) =3 = 7,815 m, 4π ρ 4π (1018 kg/m3 )

or about 8 km.

Review: A neutron star with the mass of the Sun ­w ill be smaller than a “normal” neutron star having a radius of 10 km. Or­ga­nized races in the United States have lengths of 5K, 10K, and 12K. Fast runners could run the dia­meter of the Sun-­as-­a-­neutron star in about 30 minutes.

Using the Web 46. Students go to the Chandra X-­ray Observatory’s “Variable Stars” web page and do the two activities on variable stars. Students must estimate the magnitudes of a star in each activity based on the stated mag­ nitudes of other stars in the field for 18 images for Activity 1 and 7 images in Activity 2, where the standard stars and the target star are not marked, but must be identified each time from the marked first image.

Answers w ­ ill vary.

47. Students go to the International Astronomical Union’s “List of 2015 Supernovae” and pick a few to analyze.

Answers w ­ ill vary.

134 ◆ Chapter 17 Evolution of High-­Mass Stars 48. The Zwicky Transient Fa­cil­i­t y is a time-­domain survey that ­w ill use a camera with a large, 47 square degrees, field of view. It ­w ill scan more than 3,750 square degrees an hour to discover young supernovae nightly as well as search for rare and exotic transients.

How does it improve on the Palomar Transient Factory? It has a larger field of view and faster readout electronics, an order of magnitude faster than the Palomar Transient Factory. Scientists hope to learn how and why stars explode as SNe supernovae; what the progenitors of t­ hese types are; what the characteristics are of the circumstellar material. As of autumn 2018, no science had been reported yet. The survey phase began in February 2018, so data may have still been in the analy­sis stage.

49. The GAIA mission contributing to the study of variable stars ­w ill survey more than 1.6 billion sources down to ~21 magnitude. It ­w ill focus on pulsating, eruptive, and cataclysmic variables, as well as on stars exhibiting variability due to rotation and eclipses (information obtained from https://­arxiv​.­org​/­abs​/­1804​.­09382). It contributes to the study of novae and supernovae ­because it goes over the same patches of sky again and again and can detect changes, especially of objects that ­were not ­there before or that had dramatically increased in brightness. As of autumn, 2018, ­there ­were quite a few results available, so answers w ­ ill vary.

50. Students are instructed to join the “Einstein@Home” proj­ect, create an account, and join the search for gravitational waves from spinning compact objects. ­ ill vary.

Answers w

EXPLORATION This exploration goes into more details about the complexity in the steps of the CNO fusion cycle.

EXPLORATION SOLUTIONS 1. 12C and hydrogen (1H) are involved in the first reaction.

2. The proton (hydrogen) is represented by blue.

3. The blue squiggle is a gamma ray emitted from the reaction.

4. This first reaction creates the nitrogen isotope, 13N. 5. The resulting nucleus has gained a proton through fusion, and ­because the number of protons in a nucleus defines the atom, this nucleus has changed from carbon (six protons) to nitrogen (seven protons).

6. The yellow ball that appears is a spontaneous decay.

7. The yellow ball represents a positron.

8. The dashed line represents a neutrino.

9. The resulting nucleus has had a proton inside it converted (by positron decay) into a neutron. Thus, the 13 N becomes a 13C. 10. As each proton comes into the nucleus, it gains one proton, raising the mass number (the number of nucleons) by one and turning it into the next higher ele­ment on the periodic t­ able.

11. During t­ hese collisions, gamma rays are emitted.

12. The transition from 15O to 15N is a spontaneous decay.

13. This mimics the decay pro­cess when 13N decayed to 13 C. 14. The nucleus that began this reaction was 12C. 15. The top red ball has 12 nucleons while four came into the pro­cess, so the bottom nucleus has four nucleons.

16. The bottom red ball has two protons.

17. The bottom red ball has two neutrons.

18. The bottom red ball is 4He. 19. The net reaction is 41H → 4He. 20. In this pro­cess, the 12C simply acts as the catalyst, or intermediary, ­because it is returned back to its original state and cycles through another reaction.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 21: INVESTIGATING THE CRAB NEBULA AND PULSAR Students investigate a supernova remnant, finding the speed at which a feature embedded in this explosion is traveling outward. They also: • confirm which images had to be taken from a space-­ based observatory • hypothesize as to why the images taken at dif­fer­ent wavelengths have dif­fer­ent angular sizes (images have the same scale).

• calculate the speed of a distinctive “wave” from the pulsar and compare it with the speed of light.

Text reference: Section 17.3 ACTIVITY 22: DETERMINING THE AGES OF STAR CLUSTERS In this activity, students find the ages of star clusters (and the stars in them) by considering the rate at which stars fuse hydrogen to helium and by studying the color-­ magnitude diagrams of clusters. ­A fter completing the activity, students are able to:

Chapter 17 Evolution of High-­Mass Stars

• state the relationship between the color of a ­m ain-­sequence star and its main-­sequence lifetime.

• list the steps for determining the age of a cluster given the cluster’s “turnoff ” B-­V color.

• determine cluster ages using their color-­magnitude diagrams.

• summarize the complete pro­cess of how astronomers estimate the ages of star clusters.

Text reference: Section 17.4

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CHAPTER 18

Relativity and Black Holes

INSTRUCTOR NOTES Chapter 18 covers special and general relativity. Major topics include: • relative motion and inertial frames of reference, including aberration of starlight.

• special relativity: length contraction, time dilation, and relativistic mass.

• emphasis on the twin paradox.

• general relativity: mass distorts spacetime, orbital precession, gravitational lensing, gravitational redshift, and gravitational time dilation.

• equivalence princi­ple.

• black holes: formation, event horizon, spaghettification, and Hawking radiation.

DISCUSSION POINTS • In a think-­pair-­share activity, have students list as many instances as pos­si­ble where their motion—­ walking, biking, driving, snowboarding—­a ffected the observed velocity of objects near them. (LG 1) • Do relativistic effects make intuitive sense or do they run c­ ounter to our expectations? Is that an acceptable part of science? (LG 2) • We use the equations of Einstein’s theory of special relativity ­every day. Research what the global positioning system (GPS) must consider in order to work effectively.

How many uses of GPS are critical t­ oday? (LG 2) • Work with Newton’s formulation of gravity and Einstein’s consequence of mass distorting space and time.

How do the descriptions of gravity differ? Do they result in identical observations and mea­sure­ments?

(LG 3) • Einstein’s general theory of relativity predicts the existence of gravitational waves, something that was not confirmed u­ ntil very recently. How are gravitational waves created? Does Newton’s law of gravity lead to any such prediction? (LG 3) • Create a set of cards that list the stages of a 50-­solar-­mass star from its time on the main sequence ­until its collapse into a black hole. For a challenge, mix 136

your cards and exchange them with other students; then sort the sets into accurate timelines. (LG 4) • Sketch your vision of what a stellar black hole and its environment would look like. Include key properties and what we would observe from thousands of light-­ years away. (LG 4) • The chapter-­opening image asks, “How are space and time changed near a black hole?” Review ­those parts of the movie Interstellar and evaluate the film’s answer to this question. Is the movie accurate in what it shows?

PROCESS OF SCIENCE Newton’s law of gravity was able to describe how gravity acts and make accurate predictions. In the early 1900s theories started considering limits, both very large (distance, mass) and very small (quantum theory). General relativity has conclusively encompassed the extreme limits and all in between, as it must in order to be complete.

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Einstein’s theory of general relativity predicted gravitational waves, and that t­ hose waves would not experience dispersion as light does when traveling between regions of dif­fer­ent material. This latest detection came from black hole mergers extremely far away, and no dispersion effects w ­ ere found.

Chapter 18

2. What was surprising about the masses of the black holes was that they w ­ ere larger than 20 solar masses each.

3. This observation was good enough that the directions of the spins of the black holes could be inferred. The spin of one of the black holes in the former pair may not have been aligned with its orbital motion.

4. Answers ­w ill vary as detections of gravitational waves continue. For example, on November 15, 2017, LIGO Scientific Collaboration announced the observation of another binary black hole coalescence. The gravitational waves ­were observed by the twin LIGO detectors on June 8, 2017. This was the lightest black hole binary observed so far, with component masses 12 and 7 times the mass of the sun.

CHECK YOUR UNDERSTANDING SOLUTIONS 18.1. (d) The speed of light is constant for all observers.

18.2. (b) In each rest frame, the clocks ­w ill appear to run normally as mea­sured by the person in that rest frame.

However, you w ­ ill mea­ sure your friend’s clock ­r unning slowly as your friend speeds by.

18.3. (d) Mass bends space.

18.4. (c) The event horizon of a black hole scales directly with mass.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (c, a, b) White dwarfs come from low-­mass stars (0.5 − 3 MSun). Neutron stars come from medium-­ mass stars (3 − 8 MSun). Black holes come from the most massive stars (more than 8 MSun). (Used information from Chapter 16, Section 16.1. Exceptions ­w ill occur depending on the amount of mass any given star loses over its lifetime.)

2. (c) You see the fly moving at 50 km//h−7 km/h = 43 km/h.

3. (d) The speed of light is the same for all observers.

4. (c) From the observer’s (your) rest frame, the moving clock runs slowly. This means that the clock in the spaceship ­w ill have ticked by fewer than 10 seconds.

5. (c) Moving objects appear shorter in their direction of motion than they ­really are (length contraction).

6. (d) When in orbit, astronauts feel no gravity b­ ecause they are in constant f­ ree fall.

7. (d) General relativity explains gravity on all mass scales and predicts Newton’s laws when the gravita-

Relativity and Black Holes

◆ 137

tional field is small. Thus, it does not replace Newtonian physics, but it does explain it within a larger context. Thus, we also learn why Newton’s laws work, not just what they are.

2GM 8. (b) The Schwarzschild radius is given as R S = 2 c and increases linearly with mass.

9. (c) A neutron star is held up by neutron degeneracy, which can be overcome by the inward force of gravity if the mass of the neutron star is sufficiently high.

10. (a) ­W hether one is moving at classical or relativistic speeds, relative motion can be detected, provided the detector is sufficiently sensitive.

11. (c) The speed of light is exactly the same for all observers.

12. (d) Energetic protons are harder to “push around” ­because they have more inertia than a proton at rest.

Thus, they more strongly attract other masses through gravity.

13. (d) Special relativity tells us that no massive object can travel at or faster than the speed of light. Figure 18.3 shows us the ships w ­ ill actually see each other approaching at 0.8c.

14. (c) As described for the X-­ray source Cygnus X-1, the star HD 226868 is orbiting an unseen massive object.

The X-­ray emission comes from material falling onto an accretion disk.

15. (c, d, f) We see jets or jet-­l ike streams of material from pulsars, protostars, and planetary nebulae.

Thinking about the Concepts 16. Absent other information about the object of interest, the astronomer would not be able to tell the difference between a gravitational redshift and a recessional redshift. Both produce the same Doppler redshift in the object’s spectrum.

17. Both observers (you looking out the back of the ship, and a person on a slowly moving planet) ­w ill see the light traveling at exactly c, ­because the speed of light is the same for all observers.

18. A photon is a massless particle, and all massless particles can travel at the speed of light in a vacuum.

19. Twin A returns younger than twin B but not ­because of time travel. Rather, twin A aged at a slower rate than twin B b­ ecause twin A accelerated and decelerated from moving close to the speed of light. Twin A could never return before twin B was born. This would involve ­going backward in time and time flows only in the forward direction.

20. It is pos­si­ble for temporal order to be changed or even reversed, depending on the frames of reference. Consider a person standing on a platform, with two flashbulbs on ­either end, and an observer standing fairly

138 ◆ Chapter 18 Relativity and Black Holes FIGURE IM 18.1 B

Person on the platform.

A

Observer

far away (Figure IM 18.1). Imagine that initially the platform is not moving and the bulbs A and B flash si­mul­ta­neously. The light from A and B ­w ill reach the observer at the same time. Light A ­w ill reach the person on the platform in about 1/3 of the time that the light from B does, ­because of the difference in distances to them. Now imagine the platform is moving at a substantial fraction of the speed of light to the right.

At the instant that the ­middle of the platform passes the observer, both bulbs flash. The person on the plat­ ill still see the light from A before that of B form w ­because they are in the same reference frame (with the light from A still arriving in about 1/3 the time as the light from B), but the observer w ­ ill see the light from B before the light from A. This is ­because the platform is moving and the light from flashbulb A ­w ill have a ­longer distance to travel to the observer than the light from flashbulb B. 21. Moving clocks run slowly, so the person’s pulse rate ­w ill appear extremely slow to you.

22. ­Because objects moving at a substantial fraction of the speed of light undergo length contraction from an observer’s viewpoint, the density of the person in the spaceship ­w ill appear to increase. For the person in the spaceship, every­thing seems normal.

23. The astronaut is moving with the ship and is in the same reference frame, thus he w ­ ill see no length contraction. An outside viewer who is not moving at that high speed would observe length contraction.

24. Based on the fundamentals of length contraction, one knows that moving objects appear shorter to someone at rest. This leads to the answer, “If the ball appears round to you (who is at rest), then it must be elongated (that is, an oval-­or rugby-­shaped ball) on the ship.” However, soccer balls appear to be the exception. “A soccer ball is a remarkable exception: a freak of nature, it always appears to be round, no ­matter how fast it moves and in which direction.” https://­w ww​.­s pace​t imetravel​.­org​/­f ussball​/­f ussball​ .­html

25. Moving objects appear shorter in their direction of motion. If the meter stick still appears to be 1 meter long to you, then it must be oriented perpendicular to its direction of motion.

26. It is a common misconception that black holes are g iant “vacuum cleaners” that suck up every­ ­ thing around them. ­There is no reason for anyone to be concerned about a 3-­MSun black hole located a few light-­ years from Earth due to the star’s tremendous gravitational pull. However, astronomers might be concerned about the black hole perturbing the orbits of the Oort Cloud comets, sending numbers of them to the inner solar system.

27. If a star fell into a black hole, it would become increasingly redshifted by gravity. We would see it appear redder and redder, moving into the infrared, micro­ wave, and radio parts of the spectrum as it approached the event horizon. Eventually the light would be infinitely redshifted and no longer vis­i­ble as it reached the event horizon. (We have neglected to include that the star would be “spaghettified” as it approached the event horizon.)

28. The effects of relativity show themselves only in extreme circumstances, such as traveling close to the speed of light or being right next to a black hole or neutron star. In our common lives, special and general relativity are equivalent to the laws of physics that we encounter on a daily basis.

29. It is exceedingly unlikely that we w ­ ill ever have faster-­ than-­l ight travel b­ ecause special relativity tells us that nothing can travel faster than the speed of light. An object traveling at the speed of light would have an infinite amount of inertia and it would take an infinite ­ oing beyond amount of energy to get to that speed. G it is currently impossible. NASA, however, has started theoretical work on ways to circumvent the restrictions of special relativity through the warping of space.

30. If a nearby gamma-­ray burst ­were directed ­toward us, the burst of high-­energy photons could wreak havoc on Earth. Nitrogen oxides created would cool Earth.

The gamma-­ray radiation would ionize the atmosphere, destroy much of the ozone layer, and irradiate the Earth with deadly radiation. Photosynthesis of phytoplankton would be disrupted and the entire biosphere be put off-­balance.

Applying the Concepts 31. Setup: Figure 18.1b of the text book shows that when the car is moving, rain seems to approach the car, but when the car is still, the rain falls vertically. If we are to mea­sure the speed of the rain, we have to determine

Chapter 18

­whether this a­ ngle depends only on the speed of the car, or also on the speed of the rain. A thought experiment would be: What would the situation look like if the rain ­were ­stopped? The right-hand panel of Figure 18.1b shows that the rain would move horizontally toward us at the speed on our speedometer. As the rain falls faster, but we stay at constant speed, it w ­ ill appear to fall more vertically. This means ­there must be a way to determine the speed of the rain.

Solve: Suppose the rain falls at speed v and we move to the right at speed u; the ­angle θ at which it appears v to fall ­w ill be given by tan θ = . The top triangle of u Figure IM 18.2 is an isosceles right triangle. This means tan θ = 1 and θ = 45°, when we are moving at the same speed as the rain. The bottom triangle of Figure IM 18.2 shows the situation when the car has the same velocity, but the rain is falling at a higher speed. The angle θ has increased, implying that there could be a way to determine the speed of the falling rain.

Review: If the rain is not falling and we are moving, ­ ill appear to come right at us at θ = 0°. Maththe rain w ematically, for this situation, v = 0, so tan θ = 0°. Now, if instead we are ­stopped, we expect θ = 90°; being ­stopped means u = 0 and tan θ = ∞. This is identical to the situation if the rain were to have an extremely high speed as it fell; that is, as the speed of the rain increased the angle θ would approach 90°. Note, though, this is a theoretical argument because it assumes that rain is not limited to the constant acceleration due to gravity (9.81 m/s2) nor to the braking action of the atmosphere where it would reach a terminal velocity that depended on air temperature and pressure.

32. Setup: Figure 18.2b shows how rainfall appears aberrated to d­ rivers, whereas Figure 18.2 of the text book shows the same effect for starlight. Follow the arguments in the solution to question 31.

Solve: Light travels at speed c, and we move to the right at speed u, thus the a­ ngle θ at which the light c appears to fall ­will be given by tan θ = u . You would need a pro­cess for mea­sur­i ng the ­a ngle of the aberration of starlight, most likely using an optical telescope and high precision sensors and mea­sur­ing devices.

Review: See question 31. The aberration of starlight varies over the course of a year because the speed of Earth in its orbit varies. The effect is small, about 40 arc seconds over the course of a year.

33. Setup: Einstein’s formula says E = mc2 such that energy can be turned into mass.

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◆ 139

FIGURE IM 18.2

Solve: ­Because higher energy can translate into higher mass, a hot cup of coffee has more thermal energy and thus more energy that could be transformed into mass than, say, a cup of iced coffee. However, the difference in weights would be infinitesimally small.

Review: Any object with more energy has the potential to gain mass, such as a wound-up watch that uses a small spring. The energy comes from the person who did the winding.

34. Setup: In Figure 18.5, time is shown on the bottom and speed on the left as a fraction of c, so 0.6 means 0.6c. The purple line shows Newtonian (that is, incorrect) speed, whereas the pink line accounts for relativity. To convert c to km/h, recall the speed of light is 1.08 × 109 km/h.

Solve: (a) The two curves, and thus Newtonian and relativistic effects, begin to significantly differ at about 0.33c to 0.5c. (b) 0.33c is 0.33 × 1.08 × 109 km/h = 3.6 × 108 km/h. Typical ­human speeds of travel are in the hundreds of km/h (100 km/h on the highway, a few hundred in an airplane). We do not experience the effects of relativity ­because we never travel fast enough.

Review: In general, one can ignore the effects of relativity if the speed is u­ nder 0.5c, as shown in this figure.

35. Setup: Figure 18.6 shows the Lorentz ­factor gamma, γ, as a function of the fractional speed of light.

140 ◆ Chapter 18 Relativity and Black Holes Solve: One can quibble about the exact number, but γ does not deviate significantly from 1 ­until the speed is about 0.33 to 0.5c. The Lorenz factor goes to infinity as the speed of an object approaches the speed of light.

Review: In general, one can ignore the effects of relativity if the speed is ­u nder 0.5c, as shown in this figure.

36. Setup: As noted in the prob­lem, ­there are 60 arcsec in an arcmin, and 60 arcmin in a degree, or 3,600 arcsec in one degree. Th ­ ere are also 100 years in a c­ entury.

So, we want to convert 2 degrees per c­ entury to arcsec per year.

Solve: degree 3600 arcsec 1 century arcsec 2 × × = 72 .

1 degree 100 years century year This is certainly large enough to mea­sure ­because telescopes can typically mea­sure positions smaller than 0.01 arcsec.

Review: The mea­sure­ment of the perihelion of Mercury is the easy part, and its precession has been known for a long time. Explaining it was the hard part, and ­doing so has been one of the ­g reat successes of general relativity.

37. Setup: Figure 18.12 shows how the gravitational bending of light by a massive object makes the star appear to be at a dif­fer­ent position than it r­ eally is. The massive object is causing the deflection by warping space, causing light to take a curved path, as shown in Figure 18.10a.

Solve: ­Because the warp of space increases linearly with the mass of the object warping space, doubling the mass of the Sun ­w ill double the angular deflection of the star; that is, the distance between the stars would increase.

Review: Mea­sur­ing the angular deflection of light by objects is another way of mea­sur­ing the mass of the object. It can be challenging ­because we have to know where the background star was supposed to appear.

38. Setup: Working It Out 18.1 shows us that the elapsed time when moving at high speeds changes by the Lorentz ­factor γ, which equals 1.15 for a speed v = 0.5c, as stated in the prob­lem. Traveling at 0.5c, the time passed on Earth would be 20 light-­years / 0.5c = 40 years, or 80 years round-­trip. Thus, 80 / 1.15 = 69.6 years have passed for you.

Solve: Traveling at half the speed of light means 80 years would have passed on Earth, while you register the trip to take 80 / 1.15 or 69.6 years. Your twin ­w ill be over 10 years older than you.

Review: Seventy years is a long trip even considering the fast speed at which you would travel. Better get a faster ship. See the next prob­lem.

39. Setup: Students are asked to predict, based on Working It Out 18.1(perhaps a­ fter solving prob­lem 38), how much younger they would be than their twins if they then traveled 0.999 times the speed of light instead of 0.5c. Answers w ­ ill vary.

Solve: Working It Out 18.1 tells us that for v = 0.999c, Lorenz ­factor γ = 22.37. Time passed for the trip = 20 light-­years / 0.999c = 20.02 years. The round-­trip would take 40.04 years: 40.04 / 22.37 = 1.79 years for you in the spaceship. For the twin prob­lem, a ­little over 40 years passes for your twin on Earth, while less than 2 years have passed for you. Your twin would be 38 years older than you. How close did you come to your prediction?

Review: In this case, the Earth-­based twin w ­ ill be even older by the time the space traveler returns.

40. Setup: The Schwarzschild radius is given by 2G 2GM Rs = 2 . Note that 2 = 1.48 × 10 −27 m/kg. c c Solve: A 70-kg person has a Schwarzschild radius of 2 × (6.67 × 10 −11 m 3 kg −1s −2 ) RS = × 70 kg 9 × 1016 m 2 s −2 =1.48 ×10 −27 m/kg × 70 kg = 10 −25 m.

Review: This is about 8 billion times smaller than a proton.

41. Setup: The Schwarzschild radius is given by 2GM c2 Rs = 2 . Solving for mass gives M = RS . c 2G Solve: A black hole with R S = 1.5 km has a mass of 9 × 1016 m 2 s −2 c2 M = RS = 1,500 m × 2G 2 × (6.67 × 10 −11 m 3 kg −1s −2 ) = (1,500 m) × (6.76 × 1026 kg/m) = 1030 kg.

This is half the mass of the Sun. Review: This tells us that the Schwarzschild radius for the Sun is about 3 km.

42. Setup: To find the Schwarzschild radius, use 2GM Rs = 2 . c MMoon = 3.7 × 10 −8 MSun = (3.7 × 10 −8 MSun ) × (1.99 × 1030 kg/MSun ) = 7.36 × 1022 kg Solve: (a) The mass of the moon is 7.36 × 1022 kg, so its Schwarzschild radius is 2 × (6.67 × 10 −11 m 3 kg −1s −2 ) RS = × 7.36 × 1022 kg 9 × 1016 m 2 s −2 = 10 −4 m, or 0.1 mm

Chapter 18

(b) If the Moon turns into a black hole, none of the variables in the tidal force equation are changed, thus its tidal force on Earth w ­ ill not change. (c) Any collapse of an object into a black hole is expected to produce gravitational waves.

Review: Notice that a black hole has a very small radius, which allows objects to be much closer to the mass (that is, R can get very small), which can make the tidal force huge. But at ordinary distances the effects of gravity and tidal forces remain the same, even around a black hole.

43. Setup: One critical princi­ple of relativity is that the speed of light is the same for all observers (that is, in all reference frames).

Solve: ­Because the speed of light is constant for all observers, every­one sees the l­aser travel at 3 × 108 meters per second (m/s).

Review: Always remember, light travels at the same speed no m ­ atter who is emitting it, watching it, or receiving it.

44. Setup: Working It Out 18.1 tells us for that spaceship v = 0.99c, γ = 7.09. It ­w ill take 25.25 years (25 / 0.99) to reach the alien civilization from both Earth’s and the aliens’ viewpoints.

Solve: (a) A spaceship traveling at 0.99c would take 25 ly / 7.09 = 3.53 years to get ­there. The aliens’ clocks ­w ill state that it took 25.25 years for the spaceship to arrive. (b) A return ­w ill happen 50.5 years ­later for the ­people on Earth. Perhaps the next generation would be the ones to greet their parents, the astronauts.

Review: Relativistic space travel makes it easy for space explorers to visit a planet, but hard for them to tell their generation about it.

45. Setup: Planets orbiting a black hole at 0.5 AU would A3 follow Kepler’s third law, P 2 = , where the M1 − M 2 period is in years, distance in AU, and masses in MSun. Solve: We can ignore the mass of the spaceship in our use of Kepler’s third law. The equation for the mass of the black hole thus becomes A3 13 MBH = 2 = 2 = 4 MSun .

P 0.5 Review: Note that this answer to this prob­lem would be same if the black hole ­were a four-­solar-­mass star.

Using the Web Solutions 46. Answers ­w ill vary for searches relating to supernovae, gamma-­ray bursts, or stellar black holes. When two neutron stars merge, their combined mass ­w ill exceed the limit of neutron degeneracy and so the objects ­w ill collapse into a black hole.

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47. The objective of the Fermi Gamma-­ray Space Telescope mission is to study every­thing related to the huge jets that accompany black holes. Answers w ­ ill vary for the report on any recent news relating to black holes.

48. The four listed sources of gravitational waves from the LIGO website are continuous (sine wave), inspiral (frequency and amplitude increase), burst (large amplitude that quickly dampens), and stochastic (much like static). Students ­w ill also report on what’s new with the proj­ect.

49. The NuSTAR telescopic array observes at X-­ray wavelengths (energies from 3 to 79 keV). ­There have been a number of new discoveries and contributions to science, so answers ­w ill vary.

50. As you travel into the Schwarzschild black hole, light from objects in the background becomes increasingly distorted, but as the black hole fills your field of view, you see very l­ittle, time slows, and eventually every­ thing becomes black. The distortion is called gravitational lensing. Inside a black hole, all light goes straight to the center, so t­ here ­w ill be nothing to see inside the object. The Reissner-­Nordström black hole has a “wormhole,” which is a mathematical result of the charged sphere that could, in princi­ple, take one to a new place in space. However, you w ­ ill still die as you pass through the black hole due to “spaghettification.”

Moving ­toward the Sun at close to the speed of light, you see the effects of aberration, in which ­angles change so the Sun first appears to move away from you, then Doppler shift in which the Sun changes color, followed by beaming where the Sun appears brighter than it ­really is and time dilation slows time as you approach the Sun.

EXPLORATION Following Einstein’s use of thought experiments to solve prob­lems, students ­w ill imagine rolling golf balls across a big rubber sheet that ­w ill be warped by mass.

EXPLORATION SOLUTIONS 1. The balls go straight when the sheet is perfectly flat.

2. The balls go fairly straight when they are far from the bowling ball, provided the sheet is big enough; other­ wise, they are deflected by small amounts.

3. The golf balls are deflected by a large amount.

4. The golf balls hit the bowling ball.

5. Numbers 2 and 3 change depending on the speed of the golf balls, whereas number 4 does not. The faster the golf balls travel for 2 and 3, the less their paths are

142 ◆ Chapter 18 Relativity and Black Holes deflected. Similarly, the slower they move, the more they are deflected.

6. As more golf balls fall into the pit, the total mass in the pit increases, so the pit ­w ill become deeper and larger.

7. Light is like a superfast-­moving golf ball. Its path ­w ill not be deflected as much as our golf balls, but w ­ ill still be deflected.

8. The other golf ball w ­ ill make a huge dip in the sheet, and both balls ­w ill prob­ably start rolling around each other. Certainly, the golf balls and sand would move differently if rolled anew, and in the end, all the balls and sand ­w ill prob­ably end up together in a single, very deep and wide pit.

9. Close to a black hole, time and space are all highly warped/distorted, so nothing would appear as we would see it if t­ here ­were no black hole. Answers w ­ ill vary.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES ACTIVITY 23: BENT SPACE AND BLACK HOLES Students use thought experiments and the meta­phor of a rubber sheet to explore the interaction of black holes with objects that pass near them. They also: • explain the concept of bent space.

• demonstrate an understanding of how the speed of an object affects its interaction with a massive body.

• define event horizon.

• extend the concept of an event horizon to objects that travel at speeds slower than light.

• extrapolate the two-­d imensional concepts to three dimensions.

Text reference: Section 18.3

CHAPTER 19

Galaxies INSTRUCTOR NOTES Chapter 19 introduces galaxies. Major topics include: • the pro­cess through which Milky Way nebulae became separate, very distant galaxies.

• classification, structure, kinematics, and composition of spiral and elliptical galaxies.

• the cosmological distance ladder.

• the discovery and implications of Hubble’s law.

• rotation curves and dark ­matter.

• supermassive black holes and the unified model of AGN. This chapter starts with our initial 18th ­century views that the distant nebulae ­were all part of the Milky Way. While t­ here ­were a number of theories and observations that sought to answer just how big the Milky Way is and what it contains over the next ­century, it ­wasn’t ­until Edwin Hubble’s observation of a Cepheid variable in the Andromeda galaxy that we understood that ­these distant nebulae ­were truly at im­mense distances and w ­ ere galaxies similar in size to our galaxy. We like to emphasize that Edwin Hubble’s work happened just a c­ entury ago and then discuss with the students just how far our knowledge has grown since then, especially since the Hubble Space Telescope was launched in 1990.

DISCUSSION POINTS • Search for “galaxies” and bring up images. Have students look for differences between elliptical and spiral galaxies that yield clues about the motions of their stars. (LG 1) • Ask students to sketch their visualization of a hurricane as seen from space. Then compare ­those sketches to spiral galaxies. Contrast the pro­cesses that create each. Are t­ here any similarities? (LG 1) • Step-­by-­step, the construction of the distance ladder has led to our finding distances out to the edge of the vis­i­ble universe. In a think-­pair-­share activity, have students describe a “ladder” they have “climbed” that led to greater achievements, knowledge, or discoveries. (LG 2) • Evaluate the accuracy of mea­sur­ing distances via redshift against that of other distance-­measurement techniques discussed in this and previous chapters. (LG 2)

• Discuss the evidence for dark m ­ atter. Foster a debate among students about ­whether massive compact halo objects (MACHOs) or weakly interacting massive particles (WIMPs) are more likely the “stuff” that composes dark m ­ atter. (LG 3) • Does e­ very galaxy have a potential AGN inside it? Discuss what would be the observable consequences if the center of the Milky Way w ­ ere to light up as an active galactic nucleus. (LG 4) • The chapter-­opening image asks the question, “How do galaxies get their shapes?” Students could compare elliptical galaxies and come up with ideas about why some are ­spherical and some are more flattened, similar to footballs. Some elliptical galaxies show signs of dust. Similar exchanges of ideas could include normal spiral galaxies versus barred spiral galaxies. Why would some galaxies have a large, dominant bulge while ­others have a small, barely noticeable one?

Encourage development of new ideas and then reinforce t­ hose that are similar to theories developed by astronomers.

PROCESS OF SCIENCE The detection of active galaxies began in 1943 with a paper by Carl Seyfert. Th ­ ese galaxies ­were called Seyfert galaxies, and ­later classified as Seyfert I or Seyfert II. The discovery of quasars followed a few de­cades ­later, followed by blasars and the slightly dif­fer­ent BL Lac objects.

Some w ­ ere radio quiet; some, radio loud. The Pro­cess of Science for this chapter emphasizes the importance of finding a simpler, unified model that explains most, if not all, of the reasons for the dif­fer­ent classifications.

ASTROTOUR ANIMATIONS HUBBLE’S LAW This animation uses Hubble’s observed redshifts of the light from galaxies as a starting point for an exploration of the recessional velocities of the galaxies. The observation that the farther a galaxy was from us, the faster it was receding, led to Hubble’s law and the expansion of the universe.

The animation uses mega-­light-­year (Mly) instead of megaparsec (Mpc). This example ends with a Hubble constant of 143

144 ◆ Chapter 19 Galaxies 22 kilo­meters per second per mega-­light-­year (km/s/Mly).

Additional examples are given to determine the age of the universe. The animation closes with an explanation of the cosmic micro­wave background radiation as evidence for the Big Bang.

Text reference: Section 19.2 DARK ­M ATTER The animation starts with an overview of the periodic chart and the statement that only 4 ­percent of the universe is made up of this ­matter; the rest is “dark ­matter.” It then covers the evidence for the existence of this material.

A contrast is made between the velocity and distance of planets in the solar system and velocity versus distance for stars in the galaxy. The conclusion is that ­there is extra mass distributed throughout the galaxy.

Text reference: Section 19.3 ACTIVE GALACTIC NUCLEI This animation starts with the questions “What generates all the energy?” and “How can so much luminosity come from such a compact source?” The visualization zooms into a massive black hole with an accretion disk of gas and dust.

The explanation moves on to the role of friction between gas particles in the disk and the relationship to temperature as a function of distance from the black hole. The formation of jets is covered t­ oward the end, along with a summary of the AGN characteristics and unanswered questions.

Text reference: Section 19.4

ASTRONOMY IN ACTION VIDEOS SIZE OF ACTIVE GALACTIC NUCLEI Stacy Palen explains how we determine the size of active galactic nuclei. To demonstrate, Dr. Palen times students who are marching past her. By knowing their speed, the length of the student parade (size of the galactic nucleus) can be determined. Length: 1:14 Text reference: Section 19.4

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. The huge elliptical galaxies are dominating the light in this image of Abell 370 and thus seem to be the most

common. It may be, however, that spiral galaxies in this cluster are more common, we just ­can’t see them.

2. Spiral galaxies are bluer than elliptical galaxies ­because the spiral galaxies are still producing young, hot, massive stars.

3. More remote galaxies tell us about the ­earlier stages of the universe ­because the farther away the galaxies are, the farther back in time we are viewing them.

Due to the finite speed of light, the light from t­ hese galaxies show what they looked like 4 billion years ago, when the universe was only 9 billion years old.

4. Astronomers can take spectra of the arcs and see that the images all have the same redshift if they are the same galaxy.

CHECK YOUR UNDERSTANDING SOLUTIONS 19.1A (d) Another term for this is “galaxy morphology.”

19.1B (c) Spiral galaxies have abundant gas and dust, the fuel for star formation.

19.2 (c) Hubble’s law is also called Hubble’s redshift-­ distance law.

19.3 (a) Dark m ­ atter is m ­ atter that exerts gravity but is completely nonluminous.

19.4 (c) A supermassive black hole has been found at the center of nearly e­ very galaxy examined so far.

END-­OF-­CHAPTER QUESTIONS AND PROB­LEMS SOLUTIONS Test Your Understanding 1. (a) Up to 90 ­percent of a galaxy’s mass is dark ­matter.

2. (c) Kepler’s laws use orbital characteristics of stars to determine the mass interior to their orbits.

3. (c) An object cannot vary in time any faster than the time it takes for light to cross it. This gives an upper limit to the size of any variable object. λmeasured − λrest 756.3 nm − 656.3 nm = 656.3 nm λrest 100 nm = = 0.15.

656.3vnm

4. (d) Redshift z =

5. (a) Type Ia supernovae are many powers of 10 more luminous than Cepheid variables.

6. (a) All types of spirals have a bulge and disk but the bulges of barred spirals are elongated, and not necessarily spherical.

7. (d) E5. A rugby ball is ellipsoidal.

8. (a) ­There needs to be a period-­luminosity relationship.

9. (a) An E0 is s­ pherical and thus results from perfectly randomized orbits.

Chapter 19

10. (d) See Figures 19.15 and 19.16.

11. (c) Hubble’s law indicates a direct relationship between recessional velocity and distance.

12. (a) This is the definition of the Hubble constant.

13. (c) This is a restatement of our observations of flat rotation curves.

14. (d) In the unified model of AGN, the main difference between types depends on the viewing ­angle, with a smaller contribution from the amount of accretion feeding the black hole.

15. (b) One AU is roughly 8.3 light-­m inutes. Ten hours is 600 ≈ 70 AU. 600 minutes, giving 8.3 Thinking about the Concepts 16. The ­Great Debate was a meeting of the National Acad­emy of Sciences convened to decide ­whether spiral nebulae w ­ ere small clouds that w ­ ere part of the Milky Way, or large “island universes,” as Kant described them. This single question determined the scale of the entire universe: if the spiral nebulae w ­ ere just small parts of our own galaxy, it implied that the Milky Way was the entire universe, as opposed to the vast expanse that we now know it to be.

17. Cepheids found in the Andromeda Galaxy (Messier 31, M31) by Hubble ­were determined to be too far away to be part of our own Milky Way. 18. ­There is always uncertainty as to the ­actual luminosity of any given standard candle type. Thus it is very impor­tant to make many mea­sure­ments of dif­fer­ent standard candles to gain confidence in one’s results.

19. Distance-­measurement techniques are generally only applicable to a certain distance range. To calibrate a given technique, we need to in­de­pen­dently determine the distance to a number of objects first. We calibrate the Cepheid distance formula using parallax or spectroscopic parallax. Then we calibrate Type Ia supernovae using Cepheid variables found in the same galaxy. In such a way, we build up each “rung” of a distance ladder, which lets us probe to larger and larger distances.

20. A Type Ia supernova is a standard candle only if we know exactly how luminous each one is. For that, we have to know the type of its progenitor and have a similar progenitor where we are able to use a dif­fer­ent distance mea­sur­ing technique to calibrate.

21. Young, hot blue stars dominate the luminosity of the blue regions in spirals. This implies that they are regions of active star formation that contain significant gas and dust. Meanwhile, old, cool red stars (red ­g iants) dominate the luminosity of the red regions.

This implies that they are regions lacking active star

Galaxies ◆

145

formation, though gas and dust may or may not be prevalent. Note that this does not imply that regions of space that are red are only full of old stars. For example, HII regions are also red.

22. The bulge of a spiral has an older population of stars and the stars are on randomized orbits, which mimic on a small scale the structure of an elliptical galaxy.

23. A quasar is an enormously power­ful engine at the center of a galaxy, powered by huge amounts of mass falling into a supermassive black hole. Quasars shine with the luminosity of a trillion to a thousand trillion (1012 to 1015) Suns, much more luminous than a galaxy with 1011 Sun-­l ike stars.

24. Quasars appear to be associated with the mergers and interactions between galaxies. B ­ecause they are observed at ­g reat distance, and thus back to an ­earlier time, ­these mergers must have been much more prevalent in the past when the universe was younger and the distances between galaxies smaller.

25. If we look at their spectra, normal galaxies have absorption spectra from the light of its billions of stars, with some emission from massive stars and star-­ forming regions. The nucleus of a “normal” galaxy is quiescent, suggesting that the supermassive black hole is dormant. A galaxy with an AGN exhibits an x-­ray emission spectrum from its nucleus in addition to the stellar absorption spectrum. Bipolar jets of hot gas may also be pre­sent, spiraling out along the magnetic axis of the galaxy and emitting at radio frequencies.

26. AGNs are about the size of our Solar System. We know this b­ ecause the time variability of an object constrains its size. An object cannot be much larger than the timescales of variability, other­w ise that signal would be washed out.

27. An AGN has an accretion disk fed by several solar masses ­every year. As material moves inward ­toward the supermassive black hole, conversion of gravitational energy heats the accretion disk to hundreds of thousands of kelvins, causing it to glow brightly in vis­i­ble, ultraviolet, and X-­ray light. Conversion of gravitational energy to thermal energy as material falls onto the accretion disk also is a source of energetic emission. As much as 20 ­percent of the mass of infalling material around a supermassive black hole is converted to luminous energy.

28. To detect supermassive black holes at the center of galaxies, we need to see e­ ither their gravitational influence on their immediate surroundings or catch them in the act of “feeding”; that is, having a source of ­matter falling onto the black hole. ­These black holes reveal their presence. Black holes in “normal” galaxies

146 ◆ Chapter 19 Galaxies are quiescent. The presence of black holes for them would be inferred from observations of a concentration of mass at the centers, central regions that are much more luminous than what can be explained by stars alone, and large Doppler shifts of material at the centers. We see AGN galaxies at huge distances, but the other observations are pos­si­ble only for relatively nearby galaxies.

29. A nearby rung is needed to calibrate a more distant one; t­ here must be overlap so that we can reliably use the more distant rungs. It is impor­tant to use many mea­sure­ments of dif­fer­ent standard candles at any given distance to increase the confidence we have in our results.

30. The proximity of stars in the crowded center of ­galaxies fosters a number of tidal encounters that would disrupt planetary orbits. Even if the center of a galaxy did not have an AGN, ­there may still be an overabundance of radiation harmful to life.

Applying the Concepts 31. Setup: Figure 19.15 shows a typical rotation curve of a spiral galaxy, NGC 3198, in black. The red curve indicates the enclosed mass of normal luminous ­matter, and the blue the amount of dark ­matter mass that must be pre­sent in order to explain the rotation curve.

Solve: (a) 25,000 pc is where the rotation curve peaks. The data values are about 150 km/s and the errors are about 10 km/s in length. So, the mea­sure­ ment error is 10 km/s ± 5 km / s.

10 km/s × 100% = 7%. (b) The percentage error is 150 km/s (c) Note that the error bars still trace out the same shape, so varying the black curve within the error bars ­w ill not change the conclusion.

Review: The evidence for flat rotation curves in galaxies has been extensively studied with a variety of techniques, and the fact that the error bars are as small as they are shows that this a very reliable finding.

32. Setup: Density is, in this prob­lem, n = number/volume, so the number of galaxies in a given volume is 4 N = nV, where V is the volume of a sphere, π r 3 .

3 Note that we call n the “number density” to distinguish it from mass density ρ.

Solve: N = nV 4 = 10 −68 galaxies/m3 × π [(1010 pc) 3 × (3.08 × 1016 m/pc)]3 ≅ 1012 galaxies.

Review: One trillion galaxies is a good back-­of-­the-­ envelope number to have handy, just like a galaxy typically contains 100 billion stars. λ − λrest .

33. Setup: The redshift of a galaxy is z = measured λrest The recessional velocity (for redshifts less than one) is v = cz, and the distance is given by the Hubble relationship d = v/H0. λ − λrest 750 nm − 656.28 nm = Solve: (a) z = measured 656.28 nm λrest 93.72 nm = = 0.143.

656.28 nm (b) Using this redshift, we find a recessional velocity of v = 0.143c = 4.28 × 107 m/s. (c) Using the Hubble relationship with this velocity, 4.28 × 10 4 km/s d= = 612 Mpc. 70 km/s /Mpc Review: A galaxy with high redshift has high recessional velocity and high distance. We see in this prob­ lem that this galaxy (at about 600 megaparsecs) certainly follows this trend.

34. Setup: Hubble’s law is v = H0 d, where H0 = 70 km/s/ Mpc and redshift relates to speed by v = cz.

Solve: The quasar’s redshift gives a recession velocity of v = cz = 0.158 × 3 × 105 km/s = 4.74 × 10 4 km/s, which with Hubble’s law yields d = v/H 0 = (4.74 × 10 4 km/s)/70 km/s/Mpc = 677 Mpc. Review: This shows us that just ­because the redshift is small (for example, around 0.1 or 0.2), that does not mean the object is nearby. It is still very far away.

35. Setup: A large galaxy, such as the Andromeda Galaxy, is 10 billion times the luminosity of the Sun, or 1010LSun. Solve: 3C 273 has a luminosity of 1012 LSun, and so is 100 times more luminous than the entire Andromeda Galaxy.

Review: Quasars are extremely bright objects, as shown in this prob­lem, where 3C 273 is 100 times brighter than the entire Andromeda Galaxy. Imagine how bright it would be if the quasar ­were at the same distance as that galaxy.

36. Setup: We bring in the brightness-­ luminosity-­ L , and ratios, distance relationship, b = 4π d 2 Lquasar Lgalaxy = 2 . This leads to the distance to the 2 dquasar dgalaxy Lquasar quasar, dquasar = dgalaxy .

Lgalaxy

Chapter 19 Galaxies ◆ 147

Solve: This quasar is 1 million times brighter than our galaxy, and is 2 megaparsecs away, so dquasar = 2 Mpc

106 = 2,000 Mpc (or 1,000 times 1

farther away).

Review: Quasars are all located at very large distances, as this prob­lem demonstrates.

37. Setup: The Schwarzschild radius is given as 2GM R S = 2 . We could also use the scaling relation c that a black hole with l MSun has a Schwarzschild radius of about 3 km. ­ ill break the solution into parts, starting Solve: We w with the part containing the constants followed by the mass of the supermassive black hole (SMBH) and then the Schwarzchild radius: 2G 2(6.67 ×10 −11 m 3 kg −1s −2 ) = =1.48 ×10 −27 m/kg (3 ×108 m/s)2 c2

This yields a mass energy of : E = mc2 = (4 × 1037 kg)(3 × 108 m/s)2 = 3.6 × 1054 J. ­Because the power output is 10 41 J/s, the lifetime is: 3.6 ×1054 J 1 yr t quasar = × =1.14 ×106 yr, 10 41 J/s 3.15 ×10 7 s or about 1 million years.

Review: In cosmological time spans, a million years is very short, which shows that the engine of a quasar is very short-­lived, about as long as that of the most massive stars.

40. Setup: We use E = mc2 to convert mass into energy.

Solve: Converting half of the solar-­mass star into energy produces E = 1030 kg × (3 × 108 m/s)2 = 9 × 10 46 J. Review: The Sun puts out about 4 × 1026 J in 1 second, or about 1.3 × 1044 J over its 10-­billion-­year lifetime (3.15 × 1017 s). Thus, we see that the energy released when mass falls into a black hole is significantly greater than that released via normal stellar pro­cesses.

41. Setup: Distance, luminosity, and brightness are

M SMBH = (2.6 ×1010 M Sun )(2 ×1030 kg/M Sun ) = 5.2 ×10 40 kg M Sun )(2 ×1030 kg/M Sun ) = 5.2 ×10 40 kg R S = (1.48 ×10 −27 m/kg)(5.2 ×10 40 kg) = 7.7 ×1013 m 7.7 ×1013 m In relative terms: R S = = 514 AU. 1.5 ×1011 m/AU Review: Using our scaling relation, the Schwarzschild radius is 26 × 10 9 × 3 km = 7.8 × 1010 km = 7.8 × 1013 m, as found above. Notice that this is as large as the estimated distance to the middle of the Kuiper ­Belt.

38. Setup: An object can be no larger than the distance light travels in the time during which the brightness varies.

Solve: ­Because the object varies within an 83-­m inute period, it can be no larger than 83 light-­m inutes across. Remember that it takes sunlight about 8.3 minutes to reach Earth. Light travel time for 1 AU is 8.3 light-­ m inutes. The object is 83 light-­ m inutes across, corresponding to (83 light-­ m inutes) × (1 AU / 8.3 light-­m inutes) = 10 AU across.

Review: This is about the distance to Saturn. Given that the newspaper describes the object as “cosmological,” it might well be a new quasar.

39. Setup: We use E = mc2 to convert mass into energy ­because mass feeds the quasar. The lifetime of an object equals the amount of fuel available divided by how quickly the fuel is used.

Solve: The total amount of energy pre­ sent is 20 p­ ercent of 108 solar masses, or: Mquasar = 0.2 (108 MSun) (2 × 1030 kg/MSun = 4 × 1037 kg.

L L , giving d = . The typical 2 4π d 4π b luminosity of a Type Ia supernova is 3.7 × 1036 W (see Working It Out 19.1).

related by b =

Solve: d =

L = 4π b

3.7 × 1036 W = 1.7 × 1026 m 4π × 10 −17 W/m2

or about 5,510 Mpc. Review: The most distant Type Ia supernova observed (2018) is at z = 1.914, or about 8,000 Mpc, depending on the value of the Hubble constant used.

42. Setup: We use E = mc2 to convert mass into energy.

Solve: (a) Converting one-­tenth of Earth’s mass into energy produces: E = 0.1 × 5.97 × 1024 kg × (3 × 108 m/s)2 = 5.37 × 10 40 J. 5.37 × 10 40 J (b) This is = 1.4 × 1014 , or 140 trillion 3.85 × 1026 J times more energy.

Review: A black hole swallowing an object of Earth’s mass with 10 ­percent efficiency can give off more energy per second than even an entire galaxy of stars.

43. Setup: ­There are 3.15 × 107 seconds in 1 year and one solar mass has an equivalent energy of E = mc2 = 2 × 1030 kg × (3 × 108 m/s)2 = 1.8 × 10 47 J. 7 Solve: 2 × 10 41 J/s × 3.15 × 10 s = 6.3 × 10 48 J/yr.

1 yr Dividing this by the energy equivalent of the Sun, 1.8 × 10 47 J, we find the quasar consumes 35 solar masses per year.

Review: This may seem like a lot, but consider that a supermassive black hole has a mass of millions to billions of solar masses; thus, it needs to accrete only a

148 ◆ Chapter 19 Galaxies ing incidence telescopes with specially coated optics miniscule fraction of its total mass to maintain its and newly developed detectors that extend sensitivity average energy output. dθ s = converts to higher energies as compared to previous missions 44. Setup: The small-­angle formula 57.3° such as Chandra and XMM. A ­ fter launching into orbit from angular size θ to physical size s given the dison a small rocket, the NuSTAR telescope extends to tance d. Solving for the ­ angle in degrees gives achieve a 10-­meter focal length. The observatory w ­ ill s θ = 57.3° . The question asks for the angular extent provide a combination of sensitivity, spatial, and specd tral resolution ­factors of 10 to 100 improved over previof the jet. ous missions that have operated at t­ hese X-­ray energies.”

Solve: Solve for ­angle: A number of discoveries have been made through 3 × 10 4 pc s observations of Eta Carinae, ultraluminous X-­ ray θ = 57.3° = 57.3° = 0.1° 1.7 × 10 7 pc d sources, magnetars, power­ful X-­ray bursts, and more. (b) The angular size of the Moon is about 0.5°, making the apparent extent of the M87 jet about one-­fi fth that size, about the size of the Sea of Serenity.

EXPLORATION Review: If this jet w ­ ere bright enough, it would be This activity w ­ ill be much easier for students if the origivis­i­ble to the naked eye. nal color image (to help with the classifications) is dis45. Setup: The speed of light is 1 light-­year per year, so played for the class to review. The high-­resolution image let’s work in ­these units. can be found at https://­w ww​.­nasa​.­gov​/­images​/­content​ 5,000 ly d Solve: d = vt, leading to t = = = 5,050 yr/­2 42543main​_­hstimg ​_­20080610​_­H I​.­jpg.

0.99c 0.99 ly/yr Mapping is an effective way for students to do more than 5,000 ly d just gain a cursory overview of an image. However, the = 5,050 yr t= = 0.99c 0.99 ly/yr instructions, which ask that students lay a piece of paper Review: If the distance is 5,000 light-­years, and the over the image and number the 20 or so brightest of the larggas is moving at the speed of light, it was ejected est galaxies, may require tracing paper for students to get at 5,000 years ago. ­Here, we are moving at 99 ­percent least 20 galaxies, plus the more artistic students may spend a the speed of light, so it takes 101 ­percent of the time to lot of time on this step. We have included a negative gray travel, or 5,050 years. scale image (Figure IM 19.1), with 24 galaxies numbered for easier reference, for making copies for the students so that Using the Web Solutions they can spend more time focusing on the characteristics of 46. We observe the variations in brightness of the varieach galaxy—­shape, features, colors. The Coma Cluster is a able star. We assume that t­here is a relationship large, dense cluster and thus is dominated by elliptical between the period of ­those variations and the lumigalaxies. nosity of the star. From that, we calculate the distance to the star, which is the distance to the galaxy it is in.

47. The question asks students to go to Astronomy PicEXPLORATION SOLUTIONS ture of the Day and look at recent pictures of galaxies.

1. Very round elliptical galaxies tend to be the easiest to Answers w ­ ill vary. identify, especially b­ ecause most look orange.

48. a. Question asks students to go to the website for Gal2. Lenticular galaxies are very hard to identify, as are the axy Zoo and classify some galaxies. Answers w ­ ill vary. many very small galaxies. b. Students are to go to the website for Radio Galaxy 3. The tilt of a spiral or the perspective of an elliptical Zoo and look for supermassive black holes. Answers makes it very hard to tell if the galaxy is a spiral and ­w ill vary. determine its subtype, or if it is a very eccentric ellip49. Students are to go to the Fermi Gamma-­ray Space tical. It is especially hard to tell if the small galaxies Telescope website and search ­under “Latest News” for are elliptical or spiral. a story about dark ­matter. Answers ­w ill vary.

4. We most often agreed about elliptical galaxies the most.

50. a. Students are to go to the NASA Swift Gamma-­R ay 5. We most often disagreed on the subtype of spirals. observatory website, search u­ nder “Other News” and 6. It is easier to classify galaxies if one has (a) much larger find a story about supermassive black holes. Answers or higher-­resolution images and (b) an example cata­log ­w ill vary. of many dif­fer­ent types to which one can directly b. Students are to go to the website for NuSTAR. “The compare.

NuSTAR instrument consists of two co-­aligned graz-

Chapter 19

FIGURE IM 19.1

Table IM 19.1 Sample ­Table with Pos­si­ble Classifications and Comments Galaxy No

Type

Comments

Galaxy No

Type

Comments

1

E0

Merger?

13

SBa

Barred

2

E0

14

E1

3

Sc

15

E2

4

Sb/Sc

16

Sb

5

E0

17

SBa

6

SBb

18

E5

7

Sb/Sc

19

Sb

8

Sb

Maybe has arms?

20

Sb

9

Sb

Merger?

21

Irr

10

E0

Small

22

Sb

11

E7

23

E1

12

E2

24

Sb

Merger?

Edge-on?

Galaxies ◆ 149

150 ◆ Chapter 19 Galaxies

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES ACTIVITY 24: LIGHT TRAVEL TIME AND THE SIZE OF A QUASAR Astronomers use variability in the emission from active galaxies to determine the size of the emitting region and the total energy emitted. In this activity, students connect ­these observations to the theory that active galaxies are powered by supermassive black holes. They also: • determine the size of the emitting region of a “new” quasar based on a brief intense flare.

• calculate the energy released by the flare and compare that with the energy of the Sun. • evaluate the evidence as to ­whether it supports the presence of a supermassive black hole.

Text reference: Section 19.4

ACTIVITY 29: CALCULATING THE MASS OF THE CENTRAL OBJECT OF THE GALAXY Students use observations of stars orbiting near the center of the Milky Way to confirm the existence of a supermassive black hole as the central object. The method they use is closely connected to the methods used with orbiting objects in previous activities. They also: • demonstrate proficiency in the use of scientific notation in all calculations.

• apply Newton’s form of Kepler’s third law to infer the mass within the orbit of a centrally located star in the galaxy.

• compare the results with the mass of the Sun and the size of the Solar System.

• analyze the evidence for a supermassive black hole at the center of the Milky Way. • compare the results with t­ hose published in astronomical journals.

Text reference: Section 19.4

CHAPTER 20

The Milky Way: A Normal Spiral Galaxy INSTRUCTOR NOTES Chapter 20 is an exploration of the Milky Way. Major ­topics include: • details of the size, structure, and shape of the Milky Way. • characteristics of spiral arms and star formation.

• age and chemical composition of stars, including the pro­cess of chemical enrichment.

• structure and components of the disk and halo.

• evidence for dark m ­ atter.

• evidence for a supermassive black hole at its center.

• Local Group, formation of the Milky Way and eventual collision with Andromeda.

[AUs]), (c) as a constant versus distance (example: crack-­the-­whip while skating, where extra force in the hand grips must be applied to keep the outer skaters attached). How does the rotation curve of the Milky Way differ from ­these other curves? How is this evidence that dark m ­ atter exists in the galaxy? (LG 3) • The chapter-­opening image asks, “How do we know what the Milky Way Galaxy looks like from inside?” As part of their reading of this chapter, have students make , their locations a time line of the observations concerning the contents and structure of the Milky Way and how each new discovery became another piece of the puzzle.

PROCESS OF SCIENCE DISCUSSION POINTS • Consider the challenges of mea­sur­ing the structure of your college or university campus by sitting at one single location armed only with a telescope. Discuss why it is both expected and acceptable that our understanding of our galaxy has changed over time. (LG 1) • Most, if not all, students w ­ ill have had the experience of being stuck in a traffic jam. Have students share what happened during that time. Did they look at other ­drivers? Did they get out of their cars? Did they honk their car’s horn? ­Were vehicle accidents causing the jam?

Find out if t­ here are analogies with the active star formation occurring in spiral arms of galaxies. (LG 1) • Have students make a step-­by-­step story­board of the formation and evolution of the Milky Way by taking into account the components that we see t­ oday and including the chemical evolution of the galaxy, making note of the metal-­poor versus metal-­rich groups of stars, their locations, and how they formed. (LG 2, LG 4) • Figure 20.16 shows the rotation velocity plotted against the distance from the center of the Milky Way. Have students come up with lists of objects or events where the velocity would increase (a) as a function of distance from the center as in a solid disk (example: playground merry-­go-­round), (b) as one over the distance squared (example: orbital speed of planets compared with Earth versus distance in astronomical units

­ ere have been two major “unknown unknowns” of this Th ­century: dark ­matter and dark energy. We ­didn’t know what we ­didn’t know ­until, as this Pro­cess of Science points out, ­there came sets of observations that failed to match the accepted model. In each case, many additional observations were made in attempts to disprove the findings. Research must continue, of course, ­because just ­because we know about dark ­matter and dark energy d­ oesn’t mean we know what they are. At least now, we know what we ­don’t know.

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Scientists need to know the exact brightness (apparent magnitudes), distances, motions, and colors (surface temperatures) of more than a billion stars.

151

152 ◆ Chapter 20 The Milky Way: A Normal Spiral Galaxy 2. For stars that are very far away, astronomers need to measure the shifts in the light from the stars. Light that is blueshifted means the star is approaching; light that is redshifted means the star is receding.

3. Students will have their individual answers, but mainly we know we are looking at the center of the Milky Way when the region viewed is the thickest, brightest, and with the most dust.

4. Students are asked to find recent results from the Gaia Milky Way data. Answers will vary.

CHECK YOUR UNDERSTANDING SOLUTIONS 20.1 Answers ­w ill vary. Similarities between images of distant galaxies and ours suggest a spiral galaxy. The Milky Way contains gas and dust as well as ongoing star formation, characteristics of spiral galaxies.

Observations of 21-cm radiation emitted from neutral hydrogen clouds show the existence of spiral arms.

Motions of stars in the inner part of the galaxy indicate a substantial bar. The galaxy has a well-­defined rotation curve that flattens ­toward the vis­i­ble edge.

Fi­nally, most stars in our galactic neighborhood share the same relative motion around the galactic center.

20.2 The disk (and bulge) contains old and young stars; the halo contains only old stars.

20.3 (c) Dark ­matter is detectable only by its gravity.

20.4 Andromeda and the Milky Way are moving ­toward each other ­because of their mutual gravitational attraction.

END-­OF-­CHAPTER QUESTIONS AND PROB­ LEMS SOLUTIONS Test Your Understanding 1. (c) RR Lyrae variables in globular clusters allow us to reliably determine the distances to t­ hese clusters, which are distributed throughout the galaxy and its halo.

2. (a) It is difficult to determine the structure of something when you are embedded inside it and when dust obscures much of what you can see.

3. (c) Supernovae explosions enrich the interstellar medium, causing the abundances of heavy ele­ments to increase with time.

4. (a) The glow of synchrotron radiation from cosmic rays spiraling in the galactic magnetic field is a direct confirmation of the presence of large-­scale fields.

5. (a) Observations of stellar orbits lead to the conclusion that t­ here is a very compact and massive object at the galactic center.

6. (a) We already see tidal streams of stars from the Magellanic Clouds being stripped off by the Milky Way’s gravity.

7. (c) The distribution of globular clusters shows that we are not located at its center.

8. (b) Shapley used RR Lyrae variable stars to estimate the distance to each globular cluster. Without ­t hose stars, this distance estimate would not have been pos­si­ble.

9. (a) The flat rotation curve can be explained only by the presence of a significant amount of m ­ atter that we cannot see.

10. (b) The name is a bit of a misnomer b­ ecause ­t hese are not rays of energy but rather high-­energy particles (protons, electrons, helium, and other nuclei).

11. (c) All observational evidence suggests that we are living in a spiral with a central bar.

12. (c) The disk, specifically the spiral arms, is where the youn­gest stars are located ­because this is where star formation is the most active.

13. (d) Both disk and halo stars can have the same temperatures.

14. (d) It is very hard to detect many galaxies ­because they have “low surface brightness,” which means that their light is spread out (diffuse).

15. (d) The galactic habitable zone considers where stars could form that may host life-­bearing planets, but it does not discuss the particulars of the planetary systems themselves.

Thinking about the Concepts 16. Science is about providing the best explanation pos­si­ ble at the time, but its self-­correcting nature allows for ­those explanations to change, without judgment or repercussions, as more is learned. Shapley mea­sured the wrong distances ­because he did not know about the dust, but the conclusions he drew w ­ ere nonetheless critical b­ ecause he made the first data-­driven mea­sure­ments of our galaxy’s size and our location later on have not within it. Further refinements ­ changed t­hose basic conclusions. But without t­hose first mea­sure­ments, subsequent advancements in our understanding might not have occurred.

17. Globular clusters are distributed within a rough, large sphere, showing that ­there is a large, spheroidal halo that encloses the disk of the galaxy. The center of that spheroid is roughly 8,000 pc from the Sun, implying that we are roughly that distance from the center of the galaxy and not in the center itself.

18. Open clusters are young and therefore ­w ill generally be found near spiral arms and regions of recent star

Chapter 20

formation. Globular clusters are found primarily in the galactic halo.

19. Star formation is generally more active in the denser, inner part of the Milky Way than in the outer parts. If the gas in the inner part of the galaxy has under­gone more nuclear pro­cessing, then it is more metal rich than the gas in the outer galaxy; ­ ill be hence, stars in the inner part of the galaxy w more metal rich.

20. Doppler shifts of 21-cm radiation emitted by neutral hydrogen clouds show material moving around the center of the galaxy. This is shown in Figure 20.15.

Note that dust obscures our view of the galaxy in the vis­i­ble and ultraviolet regions of the electromagnetic spectrum. However, it is transparent to the longer 21-cm radiation emitted by neutral hydrogen clouds.

This makes 21-cm radiation a useful tool for peering through the dust to see the galaxy’s structure.

21. Similar to most other spiral galaxies, the Milky Way clearly has a flat rotation curve as observed in the orbits of stars and gas at its edge. These observations imply that up to 90 ­percent of the mass of our galaxy must be dark ­matter.

22. Halo stars have a much lower metal content than disk stars, and they are not moving with the disk; instead, halo stars are moving up and down through the disk on spheroidal orbits around the galactic center. Their ­ ill be much greater than t­ hose mea­sured velocities w of stars orbiting within the disk.

23. Synchrotron radiation c­ auses the disk of the galaxy to glow, and it comes from cosmic rays, mostly electrons, spiraling through the galaxy’s magnetic field.

24. ­There is far too much dust along the path to the galactic center to see optical light from that region. Whereas dust is opaque to optical and UV light, it is transparent to radio, X-­ray, and infrared; hence, we can (and do) observe this region in t­ hese wavelengths.

25. Sgr A* is believed to be the massive object at the very center of the galaxy. It was first detected as a strong radio source and ­later in X-­rays.

26. The Doppler shifts of stars near the galactic center are very large. Plus, we use the proper motions of stars observed at infrared wavelengths to determine their orbits and directly determine the mass of the central object via Kepler's laws. These observations imply that ­there must be a tremendous mass packed into a small volume in the galactic center. Gamma ray, X-­ray, and radio emissions (particularly synchrotron radiation) emanating from the galactic center are also black hole signatures. Compared with the supermassive black holes of AGN, of order billions of solar masses, the one in our galaxy is small.

The Milky Way: A Normal Spiral Galaxy ◆ 153

27. The Large and Small Magellanic Clouds are classified as dwarf satellite galaxies of the Milky Way. They are the remnants of dwarf galaxies that crashed through the galactic disk, stripping some of our galaxy’s gas and dust in the pro­cess. Besides being at a comparable distance from our galactic center (which is why they look like small clouds), they also look spectroscopically similar to the Milky Way b­ ecause of the gas and dust they “stole” during previous passages through the galactic disk.

28. Galactic satellites are the remnants of the small protogalaxies that first formed in our dark m ­ atter halo. Most of them coalesced to form the Milky Way; however, ­there are a number of them that remain in orbit around our galaxy. Many satellite galaxies have low surface brightness, so they are hard to detect against the foreground stars of our galaxy itself. Many are also located on the other side of the galaxy, and b­ ecause of the same sources of opacity that make it impossible to see the galactic center, t­ hese dwarf galaxies cannot be detected ­because their stars ­can’t be seen.

29. (a) If the Sun ­were located near the galactic center, the night sky would be extremely bright b­ ecause of the very high density of stars surrounding the Sun. (b) If the Sun ­were located near the center of a large globular cluster, both the day and night skies would be filled with thousands of stars. However, most of t­ hese stars would be Sun-­sized and smaller. As a result, we would not be bathed with the intense radiation that we would face at the galactic center. (c) Inside a molecular cloud, no stars would be vis­i­ble in the background. We would see the Sun dimmed by the thick dust, and the dense fog might hide even the planets unless the solar wind blows dust away from the immediate neighborhood.

Meanwhile, the sky would glow a dull red b­ ecause the cloud would likely be heated by nearby stars and protostars.

30. A galactic habitable zone is the region of a galaxy in which we believe stars capable of hosting life would be found. This excludes regions of rapid star formation (­because of both the UV light and frequent supernovae), close stellar encounters, and regions that do not have enough heavy ele­ments to form rocky planets.

Applying the Concepts 31. Setup: Figure 20.7 shows the disk and halo of our galaxy. Note that the halo is listed as about 90 kpc in dia­ meter, and the disk of the galaxy is shown at 30 kpc.

Solve: The disk is about one-­third the size of the halo.

Review: In general, the disk of a galaxy is much smaller than its halo.

154 ◆ Chapter 20 The Milky Way: A Normal Spiral Galaxy

revolution. There are 180 seconds in 3 minutes, 32. Setup: Figure 20.16 shows the rotation curve for our meaning each rider goes around 180 s / 15 s/rev = 12 galaxy, where the y-­a xis shows orbital speeds in kilo­ revolutions. meters per second, and the x-­a xis shows distances in 35. Setup: To solve this prob­lem, we could use the inforkiloparsecs. The period of a circular orbit is equal to mation given to determine the orbital velocity of the the distance traveled divided by the orbital speed, or 2π r Sun and then apply this to the proposed globular clusP= . ter (GC) to find its period. Alternatively, we could v consider using ratios b­ ecause both objects ­w ill have Solve: (a) At 6 kpc (6,000 pc), a star is orbiting at the same orbital speed (­a fter all, the rotation curve about 240 km/s. is flat). The period of orbit is equal to the distance (b) Converting parsecs to kilo­meters: r = 6,000 pc 13 17 2π r × 3.086 × 10 km/pc = 1.85 × 10 km. The time to . traveled divided by the orbital speed, or P = 17 2π (1.85 × 10 km) 1 yr v 8 orbit once is the period, P = = 1.54 × 10objects × yr.

7 B10 ecause both have the same velocity, 3.15 ×­ 240 km/s s rSun r 1.85 × 1017 km) 1 yr = GC .

8 × = 1.54 × 10 yr.

It takes about 154 milPSun PGC 240 km/s 3.15 × 10 7 s lion years for a disk star to orbit once.

Solve: (a) Solving our equation for the orbit of the Review: Our Sun takes ~230 million years to orbit, globular cluster (GC), its period ­w ill be so it makes sense that a star that is closer to the galacr 8,300pc + 9,000 pc PGC = PSun GC = 230 Myr × tic center ­w ill complete one orbit in less time since it rSun 8,300 pc has a shorter distance to travel even if its orbital veloc≅ 480 Myr. ity is the same as the Sun’s.

33. Setup: Figure 20.16 shows the rotation curve for our galaxy. At a distance of 50 kpc, the Magellanic Cloud would follow a circular orbit at a speed of about 175 km/s. The time to complete one orbit is the 2π r period, P = . v Solve: Converting to kilo­meters, 50 kpc is 50,000 pc, and (50,000 pc) × (3.086 × 1013 km/pc) = 1.54 × 1018 km.

Solving for the period, 1.54 × 1018 km 2π r = 2π = 5.53 × 1016 s 175 km/s v 1 yr (5.53 × 1016 s) × ≅ 1.8 × 10 9 yr 3.15 × 10 7 s

P=

(b) The total number of orbits ­ w ill be 10 1.3 × 10 yr =27 orbits.

4.8 × 108 yr/orbit Review: In prob­lem 34, we found the Sun has orbited the galaxy about 20 times in the last ~5 billion years.

The globular has made 1.35 times as many orbits in 2.6 times the time.

36. Setup: Use the distance-­ luminosity relation: L b= . Since the luminosities of the RR Lyrae 4π d 2 d2 bstar stars are the same (or nearly so), we use GC = 2 dstar bGC and solve for dGC.

bstar 1.6 × 105 Review: At this rate, the Magellanic Cloud has comSolve: (a) dGC = dstar = 230 pc = 92,000 pc pleted about seven orbits since the Milky Way was bGC 1 5 bstar 1.6 × 10 formed. = 92,000 pc dGC = dstar = 230 pc 34. Setup: To compute the number of orbits of the Sun, bGC 1 (b) The vis­i­ble disk is about 30,000 pc in dia­meter, we note that if one trip takes x years, then n trips takes and the halo is at least three times larger than the vis­i­ n times as long. The number of trips made in a given ble disk. Globular clusters are part of the halo, and it time T is just the ratio of the two. would appear that this cluster is at its outer limits.

Solve: The number of trips the Sun has made is Review: Looking at Figure 20.7, we find that the halo 4.6 × 10 9 yr extends to about this distance. = 20 trips.

2.30 × 108 yr/trip 37. Setup: See Figure 20.18.

Review: Although the Sun’s orbital period seems very slow, we have circled the galaxy about 20 times.

We can use similar logic to find out how many times a rider completes a circle on a merry-go-round if each ride lasted 3 minutes and it takes 15 seconds for one

Solve: The stellar orbits around the galactic center show that t­ here is an object of around 4 million solar masses residing ­there, but also that this object must be smaller than 6,000 AU. If one imagines 4 million stars in a dense cluster, ­those stars would be interacting

Chapter 20

gravitationally, and many would be ejected from the cluster within a few tens of thousands of years. The object could not be a star cluster. The only known reasonable way to compact that much mass into that small of a space is in a black hole. The center of the galaxy also has superheated gas that produces X-­rays, gamma-­ray-­emitting ­bubbles, as well as synchrotron radiation at radio wavelengths. The energetic events such as these are plausibly explained by a supermassive black hole.

Review: Note that in some re­spects this is a pro­cess of deduction and exclusion rather than one of direct proof. Science often works that way.

38. Setup: Terrestrial planets form out of heavy material (ele­ments other than H and He), so they w ­ ill be orbiting only around stars that contained an appreciable amount of that material as they w ­ ere forming.

Solve: Terrestrial planet formation requires star formation from enriched gas, which is mostly in the inner disk of the galaxy. Planet formation is prob­ ably less likely in the outer disk and certainly unlikely in the halo stars, which are very metal poor.

Review: From this argument, it is unlikely to find planets in the halo or halo objects, such as old globular clusters. However, we expected that nearby stars, which formed around the same time as the Sun, could host many planets, and, indeed, thousands have been detected.

39. Setup: For this prob­lem, we ­w ill use E = mc2 and solve it for mass, m = E/c2 .

Solve: (a) Using E = mc2 to solve for the rest energy of a proton, we find E = 1.7 × 10 −27 kg × (3 × 108 m/s)2 = 1.53 × 10 −10 J. (b) The relativistic mass of the fast-­moving proton with energy of 100 J is: m=

100 J = 1.11 × 10 −15 kg (3 × 108 )2

The Milky Way: A Normal Spiral Galaxy ◆ 155

d = vt = [1 − (1 × 10 −24)](3 × 108 m/s)(3.15 × 1015 s) d = [1 − (1 × 10 −24)](9.5 × 1023 m) = (9.5 × 1023 m​ − 0.95 m).

Although the photon has traveled the full (9.5 × 1023 m), the cosmic ray was 0.95 m (about a meter) b­ ehind at the end of the 100 million years.

Review: For all practical purposes, ultra-­h igh-­energy cosmic rays and photons travel the same speed and arrive at the same time. However, unlike the photon, this particle has enough energy to create billions of new particles when it collides with, for example, an atom in our upper atmosphere.

41. Setup: Kepler’s third law says P2 = A3/(M + m) where m is orbiting M, mass is in solar masses, distance is in AU, and period is in years. ­Because the black hole is far more massive than a star, we can ignore m and solve for the period, P. As the prob­lem states, 0.02 light-­years equals 1.9 × 1014 meters, or 1,267 AU. The mass of the black hole is 5 million MSun. Solve: 0.02 ly is 1,267 AU, so Kepler’s law gives P = 1,267 3/5 × 106 = 20 yrs.

Review: This star orbits so quickly that we can see it complete many orbits during a ­human lifetime.

42. Setup: Newton’s version of Kepler’s third law states A3 4π 2 . By using the full form, we ­will G (M + m) end up not having to do awkward unit conversions as well as having practice with Newton’s version. We again ignore m. The mass of the supermassive black hole is given as 8 × 1036 kg, and the radius is 0.0131 light-­years, or 1.24 × 1014 meters. We will find the circumference of the orbit in order to find the speed of the star.

Solve: P=

P=

a3 4π 2 G (M + m)

=

(1.24 × 1014 m)3 4π 2 6.67 × 10 −11 m3/ kg /s2 8 × 1036 kg

1.11 × 10 −15 kg = 6.54 × 1011 −27 1.7 × 10 kg −15 = 1.41 × 1017 s2 = 3.75 × 108 s 1.11 × 10 kg 11 = 6.54 × 10 or 654 billion times more massive than C = 2π R = 2π (1.24 × 1014 m) = 7.79 × 1014 m 1.7 × 10 −27 kg its rest mass. d 7.79 × 1014 m Review: Fast-­moving particles gain so much energy ≅ 2.1 × 106 m/s v= = 8 3.75 × 10 s t that when they collide with another particle, the energy goes into creating an entire zoo of new parti= 2.1 × 103 km/s cles. This is the princi­ple ­behind particle accelerators.

Review: 2,100 km/s is a very high speed, about 40. Setup: This prob­lem essentially requires us to com0.7 p­ ercent the speed of light. pute distance d = vt, where v is the speed and t is the 43. Setup: A black hole has a Schwarzschild radius time.

M 36 R S = 2G 2 , where M = 8 × 10 kg for this prob­lem.

Solve: 100 million years is 3.15 × 1015 s, so in this c time, the cosmic ray has traveled (c) The relativistic proton is

156 ◆ Chapter 20 The Milky Way: A Normal Spiral Galaxy M = 2 × (6.67 × 10 −11 m 3 kg −1s −2 ) c2 8 × 1036 kg × = 1.19 × 1010 m, (3 × 108 m/s)2 or about 0.08 AU. Review: Note that this supermassive black hole has a radius much smaller than even the orbit of Mercury (0.387 AU). That is tiny.

44. Setup: Use the equation for circular velocity, v 2r M . vcirc = G , and solve for the mass, M = G r v 2r (2 × 106 m/s)2 × (1.24 × 1014 m) Solve: M = = 6.67 × 10 −11 m 3 kg −1s −2 G = 7.44 × 1036 kg, or about 3.7 million solar masses.

Review: We see that stars orbiting at a few thousand kilo­meters per second are prob­ably orbiting compact objects with millions of solar masses—­that is, supermassive black holes.

45. Setup: Volume of a torus is equal to V = (πr 2) (2πR), where, for this prob­lem, R = 8 kpc, and r = 1 kpc. We are looking for the volume of the habitable zone (HZ) of the Milky Way. Solve: V HZ = (πr 2) (2πR) = (π × 1kpc2) (2π × 8 kpc) = 158 kpc3. If we assume that the disk extends to a radius of 15 kpc (dia­meter of 30 kpc) and has a thickness of 1 kpc, then the total volume of the disk is Vdisk = πr2h = π × (15 kpc)2 × 1 kpc = 707 kpc3. The habit158 kpc 3 able zone contains = 0.22, or 22 ­percent of 707 kpc 3 the total disk.

Review: Even if we think that stars that host habitable planets can form only in ­these regions, ­there is still a reasonably large percentage of the Milky Way in which life could form.

Solve: R S = 2G

Using the Web Solutions 46. A natu­ral lightscape refers to “resources and values that exist in the absence of human-­caused light at night.” It is increasingly rare for ­people to see the Milky Way ­because of a combination of pollutants in the air and urbanization that produces light at night even in areas that are remote, such as many national parks. The National Parks Ser­v ice considers the natu­ ral landscape at night (including the sky) to be just as impor­tant of a resource to protect and trea­sure as the landforms of the parks themselves.

47. (a) Th ­ ere is a change in wavelength of the selected pictures of the APOD movie b­ ecause we can only see into the center of the galaxy in the infrared. Answers w ­ ill vary on recent images of the Milky Way and w ­ hether

you can see the Milky Way from your location. (b) The galactic center can only be seen in X-­rays and infrared ­because ­these wavelengths are needed to peer through all the dust. Answers ­w ill vary on reports of images from the Chandra X-­ray Observatory.

48. Rapidly moving stars in a small area have such short periods ­because of a large, massive central object. A combination of the mass of the central object and the small size of the orbit constrains the object to be a black hole. Answers w ­ ill vary on new results from the two research groups.

49. Students w ­ ill perform the Milky Way Proj­ect exercise at the Zooniverse to find galactic b­ ubbles. Reported results ­w ill vary.

50. The video simulation given in the link in the textbook is no longer available, and the site is not being maintained.

The simulation or one similar to it can be found at https://­s citechdaily​. ­c om​/­s imulation​- ­s hows​-­w hat​ -­h appens​-­w hen​-­t he​-­m ilky​-­w ay​-­a nd​-­a ndromeda​ -­galaxies​-­collide as well as on YouTube. The video that involves the Triangulum Galaxy in the interactions shows the results of simulations that extend further into the ­future, where the merger of the two large galaxies results in an elliptical galaxy, and the Triangulum orbits the two. A more recent link is found at https://­svs​.­gsfc​ .­nasa​.­gov​/­30955. The Andromeda Galaxy is more massive than the Milky Way, although “eating” may not be the best word for describing the merger.

EXPLORATION The Exploration for this chapter has students investigate Harlow Shapley’s work on the distances to globular clusters. His work using RR Lyrae stars is covered in Activity 25 of Learning Astronomy by D ­ oing Astronomy, where students check out the distances to two clusters using RR Lyrae stars and we plot the majority of the globular clusters for their analyses.

EXPLORATION SOLUTIONS Following is (Figure IM 20.1) a plot of all the clusters listed in ­Table 20.1, similar to Figure 20.26.

1. The geometric center of t­hese clusters appears to be at a distance between 8 and 10 kpc from the Sun. 2. The geometric center of ­t hese clusters is at a galactic longitude of 0 (or 360) degrees.

3. If the Sun w ­ ere at the center of the galaxy, then the clusters would be distributed randomly around the center of the graph. Th ­ ese clusters are not centered on the Sun but are offset, showing that the Sun is not at the center of the globular cluster distribution and therefore not at the center of the Milky Way.

Chapter 20

The Milky Way: A Normal Spiral Galaxy ◆ 157

FIGURE IM 20.1

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 25: RR LYRAE STARS AND THE ­DISTANCE TO THE CENTER OF THE GALAXY This activity applies to topics covered in Chapter 14. We are suggesting it ­here as well ­because it is impor­tant to our discovering the structure of the galaxy. Students use observations of RR Lyrae variable stars to determine the distances to globular clusters and, from ­those, find out where they are located in the Milky Way. They determine the size of the Milky Way using a method similar to that done originally. They also: • determine the distances to two globular clusters using RR Lyrae variables found in them.

• estimate the distance to the center of the Milky Way based on the distribution of 130 globular clusters.

• relate the contributions RR Lyrae variable stars have made to our knowledge of the Milky Way and our location in it.

Text reference: Section 20.1, Exploration

ACTIVITY 29: CALCULATING THE MASS OF THE CENTRAL OBJECT OF THE GALAXY Students use observations of stars orbiting near the center of the Milky Way to confirm the existence of a supermassive black hole as the central object. The method they use is closely connected to the methods used with orbiting objects in previous activities. They also: • demonstrate proficiency in the use of scientific notation in all calculations.

• apply Newton’s form of Kepler’s third law to infer the mass within the orbit of a centrally located star in the galaxy.

• compare the results with the mass of the Sun and the size of the Solar System.

• analyze the evidence for a supermassive black hole at the center of the Milky Way. • compare results with ­those published in astronomical journals.

Text reference: Section 20.3

CHAPTER 21

The Expanding Universe INSTRUCTOR NOTES Chapter 21 covers the Big Bang theory. Major topics include: • the cosmological princi­ple (homogeneity and isotropy).

• why the universe has no center.

• how Hubble’s law provided the foundations of the Big Bang theory.

• Hubble flow, redshift, and the expansion of spacetime.

• the cosmic micro­wave background (CMB).

then compare t­ hose answers with the meaning cosmologists use. (all LGs) • The chapter-­opening image asks the question, “What is the evidence for the Big Bang?” It has been less than a ­century (March 1929) since Hubble published the results of his observations of galaxies. He found that the farther away a galaxy was from us, the faster it was receding. What did that mean for our understanding of the universe? How has our knowledge “expanded” since that time?

DISCUSSION POINTS

PROCESS OF SCIENCE

• Ask students to come up with their own experiences of environments that are homogenous and/or isotropic.

Do we expect to find ­these in nature? (LG 1) • The Hubble constant has been a value of considerable and consistent debate, with values ranging from 30 to 500 km/s/Mpc. Why has a consistent mea­sure­ment of this value been elusive? Can personal bias affect its mea­sure­ment? Observational biases? (LG 2) • Students can find it difficult to comprehend the meaning of the Big Bang. The concept that space is stretching/expanding (Figure 21.5) and the effects that has on light waves (Figures 21.7 and 21.9) and on the background blackbody energy of the CMB (Figure 21.12) have been substantiated by Hubble’s law and the observations of the cosmic micro­wave background (Figure 21.13). Connect all ­these figures with Wien’s law to provoke discussion among students about the properties of the Big Bang and its observable consequences. (LG 3) • The information contained in Figure 21.14 represents a critical test of the Big Bang theory. As a think-­pair-­ share writing exercise, have students study the figure and the connected text material, and summarize their interpretation of what it means. Students can then check with a neighbor and form small teams to put a final summary together. A ­ fter pre­sen­ta­tions, the class may vote on the best summary. (LG 4) • This chapter introduces a number of words and phrases that have quite dif­fer­ent meanings from our everyday use: Hubble time, Hubble flow, Big Bang, homogeneous, isotropic, recombination, constant, look-­back time, scale ­factor, dark m ­ atter, dark energy. Have students summarize what ­these words or phrases mean to them and

The Pro­cess of Science for this chapter ends with the statement, “Even the most brilliant scientists must adjust when the data contradict their conclusions.” We could reflect on the work of Johannes Kepler and his need to drop all preconceptions and follow what the data ­were “telling” him.

This led to our understanding of orbits. Einstein thought that the universe needed to be static and inserted an extra term in his equations to explain why the universe ­hadn’t collapsed due to gravity. Hubble eliminated the need for Einstein’s extra term when he published the data that showed that the universe was expanding. It is in­ter­est­ing that Einstein’s extra term, his “biggest blunder,” has now been reinstated by researchers to explain dark energy. ­Will ­these scientists need to adjust when ­f uture data contradict their conclusions?

158

ASTROTOUR ANIMATIONS HUBBLE’S LAW This animation uses Hubble’s observed redshifts of the light from galaxies as a starting point for an exploration of the recessional velocities of the galaxies. The observation that the farther a galaxy was from us, the faster it was receding, led to Hubble’s law and the expansion of the universe.

The animation uses mega-­light-­year (Mly) instead of megaparsec (Mpc). This example ends with a Hubble constant of 22 kilo­meters per second per mega-­light-­year (km/s/Mly).

Additional examples are given in order to determine the age of the universe. The animation closes with an explanation of the CMB radiation as evidence for the Big Bang.

Text reference: Section 21.1

Chapter 21

BIG BANG NUCLEOSYNTHESIS This animation pre­sents the fusion stages of Big Bang nucleosynthesis, including why ele­ments heavier than helium did not form.

Text reference: Origins

ASTRONOMY IN ACTION VIDEOS EXPANDING BALLOON UNIVERSE Stacy Palen demonstrates the expansion of the universe using the balloon universe, starting with a half-­inflated balloon with a series of star stickers on it. Using string, Dr. Palen shows how t­ hose stars move apart by blowing up the balloon and then comparing the first and second mea­sure­ments. She also covers how the speed of expansion could be determined and how this model expands to the real universe. Length: 4:33.

Text reference: Section 21.1 OBSERVABLE VERSUS ­ACTUAL UNIVERSE In the Observable versus A ­ ctual Universe video, Stacy Palen starts with a dense, hot, opaque universe and then discusses why the universe became transparent. Dr. Palen then relates light travel time to a meter stick by shrinking the speed of light down to understandable numbers. If light traveled at 1 meter per second, then a­ fter 1 meter, we would see 1 meter away; ­after 2 seconds, 2 meters away, and so on. Length:1:44 Text reference: Section 21.2

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Before the discovery of the CMB, evidence of the Big Bang came in many forms, including Hubble’s redshift-­distance relationship and stellar chemical abundances.

2. Theoretical investigations of a Big Bang scenario predicted the existence of a background radiation between 2 K and 5 K; the discovery of the CMB confirmed this prediction.

3. Answers ­w ill vary. “Thermal echo of the universe’s explosive birth” is perhaps a wordy and overly technical description b­ecause it implies that the reader understands both thermal and echo; it is perhaps misleading as well, as it implies the event was an explosion. One generally views an explosion as traveling

The Expanding Universe ◆ 159

into the space around it, whereas the Big Bang was an explosion of space itself. The term echo implies that the explosion has bounced off of something back to us.

4. Bell Labs was actually supporting research into background micro­wave signals for the practical purpose of using such signals to determine the distances to satellites. The discovery of the CMB was accidental. However, Bell Labs has a long history of supporting scientific exploration and advancement b­ ecause, as the cliché goes, “a rising tide lifts all boats.” Scientific advances almost always benefit the public, even if that benefit is not clearly foreseeable at the time.

CHECK YOUR UNDERSTANDING SOLUTIONS 21.1 (b, a) Isotropy is uniformity in direction; homogeneity is uniformity in location.

21.2 (d) The Big Bang happened everywhere b­ ecause it was the beginning of the entire universe.

21.3A (c) The scale f­ actor shows how distances change in time.

21.3B (b) Scale ­factor and redshift are related by 1 RU = .

1+ z 21.4 (a) Temperature in the universe cools over time, so a cool blackbody t­oday would have been very hot at early times.

END-­OF-­CHAPTER QUESTIONS AND PROB­ LEMS SOLUTIONS Test Your Understanding 1. (b) Homogeneity means the distribution of ­matter is uniform on the largest scales.

2. (d) Isotropy means the distribution of ­matter looks the same from all viewing directions.

3. (c) Galaxies are so far away that we cannot resolve individual objects.

4. (d) Although the Milky Way appears to be at the center of the universe, an observer on any other galaxy would see the same ­thing, proving that we cannot be at its center.

5. (d) Galaxies are not literally moving at speeds greater than light; it just appears so ­because of the expansion of space.

6. (b, d) Although a, c, and e are explained by the Big Bang theory, they w ­ ere not predictions that it made; rather, they w ­ ere the basis from which it was developed.

160 ◆ Chapter 21 The Expanding Universe 7. (a) The “Hubble time” is the slope of the Hubble law and is an estimate of the age of the universe.

8. All answers apply.

9. (a) If repeated mea­sure­ments contradict a prediction from a theory, the theory is wrong.

10. (c) Although (a) and (b) are true, they are not part of the cosmological princi­ple.

11. (d) Our galaxy is embedded within space; that is, it is expanding away from other galaxies. However, our galaxy itself is not expanding ­because it is bound together by gravity.

12. (c) The scale ­factor is a mea­sure of how much a given distance has changed over time due to Hubble expansion.

13. (d) Although we call it a “bang,” the Big Bang was less of an explosion and more the first moment of the universe.

14. (b) Isotropy is when something looks the same in all directions.

15. (e) Carbon was too massive to have been created during Big Bang nucleosynthesis; the repulsion of the protons was greater than the thermal energy available at the time.

Thinking about the Concepts 16. (a) Yes, b­ ecause every­thing looks the same in all directions. (b) Yes, b­ ecause every­thing would look the same no ­matter where you w ­ ere standing, as long as you w ­ ere in the thick, dense fog. However, if you moved much from the center, you may see a lifting of the fog.

17. Galaxies are fixed in space, but space itself is expanding. Imagine space filled with a mesh or grid. The galaxies are not moving within the grid, but the space between gridlines gets larger. See Figure 21.5.

18. According to the cosmological princi­ple, an astronomer in another galaxy located anywhere in the universe would see all other galaxies moving away with the same redshift-­d istance law as we observe.

19. Quite the opposite, Hubble’s law implies that ­there is no center of the universe ­because all galaxies are moving away from each other. As a result, each observer thinks they are at the center of the universe, meaning ­there is no center at all.

20. The scale f­ actor tells us how the distance between galaxies has changed over time. If we imagine two galaxies as being embedded in space and space has expanded over time, then the scale f­ actor tells us the expansion rate of the universe.

21. The expansion rate of the universe does not make the Sun or the Milky Way bigger, b­ecause ­these are objects that are held together by their own self-­g ravity

or the gravitational attraction of their contents. The “Hubble flow” of expansion only happens on the largest scales, in which gravitational attraction is weaker than the expansion rate.

22. Answers ­w ill vary. The question asks for students’ viewpoints and opinions.

23. Verified predictions of the Big Bang include the chemical abundances of stars and the interstellar medium, the presence of the cosmic micro­wave background, inhomogeneities, and fluctuations within the CMB. Hubble’s redshift-­d istance relationship was the impetus for the development of the theory.

24. Regardless of your position in the universe, all distant galaxies recede from you as the universe expands. If you claim that a distant galaxy is moving rapidly away from you, an observer in that distant galaxy would say that you are actually rapidly moving away from him or her. In other words, ­every observer in the universe perceives himself or herself to be at the center of the universe’s expansion. Another answer is that the Big Bang took place everywhere; every­one thinks they are at the center.

25. As long as recessional velocities remain well below the speed of light, the more complex relativistic equation reduces to the ­simple form of vr = cz. As recessional velocities approach the speed of light, the ­simple approximation breaks down, and we must use the more complete relativistic version of the relationship and consider the scale ­factor of the universe.

According to Hubble’s law, the most distant galaxies with the highest redshifts have recessional velocities approaching the speed of light.

26. The fact that the CMB displays a blackbody spectrum ­today indicates that the universe was a blackbody at early times when the spectrum was formed; that is, dense and opaque. This is what the Big Bang theory predicted, and this observation helped to rule out the steady-­state theory.

27. The tiny variations in the CMB are the microscopic fluctuations of structure that grew over time into the large-­scale structure we observe and live in ­today.

Studying the properties of ­those fluctuations gives us very strong tests of the Big Bang theory.

28. The observed abundances of isotopes confirm the predictions of Big Bang nucleosynthesis, an impor­ tant test of the Big Bang theory. Comparing current abundances with models of isotope formation in the Big Bang helps pin down the density of the early universe. By comparing the abundances of isotopes such as 2H (deuterium) and 3He found in the universe ­today with predictions in dif­fer­ent models of how abundant isotopes should be when formed at dif­fer­ent

Chapter 21

densities, cosmologists can find the density of normal ­ atter. m 29. Nuclear fusion requires high temperatures to overcome the repulsion of protons from each other, and high densities so that many collisions happen in a given amount of time. As nuclei grow larger, that repulsion increases; thus, higher temperatures and densities are needed to build nuclei into heavier ele­ ments. However, the universe expanded and cooled as it built He from H, and thus conditions ­were never appropriate for the He to fuse into heavier ele­ments.

30. A star converts only a percentage of its mass of hydrogen into helium over its lifetime, and much of that helium is never released back into space b­ ecause it is locked up in white dwarfs, neutron stars, or black holes. Stellar evolution therefore tells us that the amount of helium ­today is larger than could have been produced inside stars and must therefore have been “primordial” in nature (produced before stars).

Applying the Concepts 31. Setup: The area of a square of size s is s2 and the volume of a cube is s3. We can use t­ hese scaling relationships to answer the questions.

Solve: If a square doubles in size, its area increases by 22 = 4. The volume of a cube whose side has doubled in size increases by 23 = 8.

Review: If we divide the volume by the area, we should recover the f­actor of the change. That is, V/A = s. If the size doubles, the ratio doubles.

32. Setup: Error bars are shown for individual data points, but are they used when one shows a range of values? If a range of values is shown, does the thickness of the lines or range indicate the uncertainty?

Solve: In Figure 21.12, individual data from the COBE mea­sure­ments are shown. The mea­sure­ments ­were so precise that the uncertainties lie within the size of the data points. Although Figure 21.12 does not show the tiny mea­sure­ment errors, the fact that the observed spectrum so closely matches the one predicted by the theory in itself is a strong statement of confidence.

Review: Error bars are not the only way to show uncertainty or to gauge ­whether results are significant.

33. Setup: Observations are marked “observed” in Figure 21.14, whereas curves and regions that have comments like “would have” or “if the” tells us that ­these are predictions.

Solve: (a) The vertical yellow bar and slanted lines/ curves show predictions of theory.

The Expanding Universe ◆ 161

(b) The horizontal pastel bars and vertical black line show observation. (c) The observations and predictions match up exquisitely. For example, the predicted helium abundance and density cross the observational region where the ­actual density and abundances are shown.

Review: A strong test of a theory is when the observations and predictions from a number of dif­fer­ent tests all match up.

34. Setup: This is an exercise in unit conversion and all the necessary conversion ­factors have been given.

Solve: Hubble’s constant is assumed to be 80 km/s/ Mpc. We start with taking the inverse and then ­doing the unit conversions.

1 1 = H0 80km/s/Mpc 106 pc 3.09 × 1013 km 1 yr 1 1 = 1 pc 3.16 × 10 7 s H0 80 km/s/Mpc 1 Mpc 10 = 1.22 × 10 yr,

or around 12 billion years If the Hubble constant increases to 80 km/s/Mpc, the age w ­ ill decrease to 12 billion years.

Review: The universe is approximately 13.8 billion years old based on the current value for the Hubble constant.

35. Setup: This is an exercise in unit conversion, and all the necessary conversion ­factors have been given.

Solve: For a Hubble constant of 50 km/s/Mpc and then for 100 km/s/Mpc, the Hubble time is 106 pc 3.09 × 1013 km 1 yr 1 1 = 1 pc 3.16 × 10 7 s H0 50 km/s /Mpc 1 Mpc 10 = 1.96 × 10 yr

106 pc 3.09 × 1013 km 1 1 = 1 pc H0 100 km/s /Mpc 1 Mpc 1 yr × = 9.8 × 10 9 yr 3.16 × 10 7 s The age of the universe for a value of 100 km/s/Mpc is in direct conflict with the ages of the oldest stars in globular clusters, ages thought to be over 12 billion years based on observations. ­There is no conflict with the ages of rocks on Earth ­because the Sun and its planets w ­ ere formed less than 5 billion years ago.

Review: Note that this is a very large age range (a ­factor of 2), which explains why ­there was g­ reat uncertainty about many of the specific details of the universe’s evolution. This also shows how hard it was to make ­these mea­sure­ments.

1 relates the expan36. Setup: The scale f­ actor RU = 1+ z sion of the universe (size) to the observable redshift.

162 ◆ Chapter 21 The Expanding Universe Solve: A redshift of 2 corresponds to a scale f­ actor of one-­third; thus, the universe has expanded by a ­factor of 3 since the light was emitted.

Review: This implies that the universe was infinitely small (RU → 0) when the universe first began (z → ∞).

1 37. Setup: The scale f­ actor RU = relates the expan1+ z sion of the universe (size) to the observable redshift.

Solve: A redshift of 8 corresponds to a scale f­ actor of one-­ninth; thus, the universe has expanded by a ­factor of 9 since the light was emitted.

Review: This implies that the universe was infinitely small (RU → 0) when the universe first began (z → ∞). Δλ . To find the size 38. Setup: Redshift is defined as z = λ of the universe at the time this light was emitted, use 1 the scale ­factor RU = .

1+ z Δλ 547.2 −121.6 Solve: The redshift is z = = = 3.5.

121.6 λ At the time this light was emitted, the universe was 1 RU = = 0.22 or 22 p­ ercent of its pre­sent size.

1 + 3.5 Review: Recall that the universe was smaller at higher redshift.

39. Setup: Hubble’s law states that v = H0 d, which we can v 1 . The scale ­factor RU = rearrange to give d = .

H0 1+ z 287,000 km/s v = = 4,100 Mpc H0 70 km/s/Mpc (b) 4,100 Mpc × 3.26 Mly/Mpc = 13.4 Mly, so the look-­back time is 13.4 billion years. If we assume a Hubble time look-­back time of 13.8 billion years, then the quasar formed 13.8–13.4 billion years = 0.4 billion years or ~400 million years ­a fter the Big Bang. (c) The scale f­actor of the universe at that time was 1 1 RU = = = 0.147. It was around 1 + z 1 + 5.82 15 ­percent of its current size.

Review: Note that this galaxy was not actually around only 400 million years a­ fter the Big Bang. At large distances, the simplistic treatment we have used ­here is not appropriate. The ­actual age is around 900 million years, as computed by a cosmological calculator such as the one found at http://­astro​.­ucla​.­edu​ /­~wright​/­CosmoCalc​.­html.

2.9 × 106 nm K .

40. Setup: Wien’s law is T = λpeak Solve: (a) d =

Solve: Reading off of Figure 21.12, the peak of the spectrum occurs just past 1 mm, around 1.05 mm (mea­sur­ing to the center of the data point). Using this value in Wien’s law, we find 6 6 2.9 × 10 nm K 2.9 × 10 nm K T= = = 2.76 K. λpeak 1.05 × 106 nm The ­actual temperature of the CMB is 2.73 K, so this is very close.

Review: We could have done much better if we had a more accurate spectrum. An Internet search turned up a value of 2.72548 ± 0.00057 K based on values from a number of sources.

41. Setup: For speeds that are not “too fast” (that is, nonrelativistic, or less than about 10 ­percent the speed of light), the redshift formula can also be written as v z = r . If it turns out that z is negative, then z is a bluec shift rather than a redshift. v −368 km/s Solve: z = r = = − 0.00123 c 3 × 105 km/s Review: This is a very small blueshift b­ ecause we are dealing with a speed that is very small compared with that of light.

42. Setup: Dividing the mass density (kg/m3) by the mass of a hydrogen atom (kg) yields the number of atoms per cubic meter.

4 × 10 −28 kg/m3 Solve: = 0.24 hydrogen atoms per 1.66 × 10 −27 kg cubic meter.

Review: This is about one atom each per 4 cubic meters, which is about the volume of a small shower stall. The average density of the universe is very low.

43. Setup: To compute the ratio of the density of the universe to that of air on Earth, simply divide the two.

1.2 kg/m3 Solve: = 3 × 1027.

4 × 10 −28 kg/m3 Review: Earth’s atmosphere is 3 billion billion billion times denser than the average density of the universe (3,000,000,000,000,000,000,000,000,000).

1 and volume scales 44. Setup: The scale ­factor RU = 1+ z with size cubed.

Solve: At a redshift of z = 10, the scale ­factor w ­ ill be 1/11, so the universe would be 113 = 1,331 times denser than ­today, for a total of 5.3 × 10 −25 kg/m3.

Review: Redshift of 10 is very high and very far back in time (RU = 9 ­percent of the current size, universe was about 600 million years old), but even then densities ­were very low compared to ordinary m ­ atter. We see

Chapter 21

that the universe was very dense for only a short period of time.

1 45. Setup: The scale ­factor RU = gives the relative 1+ z size of the universe at any redshift z when light was emitted.

Solve: In the case of when the CMB was emitted, the 1 = 0.000999 or about 0.001 scale ­factor RU = 1 + 1000 the size of the universe t­ oday.

Review: The universe was certainly much smaller at the time the CMB was emitted than ­today.

Using the Web Solutions 46. Einstein (and many of his contemporaries) believed the universe was in “steady state,” eternal and unchanging. Lemaître was one of the first to propose a definite beginning to the universe. The steady-­state theory proposes that the universe has always looked as it looks now and that t­ here is a steady creation of new m ­ atter to keep the overall density constant. ­There ­were a number of observations that called the steady state theory into question. Certainly the observations of Edwin Hubble, but most decidedly, the prediction and discovery of the cosmic micro­wave background.

47. The age of the universe, around 13.7–13.8 Gyr, is much greater than the age of our solar system (about 4.5–5.0 Gyr) and also older than the oldest globular clusters (around 12–13 Gyr). The determinations for the age of the universe change as more data are obtained or previous data are reanalyzed. Based on the 2012 data release by the Planck satellite, the age is 13.772 Gyr; based on the 2013 data release, it is 13.82 Gyr; based on the best fit to 2015 data alone, it is 13.813 ± 0.038 Gyr. 48. In his talk, Dr. Wilson judges the discovery of the CMB as his most impor­tant discovery, but he never explic­itly states why. However, as he notes, this discovery brought cosmology to the forefront of the world’s scientific attention and was quite possibly the greatest step science has made t­oward answering the question, How did we get ­here? He and his colleague, Dr. Penzias, identified the source of this radiation somewhat by happenstance when a mutual friend connected their discovery with work done by researchers at Prince­ton (led by Dr. Dicke) who had predicted the existence of this signal.

49. Answers could vary. ­There was a new study (2018) of the polarization of the CMB that aimed to solve some of the anomalies pre­sent in the micro­wave results.

“Now we ­really are confident that we can retrieve a

The Expanding Universe ◆ 163

cosmological model based on solely on temperature, solely on polarization, and based on both temperature and polarization. And they all match,” says Reno Mandolesi, principal investigator of the LFI instrument on Planck at the University of Ferrara, Italy. Basically, all par­ameters for the beginning of the universe and its evolution have been determined within reasonable uncertainties. Based on the 2013 data release by the Planck mission, the age of the universe is 13.82 Gyr; based on the best fit to 2015 data alone, it is 13.813 ± 0.038 Gyr. The Hubble constant, 67.4 km/s/ Mpc, has an uncertainty of less than 1 ­percent.

50. Students complete the Hubble law exercise, in which they mea­sure the angular size and recessional velocity of many galaxies and plot t­hese to mea­sure the Hubble constant.

EXPLORATION In this Exploration, students use the surface of a balloon to get a feel for how an “expansion” changes distances between objects. Throughout this Exploration, students are encouraged to think of the surface of the balloon as a two-­d imensional object, much as the surface of Earth is a two-­d imensional object for most ­people. The average person can move east or west, or north or south, but moving into Earth and out to space are not options. This activity is a hit with all ages, and ­really does give a good visualization of the expansion of the universe.

EXPLORATION SOLUTIONS 1. Graphs should trend upward at a roughly constant slope, although ­there ­w ill prob­ably be a lot of scatter.

2. The reference galaxy does not move.

3. The trend line would not change if a dif­fer­ent galaxy ­were picked; it is still the same balloon universe.

4. “More stuff” means it was harder for the elastic to expand.

This mimics the extra mass in some regions that created clusters of galaxies, which opposed the expansion.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES ACTIVITY 26: FINDING THE EXPANSION RATE AND THE AGE OF THE UNIVERSE Students analyze the distances to galaxies and the velocities at which they are receding from us. They also:

164 ◆ Chapter 21 The Expanding Universe • determine a value for the Hubble constant through analy­sis of mea­sure­ments of the angular sizes and redshifts of galaxies.

• estimate the corresponding age of the universe and compare it with the age of the Sun and the Milky Way.

• explain why Hubble’s law implies that the universe is expanding.

• summarize how our view of the universe has changed as the value of the Hubble constant has improved.

Text reference: Section 21.2

CHAPTER 22

Cosmology INSTRUCTOR NOTES Chapter 22 covers the physics of the Big Bang and the evolution of the universe. Major topics include: • the critical density and three pos­si­ble fates of the universe.

• the cosmological constant and evidence for dark energy.

• the geometry of the universe: age, fate, and shape.

• how inflation solves the flatness and horizon prob­lems.

• the physics of the first few seconds of the Big Bang.

• the ­g rand unified theory and a theory of every­thing.

• superstring theory and multiverses.

In the introductory astronomy curriculum, cosmology can be an opaque subject for nonscience students, which is problematic ­because it is also the subject that they look forward to the most. Part of this is due to the scope: we cover all of time and all of space, from unimaginably small zillionths of a second in time to billions of light years in space. Our advice: go slow and tie it all together.

DISCUSSION POINTS • Have students carefully review Figure 22.1 and Fig­ ure 22.3. How do they differ? Why do they differ?

Work this into a think-­pair-­share activity and have stu­ dents share which cosmological model seems most accurate according to ­today’s data. (LG 1) • This chapter opens with a discussion of how gravity affects the expansion of the universe; then we learn that dark energy is working against gravity. Similarly, Einstein first proposed, then retracted the cosmologi­ cal constant, and it is now favored again. Do ­these evolving views represent a weakness or strength in the pro­cess of science? (LG 2) • The concept of inflation has been put forward as a way to solve the flatness and horizon prob­lems in cosmol­ ogy. First, explain what is meant by the flatness and horizon prob­lems. Why are t­ hese considered “prob­ lems”? Inflation calls for two points in space to travel apart at velocities greatly exceeding the speed of light.

Discuss how inflation can happen without violating the theory of relativity, and how it solves ­those two prob­lems. (LG 3)

• Discussing the universe at its very earliest times requires imagining states that are unknown to current physics. Should t­ hese be considered scientific theories, or scientific ideas? (LG 4) • Students have learned that stars derive their luminos­ ity mainly from nuclear fusion of hydrogen into helium. Use Big Bang nucleosynthesis to explain why hydrogen and helium are the two most abundant ele­ ments in the universe. Have students create 18 × 24-­ inch posters that outline the life of a proton from the era of Big Bang nucleosynthesis to being part of an iron atom in their blood. (LG 4) • List the major stages and eras shown in Figure 22.13 on a sheet of paper, using a font that is large enough to allow for cutting the list into strips. Make enough cop­ ies to give to teams or individuals. Cut them into slices, mix the slices up, and place them in envelopes to dis­ tribute. Have the students work in a think-­pair-­share activity to place the strips in the proper chronological order. Discuss how the a­ ctual order makes sense when considering an expanding, cooling gas. Which events are related to the forces that operate in the modern uni­ verse? Which events are still occurring t­ oday? (LG 4) • The chapter-­opening image asks the question, “What ­w ill be the fate of the universe?” The discussion ­here can be an extension of one suggested above: Have students carefully review Figure 22.1 and Figure 22.3. How do they differ? Why do they differ? Work this into a think-­ pair-­share activity and have students share which cos­ mological model seems most accurate according to ­today’s data. Extend that discussion into what the fate of the universe ­w ill be if the model is correct.

PROCESS OF SCIENCE Einstein called the use of a cosmological constant, a fudge ­factor in his equations, his “biggest blunder.” A l­ittle over 40 years ­after his death, observations of Type Ia supernovae indicated an accelerating universe, and the cosmological constant has been revived as one of the driving forces for the acceleration. The interpretation has changed, and ­there are some characteristics of dark energy that cannot be explained by such a constant. The pro­cess of science w ­ ill continue as we devise ways to discover just what dark energy is.

165

166 ◆ Chapter 22 Cosmology

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Baryons are what is missing.

2. WHIM stands for warm-­hot interstellar medium. It may have been detected as a dim outline in the cos­ mic micro­wave background, indicating that radiation has been scattered.

3. The temperature of WHIM may be too cold for X-­ray emission to occur.

4. When two dif­fer­ent teams confirm a similar result, it gives added credence to the theory.

CHECK YOUR UNDERSTANDING SOLUTIONS 22.1 (c) If t­ here is enough mass to overcome gravity, the universe ­ w ill collapse. The critical density is the amount of mass needed for that to happen.

22.2 (c) At the largest distances, objects are moving away from us faster than Hubble’s law predicts.

22.3 Inflation solves the horizon and flatness prob­lems.

22.4 (d) According to our theories, the antimatter anni­ hilated with ­matter, but a small asymmetry of ­matter existed, leaving the atoms that fill our universe ­today.

22.5 (a) At pre­sent, t­ here is no way to test and nullify the idea of parallel universes, so it remains a hypothesis.

END-­OF-­CHAPTER QUESTIONS AND PROB­ LEMS SOLUTIONS Test Your Understanding 1. (d) The density of the universe (Ω) dictates w ­ hether it ­w ill be open, closed, or flat.

2. (c) “Dark energy” is the name that has been given to the “stuff” that ­causes acceleration. The cosmological constant is the mathematical treatment of that effect.

3. (a) The CMB is uniform to about 1 part in 100,000.

4. (a) String theory has not yet made testable predic­ tions, which leaves it as a hypothesis, or more accu­ rately, just an “idea.”

5. (a, f, j, b, e, g, d, i, h, c) This is shown in Figure 22.13, which gives a graphical chronology of the universe.

6. (a) Prior to the epoch at which the CMB formed (“decoupling”), m ­ atter and radiation w ­ ere in equi­ librium, which means that light could not travel very far before hitting a particle. This made the universe opaque.

7. (c) The critical density is the density at which the uni­ verse changes from open to closed.

8. (b, d) The electromagnetic force and the weak nuclear force combine to make the electroweak force.

9. (b) See Figure 22.7, which shows triangles and cir­ cumferences for three dif­fer­ent geometries.

10. (b) Protons ­were the first to form; ­these ­were built up during Big Bang nucleosynthesis into nuclei. Much ­later, the universe cooled enough for electrons to recombine with the nuclei to form neutral atoms.

11. (d) Quarks are the particles out of which baryons (pro­ tons, neutrons) and mesons (quarks + antiquarks) are made.

12. (b, e) The density of the universe is consistent with Ωm + Ω Λ = 1, which means it is flat, even though dark energy seems to dominate, which means the expan­ sion of the universe is accelerating.

13. (a) This is the pro­cess of pair production in reverse.

14. (b) High-­redshift supernovae show us that very distant objects are moving away from us faster than predicted by the Hubble law predictions.

15. (b) ­ There are many statements of the anthropic princi­ple, but the essence is that every­thing happened such that we are h­ ere to see it ­today.

Thinking about the Concepts 16. To make the universe collapse, the average density of ­matter would have to be higher than the critical den­ sity, and the forces of gravity from this ­matter would have to overpower that of any dark energy, which would cause the universe to be pushed apart.

17. When we look at distant galaxies and observe Type Ia supernovae, we see that their recessional velocities increase with distance faster than the linear rela­ tionship of Hubble’s law, suggesting that ­t here is a ­nonzero cosmological constant. Einstein brought in the cosmological constant to explain why the uni­ verse d­ idn’t collapse in on itself.

18. Dark energy is vacuum energy—­that is, a repulsive force that results from the energy pre­sent in totally empty space.

Chapter 22

19. Just as our planet, galaxy, and local group of galaxies are not being torn apart by the expansion of space, so too are they stable against the repulsive force of dark energy: gravity holds them together.

20. Hubble time does not take into account the changing dynamics that result from having dif­fer­ent cosmolog­ ical densities and a cosmological constant (dark energy). The Hubble constant may vary with time.

21. For the universe to be exactly flat ­today, it needed to be exactly flat for all time, even at the earliest points in its evolution. It is too close to being exactly flat to have happened by chance. Some condition in the early universe must have caused Ωm + Ω Λ = 1 out to at least the tenth decimal place; but, what was it?

22. ­Matter and information did not travel faster than light; space itself expanded so fast that the distances between points in space expanded faster than at the speed of light. ­There was no causal connection between ­these points.

23. The early universe was hot and dense, which are situa­ tions that are replicated (or that we try to replicate) within the realm of high-­energy physics. At t­hese high temperatures and densities, all sorts of exotic forms of ­matter can exist, which may have had sub­ stantial influences on how the early universe evolved.

24. (a) Gravity fights against expansion, so if gravity changes with time, the universe might close on itself or accelerate faster than predicted. (b) Electric force holds or repels charged particles. If this changed with time, the structure of atoms and pro­cesses of fusion and fis­ sion would be very dif­fer­ent from ­today. For example, maybe no neutral atoms would form (electric force too weak) or no nuclei would form (electric repulsion too strong).

25. A theory may be incomplete, but that does not mean it is inaccurate. Instead, ­there is just more to be learned.

However, in learning more, one might find a better theory that replaces the current, incomplete one. At pre­sent, the standard model is the most successful theory of particle physics, so the pieces that are miss­ ing are considered to be small compared to the total body of predictions that the theory makes.

26. If a photon contains more energy than the combined mass energy of a particle and its antiparticle, then the photon can become a particle-­antiparticle pair that ent directions, conserving mass flies off in dif­fer­ energy and momentum.

27. The Planck era was the earliest time of the early uni­ verse, when all the laws of physics that we know ­were combined in a single way. Other scientists define the Planck era as the early time of the universe during which we cannot explain what was ­going on ­because energies ­were too high and timescales ­were too small.

Cosmology ◆ 167

28. ­Grand unified theories (GUTs) attempt to unite the electroweak and strong nuclear forces. Theories of every­thing (TOEs) attempt to unify suitable GUTs with gravity. One basic distinction is that GUT forces are inherently quantum mechanical in nature, whereas gravity is described with general relativity, which is not yet quantum mechanical.

29. Superstring theory was developed as a pos­si­ble way to reconcile general relativity and quantum mechanics; however, it is impossible at the pre­sent time to test its proposals. A theory must be testable and disprovable.

Currently, it may be best described as an “idea” or a “philosophy.”

30. Answers w ­ ill vary. As one example, if the nuclear force ­were stronger in a parallel universe, then heavier nuclei than helium might have been created during the Big Bang. If the gravitational constant w ­ ere weaker, then structure would not have formed. If it ­were stronger, ­there would be just humongous structures left, and the universe would be closed. If the cosmological constant ­were stronger, any structures would be infinitely far apart.

Applying the Concepts 31. Setup: Logarithmic axes increase by ­factors of 10 (number spacing is not uniform), whereas linear ones increase by single values.

Solve: (a) There are two vertical axes. The left is loga­ rithmic (although not in powers of 10), the right is lin­ ear. (b) The top label is linear. (c) The bottom axis is logarithmic. (d) The relationship between relative brightness and time is not linear, but the fainter some­ thing is, the older it is.

Review: Logarithmic axes are a very useful way to pack a large range of numerical information into a single graph.

32. Setup: Acceleration means that the expansion becomes greater, faster; that is, the curve ­will become more verti­ cal with time.

Solve: (a) The solid orange curve trends t­ oward hori­ zontal, whereas the blue dashed curve trends ­toward vertical. (b) The solid orange curve shows a universe that expands more slowly with time; this is decelera­ tion. The blue dashed curve shows the universe expanding more quickly with time; this is accelera­ tion. (c) A straight line would indicate no change in the scale f­ actor of the universe, meaning it was neither accelerating nor decelerating.

Review: Models that study the past can fit the ­f uture of the universe.

168 ◆ Chapter 22 Cosmology 33. Setup: For data to be consistent with a model, the pre­ 37. Setup: Uncertainty in one variable yields a direct diction (the curves in this case) has to pass through uncertainty in the next if ­there is a ­simple linear or inverse relationship. We use the information in Work­ most of the points, or t­ here should be very l­ ittle scatter between the points and the line. ing It Out 21.1 to find the age of the universe.

Solve: (a) The right-­hand side of the graph is the f­ uture.

Solve: ­Because the Hubble constant is good to 4 ­percent, so is the age of the universe (roughly). With ­Because it has not happened yet, ­there can be no data. (b) The red, orange, and green models are excluded. (c) H0 = 70 km/s/Mpc, the universe has an age of 14 Gyr; a 4 ­percent margin of error yields minimum and Data go back roughly 9 billion years. (d) 9/13.8 = 0.65, maximum ages of 13.4 and 14.6 Gyr, respectively. or about two-­thirds the age of the universe.

Review: The research continues. Measurements from a Review: It would be helpful to get additional data for variety of observations give a value of 73.5 km/s/Mpc, Type Ia supernovae having relative brightness less than with an uncertainty of only two percent. Analysis 0.001 where the models diverge, to the extent that bet­ of data from the Planck mission give a value of ter distinction among the models would be pos­si­ble.

67.4 km/s/Mpc with an uncertainty of less than one 34. Setup: Logarithmic axes increase by ­factors of 10 percent. It seems we still have work to do. (number spacing is not uniform), whereas linear ones ­ ill use the critical density calcu­ 38. Setup: We w increase by single values. Inflation occurred during lated in Working It Out 22.1 of 9.5 × 10 −27 kg / m3. the time period indicated by the orange bar.

From Chapter 20, we learned that the total mass of Solve: Figure 22.8 has logarithmic axes. During infla­ 37 in size.

At the the Milky Way is between 1.0 × 1012 M Sun and tion, the universe increased about 10 beginning of the inflationary period, the universe was 1.5 × 1012 M Sun. We ­w ill pick the more massive esti­ 37 mate. Getting an estimate of the volume of the galaxy about 10 times smaller than in the noninflationary model. It expanded “from ten-­trillionths the size of the ­ ill use the rotation curve of Fig­ is a bit trickier. We w nucleus of an atom to a region about 1 meter across.” ure 20.16 and set the radius of the MW that includes Review: We do not yet have ­actual data to support most of the mass, including dark ­matter, at 15 kpc. this graph. If what is shown is even somewhat close to We assume that its mass is distributed roughly what ­really happened between 10 −35 and 10 −33 sec­ ­spherical out to that radius to make the calculations onds ­a fter the Big Bang, the universe truly did inflate simpler. The average density ­w ill be its mass divided exponentially in size. by its volume.

35. Setup: Review Chapter 1, which discusses the pro­ Solve: For the density of the Milky Way, we divide its cess of science. mass by its volume.

Solve: Given the excellent agreement between data from mass: 1.5 × 1012 MSun × 2 × 1030 kg / MSun = 3.0 × 1042 kg a large array of dif­fer­ent experiments, astronomers would 4π volume: (1.5 × 10 4 pc × 3.09 × 1016 m/pc)3 ≈ 4.0 × 1062 m 3 prob­ably expect that new observations that propose radi­ 3 4π studied4to ensure cally dif­fer­ent values must be carefully 16 3 (1.5 × 10 pc × 3.09 × 10 m/pc) ≈ 4.0 × 1062 m 3 ­ istakes or errors3in the mea­sure­ments. that t­ here are no m 3.0 × 10 42 kg = 7.0 × 10 −21 kg/m3 .

If t­hese data prove to be robust, then all the data taken average density: ρaverage = 62 3 4.0 × 10 m ­until now must be studied, and if all are robust, the theory 3.0 × 10 42 kg ρaverage = = 7.0 × 10 −21 kg/m3 . of the Big Bang must be considered flawed.

4.0 × 1062 m 3 Compared to the critical density of the universe: Review: The strength of science is that it accommo­ dates new, reliable findings and does not reject them, 7.0 × 10 −21 kg/m3 ≈ 8 × 105. The average density of no m ­ atter how controversial the findings are.

9.5 × 10 −27 kg/m3 36. Setup: Logarithmic axes increase by ­factors of 10 the Milky Way is about 800,000 times as ­g reat. (number spacing is not uniform), whereas linear ones Review: If we use Figure 20.12 and assume a radius increase by single values. of the Milky Way of 200,000 pc, then we get an aver­ Solve: (a) The time axis is approximately logarithmic. age density that is roughly 300 times that of the criti­ (b) Density drops from 1099 to 10 −26, or 125 o­ rders of cal density of the universe. One ­thing ­these estimates magnitude. (c) Temperature drops from 1032 to 2.73, ­ atter in the universe -­baryonic and show is that m or about 31 o­ rders of magnitude. dark - is clumpy.

Review: Over 14 billion years (give or take), the uni­ 39. Setup: As given in the prob­lem, the scale ­factor can verse has exhibited dramatic decreases in tempera­ ρ ture and density. Note that most of that happened be solved for RU = 3 o . ­a fter the first few hundred thousand years. ρ

Chapter 22

Cosmology ◆ 169

Review: Remember that although the energy per atom might be small, ­there are about 6 × 1023 atoms in ρo 3 × 10 −28 kg/m3 −10 RU = 3 =3 = 6.3 × 10 . a gram of hydrogen, so 1 gram’s annihilation results ρ 1.2kg/m3 in a huge amount of energy.

Review: This means the universe was less than a bil­ 43. Setup: We need to know how many protons are in the lionth of its current size. detector. ­Water has a molar mass of 18 grams per mole.

40. Setup: ­ Because ­ there are two particles and two ­Because ­there are 6.02 × 1023 molecules per mole, ­there are roughly 3.3 × 1022 molecules in 1 gram of ­water. gamma rays, each gamma ray is formed from the com­ With a total capacity of 20 million kilograms, or 20 bil­ plete conversion of one particle’s mass into energy, so lion grams, ­there are about 6.7 × 1032 ­water molecules we use Einstein’s mass-­energy formula, E = mc2 .

Solve: E = mc 2 = (1.67 × 10 −27 kg)(3 × 10 8 m/s) 2 ​ in the detector. ­Water is made of two hydrogen atoms −10 = 1.5 × 10 J. This is the energy of each gamma ray and one oxygen atom, which have a total of 1 + 1 + 6 = 8 created. protons, so the total number of protons in Super-­ Review: Although this might not seem like a lot of Kamiokande is around 5.4 × 1033 protons.

Solve: No decays in 5 years limits the lifetime to energy, consider that one gram of hydrogen contains 23 around 5 × 5.4 × 1033 = 2.7 × 1034 years.

6 × 10 protons, so its annihilation would result in 14 about 10 J being given off.

Review: This prob­lem is oversimplified and the real 41. Setup: At 2.7 K, using Wien’s law, each photon has a limit is around 1033 years, but it still shows this par­tic­ u­lar GUT theory may be ruled out. Statistically, we 2,900 µ K 2,900 µ K need a much longer baseline of continuous operation peak wavelength λpeak = = ≈ 1 mm.

T 2.72 K than just 5 years. −34 2 8 44.

Setup: The circumference of a circle equals 2πr in flat (6.63 × 10 m kg/s)(3 × 10 m/s) hc We use E = = space; it is more than this value in open space and is 0.001 m λ less in closed space. = 2 × 10−22 J, Solve: A gravitational well is a closed space, so the cir­ and find that the CMB photon has an energy of about cumference is less than 2πr.

2 × 10 −22 J. Knowing how many photons ­there are and Review: This is another way to think about gravitational E = mc2 , we can compute the total energy and then time dilation, ­because the planet close to the star travels equivalent mass pre­sent in the CMB. a shorter distance than expected for a flat universe.

Solve: If t­ here are 500 million CMB photons for each 45. Setup: This is an exercise in comparing the order of H atom, then the total energy in ­these 500 million magnitude of two numbers. photons is (5 × 108 photons)(2 × 10 −22 J/photon) ​ Solve: The ratio between Earth’s atmosphere density = ​10 −13 J. This corresponds to an equivalent mass of 1.2 kg/m3 10 −13 J E and that of the universe: = 1.2 × 1026 −30 −27 3 m= 2 = = 10 kg. ­ Because the 9.9 × 10 kg/m 16 2 2 9 × 10 m /s c 1.2 kg/m3 = 1.2 × 1026 times denser than the average density of −27 −27 mass of a proton is about 10 kilograms, we 9.9see × 10that kg/m3 the universe. This is a ­factor of 120,000,000,000,000 the CMB contains about 1,000 times less equivalent ,000,000,000,000. mass than all the hydrogen atoms in the universe.

Review: The density of ­water is 1,000 kg/m3. The dens­ We can ignore it when considering the density of the est material on Earth is osmium at 22,500 kg/m3. The universe. least dense material on Earth is aerographite at a mere Review: It is impor­tant to note that even though 0.2 kg/m3. For Earth, the rec­ord vacuum, made at ­there are about 500 million times more CMB pho­ CERN, is around 1,000 atoms/cm3, or 1,000,000 tons than atoms, their total energy is insignificant atoms/m3. If we ­were to assume that ­those atoms ­were compared to that mass, b­ ecause each photon’s energy hydrogen, then the density would be about 2 × 10-21 kg/ is so low. m3, or still a million times the density of the universe.

42. Setup: Rather than worry about how much energy is Solve: Using the values given in the text,

given off per atom, simply use E = mc2 where the mass is 2 grams.

Solve: (a) E = 0.002 kg × 9 × 1016 m2/s2 = 1.8 × 1014 J. (b) This is approximately 4 ­percent of the energy released in the detonation of a 1-­megaton hydrogen bomb.

Using the Web Solutions 46. The dark-­energy survey ­w ill use optical imaging to study four probes of dark energy: Type Ia supernovae, baryon acoustic oscillations, galaxy clustering, and

170 ◆ Chapter 22 Cosmology weak gravitational lensing. Th ­ ese ­w ill study accelera­ tion of the universe, clustering statistics to constrain the history of the cosmic expansion rate, the growth rate of structure in the universe, and the clumping/ distribution of dark ­matter. Answers ­w ill vary on news, the proj­ect status, and any results.

47. Particles may have had no mass when they formed in the Big Bang. The Higgs boson was responsible for the transfer of mass to particles as they interacted with the Higgs field. Answers ­w ill vary on what is new in the search for this boson.

48. Exotic decays are better found with h­uman eyes ­because the Large Hadron Collider’s computer alg­ orithms ­were not written to find off-­center vertices.

Answers w ­ ill vary for the marking up of a few images and keeping rec­ords for homework.

49. Answers w ­ ill vary. Report on arguments for and against multiverses.

50. Answers ­w ill vary. Discuss ­whether the arguments made in the series are compelling and w ­ hether the science is explained in a way that is understandable to you.

EXPLORATION This exploration is an excellent introduction to particle physics and how we determine the characteristics of par­ ticles based on their decays.

EXPLORATION SOLUTIONS 1. ­There are prob­ably 60–100 dif­fer­ent particle tracks.

2. The magnetic field pointed perpendicular to the page.

3. Answers w ­ ill vary ­because they identify the types of tracks in Figure 22.15.

4. A charged particle with a nearly straight track was prob­ably both high energy (high speed) and massive.

5. (a) The looser spiral indicates a faster electron. (b) The looser spiral indicates a higher energy.

6. (a) A faint track indicates a faster particle. (b) A tight spiral indicates a slower speed and thus it should be brighter. (c) If a looser track is also brighter, the particle was traveling slowly; therefore, it must have had a higher mass than the particle in the tight track, and thus may not have been made by the same type of particle.

7. If both a proton and electron’s tracks curve by a very small amount, then the lower-­mass particle (electron) must be traveling much faster than the higher-­mass one (proton).

8. A neutral particle could decay into a charged particle, which could then decay back into a neutral particle, to produce a track with a definite beginning and end.

9. (a) If t­ here is no trace away from the track at the location of a kink then the emitted particle is neutral. (b) B ­ ecause a particle was emitted, mass and energy w ­ ere taken away from the original particle. (c) Answers ­will vary.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES Activity 27: A Cosmic Calendar Students or­ga­nize the natu­ral history of the universe onto a calendar representing 1 Earth year to gain a feel for the relative magnitudes of the timescales involved. During the activity they ­w ill also: • calculate relative timescales.

• identify impor­tant events in natu­ral history.

• choose appropriate time steps at dif­fer­ent epochs.

Text reference: Section 22.4

CHAPTER 23

Large-­Scale Structure in the Universe INSTRUCTOR NOTES Chapter 23 covers how the universe has evolved from the Big Bang to t­ oday. Major topics include: • types of structures: galaxy groups, clusters, superclusters.

• the role of gravity and dark ­matter.

• hot m ­ atter versus cold dark ­matter.

• the Dark Ages and the first generation of stars and galaxies.

• parallels between galaxy and star formation.

• cosmological evolution of galaxies.

• comparison of observations with simulations of structure.

As a wrap-up to this chapter, the textbook notes in Section 23.3 to compare galaxy formation to star formation.

This is a perfect example of how the same laws and princi­ples of physics can be used to understand ­things on vastly dif­fer­ ent scales of size and time. ­Were it not for this universality of physics, a class such as ours would not even be pos­si­ble.

DISCUSSION POINTS • Have students address a postcard to a beginning astronomy student who lives on a planet orbiting a star in the Andromeda Galaxy. Describe to the beginner the distributions of galaxies in the universe. Students should use their complete cosmic address for the return address. Have students share their descriptions and ask for peer reviews. (LG 1) • Ask students to create 18 × 24-­inch posters that show the distribution of galaxies in the universe. They should include the Local Group and Virgo Supercluster, and extend their sketches to the largest structures and voids, as they understand it. Structures should be labeled. Perhaps do an informal vote on which posters are the most instructive. (LG 1) • Compare images of the CMB fluctuations (see ­Figure 21.13) with ­those of large-­scale structure observed ­today (Figure 23.21). Ask students to enumerate the steps through which the former evolved into the ­later and then check their lists with ­others. (LG 2) • This chapter pre­sents the results of a number of simulations (Figures 23.9 and 23.19 through 23.21). Discuss

why simulations are useful and explore their limitations in testing scientific theories. (LG 2) • When comparing galaxy and star formation ­there are many parallels, but what are some distinct differences?

For example: Can star and planet formation be considered analogous? Is a halo like an Oort cloud? Does the center of a galaxy exert as much influence as the central star of a solar system? (LG 3) • Direct a think-­pair-­share activity based on the idea that as we look farther away in distance we are looking further back in time and can observe how galaxies change over billions of years. Think of at least one analogy when we consider life on Earth. (LG 4) • The chapter-­opening image is of the Hubble Ultra Deep Field 2014, combining ultraviolet, vis­i­ble, and infrared light observations. The question asked is, “What does this image tell us about galaxy evolution?” From http://­ hubblesite​.­org​/­news​_­release​/­news​/­2018​-­35: “The image straddles the gap between the very distant galaxies, which can only be viewed in infrared light, and closer galaxies, which can be seen across a broad spectrum. The light from distant star-­forming regions in remote galaxies started out as ultraviolet. However, the expansion of the universe has shifted the light into infrared wavelengths.

By comparing images of star formation in the distant and nearby universe, astronomers glean a better understanding of how nearby galaxies grew from small clumps of hot, young stars long ago.” That’s one insight into galaxy evolution. What other differences or similarities do students notice? What additional information do they need in order to know the ages of the galaxies based on look-­ back time? How many galaxies can be classified into the basic groups?

PROCESS OF SCIENCE We may not know what dark m ­ atter is, but observations, models, and other evidence point to the fact that it does exist. Certainly the challenge for ­today and the ­f uture is finding out just what dark m ­ atter is.

ASTROTOUR ANIMATIONS None for this chapter.

171

172 ◆ Chapter 23 Large-­Scale Structure in the Universe

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Laniakea is about 500 million light-­years across, or about 150 Mpc. 2. It was hard to discover our supercluster ­because we are inside the object looking outside of it, which makes it hard to determine the detailed structure of the surroundings, much like the difficulty of discovering the properties of our own galaxy.

3. No, the Milky Way and Local Group sit near the far edge of the supercluster.

4. To map out more distance superclusters, astronomers need to know the component galaxies’ positions and velocities, which is challenging for distant objects.

5. Astronomers set the initial bound­aries of dif­fer­ent clusters of galaxies based on finding the center of mass of clusters and superclusters. For Laniakea, the boundary was set based on the overall flow of galaxies through space. For the Local Group, the flow is ­toward the ­Great Attractor.

CHECK YOUR UNDERSTANDING 23.1 (c, b, d, a) 23.2 (b) The distribution of dark m ­ atter in the early ­universe caused the large-­scale structure we see ­today.

23.3 (c) The earliest generation of stars had no heavy ele­ ments at all.

23.4 (b) At high redshift, t­ here w ­ ere more irregular galaxies, indicating the spirals took time to grow.

END-­OF-­CHAPTER QUESTIONS AND PROB­ LEMS SOLUTIONS Test Your Understanding 1. (g, b, a, c, e, f, d) See Section 23.1.

2. (a) Gravity is the dominant force in all pro­cesses of formation of large astronomical objects (planets, stars, galaxies, ­etc.).

3. (b) Sometimes called the “bottom-up” building scenario, small ­things combine to form bigger ones, whether planets from dust or galaxies from star ­ clusters.

4. (b) Hot and cold dark m ­ atter particles move at dif­fer­ ent speeds (have dif­fer­ent temperatures). The theorized particles react differently to gravitational fields.

Cold dark ­matter clumps more easily. The similarity between them is that neither emits radiation.

5. (c) This is done through weak lensing of distant galaxies by foreground clusters.

6. (a) We expect the universe to expand forever and all ­matter to eventually decay.

7. (c) Clusters have more galaxies (and therefore mass) than groups.

8. (d) As long as the galaxy is moving with the Hubble flow, redshift gives recessional velocity by the Doppler equation and distance by the Hubble law.

9. (a) It is the gravitational instabilities—at vastly dif­ fer­ent scales—­that lead to galaxy and star formation.

10. (d) Dark ­matter particles do not interact electromagnetically, so they cannot easily get rid of their energy by, for example, collisions or radiation. Eventually the motion of the dark m ­ atter material balances gravity.

11. (d) Observational evidence supports that elliptical galaxies result from the merging of two or more spiral galaxies.

12. (d) During the Dark Ages, photons could not travel freely, but instead traveled a very short distance before hitting a f­ ree electron.

13. (a) For ­there to be no stars with zero heavy ele­ments observed, such first-­generation stars must have been of high mass and exploded soon a­ fter they formed.

14. (d) With the first generation of stars came intense radiation from the high-­mass stars and subsequent supernovae, which re-­ionized much of the gas in the universe.

15. (a, c, b) We know from cosmological studies that around 70 ­percent of the universe is dark energy, around 25 ­percent is dark ­matter, and only 4–5 ­percent is ordinary ­matter, including us.

Thinking about the Concepts 16. If we could see the early universe when galaxies ­were first forming, it would be smaller and more active. Th ­ ere was a large amount of hot gas but ­little dust. Protogalaxies ­were forming the first generations of stars, and supernovae ­were frequent as the earliest massive stars ended their brief lives. Gas was still collapsing into galactic disks, so ­there ­were few (if any) large galaxies with regular structure. Instead, smaller, irregularly ­shaped protogalaxies dominated the vis­i­ble universe. Many young galaxies ­were merging with each other, making irregular galaxies a common sight. Supermassive black holes ­were forming, and galactic centers w ­ ere much more luminous across the entire electromagnetic spectrum.

Chapter 23

17. If voids ­were filled with considerable dark ­matter, then we would expect to see gravitational lensing of galaxies b­ehind t­hose voids. We would also have expected galaxies to have formed inside the dark ­matter voids. It is much more likely that voids are fairly empty of any mass.

18. In the absence of much luminous ­matter, we could detect a dark ­matter galaxy only from the gravitational influence it exerts on its surroundings. If t­ here w ­ ere a luminous companion galaxy that e­ither orbited or merged with a dark ­matter galaxy, we could detect it. If a dark ­matter galaxy passed in front of a luminous background object, we could infer its presence from gravitational lensing.

19. Both stars and galaxies form due to gravitational instabilities. A gas cloud collapses to form a star, or many stars, and each protostar spins itself up such that an accretion disk feeds the central object. Similarly, a huge ensemble of gas can collapse and spin up to form a disk-­like protogalaxy. However, once stars begin forming in a galaxy, that formation pro­cess provides feedback that can change how the galaxy continues to form. Also, galaxy formation is sensitive to the distribution of masses around it that might lead to mergers. The pro­cesses certainly operate on vastly dif­f er­ent scales.

20. WMAP and the Planck observatory observations revealed quantum-­ sized variations in the CMB. ­Those tiny variations reflect the imprinted structure of the early universe at the time of inflation, providing the “clumps” or “seeds” from which galaxies and collections of galaxies grew.

21. The density of normal m ­ atter in the universe is too low to trigger any collapse into galaxies. Without dark ­matter to boost the local mass density in cluster-­size regions, all of the normal ­matter in the universe would still reside in the form of hydrogen and helium gas clouds.

22. The correct sequence is (a). We see evidence for merging events, both in our galaxy (with small dwarfs) and in larger galaxies, which supports the notion that small t­ hings formed first and agglomerated to make larger ones. We also find evidence in the very distant (young) galaxies being much smaller and less massive than ­those observed t­ oday.

23. ­Because inhomogeneities in the density of normal ­matter w ­ ere never strong enough to cause the structure seen t­ oday, t­ here had to be another source of gravity to cause ­those seeds of structure seen in the CMB to grow into the ­actual structure we now observe.

24. A supercluster is composed of many clusters, spanning sizes up to many hundreds of Mpc and with much more

Large-­Scale Structure in the Universe ◆ 173

mass. Also, clusters are small enough that the galaxies are orbiting each other uniformly, whereas a supercluster has not had time to “relax” into this state. We are part of the Local Group that has too few large galaxies to be called a cluster. We are part of the “local supercluster” that contains the Virgo Cluster. It is the local supercluster that is affecting our motion through space.

25. Just as a cloud of gas spins up and flattens as it collapses into a star due to the conservation of angular momentic­ u­ lar, tum, so does a protogalactic cloud. In par­ uneven gravity from all the masses around the protogalaxy can help to give it a ­little spin at the beginning, which is amplified as the cloud collapses into a galaxy.

26. Observational signs of dark m ­ atter include the rotation curves of galaxies; the velocities of galaxies in groups that are higher than can be caused by vis­i­ble mass; the high temperature of intragalactic gas in clusters; and gravitational lensing. The observational evidence of the effect of something unseen was so overwhelming that theories needed to be revised. Scientists cannot ignore the data.

27. As we look back in time, if the merger scenario for the formation of elliptical galaxies is correct, we should see more spiral galaxies than elliptical galaxies. As we look back even farther in distance and time, we should see galaxies that have not yet formed spiral arms, and galaxies should have a lot of massive stars in them b­ ecause they have just formed and are not yet old enough to have exploded. This is indeed what is observed.

28. Hot dark m ­ atter could not be gravitationally bound enough to form the halos needed to contain galaxies and to explain galactic rotation curves.

29. Fluctuations in the mass distribution caused mass to flow ­toward them, and as mass attracts even more mass, t­ hese small over-­densities soon grew into large halos. ­Because mass was being pulled from the general distribution and into ­these over-­densities, the result was a foamy structure of voids surrounded by filaments and walls. Inside t­hese halos, galaxies collapsed and formed and ­were then bound into the groups and clusters seen ­today.

30. The nuclear forces are too short-­ranged to work over cosmic distances, and electromagnetic forces only work for charged particles; most ­matter in the universe is neutral. This leaves only gravity.

Applying the Concepts 31. Setup: Figure 23.3a gives us redshift on the top side of the “pie” wedge. We can compute recessional velocity from this using v = cz, where z is the redshift. The wall galaxies have a redshift around 0.035.

174 ◆ Chapter 23 Large-­Scale Structure in the Universe variables, putting together the star formation history Solve: v = 3 × 10 8 m/s × 0.035 = 1.05 × 10 7 m/s of a young galaxy is hard to model. = 1.05 ​× 10 4 km/s.

Review: This explains why galaxies are a dif­fer­ent Review: The observed objects of highest redshift subject of study than stars, even though galaxies are tend to lie around z = 6 and above. This figure shows a mostly made of stars. slice of the universe only out to redshift 0.05, so we 35. Setup: For the universe to be closed, we need a critisee that structure is still small on cosmic scales. cal mass density of 9.5 × 10 −27 kg/m3. Knowing that 32. Setup: The universe is homogeneous and isotropic on 5 ­percent of the universe is neutrinos and that their the largest scales, whereas Figure 23.3 shows strucnumber density is 300 million per cubic meter, we can tures that are clearly not ­either. However, remember directly estimate their mass by dividing the approprithat if the w ­ hole universe looks like this single slice, ate mass density by the number density. then any direction and position w ­ ill appear the same Solve: The neutrino mass density is around to any observer.

0.05 × 9.5 × 10 −27 kg/m3 = 4.75 × 10 −28 kg/m3, so the Solve: The ­Great Attractor may be massive, but it is still local. On the largest scales, we do not see prefer4.75 × 10 −28 kg/m3 mass of one neutrino is around = 1.6 × 10 −36 k ential or inhomogeneous structure. We see clusters 3.0 × 108ν /m3 −28 3 10 kg/m that comprise the large-­scale filaments in e­4.75 very×direc= 1.6 × 10 −36 kg. tion and at e­ very distance.

3.0 × 108ν /m3 Review: This is about one-­millionth the mass of an Review: Structures such as clusters and walls are still electron, truly a tiny (almost negligible) mass, in rough small on the cosmic scale and do not violate the expecagreement with current published mea­sure­ments. tation that the universe is homogeneous and isotropic.

36. Setup: This question can be answered using o­ rders 33. Setup: For particles to give up their energy, they must of magnitude. We start with the mass of our galaxy ­either radiate or collide with other particles. being about 1012 solar masses, including dark ­matter.

Solve: Dark ­matter particles do not interact electroWe assume a group of galaxies has a handful of large magnetically, so they cannot emit radiation, and colligalaxies, some small ones, and dark ­matter; a cluster sions are very rare (most “collisions” are not literal of galaxies contains a handful of groups plus dark impacts but the interactions of the electromagnetic m ­ atter; and a supercluster is a handful of galaxy clusfields of the two particles); therefore, t­here is no ters plus dark ­matter. straightforward mechanism for dark ­matter to cool. On Solve: Using an order-­of-­magnitude approach, we find the other hand, ordinary ­matter readily loses energy that the approximate mass of an average (a) group of through emission pro­cesses (lines and blackbody) as galaxies would be 1013 solar masses; (b) cluster of galwell as via collisions; thus, ordinary m ­ atter can collapse axies would be 1014 − 1015 solar masses; and (c) superto the tiny sizes needed to form stars. cluster of galaxies would be 1016 − 1018 solar masses.

Review: Note that when dark ­matter clouds “heat up” Review: Note that the total masses are about 10 times during collapse into galaxies and clusters, t­ hese parlarger than the mass of a single galaxy ­because dark ticles still are gravitationally bound and are not conm ­ atter can be up to 90 ­percent of the mass of a given sidered “hot dark m ­ atter.” structure.

34. Setup: Stellar evolution tells us how a star evolves 37.

Setup: The prob­lem states that black hole decay rate once it is born. The star formation history is the pro­ scales t ∝ M3, so to compare the lifetimes for two cess and feedback from continued star formation over black holes, compute a ratio. a long period.

M3 t Solve: Stellar evolution is quite dif­fer­ent from star Solve: 1 = 23 , or the supermassive black hole evapt2 M1 formation history. A single star is fairly easy to follow along its life ­because ­there are just a few ­simple princi­ orates in ples that dictate how the star evolves. Star formation M 23 (3 × 106 MSun)3 t = t = t2 = 1018 t2 more time. histories, on the other hand, depend on how much gas 1 2 M13 (3 M Sun )3 and dust t­ here was, where it was, how many heavy ele­ Review: The evaporation time for a one-­solar-­mass ments w ­ ere in it, how many massive versus low-­mass black hole is 2 × 1067 years, which is already much lonstars ­were made, how supernovae affected the surger than the current age of the universe. A black hole 3 rounding space, and many additional variables. This million times that mass would take (3 × 106)3 × 2 × 1067 is not a failure of scientific theories. ­Because the pro­ years = 5 × 1086 years. This lifetime fits within the cess of star formation depends on so many dif­fer­ent

Chapter 23

black hole era introduced in The Five Ages of the Universe by Fred Adams and Gregory Laughlin, 1999.

38. Setup: Elliptical galaxies appear yellow orange, and we know the stars are old, so how old are typical yellow-­orange main-­sequence stars?

Solve: Based on their colors, elliptical galaxies must be composed of low-­mass stars like the Sun. This gives them an age of up to 10 billion years. If galaxies first formed about 1 billion years ­a fter the Big Bang, then ­these assumptions suggest t­ here ­were 4 billion to 9 billion years for the merging of spirals to take place.

Review: This is a significant fraction of the lifetime of the universe, which suggests that all of the spiral galaxies that ­were ­going to form already have, and now we are simply watching them merge with other galaxies and evolve.

39. Setup: For part (a), we might assume that a large redshift should makes galaxies look redder; however, the colors in this image have been adjusted to relate star formation in very distant galaxies to more current star formation. Furthermore, while galaxies that look smaller in the image should indicate that they are farther, the sizes of the earliest galaxies w ­ ere inherently smaller. We note that the vast majority of the misshapen, irregular, and small galaxies are colored blue.

We could reasonably use t­ hose characteristics to identify galaxies that might be at higher redshifts. For part (b), count the number of galaxies in a given area of the image, assume it is typical of any point in space, and extrapolate to the observable universe.

Solve: (a) The highest-­ redshift galaxies ­ w ill look small, blue, irregular, and may cover only a few pixels. (b) Using the cosmological princi­ple, we ­w ill assume that this patch of space is typical of any other patch, so that the average number of galaxies ­here is the same as any patch of the same size. Estimating that ­there are about 10,000 galaxies in the HUDF, ­there are 10 million times 10,000, or 100 billion galaxies in the observable universe.

Review: Section 23.1 says t­ here are several hundred billion galaxies in the observable universe. We could also do a more stringent calculation. The HUDF is 11.5 square arcminutes in angular size. The entire celestial sphere is 148,510,800 square arcminutes in angular size. It would take 12,913,983 HUDF images to cover that, for a total of just over 129 billion galaxies based on our estimate of 10,000 galaxies per 11.5 square arcminutes. Both comparisons are consistent with our estimate.

40. Setup: For this exercise, let’s just count galaxies in the red and blue areas. B ­ ecause it is hard to know what is a

Large-­Scale Structure in the Universe ◆ 175

galaxy and what is a star, we ­w ill call a galaxy anything that is not a small dot or smudge.

Solve: In the top clusters, we count about 35 galaxies.

In the bottom clusters, we count about 80 galaxies.

Review: The Bullet Cluster is especially impor­tant to studies of dark ­matter, as the collision of the clusters created a separation between the dark m ­ atter and ordinary ­matter. Studies suggest that the ratio of dark to normal m ­ atter is at least 10 to 1.

41. Setup: If the initial fluctuations ­were 10 times larger, gravity in t­ hose regions would have been significantly stronger and the universe would be filled with much larger, higher density regions. If the initial fluctuations ­were 10 times smaller, the gravity would have been weaker and the large-­scale structure would be made of smaller, less dense regions.

Solve: Larger initial fluctuations leading to larger concentrations of mass and stronger gravity would have prob­ably accelerated the formation pro­cess of structure. If all galaxies formed ­earlier, ­there might be far fewer spirals than ­there are ­today ­because many more would have collided to become elliptical galax­ ere ies. On the other hand, if structure formation w slowed due to the initial fluctuations being 10 times smaller, ­there might be far fewer elliptical galaxies ­because the density of the galaxies would have been smaller and spirals would not have had time to merge and form ellipticals.

Review: One could also pose this question as: What would happen if the strength of gravity changed?

42. Setup: Figure 23.16 shows the star formation rate as a function of redshift.

Solve: The early universe is at high redshift, meaning far to the right in Figure 23.16; t­ oday is at the left. The star formation rate now is about 0.01 MSun/yr Mpc3, whereas at the peak it was around 0.2 MSun/yr Mpc3.

This shows that the star formation rate was about 20 times higher at its peak.

Review: It stands to reason that rates of star formation had to be higher in the past, ­because many stars in our galaxy had to be formed to generate the heavy ele­ments needed to make Earth.

43. Setup: The Oxford En­glish Dictionary defines simulation as “the production of a computer model of something, especially for the purpose of study.” One should remember that a simulation is not a test of real­ity but a test of ­whether our physical theory produces something consistent with real­ity.

Solve: Although the panels differ in their ­actual structure, both show filaments and voids of roughly the same size and frequency. As such, this simulation

176 ◆ Chapter 23 Large-­Scale Structure in the Universe appears to show that the physical model is consistent with real­ity.

Review: You should be suspect of a simulation that exactly reproduces real­ity, b­ ecause then you have to question the assumptions that went into producing it.

If this happened, then most likely the par­ameters ­were set reproduce the exact positions and sizes, which would be poor science.

44. Setup: To answer this question, use Figure 23.16 and compare redshift on the bottom with age on the top.

The dropout of a spectrum occurs at 121.6 nm, and λ the redshifted dropout occurs at 1 + z = obs (see λemit Working It Out 23.2), or, solving for observed wavelength, λ obs = 121.6 nm (1 + z).

Solve: At redshift 0.5, the universe was about 9 Gyr old. At z = 4, the universe was about 1.5 years old. At z = 10, the universe was about 0.4 Gyr old. The “dropout” of a spectrum at redshift (z = 0.5) would be seen at 182 nm (UV wavelengths). Similarly, at redshift 4, the dropout is at 608 nm (vis­i­ble wavelengths); and at redshift 10, it is at 1,338 nm (IR wavelengths). The first dropout can be observed only from space, and the dropout at 608 nm could be observed from the ground. The dropout at 1,338 nm would need an observatory on a high mountain or a space satellite to best observe it.

Review: We can confirm the prob­lem with ­Table 23.1, which lists age versus redshift. At redshift 0.5, the universe was 8.6 Gyr old; at redshift 4, it was 1.6 Gyr, and at redshift 10, it was 480 million years. We could have worked this prob­lem from ­either source.

45. Setup: To estimate the core mass of a cluster, we use the distance and velocity of an orbiting object as shown in Working It Out 23.1. Note that we must convert Mpc to m: 4Mpc × 3.1 × 1022 m/Mpc = 1.24 × 1023 m. v 2r (8 × 105 m/s )2 (1.24 × 1023 m) = 6.67 × 10 −11 m 3/(kg s2 ) G = 1.2 × 10 45 kg.

We can convert this to solar masses: 1 MSun 1.2 × 10 45 kg × = 5.9 × 1014 MSun .

2 × 1030 kg This is over 590 trillion suns.

Review: Compare this to Working It Out 23.1, in which a slightly closer galaxy orbiting at a slightly higher speed yielded roughly the same amount, which is typical of cluster masses.

Solve: Mcore =

Using the Web Solutions 46. Scientists are using ­these maps not only for public outreach but also b­ ecause ­these data help retrace the history of the universe. Goals of the proj­ects include determining how much of the universe is made of dark ­matter, the compositions of the nearest 500,000 stars, and the scale of baryonic acoustic oscillations from early times. The fly-­through of the survey provides a three-­d imensional visualization of the ­actual structure of the ­matter distribution. Updates given by students on the new results from the SDSS and BOSS proj­ects ­w ill vary.

47. The science goals of the dark energy survey, DES, are to image 5,000 deg2 of the southern sky to probe the origin of the accelerating universe. Observations w ­ ill study type Ia supernovae, baryon acoustic oscillations, galaxy clustering, and weak lensing. Reports on current results ­w ill vary.

48. The most distant object observed to date is 2016-­GN-­ z11, z = 11.09, a galaxy observed by the HST (although new observations may replace its current rec­ord as being the farthest). Cosmological and galaxy-­ evolution models make predictions about the universe at early times. We have to make observations of objects at ­those times (that is, at high redshifts like z = 6) to confirm or deny t­ hose predictions. We need to understand the properties of ­those objects and how common they are compared with objects that are closer and older, such as ­those at a redshift of z = 2.

49. The ALMA Observatory is an array of radio telescopes for observing the universe in millimeter and below-­m illimeter wavelengths. It is ­going to be the largest telescope at ­these wavelengths, and ­because it is located at 5,000 meters in elevation, it ­w ill suffer very l­ittle from terrestrial sources of noise. The array ­w ill study the Dark Ages by observing extremely high-­redshift galaxies whose optical emissions have been redshifted into the radio regime. The Deep Field ­w ill therefore discover many times more high-­redshift galaxies (z is greater than 1.5) than ­were discovered with the Hubble Deep Field ­because the Deep Field is looking for the extremely high-­redshifted emission that the Hubble Deep Field could not see. Answers ­w ill vary about news and new reports.

50. Students are asked to go to the James Webb Space Telescope website and provide an update on its status and w ­ hether any results are available. NASA expects the launch date to be in 2020, and so it is a bit early at this time to have any results.

Chapter 23

EXPLORATION The Exploration for this chapter asks students to put together the pieces of how they came to be sitting in a chair, reading the pages of the textbook. Basically, the students are given the five major points and are asked to fill in the details.

EXPLORATION SOLUTIONS 1. The proton was formed about 100 seconds ­a fter the Big Bang, when three quarks came together and w ­ ere not subsequently split apart by the high-­energy photons that w ­ ere also pre­sent.

2. That proton was part of a large cloud of other H and He atoms, which gravitationally collapsed due to instability to form the first clusters and stars.

3. As a C-­atom, that proton would have been expelled in the stellar winds from a 4 MSun star to become a planetary nebula before returning to the interstellar medium.

4. That C-­atom, incorporated into a molecular cloud, would collapse ­under the self-­g ravity of the cloud to form a protostar and accretion disk. Angular momentum of the atom kept it spinning around the central protostar, whereas gravity caused the cloud to collapse. Then the atom stuck to another atom through electromagnetic forces to form the beginning of a dust grain that would grow into a planet.

5. Answers w ­ ill vary considerably. In broad brushstrokes, the proton is formed about 3 minutes ­after the Big Bang, becomes part of a protogalactic clump a­ fter a few million years, becomes a star a­fter maybe a billion years, is ejected in a supernova as a carbon atom about 8 billion years ago, and becomes part of our Solar System about 5 billion years ago. The story then follows the formation of the Earth: its cooling, oceans forming, primitive life appearing, life becoming more complex and branching to vari­ous forms. Millions of years ago, ­there was an ancestor whose lineage became more complex, especially in the development of a brain that

Large-­Scale Structure in the Universe ◆

177

could reason and use tools and develop language. ­Humans spread out over Earth, populating areas that ­were suitable for their survival. As part of that survival, they had c­ hildren that passed their DNA to multiple generations. One of t­hose generations contained the student who is writing this summary.

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE LECTURE ACTIVITIES ACTIVITY 27: A COSMIC CALENDAR Students or­ga­nize the natu­ral history of the universe into a calendar representing 1 Earth year to gain a feel for the relative magnitudes of the timescales involved. During the activity they ­w ill also: • calculate relative timescales.

• identify impor­tant events in natu­ral history.

• choose appropriate time steps at dif­fer­ent epochs.

Text reference: Exploration ACTIVITY 28: THE HUBBLE DEEP FIELD NORTH Students use the Hubble Deep Field North to explore the evolution and distribution of galaxies in the universe. As they work through the activity they: • analyze the galaxies in and the structure of the Hubble Deep Field North.

• identify Hubble types and relate them to colors of galaxies.

• relate the colors of galaxies to the pos­si­ble dominant stellar populations.

• summarize the importance of the Hubble Deep Field North to our understanding of the evolution of galaxies.

Text reference: Section 23.1 and 23.4

CHAPTER 24

Life INSTRUCTOR NOTES Chapter 24 summarizes our search for life beyond Earth.

Major topics include: • origins and evolution of life on Earth.

• evolution as a mechanism of change resulting in increasing complexity.

• biological and physical expectations for the presence of life elsewhere in the Solar System.

• the concept of habitable zones.

• current and previous searches for signs of intelligent life.

• expectations for the fate of life on Earth.

A nontrivial part of our jobs is to engender positive attitudes t­oward science in our courses, as t­oday’s students ­ ill help dictate are tomorrow’s taxpayers and voters who w the extent to which public money funds the sciences.

NASA has been extremely successful in securing funding through its Cosmic Origins program, and ­these public investments have paid off. The search for life beyond Earth makes astronomical research personal for each of us; thus, covering the material in this chapter is in no small way an impor­tant part of public relations between the astronomical community and general public. The last chapter of any text is an easy one to skip, given the time constraints of our courses, but we highly encourage you to devote at least one class lesson to how we are looking for life beyond Earth, both in our own Solar System and out in the far reaches of our galaxy.

DISCUSSION POINTS • The material in this chapter could bring out contradictions in students’ beliefs and backgrounds, with the possibility of some quite determinedly rejecting the w ­ hole concept of life beginning more than 3 billion years ago.

Find a good, short video on the Internet that accurately covers the topics of Section 24.1 of the text. Discuss the scientific evidence we have ­today for this scenario and what questions are still unresolved. (LG 1) • Self-­replicating chemical reactions become more complex through a very inefficient pro­cess of mutation and natu­ral se­lection. Connect this with the fact that the rise of complex life forms on Earth did not start ­until 178

about 540 million years ago. Discuss the implications of this inefficient biological evolution for the search for life in the Milky Way. (LG 2) • As part of a think-­pair-­share activity, list the requirements for life as we know it. Then, use what we know about the history of the Solar System and list the pos­si­ ble places in our Solar System where life (as we envision it) could have originated around the Sun. (LG 3) • Figure 24.18 illustrates the distances and extent of habitable zones around dif­fer­ent types of stars. Discuss the advantages and disadvantages of living around stars of dif­fer­ent spectral types. Consider spectral types O, A, G, and M. (LG 3) • A fun way to involve all students in a class when discussing the Drake equation is to have them each take out a piece of paper and then, for each f­ actor, have them tear the paper into the fraction that they think the ­factor’s value is. Optimistic students ­w ill end up with a larger piece of the original sheet, whereas pessimistic students w ­ ill end up with a tiny piece, if any.

(LG 3 and LG 4) • Discuss each of the seven ­factors than enter into the Drake equation. Is ­t here any f­ actor missing? What can be done to decrease the uncertainty in estimating any of the f­ actors? For which ­factors of the Drake equation has the discovery of more than 3,500 extrasolar planets decreased the uncertainty in our estimations? (LG 4) • The chapter-­opening image asks the age-­old question, “Are we alone in the universe?” Bring in the latest news about exoplanet discoveries and the latest list of exoplanets that exist in their stars’ habitable zones. This topic should be g­ reat for a 2–3-­m inute writing exercise, with students volunteering their summary to the class.

PROCESS OF SCIENCE Astrobiology may just be the scientist’s ultimate dream.

This interdisciplinary field of study links astronomy with geology, biology, chemistry, physics, and atmospheric physics. It also includes oceanography. The interconnectedness of all t­ hese subjects provides a comprehensive and rewarding view of our universe.

Chapter 24 Life ◆ 179

ASTROTOUR ANIMATIONS None for this chapter.

ASTRONOMY IN ACTION VIDEOS None for this chapter.

INTERACTIVE SIMULATIONS None for this chapter.

TEACHING READING ASTRONOMY NEWS 1. Media coverage of SETI results tends to overemphasize how the discoveries are unique and thus possibly indicative of alien civilizations. Even a hint of this conclusion gets amplified through social media, impeding knowledge of the ­actual scientific implications.

2. Answers ­w ill vary for this question, as students need to “fit” the suggestions for journalists with what they have learned about the pro­cess of science. At the most basic level, when a new observation is made, it must be tested and tested again so that the results are deemed reproducible.

3. Students are asked to think of an article they have recently seen on evidence for alien life. Answers ­w ill vary.

4. Students are asked what they think about how p­ eople ­w ill react to an ­actual discovery of alien life. Answers ­w ill vary.

CHECK YOUR UNDERSTANDING 24.1 (d) Extremo-­ means “extremes” and -­phile means “loving.”

24.2 (a) Carbon is highly reactive, which makes it favorable for life.

24.3A (d) Uranus has no surface on which to land, no liquid ­water, no source of warm internal heat, and an extremely low temperature.

24.3B (a) As we learned in stellar evolution, the mass and age of a star determine its luminosity and temperature, which in turn determine the distances about the star in which liquid ­water could exist.

24.4 (b) The Drake equation is unlike other equations found in this text b­ ecause it is not used to calculate an exact quantity, but rather to provide guidelines about probabilities.

END-­OF-­CHAPTER QUESTIONS AND PROB­ LEMS SOLUTIONS Test Your Understanding 1. All are correct.

2. (b) Our easiest way to look for life beyond Earth is to look in locations that have properties that support life as we know it; w ­ ater is one of the universal needs of all life on Earth.

3. (c) The Urey-­M iller experiment showed that energy input can create the building blocks of proteins out of commonly found materials.

4. (c) On its own, a mutation is just a change in DNA. ­W hether it is beneficial, detrimental, or inconsequential depends on how that mutation plays out in the organism’s life.

5. All of the items listed are needed for life to form.

6. (a, c) The conditions given mean that a system w ­ ill change over time, and life on Earth has become more complex through changes.

7. (d) Another way to put this is that absence of evidence is not evidence of absence.

8. (c) The “explosion” was of biodiversity during the Cambrian geological period.

9. (d) A habitable zone is the region around a star where ­water (the solvent most useful for life) can be in a liquid state on the planet.

10. (c) Prokaryotes do not have cell nuclei.

11. (b) Thermo-­means “heat” and -­phile means “loving.”

12. (b) Currently, we are only able to send robots to other regions of the Solar System to probe t­ hose places for signs of current or former life.

13. (c) For a planet or moon to host life, we hypothesize that it must experience a relatively steady environment for a long time (billions of years). O and B stars certainly expire too soon to meet this criterion.

14. (a) The first ­simple forms of life, converting carbon dioxide to oxygen, appeared billions of years ago.

15. (d) This is the scientific definition of theory.

Thinking about the Concepts 16. As carbon-­based life forms, all forms of life on Earth that we know of have and use DNA and RNA. Furthermore, the evolution of life depends on changes in DNA and RNA being propagated from parent to child.

17. Amino acids, the building blocks of DNA, ­were prob­ ably built from simpler ele­ments and molecules, using energy from the Sun or lightning as a catalyst. They may also have been brought to Earth by comets and carbonaceous chondrite meteorites.

180 ◆ Chapter 24 Life 18. Many early life forms ­were prob­ably extremophiles, living in very hot, acidic, or alkaline conditions.

19. Answers ­ w ill vary with this Pro­ cess of Science question.

20. Cyanobacteria slowly broke CO2 molecules apart to form O2 . This required about 4 billion years to produce t­ oday’s oxygen levels.

21. The Cambrian explosion was a burst of biodiversity that happened around 500 million years ago. At pre­ sent, it is unclear what caused it. Some ideas include a decrease in predators, a warming of Earth, or the oxygen levels reaching high enough levels and forming an adequate ozone layer for modern life to evolve.

22. Plants and forests use photosynthesis, which relies on CO2 and which was abundant far e­ arlier than oxygen, thus explaining why flora appeared first on Earth. Once the plants converted a reasonable amount of the atmosphere to O2, large animal life could evolve and flourish.

23. The region of space around a star where we think life could form in Earth-­l ike conditions is called the “habitable zone.” In broad terms, the bound­ aries are defined by the distances at which ­water would be a solid (too cold) or gas (too hot), but the details also have to do with the energy output from the star, as well as the properties of the planet, such as albedo and the strength of the green­house effect.

24. For life to arise, we believe ­there needed to be a reasonably stable environment, abundant building blocks, a “solvent” in which the chemistry could work, an energy source, and a lot of time.

25. Evolution is certainly underway ­today. If evolution follows natu­ral se­lection, then our evolution should allow us to live better in our current or ­f uture circumstances. The possibilities are endless ­here, but it is a good exercise for the student to think about it.

26. All of the hydrogen and most of the helium are from the Big Bang. All the other atoms in our bodies are from stellar giant-­branch winds and supernovae.

27. Absence of evidence is not evidence of absence. Just ­because two probes, looking for life in two isolated regions, did not find the signs of life does not prove or disprove the presence of life on Mars.

28. If we find that Earth is the only planet in the Solar System capable of supporting life, then f 1 (the fraction of worlds on which life arises), fc (the fraction of living worlds on which intelligence arises), and L (the likelihood that an advanced civilization exists t­oday) are all likely to be small. If we do find nonterrestrial life elsewhere in the Solar System, then f 1 must be large, forcing both fc and L to be extremely small if intelligence is rare.

29. The second law of thermodynamics applies only to a closed system. Living organisms do not violate the second law ­because the Sun is constantly pouring energy onto Earth’s surface. Therefore, the order that is generated in living creatures comes at the expense of even greater disorder in the Sun. 30. Main-­sequence stars slowly become more luminous over time. In a few billion years, the increase in the Sun’s luminosity ­w ill cause the oceans to evaporate and scorch all of Earth.

Applying the Concepts 31. Setup: The Drake equation reads N = R* × fp × ne × f l × fi ​ × fc × L, where R* is the star formation rate, fp is the ­fraction that forms planetary systems, ne is the fraction of places per system that could host life, f l is the fraction of places per system that do support life, fi is the fraction of ­those places where life becomes intelligent, fc is the fraction of that life that forms civilization, and L is the number of years that advanced civilization exists.

Solve: Results from the Kepler Mission are still being published. So far, the mission has placed limits on the number of stars with planets, the number of planets around ­those stars, and the distances t­ hose planets are from their stars. Thus, this mission should help to constrain fp and perhaps ne, if one makes assumptions about the habitable zone around each star. If most stars have planets with habitable zones, then fp ­will be close to 1, which w ­ ill increase N. Launched on April 18, 2018, the Transiting Exoplanet Survey Satellite (TESS) w ­ ill continue the work of the Kepler Mission, covering an area 400 times larger.

Review: The Kepler may not have discovered all the planets around a given star (or any planets for some stars); thus, its sample is incomplete. TESS w ­ ill cover more area of the sky than Kepler did and ­w ill have the capability to refine mea­sure­ments of planet masses, sizes, densities, and atmospheric properties. This mission should also help to constrain f p and ne, if one makes assumptions about the habitable zone around each star.

32. Setup: Figure 24.22 is the famous radio signal sent from Arecibo in 1974. The top row, we are told, represents the numbers 1 through 10, where the first three rows are the numbers and the bottom row is just a placeholder to show us that a number is being given ­ t here. The numerical-­ counting scheme is binary.

Solve: Each block is composed of two columns of four rows with each space representing a power of 2.

Chapter 24 Life ◆

181

FIGURE IM 24.1

The top three rows represent the binary numbers that form the base 10 number, with the bottom row a placeholder. We normally think of counting in binary as a horizontal task, where the message has the numbers expressed vertically. To compose a given number, black out all the digits that should not be added.

The bottom row ­w ill indicate the least significant number.

Review: Binary is the language of computers, which use “bits” of ­either zero or one (on or off, respectively) to designate numbers.

33. Setup: Replication means doubling. If we have one organism that replicates in the first second, we w ­ ill have two; the next second we have four, then eight, and so forth, such that the total population is 2t, where t is the number of seconds.

Solve: Using the formula above, t­here ­ w ill be 10 = 1,024 organisms.

2 Review: Replication may start slowly, but this growth accelerates quite rapidly. This is why we call it exponential growth.

34. Setup: Replication means doubling. If we have one organism that replicates in the first second, we w ­ ill have two; the next second we have four, then eight, and so forth, such that the total population is 2t, where t is the number of seconds.

Solve: We wish to know how many seconds must elapse to reach a population of 1,024. Thus, our formula is 1,024 = 2t, which we can solve ­either with logarithms or by inspection. By inspection, the answer is 10 seconds. When using logarithms, we take the log of both sides. The answer is: log10 (1,024) = tlog10 (2) log 10 (1,024) that leads to t = = 10. log 10 (2) Review: Replication may start slowly, but this growth accelerates quite rapidly. This is why we call it exponential growth.

35. Setup: According to the prob­lem, the doubling time is roughly 70/p, where p is the percentage rate of increase per year.

Solve: If the ­human population increases 1 ­percent per year, then the doubling time is about 70/1 = 70 years. If the population doubles in 70 years, then it w ­ ill qua­dru­ ple in 140 years.

Review: The United Nations estimated the world population in October 2018 to be 7.7 billion. Thus, by 2088 we expect about 15 billion ­people on Earth and 30 billion p­ eople by 2158.

36. Setup: From Working It Out 24.1, at 1 second, t­ here is one particle. At 2 seconds, two particles. At 3 seconds, four particles, and at n seconds t­ here are 2n−1 particles.

Solve: Figure IM 24.2 shows linear graph of the growth, followed by a logarithmic-­linear plot shown in Figure IM 24.3.

Review: Exponential growth proceeds slowly at first but then hits a population explosion where the population grows into millions of billions ­a fter a minute.

The linear plot shows this quite clearly.

FIGURE IM 24.2

182 ◆ Chapter 24 Life FIGURE IM 24.3

37. Setup: For a self-­replicating object, at first t­ here is one particle. A ­ fter one doubling time, two particles. At three doubling times, four particles, and so on, such that at n doubling times ­there are 2n particles. If you start with P0 particles, then you end with P0 × 2n particles.

Solve: We need to first convert hours to minutes: 12 hours × 60 minutes per hour = 720 minutes. For the number of doubling times, 720 minutes / 20 min­ ere ­w ill be 10 × 236 = 6.87 × 1011 or 687 utes = 36. Th billion bacteria in your body.

Review: The body does not respond instantly to an infection ­because the bacteria (or virus) have to replicate to the point that they exist in large enough numbers for the body’s immune system to react.

38. Setup: One generation doubles the number of molecules, so n generations make 2n molecules.

Solve: If the probability of a mutation is 1 in 100,000, then we need 100,000 molecules to be made to get one a mutation, on average. B ­ ecause 217 = 131,072, it would take about 17 generations.

Review: For some organisms, this can be less than a day (consider yeast, for example), which shows just how quickly mutations can occur.

39. Setup: Answers w ­ ill vary considerably. The Drake equation reads N = R* × fp × ne × f l × fi × fc × L. Let’s take optimistic values from the text: R* = 7, fp = 1, ne = 2, f l = 1, fi = 1, fc = 1, L = 106. Pessimistic values could be R* = 7, fp = 0.5, ne = 1, f l = 0.1, fi = 0.1, fc = 0.1, L = 103.

Solve: The product of the optimistic f­ actors is N = 14 million; that is, t­ here should be 14 million advanced civilizations. The product of the pessimistic ­factors is N = 3; that is, t­ here should be about three advanced civilizations in our galaxy at this time.

Review: With optimistic values, our galaxy ­w ill be teeming with life, although ­because ­there are bil-

lions of stars, this means that civilization is still fairly rare, with only 1 in 20 to 100 stars having the possibility of hosting life. The pessimistic values would suggest we are the only advanced civilization in our galaxy.

40. Setup: The Drake equation reads N = R* × fp × ne × f l × fi × fc × L, where L is the lifetime of a civilization.

Solve: ­Because N depends linearly on L, N changes with L. Using optimistic and pessimistic values of L from prob­lem 39, L ranges from a million to a thousand; thus, we see that N can easily vary by three ­orders of magnitude based just on the uncertainty in this one pa­ram­e­ter alone. Short lifetimes make the chances of contacting extraterrestrials quite small ­because we both need to be alive at optimal times so one can transmit and the other can receive.

Review: Remember that we have only been transmitting and listening for about 100 years, so we must bear in mind that communicating with extraterrestrials requires an enormous congruence in timing.

41. Setup: In Kelvin, w ­ ater freezes at 273 K and boils at 373 K. Solve: The annotated figure is given in Figure IM 24.4.

From this figure, no planet is predicted to lie inside the habitable zone based on its equilibrium temperatures. Mea­sur­able temperatures show that Earth falls solidly inside the freeze and boil lines, whereas all other terrestrial planets are e­ ither fully or partly outside; that is, Mars is far too cold, Venus is far too hot, and Mercury has too much of a temperature swing in both directions (although for Mercury t­ here may be instances or places where the temperature lies within the freeze-­boil range).

Review: The notion of a habitable zone is very fluid, ­because it does not account for axial tilts, planet rotation, albedo, or the green­house effect.

42. Setup: The habitable zone is where the average temperature of a planet from its star is between the freezing and boiling points of ­water, 273 K and 373 K, respectively.

Solve: Mercury’s average temperature is hotter than boiling, and its temperature range is far above and below this range. Thus, only a small part of the planet may be inside this temperature range at a given time.

With wild temperature swings such as t­hese, one would not consider Mercury to be inside the habitable zone.

Review: Being in the habitable zone means not only having an average planetary temperature that is between boiling and freezing but also that the excursions away from this average are not too large.

Chapter 24 Life ◆ 183

FIGURE IM 24.4

43. Setup: To relate period to distance of an orbit around a Sun-­like star, use Kepler’s third law, P2 = A3, where period P is in years, and A is in astronomical units (AU).

Solve: For this prob­lem, the period is 87 days, which is 87 days × (1 year/365 days) = 0.238 year. This yields a semimajor axis of A = P2/3 = (0.238)2/3 = 0.38 AU. This is basically at the location of Mercury, which is too close to its star to be within the habitable zone for a Sun-­l ike star.

Review: Finding a planet is just the first part of our search for life. Determining the habitable zone around each star is the next step in the search for life.

44. Setup: When considering where to send a signal to search for life, does it make sense to target a single star, or a huge number at once?

Solve: Being a globular cluster with hundreds of thousands of stars, M13 provided a single “target” with many solar-­type stars, which we hoped would maximize our chances of finding life when we beamed out our radio signal in 1974.

Review: Unfortunately, M13 is metal poor and about 11 billion years old, so it is not certain w ­ hether planets with sufficient heavy ele­ments to form life would exist ­there in large numbers, and, if life did evolve, ­whether it would still exist ­today. The cluster is also over 22,000 light-­years away, and so any return message would arrive far in the ­f uture.

45. Setup: Recall that the Solar System formed about 5 billion years ago, so the life forms that did the seeding must be older than this at a minimum. But to get into the protosolar nebula and into the comets and asteroids and meteoroids, they had to have formed elsewhere and then traveled through space, which prob­ably required the explosion of a nearby star.

Solve: ­Simple life on Earth required a billion or more years to form, so t­ hese seeds prob­ably took a similar length of time. Knowing that more massive stars than our Sun live shorter life spans but also are much hotter and more luminous (too much UV radiation could be good or bad for life), we might propose that the star needs to have a mass similar to the Sun—­that is, 2 solar masses or so. This means the star formed about 8 billion to 10 billion years ago.

Review: ­Were t­here enough metals in the protosolar nebula to provide life-­building chemicals in its comets, asteroids, and meteoroids and seed Earth with life?

Using the Web Solutions 46. (a) Answers ­w ill vary. Students are asked to report on new findings on extremophiles. Astronomers care about extremophiles ­because ­these are forms of life that evolved in hostile circumstances, such as t­hose expected around other planets and moons. (b) Answers ­w ill vary. Discuss recent topics of interest from the “Life, Unbounded” blog. Include ­whether this is based on new data or a new theory.

47. (a) Curiosity includes a host of research apparatus to search for fossilized evidence for life on Mars. Astrobiologists are very interested in ­whether life evolved, to what stage it did so, and w ­ hether it could still be pre­sent in some form. Answers ­w ill vary on recent results. The InSight Mars lander was launched on May 5, 2018, and is on its way to Mars. It landed on November 26, 2018. Students should click on the “News & Events” link for updates. The Mars 2020 rover mission is currently scheduled for a launch in July/August 2020. (b) MAVEN is exploring Mars’s

184 ◆ Chapter 24 Life upper atmosphere, including how molecules like ­water and carbon dioxide have been lost in space. This provides impor­tant clues about when and how Mars became geologically dead. Answers ­w ill vary on recent news.

48. Answers w ­ ill vary with time. Th ­ ere is a formal, rather complicated, definition of what conditions are necessary for a region around a star to be labeled the habitable zone: “The HZ bound­aries change as a function of planetary mass and the amount of background N2 gas.

The conservative HZ limits for more massive planets should be wider than t­ hose for low-­mass planets if the atmospheric column depth scales with planet radius.”

At the end of 2018, the exoplanet cata­log listed zero exoplanets that ­were confirmed to be habitable, and 55 exoplanets that are potentially habitable. The total number of exoplanets confirmed as of late 2018 is 3,875. ­These numbers yield a lower limit of 1 ­percent of exoplanets being potentially habitable, which would be the fraction f l in the Drake equation.

49. (a) Gaia’s scientific mission is to produce a three-­ dimensional chart of our galaxy by making radial velocity mea­sure­ments of about 1 billion stars. This is expected to reveal tens of thousands of exoplanets ­because a star’s radial velocity changes reveal the gravitational tugs of its planets. Answers ­w ill vary on recent results. (b) The Transiting Exoplanet Survey Satellite (TESS) was launched on April 18, 2018. Its mission is to survey the brightest stars near Earth for transiting exoplanets over a two-­year period. Answers ­w ill vary about its discoveries. (c) The James Webb Space Telescope (JWST) is scheduled to be launched in 2021. It w ­ ill look for dimming of light from a star via planet transits. It ­w ill do infrared spectroscopy of the planet’s atmosphere. No results are available yet.

50. “The Breakthrough Initiatives are a program of scientific and technological exploration, probing the big questions of life in the universe: Are we alone? Are ­there habitable worlds in our galactic neighborhood?

Can we make the g­ reat leap to the stars? And can we think and act together—as one world in the cosmos?”

As of late 2018, ­there ­were a lot of status updates and news about observations at vari­ous observatories, radio telescopes, and other facilities. This should be an in­ter­est­ing research proj­ect for the students.

EXPLORATION The answers to Fermi prob­lems always vary considerably.

The solutions presented ­here are only one means to address each prob­lem and are meant to be instructive rather than exhaustive in method and accuracy. One can

always look up current values on the Internet or in a book.

We use a combination of logic and “lookup” for the following prob­lems.

EXPLORATION SOLUTIONS 1. From http://­w ww​.­worldometers​.­info​/­world​-­popula​ tion​/­world​-­population​-­by​-­year/ we find that between 2017 and 2018, the world population increased by 83 million ­people. We w ­ ill assume that the average mass (weight) of all the ­people on Earth is about 65 kilograms (kg) (~145 pounds). This means the mass of the ­human population on Earth went up by about 5.4 billion kg (12 billion pounds).

2. A ­horse is much larger and more active than a person.

Adults need about 2,000 calories per day. B ­ ecause a ­horse does a lot more work, we w ­ ill guess that a 1,000-­ pound h­ orse needs around 20,000 calories per day. If a ­horse lives 20 years, or 7,000 days, then it needs roughly 20,000 × 7,000 = 140 million calories over its life.

3. The average per capita potato consumption in the United States (all forms that potatoes might come in) is about 120 pounds per year. ­There are about 310 million p­ eople in the United States, making the total consumption 3.1 × 108 × 120 = 37 billion pounds of potatoes per year.

4. Let’s assume the average h­ uman weighs 100 pounds, which is made up entirely of cells. If we assume a cell is pure ­water (the body is mostly ­water) and is on average a sphere of radius 10 μm, then a cell has a volume of 4.2 × 10 −15 m3. ­Water has a density of 1,000 kg/m3, ­ ere are about 2 so a cell weighs about 4.2 × 10 −12 kg. Th pounds per kilogram, so a cell weighs about 9 × 10 −12 pounds. Let’s round that up to 10 −11 pounds per cell.

Then, dividing the average weight of a person by the weight of a cell yields 100/10 −11 = 1013 cells. Smithsonian​.­com says, “­There are 37.2 trillion cells in your body.” We ­aren’t that far off.

5. Let’s forget inflation for the moment and assume that each of us ­w ill work for 50 years, earning an average of $75,000 per year, for a total of about $3.75 million over a lifetime. For this prob­lem, we assume a 40-­hour workweek at 50 weeks per year, meaning we work 2,000 hours per year, or 100,000 hours in our lives.

Each of us would have earned about $37.50 per hour.

6. Per the Environmental Protection Agency (EPA), each person in the United States produces, on average, 4.5 pounds of trash per day (this includes industry and business) for a total of 4.5 × 365 × 3.1 × 108 = 510 billion pounds per year.

7. Per Wikipedia, hair grows about 0.5 inches per month. That works out to be 0.042 feet per 30-­day

Chapter 24 Life ◆ 185

month. ­There are 720 hours per 30-­day month, so this means that hair grows about 5 × 10 −5 feet per hour.

8. Per National Geographic, 2010: If ­every person on Earth stood shoulder to shoulder, we would all fit into Los Angeles.

9. Earth is 70 ­ percent ­ water, so let’s assume only 30 ­percent of the land mass is usable. Earth has a radius of about 6,400 km, or a surface area of 4π × (6,400,000 m)2 = 5 × 1014, of which 30 ­percent is usable, or about 1014 m2 of land (rounded down).

Then, if each person used 1 square meter, 1014 ­people, or 100 trillion, could live on Earth.

10. To estimate how much carbon dioxide a car gives off per year, we should find out how many gallons of gas are used per year and how much carbon dioxide is produced per gallon. A typical lease for a car is 3 years for 36,000 miles, so let’s assume a person drives 12,000 miles per year, and the car makes 20 miles per gallon (for each efficient car ­there is a gas-­g uzzling SUV).

Thus, a person uses about 600 gallons of gas per year. A gallon of ­water weighs about 8 pounds and gas is a hydrocarbon, but gas burns oxygen. Carbon and oxygen have about the same atomic weight, so burning gas ­w ill use 1 pound of carbon for each 2 pounds of oxygen. ­ ill guess that of 1 gallon of gas, 5 pounds are carWe w bon, so a total of 15 pounds of CO2 are released per gallon of gas. (The EPA lists 19.4 pounds of CO2 per gallon, by the way). This means the car releases about 600 × 15 = 9,000 pounds of CO2 in a year. The EPA estimates that an average passenger car releases 5.48 metric tons of CO2 per year, or 12,000 pounds per year.

11. Earth may be 70 ­percent ­water on the surface, but it is mostly rock and iron. ­Water has a density of 1,000 kg/ m3, and iron is about 5,500 kg/m3. We assume an average density of 5,000 kg/m3. Earth’s radius is 6,400 km, so its volume is 4/3π × (6,400,000 m)3 = 1021 m3 for a total mass of about 5,000 kg/m3 × 1021 m3 = 5 × 1024 kg. The ­actual mass is 5.9 × 1024 kg.

12. For the year 2017, AAA estimated that it cost, on average, about $8,500 a year to own and operate a vehicle. Besides any monthly payment, the study also includes depreciation, maintenance, repairs, and fuel costs. The estimate for annual costs runs from $6,400 for a small sedan to $10,000 for a pickup truck (http://­ newsroom​.­aaa​.­com​/­auto​/­your​-­driving​-­costs​/­).

LEARNING ASTRONOMY BY ­DOING ASTRONOMY: COLLABORATIVE ­LECTURE ACTIVITIES ACTIVITY 30: HABITABLE WORLDS Students explore the properties of known extrasolar planets ­ hether they might support life. They also: to determine w • describe physical and orbital characteristics of a number of extrasolar planets.

• compare characteristics of extrasolar planets to the planets in the Solar System.

• apply the definition of habitable zone to determine habitability of extrasolar planets.

• summarize what life might be like on e­ ither an exoplanet or a moon of an exoplanet.

Text reference: Section 24.3

PA RT I I :

Answers to Starry Night Workbook Exercises

133

The Norton Starry Night Workbook: Exercise Summaries and Activity Answers EXERCISE 1: THE CELESTIAL SPHERE This exercise illustrates the daily motion of the celestial sphere. It animates the apparent motion of the sky as seen from Earth’s North and South poles.

Students will learn the most basic operations of Starry Night: selecting objects to view, changing the orientation of their gaze, and altering the flow of time.

ANSWER KEY Activity 1: Directions on the Sky 1. In this view, west is toward the right, and east is toward the left.

2. From the North Pole, the apparent motion is parallel to the horizon.

Activity 2: Direction of Rotation 3. From the North Pole, the apparent motion of the sky is counterclockwise.

4. Polaris is located in the constellation of Ursa Minor.

Activity 3: View from the South Pole 5. From the South Pole, the apparent motion of the sky is clockwise.

6. The south celestial pole is in the constellation Octans.

7. Stars at the North and South poles appear to move in opposite directions. This is because the point of view of the observer is inverted, so that a counterclockwise motion as seen at the North Pole appears clockwise as seen at the South Pole.

EXERCISE 2: EARTH’S ROTATION PERIOD Here, the students will become familiar with the appearance of the celestial sphere from an intermediate latitude.

The instructor may wish to direct the students to pick a particular viewing location for this exercise.

The students will determine Earth’s rotation period by finding the average time between meridian crossings for a star. The exact value of Earth’s rotation period (23 hours 56 minutes 4 seconds), is called the sidereal day. Many 188 134

students are surprised that the rotation period is not a full 24 hours. A number of the exercises for Starry Night require that the students observe in intervals that are multiples of 1 sidereal day, so it is important that the students learn this concept.

ANSWER KEY Activity 1: Circumpolar Constellations 1. From typical latitudes in the continental United States, the circumpolar stars are in the constellations of Camelopardalis, Cassiopeia, Cepheus, Draco, Ursa Major, and Ursa Minor. You might want to point out to students that all the constellations in the Northern Hemisphere are circumpolar as seen from the North Pole. This fact was demonstrated in the exercise on the celestial sphere.

Activity 2: Rising and Setting Constellations 2. The answer will depend on the constellation selected. Small constellations take about an hour to rise, whereas larger ones (Orion, Pegasus) may take 3 hours or more.

Activity 3: Earth’s Rotation Period 3. The students are asked to determine the interval between meridian crossings a few times and average the answer to smooth out any inaccuracies in their measurements.

4. The length of the sidereal day is 23 hours 56 minutes 4 seconds.

5. The average amount by which stars cross the meridian earlier each day is 24 hours minus the length of the sidereal day, or 3 minutes 56 seconds.

6. A star that crosses the meridian at midnight tonight would cross 4 minutes earlier tomorrow night, at 11:56 p.m. Thirty days later, the star would cross the meridian 120 minutes earlier (30 3 4), at 10 p.m. This implies that any particular star will continue to transit earlier and earlier until it is eventually only above the horizon during the day. But also, the star would again

The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 135 The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 189

continue to transit earlier until a year later, when it would again transit at midnight.

EXERCISE 3: MOTION OF THE SUN ALONG THE ECLIPTIC This exercise demonstrates that the Sun moves eastward along the ecliptic each day, and asks the students to note which constellation the Sun appears in at different times of the year. The exercise reinforces the concept of Earth’s changing position in its orbit around the Sun, causing constant changes in the apparent position of the Sun in the celestial sphere as viewed from Earth.

Most students know something of popular astrology, if only their birth sign. The students will compare the actual direction of the Sun with the locations listed on many calendars and see that the dates of the astrological signs have their origins more than 2,000 years ago.

ANSWER KEY Activity 1: The Sun and the Zodiac 1. Dates on which the Sun entered the modern constellation boundaries: Constellation

Date Sun Enters

Days Spent

Gemini

6/22/2007

29

Cancer

7/21/2007

21

Leo

8/11/2007

37

Virgo

9/17/2007

45

Libra

11/01/2007

22

Scorpius

11/23/2007

7

Ophiuchus

11/30/2007

19

Sagittarius

12/19/2007

32

Capricornus

1/20/2008

28

Aquarius

2/17/2008

24

Pisces

3/12/2008

38

Aires

4/19/2008

25

Taurus

5/14/2008

38

Gemini

6/21/2008

—

Allow 6 1 day for the student estimates.

Activity 3: Dates in the Distant Past 3. The approximate year that the Sun entered Gemini on May 21 is 900 BC (allow ±1 century).

4. The Sun only appears to pass through the zodiac constellations because the ecliptic plane of Earth’s orbit around the Sun lines up with these constellations. The Sun never passes through Ursa Major because that constellation is not in the ecliptic plane.

EXERCISE 4: MOTION OF THE MOON This exercise asks the students to measure the synodic and sidereal periods of the Moon and to compare their measurements to the values quoted in the reading.

The instructor may wish to stress that any set of measurements will be accurate only to a particular level. The students should be encouraged to be honest about their measurements and not to try to land on the stated periods.

By comparing their values to the actual ones, the students will get a feel for the accuracy of their measurements. There are more accurate ways of measuring the synodic and sidereal periods using Starry Night, but this exercise is designed to produce answers accurate only to about a quarter of a day.

ANSWER KEY Activity 1: Time of Moonrise on Successive Nights 1. The exact times will depend on the viewing location.

The instructor may wish to specify a particular viewing location so that all students come up with the same times.

2. The average time interval will be about 50 minutes.

Activity 2: The Moon’s Sidereal Period 3. The students should derive numbers between 27 and 28 days. In the exercise, they work in steps of 1 day, so the fraction of a day over 27 that they derive will depend on how they interpolate between positions of the Moon on successive nights.

4. The length of the sidereal month is 27.3 days.

Activity 2: Astrological Dates

Activity 3: The Moon’s Synodic Period

2. Students will notice that (a) the current constellation boundaries mean that the Sun spends different intervals in the various constellations; (b) there is a difference of about 1 month between the traditional dates and the current dates; and (c) the Sun spends almost 3 weeks in Ophiuchus, which is not one of the traditional zodiacal constellations.

5. As in answer 3, the exact value will depend on how the students interpolate.

6. The length of the synodic month is 29.5 days.

7. The difference between the Moon’s sidereal and synodic periods is about a little more than 2 days. The synodic period is longer because as the Moon orbits Earth, the Earth-Moon system is moving around the

136 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 190 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers Sun. It takes 27.32 days for the Moon to complete one orbit around Earth (sidereal period), but in that time, Earth has moved to a different position in its orbit around the Sun. The Moon must therefore move a little farther in its orbit to line up with the Sun again, making this synodic period 29.5 days.

EXERCISE 5: EARTH AND MOON PHASES This exercise explores the relationship between the phases of the Moon as seen from Earth and the phases (shading) of Earth as seen from the Moon. Students will predict

Date and Time

Moon Phase (shadow = dark)

Phase Description

Activity 1: Moon Phases Activity 2: Earth Phases 1-3.

4. The shading (phases) of Earth appears to be opposite or the reverse of the Moon phase shading. A New Moon is a full Earth for the same date and time. Similarly, a waxing

Earth Shading (shadow = dark)

New Moon

Full Earth

Waxing Crescent Moon

Jan. 13, 2016 22:06:06 UT

Waning Gibbous Earth

3B

3A Jan. 16, 2016 21:54:18 UT

Last/3rd Quarter Earth

1st Quarter Moon

4B

4A Waxing Gibbous Moon

Jan. 19, 2016 21:42:30 UT

Waning Crescent Earth

5B

5A Jan. 23, 2016 21:26:47 UT

Full Moon

New Earth

6B

6A Waning Gibbous Moon

Jan. 27, 2016 21:11:03 UT

Waxing Crescent Earth

7B

7A Last/3rd Quarter Moon

Jan. 31, 2016 20:55:19 UT

Shading Description

2B

2A

1st Quarter Earth

8B

8A Feb. 4, 2016 20:39:36 UT

ANSWER KEY

1B

1A Jan. 9, 2016 22:21:49 UT

the best conditions to observe phases of the Earth from the Moon.

Waning Crescent Moon

Waxing Gibbous Earth

The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 137 The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 191

crescent Moon phase is accompanied by a complementary waning gibbous Earth phase. The first quarter Moon features a last/third quarter Earth. A waning crescent Moon is opposed by a waxing gibbous Earth, and so on.

5. To see Earth completely, it should be in its full phase, which corresponds to the new Moon phase.

6. It depends on how well the observers can be protected from the extreme cold on the surface of the Moon, in the total darkness of a new Moon. Observing might not be practical during the extreme cold of the new Moon phase. Since there is no atmosphere on the Moon, heat is conducted only by radiation and conduction.

Date and Time

Moon Phase (shadow = dark)

Phase Description

New Moon

Jan. 13, 2016 22:06:06 UT

Full Earth

Waxing Crescent Moon

Waning Gibbous Earth

3B

3A Jan. 16, 2016 21:54:18 UT

Last/3rd Quarter Earth

1st Quarter Moon

4B

4A Jan. 19, 2016 21:42:30 UT

Waxing Gibbous Moon

Waning Crescent Earth

5B

5A Jan. 23, 2016 21:26:47 UT

Full Moon

New Earth

6B

6A Jan. 27, 2016 21:11:03 UT

Waning Gibbous Moon

Waxing Crescent Earth

7B

7A Jan. 31, 2016 20:55:19 UT

Shading Description

2B

2A

Last/3rd Quarter Moon

1st Quarter Earth

8B

8A Feb. 4, 2016 20:39:36 UT

Earth Shading (shadow = dark) 1B

1A Jan. 9, 2016 22:21:49 UT

7.

Furthermore, Bruce Crater is near the lunar equator, so the surface experiences the most extreme high temperatures during the lunar day time such as when there are waxing and waning as phases. The temperature can rise to almost 390K (water boils at 373K) during the daytime, and sink to as low as 95K (water freezes at 273K) in the nighttime. Selecting a phase when the most moderate temperatures occur, around dawn as the Sun first rises for the lunar day, which is a little over 27 Earth days long, might be the best time to observe. This procedure was used during the Apollo lunar landings from 1969 to 1972.

Waning Crescent Moon

Waxing Gibbous Earth

138 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 192 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers

EXERCISE 6: SUNRISE ON MARS

ANSWER KEY

In this exercise, students will observe and document sunrises and sunsets along with related objects from the surface of mars. Students will also predict the seasons based on the shifting location of the sun along the Martian horizon.

Activity 1: Sunrises from Mars

1/ 9/13/2015 9/14/2015

Sunrise time Stopped time 22:0:47/ 0:05:03 UT

List other objects observed near the Sun along the sunrise path to the horizon

Sketch of Sun and any objects at sunrise (show and label horizon with directions) 30˚

Earth Venus Mercury

V Sun

75˚ 80˚

60˚ 2/ 3/13/2016

18:06:30 UT/ 20:06:57 UT

E Me

Observation step/ Date

Sunrise Table:

Me

10˚

E

0˚

90˚

100˚ 30˚

Mercury Venus

10˚ E 60˚ 3/ 9/13/2016

16:59:37/ 19:04:38 UT

80˚

30˚ 10˚

80˚

14:23:27/ 16:25:18 UT

10:48:36/ 12:52:17 UT

90˚ 100˚

0˚ 108˚ 120˚ 30˚

Jupiter Earth Venus Saturn

10˚ SE 100˚

5/ 9/13/2017

100˚

Earth Mercury Venus E

4/ 3/13/2017

0˚

90˚

65˚

103˚ 120˚

0˚ 140˚ 30˚

Jupiter Saturn Mercury Venus

10˚ E 60˚

65˚

90˚ 80˚

0˚ 100˚

The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 139 The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 193

Sunset Table:

Observation step/ Date 1/ 9/13/2015 9/14/2015

Sunset time Stopped time 10:31:17/ 12:31:29

List other objects observed near the Sun along the sunset path to the horizon

Sketch of Sun and any objects at sunrise (show and label horizon with directions)

Venus Earth Mercury

V E

30˚ 10˚ NW Me 288˚ 280˚ 300˚

2/ 3/13/2016

7:28:26/ 9:29:02 UT

0˚ 320˚ 30˚

Earth Mercury Venus

10˚ NW 0˚ 280˚

3/ 9/13/2016

04:39:16/ 6:34:30 UT

292˚ 300˚

320˚ 30˚

Earth Jupiter Mercury

10˚ W 240˚

4/ 3/14/17

2:13:29/ 4:12:50 UT

252˚ 260˚

270˚

0˚ 280˚ 30˚

Venus Earth

10˚ W 240˚ 5/ 9/14/17

0:12:27/ 2:28:05 UT

252˚ 260˚

270˚

280˚ 30˚

V

Venus Mercury Earth

0˚

Me

10˚

E NW 280˚

1. They are the planets rising or setting. The names of these may be seen by rolling the cursor over the objects or by turning on the planet labels.

293˚ 300˚

0˚ 320˚

140 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 194 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers Activity 2: Predicting Seasons from Sunrise/-Set Data 2.

Difference a. Summer Solstice

1/3/2016

-------------------

b. Autumnal Equinox

7/4/2016

6 months, 1 day

c. Winter Solstice

11/28/2016

4 months, 24 days

d. Spring Equinox

5/5/2017

5 months, 7 days

e. Summer Solstice

11/20/2017 Total

6 months, 15 days 22.6 months

3. Northern Hemisphere.

4. The greater eccentricity of the orbit of Mars causes the seasons to be unequal.

EXERCISE 7: PRECESSION Here we focus on the precession of Earth’s rotation axis.

The students will measure the angular separation between the north celestial pole (NCP) and Polaris at different epochs and will see that the current separation is close to the smallest possible value.

The text emphasizes the effect of precession on the times of the spring and fall equinoxes; we touch on that concept in Exercise 3, on the motion of the Sun along the ecliptic.

ANSWER KEY Activity 1: The Distance Between Polaris and the North Celestial Pole 1. The separation in 2007 is about 0 degrees 42 minutes (or 0° 429). Allow an error of a few minutes, as the students may land on slightly different positions near the north celestial pole.

Activity 2: When Polaris is Closest to the Ncp 2. The minimum separation will occur about the year 2100 and will be approximately 0° 289.

3. This is about 67 percent of the current value.

Activity 3: The Separation in the Past 4. In the year AD 1, the separation was almost 12° (about 11° 459)!

5. The following table shows a few bright stars that were closer than Polaris to the north celestial pole in the year AD 1. The last column displays the separation in degrees and minutes. If they ignore the instructions to examine the stars in the stick figures for Ursa Minor

and Draco, some students may zoom in and identify faint stars that are much closer to the NCP. Constellation

Star

Separation

Ursa Minor

Kochab

8° 209

Zeta Ursae Minoris

8° 329

Kappa Draconis

9° 109

Draco

6. Though Polaris was about the same distance from the NCP as Kochab at this time, Polaris is the brighter star.

7. The change in the position of the north celestial pole affects all stars, not just Polaris. The stars stay in the same positions relative to each other, so the constellation patterns themselves are not affected, but if the NCP moves away from Polaris, it must move closer to other stars.

EXERCISE 8: KEPLER’S LAWS This exercise reinforces a student’s knowledge of Kepler’s laws by creating an asteroid with a semimajor axis of 1 AU, but with a very elongated orbit. In the first part, the students will calculate the orbital period and then verify it by inspection. They will see that the period is independent of the eccentricity.

The elongated orbit for the asteroid makes Kepler’s second law more apparent than it would be for the orbits of the major planets, which are nearly circular.

This exercise views the orbits of Earth and the asteroid from a position above Earth’s ecliptic pole. This helps illustrate that Earth’s distance from the Sun does not vary much during the year. Many students get the opposite impression from figures that show Earth’s orbit from the side, which make it appear much more eccentric, thereby causing confusion when they learn the causes of the seasons.

ANSWER KEY Activity 1: Kepler’s Third Law 1. The length of the semimajor axis is a = 1 AU. From Kepler’s harmonic law, P2 = a3, so P = 1 year.

2. The orbital period is independent of the eccentricity e.

This sometimes surprises students because a circular orbit looks so different from a highly elongated orbit.

3. With a = 4 AU, a3 = 64, so P = 8 years.

Activity 2: The Period of X’s Orbit 4. The period determined by measurement should be very close to 1 year, in agreement with the value from Kepler’s harmonic law.

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5. The asteroid X orbits fastest at perihelion.

6. Kepler’s law of equal areas describes the speed of motion. It states that a line connecting the Sun to an object sweeps out equal areas in equal times.

Activity 3: Inside and Outside 7. The asteroid X spends about 3 months inside Earth’s orbit. If the students are clever with the zoom tool and the time flow rate, they might show that the time is about 3 months 2 days. The asteroid spends about 9 months outside Earth’s orbit.

Activity 4: Kepler’s Second Law 8. The distance at perihelion is 0.25 AU. The distance at aphelion is 1.75 AU. 9. The semimajor axis for a = 2 AU is twice the value of a, or 2 AU. This agrees with the measured values in questions 9 and 10, because these values add to the longest dimension across the orbit (that is, the length of the major axis).

10. Asteroids that may cross Earth’s orbit typically have very eccentric orbits. Kepler’s second law says that eccentric orbits have greater variation in orbital speeds.

The asteroids would move more slowly when they were farther away from the Sun, spending more of their time outside of Earth’s orbit. Searches for these types of asteroids should be focused away from the Sun, outside of Earth’s orbit.

EXERCISE 9: FLYING TO MARS In this exercise, the students create an asteroid with an orbit that carries it from the vicinity of Earth to near Mars.

The asteroid imitates a spacecraft, particularly the two Mars exploration rovers Spirit and Opportunity, which were launched on June 10 and July 7, 2003, respectively, arriving on Mars in January 2004. They followed an orbit that was similar to the ones created in this exercise.

Some students may be familiar with the phrase “launch window,” which is an interval of time during which a spacecraft must be launched in order to reach its destination.

Here, the students will learn that it is rather complicated to put a spacecraft into the right orbit to arrive at another planet on time. They will need to use their judgment about whether the orbit they create gets the spacecraft close enough to Mars. The instructions guide the students in creating a Hohmann transfer orbit, named after its discoverer (1925). Such an orbit usually minimizes the amount of fuel needed to reach the other planet.

The complications are greater, naturally, if the spacecraft has a human crew. The last part of the exercise asks students to think of trade-offs involved in crewed spaceflight. They simply cannot fly past Mars and expect Earth to be there to receive them!

ANSWER KEY Activity 1: The Transfer Orbit 1. e = 0.21.

2. a =1.28.

3. w = 269°.

4. L = 217°.

These values are approximate. Students should get values that are within ±0.02 for e and a, and within ±5° for w and L. Activity 2: Time to Fly to Mars 5. The spacecraft arrives near Mars around January 2004.

Activity 3: Coming Home 6. The spacecraft returns to Earth in late November 2004.

Allow about 1 month leeway, as students may pick different orbits. Earth is not there at the time; it is on the opposite side of the Sun. 7. The orbital period is approximately 1.4 years. The exact value depends on the value of the semimajor axis of the orbit that students pick.

8. There are two choices: (a) fly back on a different orbit, which requires more fuel; or (b) wait on Mars for several months until Earth would be in the right place for a return.

EXERCISE 10: THE MOONS OF JUPITER January 2010 was the 400th anniversary of Galileo’s first observations of Jupiter through his simple telescope. This exercise illustrates the observations that Galileo made when seeing Jupiter for the first time to show some of the challenges he faced in identifying the four large moons and their patterns. The students will use their observations of the moons to deduce the patterns of their motions with regard to distance from Jupiter, orbital speeds, and orbital periods.

ANSWER KEY Activity 1: Galileo’s Observations 1. Students will be observing the orientation of the Galilean moons and recording them in the table.

142 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 196 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers Date

Time

Sketch

1/07/1610

8:00 p.m.

* (2 moons) *

determine the date of both maximum tilt and the next ring plane crossing.

O

*

O

***

1/08/1610

8:00 p.m.

*

1/10/1610

8:00 p.m.

*

1/11/1610

8:00 p.m.

* *

O

1/12/1610

8:00 p.m.

* * *

O

*

1/13/1610

8:00 p.m.

*

O

***

1/14/1610

8:00 p.m.

(2 moons) *

O

* *

O * (2 moons)

**

2. Galileo may have only noticed three moons at first for two possible reasons: (a) in three of the observations ( January 7, 11, and 14), two moons were so close together that they appeared as one; and (b) in the earlier observations, one moon was quite far from Jupiter and may have been mistaken for a background star.

Activity 2: The Orbits of Jupiter’s Moons 3. When the moons are shown in motion, you can see that they pass alternately in front of and behind Jupiter.

Also, the moons seem to slow down when approaching their apparent farthest point from Jupiter, meaning that they are moving either more directly toward or away from us, rather than directly across our field of view, as when they are apparently closer to Jupiter.

4. Io stays the closest to Jupiter. Callisto moves the farthest from Jupiter.

5. Io orbits with the fastest speed. Callisto orbits with the slowest speed.

6. The elongation dates and times will vary widely among students, but the orbital periods should all be consistent. The period of Io is about 1 day 18 hours; Europa is about 3 days 12 hours; Ganymede is about 7 days 4 hours; and Callisto is about 16 days 18 hours.

7. The farther a moon is from Jupiter, the slower it moves in its orbit and the longer it takes to complete one orbital period.

EXERCISE 11: THE RINGS OF SATURN Starting in 2009 and for the following few years, Saturn’s rings were inclined at a very small angle. If students observed Saturn through a telescope during these years, they may have wondered why the rings were not as spectacular as they typically look in photographs. This exercise has students investigate the changing orientation of Saturn’s rings caused by the tilt of the planet and its orientation relative to Earth’s orbit. Specifically, they will be guided through the ring plane crossing of 2009 and then continue into the future to watch the changing orientation of the rings and

ANSWER KEY Activity 1: Finding Saturn’s Ring Plane Crossing 1. On January 1, 2009, the rings were very close to edge-on, so students would not have had the very obvious view of the rings they might have expected.

When the rings are seen in this orientation, they are much less bright than they might be otherwise. This can make it easier to see dimmer objects such as small moons nearby.

2. On September 4, 2009 (plus or minus a day), the rings appeared edge-on. Because they were not visible, the effect from question 1 was enhanced, allowing even dimmer and smaller objects to be most easily seen at that time.

3. The rings must be extremely thin. Even when zoomed in to maximum, the ring still barely appears as a faint line across the planet.

Activity 2: Saturn’s Rings at Maximum Inclination 4. Saturn appears to grow larger and smaller as Earth’s orbit around the Sun brings us closer to and farther from Saturn. The size can be seen to fluctuate over a period of 1 year. When Earth is between Saturn and the Sun, we are at our closest, and Saturn appears largest. When Earth is on the opposite side of the Sun from Saturn, we are at our farthest, and Saturn appears smallest.

5. This answer will be much harder for the students to gauge accurately. Sometime in 2016 to early 2017 would be acceptable.

6. The diameter of the (visible) rings is more than twice the diameter of the planet.

7. The year Saturn’s rings were seen edge-on was 2009 and the year of maximum tilt was found to be 2016 or 2017. Assuming that the timing of the next edge-on view of the rings follows a pattern, then a student might reasonably predict that the next occurrence would be 7 or 8 years later, sometime between 2023 and 2025.

Activity 3: Finding the Next Ring Plane Crossing 8. The rings will next appear edge-on on March 23, 2025.

9. The interval between ring plane crossings was about 15 years 6 months. Using the interval between the first crossing and maximum tilt to be halfway (about

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7 years), the student might have predicted the next crossing as sometime in 2023. The book does state that the ring plane crossings take place every 15 years, but there is not one precise interval because of other effects in the planets’ orbits.

EXERCISE 12: PLUTO AND KUIPER BELT OBJECTS In 2006, the International Astronomical Union adopted new guidelines for use of the term planet, and Pluto was no longer on the list. This change was widely reported in the popular press; less well covered was the long-known connection between Pluto and other distant, icy objects in the outer Solar System. These Kuiper Belt objects (KBOs) are being discovered regularly, and each new one shows that Pluto is related to the KBOs and not the giant gas planets.

In this exercise, students will show that Pluto’s orbit is similar to those of the largest-known KBOs by comparing the orbital eccentricity and inclination angles.

ANSWER KEY Activity 1: Perihelion and Aphelion of Neptune and Pluto 1. Pluto was at perihelion around August 1989.

2. Its distance at perihelion was 29.656 AU. 3. The distance of Neptune at the same time was 30.2 AU. Activity 2: Orbits in the Outer Solar System 4. Pluto’s orbit is more like that of the KBOs. It has an eccentricity that is very different from zero and a large inclination. The following table includes data for Neptune and Pluto: Object

Eccentricity

Inclination

Neptune

0.01

1.7°

Pluto

0.25

17°

1998 WA31

0.43

9.5°

2003 EL61

0.19

28°

2004 XR190

0.08

46°

2005 FY9

0.16

29°

5. The classical planets are all confined to be very close to the ecliptic plane of the Solar System. However, Pluto and other similar objects have eccentric orbits that are inclined widely from this plane, and so could be found in almost any part of the sky.

EXERCISE 13: ASTEROIDS This exercise explains some of the characteristics of asteroid orbits. The students first examine the orbits of four large asteroids in the main belt between Mars and Jupiter and then look at the orbit of the near-Earth asteroid Bacchus.

ANSWER KEY Activity 1: Kepler’s Laws Revisited 1. Mercury and Venus orbit inside Earth’s orbit.

2. Kepler’s third law: P2 = a3, where P is the period and a is the length of the semimajor axis.

Activity 2: Orientation of the Orbital Planes of the Asteroids 3. The orbits range from nearly circular to rather elliptical. Ceres and Vesta have eccentricities that are about the same as Mars (0.09); Pallas’s orbit is quite elliptical.

4. The asteroid orbits have a range of inclinations from the plane of the Solar System.

Activity 3: Asteroids that Pass Near Earth 5–6. Bacchus did not strike Earth because its orbit is inclined to Earth’s orbital plane.

7. The asteroids are much smaller than planets, so their orbits are more easily influenced through gravitational interactions with the planets, causing the smaller asteroids to be flung into more eccentric and inclined orbits.

EXERCISE 14: THE MAGNITUDE SCALE AND DISTANCES The magnitude scale describing the brightness for stars typically needs careful explanation because of its nonintuitive, reversed scale. Students need to be reminded that smaller positive and larger negative numbers are progressively brighter stars. The method of estimating distances using absolute and apparent magnitude comparison is a very useful tool. It can be done mathematically using the distance modulus formula, but to get just a rough estimate, this comparison method works very well and stresses the student’s reasoning skills rather than math. If a star’s absolute and apparent magnitudes are equal, then the star must be at 10 pc. If the apparent magnitude is smaller (brighter) than the absolute magnitude, then the

144 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 198 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers star must be closer than 10 pc. If the apparent magnitude is larger (dimmer) than the absolute magnitude, then the star must be farther than 10 pc, and the greater the difference, the greater the distance.

Star

The stars of Hercules were chosen to complement the H-R diagram exercise in which they were used previously and because they represent a good range of brightness, luminosities, and distances.

Apparent Magnitude

Absolute Magnitude

Brightness Ranking

Luminosity Ranking

Epsilon Herculis

3.90

0.41

11 (tie)

12

Eta Herculis

3.46

0.78

6

14

Gamma Herculis

3.71

–0.18

8

10

Iota Herculis

3.81

–2.11

9 (tie)

4

Kornephoros

2.75

–0.53

1 (tie)

9

Lambda Herculis

4.40

–0.87

15

7

Mu Herculis

3.40

3.78

5

16

Omicron Herculis

3.81

–1.34

9 (tie)

5 (tie)

Phi Herculis

4.21

–0.02

14

11

Pi Herculis

3.15

–2.12

4

3

Rasalgethi

2.75

–2.60

1 (tie)

2

Rho Herculis

4.12

–1.34

12

5 (tie)

Sarin

3.09

1.18

3

15

Sigma Herculis

4.16

–0.66

13

8

Tau Herculis

3.90

–1.03

11 (tie)

6

Theta Herculis

3.84

–2.73

10

1

Xi Herculis

3.68

0.59

7

13

Zeta Herculis

2.87

unknown

2

unknown

ANSWER KEY Activity 1: Apparent and Absolute Magnitudes of Stars in Hercules 1. Apparent and absolute magnitudes of the stars.

Activity 2: Using the Magnitude Scale 2–3. Brightness and luminosity rankings are given in the previous table.

4. The brightest stars are not necessarily the most luminous. The stars may be bright because they are closer.

Activity 3: Comparing Distances Using Magnitudes 5. There are three pairs of stars with the same apparent magnitude: Rasalgethi and Kornephoros; Iota and Omicron; and Tau and Epsilon. For each pair, the star with the smaller absolute magnitude is the one that is farther away (Rasalgethi, Iota, and Tau). A star with a smaller absolute magnitude is a more luminous star.

For a more luminous star to look the same brightness

as a less luminous star, the more luminous star must be farther away.

6. Omicron and Rho have the same absolute magnitude, which means they have the same luminosity.

If they were side by side at the same distance, they would appear the same brightness. Because Rho has a larger apparent magnitude (it looks dimmer), it must be farther away.

7. The brightest star is not necessarily the closest.

Brightness depends on both distance and luminosity. A very bright star could be far away but very luminous. Remember that two stars are tied for the greatest brightness. They cannot be the same distance away because they have different absolute magnitudes.

8. Absolute magnitude is a measure of how bright a star would look if it were 10 pc (32 light-years) away. If a star’s apparent magnitude is less than (or brighter than) its absolute magnitude, this means it is closer than 10 pc. Mu Herculis is the only star in the list that has a smaller apparent magnitude than absolute magnitude, so it must be the closest.

The rest of the stars all have apparent magnitudes greater than their absolute magnitudes, which means they are all farther away than 10 pc. The greater the

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difference in these numbers, the greater the distance of the star. Theta Herculis has the greatest difference and is therefore the farthest away.

EXERCISE 15: STARS AND THE H-R DIAGRAM In this exercise, students gather data on luminosity and temperature for two groups of stars and transfer those data to an H-R diagram. The first group consists of bright stars selected from the constellation Hercules. They will learn that most of the bright stars are very luminous and are typically red giants and supergiants. The second group consists of stars that are located near the Sun. Most of these stars are main-sequence stars that are cooler and less luminous than the Sun. Selection is an important idea here. The result of an experiment can depend on how the data are gathered. Picking the bright stars in the sky is not the best way to learn about typical stars: the stars are bright either because they are luminous or because they are close to the Sun. To determine which stars are the most common, we have to sample all stars in the local volume of space and assume that the results from that sample are typical.

Activity 2: Luminosities and Temperatures of Nearby Stars 2. Data for nearby stars:

Star

Luminosity

Temperature (K)

Sun

1

5,770

HIP71683

1.8

5,715

HIP71681

0.61

5,075

HIP54035

0.03

3,483

HIP32349

27

8,858

HIP16537

0.38

5,136

HIP37279

7.5

6,800

HIP108870

0.26

4,604

HIP439

0.03

3,572

Activity 3: The H-R Diagram 3. Bright stars in Hercules are shown as filled points and nearby stars are shown as open points. The solid line is the main sequence.

ANSWER KEY

10,000

Activity 1: Luminosities and Temperatures of Bright Stars in Hercules 1,000

1. Data for Hercules:

Epsilon Herculis

Luminosity 71

9,008

Eta Herculis

57

5,025

Gamma Herculis

103

7,389

Iota Herculis

2,015

15,551

Kornephoros

200

4,928

Lambda Herculis

883

3,635

3

5,574

Omicron Herculis

352

9,008

Phi Herculis

109

9,262

Pi Herculis

2,823

3,628

Rasalgethi

2,022

4,303

Rho Herculis

350

8,969

Sarin

32

8,473

Sigma Herculis

188

8,980

Tau Herculis

566

13,789

Theta Herculis

3,741

3,831

70

4,965

unknown

5,701

Mu Herculis

Xi Herculis Zeta Herculis

100

Temperature (K) Luminosity (Suns)

Star

10

1

.1

.01 15,000

12,000

9,000

6,000

3,000

Temperature (Kelvin)

4. Most bright stars are located away from the main sequence and are giants or supergiants. The nearby stars are mostly located near the main sequence.

146 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 200 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 5. The nearby stars are more representative of “average” stars. The brighter stars are mostly very luminous stars that are rarer, as can be seen by the fact that there are none near the Sun. Cool, low-luminosity stars are very common, as can be seen by the fact that the Sun is surrounded by them.

EXERCISE 16: NEBULAE: THE BIRTH AND DEATH OF STARS The birth and death of stars both take place in nebulae. In this exercise, students are investigating the differences in the appearance, shape, and size of two types of nebulae.

These nebulae often provide some of the most beautiful images in astronomy. By investigating these images more closely, students will be looking for patterns that will help them tell these types of nebulae apart in the future.

ANSWER KEY Activity 1: Emission Nebulae 1. Emission nebulae generally look pink because they are all composed of mainly the same gas, hydrogen.

2. Emission nebulae generally encompass many stars.

3. Emission nebulae are mostly found along the plane of the Milky Way. Activity 2: Planetary Nebulae 4. Planetary nebulae are roughly round or double-lobed in shape.

5. Each planetary nebula contains one star at the center.

6. Planetary nebulae are smaller than emission nebulae.

You must zoom in further to see the shape, and they only encompass one star, whereas emission nebulae encompass many.

7. Planetary nebulae have a much wider variety of colors because they are composed of a larger fraction of elements other than hydrogen.

8. At the low magnification of early telescopes, planetary nebulae typically looked like small fuzzy circles, which is similar to how planets appeared. They were named because of this similarity, even though they were later found to have nothing to do with planets.

EXERCISE 17: PULSARS AND SUPERNOVA REMNANTS Massive stars die violent deaths. This exercise investigates the remnants these supernova explosions leave behind.

The supernova identification portion may be challenging

because of the nature of the images used. All Type II (core collapse) supernovae should have a neutron star or black hole remaining in the center, but sometimes these are hard to detect. If a pulsar is created, then it may not even be detected if the poles do not point near the direction of Earth. If the star is hard to detect, it may be mistaken for a Type Ia (white dwarf explosion) supernova, in which the central star is completely destroyed. The determination is done more accurately by looking at the spectrum of the supernova, but this activity gives a basic idea of what astronomers are looking for.

ANSWER KEY Activity 1: Pulsars 1. Most pulsars are near the cloud of the Milky Way and in the general direction of the center of the galaxy. This cloud represents where most of the stars of the galaxy are distributed. As pulsars are the remnants of dead stars, they should be found where the concentration of stars is located.

2. The constellation Aquila appears to have the greatest amount of pulsars. This constellation is just to one side of the center of the galaxy. When we look in this direction, we are looking into the largest concentration of stars in our galaxy.

3. The period of the Crab Nebula pulsar is 0.0334 seconds. This works out to about 30 times per second.

Activity 2: Supernova Remnants 4. Table of supernova remnants and justifications: Supernova Remnant

Type Ia or II

Crab Nebula

Type II

There is a pulsar remaining in the center

Tycho

Type Ia

No central star

Cassiopeia A

Type II

Star faintly visible in the center

Vela Pulsar

Type II

Pulsar in the center

B1509-58

Type II

The brightness at the center is a pulsar

Cygnus Loop

Type Ia

No central star

Justification

Cassiopeia A is a Type II, but students will likely classify it as a Type Ia, as the neutron star remaining in the center is very dim and not actually shown in the program.

B1509-58 is a Type II as well. The bright center of the image is caused by a young, energetic pulsar, but this pulsar is not marked in Starry Night.

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5. The Cygnus Loop is much larger in diameter than the other remnants. This means it has been expanding for much longer and is older than the other remnants.

6. Not all supernova remnants have a star remaining in their centers. Type Ia supernovae are the result of a white dwarf destroying itself, leaving only the debris cloud behind.

7. Not all pulsars have supernova remnants surrounding them. All pulsars originally must have been surrounded by a debris cloud, but this cloud expands over time and will eventually dissipate into space as it ages, leaving the pulsar behind.

EXERCISE 18: GALAXY CLASSIFICATION Classifying galaxies within the elliptical or spiral sequence is somewhat subjective depending on how a student interprets the shape. There is no set division between a spiral Sb and Sc for example, so there could be a wide variety of answers to some of the galaxy classifications. Students should be encouraged to make use of a Hubble tuning fork diagram. The answer table provided for Activity 4 gives the most commonly accepted classifications, which in some cases like M64 (Sa/b) indicate that there is not necessarily agreement even among professional astronomers.

ANSWER KEY Activity 1: Spiral Galaxies 1. M101 looks spiral in shape, with dusty arms spiraling into a more compact, brighter, yellowish-orange center.

2. The stars in the two different areas are of different temperatures. The light of hotter, bluer stars dominates the arms, whereas cooler, orange to red stars dominate the central area.

3. M31 appears to be the same basic type of galaxy, but viewed at more of an angle.

4. The view of NGC 4565 shows a spiral galaxy completely on edge. In this view, the arms are shown to be very thin compared to their diameter.

5. This question is a bit subjective, but anything less than one-tenth should be acceptable.

Activity 2: Elliptical Galaxies 6. Elliptical galaxies are generally round in shape, without any other distinguishing features.

7. This galaxy appears more yellowish-orange, which means that cooler stars are dominating the light.

Activity 3: Irregular Galaxies 8. The LMC has no similarity in shape but does appear mostly bluish, similar to the arms of a spiral galaxy.

Activity 4: Galaxy Classification 9. Table of galaxies in the Messier catalog: Galaxy

Galaxy Type

LMC

Irr

M87

E0

M101

Sc

M32

E0

M51

Sb/c

M59

E3

M60

E1

M63

SBb/c

M64

Sa/b

M65

Sa/b

M74

Sc

M82

Irr or Sa

Galaxy

Galaxy Type

M85

S0

M90

Sb

M104

Sa

M105

E0

SMC

Irr

10. We see our galaxy as a cloud surrounding us on all sides, with a slightly greater density in one particular direction. This is what would be expected if you were inside of a disk of stars that had a more concentrated center, like a spiral galaxy.

EXERCISE 19: QUASARS AND ACTIVE GALAXIES In this exercise, students will use skills similar to the magnitudes exercise to estimate relative distance using apparent and absolute magnitudes. Quasars are some of the most luminous objects in the universe, yet none of them are apparently very bright. This tells us that the distances to all quasars must be huge. As all quasars are very far away, this also means that they existed earlier in the history of the universe. No quasars and few active galactic nuclei are seen to exist near the Milky Way. At one time, the universe must have been a very active place, perhaps due to more galaxy interactions.

148 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 202 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers

ANSWER KEY Activity 1: Quasars 1. Quasars appear to lie mainly above and below the dusty plane of the Milky Way. This implies that the quasars are outside of our galaxy.

2. As there are several quasars for the students to choose from, there will be a variety of different answers. Apparent magnitudes range from 14.1 to 15.11 and absolute magnitudes range from –23.5 to –29.6. The averages for these magnitudes will fall somewhere inside these ranges.

Redshifts are mostly around 0.1, with one at 1.178.

3. If these quasars were located at 10 pc from our Solar System, they would look roughly as bright as the Sun does in our daytime sky (some would be a little dimmer; two would be even brighter).

4. These quasars have a very large difference between their apparent and absolute magnitudes. Their absolute magnitudes are all large negative numbers, which means they are highly luminous. The apparent magnitudes are all larger positive numbers, which means they are very dim. For a highly luminous object to appear very dim, it must be very far away. The difference between the absolute and apparent magnitudes is much larger than the example given for M13, so the quasars must be much farther away than 25,000 light-years.

Activity 2: Active Galaxies 5. As in question 2, the answers for the active galaxy information may vary among students. Apparent magnitudes will fall somewhere between 11 and 15, and absolute magnitudes will range from –17 to –21, with averages for both falling somewhere inside these ranges. Redshifts will be between 0.001 and 0.01.

6. Active galaxies are less intrinsically bright than quasars.

7. The difference between the apparent and absolute magnitudes for active galaxies is large, but smaller than it is for quasars; therefore, active galaxies must be closer than quasars.

8. Quasars generally have greater redshifts. According to Hubble’s law, the greater the redshift, the greater the distance. This agrees with the above estimate of quasars being more distant than active galaxies, which have smaller redshifts.

EXERCISE 20: VIEWS OF THE MILKY WAY By viewing the Milky Way in all possible types of radiation, we learn much more about its true structure. Visible light is an especially poor choice for viewing the galaxy because it is easily blocked by the large amounts of dust that exist within the galactic disk. Other wavelengths of light, such as infrared, easily penetrate this dust to reveal much more of

the galaxy. This is especially helpful when probing the nature of the very center of the galaxy, which is hidden behind copious amounts of dust, gas, and other stars.

ANSWER KEY Activity 1: The Milky Way in Visible Light 1. The Milky Way appears to be a diffuse cloudy band that encircles the Solar System. It appears brighter and denser in one particular direction.

2. The fact that the cloud appears to be confined to one common plane means that the overall shape must be flat in one dimension and is likely circular like our Solar System. That this surrounds us on all sides means that we are in this plane as well. The direction that looks brighter is likely the center of this shape, meaning we are not at the center.

3. The center of the galaxy appears to be in the direction of Sagittarius/Scorpius, as this direction is the brightest and densest.

4. The plane of the Milky Way is at a large angle to the plane of the Solar System.

Activity 2: The Center of the Milky Way in Other Wavelengths 5. IR radiation is brightest toward Sagittarius/Scorpius.

This matches the earlier observation in visible light.

6. Gamma, X-rays (diffuse), near-IR, molecular hydrogen, radio (408 MHz).

7. Visible light (too many clouds), atomic hydrogen.

8. The conditions at the center of our galaxy are likely very intense because of the high-energy radiation we see from that direction. But it is also cloudier in that direction, which blocks some types of radiation (such as visible light).

Activity 3: Other Parts of the Milky Way 9. Some possible X-ray sources are the Crab Nebula (Taurus), Cassiopeia A (Cassiopeia), Tycho (Cassiopeia), Vela Pulsar (Vela), and Eta Carina (Carina).

10. Most of these objects also emit considerable amounts of gamma rays and radio (408 MHz). This tells us that these objects must be intensely energetic and hot.

11. Supernova remnants are concentrated along the plane of the Milky Way, mostly in the direction of the center of the galaxy, in the region of most intense gamma-ray emission. Because the density of stars is greater in the direction of the galactic center, supernovae are more likely to happen in this direction, so we should see both the high-energy radiation from them as well as the remnants they leave behind.

The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 149 The Norton Starry Night Workbook: Exercise Summaries and Activity Answers ◆ 203

EXERCISE 21: GLOBULAR CLUSTERS

Activity 3: The Distance to Galactic Center

This exercise re-creates how Harlow Shapley determined the location of the center of our galaxy in 1915. The globular clusters used to make the plot were chosen to give an equal number above and below the galactic plane. Also, the longitudes of the chosen clusters are within roughly 10° of the longitude of the accepted galactic center. While this is not a truly random assortment, the point is still made.

A brief explanation of how to plot points on the polar graph will likely be needed, although two clusters are plotted for the students on their answer sheet. There will be some variation in location of points and where students choose the center to be, but the results should be fairly consistent.

5. Plot of globular cluster locations relative to the Sun. 6. The apparent center should be roughly between 25,000 and 30,000 light-years from the Sun. 7. The distance given in the textbook is 27,000 light-years.

75° 60°

45°

ANSWER KEY

30°

Activity 1: The Distribution of Globular Clusters 1. Globular clusters are mostly located near the plane of the Milky Way, with almost all concentrated in the direction of the apparent center of the galaxy.

2. Globular clusters are most numerous around the constellations Sagittarius and Scorpius.

3. The Milky Way appears larger and brighter in this direction.

15°

0°

Sun

−15°

Activity 2: The Distances to Globular Clusters 4. Table of globular cluster data:

−30°

Cluster

Distance (ly)

Latitude (°)

NGC 7099

26,000

–46

NGC 5897

41,000

30

NGC 5904

24,000

46

NGC 6093

32,000

19

NGC 6121

7,000

15

NGC 6144

33,000

15

NGC 6171

20,000

23

NGC 6235

32,000

13

NGC 6284

47,000

9

NGC 6316

35,000

5

NGC 6333

26,000

10

NGC 6453

36,000

–3

NGC 6522

25,000

–3

NGC 6544

8,000

–2

NGC 6652

31,000

–11

NGC 6681

29,000

–12

NGC 6723

28,000

–17

NGC 6809

17,000

–23

NGC 6864

61,000

–25

−45°

−60° −75°

EXERCISE 22: THE NEIGHBORHOOD OF THE SUN One of the most fruitful ways of finding stars near the Sun is to look for stars with a large proper motion. The proper (or tangential) motion is the rate of change of the position of a star. Stars can have a large proper motion if they are moving rapidly through space or if they are located close to the Sun. The text emphasizes the importance of high-velocity stars as tracers of the Milky Way’s halo. On average, halo stars have high velocities relative to the Sun. Samples of high-velocity stars pick out many halo stars, but they also pick out a few nearby stars in the Milky Way’s disk. The examples in this exercise are all disk stars.

150 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers 204 ◆ The Norton Starry Night Workbook: Exercise Summaries and Activity Answers

ANSWER KEY

ANSWER KEY

Activity 1: Finding Nearby Stars

Activity 1: Stepping to the Nearest Stars

1. Rigil Kentaurus is a nearby star because it has a large motion. The rate of motion is given by v/D, where D is the distance and v the tangential velocity.

2. Rigil Kentaurus is about 4 light-years from our Sun. 3. The answer will depend on what stars were picked.

The students will generally find that apparently stationary stars will have distances that are much larger than 4 light-years, the distance to Rigil Kentaurus.

1. The current location is 500 AU from Earth. Even the nearest stars are (300,000 AU)/(500 AU) = 600 times farther away.

2. At this distance, Neptune’s orbit is visible, but small.

Earth’s orbit is too close to the Sun to be visible unless we zoom in to high magnification.

3. If the students notice any stars moving, they will pick ones with distances less than about 8 light-years.

Activity 2: The Fastest-Moving Star

Activity 2: Leaving the Milky Way

4. Barnard’s star is currently in the constellation Ophiuchus.

5. Barnard’s star will be in Hercules about 3,400 years from now.

6–7. Barnard’s star is 5.9 light-years away from the Sun. Students might reasonably conclude that Barnard’s star is closer than Rigil Kentaurus because it has a higher tangential motion.

8. Tangential velocity is the other factor besides distance that determines a star’s apparent motion.

Barnard’s star has a higher tangential velocity than Rigil Kentaurus.

4. The Milky Way is a spiral galaxy. It has a central bulge surrounded by a flat disk containing spiral arms.

5. There are several possibilities other than the example Fornax. Others that might be sighted include the Ursa Minor dwarf galaxy, the Sagittarius dwarf elliptical galaxy, Sextans, UGC, and possibly the Large Magellanic Cloud and the Small Magellanic Cloud.

EXERCISE 23: BEYOND THE MILKY WAY Here, students leave the Solar System and head out into deep space. In imitation of the movie Powers of Ten, the distance will jump by factors of 100 at each step. On occasion, students will pause to look around.

It is worth stressing just how fast distances increase with geometric jumps. Only a few jumps take us from the small to the large. Intervals of a factor of 100 correspond to half the width of a fingertip, one ordinary step, the length of a football field, a 10-km footrace, a three-hour car trip, and a third of the distance to the Moon.

At the end of the exercise, the viewpoint is located 200 Mly from Earth. As large as this distance seems, it is still only about 1.5 percent of the way to the most distant objects in the universe.

Activity 3: Distant Galaxies 6. The galaxies are not uniformly distributed. There are many dense clusters and several areas with fewer galaxies. Students may notice the zone of avoidance along the plane of the Milky Way. There are undoubtedly many galaxies there, but the catalogs do not include them because their light is blocked by dust in the plane of the Milky Way. 7. We assume that the galaxies are distributed homogeneously and isotropically; that is, they are uniform in all places and in all directions. If this assumption is true, it applies only on much larger-distance scales than we can see in Starry Night.

8. The recession velocity is given by v = Hd, where d is in units of Mly. For d = 200 Mly, the velocity is 4,400 km/s.

Credits Figure IM 12.1: Larry Marschall and Christy Zuidema, Gettysburg College Figure IM 19.1: NASA, ESA, and the Hubble Heritage Team (STScI/AURA) Figure IM 24.4: NASA