Solution Manual For Pathways to Astronomy, 2025 Release Stephen Schneider and Thomas Arny Chapter 1-

Solution Manual For Pathways to Astronomy, 2025 Release Stephen Schneider and Thomas Arny Chapter 1-86

Solutions to End of the Unit Quantitative Problems Part I: The Cosmic Landscape Unit 1 10.

If you use a volleyball as a model of Earth, how big would 1 kilometer be on it?

Assume a volleyball has a circumference of 68 cm.

10. (0.0017 cm) Use the value of 40,000 km for the circumference of the Earth given in the chapter, and create a scale factor for the model: Circumference (Volleyball) / Circumference (Earth) = 68 cm / 40,000 km = 0.0017 cm/km This means 1 km on Earth would be represented by just 0.0017 cm on the volleyball, or just 0.017 mm. This is about the diameter of extremely fine hair.

11.

What would be the circumference and diameter (circumference = π × diameter) of a ball that would represent the Moon if Earth were a volleyball? What kind of ball or object matches this size? (Note: For this and the following questions you will probably need to look up the sizes and distances of astronomical objects in the set of tables in the Appendix.)

11. There are several ways to solve this problem. One easy way is to compare the Earth’s radius or diameter to the Moon’s radius or diameter. The chapter notes the Moon’s diameter is about ¼ the Earth’s. Since C = π × D, then the circumference is also ¼ the Earth’s, so in the model where Earth is a volleyball, the Moon must have a circumference 0.25 × 68 cm = 17 cm. The Moon’s diameter is then C/π = 17 cm / π = 5.4 cm. This is smaller than a tennis ball.

A slightly more accurate calculation using the values of the Earth and Moon’s size from the Appendix (the Moon’s radius is 1738 km; the Earth is 6378) gives the Moon’s radius is 0.273 times the Earth’s; therefore the diameter and circumference are also 0.273 times the Earth’s; this gives a circumference for the model of the Moon of 18.6 cm and a diameter of 5.9 cm.

Either way, the model of the Moon would be a little smaller than a tennis ball or about the size of a small orange.

10. If Earth were a volleyball, what would be the diameter of the Sun? What object matches this size?

12. (About 24 m, the size of a house). The volleyball is about 22 cm across (use Diameter = Circumference/π = 68 cm /π = 21.6 cm). The diameter of the Sun is about 109 times that of Earth (using the ratio of its radius to the radius of Earth RSun/REarth = 696,000 km/6378 km = 109) so the model of the Sun would be 22 cm × 109 = 2398 cm = 23.98 m, or about 24 meters. This

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would be roughly the size of a house. (Also okay to use RSun/REarth ~ 100 as given in chapter, for 2200 cm).

11. How many astronomical units away is the Moon from Earth?

13. (0.0026 AU) An AU is approximately 1.5 × 108 km, while the Moon is d = 384,400 km away from Earth (Appendix Value). Performing a standard unit conversion gives a distance of d = (384,400 km) × (1 AU/1.5 × 108 km) = 0.0026 AU. This is about 1/400th of an AU, or put another way, the Moon is about 400 times closer to Earth than the Sun. 12. During the 1960s and 1970s, the Apollo spacecraft took humans to the Moon in three days. Traveling to Mars requires a trip of about 2 astronomical units in total. How long would this trip take, traveling at the same speed as to the Moon?

14. (6.4 years). The Moon is 384,400 km away from Earth (Appendix value). A three-day (72-hour) flight gives an average speed of 384,400 km/3 day ~ 128,000 km/day = 1.28 × 105 km/day. A trip to Mars on a trajectory 2 AU long would cover a distance 2 × (1.5 × 108 km) = 3.0 × 108 km. At the speed of an Apollo spacecraft, a trip to Mars would therefore take: time = distance/speed = 3.0 × 108 km / (1.28 × 105 km/day) = 2.34 × 103 days This is about 6.4 years to reach Mars. (Note: if using the rounded value of 400,000 km distance to the Moon in the chapter, the speed is about 130,000 km/day, which yields about 6 yr.)

13. Using the same assumptions as in the previous question, how long would it take to travel to Pluto, about 40 astronomical units away?

15. (128 years). Based on the speed calculated in the last problem, a 2 AU flight took 6.4 years.

Pluto is 20 times farther away, so the flight would take 20 times longer: 20 × 6.4 yr = 128 yr.

Unit 2 10. If the Milky Way were the size of a nickel (about 2 centimeters in diameter): (a) How big would the Local Group be? (b) How big would the Local Supercluster be? (c) How big would the visible universe be? (The data in Table 2.1 may help you here.)

10. (a) 60 cm; (b) 20 m; (c) 5.5 km.

Create a model scale relating the size of a nickel to the size of the Milky Way: Diameter(nickel) / Diameter (Milky Way) = 2 cm / 100,000 ly = 2 × 10−5 cm/ly.

Using the approximate radius sizes in table 2.1, and multiplying by 2 to get diameters: a) The Local Group is about 3 million ly in diameter, which in the model is:

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3 × 106 ly × 2 × 10−5 cm/ly = 60 cm. 60 cm is about the size of a beach ball. b) The Local Supercluster is 50 million ly × 2 = 100 million ly in diameter, scaled to the model this is 100 × 106 ly × 2 × 10−5 cm/ly = 2000 cm = 20 m. 20 m is about the size of a house. c) The visible universe would be about 13.7 billion ly × 2 = 27.4 billion ly in diameter, which in the model is: 27.4 × 109 ly × 2 × 10−5 cm/ly = 5.5 × 105 cm = 5500 m = 5.5 km 5.5 km is about the size of a town; vertically, it is more than 18,000 feet, or the height of many mountains in the Andes or Himalayas.

11. If we detected radio signals (which travel at the speed of light) of intelligent origin from the Andromeda Galaxy (M31) and immediately responded with a radio message of our own, how long would we have to wait for a reply?

11. (5 million years) Radio waves, like all forms of electromagnetic radiation, travel at the speed of light. We’d have to wait for the message to get to Andromeda, and for the response to make it back to us—twice the distance (2.5 million light years), divided by the speed of light. Because light travels 1 light year in a year, so: time = distance / speed = (2 × 2.5 million ly) / (1 ly/yr) = 5 million years.

There is an introduction to solving distance-velocity-time problems at the beginning of the appendix.

12. How long does it take light to travel from the Sun to Earth?

12. (About 8.3 minutes). An AU is approximately 1.5 × 108 km, and the speed of light is about 3.0 × 105 km/sec. Therefore, the time it takes light to travel from the Sun to Earth is time = distance / speed = 1.5 × 108 km / 3.0 × 105 km/sec = 500 sec, which is about 8.3 minutes.

13. If the Milky Way is moving away from the Virgo Cluster at 1000 kilometers per second, how long does it take for the distance between them to increase by 1 light-year?

13. (About 300 years). This is another distance-velocity-time problem: distance = 1 ly = 9.46 × 1015 m = 9.46 × 1012 km; Velocity = 1000 km/sec.

Therefore, we find the time for the distance to increase by 1 light-year is: time = distance/Velocity = 9.46 × 1012 km / (1000 km/sec) = 9.46 × 109 sec.

Converting this to years: 9.46 × 109 sec × 1hr/(3600 sec) × 1 day / (24 hr) × 1 yr/(365.25 day) = 300 yr.

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14. The Milky Way is moving toward the larger galaxy M31 at about 100 kilometers per second. M31 is about 2.5 million light-years away. How long will it take before the Milky Way collides with M31 if it continues at this speed?

14. (7.5 × 109 yr). This problem is like the last one, but with d = 2.5 × 106 ly and V = 100 km/sec: t = (2.5×106 ly × 9.46 × 1012 km/ly) / (100 km/sec) = 2.37 × 1017 sec.

Converting this to years gives: 1.89 × 1017 sec × 1hr/(3600 sec) × 1 day / (24 hr) × 1 yr/(365.25 day) = 7.5 × 109 yr.

So the Milky Way and Andromeda will “collide” in 7.5 billion years they continued constantly at that speed. But since the force of gravity increases in strength significantly as distances between objects become smaller, the Milky Way will accelerate towards Andromeda faster and faster, and collide in less than 5 billion years.

15. Some distant galaxies seen by the Hubble Space Telescope are 12 billion light-years away.

Using the scale in Figure 2.4 (where the Milky Way is about the size of a large city), how far would that be in the model 15. (0.08 AU). At a scale of Figure 2.4, 1 light year = 1 meter, so 12 billion light years is equivalent to 12 billion meters = 12 × 109 m = 12 × 106 km = 12 million km. So, on the scale where the Milky Way is about the size of a large metropolitan area, the most distant galaxies we can see are about 30 times the Moon’s distance *12 × 106 km/384,000 km = 31], or about a tenth of an AU distant: 12 × 109 m/(1.5 × 1011 m/AU) = 0.08 AU. Unit 3 9. The radius of the Sun is 7 × 105 kilometers, and that of Earth is about 6.4 × 103 kilometers. Show that the Sun’s radius is approximately 100 times Earth’s radius.

9. The ratio of the Sun’s radius to the Earth’s radius is 7  105 km / (6.4  103 km) = 7/6.4  105−3 ≈ 1  102 or about 100 times larger.

10. Hypergiant stars can have radii as large as 1500 times that of the Sun. Estimate the radius of a hypergiant star in kilometers. How many AU is this equivalent to?

10. (1.05 × 109 km = 7.0 AU) From Table 3.3 a solar radius is 6.97 × 108 m, or measured in kilometers 6.97 × 105 km. The radius of a hypergiant star would be about: 1500 × 6.97 × 105 km = 1.05 × 109 km.

An astronomical unit is 1.50 × 108 km making the radius of the hypergiant star: (1.05 × 109 km) × (1 AU/1.50 × 108 km) = 6.97 AU ≈ 7.0 AU, 4

which is larger than Jupiter’s orbit!

11. Using scientific notation, show that it takes sunlight about 8 ½ minutes to reach Earth from the Sun. 11. The light travel time to Earth from the Sun can be calculated from the distance-speed-time formula t = d/V, where V = c = 3.0 × 105 km sec−1, and d = 1 AU = 1.496 × 108 km: t = 1.496 × 108 km/ (3.0 × 105 km sec−1) = (1.5/3.0) × 108−5 sec = 0.50 × 103 sec = 5.0 × 102 sec Next we convert this to minutes, also using scientific notation: t = 5.0 × 101 sec × 1 min / (6 × 101 sec) = 50/6 min = 8.3 min (about 8 1/2 min).

12. How many 1-kiloton bombs would need to be exploded to produce 3.86 × 10 26 joules, the amount of energy emitted by the Sun in 1 second?

12. (9.23 × 1013 1-kiloton bombs per second).

The Sun’s luminosity is 3.86 × 1026 W = 3.86 × 1026 J/sec. Divide the by the energy output of a 9.23 kiloton bomb (from Table 3.3) to get the equivalent number of these explosions each second: (3.86 × 1026 J/sec) / (4.18 × 1012 J/kt) = 9.23 × 1013 kt per second.

13. If I want to work out the distance to the Andromeda Galaxy, do I need to be concerned about whether the distance is from the center of the Milky Way or from the Sun’s location?

Why or why not?

13. The Sun is 25,000 light years (2.5 × 104 = 0.025 × 106 = 0.025 million light years) from the center of the Milky Way galaxy. The Andromeda galaxy is 2.5 million light years away–or about 100 times farther. In other words, the distance to the center of the Milky Way is only 1% of the distance to the Andromeda galaxy.

Even if the Andromeda galaxy was located in the exact opposite direction from the center of the Milky Way (and it isn’t), its distance would be 2.5 × 106 ly 0.025 × 106 ly = 2.475 × 106 ly and rounded up to just two digits of precision, this is still ≈ 2.5 × 106 ly. So measuring from the center of the Milky Way or from our location would not affect the distance estimate to within the two-digit precision of the distance of the Andromeda galaxy.

14. Suppose two galaxies move away from each other at a speed of 6000 km/sec and are 300 million (3 × 108) light-years apart. If their speed has remained constant, how long has it taken them to move that far apart? Express your answer in years.

14. (1.50 × 1010 years). The time for the galaxies to move this distance is given by the distancespeed-time equation: t = d/V = (3 × 108 ly)/(6000 km/sec) = (3 × 108 ly × 9.46 × 1012 km/ly)/(6000 km/sec) = 4.73 × 1017 sec = 4.73 × 1017 sec × 1y/(3.16 × 107 sec) = 1.50 × 1010 years.

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This time of 15 billion years is longer than the present age of our universe, but it assumes the galaxies have moved at a constant speed.

15. The equation E = m × c2 tells us how much energy, E, is stored in a mass of m. Use this equation to calculate the energy stored in a 3.10-gram penny.

15. (2.79 × 1014 J). To get the units to work our properly in the MKS system, the 3.10 gram mass of the penny must be expressed in kilograms: E = m c2 = (3.10 × 10‒3 kg) × (3.00 × 108 m/sec)2 = 2.79 × 1014 kg m2/sec2 = 2.79 × 1014 J Note: this means the conversion of a penny to pure energy is equivalent to a 66.7 kiloton bomb.

16. Using scientific notation, evaluate (1.4 × 109)3 ∕(9.3 × 108 )2.

16. Using scientific notation: (1.4 × 109)3 / (9.3 × 108)2 = (1.43 × 109 × 3) / (9.32 × 108 × 2) = 2.74 × 1027/ 86.5 × 1016 = 2.74/86.5 × 1027 − 16 = 0.0317 × 1011 = 3.2 × 109 Note: as you do the calculation, keep at least one extra digit in all of the intermediate values, and only round off at the end. Because the original numbers (1.4 × 109 and 9.3 × 108 ) are shown with two digits of precision, we round the final answer up and record it also with just two digits.

17. Napoleon Bonaparte is often said to have been short, but contemporaries described him as being of average or slightly above-average height for his time. This mistake was apparently made because the height reported in Paris feet was misinterpreted. Bonaparte was reported to be about 5 feet 2 inches tall at his autopsy. If these were in Paris units, 6.63% longer than English inches and feet, how tall was he actually in (a) English units, and (b) metric units.

17. (a) 5 ft 6 in; (b) 1.68 m) Denote measurements made in Paris with (P). So Bonaparte’s height of 5 ft 2 in (P) would be 5 × 12 + 2 = 62 in (P). Since the Parisian units are 6.63% larger, 1 Parisian inch = 1.0663 English inches. To convert this: 62 in(P) × 1.0663 in / 1 in(P) = 66 in = (5 × 12 + 6) in Thus in the English system, Bonaparte was 5 ft 6 in tall. (b) To convert to the Metric system from the English measurement of 66 in, use the fact that 1 in = 2.54 cm and calculate: 66 in × (2.54 cm / 1 in) = 168 cm = 1.68 m.

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Unit 4 11. If a proton were the size of a large apple (about 10 cm in diameter), how large would a hydrogen atom be?

11. A hydrogen atom is 10−10 m in diameter while a proton is about 10−15 m, so the atom is about 105 times larger than the size of the proton. If the proton were blown up in size of a large apple (10 cm), then the size of the atom would become (10 × 10−2 m) × 105 = 104 m = 10 km, about the size of a city.

12. Imagine that Earth and the Moon were both positively charged by an amount proportional to their masses, so that their electrical repulsion canceled out their gravitational attraction. Using the relative strengths of gravity and the electromagnetic force, how much mass in Earth would have to be positively charged?

12. (6 × 10−12 kg). The force of gravity is 10−36 times weaker than the force of electromagnetism (see Table 4.1 in chapter 4). So only 10−36 as much matter is needed for the electromagnetic force as for the gravitational force, for them to be equal. The mass of the Earth is 5.974 × 10 24 kg, so only ~ 6 × 1024 kg × 1036 = 6 × 10−12 kg of mass needs to be charged! That is just 6 billionth of a gram! And only about 1/80th of this value would be needed for the Moon.

Note that this is about the mass of a single cell in your body, so if all the electrons could somehow disappear from a single cell on Earth, and similarly from an even smaller single cell on the Moon, the electrical repulsion from those two cells would rival the whole Earth’s attractive gravitational pull.

13. Explain at a fundamental level what is happening when you take clothes out of a dryer and find them clinging to each other. What do you think is happening when you hear a crackling sound as you pull a sock and shirt apart?

13. While the atoms in your clothes typically are electrically neutral, meaning that they have an equal number of positive and negative charges, friction between tumbling clothes can allow certain materials to capture electrons from other materials. When this happens, one material becomes more negative and the other more positive. This is what we mean when we say something has a static charge. Since opposite charges attract, a sock that has gained a negative charge in this process is attracted to and clings to a sweater that has developed a positive charge. The crackling sound is that of charges jumping from one fabric to the other in tiny sparks as you pull them apart (you may be able to see these in the dark).

Under humid conditions, surfaces become covered with water molecules that are conductive and allow electrical charges to flow away relatively easily, so large charges rarely build up.

That’s why the static charge builds up on clothes in the dryer and not in the wash. Dryer sheets often also contain chemicals that are conductive and coat the clothes, thereby reducing the ability of tumbled clothes to build up a large electric charge.

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14. It is estimated that a neutrino has to pass within a distance of about 10 –24 meters of a quark for a high likelihood of interaction. If an atom has a radius of 10–10 meters and 12 nuclear particles in its nucleus, what is the probability that a neutrino passing through the atom will undergo an interaction? (Hint: Think of the atom as a target and the area around each quark as a bull’s-eye.)

14. The probability that a neutrino passing through an atom will interact is about one out of 1027! Explanation: Assume that the atom is a sphere and the 12 nuclear particles are each spheres with circular cross-sections like circular targets. Using the given radii, calculate the cross-sectional areas for the atom as a whole (A1) and for the nuclear particles (A2). Each area = π × r2, so the ratio of areas is: 12 × A2 / A1 = 12 × π × (10−24m)2 / (π × (10−10m)2 ) = 12 × (10−24)2 / (10−10)2 Note that the pi’s (and meters) cancel out, so the fraction of the area that the neutrino might hit is = 12 × 10−48 / 10−20 = 1.2 × 10−27.

Thus, a neutrino might be expected to pass through something in the order of 10 27 atoms before there is a strong likelihood of it interacting! This means that a neutrino has a high probability of passing through a solid layer of material that is even a light year (about 10 16 meters) thick, since that’s “only” 1026 atoms thick.

Unit 5 10. What are your latitude and longitude? What are the latitude and longitude toward your nadir at the point on Earth exactly opposite you? What geographic features are located there?

10. (Adjust answer to match your location). Method: If your latitude and longitude are 40 N and 100 W (at the southern edge of the state of Nebraska, in the United States), then the points on the opposite side of the Earth would have to be somewhere on 40 S (reflected through the center of the Earth). The longitude is slightly trickier, because it is 180 away on the other side of the Prime Meridian, so it is (180−100) E = 80 E. This location (40S,80E) is in the middle of the Indian Ocean. [Look at a globe or a 360 protractor to understand why it’s (180−value) in the opposite hemisphere.]

11. If your observatory is located at a latitude of 45°N and a longitude of 90°W, what range of declinations are visible from your observatory? If you wanted to see the most stars, at what latitude should you build your observatory?

11. (From‒45° up to +90° declinations). At a latitude of 45° N, stars at declination +45° pass through your zenith, so the horizon blocks off stars more than 90° north or south of that. You can therefore observe declinations from‒45° and up to +90°. To see the most stars you would want to build your observatory on the equator, and then you would be able to observe stars across all the declinations.

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12. The constellations Ursa Major (the Great Bear), in the Northern Hemisphere, and Crux (the Southern Cross), in the Southern Hemisphere, can be used to locate the north and south celestial poles, respectively. Using the star charts at the end of the book, can you show how this is done?

12. In the Northern hemisphere, the end of the “bowl” of the Big Dipper “points” at the north celestial pole and the north star, Polaris (itself at the handle end of Ursa Minor or the Little Dipper). Continuing the line made by Merak and Dubhe (going “up” out of the bowl) leads the eye to Polaris about 30° away.

In the Southern hemisphere, the two bright stars on the longer arm of the “Southern Cross” (the constellation Crux) also point toward the south pole about 30° away, however there are no naked-eye stars close to the pole, so finding the position of the celestial pole is trickier. The distance from Acrux (the brightest star in Crux) to the pole is about four times the distance between Gamma Crucis and Acrux, or about 2/3rds the distance from Acrux to Beta Hydri, the first relatively bright star encountered along the line on the other side of the pole. The south celestial pole might also be pictured as the missing point needed to make a trapezoid out of the three brightest stars in the constellation Octans.

13. Buenos Aires and Sydney are both at latitude 34°S, but Buenos Aires is at longitude 58°W and Sydney at 151°E. Using the star chart at the back of the book, find a bright star that passes through the zenith of Buenos Aires and Sydney. How many hours after passing through the zenith in Buenos Aires will it pass through the zenith in Sydney?

13. At 34°S, a star with a declination of‒34° will pass through the zenith. Some examples you can find in the charts in the back of the book are  Scorpii (Wei),  Sagittarii (Kaus Australis), or  Columbae (Phaet). As the Earth turns, the star will pass over Buenos Aires at 58° W, the rest of South America, the Pacific Ocean, the 180° line of longitude separating the western and eastern hemispheres, and then on to Sydney, Australia at 151° E. The separation between these cities and longitude 180° is: Buenos Aires west to 180° longitude line = 180°58° = 122° 180° longitude line west to Sydney = 180°151° = 29° The total change in longitude is therefore 122° + 29° = 151°. Since the Earth turns 15° per hour, it will take 151°/(15°/hr) or just a bit over 10 hours between the star being overhead in Buenos Aires and being overhead in Sydney.

14. How far is the celestial equator from your zenith if your latitude is 31°N? 15°S?

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14. (31 south of your zenith; 15 north of your zenith). At 31 N, a star at declination +31 will pass through your zenith, so the celestial equator is 31 south of your zenith. (This is illustrated as angle A in the figure at right for latitude 42 N.) Likewise, at 15 S, the celestial equator is 15 north of your zenith.

15. Using a compass and protractor, draw a diagram of Earth. Mark the poles, the equator, and your own latitude. With a ruler, find the line that is “tangent” to your location—that is, the line should touch Earth’s surface at your location while matching the surface at this point.

This tangent line represents your horizon. Draw a line extending from pole to pole to indicate Earth’s axis, and extend both this line and your horizon line until they cross. Show that your latitude is the same as the angle of the celestial pole with respect to your horizon.

15. The drawing should look like this: (Northern hemisphere location; reverse NP and SP for Southern hemisphere.).

Note that this geometric demonstration makes use of several principles: a line (in this case the horizon) crossing parallel lines makes the same angle A to each, the sum of the angles in a triangle is 180, and the tangent line to a point on a circle is at a 90 angle to the radius line to that point.

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Some students are concerned that the different sightlines to the pole star are not exactly parallel for a star that is at a finite distance. While true for any physical star, the difference is extremely small, because stars are so far away. For example, even for the star Polaris, which is about 50 times larger than the Sun, but more than 400 light years away, the difference in sightlines from being parallel is thousands of times smaller than the seemingly pointlike appearance of Polaris in the night sky.

16. If you are observing the sky from a latitude of 39°N, what range of declinations do circumpolar stars fall within?

16. (All stars north of declination +51). From a latitude of 39 N, the north celestial pole is 39 above the northern horizon. Therefore, stars within 39 of the pole (which is at declination +90) never set. This makes all stars north of declination (+90–39) = +51 circumpolar.

Unit 6 12. Suppose that instead of Earth’s axis being tilted at 23.5°, it was not tilted at all. Where would the tropics and arctic regions be? How would seasons be different?

12. If the Earth’s axis had no tilt instead of 23.5, the tropic and arctic regions would shrink to their smallest possible size. In the tropics the Sun passes directly overhead at least once during the year. With 0 tilt, the Sun would always be directly over just the equator. In arctic regions the Sun remains below the horizon for more than 24 hours during some parts of the year, but with 0 tilt, observers at the north and south pole would find that the Sun always remained right on the horizon. Without a tilt there would not be the seasonal changes we experience as the angle of sunlight to the ground changes.

13. Suppose Earth’s axis were tilted by 50° instead of 23.5°. Where would the tropics and arctic regions be? How would seasons be different?

13. If the Earth’s axis had a 50 tilt, geographically the tropics would extend to 50 N and S latitudes. The arctic regions would reach all the way to 40 N and S latitudes, giving a 10 overlap between the two. If you lived in the overlap region, there would be times of the year when the Sun would pass straight overhead as in the tropics, but half a year later, there would be days when the Sun never rose. This would lead to much greater variations in seasonal temperatures than we experience on Earth.

14. Suppose Earth’s axis were tilted by 90° instead of 23.5°. Where would the tropics and arctic regions be? How would the seasons be different?

14. If the Earth’s axis was tilted by 90, the Sun would appear to lie directly over the North Pole once a year, then move down toward the equator and eventually lie directly over the South Pole 11

6 months later before moving back to the North Pole over the course of a year. In this situation (which is similar to what the planet Uranus experiences), there would be no tropics or arctic regions, or you might say they were both everywhere. The polar regions would sometimes be warmer than the equator, and the seasons would be extreme, varying between “North Pole hot and in constant sunlight, equator cool with the Sun on the northern horizon all day long, South Pole dark for months and extremely cold,” “equator warm with 12 hour days and nights, and both poles cool” and “South Pole hot and in constant sunlight, equator cool, North Pole cold.”

15. Describe the daily motion of the Sun that you would see on the solstices and the equinoxes if you were observing from the Arctic Circle, at a latitude of 66.5°N. 15. On the Arctic Circle (latitude 66.5 N), it remains true on the equinoxes that night and day would both be about 12 hours long, and the Sun rises due east and sets due west. The Sun would not get very high in the sky, reaching a maximum altitude of 23.5 above the southern horizon at noon. On the December solstice, the Sun would barely skim the southern horizon around noon, with an hour or two of twilight before and after. It would be fully dark from about mid-afternoon to mid-morning the following day. On the June Solstice, the Sun would be up 24 hours, reaching a height of 47 degrees in the South at noon, and just flirting with the northern horizon at midnight.

16. If you wished to observe a star with a right ascension of 12 h, what would be the best time of year to observe it? What would be the best time to observe a star with a right ascension of 6 h? (You may find Unit 5.5 helpful in answering this problem.)

16. (For a star with R.A. = 12 h, around March 20; for a star with R.A. = 6 h, around Dec. 21) For the best opportunity to observe a star with a right ascension (R.A.) of 12 h, you would wish the Sun to be as far away from it as possible, at the opposite side of the sky, so that when the star is above the horizon and visible, the Sun would be below. This would best occur with the Sun at R.A. = 0 h (see also Unit 5.5). This would occur near March 20 (although much of the year would also be acceptable, as well). For the star at R.A. = 6 h, it’s best if the Sun was again 12 hours away, at R.A. = 18 h. This happens ¾ of the year after March 21–around the winter solstice (Dec. 21).

Unit 7 10. Philadelphia is 5° east of Pittsburgh. In the early 1800s, both cities synchronized their clocks with apparent noon—that is, all the local clocks read noon when the Sun passed through the meridian. By how many minutes did clocks in the two cities differ from one other?

10. To find how the 5 longitude difference affects time measurement, we need to know how much time it takes the Earth to turn each degree. Since the Earth takes 24 hr to rotate 360, the time per degree is: 24 hr / 360 = 1 hr / 15 = 60 min / 15 = 4 min / 1.

Therefore, back in the days when clocks were set using a sundial, Philadelphia’s clocks were 5 × 4 min/1 = 20 minutes ahead the clocks in Pittsburgh.

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11. If Russia had all clocks set to Moscow time (time zone +3 in Figure 7.4) instead of using many time zones, what time would the Sun rise at the easternmost tip of Siberia on the equinoxes? Examining Figure 7.4, what places in the world have the Sun crossing the meridian earliest according to local time? What places have it latest?

11. (Around 8 pm the previous evening). Siberia is in the +12 h time zone; Moscow is in the +3 h. They appear to be 9 h apart, but are actually on opposite ends of their time zones—based on longitude, Siberia is nearly at what would be +13 h (and in fact Moscow is very close to the start of +3 h, although it is not obvious on the map). If the Sun rose at 6 am local time in Siberia, the Moscow time would be 10 h earlier, 8 pm the previous evening.

For the second part of the question, local noon moves from east to west across the map (or right to left; it’s noon in England before it’s noon in America). Time zones are accurate to local solar time in the middle of the time zone—the middle of “0” runs through Greenwich, England (except for political zig-zags). So if one is at the eastern (or right side) edge of a time zone, the Sun crosses the meridian (local solar noon) at 11:30 am, half an hour before the clock reads noon. In the middle, it crosses at the same time, and at the western edge (left on map) it crosses the meridian (local solar noon) when the clock reads 12:30 pm. Looking carefully at the map, in some places the time zones have been moved from being centered on the lines of longitude, and stretch east and west over other time zones. All of Greenland is3 h, but the easternmost edge is nearly in the middle of the1 h time zone based strictly on longitude. That means that the Sun will cross the meridian there 1 hour after it crosses the meridian in Greenwich, but the clock will be set to 3 hours earlier than Greenwich—it will be “noon” at 10:00 am!

This is one of the places where the Sun crosses the meridian at the earliest local (clock) time.

11. (cont.) In China, there is one of the latest crossings. All of China, for example, is in the +8 h time zone, but the westernmost edge of China is actually in line with the middle of the +5 h time zone based on longitude alone. Therefore, in Western China, the clock will read 3 h later than local solar time (noon); it will read 3 pm when the Sun crosses the meridian!

12. One city is at a longitude of 21°E. A second city is at 104°W. How many hours apart will a star cross the meridian in the first city and then in the second? If the time zone of each city is based on the closest longitude that is a multiple of 15°, and the star passes overhead in the first city exactly at midnight, at what time (by the clock) will it pass over the second city?

12. For cities at 21 E (Warsaw in Poland would be an example) and 104 W (near Denver, Colorado would be close), the longitude difference = 21 + 104 = 125. A star that passes overhead in the eastern city first, would move westward as the Earth turns towards the East, and next goes overhead at 0 longitude, and then continue on to the western city. The amount of time between meridian crossings = 125 × 24 hours/360 = 8.267 hours = 8 hours, 16 minutes.

The eastern city is +21/(15/hr) = +1.4 hr ahead of the prime meridian, so it would be located in +1 hr time zone. The second city is located104/(15/h) = −6.9 hr behind the prime meridian, so it is in the‒7 hr time zone. The clock near Denver at 104W is set 8 h earlier than the clock in 13

Poland at the 21E, so when it is midnight at 21E (and the star is overhead there), it will be take 8 more hours to reach midnight at 104ut the star will not yet be exactly overhead yet, because only 8 hr have passed, and the star won’t quite be overhead for another 16 minutes (part 1). The star will be overhead at 104W 16 minutes past midnight, at 12:16 am. When that happens, it would already be 8:16 am in Poland; the Sun would be up and the day started!

13. Suppose Earth’s spin slowed down until there were just 180 days in a year. Compare the length of a sidereal and a solar day in this new situation. (Do not redefine units of time—just express them in terms of our current hours, minutes, and seconds.)

13. (Sidereal day = 48.4 hr; solar day = 48.7 hr). If Earth’s spin slowed down, a day would be longer, but our planet would still orbit the Sun in the same amount of time. At present there are 8766 hr in a year (365.25 day/year × 24 hr/day). This same total number of hours divided into 180 solar days is 8766 hr / 180 day = 48.7 hr/day, or 48 hours 42 minutes. So that is the length of a “solar day” if our planet’s spin was slower.

13. (cont.) During each of these new solar days, the Earth would still move 2° in its orbit, so that it completed a full revolution of 360° around the Sun in 180 “new” days. This means that the Earth must be rotating an extra 2° compared to the stars during each solar day. Therefore the Earth takes 48.7 hr to rotate 362° relative to the stars. The length of the sidereal day, when the same star would cross the meridian, is then 360° × (48.7 hr / 362°) = 48.4 hr.

14. It is thought that the angle of Earth’s axis varies by a few degrees over tens of thousands of years. If the angle were 20° instead of 23.5°, what would be the angle of the noontime Sun from the horizon at your own latitude on the solstices?

14. The Sun would be 3.5 lower in summer, 3.5 higher in winter. For example, in Lafayette, Colorado (exactly 40 N latitude), on the June solstice the Sun rises to 73.5 above the southern horizon; on the December solstice, it rises to just 26.5. If the axial tilt changed to 20, the Sun would get only 70 high as a June maximum, and 30 high for the December maximum.

15. What is your longitude, and what is the longitude of the center of your time zone? (This should be a multiple of 15°.) Calculate the time at which the Sun should, on average, cross your own meridian, assuming that the Sun crosses the meridian at exactly 12:00 at the center of your time zone.

15. Example: Dodge City, KS, is located at almost exactly 100 West longitude in the Central Time Zone. This time zone is defined for 90 West Latitude. Therefore, Dodge City is 10 West of the 90 West longitude. The Earth turns at about 15 an hour, so 10 represents about 10/15 of an hour, or 40 minutes. So if the Sun crossed the meridian at 12 noon at 90 West longitude, it would cross the meridian at about 12:40 p.m. in Dodge City. (Note this is more than the maximum 30 minute time difference that time zones were initially designed to encompass because the edges of the Central Time Zone have been adjusted to match political boundaries).

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Unit 8 13. How long does someone have to wait from the time of a third-quarter moon until the next full moon?

13. (About 22 days). If the Moon is currently at third quarter (day 22 of the lunar month).

You have to wait for ¾ of a lunar month (approximately 22 days) until the next full moon.

14. Calculate how many minutes later the Moon rises each day on average.

14. (About 49 minutes later). Since the Moon is in the same phase and rises at the same local time once every 29.5 days (on average), the Moon must rise 1/29.5 of a day later each day (on average). This results in a shift of 24 hr / 29.5 = 0.814 hr × 60 min/hr = 49 min later each day.

Note that this will not be exactly correct primarily because of two factors. First, the Moon changes declination each day, so just as the Sun rises earlier in summer than winter, moonrise will be earlier when the Moon is farther north (for a northern observer). Second, the Moon’s orbit is quite elliptical, so it speeds up and slows down in its orbit over the course of the month (Unit 12.2).

15. Show that the ratio of the Moon’s diameter to the Sun’s diameter is very similar to the ratio of the Moon’s distance from Earth to the Sun’s distance from Earth. (Sizes and distances needed to solve this are in Appendix Tables 1 and 7.)

15. The ratio of the Moon’s and Sun’s diameters is (2 × 1738 km) / (2 × 6.96 × 105 km) = 0.00250. In other words, the Sun is about 400 times larger in diameter than the Moon. The ratio of Moon’s and Sun’s average distances: (384.4 × 103 km)/(1.496 × 108 km) = 0.00257. So the Moon is also about 400 times closer than the Sun. This coincidence means the Sun and Moon are seen to be about the same angular size in our sky.

16. A total solar eclipse occurs over Africa on August 2, 2027. Based on the 6585-day, 8-hour time between repetitions of similar eclipses, when and where will the next similar total eclipse occur?

16. (August 12, 2045 across the Gulf of Mexico and the western U.S.) A very similar total solar eclipse to the August 2, 2027 eclipse occurs a saros later. Looking at the map of eclipse shadows, we see that this eclipse follows the northern coast of Africa, around 30 degrees latitude. First, we estimate the location of the eclipse. The added 8 hours means that the eclipse shadow falls 1/3 of a day, or 120°, further west than Africa on the Earth.

This places it near the Gulf of Mexico and up into the USA. 18 years is 18*365 = 6570 days (not accounting for leap years), so 6585 days is 18 years, 15 days. Between 2027 and (2027 + 18 = 2045) are 5 leap years, (2028, 2032, 2036, 2040, 2044), so we need to subtract a day for each of these. That leaves us with 18 years, 10 days, which we add to August 2, 2027, giving us the date August 12, 2045. More information about this eclipse can be found at the NASA Goddard Spaceflight Center’s Eclipse Web Site, http://eclipse.gsfc.nasa.gov/SEgoogle/SEgoogle2001/SE2045Aug12Tgoogle.html 15

17. If Figure 8.12 A were expanded to include the Sun at the same scale, how big would the Sun be? How far away would it be? (Sizes and distances needed to solve this are in Appendix Tables 1 and 7.

17. (About 22-23 cm in diameter, and 24-25 m away). First, we need the scale of the diagram.

In Figure 8.12B, the Earth’s diameter is almost exactly 2 mm, and its radius therefore 1 mm.

The actual radius of the Earth in Appendix Table 1 is 6378 km, so the scale of the image is 1 mm = 6378 km. You can also determine the scale by measuring the distance to the Moon, which measures about 63 mm; if we assume it’s at its average distance of 384,400 km from Earth given in Appendix Table 7 this gives a scale of 384,400 km/63 mm = 6102 km/mm. (These numbers are very close, so the sizes and distances are about at the same scale in the drawing, but there is some uncertainty given the size of the Figure, and we don’t know for sure what the Moon’s distance is in the drawing. An answer based on either number should be acceptable and both are about the same.)

From Appendix Table 1, the radius of the Sun is 6.96 × 10 8 m = 6.96 × 105 km, and it is at a distance of 1.5 × 1011 m = 1.5 × 108 km from the Earth. With the Earth-determined scale, for the drawing the Sun would be: 6.96 × 105 km × 1 mm/6378 km = 109 mm in radius or 218 mm (21.8 cm) in diameter, and you would have to locate it 1.5 × 108 km × 1 mm/6378 km = 23,518 mm = 23.5 meters away! (For the Moon distance-determined scale, the Sun would be 228 mm in diameter and 24.6 m away.)

Unit 9 10. If the same year-numbering systems were used today as were used in Europe before 532 c.e., what year would we currently be in?

10. (The current year + 753.) The system in Europe before 532 C.E. was based on the founding of Rome in 753 B.C.E, and the year 532 corresponds to the year 753 + 532 = 1285 using that system. So the present year plus 753 will give the answer. The other system mentioned in the text is based on when Diocletian became emperor of Rome; in that system, 532 C.E. is year 248; so there is a difference of 532 – 248 = 284 years from the present C.E. year. For example, 2015 C.E. gives 2015 + 753 = 2768 since the founding of Rome, and 2015 – 248 = 1767 in the reign of Diocletian.

11. What year will 2020 c.e. be in the Chinese, Islamic, and Jewish calendar systems?

11. In September 2000, the Jewish calendar started year 5761 with Rosh Hashanah; in 2020 it was the year 5780 before Rosh Hashanah and 5781 after it.

From Figure 9.2, the Islamic year 1426 began on February 10, 2005. Since the Islamic year is 11 days shorter on average, after 15 years, the Islamic year 1441 began about 11 × 20 = 220 days (or about 7 months) earlier on the Gregorian calendar, placing the first day of 1441 at the end of June 2019. This means the year 2020 C.E. began with 1441 and ended in 1442 of the Islamic 16

calendar. Note that the Islamic calendar picks up an extra year compared to the common calendar once every 33 or 34 years. The extra year here was gained in 2008, which saw the beginning of two Islamic years.

In the Chinese system, the 79th sixty-year cycle began in early 1984 C.E. and will end in early 2044 C.E.. That means in early 1984 C.E., 78 cycles or 78 × 60 = 4680 years were completed, and the year 4681 began. There are 36 years between 2020 and 1984 C.E., which means that in early 2020 C.E. the Chinese year 4681 + 36 = 4717 began. Note that some authorities begin the count a year earlier (which would make it the Chinese year 4622), but in either case it is agreed to be the year of the Metal Rat. 12. Suppose a 12-month lunar calendar year (such as in the Islamic calendar) began on January 1. How many solar years would pass before the lunar calendar year would again begin within one week of January 1? How many lunar calendar years would have elapsed?

12. (32 solar years, or 33 lunar calendar years). A lunar calendar year is shorter than a solar year by 11 days, so each year the “lunar new year” begins 11 days earlier on average in the common solar calendar. For example, if the first day of the lunar calendar was Jan 1, then 12 lunar months would be completed around Dec 21 later that year, and by about Dec 10 the following year. We would have to wait a number of years until the missing time accumulated to a full year. Since 365.25/11 = 33.2, 33 lunar calendar years should accumulate the necessary difference.

Checking this, after 33 lunar calendar years the date will have shifted earlier by 33 × 11 = 363 days—this is less than 7 days short of 365.25 days, as expected. One fewer solar years has gone by in the same interval.

Note that in a more detailed calculation with a lunar month of 29.53 days, a lunar calendar year adds up to 354.36 days. Compared to a more precise value of the solar year of 365.24 days, the difference is actually 10.88 days. The ratio 365.24/10.88 = 33.57, and in fact depending on the number of leap years spanned in the common calendar over the intervening years, there may be 33 or 34 lunar calendar years corresponding to 32 or 33 solar years to get the closest match.

13. The first day of the month in the Islamic calendar begins when the crescent moon is first spotted. Usually, the earliest this is possible is about 16 hours after the new moon. How many degrees has the Moon moved since it was new at this time? Approximately how long after the Sun sets will the Moon set if it is just 16 hours past new?

13. (32 minutes). The Moon takes 29.5 days to move 360 relative to the Sun. Since 29.5 days × 24 hr/day = 708 hr, the Moon moves 360 / 708 hr  0.5/hr relative to the Sun. Therefore 16 hours after the new moon, the Moon will have moved about 16 hr × 0.5/hr = 8 from the Sun, and because the sky turns at 15/hr, it will set 8/(15/hr) = 0.53 hr = 32 minutes later. (Note that these results are approximate since the Moon may be off the ecliptic by several degrees, and moonset timing depends on the angle between Sun and Moon relative to the horizon.)

14. Under the Julian calendar system, the year averages to 365.25 days after 4 years. In the Gregorian system, determine the average length of the year after 400 years. If a country

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needed to convert from the Julian to the Gregorian calendar now, how many days would it need to remove from the calendar?

14. (13 days). In the Julian calendar, there are 365.25 d/y × 400 y = 146100 days in 400 years.

The Gregorian calendar removes leap years in centuries not divisible by 400. In 400 years, there are three centuries not divisible by 400, so the Gregorian calendar would have three fewer days, or 146097 days, and average 365.2425 days per year. To correct the calendar today, we would have to remove the 10 days needed in 1582, plus the three extra days that would have been in a Julian calendar since then for the leap years in 1700, 1800, and 1900, for 13 days total.

15. Suppose the year were 365.170 days long. How would you design a pattern of leap years to make the average year length exactly match this value within a span of 1000 years?

15. If a year had 365.170 days, over 1000 years there would be 365.170 × 1000 = 365,170 days.

If all 1000 years had exactly 365 days, there would be 365,000 days leaving an extra 170 days.

These can be made up with 170 leap years of 366 days each, and the other 830 (= 1000 – 170) would have 365 days. Since 1000/170 = 5.88, a leap year is needed every five or six years. The question is how to distribute these years with an extra day that is easy to remember. If there was a leap year every five years, there would be 200 in 1000 years, which is 30 too many. If every six years, there would only be 166 of them, four too few. One scheme that would work would be a leap year every five years, skipping 3 each century—for example, every five years except for years that end in 25, 50, and 75.

Unit 10 10. How large would Mars appear in angular size when it is closest to Earth? At its closest, the separation between Earth and Mars is about 55 million km, and Mars’ diameter is 6740 km. Jupiter is about 588 million km from Earth at its closest. Given that Jupiter has a physical diameter of 140,000 km, what is the largest possible angular size of Jupiter?

(Refer back to Unit 10.4 for the angular size formula.)

10. (Mars: 0.007°; Jupiter: 0.014°). Use the angle-distance equation: α = 57.3° × diameter/distance = 57.3° × (6740 km / 55 × 106 km) = 0.007° = 25 arc seconds At the smallest separation between Earth and Jupiter of 588 million km, when both are on the same side of the solar system and Jupiter is opposite the Sun, Jupiter (diameter = 140,000 km) has an angular diameter of: α = 57.3° × diameter/distance = 57.3° × 1.4 × 105 / 5.88 × 108 = 0.0136° = 50 arc seconds This means Jupiter would appear almost twice as large as Mars in a telescope, even though it is more than 10 times further away.

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11. Suppose you are an alien living on the fictitious warlike planet Myrmidon and you want to measure its size. The Sun is shining directly down a missile silo 1000 miles to your south, while at your location the Sun is 36° from straight overhead. What is the circumference of Myrmidon? What is its radius?

11. (1592 miles). Use Eratosthenes’ method to find the size of the planet Myrmidon. From the observations we know that 1000 miles on the surface spans an angle of 36°, so Myrmidon’s circumference must satisfy the equation: circumference/1000 miles = 360°/36° = 10.

Hence, circumference = 10 × 1000 miles = 10,000 miles. And since circumference = 2πR, we find its radius R = 10,000 miles / 2π = 1592 miles, or about ¾ the size of our Moon.

12. A cloud directly above you is about 10° across. From the weather report you know that the cloud is at an altitude of 2200 m. How wide is the cloud?

12. (380 m). The size of a cloud that is 10° across and 2200 m away is: ℓ = d × α/57.3° = 2200 m × 10°/57.3° = 380 m.

13. The Andromeda Galaxy, M31, is in many ways similar to our own galaxy but slightly larger.

The linear diameter of the Andromeda Galaxy along its longest axis is 140,000 lightyears, but from our perspective, the Andromeda Galaxy has a maximum angular diameter of 3.18°. How far away is the Andromeda Galaxy?

13. (2.5 Million ly). From the linear and angular diameters of the Andromeda galaxy, we can find its distance: d = ℓ × (57.3°/α) = (140,000 light-years) × (57.3°/3.18°) = 2.5 × 106 light-years This “nearby” galaxy is about 2.5 million light-years distant.

14. Suppose you see a person whom you know is 1.8 m (6.0 ft) tall, standing at a distance where he appears to have an angular height of 2°. How far away is he?

14. (About 52 m away). A person 1.8 m tall who has an angular size of 2° must be standing at a distance: d = ℓ × (57.3°/α) = 1.8 m × 57.3°/2° = 51.6 m For most people, when held at arm’s length, the width of a thumb has an angular size of about 2 degrees. So if you see a person standing about 50 meters away, they would appear to you to be the same angular size as your thumb on your outstretched arm.

15. Suppose the Moon is exactly 0.500° wide when it is near the horizon. If later in the night you saw it straight overhead, what would its angular diameter be? (Assume Earth’s radius is 6378 km and the Moon’s distance at the first observation is 398,000 km.)

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15. If the Moon has an angular size of 0.500° when it is near the horizon, it will be larger when it is overhead later that night because it will be slightly closer to you—by the size of Earth’s radius. The ratio of the distances is 398,000 km / (398,000 km–6378 km) = 398,000/391622 = 1.0163. Since you are closer, and angular size-distance is an inverse relationship, the Moon will appear bigger by that factor and have an angular size of 0.500° × 1.0163 = 0.508°. (Note that this is the opposite of what most people expect to be true because of the Moon illusion.)

16. Aristarchus’s critics argued that parallax was not observed and therefore Earth could not be in motion. In fact we know now that the stars are very far away and parallax motion is very small. Let’s examine this by estimating the change in angular size of the angular separation between two stars 10 light-years away from the Sun. The two stars are 3 lightyears apart. Calculate the angular separation of the two stars when Earth is closest to them in its orbit and when it is farthest away, on the opposite point on its orbit. If human vision (without the aid of a telescope) can at best distinguish a difference of about 0.02°, would the change in angular separation be detectable to the naked eye?

16. (No, the change is not detectable to an unaided eye). If two stars are 10 ly from the Sun, then their distance from the Earth can vary by plus or minus 1 AU as the Earth orbits the Sun. Expressing light years in AU: 1 ly = 63,240 AU. (from Appendix) so the distance varies from 632,401 AU to 632,399 AU. If the stars are 3 ly = 189,720 AU apart, we can calculate their angular separation from the formula: Closest: α = 57.3 × ℓ / d = 57.3 × 189,720 AU / 632,399 AU = 17.1900272 Farthest: α = 57.3 × ℓ / d = 57.3 × 189,720 AU / 632,401 AU = 17.1899728 The difference is 0.000054°, much smaller than the 0.02° difference the eye can distinguish.

Unit 11 10. If we discover a new planet in the Solar System and observe that it takes about 366 days between oppositions, what could we say about it?

10. If a planet was in opposition once every 366 days, that would imply it shifts by less than about 1° in its orbit each year, indicating that it must take more than 360 years to complete an orbit. (If you use 365.25 days as the length of the year, you might estimate an orbital period as large as 487 years.) Given that this orbit is slower than any of the other planets, the new planet must lie in a very large, slow orbit, far from the Sun. 11. On average, Mars spends about 72 days in “retrograde motion” when observers on Earth see it moving temporarily westward along the ecliptic. If Mars’ average orbital speed is 27.7 km/second, how far does Mars actually move in its orbit during that time? How far does Earth move in that time, if its average speed is 29.8 km/sec?

11. Use distance = velocity  time. In 72 days, Mars would move: 20

(27.7 km /sec)´ 72 days´

24 hours 3600 seconds ´ = 1.72´ 108 km day hour

In the same time, Earth would move farther since it is moving around the Sun faster:

(29.8 km /sec)´ 72 days´

24 hours 3600 seconds ´ = 1.85´ 108 km day hour

Because Earth moves faster around the Sun, it “catches up and laps” the outer planets, and it is when that happens that we see those planets move in retrograde motion. And because Earth’s orbit is closer to the Sun, moving farther on a smaller orbit means Mars will appear to shift its location in front of the background stars even more.

12. When Mars is close to conjunction, within an angle of about 5° of the Sun, we lose radio contact with our spacecraft there. Draw a scale diagram of the orbits of Earth and Mars and, given that Earth moves about 1° per day in its orbit and Mars about 0.5° per day in its orbit, estimate approximately how long this period of communication problems lasts.

12. (About 32-36 days) Starting from when Mars is in conjunction, the Earth’s and Mars’s motions on opposite sides of the Sun partly cancel out, keeping Mars hidden longer than if only one of them was moving. Earth’s more rapid motion eventually overtakes Mars bringing it back into view. The diagram shows a scale construction of the two orbits (with Mars’s orbit 1.52 times larger than Earth’s and ignoring the ellipticity.)

By testing several angles with a protractor, shifting Earth by 1° per day and Mars ½° per day in their orbits around the Sun, it appears that Mars reaches an angle of ~5° from the Sun (as seen from Earth) about 1618 days after conjunction. In total, then Mars is difficult to contact for about 32 to 36 days, counting the time before and after conjunction.

12. (cont.)

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13. Using a protractor, draw two lines that make an angle of 46° relative to each other, matching Venus’s average greatest elongation. At a point on one of the lines 10 cm from where the lines intersect, place a compass and draw the largest circle you can that just touches the other line. What is the radius of this circle? Explain how this geometric construction relates to the size of Venus’s orbit.

13. (The radius is 0.72 AU, representing Venus’ distance from the Sun).

First, we draw a line from Venus to Earth as shown in the diagram. Then 46 off of that, we draw a 10 cm line from Earth towards the Sun. Starting at the end of that line, we draw the circle that represents the orbit of Venus, so that it just touches our first line. We now have a line tangent to a circle, so the angle between our Venus-Earth line and the Venus-Sun line (D) is 90.

13. (cont.) The length of D divided by the length of 1 AU measured in the figure gives the value of D, the orbital distance of Venus, in AU. Alternatively, we can use the law of sines and the right triangle we have just created to solve for D. The right angle is at Venus when Venus

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appears to be farthest from the Sun (at greatest elongation). If the angle between Venus and the Sun is 46 as seen from Earth, then from trigonometry: sin (46) = Venus Distance / 1.0 A = D / 1 AU. So D = 1 AU × sin(46) = 0.72 AU.

14. Make a diagram similar to Figure 11.7 B that shows retrograde motion for a planet closer to the Sun than Earth. Does retrograde motion occur near superior conjunction, inferior conjunction, or greatest elongation? What would an observer on this planet see of Earth’s motion over this same time?

14. (Inferior conjunction; an observer would see Earth in retrograde). The retrograde motion for a planet closer to the Sun occurs around inferior conjunction as the planet passes Earth between the Earth and the Sun. An observer on this planet during this time would also see the Earth, near opposition, going through retrograde motion.

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15. Mars spends, on average, about 106 days (15.4% of its orbit) moving from opposition to quadrature. Make a geometric construction like Figure 11.11 (or use trigonometry) to show that Mars’s orbit is about 1.52 times the size of Earth’s 15. Mars spends, on average, about 106 days (15.4% of its orbit) moving from opposition to quadrature. This is illustrated in the figure below. To draw this, we need to know how much of their orbits the planets move through. In the 106 days from opposition to quadrature, the Earth travels 106/365 × 360 = 104.5 along its orbit. During the same time, Mars travels 106/687 × 360 = 55.5 along its orbit. Mars must be 1.52 times farther from the Sun than the Earth along the line at 55.5° in order for Mars to be 90° away from the Sun when viewed from Earth. If Mars were anywhere else along that line from the Sun, the angle at Earth would be different.

15. (cont.)

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Using trigonometry it is possible to solve for the size of Mars’s orbit more directly. The difference between the angles the two planets move is = 104.5°–55.5° = 49° as shown in the diagram. The distance from the Sun to Mars is the hypotenuse of the right triangle, therefore cos() = 1 AU / (Mars distance from Sun). Since = 49°, we get Mars’s distance from the Sun equal to 1 AU / cos(49°) = 1.52 AU. The eccentricity of Mars’s orbit is sufficient that the results of this calculation vary depending on the portion of the orbit where Mars is traveling—and neither Mars nor the Earth travels at constant speed, so the time between opposition and quadrature can vary significantly.

Unit 12 10. Ceres has an orbital period of 4.60 years. What is its semimajor axis? Where in the Solar System does this place Ceres?

10. (2.8 AU, in the asteroid belt). From Ceres’ orbital period of 4.6 yr, use Kepler’s 3rd Law: 3 = P2 to solve for the semimajor axis. For P= 4.6 yr, we find 3 = 4.62 = 21.2, so taking the cube root, we find a = 2.8 AU. This is between the orbits of Mars and Jupiter, placing Ceres in the asteroid belt.

11. Sedna, an object in the outer Solar System, has a semimajor axis of 526 AU. What is its orbital period? How does this compare to Pluto’s orbital period of 248.1 years?

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11. From Sedna’s semimajor axis of  = 526 AU, we can use Kepler’s 3rd Law to solve for the period. P2 = 3 = (526)3 = 1.4553 × 108, so taking the square root we find P = 12,064 years. The ratio of Sedna’s orbit to that of Pluto = 12,064/248.1 = 48.6, so the orbital period of Sedna is almost 50 times greater than Pluto’s period.

12. Suppose an asteroid is discovered with an orbit that brings it as close as 0.5 AU from the Sun and as far away as 5.5 AU. What is the semimajor axis of its orbit? What is its orbital period?

12. (a= 3.0 AU and P = 5.2 years). The semimajor axis is half the sum of the closest and farthest distance from the Sun, so for this asteroid the semimajor axis a = (0.5 AU + 5.5 AU)/2 = 3.0 AU. We can solve for the asteroid’s period using Kepler’s 3rd Law P2 = a 3 = 33 = 27, so taking the square root we find P = 5.2 years.

13. If a comet orbits the Sun and reaches 1 AU at its closest approach to the Sun, and its orbital period is 27 years, what is its maximum distance from the Sun? 13. A comet with an orbital period of 27 years has a semimajor axis that can be found from Kepler’s 3rd Law: 3 = P2 = 272 = 729; taking the cube root, we find  = 9 AU. Since the semimajor axis is the average of the maximum and minimum distance from the Sun, we know that (max + min)/2 =  or (max + min) = 2 = 18 AU. Therefore, if the minimum distance to the Sun is 1 AU, the maximum distance from the Sun must be 18 AU – 1 AU = 17 AU. 14. Calculate the speed (in km/sec) of an asteroid in a circular orbit at 4 AU by dividing the orbit’s circumference by its period.

14. (15 km/sec). For an asteroid orbit with a = 4 AU, the period is given by Kepler’s 3rd Law as P = 43/2 = 8 yr. A circular orbit of R = 4 AU has a circumference = 2πR. The speed of the asteroid is then found from: v = circumference / period = 2πR / P = 2π(4 AU * 1.5 × 108 km/AU) / (8 yr * 60 sec/min * 60 min/hr * 24 hr/day * 365 day/yr) = 15 km/sec.

15. A comet is discovered with an orbit that extends from Venus’s orbit to Neptune’s orbit.

Use Kepler’s second law to estimate how many times faster the comet is moving when it is closest to the Sun compared to when it is farthest from the Sun. (Hint: The area swept out over a short period of time when the planet is at these two extremes can be approximated by a triangle, and the area of a triangle equals one-half of its base times its height.

15. (41.5 times faster.) The ratio of distances at Neptune’s versus Venus’s distance is (30.0 AU/0.723 AU) = 41.5. When the comet is closest to the Sun, its motion over a fixed amount of time can be pictured as the base of a triangle with the height of the triangle equal to

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the distance to the Sun. When the comet is farthest from the Sun, we can imagine the same kind of triangle, but it now has a much greater “height” so the base of the triangle must be proportionately shorter. Thus, the ratio of speeds is inversely related to the ratio of the distances. Therefore, the comet travels 41.5 times faster when it is closest to the Sun than when it is farthest.

Unit 13 10. Use your hand and fingers to estimate the angular size of at least three constellations.

Sketch the constellations to scale with your measurements.

10. The main part of Orion, for example, is about 20° tall and 10° wide. (Sizes can be estimated from sky charts, but there is a lot of distortion at the edges because of the spherical projection.)

11. Sailors have “handy” rules for estimating the time until sunset. Approximately how many minutes before sunset is the Sun “one finger” above the horizon at the equator? At 45°N or S latitude?

11. Typically, the width of a finger is similar to the width of a thumbnail, which is about 2° wide when held at arm’s length. The Earth turns at about 1 degree every 4 minutes, so if the Sun were “one finger” above the western horizon at the equator, sunset would be in about 8 minutes. (The little finger might be a little less, the middle finger a little more.) At about 45° latitude, the Sun sets at a 45° angle, and hence its path to the horizon is longer. To figure out the length of the path, you can sketch the path and measure the difference: height of Sun above horizon =h

Path of Sun ~ 1.4 × h

The length of the Sun’s path may also be recognized as the hypotenuse of a 45°-45°-90° triangle, which is 2 times the length of a side. (Or using trigonometry, we find that the longer distance is h/sin(45°) = h/0.707 = 1.41 h, where h is the vertical height of the Sun above the horizon.) A finger’s width above the horizon translates to a motion of the Sun by about 2° × 1.4 = 2.8°, so the Sun will take about 8 min × 1.4  11 minutes per finger before it sets at 45° latitude.

12. Demonstrate through the use of a drawing that the altitude of the North Star, as seen by anyone in the Northern Hemisphere, is equal to the observer’s latitude.

12. The picture at right shows Earth as viewed in profile and above the equator. By inspection of the figure we note that the two arrows pointing toward the North Star are parallel lines. This 27

implies that the angles θ and ϕ are equal. Moreover, since the latitude angle + θ = 90° = Altitude + ϕ it follows that the altitude must also equal the latitude angle.

As a side note, students are often reluctant to accept that lines toward the North Star from different parts of the Earth are effectively parallel.

12. (cont.) For some it is helpful to remind them that Polaris is far bigger than the entire Earth.

In fact it is about 50 times larger than the Sun, so the difference in direction from different spots on the Earth is a tiny fraction of the point of light we see in the night sky.

13. Calculate a precise value for how many degrees the Moon moves each day, keeping in mind that to complete one lunar month of 29.53 days, the Moon must once again align with the Sun, which also has shifted along the ecliptic during that time.

13. (13.18°/day). The Moon revolves (360°/29.53 days = 12.19°/day) around the Earth each day, and during that day the Earth has revolved (360°/365.25 days = 0.99°/day) around the Sun in the same direction, so the Moon revolves 12.19°/day + 0.99°/day = 13.18°/day relative to the stars. (Note this is the same answer as you get dividing 360° by the sidereal lunar month of 27.3 days).

14. Even at greatest elongation, Mercury will not always be visible high above the horizon after sunset or before sunrise. This is primarily because the angle between the ecliptic and the horizon is shallow sometimes during the year. In that case, Mercury is far from the Sun but still close to the horizon. When the ecliptic is more nearly perpendicular to the horizon in your viewing location, Mercury will have a higher altitude. By examining the foldout star chart, estimate at what times of year the ecliptic will be at the best angle for you to observe Mercury in the evening sky. Is there an upcoming greatest elongation that is favorable for you?

14. The angle of the horizon at sunset depends on the observer’s latitude. On the equator the horizon is perpendicular to the celestial equator (running north-south on the celestial sphere), so on the equator the ecliptic runs most nearly perpendicular to the horizon at the solstices.

At other latitudes, the horizon at sunset is tilted from north-south by an angle equal to the latitude. This is illustrated below for latitude 40°N, with the grayed-out area beneath the horizon for sunset on the March equinox (left) and the September equinox (right).

As can be seen in the illustrations, the path of the ecliptic is most nearly perpendicular to the horizon in the evening sky near the beginning of spring, which is when it would be easiest to see Mercury as an evening star. The pre-dawn horizon is tilted 40° in the other direction, so Mercury is easiest to see as a morning star near the beginning of autumn for northern observers. These answers would be reversed for an observer at mid-southern-hemisphere latitudes.

14. (cont.)

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15. If you mark the position of the shadow cast by the top of a flagpole at the same clock time every day throughout the year, you will find that the marks trace out a figure-eight pattern. Explain why this happens by referring to the equation of time.

15. The Equation of Time shows the offset between sundial time and mean solar time. These are not the same because of the Earth’s changing speed in its elliptical orbit and the tilt of the Earth’s axis (as discussed in Unit 13.5). If we measure the shadow cast by the top of the flagpole once every 24 hours as measured by a standard clock, the Sun will be east of its mean position when the sundial is behind clock time, or west when the sundial is ahead as given by the Equation of Time. The shadow therefore reaches farthest to the west in February and July, and farthest to the east in May and November. Because the Sun also moves north and south of the celestial equator during the course of the year, so the east-west motions over each half year produce an S-shaped pattern that join to make a figure-8 shaped pattern.

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Alternate answer for Unit 9 #11 if you change the manuscript question to ask about 2025: 11. What year will 2025 c.e. be in the Chinese, Islamic, and Jewish calendar systems?

11. In September 2000, the Jewish calendar started the year 5761 with Rosh Hashanah; in 2025 it will be the year 5785 before Rosh Hashanah and 5786 after it.

From Figure 9.2, the Islamic year 1426 began on February 10, 2005. Since the Islamic year is 11 days shorter on average, after 20 years, the Islamic year 1446 will begin about 11 × 20 = 220 days (or about 7 months) earlier on the Gregorian calendar, placing the first day of 1446 at the end of June 2024. This means the year 2025 C.E. will begin with 1446 and end in 1447 of the Islamic calendar. Note that the Islamic calendar picks up an extra year compared to the common calendar once every 33 or 34 years. The extra year here was gained in 2008, which saw the beginning of two Islamic years.

In the Chinese system, the 79th sixty-year cycle began in early 1984 C.E. and will end in early 2044 C.E.. That means in early 1984 C.E., 78 cycles or 78 × 60 = 4680 years were completed, and the year 4681 began. There are 41 years between 2025 and 1984 C.E., which means that in early 2025 C.E. the Chinese year 4681 + 41 = 4722 will begin. Note that some authorities begin the count a year earlier (which would make it the Chinese year 4623), but in either case it is agreed to be the year of the Metal Rat.

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Solutions to End of the Unit Problems Part II: Probing Matter, Light, and Their Interactions Unit 14 12. Picture a soccer ball rolling across the ground at 1 meter per second (1 m/sec).

One player deflects the ball so that it turns by a small angle but still moves at the same speed.

Then a second player brings the ball to a stop. Draw a diagram of this with arrows indicating the relative direction and magnitudes of the velocities of the ball and of the forces that the two players apply to the ball.

12. To have little effect on the speed yet change the direction of the ball, the force should be applied approximately perpendicular to the motion but it does not need to be a large force to deflect it over a small angle. The length of the arrow represents the magnitude of the force applied, and if the ball continues moving at the same speed, it won’t take a large force. To stop the ball entirely requires a bigger force–shown with a larger arrow—opposite the direction of motion as shown in the diagram below.

13. Two books, one a thin ¼-kg paperback and the other a thick 1-kg dictionary, are placed side-by-side on the same table. Is the force by the table on each book the same (because it is the same surface) or different (for some other reason)? Explain your answer.

13. Since the books are not accelerating, there is no net force! The (normal) force by the table upwards on each of the books must balance the gravitational force pulling them down. Since the dictionary has a 4 times greater mass, it experiences a gravitational force downward 4 times greater, and so the table must equally exert a 4 times greater force upwards on this book to balance the force of gravity. The same table exerts a smaller force on the lighter paperback.

13. (cont.) Note that to many students this seems counterintuitive partly because they picture the table top as an absolutely rigid geometric surface. It becomes somewhat more intuitive when thinking of the surface of a trampoline, where the deflection of the surface is visible, and it may help students if they realize that the table surface actually does deflect in response to the books.

14. In an amusement park ride called The Rotor, you are spun in a cylinder and then held against the wall as a result of the spin when the floor drops away. People sometimes describe that effect as being due to “centrifugal force.” What is really holding you against the wall of 31

the spinning cylinder? Draw a diagram showing the direction of all forces being applied to you. What force is keeping you from sliding down the wall, countering the force of gravity?

How big is that force?

14. Centrifugal “force” is not really a force at all, the effect of inertia (see Newton’s First Law) resisting the change in motion imposed by the spinning ride. (You do experience centripetal force, from the wall, pushing you towards the center of ride, turning you as you spin.) The frictional force the wall acts upwards, and is keeping you from sliding down. It must be equal to the force of gravity pulling you down, if you stay stuck against wall.

but

the of

the

15. Consider the situation shown in Figure 14.5. If the mass of the ball were twice as large, do you think this would require a greater or lesser force applied by the string? What if the same mass were swung around twice as fast? (A qualitative result is sufficient for this problem, but try to guess how much bigger by performing the experiment if possible.)

15. When we swing any object around, experience tells us that it requires a greater force if the object is more massive or we swing it faster. We might expect that the force required to swing around a ball twice as massive will require twice as much force on the string. Swinging a ball around twice as fast has two effects: (1) the velocity is twice as big, requiring a larger force to deflect the ball by any given angle, and (2) the ball takes half as long to complete a circuit, so the deflections occur in half as much time.

15. (cont.) To find force needed to produce the necessary change in velocity per time interval, the total must then be 2 divided by ½, or a factor of 4 when the speed is twice as large.

(The relationship between force and change in velocity will be discussed in Unit 15.)

A more rigorous treatment (see Unit 17): the centripetal force needed to keep a mass, m, spinning at speed, v, at a radius, r, is Fc = m*v2/r. Therefore, spinning a larger mass (2 m) requires: (2 m)* v2/r = 2 Fc. Spinning a mass m at speed 2 v requires m*(2 v)2/r = 4 Fc. Unit 15 10. After driving behind a car that is traveling at 90 km/hr, you decide to pass the car by changing your speed to 110 km/hr. What was your average acceleration in m/sec2 if it takes 3 seconds to change your speed?

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10. (1.85 m/sec2). First convert the two speeds to m/sec: 110 kph = 110 km/hr × 1000 m/km × 1 hr/3600 sec = 110000 m/3600 sec = 30.6 m/sec 90 kph = 90 km/hr × 1000 m/km × 1 hr/3600 sec = 90000 m/3600 sec = 25 m/sec and now calculate the average acceleration: a = ∆V/∆t = (30.6 m/sec–25 m/sec) / (3 sec) = (5.56 m/sec) / 3 sec = 1.85 m/sec2 Note that in calculating the acceleration, we didn’t round off the intermediate values; 110 km/hr is actually 30.556 m/sec, and we used that value to get the difference in velocities and the resulting acceleration. If you rounded off 30.556 to 31 m/sec, and used that value in the calculation, you would get a = 2 m/sec2.

11. In metric units, forces are measured in “newtons,” which can also be written as kilograms × meters/second2. What is the acceleration experienced by a 100-kg mass if a 100-newton force is applied to it? What is the acceleration experienced by a 200-kg mass if the same force is applied to it?

11. From Newton’s 2nd law, F = m a or a =

F , so for a 100 N force applied to a m

100 kg mass: a = (100 N)/100 kg = 100 kg m/sec2 /100 kg = 1 m/sec2 200 kg mass: a = 100 kg m/sec2 /200 kg = 1/2 m/sec2) 12. A force of 1 newton acting for 1 second on a 1-kilogram-mass object will change the object’s velocity by 1 m/sec. For example, suppose an object with a mass of 20 kg is moving at a rate of 6 m/sec in a straight line. A force of 30 newtons acting opposite the direction of motion of the object for 4 sec will bring the object to a halt. a. What would happen if the force continued after 4 sec? b. What would be the resulting motion of the object if the force acts in the opposite direction? c. What if the same force acted perpendicular to the motion?

12. (a) the object reverses direction; (b) the object accelerates; (c) the object changes direction but continues at the same speed.

We are told a 20 kg object initially moving at 6 m/sec can be stopped by a 30 N force acting for 4 sec in the direction opposite the motion. a) If the same force continued to act opposite the original direction of motion for more than 4 sec, after stopping, the object would start to move in the other direction (back the way it came) faster and faster.

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b) If instead the force was in the same direction as the motion, the object would speed up in the original direction. (Eventually going twice as fast as before, if the force acted for four seconds). c) If the force is applied perpendicular to the motion, the object would move at an angle between the original direction and the direction of the new force (moving at about 45 between, if the new force acted for four seconds).

13. How long should a 4-newton force be applied to a 6-kg bowling ball to get the ball rolling at 5 m/sec? If a greater force is applied, does it take less or more time to reach the same velocity? Explain.

13. (7.5 seconds; less time with a greater force). To get a 6 kg bowling ball moving at 5 m/sec, a 4-newton force would have to be applied for a time such that: a = ∆V/∆t = (5 m/sec)/∆t = F/m = 4 N / 6 kg = 0.67 m/sec2.

Solving for the time, we have ∆t = ∆V /a = (5 m/sec) / (0.67 m/sec2) = 7.5 sec.

A greater force will cause a greater change in speed each second, so the ball will reach the same speed more quickly.

14. If the bowling ball from the previous problem is now rolling uniformly at 5 m/sec, describe the direction a force must be applied to get the ball to stop moving. If you want the ball to be at rest in 1 sec, what magnitude of a net force must be applied?

14. (Opposite to the direction of motion; 30 N). To get the same bowling ball to slow back down, the force must be applied in the opposite direction that the ball is rolling. To get its 5 m/sec speed to a stop in 1 sec corresponds to an acceleration of: a = ΔV/Δt = (–5 m/sec) / (1 sec) = –5 m/sec2.

The minus sign reminds us that the acceleration is in the direction opposite the initial velocity.

To get a 6 kg object to accelerate at this rate, a force of: F = m × a = (6 kg) × (5 m/sec2) = 30 newtons must be applied.

15. Suppose that a 1000-kg rocket’s thrusters give it an acceleration of 20 m/sec2 when it is launched from Earth’s surface. If Earth’s gravity applies a downward force of approximately 10,000 newtons on the rocket, what is the force of the thrusters on the rocket? If the same rocket were launched from the surface of another planet using the same thrust, would the acceleration of the rocket be greater or less than the acceleration experienced by the rocket taking off from Earth? Explain your answer.

15. (30,000 N). The net force Fnet = ma = 1000 kg × 20 m/s2 = 20,000 N = Frocket – Fgravity.

The force of the thrusters, then, is Frocket = 20,000 N + Fgravity = 20,000 N + 10,000 N = 30,000 N. 15. (cont.) Note that we had to add the value of gravitational force to the thruster force because the rocket’s thrusters have to overcome the force of gravity. If this same rocket were launched from a 34

planet with a smaller surface gravitational force Fplanet it would have a greater acceleration. The thrust force Frocket wouldn’t change, but Frocket–Fplanet would be a larger number, so the net force would be larger. Since the mass stays the same and a = F/m, the acceleration would also be larger than calculated for the Earth. If launched from a planet with a greater surface gravitational force, it would have a smaller acceleration.

Unit 16 10. Given that the mass of Mars is 6.4 × 1023 kg, how influential (gravitationally speaking) is Mars on you when the red planet is at its smallest separation from Earth, which is 56 million km? How does this value compare to the gravitational force between you and a person standing 1 m away?

10. (About 10‒9 times less than Earth’s gravitational force; about twice as much as a person standing nearby). The gravitational force exerted by Mars (MMars = 6.4 × 1023 kg) when it is at a distance d = 56 × 106 km = 5.6 × 1010 m is FMars = GMMarsm/d2, where m is your own mass. Since your weight on the surface of Earth is FEarth = GMEarthm/REarth2, where MEarth = 5.97 × 1024 kg and REarth = 6.37 × 106 m are the Earth’s mass and radius, the effect of Mars on you compared to the Earth is: FMars/FEarth = (GMMars m/d2)/(GMEarth m/REarth2) = (MMars/MEarth) ×(REarth/d)2 = (6.4 × 1023 kg/5.97 × 1024 kg) × (6.37 × 106 m/5.6 × 1010 m)2 = 0.107 × (1.14 × 10‒4)2 = 1.4 × 10‒9 So Mars influences your weight by about one part in a billion. By comparison, the force from a person of let’s assume 100 kg standing one meter away is: FMars/FPerson = (GMMars m/d2)/(GMPerson m/1 m2) = (MMars/100 kg) × (1 m/d)2 = (6.4 × 1023 kg/100 kg) × (1 m/5.6 × 1010 m)2 = 6.4 × 1021 × (1.79 × 10‒11)2 = 2.0 So Mars has just twice the gravitational pull of a person standing nearby, and if the person moved closer, they would have a larger gravitational pull than Mars.

11. If an object weighs 180 newtons when measured on Earth, what would its weight be if it were measured on a planet with twice the mass of Earth and a radius twice the radius of Earth?

11. (90 N). The surface gravity of a planet is g = GM/R2, where M is the mass of the planet and R is its radius. If a planet had twice the mass of Earth, it would double the surface gravity since it is proportional to the mass; doubling the planet’s radius will decrease it by a factor of four, since it is inversely proportional to the radius squared; the net effect of doubling both is 2/4 = ½ the gravity, and the weight is 90 N. Explicitly, if we look at the planet compared to the Earth, gplanet/gEarth = (GMplanet / Rplanet2)/(GMEarth / REarth2) = (Mplanet / Rplanet2)/(MEarth / REarth2)

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= ([2 × MEarth]/Rplanet2)/(MEarth / REarth2) = ([2 × MEarth]/[2 × REarth 2)/(MEarth / REarth2) = [2 × MEarth / MEarth]/[REarth2/(22 × REarth2)] = 2/4 The force of gravity on the surface of the other planet is equal to ½ the force of gravity on the Earth, so the weight of the object would have a magnitude of 90 N. 12. The Moon’s radius is 1.74 × 106 m; its mass is 7.35 × 1022 kg. Find the surface gravity on the Moon from these values.

12. (1.62 m/s2). The surface gravity is g = G × M/R2. Substituting in 7.35 × 1022 kg for M (mass of the Moon), 1.74 × 106 m for R (radius of the Moon), and G = 6.67 × 10‒11 N m2/kg2, we get gmoon = 1.62 m/s2, or about 16.5 percent that of Earth.

13. The Sun’s radius is 6.97 × 108 m, while its mass is 1.99 × 1030 kg. Using these values, find the surface gravity on the Sun. Without changing its mass, what would the Sun’s radius have to change to in order to have the same surface gravity as Earth?

13. (3.7 × 109 m). The surface gravity of the Sun is: gsun = G × M⊙ /R⊙ = (6.67×10‒11 N m2/kg2) × (1.99 × 1030 kg) / (6.97 × 108 m)2 = 273 m/sec2.

If we rearrange the surface gravity expression, but use the gravitational acceleration of Earth, we quickly arrive at the desired result, R = (G M⊙/ gEarth)1/2 = ((6.67×10‒11 N m2/kg2) × (1.99 × 1030 kg) / (9.8 m/s2))1/2 = 3.7 × 109 m As expected, the Sun would have to become much larger in size to reduce its surface gravity to match that of Earth’s.

14. The mass of the planet Jupiter is 1.90 × 1027 kg, and the minimum distance from us (on Earth) to Jupiter is 5.88 × 1011 m. The mass of a loaded 18-wheeler is 35,000 kg. a. What is the gravitational force of Jupiter on you? b. What is the force of the 18-wheeler on you if you are standing 2 meters away from its center of mass?

14. (a) 2.6 × 10‒5 N for a 70 kg person; (b) 4.1 × 10‒5 N. We can find the force of Jupiter’s gravity on a person from Newton’s Law of Gravity: F = G × Mjupiter × Myou / d2 = (6.67 × 10‒11 N m2/kg2) × (1.9 × 1027 kg) × (70 kg) / (5.88 × 1011 m)2 = 2.6 × 10‒5 N Here we used a typical mass of a person of 70 kg and the values for Jupiter supplied in the problem. (b) The gravitational force between a person and a 35,000 kg truck 2 meters away is F = G × Mtruck × Myou / d2 = (6.67 × 10‒11 N m2/kg2) × (35,000 kg) × (70 kg) / (2 m)2 = 4.1 × 10‒5 N 36

The truck’s pull is more than one and a half times as large as Jupiter’s.

15. From Appendix Table 5, Uranus has a mass of 8.68 × 1025 kg and a radius of 25,559 km at its cloud tops. What is the acceleration of gravity you would feel flying in a plane there?

15. (8.83 m/s2). The surface gravity at the cloud-tops of Uranus is given by g = GM/R2. Using the values from the appendix for Uranus, we find: g = (6.67 × 10‒11 N m2 kg‒2) × (8.68 × 1025 kg)/(2.56 × 107 m)2 = 8.83 m/s2.

This is just a bit less than the surface gravity on Earth.

Unit 17 9. Given that Neptune is about 30 AU from the Sun, how many years does it take to complete one orbit around the Sun? Approximately how many orbits has Neptune completed since its discovery in 1846?

9. (164 years; barely one orbit). From Kepler’s third law, a3 = P2 so P = (303)½ = 27000½ = 164 years. The year 2014 is 168 years after the discovery of Neptune, meaning the planet has just barely completed an orbit since its discovery.

10. Derive the mass of Earth from the Moon’s orbital period of 27.3 days and average orbital distance of 3.84 × 108 m.

10. We can solve for the sum of the Earth’s and Moon’s masses using the Newton’s version of Kepler’s 3rd Law: MEarth + MMoon = a3/P2 where a is in AU and P in years and the masses would be in terms of the Mass of the Sun. Since the Moon’s mass is much smaller than the Earth’s, the result will approximately be the Earth’s mass (in solar masses). P is 27.3 days, which is approximately 27.3/365, or 7.5 × 10‒2 years. The radius of the orbit (a) is 3.84 × 108 m, which is about (3.84 × 108 m)/(1.5 × 1011 m/AU) = 2.56 × 10‒3 AU. So: MEarth = (2.56 × 10‒3)3 / (7.5 × 10‒2)2 = 2.98 × 10‒6 Msun.

The Sun’s mass is 2 × 1030 kg, so the Earth’s mass is (2 × 1030) × 2.98 × 10‒6 kg or about 5.97 × 1024 kg, which is correct.

11. Communication satellites are typically placed in what are called geostationary orbits.

These are orbits where the orbital period of the satellite matches Earth’s rotational period, allowing the satellite to stay over a fixed point on Earth. Calculate the altitude (distance above Earth’s surface) of a geostationary orbit and compare this value to the altitude of the International Space Station, which is about 360 km.

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11. (35,700 km, which is much larger than the altitude of the ISS). To solve for the orbit of a geostationary satellite we can start by rearranging the expression M = (4π2/G) × (d3/P2) to solve for the size of the orbit: d = (G × M × P2/ 4π2)1/3 which represents the orbital separation between the center of Earth and the satellite. In the above M is Earth’s mass (= 5.97 × 1024 kg according to Appendix table 5) and P is the orbital period for the communications satellite, which equals 24 hours = (24 hr) × (60 min/hr) × (60 sec/min) = 86,400 seconds. Plugging in numbers leads to d = ((6.67 × 10‒11 N m2/kg2) × (5.97×1024 kg) × (86400 seconds)2 / 4 π2)1/3 = 4.22 × 107 m = 4.22 × 104 km = 42,200 km.

Earth’s radius (from Appendix table 5) is 6,480 km. Subtracting this value from our value of d gives a geostationary orbit altitude of 42,200 km–6,480 km = 35,700 km. This is much larger than the 360 km altitude for the International Space Station. Because it orbits much closer to Earth, the ISS also orbits much faster than a geostationary satellite. The ISS makes a complete orbit in about 90 minutes.

12. Two identical stars orbit each other with a period of 62.0 years at a distance of 25.0 AU. What are the masses of the stars?

12. (2.03 M⊙ each). From the equation MA + MB = a3AU / P2yr, the sum of the masses will be (25.0 AU)3/(62.0 yrs)2, or 4.06 Solar masses (total). Since the stars are identical, the masses of the individual stars are 2.03 solar masses each.

13. The Sun orbits the center of the Milky Way Galaxy at a speed of about 220 km/sec at a distance from the center of about 2.6 × 1020 meters (about 28,000 light-years). What is the mass of the Milky Way inside the Sun’s orbit? (We are assuming the Galaxy can be treated as a sphere of matter. This is not exactly correct, but detailed calculations give almost the same answer.)

13. (1011 M⊙ )We can solve for amount of the Milky Way’s mass inside the Sun’s orbit using the formula for mass as a function of orbital radius and speed: MMW = d × V 2/G where d is the distance from the center and V is the orbital speed. Therefore: MMW = d × V 2/G = 2.6 × 1020 m × (2.2 × 105 m/sec)2 / (6.67 × 10‒11 N m2/kg2) = 1.9 ×1041 kg ≈ 1011 MSun The conversion to solar masses is based on the fact that the Sun’s mass M⊙ = 1.99 × 1030 kg.

14. The orbital speed of stars orbiting 14,000, 56,000, and 84,000 light-years from the center of the Milky Way are also found to be 220 km/sec, as in the previous problem. What is the mass out to each of these radii? If all the mass had been concentrated in the very center of the Milky Way, what orbital speeds would you have expected at these distances from the center (assuming the speed was 220 km/sec at the Sun’s distance)?

14. (5 × 1010 M⊙, 2 × 1011 M⊙, and 3 × 1011 M⊙; 311 km/s, 156 km/s, 127 km/s)

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This problem notes that the rotation velocity of the Galaxy remains the same at ½, 2, and 3 times the Sun’s orbital radius. Since everything but the distance remains the same as in the calculation for problem 12, the mass changes in direct proportion to the distance from the center. As a result, interior to ½ the Sun’s orbital radius must be ½ the mass interior to the Sun’s orbit, or 5 × 1010 MSun. Likewise the mass at 2 and 3 times the Sun’s distance from the center, the enclosed mass must be 2 × 1011 MSun and 3 × 1011 MSun respectively.

If the mass interior to these distances had remained the same (i.e., all or nearly all the mass was concentrated at the center of the Milky Way), we can solve for the orbital velocity at these distances by rearranging the equation relating mass to distance and rotation velocity: æ Gö V 2 = MMW × G / d or V = ççM ´ ÷ ÷ ÷ çè dø

Therefore, the orbital velocity would decline in proportion to

1 d

(often called a

“Keplerian decline”) and at ½ the Sun’s orbital distance the orbital velocity would have been 1 1 2

=

2 times larger (220 km/sec ×

velocity would have been is distance,

1 3

1 2

2 = 311 km/sec). At twice the Sun’s distance the orbital

≈ 71% as large (or about 156 km/sec), and at 3 times the Sun’s

≈ 58% as large (or about 127 km/sec).

15. Jupiter’s orbital period is 11.8622 years and its semimajor axis is 5.203 AU. Show that this implies a mass for Jupiter of approximately 1/1000 the Sun’s mass. If Jupiter had a completely negligible mass, but orbited at the same distance from the Sun, what would its orbital period be?

15. (11.8681 years). Using the form of Kepler’s Third Law for two bodies, we can find the mass of Jupiter in terms of the Mass of the Sun by solving: MJ = a3/P2 – MS = (5.203 AU)3/(11.8622 years)2 – 1.00000 = 9.93 × 10‒4 or approximately 1/1000. If Jupiter’s mass was negligible, its orbital period for the same orbital distance would be P = (a3/MS)1/2 = ((5.203 AU)3/1.00000)1/2 = 11.8681 years, compared to the value of 11.8622 years above.

16. In searching for planets orbiting other stars, you notice a star that looks just like the Sun (and presumably has the same mass as the Sun) whose position is slightly shifting back and forth due to an unseen orbiting planet. It is determined that the star is moving in a circular orbit with a period of 14 days. What is the semimajor axis for the planet? Compare this value to Mercury’s semimajor axis of 0.387 AU. 16. (0.11 AU) Newton’s version of Kepler’s third law states that MA + MB = a3AU / P2yr where the masses are expressed in solar masses. If we make the safe assumption that the star’s mass, MA, 39

is much larger than the planet’s mass, MB, we can neglect the planet’s mass. Doing so allows us to rearrange Kepler’s third law to arrive at the semimajor axis: a = (P2yr MA)1/3 = ((14 days/ 365 days/year)2 × (1 MSun))2/3 = 0.11 AU. We are required to convert the orbital period from days to years to use the given form of Kepler’s third law. Lastly, by inspection of figure 17.2 we see that the semimajor axis value just calculated actually represents the total separation between the star and the planet. However, under the assumption that the planet is much less massive than the star, this value also represents the semimajor axis as we normally ascribe to planets in our Solar System.

By comparison, this planet is much closer to its parent star than Mercury is to our Sun. Unit 18 10. Calculate the orbital speed for a satellite 1000 km above Earth’s surface, using the fact that M⊕ = 5.97 × 1024 kg and R⊕ = 6.37 × 106 m.

10. (7350 m/s) To find the orbital velocity for a satellite 1000 km above the Earth’s surface, we need to add this distance to the radius of the Earth (6370 km) to find the orbital radius, then:

V=

GM R

=

æ 6.67´ 10- 11 N m 2kg- 2 5.97´ 1024 kg ö ( )÷÷÷ çç = 7350 m/s çç ÷ ÷ ççè ÷ (6.37´ 106 m + 1.0´ 106 m ) ø

(

)

11. How small would the Sun (M⊙ = 1.99 × 1030 kg) have to shrink to make its escape velocity equal to the speed of light (c = 3.0 × 108 m/sec)?

11. (About 3 km in radius). Starting with the expression for the escape velocity, Vesc =

2 G M /R , we’ll first rearrange for the radius, R = 2 GM /Vesc2, and then substitute in

numbers for the Sun and an escape velocity equal to the speed of light: R = 2 (6.67 × 10‒11 N m2/kg2)(1.99 × 1030 kg) / (3.0 × 108 m/sec)2 = 2950 m = 2.95 km.

In words, this means for the Sun to act like a Black Hole, it would have to be compressed to the size of a small city, about 6 km or less than 4 miles across.

12. On average, the International Space Station is 360 km above the surface of Earth. By what percent is the escape velocity from Earth reduced by launching from the ISS rather than the surface of Earth?

12. (3% less). The ratio of the escape velocity from the ISS to Earth is VISS/VE =

2 G M E /RISS / 2 G M E /R E =

RE /RISS

40

=

RE /(RE + 370 km ) = 6370 km /(6370 km + 370 km )= 0.97.

This means that there is a just a 3% reduction in the escape velocity from Earth by launching from the ISS. (Note, however, that by launching from the ISS in the direction of its orbit, the additional velocity required to reach escape velocity is considerably smaller, since the ISS is already moving at 17,100 miles per hour (27,600 km/hour). So launching from the ISS requires only about 30% of the escape velocity required when launching from the Earth’s surface.)

13. The escape velocity of the Sun can be considered to be the escape velocity of the Solar System. a. Calculate the escape velocity of the Sun at its surface. b. Calculate the escape velocity of the Sun at the orbit of Earth.

13. (a) 617 km/s; (b) 42.1 km/s a) The escape velocity is Vesc =

2G M

, where R is the radius of the Sun. Substituting data R from appendix table 1, the escape velocity from the surface of the Sun is

Vesc =

æ2 6.67´ 10- 11N m 2kg- 2 1.99´ 1030 kg ÷ ( )÷÷ö çç = 617 km/sec. çç ÷ ÷ ççè ÷ (6.96´ 108m ) ø

(

b) Again, the escape velocity is Vesc =

)

2G M

, but now the radius is the Earth’s orbit.

R Substituting data from appendix table 1, the escape velocity is

Vesc =

æ2 6.67´ 10- 11N m 2kg- 2 1.99´ 1030 kg ÷ ( )÷÷ö çç = 42.1 km/s. çç ÷ ÷ ççè ÷ (1.496´ 1011m ) ø

(

)

The farther out we are in the Solar System, the less it would take to escape the Sun’s pull. Note that this is much bigger than the escape velocity from the surface of the Earth itself, so sending a probe to another planet requires much more energy than just getting it into orbit around the Earth.

This problem can also be solved using the method of problem 12, and noting that the Earth is 215 times farther from the center of the Sun as the surface of the Sun. The escape velocity must be smaller in proportion to the square root of this ratio: 1/215 × 617 km/sec = 42.1 km/sec.

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14. What is the escape velocity from the surface of an asteroid with a radius of 50 km and a mass of 1.0 × 1018 kg? (These are about the values for the asteroid Hekate.) If very good pitchers can throw a fastball with a speed of 162 kph (or 101 mph), could they throw the ball so it never fell back on the asteroid?

14. (51.7 m/s; probably not). The escape speed from the asteroid is

Vesc =

2GM R

æ2(6.67´ 10- 11)(1´ 1018 )÷ ö çç ÷ ÷ = 51.7 m/sec. çç ÷ ÷ ÷ çè (50´ 103) ø

A ball thrown at 100 miles per hour = 162 km/hr × 1 hr/3600 sec × 1000 m/1 km = 45 m/sec would be slightly too slow for the ball to escape the asteroid’s gravitational pull. The pitcher would be able to put the ball in orbit though.

15. What is the escape velocity from a galaxy at a radius of 50,000 ly if the mass of the galaxy is 1012 times the Sun’s mass?

15. (750 km/s). Using Vesc =

2 G M /R ,with R = 50,000 ly = 4.73 × 1020 m, and the mass

Mgalaxy = 1012Msun = 1.99 × 1042 kg: Vesc =

2´ 6.67´ 10- 11m 3 /sec2 /kg´ 1.99´ 1042 kg)/4.73´ 1020 m = 7.5´ 105 m /sec

This is an escape velocity of about 750 km/sec.

16. Which body has a larger escape velocity, Mars or Saturn?

MMars = 0.1 M⊕ RMars = 0.5 R⊕ MSaturn = 95 M⊕ RSaturn = 9.4 R⊕ 16. (Saturn has a higher escape velocity). In the formula Vesc =

2 G M /R ,the 2 G is constant,

so it drops out of the comparison. The given masses and radii are in Earth units, so we can easily compare Saturn and Mars to Earth and then to each other. For Saturn, Vesc µ 95/9.4 = 10.1 = 3.2 times Earth’s escape velocity; for Mars Vesc µ

0.1/0.5 =

M /R µ 0.2 =

0.45 times Earth’s escape velocity. Thus, Saturn has the higher escape velocity, 3.2/0.45 ≈ 7 times as much as Mars.

Alternatively, directly calculate the ratio: VSat/VMars = =

2 G M Sat /RSat /

2 G M M ars /RM ars

(M Sat /M M ars)(RM ars /RSat)

= (95/0.1)(0.5/9.4)=

50.53 = 7.1 42

Unit 19 10. At what times would you expect high and low tides when the Moon is full? Draw a diagram to support your result.

10. Roughly midnight and noon for high tides; 6 a.m. and 6 p.m. for low tides. This is the arrangement of the Sun, Earth, Moon, and tides when the Moon is full. (Students should indicate times in the drawing). The Earth turns counterclockwise seen from above the North Pole, so it is 6am at the top of the Earth:

11. Estimate the tidal force between your feet and your head when you are standing on Earth. Assume that both your feet and head have a mass of approximately 5 kg and that you are 2 m tall.

11. (About 3.1 × 10‒5 N). The tidal force is the difference in the gravitational force experienced at two separate locations, Ftidal = Ffeet – Fhead = G × ME × Mfeet/(REarth)2 ‒ G × ME × Mhead/(REarth+ h)2 = G × ME × (Mfeet/(REarth)2 ‒ Mhead/(REarth+ h)2 = (6.67 × 10‒11 N m2 kg‒2) × (5.97 × 1024 kg) × [5 kg / (6.378 × 106 m)2 ‒ 5 kg / (6.378 × 106 m + 2 m)2] = 3.1 × 10‒5 N Note that some calculators may not have enough digits of precision when subtracting the two very similar values. Computer calculator programs and phone apps generally do.

12. Carry out the same calculation as in Section 19.2 for the tidal force on the far side of Earth. Show that this tidal force has approximately the same strength as the tidal force on the near side but points in the opposite direction.

12. From the text, the average force of gravity the Moon exerts on a kilogram of water is 3.33 × 10‒5 N. On the far side, we set up the equation, 12. (cont.)

Fgrav,far = GMm/(d + REarth)2 = 6.67 × 10‒11 N m2/kg2 × 7.35 × 1022 kg × 1 kg / (3.84 × 108 m + 6.37 × 106m)2 = 3.217 × 10‒5 N. As in the text, take the difference to find the tidal force. On the near side, in the book, we found:

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Near – average = 3.44 × 10‒5 N ‒ 3.33 × 10‒5 N = 0.11 × 10‒5 N = 1.1 × 10‒6 N Doing the same thing, Far – average = 3.217 × 10‒5 N ‒ 3.33 × 10‒5 N = ‒ 0.113 × 10‒5 N = ‒1.1 × 10‒6 N The minus sign indicates that the net force is in the opposite direction (the far side force is less than the average, whereas the near side force was more, so the tidal forces both point away from the Earth’s surface, in opposite directions to each other).

13. Copy Figure 19.3 and add a point E on the side of Earth opposite point A. Draw a vector of

appropriate length and direction to represent the force at this point, and show that after subtracting off the average force, the tidal force is in the direction toward the bulge on the side of Earth opposite the Moon.

13. Note the force of gravity towards the Moon at E is smaller than the force at C, so when combined with –C the net force is away from the Earth and Moon and towards the tidal bulge.

14. If the Moon were massive enough, it could pull a kilogram of water off of Earth’s surface with its tidal forces. How massive would it have to be for this to happen?

14. (6.37 × 1029 kg). For the Moon to pull one kilogram of water off the Earth's surface, it would have to be able to produce a tidal force equal to the weight of the water, 9.8 N. The tidal force is given by: FM,near – FM = GMm/(d‒RE)2 – GMm/d2 = GMmoonmwater × (1/(d‒RE)2 –1/d2) = (6.67 × 10‒11 N m2 kg‒2) × Mmoon × 1 kg × (1/(3.84 × 108 m – 6.37 × 106 m)2 ‒ 1/(3.84 × 108 m)2) = Mmoon × 1.540 ×10‒29 N/kg If this force is equal to 9.8 N, Mmoon × 1.540 × 10‒29 N/kg = 9.8 N 44

Mmoon = (9.8 N / 1.540 × 10‒29N) kg = 6.37 × 1029 kg.

This is 8.6 million times its current mass—approaching the mass of the Sun. 15. Demonstrate that the Sun’s tidal effects are approximately half that of the Moon’s by finding the tidal force on 1 kg of water located on Earth. The Sun’s mass is 1.99 × 10 30 kg, and the distance between the Sun and Earth is 1.50 × 1011 m.

15. Tidal force of the Sun on 1 kg water at Earth's distance is the difference of the force on the near side and the force at the average distance.

Fnear = GMm/(d‒REarth))2 = 6.67 × 10‒11 N m2 kg‒2 × 1.99 × 1030 kg × 1 kg / (1.5 × 1011 m – 6.37 × 106 m)2 = 5.8997 × 10‒3 N Favg = GMm/d2 = 6.67 × 10‒11 N m2 kg‒2 = 6.67 × 10‒11 N m2 kg‒2 × 1.99 × 1030 kg × 1 kg / (1.5 × 1011 m)2 = 5.8992 × 10‒3 N The difference here is the tidal force, 0.0005 × 10‒3 N = 5 × 10‒7 N. From section 19.2 we know that the tidal effects on the same 1 kg of water by the Moon is 1.1 × 10‒6 newton. Thus the Sun’s tidal effect is a little under half the value we found for the Moon.

16. Calculate the tidal force of Mars on 1 kg of water on Earth. The mass of Mars is 6.42 × 1023 kg, and the minimum distance between Earth and Mars is 7.78 × 10 10 m.

16. (0.001 × 10–9 N). This can be worked out as in problem 15, replacing the distance and mass of the Sun with those of Mars. The gravitational force of Mars on a kilogram of water on the near side of the planet is 7.077 × 10–9 N. The average gravitational force of Mars on a kilogram of water would be 7.076 × 10–9 N. (On the far side of the planet, it is 7.075 × 10–9 N). The tidal force is about 0.001 × 10–9 N, or an inconceivably tiny 1 × 10–12 N! Unit 20 11. Some professional baseball pitchers can throw a 0.145-kg baseball at 160 km/hr (about 100 mph). What is the kinetic energy of the ball? Using Earth’s acceleration of gravity (9.8 m/sec2), for how many seconds would it have to fall to reach this speed?

11. (143 J; 4.5 seconds). First convert the speed from km/hr to m/sec V = (160 km/hr) × (1000 m/1 km) × (1 hr/60 min) × (1 min/60 sec) = 44.4 m/sec The kinetic energy is EK = ½ mV 2 = ½ (0.145 kg) × (44.4 m/sec)2 = 143 joules.

45

The Earth’s gravitational acceleration is a = 9.8 m/sec2. We can use the relation between acceleration and velocity to determine the speed of a falling ball after some amount of time, t: V = V0 + (a × t) Rearranging to solve for t, and assuming the ball stars with V0 = 0 m/s: t = (V – V0) / a = (44.4 m/s – 0 m/s) / (9.8 m/s2) = 4.5 sec A pitcher’s arm can accelerate the ball to this speed much faster than gravity.

12. What kinetic energy must a 10,000-kg rocket have to escape from the surface of Earth?

12. (6.24 × 1011 J). Starting from the idea on the surface of the Earth the total energy is EG + EK = 0 we see that all that needs to be calculated is the gravitational potential energy of the rocket sitting on the surface the Earth, EK = ‒ EG = ‒ (‒G × m × ME/RE) = (6.67 × 10‒11 N m2 kg‒2) × (10,000 kg) × (5.97 × 1024 kg) / (6.378 × 106 m) = 6.24 × 1011 joules 13. Show that if the gravitational potential energy and kinetic energy add up to zero for an object on Earth’s surface, you can derive the escape velocity formula given in Unit 18.

13. If EG + EK = 0, then EK = ‒ EG. Thus, ½ mV 2 = GmM/R. Divide each side by m and multiply by 2 to get V 2 = 2 GM/R. Taking the square root of each side yields the escape velocity formula: Vesc =

2 GM /R .

14. Suppose that a comet with a mass of 109 kg strikes Earth at a velocity of 40 km/sec. How much energy would be released by the impact? Convert this energy into the equivalent of “kilotons of TNT” (kt), where 1 kt ≈ 4.2 × 1012 joules.

14. (8 × 1017 J; 1.9 × 105 kt.)

The kinetic energy of a comet with speed V = 40,000 m/sec and mass m = 109 kg is: EK = ½ mV 2= ½ × 109 kg × (4 × 104 m/s)2 = ½ ×16 × 108+9 joules = 8 × 1017 J Converting this to “kilotons of TNT” we find EK = 8 × 1017 J × 1 kt / (4.2 × 1012 J) = 1.9 × 105 kt.

46

This is more than 10,000 times the energy released in the Hiroshima atomic bomb.

15. Astronomers estimate that about 100 metric tons (1 × 105 kg) of meteorites strike Earth every day. a. If the meteorites increased the mass of Earth by 1% without adding any angular momentum, how long would the day become? b. Earth’s mass is 5.97 × 1024 kg. How many years would it take to increase Earth’s mass by 1% at this rate?

15. (a) 1.01 times longer; (b) 1.64 × 1015 years. a) This is a conservation of angular momentum problem in which M × V × R = constant. If the mass is 1.01 times larger than what it had been before and the radius remains constant, then the velocity must slow to (1/1.01) or 0.99 times the original velocity to keep the product constant. The time required for the Earth to complete a rotation would equal the distance traveled (which would not change if we assume the size of the Earth does not change), divided by the velocity, which is 1/(1/1.01). Therefore, the new day would be 1/(1/1.01), or 1.01 times as long as the day is now, and that would be an additional one percent, or 864 seconds.

Note that if the radius increased slightly as well (and it might be expected to increase by about 0.33% if the Earth maintained a constant density), the velocity would decrease by this same percentage in addition to the previous amount. The time to complete a rotation would also grow longer by the same percentage again, so the overall period of rotation would be 1.0066 × 1.01 = 1.017 times longer than at present, or about 1470 seconds longer. b) At the rate r given in the problem, 1% of the Earth’s mass would fall onto the Earth in a time t given by: 0.01´ M Earth = (rrate ofdeposition )t

so

0.01´ M Earth 0.01´ 5.97´ 1024 t= = = 1.64´ 1015 years 5 rrate ofdeposition 10 kg perday This is a time much longer than the Sun is expected to live.

16. Suppose that a star collapses inward until it is reduced to one-tenth of its original size while losing no mass. (Note: We are treating the star as if it rotates like a solid, uniform ball, for the purposes of this problem.) a. How many times faster will its surface be moving? b. If initially it was spinning around once every 100 hours, how long will it take to spin around after it collapses?

16. (a) 10 times faster; (b) 1 hour.

47

(a). Since the collapsing star preserves angular momentum, M × V × R = constant. If no mass is lost, then when R drops 10 times, V must go up 10 times. This means the surface will move at ten times speed that it did originally when it collapsed to one-tenth its size. (b) The time to complete one rotation after the star collapses requires us to consider two factors. Not only is the star’s surface rotating with a 10 times larger speed, the surface is 10 times smaller, so the distance the surface covers in one rotation is 1/10th as large. Therefore the star completes a rotation in 1/100th the time. If it initially completed a rotation in 100 hours, it would take 1 hour to rotate after the collapse.

Unit 21 10. Calculate how long it takes light to travel the 150 million kilometers from Earth to the Sun. 10. To calculate the time for light to reach us from the Sun, we divide the distance by the speed of light: d/V = 1.5  108 km/(3  105 km/sec) = 5  102 sec = 500 sec = 8.3 minutes.

11. Suppose you are operating a remote-control spacecraft that is attempting to fly through the rings of Saturn, which are composed of icy chunks of rock. Calculate the round-trip communication time, which includes a downlink of an image and uplink steering command to the spacecraft. Is this a reasonable approach to navigating through Saturn’s rings?

11. (2.4 hours; probably not a reasonable approach). At closest separation Saturn is 8.6 AU from Earth (Saturn’s distance from the Sun minus Earth’s distance, 9.6 AU– 1.0 AU). In meters this distance is: d = 8.6 AU × (1.5 × 1011 m/1 AU) = 1.3 × 1012 m 11. (cont.) The round-trip communication time would then be Δt = 2 d/c = 2(1.3 × 1012 m)/(3.0 × 1012 m/sec) = 8700 sec; In hours this time is Δt = (8,700 s) × (1 min/60 sec) × (1 hr/60 min) = 2.4 hr.

This is likely to be an unreasonably long time for navigating through Saturn’s rings as the spacecraft would move thousands of kilometers during the communication round-trip time interval. Objects that the spacecraft might collide with in the intervening time would probably be too far away to detect and avoid.

12. How much brighter does the Sun appear from Earth when their separation is the smallest (0.983 AU) compared to when they are at their largest separation (1.017 AU)?

12. (About 7% brighter) The brightness of an object is given by brightness = (total light output)/(4 π d2). If we take the ratio of the brightness B of the Sun during closest to the farthest separation we find that: Bclose/Bfar

= (total light output)/(4 π dclose2) / (total light output)/(4 π dfar2) 48

= dfar2/ dclose2 = (dfar/dclose)2 = (1.017 AU/0.983 AU)2 = 1.07 This means the light we receive from the Sun is about 7% brighter when Earth is closest to the Sun in January than when Earth is farthest in July.

13. The Sun is about 10 billion times brighter than the next brightest star, Sirius, as seen from Earth. How far would we have to be from the Sun for the Sun to be as bright as Sirius? (The distance from Earth to the Sun is 1.00 AU.) 13. The brightness of a star is dependent on the distance to the star squared. If we double the distance, the star will become four times fainter. If we triple the distance, the star will become nine times fainter. Therefore, if we wish the Sun to look 10 billion (1010) times fainter, we would have to move to the square root of ten billion times farther away from the Sun. This would be 105 AU (= 100,000 AU). One light-year is equivalent to 63,241 AU, so this distance would be about 1.5 light-years away. This is well out of the Solar System that we know, and about one third of the distance to the actual nearest star (not Sirius).

14. Suppose there is an atom with three excited states with energies 1, 3, and 7 × 10−19 joules above the ground state. If an electron began in the highest energy level, determine all the different ways the electron could descend to the ground state and the energies of the photons it would emit in each case.

14. (There are four different possible paths, and photons of 1, 2, 3, 4, 6, and 7 × 10‒19 Joules are all possible emission lines we would see from this exited atom.)

The electron begins at 7 × 10‒19 Joules. For the rest of this answer, the numbers should be taken to be in units of 10‒19 Joules.

From 7, the electron can: drop to ground, and emit a photon of 7; drop to 3, and emit a photon of 4; and then take the paths from 3 below; drop to 1, and emit a photon of 6; and then take the paths from 1 below; From 3, the electron can drop to ground and emit a photon of 3; drop to 1 and emit a photon of 2; From 1, the electron can drop to ground and emit a photon of 1.

In summary the four possible emission sequences produce photons of 7; 4 then 3; 4 then 2 then 1; 6 then 1. (all × 10‒19 Joules) 15. For the atom in the preceding problem, if the electron is in the ground state, what energy photons could it absorb? What if it were in the first excited state?

15. (From ground: 1, 3, or 7 × 10‒19 J; from 1st exited state 2 or 6 × 10‒19 J).

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If the electron in the atom of the previous question was in the ground state, it could absorb a photon of 1 × 10‒19 Joules (and go to the 1st excited state), 3 × 10‒19 Joules (and go to the second), or 7 × 10‒19 Joules (and go to the third). From the 1st exited state, it could absorb a photon of 2 × 10‒19 Joules (and go to the second state) or 6 × 10‒19 Joules (and go to the third).

Unit 22 11. A typical frequency used for cell phone communication is 800 megahertz. What is its wavelength? How does the wavelength compare to the size of a standard cell phone?

11. We can rewrite c = λ × ν to solve for the wavelength: λ = c/ν = (3.0 × 108 m/sec)/(800 × 106 Hz) = 0.375 m This size is about 14 inches, which is about 3 or 4 times the height of most cell phones. But the antenna inside cell phones does not have to be this long to resonate with this signal; cell phone antennas are cleverly designed to receive multiple wavelengths of radio waves from cell towers, WiFi, Bluetooth, and other wireless devices.

12. Microwave ovens operate at a 12-cm wavelength by making water molecules oscillate at the frequency of the electromagnetic radiation. What frequency is this? Explain at the atomic level why some substances heat up and others do not. How does this differ from what happens in conventional ovens at the atomic level?

12. A 12 cm photon has frequency ν = c / λ = (3 × 1010 m/sec) / 12 cm = 2.5 × 109 Hz The only substances that will heat up will be ones that are able to absorb photons of this frequency: ones which have transitions with this energy difference. This can be electrons rising to excited states in a specific atom, or shared electrons rising to excited states in a molecule, and/or the specific vibrational modes of the atoms in a molecule (probably the most typical way the photons would be absorbed in this case.). Eventually the electrons drop back down and the energy is released not as photons but as kinetic energy of the food particles—fast moving particles are what make the food hot. In a conventional oven, kinetic energy is transferred to the particles directly by hot, fast moving air particles that collide with the food particles.

13. A hydrogen atom has a diameter of about 0.1 nanometer. What is the frequency and energy of a photon that has a wavelength equal to the diameter of the hydrogen atom?

13. (3.0 × 1018 Hz; 2.0 × 10‒15 J) We can rewrite c = λ × ν to solve for the frequency: ν = c/λ = (3.0 × 108 m/sec)/(0.1 × 10‒9 m) = 3.0 × 1018 sec−1 = 3.0 × 1018 Hz.

50

The energy of this photon is E = h × c/ λ = (6.62 × 10‒34 J sec) × (3.0 × 108 m/sec)/(0.1 × 10‒9 m) = 2.0 × 10‒15 J This is a VERY energetic photon – about 12,500 electron Volts and 4000 times more energetic than visible light.

14. What is the wavelength of a photon that has 1000 times as much energy as a photon with a wavelength of 500 nm? Would it be more dangerous to be exposed to 1 million photons with a wavelength of 500 nm, or 1000 photons with 1000 times the energy of a 500-nm photon? Why? 14. Since E is inversely proportional to wavelength, a photon that has 1000 times as much energy as another photon must have 1/1000 the wavelength. The wavelength of a photon with 1000 times the energy of a 500 nm photon, or 500 × 10−9 m = 500 nm, would be 0.5 nm, or 5 × 10−10 m.

(This can also be solved directly starting with 1000 × E from the previous question). The photon would be an X-ray photon. The 500 nm photon is greenish-blue light. In most situations, one million low energy photons like the blue light would have very little effect on you, unable to deposit enough energy in one place at one time to cause a problem. On the other hand, one very high energy X-ray photon could have a serious effect on a cell. (Of course, if there were enough low energy photons, they can still have an effect—think about the microwave question).

15. Suppose the energy of an electron in its atomic orbital is 10 −18 joule above the ground state. What wavelength of light would have photons matching this energy difference? Which band of electromagnetic radiation is this in?

15. (1.99  10‒7m) For a photon with energy E = 10‒18 J, we have E = hc/λ. Solving for the wavelength: λ = hc/E = 6.626 × 10‒34 J sec × 3 × 108 m/sec /10‒18 J = 1.99  10‒7m.

This is about 2  10‒7 m, which is in the ultraviolet range.

16. a. What is the energy (in joules) of an extremely powerful gamma ray with a wavelength of 10−7 nanometer? b. How many of these gamma-ray photons would add up to the same amount of energy as a housefly in flight, which has a kinetic energy of about one millionth of a joule as it flies along?

16. (a) 1.99 × 10‒9 J; (b) About 500 photons.

The wavelength is 10‒7 × 10‒9 m = 10‒16 m. Solving for the energy, we find: E = h × c/λ = 1.99 × 10‒25 J m /10‒16 m = 1.99 × 10‒9 J. b) If a fly has a kinetic energy of about 10‒6 joules, then about 500 of these gamma ray photons equal the energy of a fly: 51

# photons = 10‒6 Joules KE per fly / (2 × 10‒9 J/photon) = 500.

Unit 23 9. What is the temperature range for objects whose wavelength at maximum falls within the visible spectrum, 400 to 700 nm?

9. (From 7250 K to 4140 K) From Wien’s law (T = 2.9 × 106 K nm/λ max), the temperature for λmax = 700 nm is T = 2.9 × 106 K / 700 = 4,140 K, and for λmax = 400 nm: T = 2.9 × 106 K / 400 = 7,250 K. Note how the longer wavelength red light corresponds to a lower temperature object, and the shorter wavelength blue light to a hotter object.

10. A lightbulb radiates most strongly at a wavelength of about 3000 nanometers. How hot is its filament?

10. A lightbulb with λmax = 3000 nm has a temperature: T = 2.9 × 106 K nm/3000 nm = 970 K. This is almost 1300° Fahrenheit.

11. How hot would a source need to be to emit its maximum radiation in the FM range of the radio dial (100 MHz)? Why don’t common objects interfere with our radio reception?

11. (9.7 × 10‒4 K). A frequency of 100.0 MHz corresponds to a wavelength of λ = c/ν = (3.0 × 108 m/sec)/(100 × 106 Hz) = 3 m = 3 × 109 nm From Wien’s law, the temperature will be T = 2.9 × 106 K nm/λ max = 2.9 × 106 K nm/ 3 × 109 nm = 9.7 × 10‒4 K. Common objects are much hotter than this (200 to 300 K) so they actually radiate primarily at much higher energies (shorter wavelengths).

12. By measuring relic radiation left over from the Big Bang, astronomers have measured the average temperature of the Universe to be 2.73 K. At what wavelength does this background

52

radiation emit most strongly? What part of the electromagnetic spectrum does this wavelength lie in?

12. From Wien’s Law, T = (2.9 × 106 Knm /λmax), we solve for the wavelength: λmax = (2.9 × 106 Knm)/T = (2.9 × 106 Knm)/(2.73 K) = 1.1 × 106 nm = 0.0011 m At this wavelength, about 1 mm, the radiation falls within the microwave portion of the electromagnetic spectrum.

13. To forge copper, the metal must be heated to a temperature of 1200 K. How much more thermal radiation does a piece of copper produce when heated to its forging temperature as compared to room temperature?

13. (256 times more). Room temperature is about 300 K so the temperature of 1200 K copper would be 4 times as great. By the Stefan-Boltzmann Law, the radiated energy would be 44 or 256 times as much.

14. The total light an object emits depends on its temperature and its surface area. How large would a star with a surface temperature of 3000 K need to be to have the same luminosity as a star with a surface temperature of 6000 K? (Phrase your answer in terms of the size of the 6000-K star.)

14. (The cooler star would need to be 4 times the radius, and 16 times the surface area).

The total luminosity depends on temperature to the fourth power according to the StefanBoltzmann law, and as the surface area, which depends on the radius squared. If a blue and a red star have the same luminous output, We must have: Rred2Tred4 = Rblue2Tblue4.

Since we are told that Tblue = 2 Tred, we find: Rred2Tred4 = Rblue2Tblue4 Rred2Tred4 = Rblue2(2 Tred)4 Rred2 = Rblue224 =16 Rblue2 Rred = 4 Rblue This can also be solved by proportionalities, noting that since L  T4R2, then the effect of the lower temperature by a factor of 6000/3000 = 2 means the star would be ½4 =1/16 as luminous if it was the same size, and the same luminosity if it has 16 times the area. Area goes as radius squared, so the cooler star must have 4 times the radius to have 16 times the area.

15. What is the ratio of the luminosity per square meter coming from a part of the Sun that is at 5800 K versus a sunspot where the temperature is 5000 K? 15. (1.81 times). Luminosity is proportional to the size of the emitting area and the temperature to the fourth power. In the question, the area is constant (1 square meter) so the ratio is just the temperatures to the fourth power: 53

ratio = (5800 K)4/(5000 K)4 = (5800/5000)4 = (1.16)4 = 1.81.

The hotter area is 1.81 times brighter.

16. Find the temperature in degrees Celsius and Fahrenheit for (a) a star whose surface is at 10,000 kelvins, and (b) the 3-kelvin temperature of deep space far from any stars.

16. (a) TC = 9727 C; TF = 17541 F; (b) TC = ‒270 C; TF = ‒454 F. (a). If TK = 10,000 K, then TC = TK ‒ 273 so 10000 – 273 = 9727 C; and TF = 9/5 TK ‒ 459.4 so 9/5(10000) – 459.4 = 17540.6 F (b). If TK = 3 K, then: TC = 3 – 273 = ‒270 C; and TF = 9/5(3) ‒459.4 = ‒454 F. Unit 24 12. In the formula for hydrogen’s energy levels, find the energy difference between levels n = 1 and n = 2, and determine the wavelength of a photon with this energy. What part of the electromagnetic spectrum is this?

12. (1.635 × 10‒18 J; 121.7 nm, UV) The energy of level n for hydrogen is En = –2.18 × 10‒18 joule/n2, so at n = 1 the energy is E1 = –2.18  10‒18 joule, and at n = 2 it is: E2= –2.18 × 10‒18 joule/22 = –5.45 × 10‒19 joule.

The energy difference is: E2 – E1 = –5.45 × 10‒19 joule – (–2.18 × 10‒18 joule) = 1.635 × 10‒18 joule.

We use the relation between energy and wavelength to find the corresponding wavelength: λ = (h × c)/E = 1.99 × 10‒25 joulem/1.635  10‒18 joule = 1.217 × 10‒7 m = 121.7 nm, which is in the ultraviolet. (Values vary slightly from laboratory values because of rounding).

13. Using Wien’s law (Unit 23.4), find the temperature of a star that has its wavelength of maximum emission at the wavelength needed to raise an electron from the ground state to the n = 2 state.

13. (T = 23,830K) The energy and wavelength of the photon were worked out in question 12.

Using Wien’s law this corresponds to the wavelength of maximum emission for a blackbody of temperature: T = (2.9 × 106 K nm) / (121.7 nm) = 23,830 K. 14. Using the formula for hydrogen’s energy levels, calculate the energy of a photon emitted when hydrogen goes from level 143 to level 142 (a transition that has been observed by astronomers). What is the wavelength of this spectral line? What part of the electromagnetic spectrum is this?

54

14. (1.507  10‒24 J; 13.2 cm) When hydrogen transitions from level n = 143 to 142, the energy difference is: E143 – E142 = –2.18  10‒18 joule × (1/1432–1/1422) = –2.18  10‒18 joule × (–6.91  10‒7) = 1.507  10‒24 joule The wavelength corresponding to this energy is: λ = h × c/E = 1.99 × 10‒25 joulem/1.507  10‒24 joule = 0.132 m = 13.2 cm This is in the radio wavelength range.

15. Explain the physical meaning of a transition from n = 1 to n = infinity. For hydrogen, calculate the energy associated with this transition (recall that anything divided by an infinite number can be treated as 0).

15. The n = 1 to n = ∞ (infinity) transition represents the energy required to ionize the atom from its ground state, and remove the electron altogether from the atom. For hydrogen this is ΔE = (–2.18 × 10‒18 joule/∞2) ‒ (–2.18 × 10‒18 joule/12) = 0 + 2.18 × 10‒18 joule where we have used the fact that a number divided by infinity is zero. This energy is, of course, the same as the hydrogen ionization energy stated in Table 24.1.

16. From the data in Table 24.1, what wavelength photon can ionize helium? What wavelength can ionize uranium?

16. (Helium: 50.5 nm; Uranium: 201 nm) To find the wavelength of an ionizing photon, we plug the ionization energy into the relationship between energy and wavelength.

For helium the ionization energy is E = 3.940  10‒18 joule, which corresponds to wavelength: λ = h × c/E = 1.99 × 10‒25 joulem/3.940  10‒18 joule = 5.05  10‒8 m = 50.5 nm For uranium the ionization energy is E = 0.992  10‒18 joule, which corresponds to wavelength: λ = h × c/E = 1.99 × 10‒25 joulem/0.992  10‒18 joule = 2.01  10‒7 m = 201 nm Both of these are in the ultraviolet part of the electromagnetic spectrum.

55

Solutions to End of the Unit Problems

Part II cont’d.

Unit 25 10. About how fast would you have to be driving toward a red stop-light for it to appear green?

10. ( 6.4´ 107 m /s)If we take λ= 550 nm to be the wavelength of light from the green circle in the stoplight, and λ= 700 nm to be the wavelength of the red, then v=

(550nm - 700nm ) æ Dl mö c= ´ çç3´ 108 ÷ ÷= 6.4´ 107 m /s çè l 700nm s÷ ø

which means you’d need to be moving toward the light (the minus sign confirms that we are looking at a blueshift) at about 20% the speed of light (150/700 = 0.21).

11. Hydrogen produces a spectral line at 656.3 nm. If it is detected from a star at 655.5 nm, first argue if the star is approaching or receding, then calculate the radial velocity of the star.

11. ( 3.66´ 105 m /s aw ay). The observed emission line is at a wavelength (655.5 nm) that is shorter than the rest wavelength (656.3 nm). The light from this star is blueshifted, which means it is moving toward us. The actual radial velocity is

v=

(655.5 nm - 656.3 nm ) æ Dl mö c= ´ çç3´ 108 ÷ = - 3.66´ 105 m /s ÷ ÷ ç l 656.3 nm sø è

The negative value reinforces the idea that this star is approaching us.

12. Suppose that a star is rotating at a speed of 50 km∕sec at its equator. Its spectral lines will be spread out because one side of the star will be redshifted and the other side will be blueshifted. For the 656.3-nm line of hydrogen, over how wide a range in wavelengths will the spectral line be spread?

12. (0.218 nm) Start with a Doppler shift for 50 km/s:

56

æ

Δl =

lv = c

æ m ö mö ÷ ´ çç5´ 104 ÷ ÷ ÷ ÷ èç ÷ nm ø sø è = 1.09´ 10- 10 m = 0.109nm 8 3´ 10 m /s

(656.3 nm )´ ççç10- 9

This gives a shift in wavelength of 0.109 nm toward the blue at the edge of the planet rotating toward the observer, and another 0.109 nm toward the red at the limb of the planet moving away from the observer, for a total shift of 0.218 nm.

13. Hydrogen produces a spectral line at 656.3 nm. Oxygen produces a spectral line at 645.5 nm. If you detected a galaxy with a spectral line at 655.0 nm, what Doppler shift would you derive if you assumed it was produced by hydrogen? by oxygen? How could you determine which was correct?

13. (By hydrogen, −5.9 × 105 m/s; by oxygen, +4.4 × 106 m/s).

If the line is hydrogen: and the minus sign indicates the source is moving towards us (a blueshift). If the line is oxygen: v=

(655.0nm - 645.5nm ) æ Dl mö Dl c= ´ çç3´ 108 ÷ = + 4.4´ 106 m v = c ÷ ÷ çè l 656.3nm sø l =

(655.0nm - 656.3nm ) æ ç

mö ´ ç3´ 108 ÷ ÷= - 5.9´ 105 m /s çè s÷ ø

656.3nm

/s

and the plus sign indicates the source is moving towards us (a redshift).

We can determine which conclusion is correct by applying some knowledge about galaxies. If the line is hydrogen, then the galaxy is moving very fast toward us. No galaxy in the Local Group moves that fast relative to us and no galaxy outside the Local Group is moving toward us.

Therefore, the red-shifted line is oxygen and the galaxy is outside the Local Group.

14. Venus orbits the Sun at 35 km∕sec, whereas Earth orbits at 30 km∕sec. Draw a diagram of the two planets’ orbits (Venus orbits at ~0.7 AU) with vectors representing the planets’ velocities, and use it to estimate the radial velocity we would see when Venus is at its greatest elongations and when it is in conjunction with the Sun. What would be the change in wavelength for a 10-cm radar signal bounced off Venus at these locations?

14. (See Diagrams on the next page).

For there to be a Doppler shift, there must be a net radial component of Venus’s velocity vector relative to the Earth. At the point of greatest elongation, Venus is 47° from the Sun and the geometry is as shown in the diagram. Venus is moving towards the Earth at 35km/s, but there is a component of the Earth’s motion in the same direction in space (hence away from Venus).

57

From the diagram, it is clear that the component of the Earth’s motion away from Venus is 30 km/s × sin(47°) = 22 km/s. Venus is coming towards the Earth, but the Earth is moving away, so the net motion is 35 km/s + 22 km/s = 13 km/s, where the minus sign indicates the motion is towards Earth (blueshift).

For the greatest elongation shown in the diagram, the change in the wavelength is

Dl v l v (0.1 m )(- 13000 m /s) = so D l = = = - 4.3´ 10- 6 m 8 l c c 3´ 10 m /s 14. (cont.) There is a second possible elongation orientation (Venus on the other side of the Sun). In this case, Venus moves away from the Earth and Earth moves towards Venus (but not as fast) so the net motion results in a redshift. (A similar sketch will show the radial component of Earth’s velocity for this second case is 30 km/s × sin(43°) = 20 km/s and the resultant relative velocity is +15 km/s).

For the case of conjunction, when Venus is directly in line with the Sun, the Earth and Venus are traveling in the same direction—since there is no radial motion there is no Doppler shift.

58

15. At its equator, Mercury rotates at 3.0 m∕sec. If an astronomer sent a radar signal at precisely 10 cm to bounce off of Mercury, what would be the breadth of the return signal?

15. (2.0 nm). From the side of Mercury rotating towards us, the signal would be blueshifted; from the side rotating away, redshifted, and the signal would remain at 10 cm where reflected from the equator. The amount of the red and blue shift would each be:

Dl v l v (0.1m )(3m /s) = so D l = = = 1.0´ 10- 9 m = 1.0nm 8 l c c 3´ 10 m /s So the total breadth of the signal would be twice that, 2.0 nm or 2.0 × 107 cm.

16. Hydrogen produces spectral lines at radio wavelengths, notably at 21.1 cm. If a galaxy is moving away from us at 10% of the speed of light, at what wavelength will we detect this line? Convert this into a frequency.

16. (23.2 cm, 1.29 × 109 Hz). Starting with - 2 7 Dl v l v (21.1´ 10 m )(3´ 10 m /s) = so D l = = = 21.1´ 10- 3 m = 2.11cm 8 l c c 3´ 10 m /s

So Redshifted =  +  = 21.1 cm + 2.1 cm = 23.2 cm.

Frequency  = c/Redshifted = 3 × 1010 cm/s / 23.2 cm = 1.29 × 109 Hz.

Unit 26 11. For the javelin thrown in Figure 26.2, would the effect of special relativity result in a speed for the javelin slower than Galilean relativity would predict, or faster? Explain.

59

11. The measured speed would be slower than what Galilean relativity predicts. Galilean relativity would have the measured speed seen by observers for the javelin moving over the ground to be the sum of the speeds of the thrower running and the speed of the javelin with respect to the thrower. Special Relativity however takes into account changes in length and duration due to motion, and observers in the ground’s rest frame would see the javelin as both being shorter in length and taking more time to move a specific distance.

12. Voyager 1, the farthest and one of the fastest space probes, is now traveling at a speed of 16.9 km∕sec relative to the Sun. What fraction of the speed of light is Voyager 1 traveling?

Approximately what is the Lorentz factor associated with this speed? [https://voyager.jpl.nasa.gov/mission/status/#where_are_they_now] 12. (~1.000000002). Voyager 1 is traveling at (16.9 km/s)/(3.0 × 105 km/s) = 0.0056% of the speed of light. At this slow speed the Lorentz factor is strongly approximated as unity, which can be verified from calculating]

g=

1 æ v2 ö÷ çç1÷ çè c2 ø÷ ÷

=

1 æ ö çç 16.92 ÷ ÷ ÷ çç12÷ çç (3´ 105 ) ÷ ÷ è ø÷

= 1.000000002

13. If we were to say that the Lorentz factor becomes significant if it affects measured lengths and times at the 10% level (a Lorentz factor value of 1.1), how fast must the observed speed of an object be to reach this threshold?

13. (1.26 × 108 m/s). Start with:

g=

1 2

1- Vc2

Re-arranging, v 1 1 = 1- 2 = 1= 0.42 c g 1.12 The observed speed must be 0.42c, or 1.26 × 108 m/s.

14. What speed would give a Lorentz factor of about 1 million (10 6)?

14. (0.9999999999995 c). Start with:

60

g=

1 2

1- Vc2

Re-arranging, v 1 1 = 1- 2 = 1= 0.9999999999995 c g (106 )2 The object would need to travel at 0.9999999999995 times the speed of light.

15. Earth is traveling at about 30 km∕sec as it orbits the Sun. a. How much more slowly does time run on Earth than in the Sun’s rest frame? b. Why will the time be measured as slower on Earth than in the Sun’s rest frame? (Why is it not “just relative”?)

15. a) Since the Earth travels 30 km/s as it orbits the Earth, its Lorentz factor is:

g=

1 2

1- Vc2

1

=

= 1.000000005

2

1-

(30 kms )

2

(3´ 105 kms )

Time runs 1.000000005 times slower than in the Sun’s reference frame. That’s about one second slower in 6.3 years. b) Due to its orbit around the Sun, the Earth is constantly accelerating. Clocks run slower when accelerating, as described in General Relativity. (Note: This problem is concerned only with special relativistic effects of an observer in the Earth’s constantly accelerated frame of reference versus the Sun’s frame of reference. We have not accounted for gravitational time dilation effects that would occur due to the Earth’s or Sun’s gravitational pull.)

16. Suppose a 1-gram paperclip struck a spaceship at 99% of the speed of light. Special relativity indicates that the paperclip’s energy goes from γ × mc 2 and drops down to its “rest mass energy” of mc2 once it stops moving. a. How much energy is released in the collision? b. How much energy is this in kilotons of TNT? (1 kt = 4.18 × 10 12 joules) 16. (a) 5.4 × 1014 J; (b) 129 kilotons) a) At 99% the speed of light, the Lorentz factor is

61

g=

1 2

1- Vc2

=

1 1- .992

= 7.

The energy released when it stops will be the energy from moving (γmc2) less the rest energy (mc2), so the total energy released is æ mö E = g m c2 - m c2 = (g - 1)m c2 = (7- 1)(0.001kg)çç3´ 108 ÷ = 5.4´ 1014 J ÷ ÷ çè sø

b) This is about 129 kt of TNT. (5.4 × 1014 J)/(4.2 × 1012 J/kt) = 129 kt).

17. How fast must a spacecraft travel to reach the nearest star system (4.3 ly away) in 1 year of “ship time”?

17. (0.97c). Special relativity shows that if 4.3 years have passed in a stationary reference frame (t) (relative to the moving ship), while aboard the ship only 1 year has passed (t0), the trip would require a Lorentz factor of:

g=

t 4.3y = = 4.3. t0 1y

To achieve this Lorentz factor, the ship would need to travel at a speed of: Starting w ith the form ularforthe Lorentz factor, 1 g= 2 1- Vc2 Re-arranging, v 1 1 = 1- 2 = 1= 0.97c. c g (4.3)2

Unit 27 10. The discrepancy in Mercury’s orbit explained by general relativity is a shift in the orientation of the long axis of the orbital ellipse of 43 arc seconds per century. (An arc second is 1/3600 of a degree.) Mercury’s orbital period is 88 days, so how much does the orientation change after a single orbit?

10. How much does the orbital orientation change after 88 days? We first start by finding how many orbits per century Mercury makes:

62

1 orbit 365.25 days 100 years orbits ´ ´ = 415 88 days 1 year 1 century century 10. (cont.) Since Mercury’s orbit precesses by 43 arcseconds (“) per century, the precession per orbit is:

43¢¢ 1 century ´ = 0.1¢¢ perorbit 1 century 415 orbits The orientation changes by 0.1″ per orbit.

11. The escape velocity from Jupiter is about 60 km/sec, whereas for Mercury it is 4 km/sec. Find the gravitational time dilation on the surface of these planets.

11. (1.00000002 for Jupiter; 1.00000000009 for Mercury).

The time dilations are given by (a high precision calculator or algebraic approximation is needed for this): 1 1 gG = = = 1.00000002 (forJupiter) 2 2 Vesc 60 kms ) ( 1- c2 12 (3´ 105 kms )

gG =

1 1-

2 Ves c 2

c

1

=

= 1.00000000009 (forM ercury)

2

1-

(4 kms )

2

(3´ 105 kms )

12. The neutron star system discovered by Taylor and Hulse takes about 8 hours to complete an orbit, but the period is shortening by about 80 microseconds per year. If it continues shortening at this rate, calculate how much time will pass before the neutron stars collide.

12. (3.6 × 108 years). Every year, the period decreases by the same amount, and the neutron stars get closer together. How many 80 microsecond increments are there in 8 hours?

8hours´

3600sec 1year ´ = 3.6´ 108 years 1hour 80´ 10- 6 sec

13. Earth’s escape velocity is 11 km/sec. How much slower do clocks run on Earth’s surface than in deep space, far away from any massive object?

13. The time dilation at the Earth’s surface is given by (a high precision calculator or algebraic approximation is needed for this):

63

1

gG =

1-

2 Ves c 2

1

=

c

= 1.000000000672222

2

(11 kms )

1-

2

(3´ 105 kms )

14. What is the gravitational time dilation for the Sun’s escape velocity of 618 km/sec? If the frequency of light corresponding to a 500-nm photon is slowed by this amount, what wavelength will this photon have far from the Sun? 14. (500.00106090337655 nm). The time dilation at the Sun’s surface is given by (a high precision calculator or algebraic approximation is needed for this):

gG =

1 1-

2 Ves c 2

1

= 1-

c

= 1.000002121806753

2

(618 kms )

2

(3´ 105 kms )

If a 500nm photon is slowed down by γG, far from the Sun it will have a (gravitationally redshifted) wavelength of λ × γG = 500.00106090337655 nm.

15. Find the escape velocity from a neutron star with the mass of the Sun and a radius of 10 km. What is the gravitational time dilation at the surface of the neutron star?

15. (1.6 × 108 m/s; 1.19) First we need the escape velocity from the surface of the neutron star: 2GM

vesc =

R - 11m 3

(

2 6.67 × 10 =

)

/kg×s2 (1.99 × 1030 kg) 3

10´ 10 m

= 1.6´ 108

m s

15. (cont.) This is more than half the speed of light! Now, the time dilation at the neutron star’s surface is given by: gG =

1 1-

2 Ves c 2

c

=

1 2

= 1.19

(1.6´ 108 ms ) 12 (3´ 108 ms )

Time is running almost 20% slower there.

16. Suppose that the gravitational time dilation on the surface of a star is 1.1. What would be the escape velocity from the star?

64

16. (Vesc = 0.42 c)

gG = Vesc c

1

« 2

Vesc

1-

2 Ves c

1-

2

c

1 « gG

=

c2

2 Ves c

c2

= 1-

1 « g 2G

Vesc c

= 1-

1 g 2G

1 1 = 1= 0.42 2 g G 1.12

= 1-

17. What is the maximum delay between detections of a gravitational wave that passes through each of the two LIGO observatories? The observatories are separated by 3000 km and gravitational waves travel at the speed of light.

17. (10 ms). The maximum delay occurs for a gravitational wave that travels along a line that connects the two observatory sits, Δt = d/v = (3,000 km)/(3 × 105 km/s) = 0.01 s = 10 ms.

18. For a satellite in a low circular orbit around Earth, calculate and compare the Lorentz factor associated with its orbital velocity (Unit 18) with the gravitational time dilation.

18. A satellite in a low circular orbit has Vcirc = 2GM

√2 times more, Vesc = g=

1 1-

2 Vci rc 2

GM R

; the escape velocity at that orbit is

. The Lorentz factor associated with the orbital velocity is

R

; the Lorentz factor associated with time dilation, g G =

c

1 1-

2 Ves c

.

c2

18. (cont.)

So,

1

g=

2 circ 2

v c

1-

= 1-

1 = æG M ÷ ö çç ÷ çè R ÷ ø

- 1/2

æ GM ö ÷ = çç1÷ 2÷ ç è Rc ø GM 1Rc2 1

? 1+

GM 2 Rc2

c2

and 1

gG = 1-

2 esc 2

v c

=

1 = æ2 G M ö ÷ çç ÷ çç R ÷ ÷ ÷ è ø 1c2

1-

- 1/2

æ 2GM ö ÷ ÷ = ççç12 ÷ ÷ çè ÷ Rc ø 2GM

1

? 1+

GM Rc2

Rc2

Where we have used just the first two terms of a mathematical series approximation (the Binomial Approximation) for the value of 65

- 1

2 æ GM ö ÷ çç1÷ ÷ çè Rc2 ø

The Lorentz factors themselves will differ by GM/2Rc2, and gravitational time dilation will have about twice the effect of special relativistic time dilation. For the 5.6 km/s orbit shown in Figure 18.1, the effects would be 1.000000000174222 (special relativistic) and 1.000000000346722 (general relativistic), for example.

Unit 28 10. Suppose that you are using an older CCD camera that collects only 40% of the light that would otherwise be gathered. How would this affect the magnification of the image you are taking? How would this affect the brightness of the image?

10. The brightness of the image would only be 40% of the image that we otherwise would have seen. The magnification would be unchanged.

11. The Crab Nebula is about 6500 ly from Earth. In what year did the supernova event that formed the Crab Nebula occur?

11. (5445 B.C.E.) The actual Crab supernova occurred 6,500 years before the Crab supernova was first observed, 1054 C.E. Hence, 1054 – 6500 = –5446. If we work in the Gregorian calendar, then there is no year 0 and we are left with 5445 B.C.E. for the actual year that the Crab supernova occurred. This is about the time agriculture was first being established.

12. One of the 10-m-diameter Keck telescopes can observe 500-nm visible-wavelength photons. If a radio telescope observing 10-cm photons was built to be proportionally as large relative to the photons it was observing, what would its diameter be? What about a gammaray telescope observing 0.01-nm photons?

12. The ratio of the Keck diameter to photon wavelength is

10 m - 9

500´ 10 m

= 2´ 107

so the diameter is 20 million times greater than the wavelength. To find the required diameter for the radio telescope, multiply the desired wavelength (10 cm) by 20 million to get 2 × 106 meters or 2000 km. For the gamma ray telescope, multiply the 0.01 nm resolution by 20 million to get 2 × 104 m or 0.2 mm. Clearly either of these sizes are not reasonable for a single telescope or detector. At least with radio telescopes, we can combine signals from sites separated by 2000 km or more using interferometry, to achieve better resolution. To collect gamma rays (from space, since they don’t penetrate Earth’s atmosphere), NASA’s Fermi Gamma Ray Space Telescope had 16 detector “towers” each around 40 cm across – much larger than the proportional size required compared with Keck.

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13. The diameter of your eye’s pupil is about 8 mm, while the Arecibo radio telescope has a diameter of about 300 m. What wavelength of light would have the same size relative to the Arecibo radio telescope as 500-nm visible light has to your eye?

13. (1.9 cm) Create a ratio comparing the diameter of Arecibo to the diameter of your eye; solve for the wavelength of light that would be proportional: D arecibo l arecibo l arecibo l visible = so = D eye l visible D arecibo D eye

l arecibo l visible = 300 m 8´ 10- 3m

500´ 10- 9 m l arecibo = 300m ´ = 0.019m = 1.9cm 8´ 10- 3m The Arecibo telescope has the same resolution as our eye at a wavelength of around 0.019 m, or 1.9 cm.

14. To provide clear images, the surface of a telescope mirror should be smooth to within 1∕10 of a wavelength of the light being collected. What size is this for a visible-wavelength telescope? for a radio telescope? for a gamma-ray telescope? What physical objects match these sizes?

14.

For visible, with λ ~ 550 nm, you need smoothness at 55 nm (1/10th of 500 nm) (some viruses are about this size) For λ ~ cm radio, you need 1 mm (1/10 of 1 cm) (size of a pinhead) For λ ~ 21 cm radio, you need 2 cm (a nickel) For λ ~ 1 m radio, you need 10 cm (length of a pen; height of a cell phone) For gamma rays of λ ~ 0.01 nm, you would need 0.001 nm (1 × 10−12 m) tolerance. (this is larger than a proton but smaller than an atom—gamma ray telescopes don’t use mirrors!)

15. Small knots of gas in the Crab Nebula are observed to be moving away from the position of the dead star at its center. Photographs taken in 1940 and 1990 show that knots originally located 150 arc seconds from the center have moved 8 to 9 arc seconds farther out. From this information, estimate the range of dates when the explosion might have occurred.

15. (AD 1052 to 1157). The range is roughly 833 to 938 years before 1990.

15. (cont.) Over the 50 year range of observations, the knots appear to have moved at 8 arcsec arcsec = 0.16 ,and 50 yr yr

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9 arcsec arcsec = 0.18 .

50 yr yr

The time of expansion from the center would be 150 arc seconds divided by that rate. speed =

distance ,rearranged to be tim e

tim e =

distance speed

150 arcseconds = 938yr,and arcseconds 0.16 yr 150 arcseconds t= = 833yr. arcseconds 0.18 yr

t=

Most astronomers believe that the event took place in AD 1054, Earth time.

Unit 29 11. How thick are the pages of this book in nanometers? Find this by measuring the thickness of many sheets and dividing. How much bigger is this in comparison with the precision that an optical telescope operating at 500 nm must maintain?

11. (About 0.03 mm; about 625 times larger). Start by measuring the book. The ~800 pages of this book are about 2.5 cm thick.

2.5 cm = 0.025 m = 0.025 × 109 nm = 2.5 × 107 nm 11. (cont.) So the relative thickness of each page, compared to the book of 800 pages, is 2.5 × 107 nm / 800 pages = 31,250 nm/page ~ 0.03 mm The mirrors in an optical telescope need to be built to a precision of about 1/10 of a wavelength of visible light, 500 nm, or 50 nm. One page is 625 times thicker than that. Or put in another way, to achieve good resolution, an optical telescope’s mirror surface must be made so smooth that the difference between any higher or lower spots on the surface would be less than 1/625 the thickness of a single piece of paper.

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12. Small binoculars have a 1-inch aperture, while a large commercial telescope has a 50-inch aperture. How many times fainter a star should you be able to see using such a large telescope in comparison to the binoculars?

12. (About 2500 times fainter). How faint an object you can see depends on the collecting area of the telescope. A 1-inch diameter binocular tube has an aperture area of: 2

æ1inchö ÷ ÷ p ççç = 0.79 square inches per tube, ÷ ÷ çè 2 ø ÷

while a 50-inch telescope has an area of 1.96 × 103 square inches. The telescope aperture is almost 2500 times greater than that of one binocular tube, so the telescope would allow the observation of an object 2500 times fainter.

13. Compare the collecting area of one of the 36 segments of the 10.4-m Gran Telescopio Canarias to the collecting area of the 2.4-m Hubble Space Telescope.

Which is larger? By what factor?

13. (HST is larger than one segment of GTC by a factor of 1.9).

2 æ10.4 m ÷ ö ç ÷ The total collecting area of the 10.4 m GTC is p çç = 84.9m 2.The area of one of the 36 ÷ ÷ çè 2 ø ÷ segments is then: 84.9 m2/36 = 2.36 m2.

2

æ2.4 m ÷ ö ÷ The collecting area of the Hubble Space telescope is p ççç = 4.5m 2 ÷ ÷ çè 2 ÷ ø The HST collecting area is larger than one of the segments by a factor of 1.9.

14. A group of stars is observed through several different filters, and some stars are seen to be brightest when observed through either the V (530 nm) or R (600 nm) filter. Estimate the range of temperatures for these stars as follows: a. Assuming a star is brightest in the middle of the R filter, use Wien’s law (Unit 23.4) to estimate its temperature. b. If a star is brighter through the V filter, use Wien’s law to estimate the range of temperatures it might have.

14. (a) 4833 K; (b) 5472 K. The R filter is at 600 nm. Using Wien’s Law, the temperature corresponding to a wavelength at maximum of 600 nm is

2.9´ 106 K ´ nm 2.9´ 106 K ´ nm TR filter = = = 4833K l m ax 600 nm 69

TV filter =

2.9´ 106 K ´ nm 2.9´ 106 K ´ nm = = 5472K l m ax 530nm

Therefore the star that peaks in the V band is about 5472 K/4822 K = 1.13 times hotter than one that peaks in the R band.

15. Suppose you used a telescope to detect starlight that is spread fairly evenly across 300to 900-nm wavelengths. If you put a V filter in front of your detector, about how many times longer will you have to observe to collect as many photons? (Estimate this approximately, based on Figure 29.9.)

15. (About 6.7 times longer). With no filter, we would detect 100% of the photons from 300 to 900 nm. The V filter only allows photons through from about 470 nm to 640 nm, with a fairly symmetric, Gaussian-looking shape. It peaks at about 90% transmission and drops to about half that (45%) at 500 and 600 nm. Since the shape is roughly symmetric, we can approximate the total transmission of the filter as a rectangle with ~ 90% transmission over 100 nm from 500 to 600 nm—effectively lopping off the outside parts and adding them to the top:

11. (cont.) This approximation is only appropriate because the photons coming in are evenly divided over wavelengths. We can then estimate the transmission of the filter as 90% × 100 nm.

The total area from 300 to 900 nm at 100% (all the photons!) can be expressed as 100% × 600 nm. So with the filter we only see (90% × 100 nm)/(100% × 600 nm) = 9/60 of the photons (about 15%). Therefore it will take 60/9 = 6.67 times as long to collect the same number of photons.

16. The power of the light reaching Earth from a “first magnitude” star is approximately 10−6 watts (joules per second) over each square meter of a telescope’s collecting area. Assuming that this light is composed of photons of wavelength 500 nm, about how many photons are arriving in a square meter each second? How many photons enter your eye each second if your pupil is 8 mm in diameter?

16. (5 × 108 photons per second.) Each photon has an energy

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æ mö Js)´ çç3´ 108 ÷ ÷ çè s÷ ø = 4´ 10- 19 J - 9 500´ 10 m - 34

E=

hc = l

(6.626´ 10

so this means the 106 W of star light hitting each 1 m2 of the mirror corresponds to:

10- 6 J/s = 2.5´ 1012 photonspersecond 4´ 10- 19 J Your eye has an area of πr2 = (0.008 m)2 = 2 × 104 m2, 5000 times smaller than the mirror.

Your eye would receive 2.5 × 1012 / 5000 = 5 × 108 photons per second.

17. At a radio wavelength of 30 cm, the Arecibo radio telescope can collect about 3 × 10−14 watts (joules per second) of radiation from the Crab Nebula. About how many photons is it collecting each second?

17. (4.53 × 1010 photons). Each photon has an energy of æ mö Js)´ çç3´ 108 ÷ ÷ çè s÷ ø = 6.62´ 10- 25 J - 2 30´ 10 m - 34

E=

hc = l

(6.626´ 10

17. (cont.) so Arecibo can collect:

3´ 10- 14 J/s = 4.53´ 1010 photonspersecond - 25 6.62´ 10 J

Unit 30 10. An image will appear upside down when we look through a simple refracting telescope, as can be seen in Figure 30.1. Trace the light through a similar diagram to show the orientation objects would have through a Cassegrain telescope. Assume the Cassegrain’s secondary mirror is flat.

10. The image would remain inverted. The diagram would look like Figure 30.5.

11. If a reflector has a primary mirror with a diameter of 28 cm and a secondary mirror with a diameter of 9.5 cm, then what fraction of the collecting area does the secondary mirror block? What diameter telescope would have the same collecting area if no part of its aperture was blocked?

11. (11.5% of the incoming light is blocked; d = 26.3 cm).

To find the blocked fraction, take the ratio of the areas:

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2

p D s2ec /4 4p (9.5 cm ) fraction blocked = = = 0.115 or11.5% 2 p D pr /4 4p (28 cm )2 i

pD 2 p = (282 - 9.52 ) = 545 cm 2 The useful collecting area is: A = 4 4 To find the equivalent diameter for an unblocked aperture:

545 cm 2 =

pD 2 so D = 4

4´ 545 cm 2 = 26.3 cm p

12. The front surface of your eye (the cornea) does most of the focusing of light entering your eye; the back wall of the eye is like a focal plane. Draw a diagram similar to Figure 30.1 to show how your eye focuses the light rays from a distant star straight in front of you. Now draw light rays for a nearer point of light (a few times farther away from your cornea than the cornea is from the back of the eye). The light rays will be bent by the cornea by about the same angle as the distant starlight, so do they now focus closer to or farther from the cornea?

12. For the same lens—the nearer object will focus behind where the distant star focuses.

13. A nearsighted person is able to focus on nearby objects, but not distant ones. Based on the previous problem, what does this tell you about the shape of the cornea for a nearsighted person? Will a nearsighted person focus red or blue colors better?

13. Since a nearsighted person can see close objects in focus, but not farther ones, the cornea is curved too strongly or the eye is too long, or both. The more curved the cornea, the more refraction, and the more separation of the near and far focus—the farther in front of the retina distant objects will focus (we know the person can see the near objects in focus at the retina).

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Since blue light is bent more than red light by refractive lenses, it will focus closer to the lens than red light—so the nearsighted person should be able to see distant red objects better.

14. After the Palomar 5-m telescope was completed, astronomers estimated that the mass of larger telescopes would increase as the (mirror diameter)3 or (mirror diameter)4 because of the greater mass of glass and support needed to keep the entire telescope stiff. If the 10-m Keck Telescopes followed this rule, how many times more massive would they have been than the Palomar telescope?

14. They would be 8 to 16 times more massive. A 10 m Keck scope is twice the diameter of the Palomar scope, and if mass increases as the 3rd or 4th power of diameter, that is 23 or 24, which is 8 and 16. Improvements in engineering and materials over time allowed the Keck telescopes to have a mass of just ~ 300 tons, a bit over ½ the mass of Palomar (540 tons).

15. The index of refraction measures the factor by which light slows down in a substance.

The index of refraction of water is 1.33. If you sent a pulse of light through water and a pulse of light through air (index of refraction of 1.00), how much longer would it take the light to travel 1 km through the water? (The speed of light in air is 299,800 km/sec.)

15. (1.1 × 10−6 seconds). The velocity of light in air is 2.998 × 105 km/s, and the speed of light in water is depends upon the index of refraction n:

cwater =

cair 2.998´ 105 km /s = = 2.254´ 105 km /s n 1.33

The time difference, ∆t, is Δ t=

æ ö 1km 1km 1 1 ÷ = 1km çç ÷ 5 5 ÷ çè2.254´ 10 km /s 2.998´ 10 km /sø cw ater cair

Δ t= 1.1´ 10- 6 seconds

Unit 31 10. If you are trying to resolve a double star with a 10-cm-diameter telescope, should a blue filter or a red filter make it easier to tell the stars apart? What resolution do you expect with each?

10. (Blue is better; 1 arcsecond resolution for blue, 1.5 arcseconds for red).

The smallest possible separation angle a telescope can resolve is proportional to the wavelength: arcsec > 0.025 λnm/Dcm. The shorter the wavelength, the smaller the minimum 73

possible separation angle and the better the resolution. We want the smallest possible separation angle, so we want the smallest wavelength possible—the blue filter.

The resolution expected from each filter depends upon their transmission wavelength; Blue filters transmit strongest at 420 nm, Red at 600 nm (see Unit 29.3). So

a blue >

0.025´ 420 nm 10 cm

~1arcseconds

and a red >

0.025´ 600nm ~1.5arcseconds 10 cm

11. When the 2.4-m-diameter Hubble Space Telescope (HST) is observing 500-nm visible light: a. What is the smallest angle it can resolve? b. Considering that the Moon is 384,400 km away, what is the smallest object HST could resolve on the Moon (see Unit 10.4)? Could the HST resolve the Apollo landing module?

11. (a) 1.44 × 10−5 °; (b) 97 meters, and no HST cannot resolve the Apollo lander.

The minimum angle that can be resolved at a wavelength of 500 nm is:

a arcsec >

0.025l nm 0.025´ 500nm = = 0.052 arcseconds D cm 240cm

11. (cont.) In degrees, this is

a = 0.052arcseconds´

1arcm inute

1degree ´ = 1.44´ 10- 5 degrees 60arcseconds 60arcm inutes

(b) Using the angular size formula from Unit 10.4, the smallest object that the HST can resolve is: - 5

(1.44´ 10 a l= d = (3.844´ 108 m ) 57.3° 57.3°

)° = 97 m

This size is much larger than the Apollo landing module. No, HST could not resolve it.

12. What is the theoretical resolution of your eye to 500-nm light in bright light when the pupil is only 3 mm across? How many times brighter would the light have to be to get the same number of photons entering the eye as when the pupil is dilated to 8 mm?

12. The minimum angle that the eye can resolve at a wavelength of 500 nm is:

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a arcsec >

0.025l nm 0.025´ 500nm = = 42arcseconds D cm 0.3cm

The collecting area of the eye is πr2, where r is the radius of the pupil. At 3 mm, the pupil’s collecting area is π × (1.5 mm)2 = 7.1 mm2. At 8 mm, the pupil’s collecting area is π × (4 mm)2 = 50 mm2. The 8 mm pupil has a collecting area that is 50/7 = 7.1 times larger than the 3 mm one, so you would need the source to be 7 times as bright to get the same number of photons at 3 mm.

13. Find the smallest resolvable angle of the 100-m-diameter Green Bank Telescope when it is observing 20-cm-wavelength radio waves. What is its resolution when observing at 3-mm wavelength?

13. (At 20 cm,  > 500 arcseconds; at 3 mm,  > 7.5 arcseconds).

The smallest resolvable angle possible for the Green Bank Telescope at a wavelength of 20 cm is: 8

0.025l nm 0.025´ (2´ 10 nm ) a arcsec > = = 500arcseconds D cm 10,000cm 13. (cont.) The smallest resolvable angle possible for the Green Bank Telescope at a wavelength of 3 mm is: 6 0.025l nm 0.025´ (3´ 10 nm ) a arcsec > = = 7.5arcseconds D cm 10,000cm

Smaller wavelengths of light provide better resolution.

14. Suppose that for an interferometer to work properly, the electromagnetic waves must arrive “in sync” to 1% of the period of the wave. What time difference is this for a 20-cmwavelength radio wave? What time difference is this for a 500-nm visible wavelength wave?

(Recall that the period of the wave is the inverse of its frequency ν.)

14. (radio: 6.7 × 10–12 seconds; visible: 1.7 × 10–17 s).

For a 20 cm wave, the frequency is:

n=

and since the period =

c 3´ 1010 cm /s = = 1.5´ 109s- 1 = 1.5´ 109 H z l 20cm

1 , 1% of the period is 6.7 × 10–12 s. frequency

For 500 nm, the frequency is 6.0 × 1014 s and 1% of the period is 1.7 × 10–17 s.

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This calculation shows why building a radio-telescope interferometer was possible long before building an optical interferometer. Synchronizing radio waves in time collected from multiple telescopes is orders of magnitude “easier” compared to optical light.

15. The Very Large Array (VLA) consists of 27 radio dishes, each 25 m in diameter, connected together as an interferometer. Compare the resolution of a single VLA dish to the resolution of the VLA working as an interferometer, assuming its maximum baseline of 36 km and observing 7-mm microwave signals.

15. (The interferometer resolution is 1400 times better than that of a single dish).

The smallest resolvable angle possible for one 25 m dish from the VLA at a wavelength of 7 mm is: 0.025´ (7´ 106 nm ) 0.025l nm a arcsec > = = 70 arcseconds D cm 2500 cm

The smallest resolvable angle possible for the entire VLA (with a baseline of 36 km) at a wavelength of 7 mm is: 6 0.025l nm 0.025´ (7´ 10 nm ) a arcsec > = = 0.05arcseconds D cm 3.6´ 106 cm

The interferometer’s resolution is 70/0.05 = 1400 times better than that of single dish.

(This same result can be found by found by taking the ratio of effective diameters, 3.6 × 106 cm/2500 cm).

16. Suppose you wanted to build a telescope capable of resolving a planet the size of Earth that is 10 light-years distant. a. Calculate the angular diameter of Earth if it were that far away (see Unit 10.4). To see any detail on the surface, we would want to see an angular size about 10 times smaller than this. b. Calculate the diameter of a telescope that would be needed to resolve this angular size if observing at 500 nm. Is this feasible?

16. (a) 7.7 × 10−9 °; (b) 4.5 km, which is not feasible. a) We can start with the small-angle formula from Unit 10.4, and rearrange it to find the angular diameter of Earth (linear diameter of 2 × (6378 × 103 m)) from a distance of 10 light-years: 2´ (6378´ 103 m ) a l l= d soa = 57.3° = 57.3° = 7.7´ 10- 9 ° 15 57.3° d 10 ly´ 9.46´ 10 m /ly

For part b, we need this answer in arcseconds, so: 76

a = 7.7´ 10- 9 °´

60 arcm inutes 60 arcseconds ´ = 2.8´ 10- 5 arcseconds 1 degree 1 arcm inute

16. (cont.) (b) We can rearrange the formula for the minimum resolvable angle to find the necessary diameter of the telescope:

a arcsec >

D cm >

0.025l nm rearranges to D cm

0.025l nm 0.025´ 500nm = = 4.5´ 105 cm - 5 a arcsec 2.8´ 10 arcseconds

This is 4.5 km. To resolve details on the surface we’d have α ten times smaller, and D ten times bigger, so we would need a 45 km diameter optical telescope. This is not feasible.

Unit 32 11. Suppose that in the atmosphere there are pockets of air of different density that are 10 meters across. If winds move these pockets of air at a speed of 30 kph, for how long is an image through a telescope likely to remain stable? How would having many of these pockets along your line of sight change your time estimate?

11. (About 1 second). At a speed of 30 kph (= 8.3 m/s), a 10 m pocket of air will move a distance equal to its diameter in a time equal to (10 m)/(8.3 m/s) = 1.2 s. An image would be expected to remain stable for this period of time. Having many such pockets along your line of sight would reduce this time estimate significantly, as each pocket would cross along your line of sight at a random time during this interval.

12. A telescope in low orbit around Earth takes about 90 minutes to circle Earth. If the telescope remained oriented in the same way relative to Earth, by how many arc seconds would its direction change relative to the stars during a 1-second exposure? (Actually, space telescopes are stabilized by gyroscopes that maintain their orientation relative to the stars.)

12. (240 arcseconds or 4 arcminutes). An entire revolution around the Earth would cause the telescope to rotate 360° in 90 minutes, or

60 arcm inutes 60 arcseconds ´ = 1,296,000 arcseconds.

1 degree 1 arcm inute 12. (cont.) The rotation takes 90 min × 60 s/min = 5400 s of time, so the spacecraft turns at 1,296,000 arc sec/5400 s = 240 arc sec/s. In the 1-second exposure, the stars would then drift 240 arcseconds, or about 4 arc minutes. This would result in a very blurry image!

360 degrees ´

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13. If the refraction of the atmosphere causes the Sun’s position to shift up by 0.5° at the horizon, calculate how long the day would be on the equinox for someone living at the equator. Would that same day be longer or shorter for someone living at latitude 45°? Explain using a diagram.

13. It would be about 12 hours, 4 minutes at the equator and longer at 45 degrees. The Sun rises straight up at the equator, and in theory there should be exactly 12 hours of potential sunshine. The refractive effect adds extra Sun at sunrise and sunset because you can see the Sun up to 0.5° below the horizon. The Earth turns 15°/h, so the extra total 1° of Sunlight amounts to 1° / (15°/60 min) = 4 min of daylight, and the day at the equator would be 12 h 4 minutes.

At higher latitudes, such as 45 degrees North, the Sun rises at an angle less than 90 degrees relative to the horizon, so it spends a longer time within 0.5° below the horizon. (Note: as in the chapter discussion, this answer ignores the size of the Sun—about 0.5°, although that gets distorted when rising and setting—which could further increase the length of the day by up to the time to turn an extra 0.25° at sunrise and sunset).

14. Based on the graph in Figure 32.1, estimate the maximum energy and frequency for a photon that can be detected by a ground-based telescope.

14. (6.6 × 10−19 J and 1015 Hz).

From figure 32.1 we can estimate that any photons with a wavelength smaller than about 300 nm will be blocked by the atmosphere. The energy of such a photon is: æ mö Js)´ çç2.99´ 108 ÷ ÷ ÷ çè sø = 6.6´ 10- 19 J - 9 300´ 10 m - 34

E=

hc = l

(6.626´ 10

The frequency for this photon is ν = c/λ = (3.0 × 108 m/s)/(300 × 109 m) = 1015 Hz.

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15. The orbiting Chandra X-ray telescope, observing at about 0.1 nm, can collect about 2 × 10−13 watts (joules per second) of radiation from the Crab Nebula. About how many photons is it collecting each second?

15. (Chandra collects 2 × 1013 J/s of 0.1 nm photons). Each photon has energy æ mö Js)´ çç2.99´ 108 ÷ ÷ ÷ çè sø = 2´ 10- 15 J 0.1´ 10- 9 m

- 34

E=

hc = l

(6.62´ 10

So it collects (2 × 1013 J/s) / (2 × 1015 J) = 100 photons per second.

(The Crab is quite bright, many sources observed by Chandra only provide a few X-ray photons every few seconds!)

16. If the cost of electricity is $0.10 per hour to run a 1000-watt lightbulb, estimate the total annual cost for a town running 200 streetlights, each consuming 500 watts. If a reflector is placed over each lamp so that twice as much light shines downward, how much money can the town save each year by switching to lower-wattage bulbs?

16. Assuming $0.10/h/1000 W means it costs $0.05/h for a 500 W light bulb, then running 200 lights will cost 200 × $0.05/h = $10/h. If the reflector is placed on the lights, the town can get an additional 250 W of useful light from a 500W bulb. They could therefore switch to 250 W bulbs, get the same amount of useful light as a non-reflected 500 W bulb would provide, and cut their energy use and energy costs in half.

16. (cont.) The annual cost for the 500 W bulbs, if we assume 12 hours of operation per day on average, is 365 d × 12 h/d × $10/h = $43,800. If they change to the 250 W bulbs, it will only cost $21,900, saving the town $21,900, quite a good thing for the town budget.

Unit 33 11. Suppose you are using a telescope with a 100-mm (4-inch) aperture diameter and a focal length of 1000 mm (40 inches). a. What is the f-ratio? b. What focal length eyepiece will give you a magnification factor of 100?

11. (a) f-ratio = 10; (b) a focal length of 1 cm would be required for a magnification of 100. a) From the formula for the focal length, we find: focallength = (aperture diam eter)´ (f-ratio),so 1000 m m focallength f-ratio = = = 10. aperture diam eter 100 m m

So the f-ratio is 10.

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b) We need a magnification of 100, and since

focallength ofobjective rearranges to focallength ofeyepiece focallength ofobjective 1000 m m focallength ofeyepiece = = = 10 m m m agnification 100 M agnification =

For a magnification of 100 you would need an eyepiece with a focal length of 10 mm = 1 cm.

12. Suppose you are using a 20-cm (8-inch) f∕10 Schmidt-Cassegrain telescope. You have a set of eyepieces with labeled focal lengths of 6, 12, 20, and 40 mm. What magnification will each eyepiece give you? Why will images probably look a bit blurry with the 6-mm eyepiece?

12. With a 20 cm, f/10 objective, the focal length is

focallength = aperture diam eter´ fratio = 20cm ´ 10 = 200cm = 2000m m M agnification =

focallength ofobjective focallength ofeyepiece The 6 mm eyepiece will produce blurry images because at high magnification you will be magnifying distortions in the atmosphere—you will be magnifying atmospheric “seeing.”

(Without an atmosphere, you’d get clear, high magnification images).

13. Suppose you have a pair of 8 × 60 binoculars, meaning they have a magnification of 8× and an aperture 60 mm in diameter. a. Compared to your eyes with your pupils dilated to 8 mm, how many times more light do the binoculars collect? b. If the binoculars have an f-ratio of 4, what is their focal length? What is the focal length of the eyepieces?

13. (a) 56 times more light; (b) f for the binoculars = 240 mm and for the eyepieces, 30 mm. a) The collecting area of a device is πr2, where r is the radius of the aperture. At 8 mm, the pupil’s collecting area is: π × (4 mm)2 = 50 mm2.

The binocular’s collecting area is: π × (30 mm)2 = 2.8 × 103 mm2.

The binoculars have a collecting area that is

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(2.8 × 103 mm2)/(50 mm2) = 56 times as much as your eyes. b) For an f-ratio of 4, the binoculars have a focal length of 60 mm × 4 = 240 mm. If the binoculars have a magnification factor of 8, focallength ofeyepiece =

focallength ofobjective 240 m m = = 30 m m . m agnification 8

14. Similar to an eye, cameras vary f-ratio by changing the aperture. Suppose you have a camera lens labeled as having a focal length of 50 mm with f-ratios varying from f∕1.4 to f∕30. a. What is the range of aperture diameters for this camera? b. When the f-ratio is small, the “depth of field” (range of distances over which objects remain in focus) becomes smaller.

Why would the focus be likely to remain sharper when the aperture is smaller?

14. (a) 1.7 mm for f/1.4 to 36 mm for f/30; (b) light rays are more nearly parallel. (a) From the formula for the focal length, we find:

focallength = (aperture diam eter)´ (f-ratio),so focallength 50 m m aperture diam eterm axim um = = = 36 m m ,and f-ratio 1.4

focallength 50 m m = = 1.7m m . f-ratio 30

aperture diam eterm inim um =

(b) As f-ratio = focal length/aperture, and better depth of field results from a larger f-ratio, you get better depth of field—more objects in focus—with a smaller aperture and the resulting higher f-ratio. The focus remains sharper over a bigger depth of field when the aperture is smaller because the light rays entering the camera are more nearly parallel than when the aperture is wider.

15. If a camera is pointed at a star on the celestial equator and the shutter is left open for 20 minutes, what length of a star trail would show up on the resulting picture? For comparison, the angular size of the Moon is 0.5°. How many moons does the star move?

15. (The trail will be equivalent to 10 full moons).

The Earth rotates 360° every 24 hours. In 20 minutes, 1/3 of an hour, a star will make trail of angular length: (360°/24 hr) × (1/3 hr) = 5° long.

This angular distance is equivalent to 10 moons.

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Solutions to End of the Unit Problems

Part III: The Solar System

Unit 34 11. By what factor would the Sun have to shrink to be the size of a large beach ball, 1 meter in diameter? In Appendix Tables 5 and 6, look up the sizes and distances from the Sun of Jupiter, Neptune, Earth, Mercury, Pluto, and Ceres. a. What diameter would these six objects have if scaled down by the same factor? b. What would their masses be if their density stayed the same?

11. The Sun’s diameter is 1.39 × 106 km across, or 1.39 × 109 m, so it would need to be shrunk by a factor of 1.4 billion times to be the size of a 1m diameter beach ball. For simplicity, values in the table are given in km(real) and meters(model). (a) The model diameters are in bold in the table. (b) Mass = density × volume. V = 4/3π R3, and 1000 L = 1 m3. The masses for the models are in bold in the table. Shrinking the sun to a 1-m beach ball with the same density would require the ball to have a mass of about ¾ of a kilogram, or about 1.5 lbs. With this same scale, Earth would be about the size of a pea – with a mass of a bit over 2 grams.

Body

Radius (km)

Diameter (km)

Sun Jupiter Neptune Earth Mercury Pluto Ceres

6.96E + 05 71492 24764 6378 2440 1160 487

1392000 142984 49528 12756 4880 2320 974

(a) Model Diam (m)

Density kg/L

Volume (L)

(b) Model Masses (kg)

1.00000 0.10272 0.03558 0.00916 0.00351 0.00167 0.00070

1.33 1.64 5.52 5.43 2.00 2.08

0.567470448 0.02358486 0.000402925 2.25601E-05 2.42407E-06 1.79373E-07

0.754735696 0.03867917 0.002224147 0.000122501 4.84814E-06 3.73096E-07

12. What would be the distance from the Sun for each of the bodies listed in Problem 11, scaled down by the same factor that would make the Sun 1 meter in diameter?

12. Scaled distances: Jupiter = 559 m, Neptune = 3.23 km, Earth = 107 m, Mercury = 41 m, Pluto = 4.24 km, Ceres = 298 m. (The scale shrinks the real distances by 1.39 billion times; see Q11).

Body

Distance (km)

Distance (m)

Scale Distance (m)

Jupiter Neptune Earth Mercury Pluto Ceres

7.79E + 08 4.50E + 09 1.50E + 08 5.76E + 07 5.91E + 09 4.15E + 08

7.79E + 11 4.50E + 12 1.50E + 11 5.76E + 10 5.91E + 12 4.15E + 11

559 3229 107 41 4243 298

13. Saturn has a mass of 5.68 × 1026 kg and a radius of 60,300 km. Calculate Saturn’s density from these values. According to Archimedes’ principle, for something to float, it must have a density less than what it is floating on. Would Saturn float on water, which has a density of 1 kg/L? 82

13. (0.62 kg/L). We find the density (ρ) by dividing the object’s mass (M) by its volume (V).

V=

So r =

3 4 3 4 p R = p R (60,300´ 103 m ) = 9.18´ 1023 m 3 = 9.18´ 1026 L 3 3

M 5.68´ 1026 kg kg = = 0.62 26 V L 9.18´ 10 L

Saturn is less dense than water, so yes, it would float on water if a big enough bathtub could be found. (Of course, when it gets out of the tub it leaves a ring.) Note that the density calculated here uses the equatorial radius and Saturn has an oblate spheroid shape.

Consequentially, the density found by this simplified calculation is slightly lower than the value reported in Appendix Table 5.

14. Use the data in Appendix Table 5, “Physical Properties of the Planets,” to list the planets in order according to several different parameters: (i) smallest to largest radius; (ii) smallest to largest mass; (iii) most to least dense. Note that the order of some planets is changed in each case. Can you draw any approximate conclusions from these results? For example, are smaller planets always more dense? What other factors might explain why some planets are out of order with the overall trends you find?

14. The lists for i, ii, iii are provided below; some trends to note are on the following page.

Radius (Sm->Lg) Ceres Pluto Eris Mercury Mars Venus Earth Neptune Uranus Saturn Jupiter

Mass (Sm->Lg) Ceres Pluto Eris Mercury Mars Venus Earth Uranus Neptune Saturn Jupiter

Density (Lg->Sm) Earth Mercury Venus Mars Ceres Eris Pluto Neptune Jupiter Uranus Saturn

Trends: Smaller planets generally have smaller mass; the trend is bucked by Neptune, which is smaller but more massive than Uranus (but not by much). In density, the determining factor is not radius but composition—the gas giants are the least dense, the terrestrial planets the most dense. Mercury is out of place even among the rocky planets for radius and density because it is mostly made of iron. (Students may comment on other trends and planets out of order).

15. You can calculate the acceleration due to gravity at the surface of a planet using the formula: g = GM∕R2 where M and R are the mass and radius of the planet, and G is the gravitational constant (see Unit 16). How much do you weigh on Earth? Determine how much you would weigh

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on Jupiter, Mars, and Ceres. (Hint: Because your mass would not change, you can simply compare the values of g for the other planets with Earth’s 9.8 m/sec2.)

15. F = ma, and g = acceleration, so on Earth your weight is your mass times g, e.g., for someone 75 kg in mass , their weight is 75 kg × 9.81 m/s2 = 735 N (165 lbs). The easiest way is to compare weights for other planets is to use ratios so that “G” cancels (unless exercises in scientific notation are desirable). gJupiter gEarth

=

2

2 2 ö M Jupiter æ ö REar REarth ÷ 318 æ 1 ÷ 318 ç th ç ÷ ç ´ = ´ ç = ´ ç ÷= ÷ 2 ÷ ÷ ç M Earth çè RJupiter ÷ 1 è11.19ø 125.2161 RJupiter G M Earth ø

G M Jupiter

15. (cont.) gJupiter = 2.54 gEarth

gM ars 0.107 0.107 = = = 0.38 gEarth 0.5322 0.283

gCeres 0.00016 = = 0.0277 gEarth 0.0762 You would weigh 2.54 times as much on Jupiter, 38% as much on Mars, and 2.8% as much on Ceres as on Earth. For someone with a mass of 75 kg, their weight on Jupiter would then be 1867 N (419 lbs); on Mars 138 N (63 lbs); on Ceres 10 N (4.6 lbs).

Unit 35 10. Use the data from Figure 35.1 to draw a graph showing the ratio of argon atoms to potassium-40 atoms that a rock will contain over a 5.12-billion-year history. Draw a curve through the points. From your curve, estimate the age of a rock that contains half as many daughter argon atoms as the potassium isotope from which they formed. Estimate the age of a rock that contains an equal amount of the two.

10. (About 4.3 billion years old.)

The ratio of argon atoms to potassium-40 atoms for a rock over 5.12-billion years.

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#Argon/#Potassium

Argon/Potassium Ratio

Time (Billions of Years)

10. (cont.) The ratio of argon to potassium is about ½ at 3.3 billion years. A rock with an equal amount of Potassium and Argon (ratio = 1) would be about 4.3 billion years old.

11. The common form of uranium (238 U) has a half-life of 4.5 billion years. It decays through a sequence of unstable isotopes to lead (206 Pb), releasing a number of helium nuclei in the process. (Radioactive decay is the source of most of the helium in Earth’s atmosphere today.)

Find the ratio of the number of 238 U to 206 Pb atoms you would expect to find for rocks that last melted 1, 2, 3, and 4 billion years ago.

11. 1 billion years corresponds to N=

tim e interval 1 billion years = = 0.22 halflives halflife 4.5 billion years

And then 2, 3, and 4 billion years have N = 0.44, 0.67, 0.89 half lives, respectively. The fraction of 238U remaining after these times, compared to the original amount (given by (1/2) N) is 0.86, 0.74, 0.63, and 0.54, respectively. So, if one considers 100 238U atoms initially, after 1 billion years there will be 86 238U atoms and 14 206Pb atoms, for a ratio of 86/14 = 6.1. The ratios for the other times are 2.8, 1.7, and 1.2.

12. If we estimate that an average planetesimal was 10 km in diameter and had a density of 3 kg∕L, approximately how many planetesimals merged together to form Earth?

12. (About 3.7 × 109). Using the information from the problem, the mass of a single planetesimal is: M = rV = r

kg 1000L 4 3 ´ ´ pR L 3 m3

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= 3

kg 1000 L 4 ´ ´ p (5´ 103 m )3 = 1.6´ 1015 kg 3 L 3 m

From Appendix Table 5, the mass of the Earth is 5.97 × 10 24 kg. The number of planetesimals would then be: 5.97´ 1024 kg N= = 3.7´ 109 1.6´ 1015 kg 13. Silicate molecules can contain different numbers of oxygen atoms per silicon atom.

Nearly all of the oxygen in Earth is locked up in silicates. Based on the data in Table 37.1, calculate how many oxygen atoms there are per silicon atom. (Note that silicon has an atomic mass of 28, while oxygen is 16.)

13. (~3.5 Oxygen atoms per Silicon). Using the data in Table 37.1, the mass ratio of oxygen to silicon in the Earth is 30.1% to 15.1%. We need to find the ratio of the numbers of atoms. The number of atoms will be equal to the total amount of material (the percentage by mass of the Earth times the mass of the Earth) divided by the mass of a single atom.

N um ber=

(Percentby m ass)(M ass ofEarth) (A tom ic w eight)

(Note that the mass of the Earth cancels out when we calculate the ratio below. Oxygen atoms have an atomic mass of 16, and silicon of 28, so the average ratio of oxygen atoms per silicon atom is: 30.1% N um berof0 atom s Ratio = = 16 = 3.5.

N um berofSiatom s 15.1% 28

Common silicates must include 3 or 4 oxygen atoms to each silicon atom. And this is indeed the case. Check out https://en.wikipedia.org/wiki/Silicate_mineral for a table of silicates common to Earth’s crust showing the ratios of Oxygen and Silicon.

14. Suppose that the Jovian planets have the same overall composition as the Sun given in Table 35.1, so they are 0.14% iron. a. What fraction of their mass would be silicates, assuming that oxygen atoms combined 3-to1 with silicon? b. What would be the mass of rock (silicates) and iron in Jupiter compared to Earth’s mass? c. What would be the mass of rock and iron in Uranus? What terrestrial planet’s mass is this closest to?

14. (a) 0.16% silicates; (b) 5.7 × 1024 kg; (c) 2.6 × 1023 kg. (a) The Jovian mass fraction for silicates (assuming SiO3) would be:

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0.06% + (3 × (0.06%) × 16/28) = 0.16%. (b) The Jovian mass fraction for silicates plus iron is 0.14% + 0.16% = 0.30%. Given Jupiter's mass of 1.90 × 1027 kg, this is a mass of 5.7 × 1024 kg, which is comparable to the mass of Earth (5.97 × 1024 kg). (c) Given Uranus's mass of 8.68 × 1025 kg, a mass fraction of 0.30% equals 2.6 × 1023 kg, which is similar to the mass of Mercury (3.3 × 1023 kg).

Note this answer neglects including magnesium or other heavy elements in the silicates, and also the solar mass fractions may not be entirely appropriate for the gas giants, so it is only a very rough estimate of the mass of rock in these planets.

15. The conservation of angular momentum (Unit 20) says that a contracting gas cloud must spin faster as it gets smaller. a. Suppose the material orbiting today at 1 AU began 1 ly away. By what factor has its orbital radius changed? b. Material orbiting at 1 AU today (like Earth) is moving at 30 km∕sec. According to the conservation of angular momentum, how fast was its rotation speed when it was at a radius of 1 ly?

15. (a) 63,240 times smaller; (b) About 48 cm/s. (a)There are 63,240 AU in a ly, so contracting to 1 AU means the radius decreased by a factor of 63,240 compared to its original size. (b)Angular Momentum L = MVR. L is conserved (constant) so if M does not change, the change in radius is accompanied by an inversely proportional change in V. If the current orbital velocity is 30 km/sec, the velocity at one light year was (Rthen/Rnow) times less.

30 km/s / (63,240) = 4.7 × 10−4 km/s or about 48 cm/s.

This is just about 1 mile per hour. So that material would not have been moving very fast before it started collapsing inwards, speeding up as it did.

16. Planetary scientists estimate that a planet can hold on to a gas for the age of the Solar System if the average speed of molecules in the gas is less than one-sixth the escape velocity of the planet. The average speed of a molecule depends on the gas temperature and the mass of the molecule according to the following equation: Vgas = 157 m /s

temperature molecule ¢s mass

. The molecular mass in this formula is found by adding up the

atomic masses of its constituents, so hydrogen molecules (H2) have mass 2, and nitrogen molecules (N2) have mass 28. a. What is the average speed of H2 in Earth’s atmosphere if the gas temperature is 300 K? Earth’s escape velocity is 11,200 m∕sec. Can Earth hold on to H2? b. What is the average speed of N2 in Earth’s atmosphere? Can Earth maintain a nitrogen atmosphere?

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c. What is the average speed of H2 on Jupiter, where the temperature is 160 K? Jupiter’s escape velocity is 59,500 m∕sec. Can Jupiter hold on to a hydrogen atmosphere? d. If Jupiter were placed at Mercury’s distance from the Sun, where the temperature would rise to about 450 K, would it be able to hold on to its hydrogen atmosphere?

16. Substitute values into the equation given. a) Average speed of H2 in Earth’s atmosphere: 1923 m/s; This is more than Vesc/6 so the Earth cannot hold onto H2.

m tem p m 300 m = 157 = 1923 s m m olecule s 2 s Vesc/6 = 11,200 m/s /6 = 1867 m/s.

Vgas = 157

b) Average speed of N2 in Earth’s atmosphere: 514 m/s; This is less than Vesc/6 = 1867 m/s so the Earth can hold onto N2.

Vgas = 157

m tem p m 300 m = 157 = 514 s m m olecule s 28 s

c) Average speed of H2 in Jupiter’s atmosphere: Vgas = 1404 m/s; Vesc/6 = 59,500 m/s/6 ~ 10,000 m/s, so Jupiter can hold onto an H2 atmosphere.

m tem p m 160 m = 157 = 1404 s m m olecule s 2 s 16. (cont.) d) Average speed of H2 in Jupiter's atmosphere if it were at the distance of Mercury from the Sun: 2355 m/s; This is still less than Vesc/6 ~ 10,000 m/s for Jupiter, so it would still be able to hold onto the H2 atmosphere.

Vgas = 157

Vgas = 157

m tem p m 450 m = 157 = 2355 s m m olecule s 2 s

Unit 36 10. Suppose a Jupiter-size exoplanet (radius 71,500 km) passed in front of a Sun-size star (radius 696,000 km). What percentage of the star’s light would be blocked by the exoplanet?

10. (1.06%).

2

2 2 æ 71500 km ÷ ö p RPl RPl anet anet fraction = = = çç = 0.0106 = 1.06% ÷ 2 2 ÷ p RSun RSun çè696000 km ø

11. The radial velocity method for finding planets can be used to see how fast the Sun moves in response to a planet orbiting it.

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a. Using Kepler’s third law (P2 = a3, for P in years and a in AU), find the period of the orbit and its circumference if the semimajor axis is 0.05 AU, 0.5 AU, or 5 AU. b. Calculate how fast a planet would be orbiting the Sun in meters per second at each of these three distances. c. The Sun also “orbits” the point of balance between the Sun and the planet (although this point may be inside the Sun). According to Newton’s third law, the Sun’s speed around this point must be smaller than an orbiting planet’s in proportion to the ratio of their masses.

What is the Sun’s speed in each case above if the orbiting planet has Jupiter’s mass? Earth’s mass?

11. (see the table below). a) P2 = a3; P = a3/2; Circumference = 2 πa; 1 y = 3.16 × 107 s (Not all decimals shown in intermediate steps) For semimajor axis of 0.05 AU: P = (0.05)3/2 = 0.011 y = 3.53 × 105 s C = 2π (0.05 AU) = 0.314 AU = 4.7 × 1010 m For semimajor axis of 0.5 AU: P = (0.5)3/2 = 0.35 y = 1.106 × 105 s C = 2π (0.5 AU) = 3.14 AU = 4.7 × 1011 m For semimajor axis of 05 AU: P = (5)3/2 = 11.2 y = 3.54 × 108 s C = 2π (5 AU) = 31.4 AU = 4.7 × 1012 m b) The speeds are calculated from Velocity = distance/time = Circumference/Period; For a = 0.05 AU, V = C/P = 4.7 × 1010 m / (3.53 × 105 s) = 1.33 × 105 m/s.

The other values are in the table below. c) Ratio of masses: Sun/Jupiter = 1.99 × 1030 kg / (1.9 × 1027 kg) = 1047 Sun/Earth = 1.99 × 1030 kg / (5.97 × 1024 kg) = 333,333 The Sun will move 1/1047 as fast if the planet had the mass of Jupiter, and 1/333,333 as fast if it had the mass of Earth. Dividing the above velocities we get solar velocities: a (AU)

(b) V planet m/s

0.05 0.5 5

1.33E + 05 4.22E + 04 1.33E + 04

(c) - Jupiter mass Planet V Sun m/s

(c) - Earth mass planet V Sun m/s

127.395 40.286 12.739

0.400 0.127 0.040

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12. When planets orbit a star, both orbit their common “center of mass” (Unit 17), which lies at a point between them closer to the center of the more massive object in proportion to the ratio of their masses. (If the planet is one-tenth as massive as the star, the point will be 10× farther from the planet than from the star.) a. By how many AU does the Sun shift in each of the cases in Problem 11? b. If viewed from 30 light-years away, how large will its angular shift be? (Refer to the angular size formula in Unit 10.)

12. (a) see the table on the next page; (b) 1/1000th of an arc second.

How far does the Sun shift? The separation of the Sun and the planet is the orbital radius = semi-major axis = a = 0.05, 0.5 and 5 AU (as above). The center of mass lies somewhere along this distance between the Sun and planet, in proportion to their masses.

If the distance from the Sun to the center of the orbit is x, then we can express this as: MSun × x = Mplanet × (a − x) (See Unit 57). x (MSun + Mplanet) = Mplanet × a x = a × Mplanet /(MSun + Mplanet) = a × 1/(MSun/Mplanet + 1) The TOTAL shift of the Sun from one side of the orbit to the other would be 2x. b) The angular shift can be found from using the angular size – distance relation to solve for the linear size of the shift, in terms of the same units used for distance: L = d × A/57.3°, so A = L / d × 57.3° = (x) / d × 57.3° = (x) / (30 ly) × 57.3° = (x) / (63,240 × 30 AU) × 57.3° Again, note that the TOTAL shift would be twice this angle, as the Sun moves from one side to the other of the center of the orbit. The table on the next page shows both values:

12. (cont.)

A (AU) 0.05 0.5 5

Jupiter-like planet x (AU) 4.77E-05 4.77E-04 4.77E-03 Earth-like planet

Angular size of x (degrees) 1.44E-09 1.44E-08 1.44E-07

Angular size of x (arc sec) 5.19E-06 5.19E-05 5.19E-04

Total Shift AU (2x) 9.54E-05 9.54E-04 9.54E-03

Total Angular Shift (2x) (arc sec) 1.04E-05 1.04E-04 1.04E-03

0.05 0.5 5

1.5E-07 1.5E-06 1.5E-05

4.53035E-12 4.53035E-11 4.53035E-10

1.63093E-08 1.63093E-07 1.63093E-06

3.00E-07 3.00E-06 3.00E-05

3.26E-08 3.26E-07 3.26E-06

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The total apparent angular shift of the Sun seen from 30 ly away, resulting from Jupiter’s orbit at 5 AU, is about 1/1000th of an arc second, which is still smaller than what the Hubble Space Telescope can resolve. And the angular shift of the Sun resulting from Earth’s orbit is a thousand times smaller still. So beyond 30 ly away, observers would need a much bigger telescope to detect the Sun’s angular shift that results from our solar system’s planets.

13. Gliese 581 is a star with a mass 31% of the Sun’s mass, and it generates only about 1.3% as much light as the Sun. Interestingly the star may be host to upwards of six exoplanets (although only three have been confirmed and another is probable). a. Use Newton’s version of Kepler’s third law to find the semimajor axis for exoplanets with orbital periods of 3.15, 5.37, 12.9, and 66.6 days. b. Compare the brightness of light received from Gliese 581 at the distances found for the four exoplanets to the brightness of light received at Earth from the Sun. (Use the inverse-square law—Unit 21.2.)

13. (See table on the next page for semimajor axes and relative brightness for each exoplanet.) a) Newton's version of Kepler's third law is (MA + MB) = a3/P2 (Unit 17). Assuming that the planet is very small mass compared to the star, MGliese 581 = a3/P2. For the exoplanet with a period of 3.15 days = 0.00863 years, which leads to a semi-major axis of: a = (MGliese 581 × P2)1/3 = (0.31 × (0.00863)2)1/3 = 0.028 AU. 13. (cont.) Following a similar calculation for the other exoplanet, the values are shown in the table below. As expected, the exoplanets are very close to their star to have such short orbital periods. (b) The brightness is proportional to luminosity and inversely proportional to distance squared.

Gliese 581 is 0.013 (1.3%) as luminous as the Sun, so the ratio of Gliese’s brightness to the Sun’s brightness at a distance of 0.028 AU is 0.013/(0.028)2 = 17. That is, the inner exoplanet sees its parent star as being 17x brighter than the Sun is from Earth. A similar calculation for the other exoplanets give ratios of 7.7, 2.4, and 0.27 times brighter. So even though Gliese 581 is smaller and much dimmer than our Sun, because its planets orbit so closely, the three closest still would receive more light than we do on Earth from our star.

Exoplanet a b c d

Period (dy) 3.15 5.37 12.9 66.6

Semi-major axis (AU) 0.028 0.041 0.073 0.22

Brightness of Gliese 581 for this planet (relative to Sunlight at Earth)

17 times brighter 7.7 times brighter 2.4 times brighter .27 times as bright

14. If Neptune originally formed at 20 AU and then migrated to its present distance of 30 AU, how much was Neptune’s orbital speed reduced during the migration?

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14. We can use Conservation of Momentum to calculate the reduction in speed during the migration. Conservation of Momentum says that:

M initVinitRinit = M finalVfinalRfinal Here, Minit is the initial mass of the planet, Vinit is its initial velocity, and Rinit is its initial distance from the center of the solar system. Mfinal, Vfinal, and Rfinal are the final mass, velocity, and distance. The mass of Neptune (presumably) did not change during its migration, so

VinitRinit = VfinalRfinal

Vfinal =

Rinit 20 A U Vinit = V = 0.667 Vinitial Rfinal 30 A U initial

Neptune’s final orbital speed is 67% of its original speed.

Unit 37 11. P waves travel at about 6 km/sec. a. At this speed, how long would it take P waves to travel to the opposite side of Earth from the center of an earthquake? b. In the iron core, the speed of P waves is approximately two times faster. Sketch a picture of the interior of Earth and estimate how much sooner the wave would reach the opposite side of Earth if the wave traveled this much faster while in the core.

11. (a) 25.5 hours; (b) ~19.2 hours. a) The equatorial diameter of Earth is 2 × 6378 = 12756 km. Travel time is distance divided by speed: t = 12756 km/(500 km/hr) or 25.5 hours. b) A scale sketch reveals the core (liquid and solid) is about half the trip through the center of the Earth. If the P-waves travel twice as fast through the core, then the total time would be ~ 25.5/2 = 12.75 hours for the mantle plus an additional 12.75/2 = 6.4 h for the core. The trip would take about 75% as long as calculated in part (a).

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12. Draw a scale model of Earth (radius 6357 km) and the liquid core (radius 3500 km). How large a region is “shadowed” from S waves? Use a protractor to estimate how many degrees away from the site of Earthquake S waves are still detected.

12. From the scale diagram below, we can measure that S-waves can be detected up to ~ 114 degrees along the Earth’s surface away from the Earthquake (in either direction).

This leaves 360 − (2 × 114) = ~ 132 degrees on the far side of the Earth in the shadow zone.

13. S waves travel slower than P waves—about 4 km/sec versus 6 km/sec. Suppose a seismometer detects P waves at 7:00 p.m. and S waves at 7:05 p.m. How far away was the earthquake? At what time did the earthquake occur?

13. (3600 km; 6:50 pm). The time difference is 5 minutes = 300 s. Set up the following equations and solve for tP and d: tP = d/vp tS = d/vS d/vP +300 s = d/vS d(1/vP − 1/vS) = −300 s

tS = tP + 300 s = d/vP +300 s

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d = −300 s / (1/6 km/s − 1/4 km/s) = 3600 km tP = 3600 km / 6 km/s = 600 s = 10 minutes, and 10 minutes before 7 pm was 6:50 pm. so the Earthquake would have occurred 3600 km away at 6:50 pm local time.

14. Suppose Earth were made of nothing but iron (at a density of 7.8 kg/L) and silicates (at a density of 3.0 kg/L). a. Calculate the percentage (by volume) of iron and silicates that would be required to give a density of 5.5 kg/L. (Hint: You know that the two volume percentages must add up to 100%, so the percentage of silicates equals 100% minus the percentage of iron.) b. The outer radius of the core is approximately 3500 km, while the overall radius of Earth is 6357 km. What percentage of Earth’s volume is occupied by the core? c. The percentage of Earth’s mass that is composed of iron is estimated to be 32% (see Unit 35.3). What might explain why this percentage is smaller than the value in (a) and larger than the percentage in (b)?

14. (a) 52% iron and 48% silicates; (b) 16.7% of the Earth’s volume.

Let x represent the fraction of iron; then 1 – x = fraction of silicates. Set up an equation relating the relative fractions of iron and silicates required to create an overall average density of 5.5 kg/L: 7.8x + 3(1 − x) = 5.5. Solve for x: (7.8 − 3) x = 5.5 − 3 = 2.5 x = 2.5 / (4.8) = 0.52, and so 1 − x = 0.48.

The Earth would be 52% iron and 48% silicates. (b) Volume goes as R3, so the core occupies (3500 km/6357 km)3 × 100% = 16.7% of the Earth's volume. (c) Material in the interior of the Earth, particularly silicates, is compressed to greater densities than 3.0 kg/L, meaning that the contribution to the density of the Earth of silicates can be greater and the total amount of iron can be less than in (a). However, iron is still substantially denser than silicon, so the mass fraction of iron is greater than the volume fraction calculated in (b).

15. a. We can get a rough estimate of how fast the rift zone in the Atlantic Ocean is moving the continents of South America and Africa apart simply by assuming that 240 million years ago Natal, Brazil, and Douala, Cameroon, occupied the same neighborhood. These cities are presently 5000 km apart. How fast would they have to be moving, on average, to reach their present separation? Express your result in cm/year. b. Another way to estimate the same quantity is by analyzing the convection cells in the mantle. If we assume that the convection cells are circular in shape with a diameter of 700 km, and that it takes 100 million years for a piece of rock to complete a trip around a cell, how fast is the rock traveling? How does this value compare to your previous result?

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15. (a) 2.1 cm/year; (b) 2.1 cm/year. (a) Use t = d/v, so: v = d/t = 5000 km/240 million yr = 5 × 108 cm/(2.4 × 108 y) = 2.1 cm/y.

So the cities and continents are not moving very fast, but over a long time, they have separated significantly. (b) Use v = d/t with d = circumference of the cell = 2r =  D, t = 100 million years: v = d/t =  D/ t = (700 km)/100 million yr = 7 × 107 cm/(1.0 × 108 y) = 2.1 cm/y This value is the same as found in part (a).

16. In 2005 Earth’s magnetic field was about 10% weaker than it was in 1845 when German mathematician Carl Friedrich Gauss first measured it. If the magnetic field continues to weaken by 10% of its current value every 160 years, approximately when will it be at half of its 1845 value? When will it be only 10% of its 1845 value? (Note that we are supposing, in this question, that after 320 years the strength of the field is 90% of 90%—or 81% of its original value.)

16. The mathematical way: (0.9)N = 0.1 for N = log(0.1)/log(0.9) = 21.8. 22 cycles, or 3520 years after 1845 is 5365. Through direct calculation: 1845 1.00

2005 0.90

2165 0.81

2325 0.73

2485 0.66

2645 0.59

2805 0.53

2965 0.48

3125 0.43

3285 0.39

3445 0.35

3605 0.31

3765 0.28

3925 0.25

4085 0.23

4245 0.21

4405 0.19

4565 0.17

4725 0.15

4885 0.14

5045 0.12

5205 0.11

5365 0.10

Unit 38 21

11. Imagine gathering all the 1.4 × 10 kg of water in the world and forming it into a sphere.

What would be the diameter of that sphere and how does it compare to the Moon’s diameter of 3400 km? (Hint: The density of water is 1 kg∕L.) 11. (~1400 km, about 2.5 times smaller than the Moon). From the definition of a density: ρ = M/V = M/ (4/3 p R3), we can solve for R: R = (3 M / (4 ))1/3 = ((3 × 1.4 × 1021 kg) / (4 p × 1000 kg/m3))1/3 = 6.94 × 105 m = 694 km.

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The diameter of the sphere is twice this value, D = 1,400 km. For comparison, Moon’s diameter almost 2.5 times bigger.

12. It is estimated that there are about 3.6 × 1016 kg of carbon (C) dissolved in the oceans. If all that combined with oxygen (O) to form carbon dioxide (CO2), how much mass would it be?

What percentage of Earth’s atmospheric oxygen would that be? (Note: Carbon has an atomic mass of 12, whereas oxygen has an atomic mass of 16.)

12. (The total mass of CO2 would be 1.3 × 1017 kg; the mass of oxygen would be 9.6 × 1016 kg; that is about 8% of the atmospheric oxygen).

For each carbon atom, we would add two oxygen atoms, and oxygen has a mass of 16 to carbon’s 12. So the mass of oxygen would be 2 × 16/12 × 3.6 ×10 16 kg = 9.6 × 1016 kg. Adding in the original carbon (3.6 × 1016 kg) the total is 1.3 × 1017 kg.

To find the percentage of the Earth’s atmospheric oxygen, the correct calculation requires us to determine the percentage of oxygen in the atmosphere by mass, using the percentages by volume given in the text in Table 38.1. The Earth’s atmosphere is mostly nitrogen gas in the form of N2 (78.1% by volume; molecular mass 2 × 14 = 28), with a substantial amount of oxygen in the form of O2 (20.9% by volume, mass 2 × 16 = 32), and then smaller amounts of other gases including argon as Ar (noble gas) (0.93%, by volume, atomic mass 1 × 40 = 40). We can estimate the fraction of the atmosphere that is oxygen by mass by: (percent by volume of oxygen × molecular mass) / (sum of (% by volume × molecular mass) for all the components).

12. (cont.) Many of the components are trace gases, and we can ignore their contribution—we can simplify the calculation to 78% N2, 21% O2, and 1% Ar. Then, the fraction of the mass that is oxygen must be: (21% × 32)/(78% × 28 + 21% × 32 + 1% × 40) = 672 / 2896 = 0.23.

Oxygen is 23% by mass—pretty close to the number by volume.

The mass of oxygen must be 0.23 × 5.1 × 1018 kg = 1.2 × 1018 kg.

So (9.6 × 1016 kg)/(1.2 × 1018 kg) ~ 0.08, about 8% of the atmospheric oxygen.

(Note: Much more carbon is locked up in the Earth’s crust in solid form.)

We can also roughly estimate the percentage—because most of the atmosphere is made up of nitrogen and oxygen, and they have nearly equal masses, the percent by mass for both elements is very close to the percent by volume. So 0.21 × 5.1 × 1018 kg = 1.1 × 1018 kg (which still gives 8%) is also a reasonable estimate for this problem.

13. Geologists estimate that the Grand Canyon was created primarily during a period of about 3 million years. Given its current depth of about 1.6 km (1 mile), on average how many centimeters deeper did the canyon become each year during this time period?

13. (0.05 cm/yr). Relate the current depth to the time it has taken to create the Grand Canyon:

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1.6 km / (3 × 106 y) = 1.6 × 105 cm/(3 × 106 y) = 0.05 cm/yr 14. The colors in the aurora borealis are due to the interactions between electrons and atoms or molecules in the atmosphere. At low densities, collisions with oxygen produce light at 630 nm. At higher densities, collisions with oxygen atoms produce light at 558 nm, and collisions with nitrogen molecules produce light at 428 nm. What colors do we associate with these wavelengths? Examine Figure 38.5 A and describe what is producing the aurora seen there.

14. Colors: 630 nm is red, 558 nm is green, and 428 nm is blue. Up toward the top is a purplish color, so maybe a combination of excited nitrogen and oxygen. The bottom of the “curtain” is greenish/yellowish so probably excited oxygen atoms.

15. Antarctica holds about 3 × 1019 kg of ice. If all of that ice melted and entered the oceans, approximately how much would sea level rise? Use the fact that the total mass of oceans is about 1.4 × 1021 kg, while the average depth of the oceans would be 2.7 km if they covered the entire Earth.

15. (57 m). The volume of Antarctic ice, when melted, would be (3 × 1019 kg/1.4 × 1021 kg) 100% = 2.1% of the volume of the oceans. Assuming that the area of the oceans remains the same, this means that the depth of the ocean would increase by 2.1%, which would be 0.021 × 2.7 km = 0.057 = 57 m.

16. As ocean water warms, the water expands in volume by about 0.014% per 1°C rise in temperature. This thermal expansion is small, but it contributes a significant part of the recent rise in sea level. The average depth of the oceans today is 3800 meters. How much has the water risen over the last century due to the 0.8°C rise in ocean temperature? If the temperature rose by 20°C, how much would sea level rise?

16. (0.32 m from the 0.8°C rise; 10.6 m if temperatures rise by 20°C).

The current volume increase is (0.014% × 0.6° C) = 8.4 × 10−3 %. Assuming that the area of the oceans remains the same, the depth of the ocean has increased by 8.4 × 10 −5 × 3800 m = 0.32 m. For a temperature change of 20° C, the volume change would be (0.014% × 20° C) = 0.28 %, which would increase ocean depth by 0.0028 × 3800m =10.6 m.

17. Air moving along with the surface of Earth takes 24 hours to circle Earth’s axis. At a latitude of 60°, the air is 3183 km from the axis and its path around the axis is 20,000 km long.

Calculate its speed in kph. Air at a latitude of 61° is 3086 km from the axis, following a path around the axis that is 19,392 km long. Suppose now that a mass of air at 60° latitude moves to 61° latitude. Use the principle of conservation of angular momentum (Unit 20.3) to estimate how fast air must move around Earth’s axis at the smaller radius there. How fast is this relative to stationary air at this location?

17. (About 50 kph faster at the higher latitude). Speed of the air at 60° latitude = 20,000 km/24 h = 833 kph. MVR = L = constant; the mass doesn’t change, so the factor of change in velocity must be inversely proportional to the factor of change in radius. The speed of the 60° air

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at 61° latitude = ((833 kph × 3183 km)/3086 km) = 859 kph. Stationary air at 61° latitude has a speed of 19,392 km/24 h = 808 kph, so that the air that has migrated from 60° latitude has a speed over 50 kph faster.

17. (cont.) This means the air moving north would also move in orbit around the surface faster than local air, and thereby would appear when looking at the globe to curve to the right. This result mathematically illustrates the Coriolis Effect, as air moving in latitude in the northern hemisphere “curve” to the right, and to the left in the southern hemisphere.

Unit 39 12. Galileo was able to estimate the height of mountains on the Moon by measuring the length of their shadows. Make a drawing of a tall mountain sitting on a flat plane viewed from the side. With a protractor, draw a line with a 10° angle to the plane that touches the mountain peak. This shows how long the shadow is when the Sun is 10° above the horizon. a. If a shadow is 5 kilometers long when the Sun is 10° above the horizon, estimate the height of the mountain. b. Estimate at what angle the Sun must be above the horizon for the shadow to be twice as long as the mountain’s height.

12. The distances on the scale diagram should produce the same answers as the trigonometry:

13. The Moon’s density is 3.3 kg∕L. Suppose that the Moon is made of just two substances: silicates with a density of 3.0 kg∕L and iron with a density of 7.9 kg∕L. a. What fraction of its mass must be iron? b. What fraction of its volume must be iron?

13. (a) mass fractions of 14.7% iron and 85.3% silicate; (b) volume fractions of 6.1% iron and 93.9% silicate.) It’s easier to do (b) first! (b) We can solve for the fraction of the volume of each element by calculating the contributions to the Moon’s density.

Avg density = (fSi × 3.0 kg/L) + (fFe × 7.9 kg/L) = 3.3 kg/L, and fFe = 1 − fSi, so 3.3 = 3 fSi + 7.9 (1 − fSi) = 3 fSi + 7.9 − 7.9 fSi 4.9 fSi = 4.6 98

The fraction of the volume that is silicon is fSi = 4.6/4.9 = 0.939 and the fraction of the volume that is iron is fFe = 1 − 0.939 = 0.061. This gives 6.1% of the Moon’s volume being iron, and 93.9% silicate.

With part (b) done, the fraction of the masses in (a) can be solved by noting if d = m/v then M Si = dSi VSi and MFe = dFe VFe and comparing each to the total mass.

MSi = dSi VSi = 3 kg/L × 0.939 VMoon MFe = dFe VFe = 7.9 kg/L × 0.061 VMoon fM,Si = MSi/MMoon = (3 kg/L × 0.939 VMoon) / (3 kg/L × 0.939 VMoon +7.9 kg/L × 0.061 VMoon) = (3 × 0.939) / (3 × 0.939 +7.9 kg/L × 0.061) = 2.817 / (2.817 + 0.48) = 0.85 fM,Fe = MFe/MMoon = 0.48 / (2.817 + 0.48) = 0.15 (Also note this is 1 – fM,Si, as expected).

This gives mass fractions of 14.7% iron and 85.3% silicate.

14. To measure the distance to the Moon, astronomers measure the round-trip time taken by pulses of light bouncing off retroreflectors placed on the Moon by Apollo astronauts. a. The Moon’s distance varies from 363,000 km to 405,000 km over its elliptical orbit. What is the round-trip time for laser pulse to travel these distances? b. The results of the experiment demonstrate that the Moon is moving away from Earth at 38 mm per year. How much longer does it take a light pulse to make a round-trip each year?

14. (a) 2.42 to 2.70 seconds; (b) 25 nanoseconds longer each year.

Find time from t = d/c. For the closest distance, t = 2 × (363,000 km)/(3.0 × 105 km/s) = 2.42 s, where the factor of 2 is need for the round-trip aspect. For the farther distance, t = d/c = 2 × (405,000 km)/(3.0 × 105 km/s) = 2.70 s b. The time for light to travel 38 mm is t = d/c = (38 × 10−3 m)/(3.0 × 108 m/s) = 1.27 × 10−10 s = 12.7 ns.

The round-trip time would therefore be about 25 ns longer each year.

15. As the Moon moves away from Earth, the radius of its orbit grows larger, although its orbital speed grows smaller. The angular momentum of the Moon is equal to the product of its mass and its orbital velocity and radius (M × V × R). Is the Moon’s orbital angular momentum growing larger or smaller? Use Newton’s version of Kepler’s third law (Unit 17) to determine the Moon’s speed as a function of its orbital radius, to show if the angular momentum is increasing or decreasing.

15. From Newton's revision of Kepler’s 3rd Law, ME + MM = a3/P2. Since the masses stay constant, we can replace them with a constant k, and call a r instead. Thus, r3 = kP2.

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We also know that P = 2r/v, so k(2r/v)2 = r3, and finally v = (k42/r). Thus, as orbital radius (r) increases, the velocity decreases proportionally to the inverse square root of the orbital radius. Since L = mvr and v decreases more slowly than r increases, the angular momentum increases as the Moon recedes.

In an isolated system, L is conserved. But this system also includes the Earth. Ignoring the very small contribution of the Moon' rotation to the overall angular momentum of the Earth-Moon system, and the unchanging radii and masses of Earth and Moon, the entire angular momentum L = LE+LM, where LE is the angular momentum of the Earth and LM that of the Moon.

If L is to be "conserved" (that is, remain constant), then if either LE or LM change, the other must change in inverse proportion. We have evidence that the Earth is slowing in its rotation, and therefore its angular momentum must be decreasing. Consequently the angular momentum of the Moon must increase.

16. Assume that an asteroid’s density is 3.0 kg∕L, and that impact craters on the Moon are 20 times the diameter of the asteroid. a. What was the diameter of the asteroid that created the 240-km-diameter Clavius crater?

Using the density of an average asteroid, calculate the mass of this asteroid. b. Calculate the kinetic energy of this impact if the speed of the asteroid was 20 km∕sec relative to the Moon.

16. (diameter = 12 km, mass = 2.7 × 1015 kg; KE = 5.4 × 1023 Joules).

Use V = 4/3πR3, 1 m3 = 1000 L, M = V, Kinetic Energy Ek = ½ Mv2 the relations and data shared in the problem: Diameter of Asteroid = diameter of crater/20; Density = 3 kg/L, and velocity v = 20 km/s = 20,000 m/s to calculate: Crater diameter (km) 240

Asteroid Diameter (m) 12000

Asteroid Volume (L) 9.04779E + 14

Asteroid Mass (kg) 2.71434E + 15

Asteroid KE (J) 5.42867E + 23

17. The Apollo 11 command module orbited the Moon at an altitude of about 110 km. Use Newton’s modification to Kepler’s third law (Unit 17.2) to calculate the module’s orbital period.

17. (About 2 hours). Kepler’s modified third law reads, M = (4 2/G) × (d3/P2). After some algebraic rearranging, and noting that the d is the distance from the center of the Moon d = 110 km + 1740 km = 1850 km this gives an orbital period of P = ((4 2/G) × (d3/M) )1/2 = ((4 2/(6.67 × 10−11 m3 kg−1 s−2)) × ((1.85 × 106 m)3/7.35 × 1022 kg))1/2 = 7,140 s  2 hours

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Unit 40 10. Astronomers think that Mercury’s scarps formed as the planet’s diameter shrank by about 5 kilometers. a. If the 3600-km diameter core shrank by 5 km, by what percentage did its volume decrease? b. If the volume of iron changes by about 30 parts per million for each kelvin of temperature change, by how much would the core’s temperature have had to change?

10. (a) 0.42%; (b) Lower by 140 K. Volume goes as the radius cubed. A decrease in radius of a factor of (3595 km/3600 km) = 0.9986 would result in a decrease in volume to: (3595/3600)3 = 0.9958 = 99.58% of the original.

The volume decreased by 0.42%. (b) The total change needed is a factor of 0.0042, which is 4.2 × 10 −3, or in parts per million, 4200/1,000,000. So: 4200 ppm/(30 ppm/K) = 140 K. The temperature dropped by 140 K. 11. In this problem, we are going to estimate the total number of craters on Mercury that have a diameter greater than 25 km. a. Given Mercury’s radius of 2440 km, what is the planet’s total surface? b. Estimate the total surface area shown in Figure 40.3 A. c. Count the number of craters in Figure 40.3 A that appear greater than 25 km in diameter.

Estimate the number of craters on Mercury that are greater than 25 km based on your results in a and b.

11. (a) 7.45 × 107 km2; (b) 7.8 × 104 km2; (c) ~ 5 larger than 25 km; (d) ~ 4800 craters greater than 25 km. a. Atot = 4  R2 = 4  (2449 km)2 = 7.45 × 107 km2 11. (cont.) b. Using the scale printed on the figure, the dimensions are approximately 260 km by 300 km giving a total area of A = (260 km) × (300 km) = 7.8 × 10 4 km2. c. There are about 5 craters with a diameter greater than 25 km. d. The density of craters larger than 25 km is approximately 5 per 7.8 × 10 4 km2.

By multiplying this value by the total surface area found in part a, we get an estimate of 4,800 craters. This calculation does assume that the cratering shown in the figure is typical of Mercury’s surface in general.

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12. The highest temperature on Mercury is about 710 K, while the highest temperature on the Moon is about 400 K. The average speed of a molecule in a gas is proportional to the square root of the temperature. The rate of molecules escaping to space depends on the fraction of their average velocity relative to the escape velocity, and Mercury’s escape velocity is about twice the Moon’s. Given these facts, which would have retained an atmosphere longer, Mercury or the Moon?

12. Mercury would have retained an atmosphere longer than the Moon. The speed of a molecule is proportional to the square root of the temperature, so the average speed of a gas on Mercury is ~ (700/400) ~ 1.3 times the speed of the same gas on the Moon. At 4.3 km/s, Mercury’s escape velocity is nearly twice that of the Moon’s (2.4 km/s).

13. Mercury’s orbit ranges from 46 to 70 million kilometers from the Sun, while Earth orbits at about 150 million kilometers. a. The Sun has a 30-arc-minute diameter viewed from Earth. What range of sizes does it have when viewed from Mercury? (Unit 10) b. At Mercury’s orbital extremes, how many times stronger is the Sun’s radiation on Mercury than on Earth? (Unit 21) 13. (a) From 64 to 98 arcminutes; (b) 4.6 times stronger.

Angular size is inversely proportional to the distance, so the Sun at Mercury appears to be: 30 arc min × 150/46 = 98 arc min, at closest approach, 30 arc min × 150/70 = 64 arc min, at the farthest part of the orbit.

13. (cont.) (b) The brightness (intensity of the Sun’s radiation) is inversely proportional to the square of the distance, so at the closest approach it will be (150/46)2 = 10.6 times stronger than at Earth, and at the farthest part of Mercury’s orbit, (150/70)2 = 4.6 times stronger.

14. How many times faster is Mercury orbiting the Sun at perihelion than at aphelion given that Mercury’s perihelion and aphelion distances are 46 and 70 million kilometers, respectively?

14. (1.5 times faster). By conservation of angular momentum (Unit 20.3) the quantity L = m × V × R is conserved. Since Mercury’s mass is not changing it must be the case that Vp × Rp = Va × Ra, or equivalently, Vp/Va = Ra/Rp. From the given values we find that Mercury is moving 70/46 = 1.5 times faster at perihelion than at aphelion.

15. The cause of the Sun occasionally appearing to move backward across Mercury’s sky is explored in this problem. Use the facts that Mercury rotates once every 58.6 Earth days relative to the stars, and takes 88.0 Earth days to revolve about the Sun. (It may also help to consult Figure 40.8 for this problem.) a. How fast would the Sun move across Mercury’s sky if Mercury were spinning at its same rate, but not revolving about the Sun? In other words, calculate the angular speed (degrees per Earth day) at which the Sun moves across the sky due to Mercury’s rotation alone.

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b. Calculate the average angular speed (degrees per Earth day) at which the Sun moves across the sky due just to Mercury’s orbit of 88 days. (This is slower than the motion of the Sun due to its rotation, which is why the Sun rises in the east and sets in the west on Mercury.) c. When Mercury is at perihelion, its orbital speed is about 25% faster than its average orbital speed. In addition, because Mercury is closer to the Sun, the angle the Sun shifts by is about 25% greater, too. What is the resulting angular speed when Mercury is at perihelion?

15. (a) 6.14/Earth Day; (b) 4.1/day; (c) 6.14/Earth Day or about 0.256/hour. a) The rate is about 6.14/Earth Day or about 0.256/hour. Divide 360 by 58.6 d for degrees per day, then divide by 24 to get degrees per hour.

15. (cont.) b) The rate is about 4.1/day or 0.17/hour. Divide 360 by 88 d, and then by 24 h/d. c) If it is 25% faster and the angle changes 25% faster, that is a 50% change, so the angular speed at perihelion is (0.17/hour) × 1.5 ≈ 0.256/hour (the same amount as (a), but resulting in the opposite apparent motion of the Sun).

Unit 41 15. Venus’s atmosphere contains about 4.6 × 1020 kg of carbon dioxide (CO2). If all of the oxygen atoms in the CO2 were combined with hydrogen to make water (H2O) instead of carbon dioxide, what would the total mass of water be? Compare this to the mass of Earth’s oceans, ~1.4 × 1021 kg. (Note: Hydrogen has an atomic mass of 1; carbon, 12; oxygen, 16.)

11. (About 3.7 × 1020 kg.) The molecular mass of CO2 is 44 (Carbon is 12, two Oxygens are 32). Of the total mass of 4.6 × 1020 kg, 32/44ths are oxygen: (32/44) × 4.6 × 1020 kg = 3.3 × 1020 kg.

In H2O, H is 2; O is 16; so the total mass of water using 3.3 × 1020 kg of Oxygen is: 3.3 × 1020 kg × (18/16) = 3.7 × 1020 kg.

This is 3.7 × 1020 kg / (1.4 × 1021 kg) = 3.7/14 ~ 0.26, just over a quarter of the water in Earth’s oceans.

12. From your answer to Problem 11, how deep an ocean would this water produce if it were the same depth everywhere on the surface? (Note that 1000 kg of water fills a volume of 1 m3.)

12. (804 m). The depth will be volume of water divided by surface area. If 1000 kg fills 1 m3, the volume is 3.7 × 1020 kg / 1000 = 3.7 × 1017 m3. The surface area is 4 πR2 = 4 π(6052 × 103 m)2 = 4.6 × 1014 m2.

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depth =

V olum e = 3.7 × 1017 m3 / 4.6 × 1014 m2 = 804 meters.

Surface A rea

It would be about 800 meters deep.

13. Kepler’s third law (Unit 12.2) describes planetary orbital sizes relative to Earth’s orbit, which is assigned a size of 1 AU. One method for finding absolute orbital sizes is to use radar ranging between planets. Use Venus’s scaled orbital size of 0.723 AU, and the observation that a radar signal takes 276 sec to make a round-trip between Earth and Venus when they are minimally separated, to find how many kilometers are in an AU. 13. (1.49 × 108 km). In 276 s light travels a distance of: d = c × t = (3.0 × 105 km/s) × (276 s) = 8.28 × 107 km.

The minimal separation between Earth and Venus is 1 AU − 0.723 AU = 0.277 AU, which is also half the distance the radar signals traveled. Therefore, it must be the case that: ½ (8.28 × 107 km) = 0.277 AU, or equivalently, 1 AU = ½ (8.28 × 107 km)/0.277 = 1.49 × 108 km.

14. Venus and Earth make their closest approach to each other (inferior conjunction, Unit 11) every 583.9 Earth days. Use a diagram like Figure 41.8 or math to show that nearly the same side of Venus faces Earth at each closest approach.

14. The time from noon to noon on Venus is about 117 Earth days, as explained in sec. 39.3.

Thus, every 117 days, the same side of Venus is facing the Sun, and the same side is facing outward to space opposite the Sun. When Earth and Venus are closest every 583.9 days, the Earth is opposite the Sun (Venus is in “inferior conjunction” as viewed from Earth, and Earth is in “opposition” as viewed from Venus.). Almost exactly 5 of Venus’s 117 day-long days fit into the 583.9 time from closest approach to closest approach (5 × 117 = 585) so the Earth always see the same side of Venus in inferior conjunction. (See diagram).

14. (cont.)

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15. Earth covers about 1° per day in its orbit about the Sun, and the solar day is slightly longer than the sidereal day (Unit 7). If Earth spun in a retrograde direction like Venus but it still had the same sidereal period (23 hr 56 min), how long would the solar day be? Use a diagram to explain your answer.

15. The solar day is slightly longer than the sidereal day now, because the Earth has to rotate a little farther to align with the Sun than to align with the stars. If the Earth had a retrograde rotation, then the solar day would be shorter than the sidereal day, as shown in the diagram below (on next page). After “noon” at (1), the same spot on Earth would turn (2) and face the Sun for “noon” again (3, 1 solar day) before it returned to the same orientation with respect to distant stars (4). (Earth aligned with both Sun & star at 1).

Unit 42 13. In Figure 42.3, suppose that the areas with negative elevations were actually oceans.

Estimate from the figure the surface area covered by the northern ocean. Multiply this by an

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average depth of 5 kilometers to get a volume. Compare your result to the 1.4 × 1018 cubic meters of Earth’s oceans.

13. (1.8 × 1017 m). Taking into account that some of the green areas are negative elevation, and projection effects on the map, the ocean in the northern hemisphere would cover at least 50% of the northern hemisphere, so perhaps 25% of the planet. Therefore: ¼ Area = ¼ × 4R2 =  (3396 km)2 = 3.6 × 107 km2.

Volume = area × depth = 3.6 × 107 km2 × 5 km × (1000 m/1 km)3 = 1.8 × 1017 m3.

This is about 1/10 the volume of Earth’s oceans (1.8 × 10 18 m3).

14. Mars’s orbit is quite elliptical, so at different times when it is in opposition, its distance from Earth can vary substantially. From the data given in this Unit, estimate the largest and smallest distance to Mars at opposition and when it is on the far side of the Sun. Assume that Earth’s orbit is circular at 1 AU. 14. (At opposition, maximum distance = 0.672 AU and minimum = 0.368 AU; at conjunction, maximum = 2.672 AU and minimum = 2.368 AU). According to the Unit Mars’ orbit varies by 45 million km, or 20%, of its distance to the Sun. The change is about 10 percent either way, so its maximum distance is 1.52 AU × 1.1 = 1.672 AU and its minimum distance to the Sun is 1.52 AU − (0.1 × 1.52) = 1.52 − .152 = 1.368 AU. Ignoring any effects from Earth’s changing distance (as noted in the question we assume Earth’s orbit is circular), at opposition, the maximum distance from Earth to Mars is 1.672 − 1 = 0.672 AU and the minimum distance is 1.368 AU − 1 AU = 0.368 AU. When Mars is on the far side of the Sun, the maximum distance would be 1 + 1.672 = 2.672 and the minimum 1 + 1.368 AU = 2.368 AU. 15. The volume of a cone is given by V = 1/3 πR2 × h, where h is the height of the cone and R is the radius of its base. Use this formula, and the dimensions shown in Figure 42.5, to estimate the volume of Olympus Mons. Compare its volume to that of Mauna Kea, which has base radius of approximately 60 km.

15. (2.5 × 106 km3; about 66 times larger). The volume for a cone is given by V =  R2 × h/3, where h is the height of the cone and R is the radius of its base. From figure 42.5, the diameter of Olympus Mons is around 600 km (R = 300 km). Its height is 26 km. Its volume is then:

V = p R2

2 26 km h = p (300 km ) = 2.5´ 106 km 3 3 3

Mauna Kea, with a base radius of around 60 km and height above the sea floor of 10,200 m has a volume of: 2

V = p (60 km )

10.2 km = 3.8´ 104 km 3 3

16. From the surface of Earth under very good conditions, we can resolve an angular size of about 1 arc second. When Mars is 0.5 AU away, calculate the linear size (in kilometers) of a 106

feature on the surface that would appear to be 1 arc second across. The Hubble Space Telescope can resolve about 0.04 arc seconds; what size on Mars does this correspond to?

16. (364 km from Earth; 15 km with HST). An object one second of arc in the scope would be about 363 km across as seen from Earth; it would be 14 km with the Hubble’s higher resolution.

Use the angle and distance formula A = 57.3 × L/d, where A is the angle involved, L is the object size and d is the distance, with both L and d in the same units. We know d and A, so this can be rearranged to find L = d × A/57.3. Convert the distances to km (0.5 AU = 7.5 × 107 km), and convert the values given for A into degrees. 1 arc sec = 1/3600; 0.04 arc sec = 1.11 × 10−5.

17. What angular size would Phobos and Deimos have as viewed from the surface of Mars?

Could either produce a total solar eclipse?

17. (Deimos about 3.4 arc minutes, and Phobos about 11.5 arc minutes). The Sun from Mars is about 20 arc minutes across, so there would be no solar eclipses (0.5 = 30 arc min from Earth × (1/1.52) = 20 arc min). Use A = 57.3 × L/d. The text lists both Deimos and Phobos as about 20 km across. Deimos is about 20,000 km from the surface of Mars; Phobos is only 6,000 km.

Plugging those numbers in, we find that Deimos is about: A = 57.3× L/d = 57.3 × 20 km/20000 km = 0.0573 = 3.44 arc minutes across, 17. (cont.) and Phobos is 3.33 times closer, and will appear 3.33 times larger, or about 11.5 arc minutes across. Both are much smaller than the Sun, so there would not be very exciting eclipses. Even Phobos would only cover about (11.5/20)2 = 33% of the Sun. 18. Earth’s atmospheric pressure drops by about a factor of 2 every 5 km higher up you go, so it is one-fourth the surface pressure at 10 km, for example. How high up in Earth’s atmosphere would you have to go to reach a pressure similar to the 0.6% atmospheric pressure on Mars? (You can do this mathematically with logarithms or plot the changing pressure with elevation and estimate the height where this pressure is reached.)

18. (37 km). If N is the number of 5 km layers required, the equation to be solved is (0.5)N = 0.006.

Taking logarithms of both sides, N = log(0.006)/log(0.5) = 7.4 layers. With each layer = 5 km, this corresponds to an altitude of 7.4 × 5 km = 37 km.

19. The average surface pressure on Mars is almost exactly equal to that below which water cannot be liquid. Review the topographic map of Mars in Figure 42.3. If you wanted to search an area that had the greatest probability of any location on Mars for liquid water today, where would you go and why?

19. The places on Mars that would have the highest atmospheric pressures are those with the lowest altitudes, such as Hellas (which has the global lowest altitude), Argyre or the bottom of Valles Marineris.

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Unit 43 14. Asteroids are often discovered on long-exposure images because they are moving relative to the stars during the exposure. They therefore leave a short trail on the image while the stars remain stationary. Suppose that an exposure 25 minutes long is made, and that this image has a “plate scale” of 0.11 arc minutes per millimeter. a. If the asteroid leaves a trail 3 mm long on the image, how many arc minutes did it move during the exposure? b. If the asteroid is about 0.5 AU away, what is its velocity through space? c. How would Earth’s motion affect this estimate? Draw a diagram to support your explanation.

14. (a) 0.33 arc minutes; (b) About 5 km/s. a) The asteroid traveled 3 mm × 0.11 arc min / mm = 0.33 arc min. (We need to be cautious about these calculations since the motion of the Earth will give a parallax effect if the exposure time is long enough, see c). b) 0.5 AU = 7.5 × 1010 m. Using the angular-size distance relation, and converting 57.3 to arc minutes, we can find the actual length it traveled: L = d × A/57.3 = d × A/3438 arc min = 7.5 × 1010 m × 0.33/3438 = 7.2 × 106 m And the time was 25 minutes = 1500 s, so v = 7.2 × 106 m / 1500 s = 4800 m/s = 4.8 km/s; at this speed the asteroid moved about 7.2 × 106 km in 25 minutes through space. c) In 25 minutes, the Earth rotates (25/60 × 15) = 6.25. As a worst case scenario, you might have been moved through an arc of nearly 700 km from this motion if you were on the equator: 6.25/360 × circumference of the Earth = 0.174 × 4.0 × 105 km) = 694 km.

Even if the asteroid were sitting still, the motion would cause the asteroid to seem to move against the background stars from parallax. 700 km is 700,000 m, so the (maximum) distance you travel from the Earth’s motion is about 10% of the distance we infer the asteroid to be traveling—depending on the orientation of your observing location, the Earth’s spin, and the asteroid’s motion, this could add to, subtract from, or affect the apparent direction of the asteroid!

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15. Given Jupiter’s orbital period of 11.9 years, use Kepler’s third law (Unit 12.2) to calculate the distances from the Sun to each Kirkwood gap.

15. (2.07 AU, 2.51 AU, 3.28 AU, and 5.21 AU). Kepler’s third law in the form a3 = P2 requires the semi-major axis to be expressed in AU and the period in years. From Kepler’s third law, the semi-major axis is simply, a = P2/3. The first Kirkwood gap is at a distance of a = (11.9/4)2/3 = 2.07 AU. Following a similar calculation for the other gaps gives distances of 2.51 AU, 3.28 AU, and 5.21 AU. 16. If an asteroid 500 km in diameter is destroyed by an impact, how many 1-km-size asteroids can result from the collision? Assume that the initial and resulting asteroids are spherical in shape, which means their volumes are given by 4/3 πR3.

16. (1.25 × 108). The total volume of the initial asteroid must equal the sum of all the volumes of the resulting asteroids, Vi = n × Vf, where n is the number of smaller asteroids. Substituting in the volumes: 4/3 π Ri3  n  4/3 Rf3, leads to n  (Ri/Rf)3  (250 km/0.5 km)3  1.25 × 108.

17. The asteroid Icarus has an elliptical orbit that carries it between 0.19 and 1.97 AU from the Sun. What is its semimajor axis? How often does it cross Earth’s orbital radius?

17. (1.08 AU; about 205 days). The semi-major axis is half the sum of the nearest and farthest points, a = (0.19 AU + 1.97 AU)/2 = 2.16 AU/2 = 1.08 AU. Since Icarus’ orbit extends farther than the Earth’s orbit, it must cross the radius of the Earth’s orbit (although not necessarily the orbit) twice per orbit, once heading sunward and once away. Use Kepler’s 3 rd Law to find the period, P = (1.08)3/2 = 1.12 y, about 410 days. The average therefore is about one half of this, or 205 days, although this does not specify the exact times between subsequent crossings.

18. The asteroid 2004 FH was discovered on March 15, 2004, by the NASA-funded LINEAR asteroid survey. It is approximately 30 meters in diameter. Through careful observation, astronomers determined its orbital period to be 270.1 days. a. What is its semimajor axis? b. If its orbit varies from 1.05 AU to 0.58 AU, could this be a potentially hazardous asteroid?

Explain your answer. c. If it has a density of 2 kg/L, what is its mass? d. If this asteroid were to strike the surface of Earth at 30 km/sec (Earth’s orbital velocity), how much kinetic energy would it have?

18. (a) 0.82 AU; (b) yes; (c) 2.8 × 107 kg; (d) 1.2 × 1016 J a) Use Kepler’s 3rd Law to find the semimajor axis, first converting period into years: (270.1 d) / (365.25 d/y) = 0.739 y; so A = P2/3 = 0.82 AU. 109

b) If its orbit crosses the plane of the Earth’s orbit, it could hit the Earth if it was at the right distance and the Earth was in the right place (or the wrong place!). c) If the asteroid is spherical, V = 4/3 R3 = 4/3  (30 m/2)3 = 1.4 × 104 m3 = 1.4 × 107 L. Then the mass = density × volume = 1.4 × 107 L × 2 kg/L = 2.8 × 107 kg. d) Kinetic Energy Ek = ½ Mv2 = ½ × 2.8 × 107 kg × (30 × 103 m/s)2 = 1.2 × 1016 J. 19. The NEAR spacecraft landed on Eros with a velocity estimated at 1.6 m/sec. Convert this speed to kph. If the spacecraft had a mass of about 800 kg, what was the kinetic energy of its impact? Discuss how much this would have affected the asteroid’s orbit.

19. (1024 J). Total Kinetic Energy Ek = ½ Mv2 = ½ × 800 kg × (1.6 m/s)2 = 1024 J; This is not a huge amount of energy, and the impact of the spacecraft would hardly affect anything at all! Perhaps more like some meteorites that get slowed down so much in Earth’s atmosphere so that they hardly make a dent in the ground. Note the spacecraft’s velocity of 1.6 m/sec × 3600 sec/hr × 1 km/1000 m = 5.8 km/hr is about 3.5 miles per hour. This is not very fast!

20. Ceres has a mass of approximately 8.7 × 1020 kg and a diameter of approximately 930 km. a. What is the acceleration of gravity on its surface (Unit 16)? b. What is the escape velocity from its surface (Unit 18)?

20. (a) g = 0.27 m/s2; (b) 500 m/s). a) To find the acceleration due to gravity on Ceres, use: G × M/R2 = 6.67 × 10−11 Nm2/kg2 × 8.7 × 1020 kg / (930/2 × 103 m)2 = 0.27 m/s2 Compared to g = 9.81 m/s2 on Earth, objects would weight 36 times less on Ceres. b) To find the escape velocity, use: Vesc

= (2 GM/R) = (2 × 6.67 × 10−11 Nm2/kg2 × 8.7 × 1020 kg / (930/2 × 103 m)) = 500 m/s.

Unit 44 15. For each terrestrial planet, use the values given in Units 37, 40, 41, and 42 to calculate what percentage of the planet’s total volume is comprised of its core. Rank the terrestrial planets according to this percentage.

15. Given the radius of the core, RC, and that of the planet, RP, the ratio of the core’s volume to the planets is VC/VP = (4/3  RC3)/(4/3  RP3) = (RC/RP)3. The radii are found in the figures referenced in the table below.

Planet Mercury Venus

Figure 40.6 41.6

RC (km) 1800 3000 110

RP (km) 2439 6052

Core Percentage 40% 12%

Earth Mars

37.5 42.2

3500 1700

6357 3376

17% 13%

16. The mass of Uranus is about 14.5 times that of Earth, yet the escape velocity for Uranus is a little less than twice that of Earth’s. Explain the seeming discrepancy.

16. Vesc = (2GM/R). Basically, Uranus’ escape velocity is only ~2x Earths because of the square root in the equation, and Uranus’ large radius compared to Earth. The radius of Uranus is almost 4 times that of Earth. Since the mass is 14.5 times as much, inside the square root in the formula, the difference between the calculation for Uranus and Earth is 14.5/4 = 3.6. But then taking the square root of that number, since 3.6 is almost 4 we get a number that’s almost 2.

17. Scientists classify objects based on various characteristics. If we rank the terrestrial planets according to their densities (mass divided by volume), we will get one relationship. If we then rank the terrestrial planets according to their uncompressed densities, we will get a slightly different relationship. Compare the rankings. What do the differences in the rankings tell us about the planets themselves?

17. The rankings based on compressed densities show the highest densities for planets closest to the Sun, and the density steadily dropping with distance from the Sun—reflecting the different density materials which could condense out of the solar nebula. Mercury is mostly metal; Mars has a lot more rocks. The compressed densities show Earth as the densest, and that Earth and Venus have much higher compressed than uncompressed densities while the Moon and Mercury hardly change. This tells us that the gravity of a more massive planet will compress the material in the planet.

18. Examine the data in Tables 44.1, 44.2, and 44.3, and consider that the average speed of a molecule in a gas with temperature T is approximately Vmol = 0.15 km/sec × (T/m)− ½ where T is the temperature in kelvin and m is the mass of the molec1ule in atomic mass units; so, m(H2) = 2, and m(CO2) = 44. About how many times smaller than the escape velocity does Vmol have to be for a planet to retain a gas?

18. (Less than ~ 1/10 the escape velocity). From the data, we can infer that the molecular velocity should be no more than about one-tenth the escape velocity for the planet to retain the gas. (A widely accepted rule is 1/6, as in the problem in Unit 35, but in this rough calculation 1/10 is more appropriate). To see this, construct a table (or spreadsheet) to compute molecular masses for various gases and compare the gas velocities to the escape velocities of the planets as shown below. In the table, the boxed entries are cases where the gases escape—no atmospheres on Mercury or the Moon, and little H2 in the Earth’s atmospheres. (Note: not all the other chemicals are present in all the other atmospheres for chemistry or composition reasons).

18. (cont.) The cut-off seems to be when the escape velocity is about 10 times the molecular velocity for the molecule to be present in significant quantities, so the molecular velocity must 111

be less than about one-tenth the escape velocity. For the gas planets, all gases can be retained. Even for molecular hydrogen, Vgas/Vesc ~ 1/20 for Uranus, a smaller and colder gas giant. By contrast, the ratio at Earth is Vgas/Vesc ~ 1/6, too high to retain much molecular hydrogen. For Mercury, with a high temperature and low gravity, not even heavy CO 2 is retained, Vgas/Vesc ~ 1/9.

(All Velocities in km/s) Gas H

Vesc/Vgas shown for clarity CO2 N2

O2

Gas Mass

2

44

56

32

Earth T Earth Vgas Vesc/Vg

287 1.80 6.23

287 0.38 29.24

287 0.34 32.98

287 0.45 24.93

Mercury T Merc Vgas Vesc/Vg

440 2.22 1.93

440 0.47 9.07

440 0.42 10.23

440 0.56 7.73

Moon T Moon Vgas Vesc/Vg

250.00 1.68 1.43

250.00 0.36 6.71

250.00 0.32 7.57

250.00 0.42 5.72

Saturn T Saturn Vgas Vesc/Vg

130 1.21 29.35

130 0.26 137.69

130 0.23 155.33

130 0.30 117.42

Uranus T Uranus Vgas Vesc/Vg

80 0.95 22.45

80 0.20 105.31

80 0.18 118.81

80 0.24 89.81

Escape Velocities (km/s) Moon Earth

Mercury

Saturn

Uranus

2.4

4.3

35.5

21.3

11.2

19. Heating by the Sun decreases with distance. For an object absorbing all sunlight, we can predict that the temperature will be 227aK where a is the distance from the Sun in AU. Find the expected temperatures at the distances of Venus, Mars, and Jupiter. What factors might cause the temperatures to differ from what this formula yields?

19. Using the equation given, the expected temperatures for Venus (0.723 AU), Mars (1.52 AU), and Jupiter (5.2 AU) would be 267, 184, and 100 K, respectively. Example: T = 227 K × 1/(a) = 227 K × 1/(1.52) = 184 K.

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The observed surface temperatures are about 737, 250, and 160K for these planets.

Atmospheric warming via the Greenhouse effect is the major reason for the higher than expected temperatures of Venus and Mars. For Jupiter, the excess warmth likely is due to heat flowing out from the interior of the planet.

Unit 45 15. Given its rate of rotation, at what speed does a point on the equator of Jupiter move?

Give your result in km/hr.

10. (About 45,000 km/hr). Distance = circumference = 2 πR = 2π  71492 km = 449197 km.

Speed = distance / time = 449197 km/9.9 h = 45,373 km/hr.

16. Given its rate of rotation, at what speed does a point on the equator of Saturn move?

Compare this speed to that of Jupiter.

11. (About 36,000 km/hr). The equatorial radius of Saturn is 60,268 km. Its day is 10.6 hours long. We need the equatorial circumference first: 2 πR = 378,675 km. Speed = distance / time = 378,675 km / 10.656 hrs = 35,724 km/hr.

Comparing this with the 45,373 km/hr speed at Jupiter’s equator due to rotation from #10 above, Jupiter has a larger circumference at its equator and its day is shorter, so the speed must be higher. (Note: 45/36 ~ 1.3. The effect is 2/3 from the bigger radius, which is 71/60 ~ 1.2, and 1/3 from the length of day, 10.6/9.9 ~ 1.1).

12. a. Jupiter is 318 times more massive and 11.2 times larger than Earth. How much greater is the surface gravity (Unit 16.3) on Jupiter compared to Earth? b. Do the same calculation for Saturn, which has a mass 95 times larger than Earth and is 9.5 times bigger.

12. (2.54 times larger; 1.05 times larger). From Unit 16.3, the surface gravity is given by g = G × M/R2. When comparing to we can write: gP/gE = (G × MP/RP2) / (G × ME/RE2) = (MP/ME) × (RE/RP)2 = (MP/ME) / (RP/RE)2.

Here the E subscript stands for Earth while the P stands for planet. a) Comparing Jupiter to Earth: gP/gE = (318) / (11.2)2 = 2.54 b) Comparing Saturn to Earth: gP/gE = (95) / (9.5)2 = 1.05 13. Compare the volume of Jupiter to the volume of Earth. How many bodies the size of Earth could fit inside Jupiter?

13. (About 1400 times larger). Volumes compare as the cube of the radius, so assume that Earth and Jupiter are perfectly spherical (they aren't really), the volumes compare as (Rj/Re) 3.

That's (11.21/1)3 = 1409. So the volume of Jupiter is about 1400 times that of Earth. (Note: If you read the problem literally, you might worry about the closest packing of spheres. In this case one could fit in only about 75% as many spheres as the simple volume ratio implies.

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14. If Saturn were compressed until it had the same density as Jupiter, what would Saturn’s new radius be? Compare this to Jupiter’s radius. If Saturn were compressed until it had the same density as Earth, what would Saturn’s new radius be?

14. (48,400 km; about 2/3 of Jupiter’s radius; about 30,000 km).

The ratio of Jupiter's density to that of Saturn is (1.33 kg/L)/(0.69 kg/L) = 1.93. Thus, the new density of Saturn would be 1.93 times the old density. Density = mass/volume. Since mass is assumed to stay the same, the only thing that changes is volume, Vnew = Vorginal/1.93. Since V= 4/3 R3, volume depends on the cube of the radius and so rnew = roriginal/(1.93)1/3 = 0.80 roriginal.

Thus, the new radius would be about 80 percent of Saturn's current radius, or about: 60268 km × 0.80 = 48,400 km.

14. (cont.) This is 48,400/71492 ~ 2/3 of Jupiter’s radius. Using Earth’s density, the ratio is: (5.52/0.69) = 8 so the ratio of radii will be: rnew = roriginal/(8)1/3 = ½ roriginal = 60268 km/2 = 30134 km.

(Note that these figures vary somewhat from the actual figures because we have simplified by assuming that Jupiter and Saturn are spheres, when in fact they are oblate spheroids.)

15. In the Sun, hydrogen makes up 90% of the atoms, corresponding to 75% of its mass. Iron and other elements with atomic weights of about 56 make up 0.001% of the atoms, while silicon and other elements with atomic weights of about 28 make up 0.01% of the atoms.

These represent the percentage of atoms, not the fraction of the mass. If Jupiter has the same proportions of atoms in its interior, what would the total mass of Jupiter’s iron- and siliconlike atoms be? Compare the masses you find to the mass of Earth.

15. (For iron-like atoms: 8.9 × 1023 kg; for silicon-like atoms: 4.5 × 1024 kg).

If Jupiter has the same composition of the Sun, then 75% of its mass is hydrogen atoms: Mass of Hydrogen = 0.75 × 1.90 × 1027 kg = 1.43 × 1027 kg Since each hydrogen atom has a mass of 1.6735 × 10−27 kg, the total number of hydrogen atoms in Jupiter is about Total # of hydrogen atoms = (1.43 × 1027 kg) / (1.6735 × 10−27 kg) = 8.54 × 1053.

Since this is 90% of all the atoms, there are (100/90 × 8.54 × 1053) = 9.5 × 1053 atoms in total.

The question notes that iron-like elements are 0.001% (10−5) of these, so we can get a total mass by multiplying this number of atoms by their atomic weight: Total mass of iron atoms = 9.5 × 1053 × 10−5 × (56 × 1.6735 × 10−27 kg) = 8.9 × 1023 kg.

Silicon-like elements (0.01% by number) are similarly calculated to have a mass of about Mass of silicon atoms = 9.5 × 1053 × 10−4 × (28 × 1.6735 × 10−27 kg) = 4.5 × 1024 kg.

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15. (cont.) The Earth’s mass is 6.0 × 1024 kg, so combined these amounts are a bit less than an Earth mass. Note, however, that these are extremely rough numbers, and in reality this calculation probably underestimates both the iron and silicon mass fractions in Jupiter—but we can conclude that we should expect something like terrestrial-sized planet amounts of iron and rock in gas giants, rather than much, much more, or much, much less.

16. Look up information on the speed, structure, and lifetime of a class 5 hurricane on Earth and compare it to the properties of the Great Red Spot. Describe the differences in their causes.

16. A Class 5 hurricane on Earth has winds greater than 155 miles per hour (or greater than about 250 km/hr). The Great Red Spot’s (GRS) speed is about 2 times that. Both hurricanes and the GRS lie above the “surface” of the planet. Hurricanes occur only during certain times of the year; the GRS may be a permanent feature of Jupiter’s atmosphere.

Unit 46 11. Find the “surface” gravity on Uranus and Neptune (Unit 16.3). How does this compare with Earth’s surface gravity?

11. (gUranus = 8.9 m/s2 which is .90 times Earth; gNeptune = 11.2 m/s2 which is 1.14 times larger).

Use g = GM/R2 . To find values relative to Earth, substitute the relative mass and radius values compared to Earth into the equation. For example, the mass of Uranus is 14.54 times that of Earth, and its radius is 4.01 times Earth’s. Substituting in gives 14.54/4.012 = 0.90 or about 90 percent that of Earth. The surface gravity of Uranus is about 0.90 × 9.81 m/s2 = 8.9 m/s2.

For Neptune, it’s 17.15/3.882 = 1.14, or 114% Earth’s, and 1.14 × 9.81 m/s2 = 11.2 m/s2.

12. If Uranus and Neptune had the same density, how many times bigger than Uranus would Neptune be?

12. Neptune would be about 1.06 times larger than Uranus in radius. Use D = M/V. From the Appendix, DU = 1.27 kg/L. DN =1.64, so Neptune’s density must decrease to 1.27/1.64 = 0.774 times its current value. Since the mass will not change, the volume must increase a factor of 1/0.774 = 1.29 = 1.64/1.27. V  R3, so the radius must increase by (1.64/1.27)1/3 = 1.09 times to 1.09 × 24764 km = 26,993 km. Compared to Uranus’ radius of 25,559 km, Neptune would be 26993/25559 = 1.06 times larger.

13. In order to match the densities of Uranus and Neptune, scientists hypothesize that their interiors are composed of three layers: (i) a dense rock-iron core; (ii) a liquid water-ammonia layer; and (iii) a hydrogen-helium gas layer. Suppose the densities of these layers average 8, 1, and 0.2 kg/liter, respectively. a. Find the density of a planet that is 10% (i), 40% (ii), and 50% (iii). b. If there were no liquid layer, find a mix of percentages of the other two layers that gives the same density as in a. c. What other constraints do astronomers have that help decide on the correct mix?

13. (a) 1.3 kg/L; (b) 14% iron-rock core and 86% hydrogen-helium gas.

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a) For planet with 10% composed of a dense rock core, 40% liquid, and 50% gas, its density would be (0.1 × 8 kg/L) + (0.4 × 1 kg/L) + (0.5 × 0.2 kg/L) = 1.3 kg/L. (b) Let x be the fraction that is rock and 1 − x be the fraction that is gas.

1.3 kg/L = (8x + 0.2(1 − x)) kg/L = (8x − 0.2x + 0.2) kg/L 1.1 = 7.8x, so x = 1.1/7.8 = 0.14.

The mixture is 14% iron-rock core and 86% hydrogen-helium gas. (c) Astronomers can assume that the solar nebula had about the same composition beyond the water line, so the ratio of iron-rich material to other materials is the same as in other parts of the solar system.

14. Given Neptune’s semimajor axis of 30 AU, use Kepler’s third law (Unit 12.2) to calculate its orbital period. How many orbits has Neptune completed since its discovery in 1846?

14. (164 years; just one). From Kepler’s third law, P2 = a3, we get an orbital period of P = a3/2 = (30)3/2 = 164 years.

Assuming a year of 2013, it has been 167 years since Neptune was discovered. Therefore, it’s barely made one trip around the Sun since its discovery.

15. Assume that Earth’s axis is tilted 90° wit respect to its orbital plane, similar to that of Uranus. How would this affect the length of Earth days? What would happen to the seasons? Describe a complete year for where you live.

15. The length of day would vary during the year and depending on your latitude (assuming Earth was still spinning the same). At all latitudes except the equator, for some part of the year (multiple days to months) the Sun would not set at all. We would still have seasons, but they would be very extreme: the Sun will end up being directly overhead for part of the year for all parts of the Earth.

Start the year with the Earth’s North Pole pointed at the Sun. For the next several months, it would be daylight at the North Pole, with the Sun directly overhead. The Sun would not set for much of the Northern hemisphere; instead it would circle in the sky (assuming the Earth was still spinning). The day would last for months. The Sun would be lower and lower in the sky as you approached the equator; the Southern hemisphere would be dark when the Sun was aligned with the North pole. The South Pole would be completely dark.

Over the first three months of the year, as the axis turned from the Sun, the equator, and then increasingly more of the Northern hemisphere would start to have nights of a few hours, and south of the equator the Sun would be visible for a few hours. At three months into the year, the Sun would be directly over the equator and appear on the horizon from both poles. There would be 12 hours of sunlight and 12 hours of sunlight. Then, the process would reverse—the North Pole would be in darkness, daylight hours shorter in the Northern hemisphere, and by six months into the year the South Pole would be directly facing the Sun and the Sun would be at the zenith there. Over the next six months, the trends would reverse. After three more months, 116

the Sun would be overhead at the equator again and the North Pole would come out of darkness; three months after that the Sun would be overhead at the North Pole.

Start with “summer” at the North Pole: For Seattle, the Sun will be 48 degrees above the northern horizon. As the year progresses, the Sun will approach the zenith (!). Once the Earth has moved 48 degrees along its orbit, it will begin to set during each rotation of the Earth. Then it will pass into the southern part of the sky. When the Sun is directly over the equator, it will be 42 degrees above the southern horizon and the daylight will be 12 hours long. Slowly the Sun will sink below the southern horizon, and it will be completely dark for part of the year, as when the Sun is above the South Pole, it is at an angle of 138 degrees from our zenith in Seattle.

Slowly twilight will come and the Sun will peek up over the southern horizon, heading toward our zenith and the North Pole.

Unit 47 11. Astronomers think that the diffuse outer ring of Saturn is replenished by material from Enceladus’s geysers that escapes from the Moon. Calculate the escape velocity from Enceladus, given its radius of 252 km and mass of 1.08 × 1018 kg.

11. (860 km/hr) From Unit 18.2, the escape velocity is given by Vesc = (2 G M/R)1/2. Using its radius of 252 km and mass of 1.08 × 1018, the escape velocity of Enceladus is Vesc

= (2 × 6.67 × 10−11 m3/s2 ∙ kg × (1.08 × 1020 kg) / (2.52 × 105 m))1/2 = 239 m/s = 860 km/hr.

12. Earth’s Moon experiences tidal forces that are gradually expanding its orbit by a few centimeters per year (Unit 19.4). Explain, using vectors, how a moon orbiting faster than its planet rotates will experience a “tidal drag” that causes its orbit to shrink.

12. In the diagram, the moon orbits and the planet spins counterclockwise, but the moon orbits faster than the planet spins. The resulting situation is the opposite of the Earth and our Moon; Earth rotates faster than the Moon orbits and so the tidal bulges lead the moon, pulling it forward in its orbit and transferring angular momentum from the planet to the moon. The Earth slows and the Moon’s orbit gets larger. Here, the bulges lag the moon, exerting the drag shown in the figure on the orbit. The moon’s orbit decreases as angular momentum is transferred from the moon to the planet—the moon is acting to speed up the rotation of the planet and the planet to slow down the motion of the moon in the orbit, towards a state of tidal lock with the bulge permanently under the moon. Actually, the moon’s velocity may speed up as result of the momentum transfer. L = MVR, and even though LMoon decreases, because the velocity of a Keplerian orbit increases more slowly than the radius decreases (see U39 QP 15) the drop in radius can be accompanied by the necessary increase in velocity for the new, smaller orbit to be stable (however the tidal effects will still be a problem).

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12. (cont.)

13. The top-right of Figure 47.14 shows a pair of shepherd moons and one of Uranus’s rings.

Assume that the moons and ring particles are orbiting counterclockwise in the image. a. Qualitatively describe the relative speeds of the moons and ring particles as they pass each other. b. If a passing moon boosts the speed of a ring particle, what will happen to the size of the ring particle’s orbit? c. If a ring particle slows when it passes a slower moon, what happens to the size of the ring particle’s orbit? d. Explain how two shepherding moons might drive ring particles into a narrow stream.

13. a) The inner moon is fastest, then ring particles, then the outer moon, according to Kepler’s laws. b) The inner moon will pull on the ring particles, causing them to accelerate, and enlarge the semi-major axis of their orbit. c) Since the outer moon is more massive than the particles, it will cause the ring particles to decelerate and this will result in a decrease in the semi-major axis of their orbit. d) The combined effect is that inner ring particles move up and outer particles move down, forming a narrow stream of particles in the middle.

14. Examine Appendix Tables 5 and 7 to determine which moons orbit inside the Roche limit around each of the giant planets. Make a list of these moons and discuss how it is possible for these moons to orbit there. Would you expect these moons to be spherical or irregularly shaped? Why? 14. The Roche limits are 2.44 times the planet’s radii: Planet Limit Interior Moons in Table 7 Jupiter – 174440 km Metis, Adrastea, Amalthea, Thebe Saturn – 147053 km Pan, Atlas, Prometheus, Pandora 118

Uranus – Neptune –

62364 km 60424 km

Cordelia, Ophelia, Bianca, Cressida Naiad, Thalassa, Despina

Objects can survive to orbit inside the Roche limit if they are not held together by their own gravity. These moons would be expected to be small and irregular, since if they were large enough for gravity to smooth them into spheres, they would be subject to destruction by the tidal forces. These bodies must be held together by the (chemical) strength of the rock or ice that makes them up.

15. Suppose a moon that was 100 kilometers in diameter broke up into particles just 1 millimeter in diameter. a. What was the surface area of the moon before it broke up? b. How many 1-mm particles could be made from the moon? c. What is the total surface area of all of the 1-mm particles you found in part b? d. What is the ratio of the surface area in part c to the surface area in part a? Explain why this ratio also represents how much sunlight would be reflected by the particles versus the original moon.

15. (a) 3.1 × 1010 m2; (b) 1024 particles; (c) 3.14 × 1021 m2; (d) 1011. (a) Surface Area = πD2 = 3.1 × 104 km2 = 3.1 × 1010 m2; (b) VMoon/Vparticle = (4/3 πR3) / (4/3 πr3) = (100 km/1 mm)3 = (105/10−3)3 = 1024 particles. (c) 1024 particles × πD2 = 1024 particles × π(10−3 m)2 = 3.14 × 1021 m2. (d) 3.14 × 1021 m2 / 3.1 × 1010 m2 = 1011. There’s about 100 billion times the surface area. Since all the particles are all spherical, and made of the same materials as the original moon, this is also the difference in how much sunlight would be reflected – the geometric effects cancel out and the albedo will be the same.

Unit 48 13. Io, Europa, and Ganymede orbit at distances from Jupiter’s center of 421,800 km, 671,100 km, and 1,070,400 km, respectively. Calculate how many times longer it takes for Europa and Ganymede to orbit Jupiter than for Io to orbit. (Hint: You do not need to know Jupiter’s mass to calculate the ratio of times.)

13. (2 and 4 times longer, respectively, for Europa and Ganymede). One form of Newton’s version of Kepler’s Third Law (see Unit 17) yields A3 = M × P2. So the ratio of periods will be: 3 AEur opa

PeriodEuropa PeriodIo

=

M Jupiter AI3o

=

3 AEur opa

AI3o

M Jupiter

for Europa, and by the same method, 4 times longer for Ganymede.

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3/2

æ671,100 km ÷ ö = çç = 2.01 ÷ ÷ èç 421,800 km ø

14. Use the modified form of Kepler’s third law (discussed in detail in Unit 17) to calculate Pluto’s mass from the orbital data for Charon given in the text. Be sure to convert the orbital period to seconds and the orbital radius to meters before putting those numbers into the formula.

14. (About 1.452 × 1022 kg). Newton’s version of Kepler’s 3rd is (M1 + M2) = a3/P2, where a should be in AU and P should be in years. Here a = 19600 km = 1.306 × 10−4 AU; P = 6.387 d = 1.75 × 10−2 years. Substituting these yields M1 + M2 = 7.27 × 10−9 solar masses × 1.99 × 1030 kg/(solar mass) = 1.45 × 1022 kg. Note that this is actually the combined Pluto-Charon mass, Pluto is about 1.31 × 1022 kg.

15. Calculate the density of Eris based on the data given in Table 48.2. (Be sure to convert kilometers to meters, and remember that there are 1000 liters per cubic meter.) Is it likely that Eris has a large iron core? Why? 15. (About 1.26 kg/liter). Assuming Charon to be spherical, V = 4/3 πR3. With r = 593 km = 5.93 × 105 m, V = 8.73 × 1017 m3 = 8.735 × 1020 L. The mass M = 1.1 × 1021 kg, and so Density = M/V = 1.26 kg/liter. Charon has little if any iron, and must be mostly ice (1 kg/L) with a little rocky material (3 kg/L).

16. Calculate the angular size of Pluto in arc seconds as seen from Earth, based on the data given in Table 48.2.

16. (Angular size = 2.27 × 10–5 degrees or 0.08 arcsec.) Use the angle-distance relation: A = L/d × 57.3° = (2320 km)/(39.5 × 1.5×108 km) × 57.3° = 2.27×10−5 ° = 0.08 arcsec.

17. From conservation of energy considerations it is possible to estimate the ejection speeds from one of Io’s volcanoes using the formula V = 2 gh where h is the maximum height above the surface the material reaches, and g is the surface gravity (Unit 16.3). If we take the maximum height as 200 km, what is the initial ejection speed? The mass and diameter for Io are given in Table 48.1.

17. (849 m/s). The surface gravity of Io is: g = G M/R2 = (6.67 × 10−11 m3/s2 ∙ kg) × (8.93 × 1022 kg) / (1.82 × 106 m)2 = 1.80 m/s2.

Using this value we get an ejection speed of: v = (2 (1.80 m/s2) × (2.00 × 105 m)) = 849 m/s.

18. Triton’s orbit is tilted about 157° with respect to Neptune’s equator, while Neptune’s equator is tilted about 28° with respect to the plane of its orbit around the Sun. a. Draw the Neptune–Triton system, marking the plane of Neptune’s orbit and setting up the tilts and inclinations correctly.

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b. Tidal forces cause Earth’s Moon to gradually spiral outward. Draw a diagram similar to the one shown in Figure 19.5 for Earth’s Moon to explain why Triton will instead spiral inward. c. Is it possible that any of the giant planets might have previously “swallowed” a moon?

Explain your reasoning.

18. (a) The relative orientations and motions work out as in the diagram below.

(b) As shown in the diagram, because of the opposite motion of Neptune’s rotation and the orbit of Triton, the tidal bulges will lag behind the moon’s motion, exerting a tidal drag on Triton (see also U47 QP 12). The net result of this drag force will be to transfer angular momentum from the Moon to Neptune, which will result in a decrease in Triton’s orbital radius (L = MVR, and the velocity will actually go up for a smaller orbit, but not as fast as R goes down). The new orbit will have the same problem, so the moon will continue to experience a drag force and eventually collide with Neptune.

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18. (cont.) (c) It’s entirely reasonable to believe that giant planets may have swallowed moons – the many ring systems around the gas giants are good evidence that moons have migrated inwards and been destroyed at the Roche limit. Certainly there are moons in retrograde orbits, which would be likely to spiral in.

Unit 49 10. Use Kepler’s third law to find the semimajor axis of Halley’s comet, given that its orbital period is 76 years.

10. (About 17.9 AU). As shown before, Kepler’s Third Law yields A = P2/3. If the Period is 76 years, then the semimajor axis (A) in AU is 762/3 or 17.9.

11. Halley’s Comet has a mass of 2.2 × 1014 kg and loses about 1∕1000 of its mass for every trip it makes through the inner Solar System. a. The first recorded observation of Halley’s comet was in 240 b.c.e. Assuming a 76-year period, how many times has the comet passed through the inner Solar System since it was first observed? b. How much mass has Halley’s comet lost since its first sighting? What percentage of the total mass does this represent?

11. (a) Almost 30 times; (b) less than 3% of its mass). a) Since 240 BCE there have been 2013 − (−239) = 2252 years (there was no year zero). Dividing this value by the orbital period for Halley’s gives the number of passages through the inner Solar System, n = 2252 yr/76 yr = 29.6. b) The total mass lost since 240 BCE is: M = (2.2 × 1014 kg) × (1/1000 passage) × (29 passages) = 6.4 × 1012 kg.

This is only (6.4 × 1012 kg) / (2.2 × 1014 kg) = 2.9% of the mass.

12. Comet McNaught’s orbit ranges from 0.17 AU out to 4100 AU. Use Kepler’s third law to determine the comet’s period.

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12. (9.28 × 104 years). The semi-major axis is half the sum of the closest and farthest distance, a = (0.17 AU + 4100 AU)/2 = 2050 AU. Kepler’s third law then gives a period of P = a 3/2 = (2050)3/2 = 9.28 × 104 years, about 93 thousand years.

13. Use data from the text to estimate the total kinetic energy, in joules, of the impact of NASA’s Deep Impact probe onto Comet Tempel 1. If the comet has a mean diameter of about 6 km, a density of about 0.8 kg/L, and is orbiting at about 30 km/sec, what is its kinetic energy?

13. (Impact energy = 1.85 × 1010 J; Kinetic energy = 4.1 × 1022 J). Deep Impact sent a 370 kg probe into the comet at a speed of 10 km/s (from the text) so the kinetic energy of the impact is: Ek = ½ mv2 = ½ (370 kg) (10,000 m/s)2 = 1.85 × 1010 J. With an average diameter of 6 km, the radius of Tempel 1 is 3 km = 3000 m. We know its density, and can determine its mass and kinetic energy: Mcomet = density × 4/3 πR3 = 0.8 kg/L × 1.1 × 1014 L = 9 × 1013 kg.

Ekcomet = ½ mv2 = ½ (9 × 1013 kg) × (30,000 m/s)2 = 4.1 × 1022 J. 14. The temperature, T, of a body decreases as the square root of its distance, d, from C onstant the Sun increases: T = . If the temperature at 1 AU is 250 K, estimate the d temperature of a comet nucleus in the Oort cloud.

14. (Temperature ~1.1 K). From the equation T = constant/√d, the temperature decreases by a factor of the square root of the change in d as d increases. If we take the distance to the Oort Cloud to be ~ 50,000 AU, that’s 50,000/1 = 50000 times further, so the temperature drops to 1/√50000 = 1/224 the value at 1 AU, or 250 K/224 = 1.1 K. That very cold!

15. As a comet orbits the Sun, its temperature changes approximately as described in the previous problem. a. Calculate the distance from the Sun at which it would reach the sublimation temperature of water (which sublimates at T = 180 K). Repeat for ammonia (130 K) and carbon dioxide (80 K). b. Based on your results in (a), describe how a comet would behave as it orbits the Sun. c. What would be the effect on the comet’s temperature if it had dark regions on its surface?

Qualitatively, how would this affect its behavior as it orbited the Sun? Explain your reasoning.

15. (a) For sublimation temperature of water, 1.9 AU, ammonia 3.7 AU, CO2, 9.7 AU. a) From #14, we know Temperature decreases a factor of the square root of the change in d as d increases, or T = constant/√d, and for d = 1 AU, T =250 K. To drop to the temperature where water sublimates at 180 K, 250/180 is a drop of 1.39 times in temperature, so the distance must increase by 1.392 = 1.9 times. The sublimation of water would be reached at 123

1.9 AU. For Ammonia, 250/130 = 1.92, and the distance must be (1.92)2 × 1 AU= 3.7 AU. For carbon dioxide, 250/80 = 3.125 and d = (3.125)2 × 1 AU = 9.7 AU. b) Based on the results from (a), a comet would sublime different compounds at different distances from the Sun (thus growing its coma and tail). As it passed a distance of 10 AU (80 K), carbon dioxide would sublime. At 4 AU (130 K), ammonia would sublime. At 2 AU (180 K), water would sublime. c) The dark material would absorb radiation better, so the comet would be a little bit warmer— at least at those locations. Materials might evaporate earlier, or more efficiently, from those regions, which could result in instability and stress on the nucleus as different regions on the comet evaporated at different rates. Regions might even break off or explosively blow out if the temperature differences were extreme enough.

Unit 50 7. Each day about 100 tons (105 kg) of meteoritic material strikes Earth. What fraction of Earth’s current mass does this represent?

7. (2 × 10−20 of Earth’s current mass). Each day, the Earth’s mass increases by 105 kg.

The fraction of Earth’s total mass that this amount equals is (105 kg)/(5.97 × 1024 kg) = 2 × 10−20.

8. Use the formula for the kinetic energy of a moving body to estimate the energy of impact of a 106-kg object hitting Earth at 30 km/sec. Express your answer in kilotons of TNT, using the conversion that 1 kiloton is about 4 × 1012 joules. Note: Be sure to convert kilometers/second to meters/second.

8. (108 kT). Using Ek = ½ mV2 with a mass m = 106 kg and velocity V = 30,000 m/s, KE = ½ (106 kg) × (30,000 m/s)2 = 4.5 × 1014J and in kilotons of TNT: KE = 4.5 × 1014J / (4.18 × 1012 J/kT) = 108 kT of TNT 9. Suppose a major impact blasted out rock from a 100-km-diameter crater that was, on average, 1 km deep. If all that material became dust that settled out evenly over the entire surface of Earth, how thick a layer would it form?

9. (About 1.54 cm). The volume of material thrown out would be roughly the volume of the crater, which is the surface area of the crater, at diameter 100 km, times the depth of one kilometer. The surface area is R2 so the volume = π (100 km/2)2 × 1 km = 7854 km3. The surface area of Earth (a sphere) is 4 R2, so with a radius of 6378 km, that surface area is about 5.11 × 108 km2. To get the depth to which material settles, divide the volume by the surface area:

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7854 km3 / (5.11 × 108 km2) = 1.54 × 10−5 km.

And 1.54 × 10−5 km × 105 cm/km = 1.54 cm.

10. There is about a 1 in 45,000 chance that the potentially hazardous asteroid named Apophis will impact Earth in 2036. The asteroid is estimated to have a diameter of 0.25 km and a mass of 2.1 × 1010 kg. Its impact velocity would be about 12.6 km/sec.

If the mass may be uncertain by a factor of 3 and the impact velocity uncertain by a factor of 2, what is the range of resultant impact energy in joules?

10. (The low end is ~ 1.4  1017 J; high end, 2.0  1019 J).

The kinetic energy Ek = 1/2 mV2 so the lowest KE will be from the lowest mass and lowest velocity (and vice versa for highest KE). Taking the uncertainty mentioned in the question to mean the mass could be as much as 3 m, or as little as m/3, and the velocity 2 V or v/2, we get a low end of the range for impact energy using the smaller values: E lower end = ½ (m/3)(v/2)2 = 1/12 Ek, and the higher end of the range of impact energies as: E high end = ½ (3 m)(2 V)2 = 12 Ek. Ek = ½ (2.1 × 1010 kg)(12600 m/s)2 = 1.67 × 1018 J, so the low end is ~ 1.4  1017 J; high end, 2.0 × 1019 J. (This is approximately 33,000 − 5 million kilotons of TNT.) 11. An impact crater’s diameter is typically 10 times the impactorʼs diameter. If there is an uncertainty in the mass of Apophis of a factor of 3, what is the uncertainty in the asteroid’s diameter? What range in crater size would this imply?

11. (Apophis diameter ranges between 173 m and 360 m; the impact crater size would then be from 1.73 km to 3.6 km.)

Using the same interpretation as in Quantitative Problem #10, the factor of 3 uncertainty in the mass means the asteroid could be 3x bigger or smaller, since volume = mass/density, and volume goes as R3, the upper limit would be (3)1/3 = 1.44 times the given radius, and the lower limit (1/3)1/3 = 1/1.44 = 0.69 times the given radius. The uncertainty in the diameter would be the same (a factor of 1.44, or the cube root of 3). The range of diameters for the 250 m asteroid would be 173 m to 360 m, and the impact crater would be 1.73 km to 3.6 km.

12. It is most likely that an asteroid will impact in an ocean rather than on land. Here is a simplified formula that gives the height of the wave (h), in meters, at a distance of 1000 km from the impact site:

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3

2

æ D ö æV ö ær ö çç ÷ ´ çç ÷ ÷ ´ ççç ÷ ÷, çè 406÷ ÷ ø çè 20÷ ø è 3÷ ø

h = 8´

where D is the asteroid diameter in meters, V is the velocity in km/sec, and ρ is the asteroid density in kg/L. Calculate the expected tsunami height in meters at 1000 km from the impact if Apophis happens to land in the middle of an ocean. How would you expect the tsunami height to vary with distance from the point of impact?

12. The height of the wave at 1000 km is, and from #10, the asteroid’s diameter ~ 0.25 km = 250 m, giving a volume (if spherical) of 8.18 × 106 m3. Its mass is estimated to be 2.1 × 1010 kg, and therefore its density is 2.6 × 103 kg/m3 = 2.6 kg/L. The impact velocity was 12.6 km/sec. Using these values, and the formula given: 3

2

æ D ö÷ æV ö÷ ær ö÷ çç ÷´ ç ÷ ´ ç ÷ çè 406ø÷ èçç20ø÷ èçç 3ø÷

h = 8´

we find that 1000 km from the impact, the height of the wave created would be just 3

= 8´

2

æ250ö÷ æ12.6ö÷ æ2.6ö÷ çç ÷´ ç ÷ ´ ç ÷ = 2.27 m çè 406ø÷ èçç 20 ø÷ èçç 3 ø÷

Not too bad considering what happens when earthquakes occur in the ocean. But, this is 1000 km away from the impact site. Since the wave will spread out like a ripple in a pond, we might expect the height to be much taller near the impact. However, it will also depend on how deep the ocean is—waves get taller as they come in to shore.

Solutions to End of the Unit Problems Part IV: Stars and Stellar Evolution Unit 51 10. The Sun’s angular size is 0.5°, and its distance is 1.5 × 10 8 km.

Use this information to calculate the Sun’s diameter.

10. (1.31 × 106 km). Using the formula L = d × (A/57.3°), where d is the known distance and A is the angle subtended, the diameter is L. With d = 1.5 × 108 km and A equal to 0.5 degrees, this is 1,308,901 km. Round this to three significant digits.

11. The central density of the Sun is about 160 kg/L. Estimate how much a chunk of this material the size of a sugar cube would weigh on Earth.

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11. (1.57 N) One side of a sugar cube is about 1 cm in length, making its volume 1 cm3 = 10−6 m3 = 10−3 L. The mass of a sugar-cubed sized chunk of the Sun’s interior is (160 kg/L) × (10−3 L) = 0.160 kg.

To get its weight we only need to multiply by the surface gravity of Earth, W = (0.160 kg) × (9.8 m/s2) = 1.57 N. 12. How much mass would the Sun have to lose each second in order to lose 10% of its mass in 10 billion years?

12. (6.3 × 1011 kg/s). There are about 3.16 × 1017 seconds in 10 billion years, and 10 percent of the Sun's mass is about 1.99 × 1029 kg. Dividing the mass by the seconds we get 6.3 × 1011 kg/s.

13. At the rate at which the Sun is losing mass in the solar wind, how many years does it take the Sun to lose as much mass as the mass of the entire Earth?

13. (126 million years). According to Sec. 51.4, the Sun loses about 1.5 million tons (1.5 × 109 kg) per second. The mass of the Earth is 5.97 × 1024 kg. Divide the Earth's mass by the mass loss rate to get the seconds = 5.97 × 1024kg / (1.5 × 109kg/s) = 3.98 × 1015 seconds or 1.26 × 108 years.

14. Wolf-Rayet stars have very strong stellar winds, losing as much as 10−5 M⊙ per year. If a typical Wolf-Rayet star is 20 M⊙, what percentage of its mass might it lose in a million years?

14. (50%). (10−5 solar masses/year × 106 years)/20 solar masses = 0.5 = 50% of their mass is lost in a million years.

15. Use Wien’s law to calculate the wavelength of highest emission for the bottom and top of the photosphere where the temperatures are 7500 K and 4500 K, respectively. What part of the electromagnetic spectrum do these wavelengths correspond to?

15. (387 nm; 644 nm). Wien’s law states that T = (2.9 × 106 K ∙ nm)/λmax which can quickly we rearranged to read λmax = (2.9 × 106 K ∙ nm)/T. For the bottom of the photosphere λmax = (2.9 × 106 K ∙ nm)/7500 K = 387 nm which, from Unit 22, is on the ultraviolet/visible light boundary of the electromagnetic spectrum. For the top of the photosphere, λmax = (2.9 × 106 K ∙ nm)/4500 K = 644 nm, which is about orange visible light.

16. If a packet of gas rising through the interior of the Sun reduces its temperature from 1.5 million K to 10,000 K and has a density decrease from 120 kg/L to 5.5 kg/L, by what factor has the pressure on it changed?

16. (About 1/3300 as much). The ideal gas law states P = Constant × T × Density. The packet has a decrease in temperature of a factor of (1.5 × 106/104) = 150. This could happen from a decrease in the pressure by the same factor since P is proportional to T. However, the density also decreases, a factor of (120/5.5) = 22 times. This would happen from a decrease in the pressure by the same factor (22) since P is also proportional to density. For both changes to happen together, the pressure must now be 1/3300 its previous value (150 × 22 = 3300).

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Unit 52 10. In chemical reactions, when energy is released, mass is lost according to Einstein’s formula, just as it is for nuclear energy. a. When a kilogram of coal is burned, it releases about 40 million joules of energy.

What amount of mass is lost? b. How many carbon atoms (mass = 2.0 × 10−26 kg) would it take to match the amount of mass that is lost?

10. (a) 4.44 × 10−10 kg; (b) 2.2 × 1016 carbon atoms. (a) E = mc2, so Mass = E/c2 = (4 × 107 kg m2 s−2)/(9 × 1016 m2 s−2) = 4.44 × 10−10 kg. (b) For the number of carbon atoms, divide the needed mass loss by the mass of a single carbon atom. Number of carbon atoms = 4.44 × 10−10 kg / (2 × 10−26 kg) = 2.2 × 1016. Note that this is simply the number of carbon atoms that would have that mass—it is not expected that the carbon atoms would necessarily actually be converted into energy—there is binding energy in the fuel (as a somewhat similar example, remember that relativistic particles have a higher mass simply because they have energy).

11. Worldwide, humans use about 4 × 1020 joules of energy each year. A typical power station generates about 109 watts of power, while the Sun’s luminosity is about 4 × 1026 watts. (Recall that 1 watt = 1 J/sec.) a. How many power stations would be needed to produce power equivalent to the Sun’s? b. How many typical power stations are needed to supply current human energy usage? c. If we could store all the energy that the Sun produces in 1 second, how long would it last humans at current use rates?

11. (a) 4 × 1017 stations; (b) 1.3 × 104 stations; (c) 106 yr (a) 4 × 1026 W / (109 W/station) = 4 × 1017 stations. (b) Human power usage: 4 × 1020 J/(3.16 × 107 s) = 1.27 × 1013 W. So 1.27 × 1013 W / (109 W/station) = 1.27 × 104 stations (c) Sun produces 4 × 1026 J in 1s. 4 × 1026 J / (4 × 1020 J/y) = 106 yr.

12. The Sun’s luminosity indicates how much energy it radiates each second. Starting from the Sun’s luminosity, 3.83 × 1026 watts, calculate how much mass must be converted to energy each second in order to maintain hydrostatic equilibrium. (Recall that 1 W = 1 J/sec.)

12. (4.3 × 109 kg/s). The conversion from mass to energy is governed by E = m × c2. Rearranging this for mass, m = E/c2, and substituting in the value for the Sun’s luminosity for the energy, gives the how much mass is converted each second: Δm/Δs = L/c2 = (3.83 × 1026 J/s)/(3.0 × 108 m/s)2 = 4.3 × 109 kg/s.

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This agrees with the value found section 52.2 but was derived from a different starting point.

13. If the Sun generates 3.83 × 1026 watts throughout its lifetime, a. how much mass will our Sun convert to energy over its entire main-sequence lifetime (1010 years)? b. what fraction is this of the mass of the Sun? c. how does this compare to the mass of Earth?

13. (a) 1.4 × 1027 kg; (b) 7 × 10−4; (c) 234 Earth masses. (a) 3.83 × 1026 J/s × 3.16 × 107 s/y × 1010 y = 1.21 × 1044 J over the Sun’s lifetime. m = E/c2 = (1.26 × 1044 J) / (3 × 108 m/s)2 = 1.35 × 1027 kg. (b) 1.35 × 1027 kg / (1.99 × 1030 kg) = 7 × 10−4, or less than 1/1000th. (c) This is (1.35 × 1027 kg)/(5.97 × 1024 kg) = 226 Earth masses. (It’s closest in mass to Jupiter at 318 Earth masses.)

14. One “kiloton” of explosive energy is 4.2 × 10 12 joules. Using Einstein’s formula, calculate the mass that is converted into energy in a 1-megaton H-bomb explosion.

14. (0.05 kg). A kiloton is 1000 tons and 1 kT = 4.2 × 1012 J. A megaton would be a million tons, or 1000 times as much as a kiloton, 1 MT = 4.2 × 1015 J. m = E/c2 = (4.2 × 1015 J) / (3 × 108 m/s)2 = 0.0467 kg ~ 0.05 kg.

15. How long does it take a neutrino produced in the center of the Sun to reach the Sun’s surface?

15. (2.3s). Since neutrinos do not interact with matter very frequently, neutrinos mostly move out of the Sun as if it were not there. They travel at just under the speed of light, so neutrinos leave the Sun in a couple of seconds: 6.96 × 108 m / (3 × 108 m/s) = 2.3 s.

Unit 53 13. If the surface of the Sun near the equator completes one rotation every 25 days, compared to a 30-day rotation near the poles, how long does it take for the surface near the equator to complete one extra rotation around the Sun compared to the poles?

After 11 years, how many extra turns would this result in?

13. (125 days; 32 extra turns).

Every rotation, the equator finishes 5 days ahead of the pole. Since the period at the equator is 25 days, after 5 rotations the equator will “lap” the pole. 5 rotations at the equator takes

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5 × 25 days = 125 days. In 11 years, there are 11 × 365.25 = 4018 days, which would be 4018/125 ~ 32 extra turns for the equator.

14. What is the speed of material ejected by a solar flare if it takes four days to reach Earth?

14. (434 km/sec). Four days = 96 hours or 345,600 seconds. The average distance between Sun and Earth is 1.5 × 108 km. Divide distance by time to get speed: 1.5 × 108 km / 345600s = 434 km/sec.

15. How long does it take the material ejected by a solar flare to reach the orbit of Neptune?

15. (1.1 × 107 s, or 0.36 years). According to the text, solar flare material moves at about 400 km/s. Neptune is at 30.05 AU, so it will take t = (30.05 × 1.5 × 108 km) / (400 km/s) = 1.1 × 107 s, which is about 1/3 of a year. (1.1 × 107 s / (3.16 × 107 s/y) = 0.36). Note that one could also apply the results from Q14 (434 km/s), and/or argue that since Neptune is 30 times as far from the Sun as Earth, if it takes 4 days for the material to reach Earth it should take 4 d × 30 = 120 d, which is about a third of a year.

16. The amount of light emitted by a hot region is proportional to the temperature to the fourth power (L ∝ T4). If a sunspot is at a temperature of 4500 K, how much less luminosity does it generate than if it were at 6000 K? 16. (Roughly 68 percent less, or 3 times less, or 1/3 the luminosity). Since L goes as T4, the cooler spot is (4500/6000)4 = (3/4)4 = 27/64 = 0.32 times as luminous. So the 4500 K sunspot is only about 32 percent as luminous as an equivalently sized 6000 K section of the photosphere.

17. The largest solar flares can release as much as 6 × 10 25 joules of energy. How does this energy compare to the total energy output of the Sun each second?

17. (About 15%) The total energy output of the Sun per unit time is the Sun’s luminosity, 4 × 1026 J/s. Therefore, a solar flare that releases 6 × 1025 J of energy is about (6 × 1025 J)/ ( 4 × 1026 J) = 0.15 = 15% of the total energy released by the Sun every second.

Unit 54 13. Using triangulation, make an approximate measurement of the distance from where you live to a neighboring building. Make a scale drawing, perhaps letting 1 cm represent 1 m (or you might let 1 inch represent 10 feet), and choose a baseline. Construct a scale triangle on your drawing and find how far apart the buildings are from one another.

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13. As an example, consider two houses which are (at least approximately) aligned on their plots as shown in the example of a possible scale diagram.

First one could sight from the corner of one house to the corner of the other. Then, the baseline might be the edge of one house, say, 8m long. Then, from the other end of the baseline one would measure the angle from the baseline to the sightline to the corner of the far house, 63° in this example. On a scale drawing, the intersection of the first and second sightlines can be measured to give the distance. Using trigonometry, for the example the distance is D = b × tan (A) = 8 m × tan 63° = 15.7 m. The house is about 16 m away. Of course, in many cases the edges of the two buildings wouldn’t be aligned. The only really important thing is to make sure that the baseline extends out at a right angle from the first sightline.

14. Procyon has a parallax of 0.284 arcsec. How far away is it in parsecs and in light-years?

14. (3.52 pc; 11.5 ly). From the parallax measurements, distance (in pc) = 1/parallax (in arcseconds) = 1/0.284 = 3.52 parsecs.

There are 3.26 light-years in a parsec, hence, d = (3.52 pc) × (3.26 ly/pc) = 11.5 ly.

15. If a star is 5,000 pc away, what should its parallax be?

15. (It would have a parallax of about .0002 arcseconds).

Parallax (in arcseconds) = 1/distance (in pc) = 1/5000 = 0.0002.

16. If a star’s parallax is 0.1 arcsec on Earth, what would it be if measured from an observatory on Mars?

16. (0.152 arcsec). The parallax angle is derived from trigonometry—distance = baseline × tan A—for very small angles. Note that for a larger baseline, the angle (or tangent of the angle for large angles) must be proportionately larger. In the parallax formula, d = constant/p where the constant depends on the baseline; likewise: p = constant/d.

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The parallax angle p is proportional to the baseline, which is proportional to the constant. The baseline for measuring parallax from Earth is proportional to 1 AU; measured from Mars, the baseline is larger: 1.52 AU. So from Mars the parallax angle would be 1.52 times larger, or 0.152 arcsec.

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17. Suppose a satellite is developed that can measure the parallax of stars with an uncertainty of 2 × 10−5 arcsec. Suppose a star is measured with a parallax of 3 × 10−5 arcsec with this satellite. What range of distances could the star be at and still fall within the range of uncertainty of the measurement?

17. (2 × 104 pc to 105 pc).

If the measurement is 3 × 10 −5 with an uncertainty of 2 × 10 −5, the minimum possible angle is 1 × 10−5, which corresponds to d = 1/p = 105 pc. The maximum parallax angle is 5 × 10 −5, which corresponds to d = 1/p = 1/(5 × 10−5) = 2 × 104 pc.

18. Barnard’s Star has a parallax of 0.545 arcsec, a proper motion of 10.3 arc seconds per year, and a radial velocity of −110 km∕sec (the negative sign means it is approaching Earth). What is the star’s overall speed through space?

18. (142 km/s). The distance to Barnard’s Star is d = 1/0.545 = 1.83 pc.

The transverse velocity is: VT = 4.74 km/s × arcsec/yr × dpc = 4.74 km/s × (10.3) × (1.83) = 89.3 km/s.

The overall speed is V = √(VR2 + VT2) = √( (110 km/s)2 + (89.3 km/s)2 ) = 142 km/s.

19. How much greater is the angle of aberration of starlight measured from Mercury compared to that measured from Earth? What would the angle of aberration of starlight be as measured from Saturn?

19. (1.6, 1/3).

The angle of aberration of starlight scales directly with orbital speed. It’s easiest to compare orbital speeds by remembering that for Keplerian orbits, the velocity goes as the square root of the radius. Mercury orbital radius is 0.387 the Earths, so it’s speed is 1/√(0.387) = 1.6 times the Earth’s, and the aberration angle will be 1.6 times as much. At 9.582 AU, Saturn moves 1/√9.582 ~ 1/3 as fast as the Earth, so the angle there would be 1/3 as much as on Earth.

Unit 55 9. The faintest visible stars deliver only about 10−14 watts to your eye. A typical visible-wavelength photon has an energy of about 3 × 10−19 joules. How many photons per second are you seeing?

9. (3.33 × 105 or 333,000 photons per second).

A watt is one joule per second, so the question is how many times will 3 × 10−19 J go into 10−14 J? Just divide: 10−14 J/s /(3 × 10−19 J/photon) = 333,333 photons per second.

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10. Starting with the Sun’s luminosity of 4 × 1026 watts, and the typical visible-wavelength photon energy stated in the previous problem, calculate how many photons per second your eye receives when you look at the Sun. Assume a pupil size of 3 mm?

10. (5.3 × 1017 photons per second) From Earth the Sun’s brightness is: B = L ꙩ/4 π d2 = (4 × 1026 watts)/(4 π (1.50 × 1011 m)2 ) = 1415 W/m2.

The amount of energy entering your eye each second is P = B × 4 π r2 = (1415 W/m2) × (4 π (0.003 m)2 ) = 0.160 W. Finally, dividing by the energy per photon gives the number of Solar photons entering your eye each second, N = (0.160 J/s)/(3 × 10−19 J) = 5.3 × 1017 photons per second. Don’t stare at the Sun! 11. Suppose star A is 36 times brighter than star B, but we have reason to believe that they have the same luminosity. a. Which is more distant? b. How many times more distant is it than the other star?

11. (a) Star B is more distant. Brightness varies as the inverse square of distance; if the two stars have the same Luminosity, the more distant star will appear dimmer. (b) If star B is 36 times dimmer than star A, it must be 6 times farther away.

12. Star C and star D are at the same distance from us, but star D is 10,000 times more luminous than star C. How do their brightness levels compare?

12. (Star D would be 10,000 brighter). Brightness and luminosity are directly proportional, so a change in one causes a proportional change in the other.

13. How do the magnitudes of stars C and D in Problem 12 compare?

13. (The magnitude of star C would be 10 more than the magnitude of stars D.) Let the magnitude and brightness of star D be m1 and B1, and the magnitude and brightness of star C be m2 and B2. Given that:

m 1 - m 2 = - 2.5 log

B1 = - 2.5 log(10,000) = - 2.5(4) = - 10 B2

the magnitude difference between the two stars is 10. Remember that a brighter object has a smaller numerical magnitude, with negative #s being even brighter. So Star D has a magnitude 10 less than star C.

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14. The star Deneb has an apparent magnitude of 1.25 and an absolute magnitude of −8.5. What two statements can you make about it, based on this data?

14. First, it is exceptionally luminous (among the most luminous stars in the sky). Second, it must be very far away from us for its apparent magnitude to be so low (it is probably around 3,000 light years away).

15. Star A has a magnitude of 3.5. Star B has a magnitude of −1.5. Which star is brightest, and by what brightness ratio?

15. (B is 100 times brighter). Star B has a more negative magnitude, and the magnitude scale has brighter objects with a smaller positive number, or even brighter objects with increasingly negative numbers. The brightness ratio is 3.5 – (–1.5) = 5 magnitudes, or 2.5125 = 100 times.

16. If you observed the Sun from Antares, what apparent magnitude would it have?

16. (About magnitude 11). Using the Appendix values, Antares is 550 ly from the Sun. 550 ly / 3.26 ly/pc = 169 pc. The Sun has an absolute magnitude (magnitude at 10 pc) of approximately 4.83. At 169 pc, it would be (10/169)2 = 0.0035 times as bright. This corresponds to a reduction in brightness of 1/0.0035 = 285 times. We need N for 2.512N = 285, so taking logs, N = log 285/log 2.512 = 6.14 magnitudes. The Sun would therefore appear to be a magnitude 11 (4.83 + 6.14 = 10.97 ~ 11) star as seen from Antares. (Alternatively, using the table in the Unit, the Sun is ~ 5, and the magnitude change ~ 6, again for 11).

17. Proxima Centauri is approximately the same distance from Earth (slightly less than 4.3 ly) as Alpha Centauri A, but it has an apparent magnitude of +11.05, compared to Alpha Centauri A’s magnitude of −0.01. How much more luminous is Alpha Centauri A than Proxima Centauri?

17. (About 26,600 times). Since both stars are roughly at the same distance, we can directly compare magnitudes. The magnitude difference is 11.05 – (–0.01) = 11.06. The luminosity difference is 2.51211.06 = 2.66 × 104 times. So Alpha Centauri A is 26,600 times more luminous that Proxima Centauri. (Note: the final number is very sensitive to the exponent. 2.51211 = 2.51 × 104.)

Unit 56 13. Suppose a star radiates most strongly at about 200 nm. How hot is it? What spectral type is it?

13. (14,500 K). Wien’s Law: T = 2.9 × 106 K nm/λmax =2.9 × 106 K nm/200 nm = 14500 K. From Table 56.1 we find that this a B spectral type star.

14. The bright southern star Alpha Centauri A radiates most strongly at about 500 nm. What is its temperature? How does this compare to the Sun’s?

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14. (5800 K). T = 2.9 × 106 K nm/ λmax, where λmax is in nanometers.

Thus, T = 2.9 × 106 K nm/500 nm = 5800 K. This is pretty much the same temperature as the Sun (Sec 56.3). Both are G2 stars.

15. Estimate the wavelength of maximum brightness for a star with dark calcium lines.

15. (~ 527 nm). From the right side of Figure 56.8 we see that G spectral type stars strongly absorb calcium. From Table 56.1 we find that G stars range in temperature from 5000 K to 6000 K. Taking the average temperature, and using Wien’s law, leads to a wavelength of maximum brightness of: λmax = 2.9 × 106 K nm/T = 2.9 × 106 K nm/ 5500 K = 527 nm, about green light.

16. If a T star has a surface temperature of 1000 K, at what wavelength should it be brightest?

16. (2900 nm or 2.6 microns). Use Wien’s law: 1000 K = 2.9 × 106 K nm/λmax so λmax = 2900 nm = 2.6 microns.

17. When our star reaches the end of its main-sequence lifetime, it will become an M-type red giant (Unit 65) with a surface temperature of approximately 3300 K. How will its wavelength of peak radiation differ from the current value?

17. (Peak wavelength will be almost twice as long (1.75 times)). The peak radiation wavelength at our star’s current temperature 5800 K = 500 nm. At 3300 K, it peaks at: λmax-nm = 2.9 × 106 K nm/T = 2.9 × 106 K nm/3300 K = 879 nm.

The peak wavelength is almost twice as long (879/500 = 1.75).

18. Suppose a star radiates most strongly at 330 nm. Would you expect to see strong absorption lines in its spectrum? Why or why not?

18. (Yes, you would see strong Balmer lines of Hydrogen).

2.9´ 106 K nm Using Wien’s law, we have T = = 8788K » 9000 K .This would be a star of 330 nm spectral type A, and would have strong hydrogen absorption lines. The star is hot enough to have completely stripped heavier elements of their electrons (or at least most of them). A temperature of 9000 K will have most of hydrogen’s electrons in the 2nd energy level, producing a strong Balmer line.

Unit 57 10. How far from the center of Earth is the center of mass between Earth (M⊕ = 5.97 × 1024 kg) and the Moon (MM = 7.34 × 1022 kg)? Is this point outside Earth?

(Hint: The Moon–Earth separation is 3.84 × 105 km.)

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10. (4680 km) Starting from ME × aE = MM × aM and using the given we arrive at: aM = (ME/MM) × aE = (5.97 × 1024 kg/734 × 1020 kg) × aE = 81 aE. Next we realize that the total Earth-Moon separation is the sum center of mass separations, a = aE + aM. Hence, a = aE + 81 aE = (1+81) aE which implies that: aE = a/(1+81) = 384 × 103 km/ (1+81) = 4680 km.

Earth’s radius is 6400 km so the center of mass is inside the Earth.

11. Two stars are in a binary system. Find their combined mass if they have an orbit with a. period, P, of 5 years and a separation, a, of 10 AU. b. period, P, of 2 years and a separation, a, of 4 AU. 11. (a) 40 solar masses (b) 16 solar masses. Use Kepler’s law: For the first example; (MA + MB) = a3/P2 = 103/52 = 40 b)

(MA + MB) = a3/P2 = 43/22 = 16

12. Two stars are orbiting each other, and the sum of their masses is 6 M⊙. It is found that star A of the pair is orbiting 3 arcsec from the center of mass, while star B is orbiting 6 arcsec from the center of mass. What are the two stars’ masses?

12. (MA = 4 solar masses; MB = 2 solar masses).

From Sec 57.3, MA × aA = MB × aB. So, 3 MA = 6 MB, which reduces to MA = 2 MB. Then we have 2 MB + MB = 6 Msun → MB = 2 Msun, and then MA = 2 MB = 4 Msun.

13. Find the distance from the center of the Sun to the center of mass between the Sun (Mꙩ = 1.99 × 1030 kg) and the following planets: a. Earth (M = 5.97 × 1024 kg; a = 1.50 × 108 km) b. Jupiter (M = 1.90 × 1027 kg; a = 7.78 × 108 km) c. Saturn (M = 5.68 × 1026 kg; a = 1.43 × 109 km) 13. The distances from the center of the Sun to the center of mass of the following planets is (a) Earth, 450 km; (b) Jupiter, 743,000 km; (c) Saturn, 408,000 km.

MSun × aSun = MPlanet × aPlanet can be re-arranged to be aSun = (MPlanetaPlanet)/MSun. All of the needed values are given in the problem. As an example, aSun = (MPlanetaPlanet)/MSun = (5.97 × 1024 kg × 1.5 × 108 km) / (1.99 × 1030 kg) = 450 km.

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14. When Jupiter and Saturn are on the same side of the Sun, is the center of mass inside or outside the Sun’s radius (R⊙ = 6.96 × 105 km)?

14. (Outside) We can directly calculate the expected distance by: MSun × aSun = MJupiter × aJupiter + MSaturn × aSaturn, which gives 1.2 × 106 km, which is larger than the Sun’s radius, 6.96 ×105 km. We could also estimate the maximum and minimum distances possible by computing the center of mass if all of Jupiter and Saturn’s combined mass was at Saturn’s distance or at Jupiter’s distance. We would expect the center of mass to be somewhere in between (and it is). For Saturn’s distance: aSun = (MPlanetaPlanet)/MSun = (2.468 × 1027 kg × 1.43 × 109 km) / (1.99 × 1030 kg) = 1.8 × 106 km.

This is outside the Sun’s radius, but Jupiter is far more massive than Saturn, so the center of mass of Jupiter and Saturn is closer to Jupiter. We need to check if all the mass was at Jupiter’s distance: aSun = (MPlanetaPlanet)/MSun = (2.468 × 1027 kg × 7.78 × 108 km)/(1.99 × 1030 kg) = 9.6 × 105 km.

This is STILL bigger than the Sun’s radius.

15. Two stars are orbiting each other, both 4.2 arcsec from their center of mass. Their orbital period is 420.3 years and their distance from Earth is 104.1 ly. Find their masses.

15. (54.5 solar masses). First of all, the two stars have the same mass, as they are both the same distance from the center of mass of the system. Their separation can be worked out from their apparent angular separation and their distance (unit 10). Their separation is 8.4 arc seconds (8.4 arcsec/3600 arcsec/° = 0.00233°). At a distance of 104.1 light years, this means an orbital separation of: d = 104.1 ly × (0.00233°)/(57.3°) = 0.00423 ly = 268 AU. The combined mass of the system is then (268 AU)3/(420.3 y)2 = 109 solar masses, so each star has a mass of 54.5 solar masses.

16. Using Kepler’s third law, we find that the combined mass of a binary system is 55.5 M⊙. The stars have separations of 6.3 arcsec and 2.5 arcsec from the center of mass, and the system is at a distance of 64 ly from Earth. Find their distances from the center of mass of the system and their individual masses.

16. (Call star A the one at 6.3 arcsec: its mass = 15.8 M⊙ and distance = 124 AU; call star B the one at 2.5 arcsec: its mass = 39.7 M⊙ and distance = 49 AU.) Find the distances of each from the center of mass from L = d × (A/57.3°) where d is the known distance and A is the angle subtended in degrees, the diameter is L in whatever units you are using for distance. This is easiest if you change the angles for the two stars from arcseconds to

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degrees by dividing by 3600. Their separations are 6.3 and 2.5 arc seconds which equals 0.00175° and 0.000694°.

LA = d × (A/57.3°) = (64 ly) × (0.00175°/(57.3°) = 0.00195 ly = 124 AU, and LB = d × (A/57.3°) = (64 ly) × (0.000694°)/(57.3°) = 0.000775 ly = 49.0 AU. Their masses can be found from M A × aA = MB × aB and MA + MB = 55.5 solar masses. Note that MA = MB × (aB/aA) so MB (aB/aA+1) = 55.5 solar masses. If B is the one at 2.5 arcsec, then it has a mass of: MB = (55.5 solar masses)/(2.5/6.3 + 1) = 39.7 solar masses and is at 49 AU. Star A at 2.5 arcsec must have a mass of 15.8 solar masses and is at 124 AU.

Unit 58 11. Alpha Centauri is actually two stars that orbit each other. Recent measurements show that Alpha Centauri A has an angular diameter of 0.0085 arcsec, while Alpha Centauri B has an angular diameter of 0.0060 arcsec. The pair are at a distance of 4.4 ly. What are their diameters, compared to the Sun’s?

11. (1.72 × 106 km for Alpha Cent A, which is about 1.24 times larger than the Sun’s diameter; 1.22 × 106 km for Alpha Cent B, or about 88 percent of the Sun’s diameter).

Using the formula L = d × (A/57.3°), where d is the known distance and A is the angle subtended in degrees, the diameter is L in whatever units you are using for d. This is easiest if you change d to km and change the angles for the two stars from arcseconds to degrees by dividing by 3600.

Thus, 0.0085 arcsec = 2.36 × 10−6 °; and 0.0060 arcseconds is 1.67 × 10−6 °. In km, the distance of the Alpha Centauri system is 4.4 ly × 9.5 × 1012 km = 4.18 × 1013 km.

For Alpha Cent A, this is L = 4.18 × 1013 km × (2.36 × 10−6/57.3) = 1.72 × 106 km.

For Alpha Cent B, L = 4.18 × 1013 km × (1.67 × 10−6/57.3) = 1.22 × 106 km.

The Sun’s diameter is (6.96 × 105 km) × 2 = 1.39 × 106 km. So the ratios for Alpha Centauri A and B relative to the Sun are, respectively: 1.72/1.39 = 1.24, and 1.22/1.39 = 0.88.

12. A star with the same color as the Sun is found to produce a luminosity 81 times larger.

What is its radius, compared to the Sun’s?

12. (9 solar radii). If the star is the same color it has the same surface temperature. From the text we have L = 4πR2 × T4 and so R =

L 2

T ´

4ps

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. Since nothing below the division bar

changes, we can say that R is proportional to (L). Since L = 81 times the luminosity of the Sun, the star must have a radius 9 times greater.

13. A star with the same radius as the Sun is found to produce a luminosity 81 times larger.

What is its surface temperature, compared to the Sun’s?

13. (3 times the Sun's surface temperature). Use the Stefan-Boltzmann relationship L = 4 πR2 × T4. The luminosity is 81 times that of the Sun with the same R, and L is proportional to T4. The fourth root of 81 is 3.

14. A certain class of stars, known as hypergiants, have extremely large radii. One particular hypergiant, VY Canis Majoris, has an estimated luminosity 270,000 times greater than the Sun’s but a surface temperature of only 0.6 times that of the Sun. What is the radius of this star? If it was located in our Solar System, which planets would be inside the star?

14. (About 1450 times larger).

From the text we have R =

L

. Since the luminosity of VY Canis Majoris is 270,000 T ´ 4ps times the Sun’s, and R goes as the square root of L, the change in L would increase R compared to the Sun by a factor of √(270,000) = 520. However, the temperature is 3/5 as much, and R goes as 1/T2. So that factor will be: 2

1/(3/5)2 = 1/(9/25) = 25/9 = 2.78.

The net effect is a star that is 520 × 2.78 = 1450 times larger than the Sun. The Sun’s radius is 7 × 105 km so the estimated radius of VY Canis Majoris is: R = 1450 × 7 × 105 km = 1.0 × 109 km = 6.78 AU. From Appendix Table 5 we see that this star is larger than Jupiter’s orbit.

15. If a star’s surface temperature is 3000 K, how much power does a square meter of its surface radiate?

15. (4.6 × 106 watts/m2). From the Unit we have energy radiated per square meter is given by  × T4, where  = 5.67 × 10−8 watts per square meter per Kelvin degree to the 4th power. So Energy/m2 = 5.67 × 10−8 watts/(m2 K4) × (3000 K)4 = 4.6 × 106 watts/m2.

16. Betelgeuse is roughly the same temperature as Proxima Centauri, the closest star to Earth aside from the Sun. However, Betelgeuse is 109 times brighter. How much larger in radius is Betelgeuse than Proxima Centauri?

16. (About 32,000 times larger radius). L  R2 T4 so 109 times the luminosity at the same temperature means a radius greater by a factor of (10 9)0.5 = 31622 ~ 32,000. So Betelgeuse has a radius 32,000 times larger than Proxima Centauri.

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17. A star is five times as luminous as the Sun and has a surface temperature of 9000 K. What is its radius, compared to that of the Sun? 17. (About the same size; or 0.93 times the size, depending on the value used for the temperature of the Sun). Since L  R2 T4, the result is very sensitive to the temperature because of the fourth power. If the surface temperature of the Sun is taken to be 5800 as in Sec. 56.3, the factor of change in T is an increase of 9000/5800 = 1.55 and the result is R = 0.93 solar. It the rounded value of 6000 K is used, the factor is 1.5 times and the R ~ solar.

Rµ

L = T4

5 = 0.93 or R µ 1.554

L = T4

5 = 0.99 » 1 1.54

18. Using the information that Betelgeuse has an angular size 0.05 arcsec and is 197 pc from Earth, show how you can calculate its diameter in kilometers.

A . For Betelguese, 57.3° d = 197 pc × 3.085678 × 1013 km = 6.08 × 1015 km.

A = 0.05 arc seconds × 1°/3600 arc seconds = 1.39 × 10−5 °

18. (1.47 × 109 km). From Unit 10, we have L = d´

Using these values, the diameter of Betelgeuse is L = 6.08 × 1015 km × 1.39 × 10−5 °/57.3° = 1.47 × 109 km.

(Note this is about 10 AU!)

Unit 59 11. Estimate the luminosity and temperature of the star Acrux by examining its vertical and horizontal position in the H-R diagram (Figure 59.3). (You should make an estimate of these values by interpolating between the values marked on each axis.) Using the values you estimated and the Stefan-Boltzmann law (Unit 58), determine the size of Acrux as compared to the Sun. 11. (About 6.4 times the size of the Sun, if 5800 K is used as the temperature of our star for the comparison.)

From Figure 59.3, noting that it’s a log scale, we can estimate that Acrux has a luminosity of about 3 × 104 solar luminosities, and a temperature of 30,000 K. The form of the StefanBoltzmann law we want is: R=

L T

2

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4ps

where R is the radius, L the luminosity, and T the temperature in Kelvin.

Note R is proportional to the square root of L and inversely proportional to the square of T. The effect from L is to increase R by (30,000) = 173 times. The temperature is 30,000 K / 5800 K ~ 5.2 times the Sun’s, so the effect of 1/T2 is to decrease R by 1/27. The combined effect is that the radius of Acrux should be about 173/27 = 6.4 times the Sun’s. Note: A student might incorrectly interpolate the log scale and assume 5 × 104, or use 6000 K for the Sun’s temperature, which would result in an answer of 8.28 times—the result is quite sensitive to the estimated values.

12. Alnitak is the brightest O-type star in the night sky, located at the eastern end of Orion’s belt. Proxima Centauri is the nearest M-type main-sequence star. Estimate the luminosities and temperatures of these two stars from their positions in the H-R diagram (Figure 59.3) as in Problem 11. Calculate the ratio of their sizes using the StefanBoltzmann law.

12. (The ratio is about 316 times). From Figure 59.3, Alnitak has a luminosity of 10 5 solar luminosities, and a temperature of 30,000 K. Proxima Centauri has a luminosity of about 10 −4 solar luminosities (perhaps 2 × 10−4, which changes the final answer slightly) and a temperature of 3,000 K. Use the Stefan-Boltzmann equation: R=

L T

2

4ps

R is proportional to the square root of L and inversely proportional to the square of T, so the ratio of their sizes is: RAlnitak / RProxima

= (LAln/LProx) × (TProx/TAln)2 = (105/10−4) × (3000/30000)2 = 109/2 × 10−2 = 102.5 = 316.

(Alnitak is 316 × bigger in radius than Proxima Centauri; it’s 224 if 2 × 10 −4 is used for the luminosity of Proxima).

13. The red supergiant Antares has a mass of 12 M⊙ and a radius of 880 R⊙. Barnard’s Star is a main sequence star with a mass of 0.14 M⊙ and a radius of 0.20 R⊙. Which star has a greater average density and by what factor?

13. (Barnard’s star is denser by a factor of ~ 109). As discussed in section 59.3, supergiants have an average density less than main sequence stars, so Barnard’s Star will have the greater average density. The factor is found from calculating:

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ρB/ρA = (MB/(4/3 π RB3)) / (MA/(4/3 π RA3)) = (MB/MA) × (RA/RB)3 = (0.14/12) × (880/0.20)3 = 9.9 × 108.

Barnard’s Star has an average density that is about a billion times larger than Antares.

14. If a main-sequence star has a luminosity of 3000 L⊙, what is its mass in relation to the Sun’s?

14. (The mass is about 10 times that of the Sun).

From the Unit, L ≈ M3.5. Here L = 3000 relative to the Sun, so we can express this as 3000 = M3.5. This can be rearranged to M = (3000)1/3.5 = 9.85. Given the uncertainty in the relationship, this can be rounded up to 10.

15. Estimate the luminosity of a main-sequence star that has a mass twice the Sun’s.

By what factor is such a star brighter than the Sun? 15. (About 11 times as bright as the Sun). From the Unit, L ≈ M3.5. If M is twice that of the Sun we have L ≈ 23.5 ≈ 11.3.

16. How massive would a main-sequence star need to be, to be 100 times more luminous than the Sun? 16. (3.73 times the Sun’s mass). From the Unit, L ≈ M3.5. So M = (100)1/3.5 = 3.73. So this star would have to be 3.73 times more massive than the Sun. 17. Use the mass–luminosity relationship and the Stefan-Boltzmann equation from Unit 58 to work out the surface temperature of a star with half the radius of the Sun and one-half the mass.

17. (~4470 K). Using the standard Mass-Luminosity relationship for main sequence stars, the luminosity would be: (0.5)3.5 = 8.8 × 10−2 solar luminosities.

The Stefan-Boltzman law is L = 4 πR2 × T4 so T4  L/R2: T4 = (8.8 × 10−2) / (1/2)2 = 0.354, so the T compared to the Sun would be (0.354)1/4 = 0.77 times the Sun’s temperature: 0.229 × 5800 K = 4470 K (or 4630K if you use 6000 K for the Sun’s temperature). The less-massive star would have a lower surface temperature.

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Unit 60 12. Examine the evolutionary path of the Sun-like star in Figure 60.6. Compare its luminosity on the main sequence to its luminosity as a red giant and as a white dwarf. How many times brighter and fainter will the Sun become than it is currently? What effects will this have on Earth?

12. (~ 1000 times brighter as a red giant, and 100 times fainter than the Sun initially as a white dwarf.)

From Figure 60.5, the Sun will have a luminosity of several hundred to about 1000 solar luminosities when it is a red giant (certainly 1000 near the turn-arounds) so it will be about 1000 times brighter as a red giant.

As a white dwarf, the luminosity will be from 1/100 the current luminosity, and down through 1/1000th to 1/10000th as much as the Sun travels down the white dwarf track (white dwarfs just slowly cool…).

The Earth will see the Sun swell to fill much of the sky, and the oceans will vaporize and the Earth’s surface bake when the Sun becomes a red giant; its atmosphere will probably boil away.

Then, what’s left of Earth will cool to temperatures far below current values once the Sun becomes a white dwarf.

13. Examine the evolutionary path of the Sun-like star in Figure 60.6. Estimate the size of the Sun when it becomes a red giant and white dwarf by referring to Unit 58. If the Sun is currently 0.5° across, how big will it appear to be when it is in its later stages of evolution?

13. (Almost 90 times larger, 43.4°). From Figure 59.4, we find the Sun as a red giant has a luminosity of 103 solar luminosities, and a temperature of ~ 3,500 K. The form of the StefanBoltzmann law we want is R = (L/(4πςT4))1/2. This can be solved directly but is probably easiest if we compare to the current Sun’s radius: RRG = Rnow ((103/1)/(3500/5800)4)1/2 = 86.8 Rnow = 6 × 1010 m 13. (cont.) (This is almost half an AU). For the white dwarf, we might select the point on the track with a luminosity of 0.01 solar luminosities and a temperature of 20,000 K, yielding RWD = 0.008 × Rnow = 6 × 106 m (about the size of Earth).

If the Sun is currently 0.5°, it will swell out to appear 43.4° across as a red giant, and contract to only 0.004° = 14.4 arc seconds in diameter as a white dwarf.

14. Imagine that a large collection of stars with a wide assortment of masses forms in a group. Draw sketches of the Hertzsprung-Russell diagram at three times: (a) soon after they formed; (b) at the end of the main-sequence life of a star similar in mass to the Sun;

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(c) at the end of the main-sequence life of a star much less massive than the Sun. Describe in words the differences between the three sketches.

14. What the sketches should show is a progression of stars moving off the Main Sequence to the giant and supergiant branches starting with the more massive stars (upper left on the Hertzsprung-Russell diagram). This should look like Figure 70.4 in Unit 70. (a) should look like the middle panel (or the first), (b) should look like the fourth panel, and (c) should look like the last panel (or a more extreme version of it).

15. The solid line that illustrates the evolutionary path of a star like the Sun in Figure 60.6 can lead to some confusion. It may give the impression that we are as likely to see stars like the Sun in their protostar phase as in their main-sequence phase. However, a star moves through some phases rapidly and others much more slowly. If instead of a line you put down one point for every 10 million years, you would get a better idea of where you would be likely to see stars like the Sun in the H-R diagram.

Lightly sketch or trace the Sun’s evolutionary path in an H-R diagram and then use the following times to estimate how many dots to put down in each portion of the Sun’s evolutionary path: protostar and pre-main-sequence phases ∼20 million years (Myr); main-sequence phase ∼10,000 Myr; first red giant phase ∼1000 Myr; yellow giant phase ∼100 Myr; second red giant phase ∼10 Myr; planetary nebula phase ∼0.01 Myr; white dwarf phase ∼3000 Myr. (The star remains a white dwarf beyond this time, but fades from visibility.)

15. The number of dots for the different phases would be 1/10 given times; Phase Protostar Main Seq. st 1 RG YG nd 2 RG PN WD

Time (Myr) 20 10000 1000 100 10 0.1 3000

Dots 2 1000 100 10 1 not even 1!

300

The diagram might look a bit like this adapted version of Figure 60.5:

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Solutions to End of the Unit Problems Part IV cont’d.

Unit 61 11. Use Wien’s law to find the wavelength of maximum radiation from a 2000-K protostar. To what part of the electromagnetic spectrum does this wavelength belong?

2.9´ 106 K nm = 1450nm 2000K From Figure 22.4 this lies in the infrared region of the electromagnetic spectrum.

11. (1450 nm, infrared). Using Wien’s law (U23), l m ax =

12. Use Figure 61.9 to estimate by what factor a pre-main-sequence star shrinks as it forms a one solar mass star.

12. In Figure 61.8 the pre-main sequence phase is denoted by the dashed red line. For a one solar mass star, the pre-main sequence phase starts at about Li = 10 L⦿and a temperature of about 4200 K. The pre-main sequence phase ends at Lf = L⦿and a temperature of 5800 K. From the Stefan-Boltzmann law, L = 4 πR2 × T4, we can get the radii, R = √(L/4 πT4). Since we’re looking for the how much the pre-main sequence star shrinks, we want the ratio: Rf/Ri

= √(Lf/4 πTf4) / √(Li/4 πTi4) = (Lf/Li)1/2 × (Ti4/Tf4)1/2 = (Lf/Li)1/2 × (Ti/Tf)2 = (1 L⦿/10 L⦿)1/2 × (4200 K/5800 K)2 = 0.17 = 17%.

The pre-main sequence star shrinks by 17% of its initial size.

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13. The jets from a T Tauri star are measured to have speeds of about 300 m/sec. At this speed, how long would it take the ejected material to travel 1 ly away from the protostar?

13. (About 1000 years). 1 light year = 9.46 × 1012 km, so t = 9.46 × 1012 km / (300 km/sec) = 3.15 × 1010 seconds, approximately 1000 years.

14. Suppose a molecular cloud begins with a mass of 1 M⊙ and has a radius of 1 ly. Imagine a dust grain at the outer edge of the cloud pulled straight inward, accelerating the entire way. When it reaches a radius of 1 million km from the center of the forming star, how fast will it be moving? (Hint: This problem is much easier to answer through the conservation of energy.)

14. (515 km/s). For a dust grain falling in nebular cloud from Rstart to Rfinish use the following values: Rstart = 1 light year = 9.4608 × 1015 m Rfinish = 106 km = 109 m M = mass of forming star = 1 solar mass = 1.99 × 1030 kg m = mass of dust grain G = 6.67 × 10−11 m3 kg−1 s−2

By Conservation of Energy, Kinetic Energy = change in Gravitational Potential Energy Since EG = −G m M/R (see Unit 20) and the particle starts from rest, we find: Ek = ½ mv2 = (PEstart – PEfinish) = −G m M/Rstart −(−G m M/Rfinish) = ½ mv2 = GmM(−1/Rstart − (−1/Rfinish)) = GmM(1/Rfinish − 1/Rfinish)) This can be rearranged, and the mass of the dust particle cancels out: v2 = 2 GM(1/Rfinish−1/Rstart) so v = √(2 GM(1/Rfinish − 1/Rstart)) = √(2 × 6.67 × 10−11 m3 kg−1 s−2 × 1.99 × 1030 kg × (1/(109 m) − 1/(9.4608  1015 m))) So v = 515,233 m/s ≈ 515 km/s.

Note that the 1/Rstart is such a tiny number (~10−16) it can even be neglected in the calculation.

15. Compare the average density of a newly formed star of mass 20 M⊙ (radius ≈ 10 R⊙) and 0.1 M⊙ (radius ≈ 0.1 R⊙) to the Sun’s density. You should find that the higher-mass star has a lower density. Relate these relative densities to the formation processes that determine the upper- and lower-mass limits for stars.

15. (0.02 times and 100 times as much).

The high mass star has a density about 0.028 g/cm2, roughly 0.02 times that of the Sun, or 2 percent; the low mass star is 142 g/cm2, or 100 times the density of the Sun. Using the spherical volume formula, 4/3 πR3 and that for density, D = Mass/Volume, then: Dnew = (factor in mass) / (factor in radius)3 Dsun.

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For the high mass star, the mass is 20 times the Sun’s and the volume will be 103 = 1000 times that of the Sun, so the density is 20/1000 = 0.02 times that of the Sun, or 0.02 × 1.4 kg/L = 0.028 kg/L. For the low mass star, the density is 0.1/(0.1)3 = 100 times that of the Sun, or 140 kg/L. The high density of low mass stars implies that they must be compressed much more than other stars to reach fusion temperatures. At even lower masses, the gravitational force cannot squeeze the matter enough to initiate fusion. Very high mass stars reach fusion temperature while they still have a low density, and the gravitational pull at the surface is relatively weak. At even higher masses the radiation pressure at the surface can become strong enough to drive off the outer layers.

16. a. If a 0.08-M⊙ brown dwarf has the same density as Jupiter (1.33 kg/L), what is its radius? b. The coldest brown dwarfs have surface temperatures near 600 K. For this brown dwarf, what is its luminosity?

16. (a) 3.1 × 108 m; (b) 2.3 × 10−5 L⊙. a) Density = Mass/Volume, so Volume = Mass/Density, and Volume scales as radius cubed, so the cube root of the factor of change in the volume will be the ratio of the low mass star’s radius to the Sun’s radius. The low mass star is 0.08 times the Sun’s mass. Jupiter’s density is 1.33 kg/L, and the Sun’s is 1.4 kg/L, so the ratio in density is 1.33/1.4 = 0.95. Putting these factors together: Vbrown dwarf = (0.08) / 0.95 = 0.084 Vsun, and so Rbrown dwarf = (0.084)1/3 = 0.44 times the radius of the Sun. 16. (cont.) Using the value for our Sun, the brown dwarf would have a radius of Rbrown dwarf = 0.44 × 6.96 × 108 m = 3.1 × 108 m.

Alternatively, compare the low mass star to Jupiter: Radius ~ (Mass/Density)1/3 ~ (Mass)1/3 for the same density.

And express the mass of the brown dwarf in terms of the mass of Jupiter: 0.08 solar masses = 1.59 × 1029 kg = 1.59 × 1029 kg/(1.90 × 1027 kg/Jupiter masses) = 83.8 Jupiter masses.

Using this value, Rbrown dwarf ~ (83.8)1/3 RJupiter = 4.376 × (7.15 × 104 km) = 3.1 × 105 km, 0.45 times the radius of the Sun.

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b) From the Stefan-Boltzmann law: L = 4 πR2 × T4 = 4 π (3.1 × 108 m)2 × (5.67 × 10−8 W/m2/K4) × (600 K)4 = 8.9 × 1021 W = 2.3 × 10−5 L⊙.

Unit 62 11. Calculate the luminosity of the 2.0-M⊙ star Sirius predicted by the mass–luminosity equation (Unit 59). Its observed luminosity is about 20 L⊙. How might you explain the difference?

11. The mass-luminosity relation, L ≈ M3.5, predicts that a 2.0 solar mass star will have a luminosity of (2.0)3.5 = 11.3 solar luminosity units. Sirius is observed to have 20 solar luminosities. In fact, the mass-luminosity relationship varies between L = M3 and L = M4 over different parts of the main sequence. For a bright star such as Sirius, the second exponent seems more accurate, giving a luminosity of (2.0)4 = 16 solar luminosity units.

11. (cont.) Also, as stars age their cores get hotter and their surface areas expand slightly. This causes an increase in luminosity. Stars that are hotter than the Sun age faster, and it is likely that Sirius, although actually younger than the Sun, is relatively nearer the end of its life, and therefore diverges more from the mass-luminosity prediction.

12. Calculate the main-sequence lifetime of a 25-M⊙ star using the mass–luminosity relationship (Unit 59) to estimate its luminosity.

12. (3 million years). Using L ≈ M3.5, we find L ≈ (25)3.5 = 78125 solar luminosities. The mainsequence lifetime would then be t = M/L × 1010 years = 24/78125 × 1010 years = 3 × 10−4 × 1010 years = 3 × 106 years = 3 million years.

13. Given that the Universe is 13.8 billion years old, if a 0.1-M⊙ star with a luminosity of 10–4 L⊙ was born in the early days of the Universe, would it have died out by now? If so, how long ago did it die out? If not, how long until it does die out?

13. (No, the star would still be shining; 1013 yr.) The expected lifetime for this star is: t = M/L × 1010 yr = (0.1/10−4) × 1010 yr = 1000 × 1010 yr = 1013 yr.

This is 10,000 billion years! Compared to the age of the Universe, this star basically will live for another 10,000 billion years. In this calculation there are two things to note: the formula for the stellar lifetime does not quite accurately apply to low mass stars. The text implies it may even underestimate the lifetime. Also, star formation did not immediately start with the birth of the Universe, but this difference in minor in the estimate. Neither of these complications takes away from the main message of the problem, red dwarf stars have extremely long lifetimes.

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14. Use the main-sequence lifetime equation and the mass–luminosity relationship to estimate the mass of a star that would have a main-sequence lifetime of 500 million years.

14. (3.3 solar masses). Using L ≈ M3.5, we use t = M/L × 1010 years = M/M3.5 × 1010 years = M−2.5 × 1010 years. This can be rearranged as M = (t/1010 years)−1/2.5. For t = 500 million years, this gives M = 3.3 solar masses.

15. According to a recent computer model of the Sun’s evolution, when the Sun was young its surface temperature was 130 K cooler (than its current value of 5800 K) and its luminosity was 0.71 L⊙. When the Sun is approaching the end of its main-sequence lifetime 5.5 billion years from now, its surface temperature will be 650 K cooler than it is currently, and its luminosity will be 2.02 L⊙. Use the Stefan-Boltzmann law (Unit 58) to estimate the Sun’s diameter at these past and future times.

15. (Young sun: 0.88 times, diameter = 1.2 × 109 m, and future sun: 1.8 times its current value, diameter = 2.5 × 109 m).

The form of the Stefan-Boltzmann law we want is R = (L/(4 πςT4))1/2. The radius goes as the square root of the luminosity and is inversely proportional to the square of the temperature.

For a surface temperature of 5800 K – 170 K = 5670 K and a luminosity of 0.71 solar luminosity units, we can compare the current and past radius: Rpast

= [(Lpast/Lnow)/(Tpast/Tnow)4]1/2 Rnow = [(0.71)/(5670/5800)4]1/2 Rnow = 0.88 Rnow = 0.88 × 6.96 × 108 m = 6.1 × 108 m, or 0.88 times the current value.

The diameter of the Sun would be 2R = 1.22 × 109 m.

For a surface temperature of 5150 K and a luminosity of 2.02 solar luminosity units, we get a radius of 1.25 × 109 m, or 1.8 times the current value = a diameter of 2.5 × 109 m.

Note the future Sun is about twice the diameter of the past Sun! 16. Suppose the Sun had a luminosity of 0.71 L⊙ when it was young. Compare the brightness of sunlight (Unit 55) on Venus (at 0.72 AU) when the Sun was young to the brightness of sunlight on Earth today.

16. (1.4 times brighter). Brightness B = Luminosity/(4 πRadius2), so B goes as L and as 1/R2.

When the Sun was young, its luminosity was 0.71 times that today; Venus is 0.72 times as far from the Sun as Earth. The brightness of sunlight on Venus when the Sun was young can be calculated as (0.71/0.722) = 1.4 times the brightness of sunlight on the Earth today.

17. Estimate the size of the core for a 10-M⊙ star. Assume the values discussed in the text: A star’s core is roughly 10% of the star’s overall mass and the core has a density of roughly 3 kg/L. Express your result in terms of solar radii.

17. (0.78 solar radii). Density ρ = M/V = M/(4/3 πR3), so the radius is:

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R = (M/ (4/3 π ρ))1/3 = ((0.10 × 10 × 6.96 × 1011 m) / (4/3 π × 3000 kg/m3))1/3 = 5.4 × 108 m = 0.78 R⊙.

Unit 63 10. Using the plots in Figures 63.3 and 63.5, estimate the luminosity and temperature of a 5solar-mass star (a) at the start of the main-sequence phase, (b) at the end, (c) while it fuses helium in its core, and (d) at its most luminous final stage.

10. The table below shows the estimated values for Luminosity and Temperature for a 5 solar mass star. Keep in mind the axes are log-scale, so a point about ½ between 1 and 10 is really 3, 2/3 along is about 5, etc. Helium fusion is the red line hairpin turn in the white dotted area of Figure 63.5.

Phase (a) at start of MS phase (b) at the end of MS (c) while it fuses helium (d) at most luminous final stage

L (Lsolar) 500 1,000 1,500 10,000

T (K) 17,000 13,000 5,100 3,500

(red, not grey, line) (end of red line)

11. When the Sun expands to be a red giant, its radius may be 100 R⊙.

What will the average density of the Sun be at that time?

11. (1.4 × 10−6 kg/L).

Volume increases as the cube of the radius, so if the red giant Sun has 100 times the current radius, it will have 1003 (106 or one million) times the current volume. Density is mass divided by volume, so assuming mass doesn’t change, the Sun’s density at red giant stage will be one millionth what it is now. Currently it is 1.4 kg/L, so in the red inflated stage, it will be 1.4 × 10−6 kg/L. 12. When the Sun becomes a red giant, its core region (containing about 20% of its mass) may contract to 0.01 R⊙. What will the density of the core be then?

12. (282,000 kg/L, about 200,000 times the present average density of the Sun). ρred giant core

= M/V = (solar mass × 0.2)/((4/3) × π × (0.01 × solar radius)3) = (1.99 × 1030 kg × 0.2)/((4/3) × π × (0.01 × 6.96 × 108m)3) = 2.82 × 108 kg/m3 = 2.82 × 105 kg/L.

The current average density is 1.4 kg/L, so this is 282,000/1.4 ~ 200,000 times more than the current average density of the Sun! 13. By how many times its current radius would the Sun have to expand before its outer atmosphere reached Earth?

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13. (Roughly 216 times). The Sun’s radius is about 6.96 × 108 km, whereas the distance between the Earth and Sun is 1 AU = 1.5 × 1011 m. So 1.5 × 1011 m /6.96 × 108 m ≈ 216.

14. Suppose a Sun-like star is just barely visible to the naked eye. If an identical star has already become a red giant, with a radius 100 times larger and a surface temperature half as large, how many times farther away must it be to look equally bright?

14. Going from a main sequence to a giant star, the luminosity increases by a factor of: Lf/Li = (4Rf2Tf4)/ (4Ri2Ti4) = (Rf/Ri)2 × (Tf/Ti)4 = (100)2 × (0.5)4 = 625.

A star’s brightness is given by B = L/4 πd2. For this problem both situations are for stars that is just barely visible, so the brightness does not change, meaning that Li/4πdi2 = Lf/4 πdf2 or, similarly, rf/ri = (Lf/Li)1/2 = (625)1/2 = 25. The red giant would still be visible if it was twenty five times farther away.

15. We observe a red giant with a luminosity 127 times the Sun’s and a wavelength of maximum radiation of 675 nm. What is its radius? (Use Wien’s law, Unit 56.2, to find its temperature.)

15. (19.5 solar radii or 1.36 × 1010 m). Using Wien’s law (Unit 56 or 23) we have 106 K nm T = 2.9´675 = 4396 K . This temperature is 4396/5800 = 0.76 times the Sun’s. Using the nm relationship in Unit 57, we have R = (L/(4 πςT4))1/2, so the radius is proportional to the square root of (L/T4). The radius of the star will be (127/0.764)1/2 = 19.5 times that of the Sun, or 19.5 × 6.96 × 108 m = 1.36 × 1010 m. (Note: you can also use the equation to solve directly for the radius by plugging in all the values).

16. Estimate the radius of the 15-M⊙ star shown in the H-R diagrams in this Unit from when it is initially a blue supergiant to when it becomes a red supergiant. (You will need to estimate luminosities and temperatures from the H-R diagram and then use the Stefan-Boltzmann equation from Unit 58 to solve for radius.)

16. (3.7 solar radii and ~870 solar radii). The form of the Stefan-Boltzmann law we want is R = (L/(4 πςT4))1/2. Since the radius is proportional to the square root of (L/T 4), when the 15 solar mass star is initially a blue supergiant, we can estimate that its luminosity is 10 4 solar luminosity units and its surface temperature is 30,000 K, which is 30,000 K/5800 K = 5.17 times that of the Sun. This implies a radius of: R = (104/5.174)1/2 = (10/5.17)2 = 3.74 times that of the Sun’s, or 2.6 × 109 m. When the star is a red supergiant, we can estimate that its luminosity is 10 5 solar luminosity units and its surface temperature is 3500 K, so its radius must be: R = (105/(3500/5800)4)1/2 = 868 solar radii or 6 × 1011 m.

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Unit 64 10. An interferometer measured the change in radius of one Cepheid as about 10% during its 10day period. The star was measured to have an average radius of 40 million km. How fast was the surface of the star moving outward, on average, if it took 5 days to expand by 10%?

10. (About 9.3 km/s). A 10 percent increase in a 40 million kilometer radius is 4 million kilometers. So the surface of the star moved outward 4 million km in 5 days. There are 432000 seconds in 5 days, so the speed was 4 × 106 km/432000 s = 9.3 km/sec.

11. The same interferometer as in problem 10 measured the change in radius of an RR Lyrae star to be 10% during its 10-hour period. If the star was measured to have a radius of 10 million km and took half a period to expand by this amount, what would be the speed of the surface of the star as it expands?

11. (5.6 km/s). The speed of the surface of the RR Lyrae star over the 5-hour half-period would be (0.1 × 10 × 106 km)/(5 × 3600 s) = 5.6 km/sec.

12. A variable star varies in luminosity and temperature throughout its cycle, according to the values listed in Table 64.1. The phase indicates the fraction of time between one maximum and the next. Graph the luminosity and temperature change, then plot the positions of the star in the H-R diagram. Connect the sequential points with a line to show the pattern of variation.

Luminosity (Solar Units)

Luminosity vs.

Phase

Phase (2 cyles shown)

Temperature (K)

12. The plots should look something like below, although only one phase cycle is required plotting two reveals the pattern more clearly:

Temperature vs. Phase

Phase (2 cycles shown)

The path on the H-R Diagram is across the instability strip (the HR diagram sketch should at least indicate the location of the main sequence). Here is an example based on a figure from the Unit, as well as a close-up of just the path of the star (as long as the axes are oriented correctly, log scale isn’t needed on the close-up).

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Luminosity (Solar Units)

H-R Diagram

Temperature (K)

13. Using the information from Table 64.1, calculate the average luminosity of the variable star at the four distinct phases listed (the phase listed as 1.00 is back around to the same phase as 0.00). Using your average luminosity, estimate the period this variable should have if it is a Cepheid from Figure 64.5.

13. (About 10 days). Assuming the star spend about equal time at each L in the table, the average luminosity is (10000 + 7800 + 6100 + 5000)Lsun/4 = 7225 Lsun. From Figure 64.5, we might estimate a period of about 10 days for this average luminosity.

14. Use the information from Table 64.1, along with the Stefan-Boltzmann law, to calculate the radius of the variable star at each phase. When is the star largest? When is it smallest?

14. This can be done by direct calculation or proportionalities, using the rearrangement of the Stefan-Boltzmann law, R = (L/(4 πT4))1/2. If using proportionality, the radius in units of the Sun’s radius will be: R = ((L/Lsun)/(T/Tsun)4)1/2, e.g. for the 0.00 phase, R = (104/(6700 K/5800 K)4)1/2 = 75 Rsun.

Phase 0.00 0.25 0.50 0.75 1.00

Luminosity 10,000 Lsun 7,800 Lsun 6,100 Lsun 5,000 Lsun 10,000 Lsun

Temperature 6700 K 5900 K 5400 K 5500 K 6700 K

T/Tsun 1.155 1.017 0.931 0.948 1.155

Radius (m) 5.28 × 1010 m 6.01 × 1010 m 6.34 × 1010 m 5.54 × 1010 m 5.28 × 1010 m

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Radius 75 Rsun 85 Rsun 90 Rsun 78 Rsun 75 Rsun

Thus, the Cepheid is largest at about the time it is coolest (phase 0.5), and smallest when it is near its hottest point (phase 0 or 1).

15. Suppose a Cepheid has a period of 5 days. Estimate its luminosity from Figure 64.5, then determine its distance if its measured brightness is 4 × 10 −12 watts/m2.

15. (about 9400 ly). From the diagram, the luminosity would be about 103 solar luminosity.

The Sun's luminosity is 4 × 1026 watts so the star is 4 × 1029. We know The star’s brightness B = 4 × 10−12 W/m2. From the brightness-distance formula, we know Distance = √(L/4 B), so the result is: D = 8.92 × 1019 m, or 8.92 × 1019 m/(9.46 × 1015 m/ly) = 9429 light years.

16. Use Figure 64.3 to estimate the range of sizes for Cepheid variables depending on their luminosity. Note that these sizes only represent averages since Cepheid variables are constantly changing in size.

16. (From ~ 8 times to more than 430 times the size of the Sun.) Stellar sizes can be estimated using the Stefan-Boltzmann law, L = 4 R2T4, or similarly, R = (L/(4 T4))1/2. From Figure 64.3 we see that the smallest Cepheids are approximately: R = (100 × 4 × 1026 W/(4 × (5.7 × 10−8 W/m2/K4) (6500 K)4))1/2 = 5.6 × 109 m = 8.1 R⦿.

The largest Cepheids are approximately: R = (105 × 4 × 1026 W/(4  × (5.7 × 10−8 W/m2/K4) (5000 K)4))1/2 = 3.0 × 1011 m = 435 R⦿.

Unit 65 11. As a dying star moves across the H-R diagram during its planetary nebula ejection phase, it maintains an almost constant luminosity as successively deeper and hotter layers inside the star are exposed. If it begins in the upper-right part of the H-R diagram as a 3000-K, 100-R⊙ red giant, use the Stefan-Boltzmann law (Unit 58.3) to calculate the central star’s radius when its surface temperature isa. 6000 K. b. 30,000 K. c. 100,000 K. 11. (a) 25 Rsun, (b) 1 Rsun, and (c) 0.09 Rsun.

Using the Stefan-Boltzmann Law and holding luminosity L constant, if temperature varies, the radius must also vary to compensate. L = 4R2T4 which can be rearranged as R2 = L/4T4. If the luminosity stays constant, the value of L/4 can be considered constant, so effectively R  (1/T2); the change in R will go as (1/change in T) squared.

11. (cont.) At T = 6000 K the temperature has doubled compared to the start (3000 K) so R reduces to (1/2)2 = ¼ the original value of 100 Rsun, or 25 Rsun.

At 30,000 K, the T is 10 times more, and R = 1/100 Rstart = 1 Rsun.

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At 100,000 K, R = 1/(100,000/3000)2 Rstart = (3000/100000)2 Rstart = (3/100)2 Rstart = 0.0009 (100 Rsun) = 0.09 Rsun.

12. How long will it take a planetary nebula shell moving at 20 km/sec to expand to a radius of one-fourth of a light-year?

12. (About 3750 years). One quarter light year is 9.46 × 1012 km/4 = 2.37 × 1012 km.

Use t = d/v = (2.37 × 1012 km) / (20 km/s) = 1.18 × 1011 s = 3750 years (using 3.16 × 107 s/y).

13. The tail of gas behind the star Mira spans about 13 ly in total. The star’s proper motion indicates that it is moving at 130 km/sec. How long ago was Mira at the position of the start of the tail? Give your answer in years.

13. (About 32,000 y). t = d/v = (13 ly) / (130 km/s) = 1/10 s ly/km = 1/10 s × 9.5 × 1012 km/km = 9.5 × 1011 s or about ~ 1012 s. At 3.16 × 107 s/y this is 31,645 y, or about 32,000 years; the exact calculation (using 9.5) gives about 30,000 years.

14. How does the luminosity of a white dwarf at a temperature of 20,000 K compare to its luminosity early on when it had a temperature of 100,000 K? What would be the wavelength of peak radiation of the white dwarf at these temperatures?

14. (For the cooled white dwarf, 0.0016 times as bright, with a peak wavelength of 145 nm, compared with the hotter white dwarf’s peak wavelength of 29 nm.)

The luminosity difference is found from using the Stefan-Boltzman law (Unit 58), L = 4 R2T4, where L is proportional to the fourth power of temperature. So the cool white dwarf has: (20,000 K/100,000 K)4 = 0.24 = 0.0016 times the luminosity 14. (cont.) of the hot white dwarf. Using Wien’s Law (Unit 23), the wavelength of peak radiation is given by λ = 2.9 × 106 K nm/T. So the cool white dwarf has a wavelength of peak radiation of 145 nm and the hot white dwarf has a wavelength of peak radiation of 29 nm.

Both of these wavelengths lie in the ultraviolet region of the electromagnetic spectrum.

15. a. Estimate the average density (in kg/L) of a planetary nebula around a solar-type star, assuming that a star like the Sun loses half its mass to a spherical nebula that expands to a light-year in diameter. b. For the same system, estimate the average density of the resulting white dwarf, assuming a radius a hundredth of the Sun’s.

15. (Average density of the planetary nebula: 2.25 × 10−21 kg/L; average density of white dwarf: 7.1 × 105 kg/L.)

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a) The volume of this sphere is given by V = 4/3 πR3, where R = 0.5 ly = 4.73 × 1015 m. Therefore, V = 4.43 × 1047 m3. The mass of this sphere (excluding the white dwarf at the center) is given by M = 0.5 solar masses = 9.945 × 1029 kg. The average density of the planetary nebula is therefore: M/V = 2.25 × 10−18 kg/m3, or 2.25 × 10−21 kg/L. b) For a radius of R = 0.01 R⦿= 0.01 × 6.96 × 108 m = 7 × 106 m, the volume is V = 1.41 × 1021 m3.

Taking the mass as the other half from part a, M = .5 solar masses = 9.945 × 10 29 kg, gives an average density for the white dwarf of: M/V = 9.945 × 1029 kg / 1.41 × 1021 m3 = 7.1 × 108 kg/m3 = 7.1 × 105 kg/L. 16. Using the result of problem 15a and the fact that the mass of a hydrogen atom is 1.7 × 10−27 kg, calculate the average number of hydrogen atoms per liter in a planetary nebula. What radius would the planetary nebula have to be so that its density matched the interstellar medium—about 1000 atoms per liter?

16. (1.32 × 106 atoms/L; ~ 5.5 ly). The previous question gives a density for the planetary nebula of 2.25 × 10−18 kg/m3. If this was hydrogen: nH = ρ/(mH) = 2.25 × 10−18 kg/m3 / (1.7 × 10−27 kg) = 1.32 × 109 atoms/m3.

16. (cont.) With 1 Liter is 10−3 m3, so that’s 1.32 × 106 atoms/L. The interstellar medium has a density of 1000 atoms/L = 103 atoms/L, which is (1.3 × 106 atoms/L / 1000 atoms/L) = 1300 times lower. Since density is inversely proportional to the volume, the volume needs to increase 1300 times for the density of the PN match the interstellar medium. Volume goes as radius cubed, so the radius needs to increase (1300) 1/3 = 11 times. The initial radius in the previous problem was a half light year, so when the cloud expands to a radius of 11 × 0.5 ly = 5.5 ly, the density will match the interstellar medium.

Unit 66 11. How fast does an accretion disk particle orbit a 1-M⊙ white dwarf when the particle is at a radius of 104 km? How does this value compare to the speed of light?

11. (0.012 c) From Unit 17.1 we have the relationship M = d × V2/G, which we can use to get the orbital speed: V = √(G × M/d) = √((6.67 × 10−11 m3 / kg∙s2) × (1.99 × 1030 kg) / (107 m)) = 3.64 × 106 m/s = 0.012c. The material is moving at about 1% the speed of light.

12. How does the acceleration of gravity (Unit 16) on the surface of a 1-M⊙ white dwarf, which has a radius of 104 km, compare to the acceleration due to gravity on the surface of Earth?

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12. (It’s about 1.33 × 106 m/s2, or 1.36 × 105 times the value of “g” on Earth).

The formula for surface gravity is g = GM/R2; g is proportional to M and inversely proportional to R2. The mass of the white dwarf is: Mwd = (1.99 × 1030 kg)/(5.97 × 1024 kg) = 333,333 times the Earth’s, but the radius is 104 km/6378 km = 1.567 times the Earths, so: gwd

= (333,333)/(1.567)2 gE = 135,750 gE = 135750 × 9.81 m/s2 = 1.33 × 106 m/s2.

This can also be calculated directly: g = GM/R2 = (6.67 × 10−11 m3 kg−1 s−2) × 1.99 × 1030 kg /(107 m)2 = 1.33 × 106 m/s2 13. What is the escape velocity (Unit 18) from the surface of a 1-M⊙ white dwarf that has a radius of 104 km?

13. (About 5200 km/s). Use the equation for escape velocity: Vesc

= (2GM/R) = [2 × 6.67 × 10−11 m3 kg−1 s−2 × 1.99 ×1030 kg /(107 m)] = 5.2 × 106 m/s = 5200 km/s.

14. The brightest star in the night sky, Sirius A, is located 8.60 ly from Earth and has a luminosity of 25.4 L⊙. What is the maximum distance at which a Type Ia supernova could occur such that it would have the same brightness as Sirius A? 14. (1.70 × 105 ly). Brightness is given by B = L/4πd2. Setting the brightness of Sirius A to a Type Ia supernova, LS/4πdSirius2 = LIa/4πdIa supernova2, leads to: dIa supernova

= (LIa supernova/LSirius)1/2 dSirius = (1010 L⦿/ 25.4 L⦿)1/2 (8.60 ly) = 1.70 × 105 ly.

This distance is well outside our galaxy but not as far as the Andromeda galaxy.

15. If the maximum luminosity of a Type Ia supernova is 10 10 L⊙, and it remains at this brightness for 15 days, estimate how long our Sun would take to emit the same amount of energy.

15. (4.11 × 108 years). The total energy output of the supernova found from: Energy output = energy output rate × time interval, or E = 1010 L⦿× 15 days.

The luminosity of the Sun is L⦿= 3.83 × 1026 W, and 1 day = 86400 s, so E = (1010 )´ (3.83´ 1026 W )´ (15)´ (86400 s)= 4.96´ 1042 J

The length of time it would take our Sun to emit the same amount of energy is:

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t=

4.96´ 1042 J = 1.30´ 1016 s= 4.11´ 108 years 26 3.83´ 10 J/s

One could also use ratios to arrive at the same answer.

16. IK Pegasi, a white dwarf/A-type main-sequence binary system 150 ly away, is a candidate to become a Type Ia supernova in the future. Assuming that it will reach a luminosity of 1010 L⊙, how bright will it appear from Earth? How many times brighter than an average star in our sky would it appear?

16. (0.15 W/m2; 7.5 million times brighter). From Unit 55, brightness is given by: B = L /4p d2 = (1010 × 3.83 × 1026 W) / (4 × (150 × 9.46 × 1015 m)2) = 0.15 W/m2.

The typical brightness of a star in our night sky is 2 × 10−8 W/m2 (see Unit 55), so this Type Ia supernova would be 7.5 million times brighter than that! (Fortunately, this is a long time in the future, and stellar orbital motion may well move it much farther away by then).

Unit 67 Note: The exact values of some answers in this unit are sensitive to whether one uses 3.83 × 1026 W for 1 solar luminosity (as in the Appendix), or the easier rounded figure of 4 × 1026 W (frequently used in the text). Expect some range on the given values below if the other value was used.

12. If every proton in the core of a massive star turns into a neutron and releases one neutrino, how many neutrinos are produced? Assume the core contains 2 M⊙, and half the mass is made of protons (M proton = 1.67 × 10−27 kg).

12. (1.2 × 1057 neutrinos). If the star is 2 solar masses and half of that mass is of protons, then the total mass of protons is one solar mass or 2 × 1030 kg. Each proton has a mass of 1.67 × 10−27 kg.

There are (2 × 1030 kg)/(1.67 × 10–27 kg/proton) = 1.2 × 1057 protons. The reaction is p + e− + Energy = n + neutrino, so if all those protons change into neutrons they will release one neutrino each for a total of 1.2 × 1057 neutrinos released.

13. Assume that 1012 neutrinos passed through every square meter of surface perpendicular to the radiation from Supernova 1987A (which was 160,000 ly away).

How many neutrinos were produced by the explosion?

13. (2.9 ×1055 neutrinos). The surface of a 160,000 light year radius sphere is 4(1.6 × 105 × 9.5 × 1015 m)2 = 4(1.52 × 1021 m)2 = 2.90 × 1043 m2. If each square meter of this surface intercepts 1012 neutrinos, the supernova must have emitted 2.90 × 1043 m2 ×1012 neutrinos/m2 = 2.9 ×1055 neutrinos.

14. How long does it take the blast wave from the center of a star undergoing a supernova explosion to reach the star’s surface if the blast wave is traveling at 10,000 km/sec and the star is about 100 times the size of the Sun? 14. (1.93 h). 100 Rsun = 6.96 × 107 km; 159

t = d/v = 6.96 × 107 km /(104 km/sec) = 6.96 × 103 sec = 1.93 hours or ~ 2 hours 15. Assuming an average collapse speed of 50,000 km/sec, how long does a star take to collapse during the supernova stage if it starts off with a surface radius of 7 AU and ends up at the surface of the neutron star core, 10 km from the center of the star?

15. (5.8 h). The time is found from t = d/v with d = 7 AU = 1.05 × 109 km and v = 50,000 km/sec. This gives a collapse time of 21,000 sec, or 5.8 hours. The ending radius (10 km) can be safely ignored because the time to travel that distance would be a fraction of a second, tiny compared to the time to travel the 7 AU. 16. The Crab Nebula is today measured to be about 10 ly across. If it exploded in 1054, what is its expansion speed in km/sec?

16. (About 3100 km/s). 10 ly = 9.5 × 1013 km. From 1054 to 2025, the time span for it to expand to 10 ly is 971 years × 3.16 × 107 s/y = 3.07 × 1010 s.

This gives v = d/t = 9.5 × 1013 km / 3.07 × 1010 s ~ 3100 km/s.

17. Betelgeuse is a red supergiant about 640 ly from Earth. Astronomers expect that within the next million years Betelgeuse will explode as a Type II supernova. Use Figure 67.4 to estimate how bright Betelgeuse will appear. Will it outshine the full Moon, which has a brightness of 3 × 10−3 watts/m2?

17. (8.3 × 10−4 W/m2; a bit dimmer than the full Moon).

Assuming the Type II supernova from Betelgeuse reaches a maximum luminosity of 10 9 L⦿then its brightness on Earth would be: B = L/4d2 = (109 × 3.8 × 1026 W)/(4 (640 ly × 9.46 × 1015 m/ly)2 ) = 8.3 × 10−4 W/m2.

17. (cont.) For the assumed luminosity, the Betelgeuse supernova will be slightly dimmer than the full Moon. However, by inspection of Figure 67.4 a luminosity of 10 9 L⦿may be an underestimate. If a slightly larger value is assumed, then it is possible the supernova will be brighter than the full Moon.

18. It is estimated that in the final stages of fusion, a giant star produces a core of iron more massive than the Sun. From the energy release given in Table 67.1, a. how much energy is this (in joules)? b. how long would it take the Sun to emit this much energy at its rate of 4 × 1026 joules/sec?

18. (a) About 6.1 × 1043 J; (b) About 4.8 billion years. a) From Table 67.1, the transition from Si to Fe happens in the last day. If the final mass is one solar mass (1.99 × 1030 kg) and the energy-mass equivalent is 0.034%, that's 6.8 × 1026 kg. Using E = mc2, that mass equals 6.1 × 1043 J. 160

b) The Sun radiates at about 4 × 1026 J/sec, so it would take about (6.1 × 1043 J)/(4 × 1026 J/s) = 1.53 × 1017 seconds to radiate that much energy. That's about 4.83 billion years, almost the current age of the Sun.

Unit 68 11. Calculate the escape velocity (Unit 18) from a white dwarf and a neutron star. Assume that each has 1.4 M⊙. Let the white dwarf’s radius be 104 km and the neutron star’s radius be 10 km.

11. (6100 km/s; 193,000 km/s). Calculate this using vesc = (2 GM/R). Escape velocity for the white dwarf is: vesc =

2M G R

=

2´ 1.4´ 1.99´ 1030 kg´ 6.67´ 10- 11m 2s- 1kg- 1 = 6.1 × 106 m/s = 6100 km/s 7 10 m

Escape velocity for the neutron star is: vesc =

2M G R

=

2´ 1.4´ 1.99´ 1030 kg´ 6.67´ 10- 11m 2s- 1kg- 1 = 1.93 × 108 m/s = 193,000 km/s 4 10 m

12. Suppose a neutron star with a radius of 10 km is spinning with a period of 0.001 sec. How fast is its equator moving? Compare this speed to the neutron star’s escape velocity calculated in the previous problem.

12. (6.3 × 107 m/s). For a neutron star of radius 10 km, the speed at the equator: V = 2 πr/t = 2 π (104m)/(0.001 s) = 6.3 × 107 m/s, which is 0.2 times the speed of light. This is less than the escape speed from problem 11 of 1.9 × 108 m/s. For a white dwarf of 104 km, the equatorial speed for this period would be (6.2 × 1010 m/s), well in excess of the speed of light and the escape velocity, but white dwarfs do not spin that fast.

13. The mass of a neutron is 1.67 × 10−27 kg. How many neutrons are in a neutron star that has 1.4 M⊙?

13. (About 1.60 × 1057 neutrons).

The Sun's mass is about 2 × 1030, and the neutron star is 1.4 times that, or about 2.8 × 10 30 kg.

Divide this by the mass of a single neutron, 1.67 × 10−27 kg, to find out how many neutrons would fit into a neutron star. So you have: 2.68 × 1030 kg / (1.67 × 10−27 kg/neutron) = 1.60 × 1057 neutrons.

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14. The volume of a neutron is about 10−45 m3. Suppose you packed the number of neutrons you found in the previous problem into a sphere of radius R. The volume of a sphere is 4/3 πR3. What is R in kilometers?

14. (Radius is about 7.3 km).

To get the volume of neutrons, multiply the number of neutrons by the volume of a single neutron, (1.60 × 1057) × (10−45 m3/neutron) = 1.60 × 1012 m3. Rearrange the volume equation to get R = (3/4 × V/)1/3 to get the radius of about 7260 meters, or about 7.3 km.

15. What is the luminosity of neutron star that has a surface temperature of 700,000 K and a radius of 10 km? Where on the H-R diagram should neutron stars be placed?

15. (1.7 × 1025 W). Use the Stefan-Boltzman law (Unit 58), L = 4 R2T4 so L = 4 (104 m)2 × (5.67 × 10−8 W/m2K) × (7 × 105 K)4 = 1.7 × 1025 W or about 4% the luminosity of the Sun. This would place the neutron star lower on the vertical scale of an H-R diagram, but its surface temperature of 700,000 K is at least 20× hotter than any normal star, which would mean the neutron star would be placed far, far to the lower left (at more than twice the width of a typical H-R plot.)

16. What is the orbital separation for a binary composed of two 1.4-M⊙ neutron stars orbiting each other once every 10 seconds? Such theoretical systems are targets for current gravitational wave detectors (Unit 27.4).

16. (9800 km). Use the modified form of Kepler’s 3rd law, a3/T2 = G (M + m)/4 2 to find the semi-major axis “a”, with the period T = 10 seconds, and M = m = 1.4 × (1.99 × 1030 kg). a3 = T2 × G (M + m)/4 2 9.41 × 1020 so a = 9800 km.

Unit 69 11. a. Calculate the Schwarzschild radius of a 3-M⊙ object. b. The orbital period and separation of the Cygnus X-1 binary system are 5.6 days and 0.2 AU, respectively. If the visible B-type supergiant has a mass of 19 M⊙, what is the mass of the unseen companion? c. What is the Schwarzschild radius of the invisible companion?

11. (a) ~ 9 km; (b) 16.6 M⦿; (c) ~49 km).

The Schwarzschild radius of a 3-solar-mass object is:

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2GM

Rs =

c2

=

2(6.67´ 10- 11m 3kg- 1s- 2 )(3´ 2´ 1030 kg) 2

æ ö çç3´ 108 m ÷ ÷ ÷ çè sø

= 8900m ~9km

11. (cont.) b) From Kepler’s third law (Unit 17.2), MA + MB = a3AU/P2yr converting 5.6 days to 0.015 years: MB = a3AU/P2yr – MA = 0.23/0.0152 – 19 = 16.6 M⦿. c) Using the mass from part (b):

Rs =

2GM 2

c

=

2(6.67´ 10- 11m 3kg- 1s- 2 )(16.6´ 2´ 1030 kg) 2

æ ö çç3´ 108 m ÷ ÷ ÷ çè sø

= 4.9´ 104 m ~49 km

12. Calculate your Schwarzschild radius. What in nature has a comparable size?

12. (About 10−25 m).

A typical adult mass might be 75 kg. Using the formula for the Schwarzschild radius: Rs = 2 GM/c2 = 2 × 6.67 × 10–11 m3 kg–1 s–2 × 75 kg / (3 × 108 m/sec)2 = 1.1 × 10−25 m.

Basically, nothing is this size except possibly fundamental particles. This is much smaller than a proton (10–15 m), but possibly in the ballpark for the size of quarks.

13. Calculate the radius of a billion-M⊙ black hole at the center of a galaxy. Calculate its density in kilograms per liter. (Reminder: There are 1000 liters in a cubic meter.) Use 4/3πRs2 as the volume of the black hole.

13. (The radius would be ~ 20 AU or 3 billion km; the density would be 0.018 kg/L).

A billion solar masses would be 2 × 1030 kg × 109 = 2 × 1039 kg. Using the Schwartzschild radius formula, Rs = 2 GM/c2 Rs = 2 × (6.67 × 10–11 m3 kg–1 s–2) × (2 × 1039 kg)/(3 × 108 m/s)2 = 3 × 1012 m.

13. (cont.) This radius is about 3 × 1012 m/ (1.5 × 1011 m/AU) = 20 AU. The volume would be (4 π/3)(3 × 1012 m)3 = 1.13 × 1038 m3 = 1.13 × 1041 L, and density  = M/V:  = (2 × 1039 kg)/(1.13 × 1041 L) = 0.018 kg/L.

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14. If a neutron star is 10% bigger than its Schwarzschild radius, how slowly would a clock run on its surface?

14. (About 3.3 times slower than far away). From the general relativity unit, 1 1 gG = = = 3.3. The time dilation factor for such a neutron star is 3.3.

Rs 1Rs 11R 1.1Rs 15. Suppose you observe two friends. One takes her spaceship to an orbit around a black hole a distance of 1.1 RS from the singularity (i.e., 10% farther away from the singularity than the event horizon is). The other moves in a straight line at 0.1 c. Which friend has the greatest time-stretching factor (as observed by you)? See Unit 26 for the special relativity Lorentz factor.

15. (The friend at the black hole as observed by you).

Gravitational Lorentz factor for friend near black hole: g G =

1Special relativity Lorentz factor: g =

1

1 Rs

=3.3.

R

1

= = 1.005.

2 2 v 1 ( 0.

1) 1c2 The friend close to the black hole experiences the greater Lorentz factor (as observed by you).

16. The evaporation time for a black hole can be calculated from τevap = 2.67 × 10–24 M3, where M is the mass of the black hole expressed in kilograms and the resulting time is in years.

Calculate the evaporation time for black holes with masses equal to (a) the Sun, (b) Earth, and (c) a medium-size asteroid, 1015 kg. Which of these black holes have an evaporation time shorter than the current age of the Universe (13.8 billion years)?

16.

a. τev = 2.67 × 10−24 × (1.99 × 1030)3 = 2.1 × 1067 years b. τev = 2.67 × 10−24 × (5.97 × 1024)3 = 5.7 × 1050 years c. τev = 2.67 × 10−24 × (1015)3 = 2.7 × 1021 years

None of these black holes have an evaporation time shorter than the current age of the Universe.

17. Neutron-density matter is close to being incompressible. Assuming that it is, we can calculate how big a mass of neutron-density matter turns into a black hole. a. Calculate the density ρ of a 1.4-M⊙ neutron star, which has a radius of 7 km. b. The mass of a neutron star is equal to its volume times its density. Using the density you found in (a), multiply it by the volume 4/3πR2 and use the Schwarzschild radius formula to determine the radius when R = RS. c. Use the radius you found in (b) to calculate the mass of this neutron star.

17. (a) 1.9 × 1015 kg/L; (b) 9200 km; (c) 6.2 × 1033 kg.

First find the volume, V = (4 π/3)(7 × 103 m)3 = 1.44 × 1012 m3 = 1.44 × 1015 L.

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Then, density =  = M/V = (1.4 × 2 × 1030 kg) / (1.44 × 1015 L) = 1.9 × 1015 kg/L. (b) Next, we want to find a radius at which RSch = Rstar if the star has the density from part (a).

This can be solved this way: Rs = 2GM/c2 = (2G/c2)×(V) = (2G/c2) × () × (4π/3 × Rs3) We’ll need to express  in kg/m3. 1.9 × 1015 kg/L = 1.9 × 1012 kg/m3. Simplifying, 1 = Rs2 × (8 πG)/(3 c2) Rs2 = (3 c2)/(8 πG) = (3 × (3 × 108 m/s)2) / (8 π × (6.67 × 10–11 m3 kg–1 s–2) × 1.9 × 1012 kg/m3) = (27 × 1016 m2/s2)/(3185 s−2) = 8.5 × 1013 m2 so Rs = 9.2 × 106 m = 9200 km.

If the neutron star had a radius of ~ 107 m = 10,000 km, it would have a radius equal to the Schwarzchild radius. (c) The mass of this object of R = 9.2 × 106 m would be M =  × V = 1.9 × 1012 kg/m3 × (4 π/3)(9.2 × 106 m)3 = 6.2 × 1033 kg.

This is about 3,100 solar masses. (Quick check: Rs for an object of 3100 solar masses ~ 3100 × 3 km = 9300 km).

Unit 70 11. The globular cluster Omega Centauri looks like a “fuzzy star” to the unaided eye, and is about 100 times less bright than the brightest star in the sky. It is approximately 16,000 light-years distant. At what distance would it be as bright as the brightest star in the sky?

At what distance would it appear as bright as the full moon, which is about 4 million times brighter than Omega Centauri at its present distance? What problems are there with this result?

11. (1600 ly; 8 ly). Brightness is inversely proportional to the square of the distance: B = L/4 πd2. To be as bright as the brightest star, the cluster would need to appear 100 times brighter than it does (B → 100 B). For B to go up 100, d2 must go down by 100, and d by 10. It needs to be 1/10 as far as it is. (100 B = L / 4 π(d2/100) = L / 4 π(d/10)2). Since it’s 16,000 ly away, the cluster would need to be 1600 ly away to be as bright as the brightest star. To be as bright as the full moon, B→4 × 106 B; d2 → d2/(4 × 106), d→ d/(4 × 106) = d/2000 = 16000 ly/ 2000 = 8 ly.

One of the problems with this result is that is presumes there are no effects from dust or gas absorbing light from the cluster. If there is, the cluster itself would have a larger Luminosity 165

than we think, and it would not need to be as close as estimated above to appear brighter to us if there was less gas or dust between us and the new location. A second problem is that far away, the cluster looks like a single fuzzy star. If it was moved much closer, we would be able to resolve the individual stars; it would no longer look like a single bright point, but instead several dimmer points grouped together. In fact, since globular clusters are ~ 100 – 200 ly in diameter, we would be inside near the center of it if it was 8 ly away.

12. We can see associations of O and B stars in the Andromeda Galaxy, 2 million ly away.

What would be the brightness of an association of 70 O stars at this distance?

12. (About 6.2 × 10−13 W/m2). A typical O star has luminosity 100,000 times greater than the Sun. This must mean that the association has a luminosity of: L = 70 × 100,000 × 4 × 1026 watts = 2.8 × 1033 watts.

Brightness (Unit 55) is given by B = L/(4 πd2), and 2 million ly is: 2 × 106 ly × (9.5 × 1015 m/ly) = 1.9 × 1022 m, so B = 2.8 × 1033 W / [4 × (1.9 × 1022 m)2] = 6.2 × 10−13 W/m2.

13. The Hyades is 150 ly away from Earth, and is 75 ly in diameter. The globular cluster Omega Centauri is 18,000 ly away from Earth, and is 200 ly in diameter. What are the angular diameters of these two objects as seen from Earth?

13. (29° and 0.64°). From Unit 10, we have L = d × A/57.3°, which we can rearrange to give A = 57.3° × L/d, where L is the diameter of the object, d is its distance and A is its angular diameter. This equation gives the following angular diameters for the two objects are: Hyades: Omega Centauri:

A = 57.3° × L/d = 57.3° × 75 ly/150 ly = 29° A = 57.3° × L/d = 57.3° × 200 ly/18,000 ly = 0.64°

(Note that only the bright central core of the Hyades is easily visible).

14. A globular cluster contains no main-sequence stars with surface temperatures higher than 5000 K and luminosities greater than 0.80 L⊙. How old is the cluster?

14. (About 11 billion years). From Unit 59, the mass luminosity relationship for Main Sequence stars is L≈M3.5. This gives a mass for the most luminous remaining star of (0.8)1/3.5 = 0.94 solar luminosities. From Unit 62, the Main Sequence age of such a star is given by t = M/L × 1010 years = 11 billion years, which would be the age of the globular cluster. One could also estimate an age on the order of 10 billion years by looking at the lifetime of 0.8 Lsun stars plotted on the H-R diagram in, e.g., Unit 62.

15. If a globular cluster contains 1 million stars in a volume with a radius of 100 ly, what is the average separation between stars? Calculate this by assuming that each star fills one millionth of the total volume of the cluster. The diameter of the sphere it occupies is then the average separation. (The volume of a sphere is 4/3 πR3.)

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15. (About 2 light years). 100 light years = 9.46 × 1014 km, and a sphere with that radius has a volume: V = ((4 π/3) × R3) = 3.55 × 1045 km3.

Divide that by 106 for the volume of one star’s section to get 3.55 × 1039 km3. Now R = (V × 3/(4 π))1/3 yields R = 9.46 × 1012 km or ~ 1 light year.

Since that is the radius of the sphere for each star, the separation is the diameter or twice that: 2 ly.

16. Estimate the escape velocity for an open cluster that contains 1000 stars and is 6 pc in radius. For this calculation, make the assumption that each star is like our Sun. How does this value compare to Earth’s escape velocity (11.2 km/sec)?

16. (~ 1.2 km/s). The escape velocity from a body of mass M at a distance R is:

vesc =

2GM R

The radius is 6 pc, which is

6pc 3.26 ly 9.5´ 1015 ´ ´ = 1.9´ 1017 m 1 1pc 1 ly The escape velocity is then: vesc 





2 6.67 1011 m3 kg 1s 2 1000  2 1030 kg 1.9 10 m 17

  1185 m ~1.2 km/s s

This is around 10 times less than the Earth’s escape velocity.

Solutions to End of the Unit Problems

Part V: Galaxies and the Universe

Unit 71 12. If the total mass of the Milky Way is 1012 M⊙, and 10% of that is in the disk, and 15% of the disk is dust and gas, how many solar masses of dust and gas are in the Galaxy?

12. (About 1.5 × 1010 MSun). The problem notes that 10% of the 1012 solar masses of material that makes up the galaxy is in the disk. Gas and dust make up 15% of the galaxy’s disk. So, mass of disk = 0.10 × 1012 solar masses = 1011 solar masses mass of gas and dust = 0.15 × mass of disk = 0.15 × 1011 solar masses 167

= 1.5 × 1010 solar masses.

13. If a star located at the center of the Galaxy goes supernova, how long does the supernova’s light take to reach the Solar System?

13. (About 26000 years). The center of the galaxy is about 8 kiloparsecs = 8 kpc from the Sun. One kpc is 1000 pc and each pc is 3.26 ly, so 1 kpc = 3260 ly, so 8 kpc = 26080 light years. Thus, the center of the galaxy is about 26,000 light years away, and light from this region takes just over 26,000 years to reach us.

14. Traveling at 60,000 kph, how long would it take us to reach the center of the Milky Way? 14. (About 4.8 × 108 years). The distance to the center of the Milky Way is 8 kpc = 26080 ly = 26080 × 9.5 × 1012 km = 2.5 × 1017 km.

At 60,000 kph, t = d/v = 2.5 × 1017 km / (60,000 km/h) = 4.2 × 1012 h.

It would take 4.2 × 1012 h = 4.8 × 108 years to cover this distance.

15. Suppose a galaxy like the Milky Way is about 2 million ly away (which is the distance to M31, the nearest galaxy similar in size to the Milky Way). What would its angular size be, given that its disk is about 100,000 ly across?

15. (About 2.9 degrees). We have A = 57.3° × L/d where A is the angle subtended in degrees, L is the linear size, and d is the distance in the same units. L = 105 light years, and d = 2 × 106 ly.

Thus, A = 57.3° × (105)/(2 × 106) = 2.865°.

16. If we made a model of the Milky Way that had the diameter of Earth, how big would Earth itself be in this shrunken model? (Assume that the Milky Way’s diameter is 35 kpc for this problem.)

16. (About 1.1 × 10−7 m, or 110 nm). Create a scale factor using the size of the Earth: Escale / MWscale = Ereal/MWreal. Earth’s diameter is 6738 km × 2 = 12756 km.

We need to convert Earth’s diameter into kpc or the MW’s into km; the latter is easier: 35 kpc × 1000 pc/kpc × 3.09 × 1013 km = 1.08 × 1018 km.

Escale = (Ereal/MWreal) × MWscale = (12756 km)/(1.08 × 1018 km) × (12756 km) = 1.5 × 10−10 km = 1.5 × 10−7 m = 110 nm.

This is smaller than the wavelength of visible light.

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17. Kapteyn used observations of the smaller shift in the position of stars that are farther away to estimate their distances. This is essentially a consequence of the angular size versus distance relationship discussed in Unit 10, as illustrated by the following: (a) If stars move at a typical speed of 20 km/sec, how far will they move in 10 years? (b) If a star is 10 pc distant, what will its angular shift be over these 10 years? (c) If a star is 100 pc distant, what will its angular shift be?

17. (a) 6.3 × 109 km; (b) 4.2 arcsec; (c) 0.42 arcsec. (a) A star with a speed of 20 km/sec moves 6.3 × 109 km in 10 years (= 3.16 × 108s). (b) This distance of 6.3 × 109 km = 2.0 × 10−4 pc. At a distance of 10 parsecs, this corresponds to an angle of A = 57.3° × L/d = 57.3° × (2.0 × 10−4 pc / 10 pc) = 1.17 × 10−3 ° = 4.2 arcsec. (c) At a distance of 100 parsecs, the angle is 10 × smaller, 0.42 arcseconds.

18. Interstellar dust limits our view into the Galactic disk to only the nearest 3 kpc. If the entire galactic disk has a radius of 15 kpc, what fraction of the disk is visible from Earth?

Recall that the area of a disc is given by A = π R2.

18. (About 4%) The ratio of the visible area AV to the total area AT is: AV/AT = π RV2 / π RT2 = (RV/RT)2 = (3 kpc/15 kpc)2 = 0.04 = 4%.

Unit 72 13. A Pop II star has a highly eccentric orbit 12 kpc in diameter. What is its orbital period?

13. (About 44 million years). Use Kepler’s 3rd law, a3 = (MA + MB) P2 with a in AU, P in years, and MA and MB in solar masses. We’ll need to convert the semi-major axis into AU. If the orbit is highly eccentric, then the semi-major axis is about ½ the 12 kpc axis: 6 kpc × 1000 pc/kpc × 206265 AU/pc = 1.24 × 109 AU. Next, we need the mass. At 12 kpc, the orbit doesn’t quite reach the edge of the galaxy, but we can assume it does for this question (we’ll learn how to estimate the mass of the galaxy more exactly in Unit 74). We can safely neglect the mass of one Pop II star compared to the mass of the galaxy, so MA + MB ~ 1012 solar masses. So: P2 = (1.24 × 109)3 / (1012) = 1.9 × 1015, and P = 4.4 × 107 y.

14. Conservation of angular momentum (Unit 20) requires that the product of the speed at which an object is orbiting times the distance from the point around which it is orbiting remain constant. If the Sun is currently orbiting at 220 km/sec, 8 kpc from the center of the

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Milky Way, how fast did the material that formed the Sun rotate if the gas began 400 kpc distant (about halfway to the next large galaxy)?

14. (4.4 km/s). Create an equation relating the angular momentum initially when the material was further from the center to what the Sun’s angular momentum is now:

L initial = L final so vinitial´ 400kpc= 220km /sec´ 8kpc Rearranging gives vinitial = (220 km/s × 8)/400 = 4.4 km/s.

15. For a Pop III star to survive the 13 billion years since the formation of the Galaxy, it can only have a maximum mass of about 0.9 M⊙. Estimate the corresponding luminosity for this mass using the mass-luminosity relationship. Why do you think it may be difficult to find Pop III stars today?

15. (~70% as bright as the Sun) Using the mass-luminosity relationship presented in Unit 59.4, L ≈ M3.5, the luminosity is approximately, L ≈ (0.9)3.5 = 0.69 L⊙. With a maximum luminosity for Pop III stars that is only 70% that of the Sun, it may be somewhat difficult to find these stars simply because they are dim.

16. Suppose that after x billion years, the Milky Way converted half of its gas into stars. In the next x billion years, it converted half of the remaining gas into stars—and so on, each x billion years. The Milky Way is about 13 billion years old, and only about 15% of its disk remains today in the form of gas. a. What value of x would give the Milky Way’s current gas fraction? (Hint: You can calculate this mathematically or make a graph of the declining gas fraction and estimate x.) b. In how many billion years will the gas fraction be just 1%?

This is a “half-life” problem, where x is the half-life we seek, and in a time t, the fraction remaining will be (1/2)t/x. For 15% remaining after 13 billion years, this becomes: 0.15 = (1/2)(13 billion y)/x then taking logs of both sides, log(0.15) = ((13 × 109 y)/x)log (1/2)

Fraction remaining

16. (a) About 4.75 billion years; (b) In about another 18.5 billion years.

Fraction vs. Number of Half Lives

Number of Half Lifes

and solving for x, x = (13 × 109 y) × log(1/2)/log(0.15) 170

x = 4.75 × 109 years Alternatively, one can set up the graph at right to find about 2.7 half-lives are needed to get down to 15%, and then divide accordingly. (b) Solve for N such that there is just 1% = 0.01 remaining: 0.01 = (1/2)N, where N = number of half-lives each of which is 4.75 × 109 years. log(0.01) = N log(0.5) → N = log(0.01)/log(0.5) = 6.64 So the total is 6.64 × 4.75 × 109 y = 31.54 billion years. Subtracting the 13 billion years that have already happened, we get 18.5 billion years to go. (Estimating based on the chart, one might get 7 half-lives and slightly larger final answer.)

17. Suppose we found a nearby Pop II star that is on a highly elliptical orbit, but it is currently near its most distant point from the center of the Milky Way in its orbit. To stars like the Sun in orbit in the disk, what would be the approximate relative speed of this Pop II star? In what direction?

17. (Zero). If the star is near its farthest point from the center of the galaxy, in general the star would be near zero speed relative to disk stars. This is somewhat analogous to a pendulum at one end of its swing as it slows down and stops before falling back. Due to the random orbital inclinations of such Pop II stars, it could be in any direction.

Unit 73 11. What is the density of a gas cloud that is 10 ly in diameter and contains 100 M⊙ of material?

11. (About 5.5 × 10–23 kg/L). Density is mass/volume = M/V. We know M = 100 solar masses = (100 × 1.99 × 1030 kg). Calculate the volume converting ly to meters: V = 4/3 × π × R3 = 4/3 × π × (10 ly × 9.5 × 1015 meters)3 = 3.59 × 1051 m3 = 3.59 × 1054 L Density = M/V = (100 × 1.99 × 1030 kg)/(3.59 × 1054 L) = 5.5 × 10–23 kg/L. 12. Given that a single hydrogen atom has a mass of 1.67 × 10–27 kg, how large would a spherically shaped interstellar cloud, with a density of 106 atoms/liter, need to be to contain a solar mass of material?

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12. (0.69 ly) From the definition of density, ρ = M/V, we quickly get the volume of a sphere that contains a solar mass worth of material: V = M/ρ = (1.99 × 1030 kg) / ((106 atoms/liter) × (1.67 × 10−27 kg/atom)) = 1.19 × 1051 liters = 1.19 × 1048 m3.

The volume of a sphere is given by 4/3 πR3, which leads to a size of: R = (3 V/4 π)1/3 = (3 × 1.19 × 1048 m3 /4 π)1/3 = 6.57 × 1015 m = 0.69 ly.

13. Convert these radio wavelengths to frequencies (see Unit 22): a. 21-cm line of atomic hydrogen b. 2.6-mm line of carbon monoxide 13. (a) 1429 MHz; (b) 1.2 × 1011 Hz (a) From Unit 22 Wavelength (λ) times frequency (ν) = the speed of light, c. This can be rearranged to yield ν = c/λ = (3 × 1010 cm/sec)/21 cm = 1.429 × 109 Hz = 1429 MHz. (The wavelength is actually closer to 21.1 cm, giving a frequency of about 1421 MHz.) (b) A 2.6 mm wavelength corresponds to a frequency of 1.2 × 1011 Hz. ν = c/λ = (3 × 108 m/sec)/(2.6 × 10−3 m) = 1.2 × 1011 Hz 14. How do the energies of photons with the wavelengths given in problem 13 compare to the wavelength of a visible photon at 500 nm?

14. To compare the energies of the photons to the wavelength of a visible photon at 500nm, we do not need to compute the actual energies. We just need to note that from E = hc/λ (Unit 22), energy is inversely proportional to wavelength. Compared to the blue 500 nm photon, the 21 cm line is: (21 cm/500 nm) = (21 × 10−2 m) /(500 × 10−9 m) = 4.2 × 105 times longer so it also has 1/(420,000) times the energy. The 2.6 mm wavelength photon is: (2.6 × 10−3 m) / (500 × 10−9 m) = 5200 times longer, so it has an energy 1/5200 times that of the blue 500 nm photon. (The problem can also be solved by direct computation, but you have to find all three energies and then divide them to see how they compare).

15. Find the ratio of the energy of a 21-cm photon to the energy of an electron shifting from the second energy level of hydrogen to the ground state, which produces a photon of 122-nm wavelength.

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15. (The 21 cm has about 5.8 × 10–7 times the energy of the 122 nm photon).

That is, the 21 cm photon is that much less energetic than the 122 nm photon. E = hc/λ (Unit 22), so the energy ratio goes as the inverse of the ratio in wavelength: (122 nm/21 cm) = 1/(2.7 million) = 5.8 × 10–7.

As for #14, this can also be solved by computing the actual energies and then comparing them.

16. On average, a hydrogen atom emits a 21-cm photon once every 15 million years.

How many 21-cm photons does a 100-M⊙ gas cloud emit per second?

16. (2.5 × 1044 photons each second).

Number of H atoms in cloud = 100 × solar mass/mass of a H atom = 100 × 1.99 × 1030 kg/(1.67 × 10–27 kg/atom) = 1.2 × 1059 atoms.

Rate of emission = 1 photon/(15 million y atom) = 1 photon/(15 × 106 y × 3.16 × 107 s/y) = 1 photon/(4.7 × 1014 s atom) Number of photons per second = Number of atoms × rate of emission = (1.2 × 1059 atom) × 1/(4.7 × 1014 s atom) = 2.5 × 1044 photons/s 17. The dust in an interstellar cloud blocks blue light in the following way: For every 1 pc light travels through the cloud, only 90% of the light continues. Thus, after 2 pc, 90% of the remaining 90%, or 81% (0.9 × 0.9 = 0.81), remains. a. You might initially expect that after 10 pc, the ten 10% reductions would have removed all of the blue light. How much blue light actually remains? b. The same cloud removes about 7% of the red light every parsec, so after 1 pc, the ratio of blue to red has dropped to 90∕93 ≈ 97% of its unreddened value. What is the ratio of blue to red after 5 pc? After 10 pc?

17. (a) (About 35 percent remains). (b) (ratio of blue to red is 0.85 = 85% after 5 pc and 0.72 = 72% after 10pc). a) This is like a half-life problem, with the appropriate formula being, Fraction remaining = (9/10)N, where N = the number of parsecs.

For part a, (9/10)10 = 0.35, or 35%. This is also easily enough directly calculated for each parsec up to 10: (after 1 pc) 0.9, (after 2 pc) 0.81, (after 3 pc) 0.73, (after 4 pc) 0.66, (after 5 pc) 0.59, (after 6 pc) 0.53, (after 7 pc) 0.48, (after 8 pc) 0.43, (after 9 pc) 0.38, and (after 10 pc) 0.35. b) It is 0.85 after 5 pc, and 0.72 after 10 pc. There are several different ways to calculate this; one is to set up the formula ratio = (90/93)N.

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(90/93)5 = 0.85 and (90/93)10 = 0.72.

Or, one can calculate the fractions of each directly, and the ratio: After Blue Red

1 pc 0.90 0.93

2 pc 0.81 0.85

3 pc 0.73 0.80

4 pc 0.66 0.74

5 pc 0.59 0.70

6 pc 0.53 0.65

7 pc 0.48 0.60

8 pc 0.43 0.56

9 pc 0.38 0.52

10 pc 0.35 0.48

So the ratio at 5 pc = .59/.70= .84, at 10 pc = 0.35/0.48 = 0.73. (These don’t match exactly because of the rounding).

Unit 74 12. If the Sun moves in a circular orbit of radius 8 kpc around the center of the Milky Way at a speed of 220 km/sec, how long does it take to complete one orbit? How many orbits has the Sun completed in the 4.5 billion years since it formed?

12. (About 222 million years, and about 20 trips around the Galaxy). Assume that the Sun has a circular orbit. Circumference = 2 πR where R = 8 kpc or 8 × 3260 light years, which converts to about 2.45 × 1017 km. Thus, C = 2 × π × 2.45 × 1017 km = 1.54 × 1018 km. At 220 km/s, it would take 1.54 × 1018 km/(2.2 × 102 km/sec) = 7.0 × 1015 s to make one complete orbit. Since there are about 365 d × 24 h/d × 60 m/h × 60 s/h =31536000 seconds in a year, the orbital time is: (7.0 × 1015 s)/(3.15 × 107 s/y) = 2.22 × 108 y, or about 222 million years.

In 4.5 billion years, this translates to 4.5/0.222  20 trips around the Galaxy.

13. If the Galaxy’s mass doubled, what would the Sun’s velocity be?

13. (It would be 1.4 times as much as now, or about 311 km/s).

According to the unit, the mass inside a radius is M = (v2 × R)/G. Mass is directly proportional to the square of the velocity. So if the mass is doubled, then v2 would also double, and 2v2 = (2 × v)2 = (1.414 v)2. The velocity would be square root of 2 times as much as now. The current velocity is about 220 km/s, so if the mass were doubled the velocity would be about 311 km/s.

14. Near the center of the Milky Way, the rotation speed climbs in proportion to the radius; so at 400 pc from the center, the speed is twice as fast as at 200 pc from the center. a. What does this imply about the mass within 400 pc versus the mass within 200 pc? b. Is there any differential rotation in this region? Explain.

14. (a) 8 times as much mass; (b) no differential rotation. a) It means that mass increases roughly four times for the increase in radius. Actually, the increase is by the cube of the radius, as here 8 = 23. Consider the formula from the unit: M<r = V2 × R/G, where V is the velocity. M<r is proportional to V2 × R. If velocity is proportional to 174

radius in this section, then M<r is proportional to R2 × R = R3. There should be 23 = 8 times as much mass inside the 400 pc orbit as the 200 pc orbit.

14. (cont.) b) At 400 pc, the circumference is twice as much as at 200 pc (C = 2 R). t = d/v = c/v, so the time (the period) of the orbit goes as R/v. But if v is proportional to R, then the time goes as R/R – the periods of the orbits are the same. As a result, in the region where velocity is proportional to radius there is no differential rotation. The increase in velocity compensates for the increased circumference, and a star at 400 pc stays aligned with a star at 200 pc – it’s like the solid wheel in the Figure 74.1 in the Unit.

15. Suppose there were no additional mass in the Milky Way beyond the distance of the Sun’s orbit. What would the rotation speed be at a radius of 16 kpc (twice as far from the center as the Sun)?

15. (About 71% as much as the Sun’s, or 156 km/s). If there were no additional mass beyond the Sun’s orbit, then in that region the speed would vary as the inverse square root of radius.

We can still use the formula from the unit: M<r = V2 × R/G. This can be rearranged as V2 = GM<r/R. In this case both G and M are constant beyond the Sun’s orbit, so we can say that V is just proportional (1/R). So if R doubled from 8 kpc to 16 kpc, we would have V proportional (1/2) = 0.707. So the speed at 16 kpc would be about 71 percent that at 8 kpc, or 220 km/s × 0.71 = 156 km/s.

16. What is the Schwarzschild radius of a black hole with the mass of the black hole thought to be at the center of the Milky Way? What is the angular size of the black hole as viewed from Earth?

16. (1.2 × 107 km; 1.0 × 10−5 arc-seconds). Use Rs = 2 GM/c2, using M the mass of the black hole of ~ 4 million times the mass of our Sun: Rs = 2 GM/c2 = 2(6.67 × 10−11 m3 kg−1 s−2)(4 × 106 × 2 × 1030 kg)/(3 × 108 m/s)2 = 1.2 × 1010 m = 1.2 × 107 km = 0.08 AU. Alternatively, one can use: RS = 3 km (M/MSun) = 3 × (4 × 106) km = 1.2 × 107 km.

From Unit 10, the linear size of an object is given by ℓ = d × (α /57.3°) which we can rearrange for the angular size, α = ℓ/d × 57.3°. Using a distance to the central black hole of 8 kpc = 1.65 × 109 AU, we get an angular size of: α = (0.08 AU)/(1.65 × 109 AU) × 57.3° = 2.79 × 10−9° = 1.0 × 10−5 arc-seconds.

17. If spiral arms are 2 kpc wide, how long does the Sun, 8 kpc from the Galactic center with an orbital speed of 220 km/sec, spend inside a spiral arm on its motion around the Milky Way? 17. (About 9 million years). The question is how long it takes the Sun to travel through 2 kpc.

Without knowing the exact orientation of the spiral arm and the Sun’s path, we can assume the

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Sun simply needs to move through 2 kpc along its orbit to get an estimate. This is simple since we have been given the Sun’s velocity.

2kpc = 2000 pc × 3.09 × 1016 m/pc = 6.18 × 1019 m = 6.18 × 1016 km.

6.18 × 1016 km / (220 km/s) = 2.8 × 1014 s = 8.9 × 106 y. (Using 3.16 × 107 s/y).

Unit 75 12. A Cepheid variable with a period of 20 days and an apparent brightness of 5.28 × 10–16 watt/m2 is observed in a nearby galaxy. Estimate the Cepheid’s luminosity from the period-luminosity relation (Figure 64.5) and convert this to watts. How far away is the host galaxy?

12. (3.83 × 1030 W and 2.5 Million ly) Figure 64.5 shows the period-luminosity for Cepheid variables. From this a Cepheid with a period of 20 days has a luminosity of 104 L⦿which is 104 × 3.83 × 1026 W = 3.83 × 1030 W. Following the calculation in this unit, the distance to the galaxy is: dg = √(LC/4 πBC) = √(3.83 × 1030 W / (4 π × 5.28 × 10−16 W/m2)) = 2.40 × 1022 m = 2.5 × 106 ly = 2.5 Mly. 13. Suppose that a galaxy’s 21-cm line of atomic hydrogen is shifted to 22 cm. a. What is the redshift of the galaxy? b. What is its recession velocity?

13. (a) z = 0.048; (b) 14,300 km/s. (a) From the definition of redshift: z = Δ λ/ λ = (22-21) cm/21 cm = 0.0476, the red shift is 0.048. (b) The recession velocity is v = zc = (0.0476) (3 × 105 km/s) = 1.43 × 104 km/s.

14. Two identical galaxies have redshifts of 0.036 and 0.072. Which has the larger angular size? What is the ratio of their apparent sizes?

14. (The galaxy at z = 0.036 is closer; the one at 0.036 appears twice as large).

We’ll need to chain together some relations. First, the recession velocity is proportional to the redshift, v = zc. Second, using Hubble’s Law, and taking the distance to the galaxies as d = v/H, the distance is proportional to the velocity, therefore it is proportional to the redshift. Finally, the angular size is A = 57.3° × L/d, so the angular size is inversely proportional to the distance, and therefore inversely proportional to the redshift (remember L is the same for the two “identical” galaxies). For two galaxies of the same size, then, the one with the smaller redshift— the one with z = 0.036—will appear larger. Since the redshift of the more distance galaxy at

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0.072 is 0.072/0.036 = 2 times larger than the redshift of the closer galaxy, the galaxy at 0.036 will appear twice as large.

15. For most of the second half of the twentieth century, measured values of Hubble’s constant ranged from about 50 to 90 km/sec per Mpc. For a galaxy with a recessional velocity of 28,000 km/sec, what range of distances would Hubble’s law return?

15. (From 560 Mpc to 311 Mpc). Following the example in the Unit, d = V/H, and assuming that H is the smaller value of 50 (km/s)/Mpc, we have: d = 28000 km s−1/(50 km s−1/Mpc) = 560 Mpc. For H of 90 (km/s)/Mpc, we have d = 28000 km s−1/(90 km s−1/Mpc) = 311 Mpc. 16. What is the approximate recession velocity of a galaxy that we are seeing as it looked 1 billion years ago?

16. (About 21500 km/s). If we see it as it was 1 billion years ago, that means that it is at 1 billion light years distance. There are 3.26 light years in a parsec, so 1 billion light years is about 3.07 × 108 pc = 307 Mpc. With a Hubble constant of H = 70 km s−1/Mpc: V = d × H = 307 Mpc × 70 km s−1/Mpc = 21490 km/s.

Unit 76 11. If a galaxy 30 kpc in diameter were at each of the following distances, what would its angular size be? (See Unit 10.4.) Express your answers in arc minutes. a. 10 Mpc; b. 1000 Mpc. 11. (a) 10 arcminutes; (b) 0.1 arc minutes.

From Unit 10, L = d × A/57.3° where L is the linear size, d is the distance and A is the angular diameter. This can be changed to A = 57.3° × L/d. L and d must be changed into the same units.

L = 30 kpc, and for (a) d = 10 Mpc = 10 × 103 kpc = 104 kpc; for (b) d = 103 Mpc = 106 kpc. (a) A = (30 kpc/104 kpc) × 57.3° = 0.17° = 10 arc minutes. (b) A = (30 kpc/106 kpc) × 57.3° = 0.0017° = 0.1 arc minutes = 6 arc seconds.

Quick check: (b) is 100 times more distant and so should be 1/100 as large.

12. If the separation between galaxies is 1 million pc (1 Mpc), how many galaxies are found in a volume of radius 1,000,000 Mpc?

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12. (about 8 × 1018 galaxies). We must assume the galaxies are evenly spaced, and divide the total volume by the volume “occupied” by each galaxy. The volume of a galaxy is 4/3 × π(Rgal)3, and the volume of the whole thing is 4/3 × π(RAll)3, so the number will be N = *4/3 × π(RAll)3+ / *4/3 × π(Rgal)3] = (RAll / Rgal)3.

The question gives the radius RAll = 1 million Mpc. The separation between galaxies is 1 Mpc, so only tricky thing is that Rgal is half the separation between galaxies, or only 1 Mpc/2 = 0.5 Mpc. So there are N = (106 Mpc/0.5 Mpc)3 = 8 × 1018 galaxies.

13. For the majority of galaxies astronomers observe, the light travel time is less than a billion years. This is only a fraction of the Sun’s age, and therefore we do not expect to find many differences in the galaxies’ properties due to their age. What would the redshift be for a galaxy that we are seeing as it was 1 billion years ago?

13. (The red-shift is about 0.072).

If we see the galaxy as it was 1 billion years ago, it is 1 billion light years away.

109 ly /(3.26 ly/pc) = 3.07 × 108 pc = 307 Mpc. Then, V = d × H = 307 Mpc × 70 km s−1/Mpc = 21490 km/s.

Since v = cz, z = v/c = 21490 km/s / (300,000 km/s) = 0.072.

14. The Hubble Deep Field covers an area of about 5 square arc minutes and contains about 3000 galaxies. If we observed the entire sky to this depth, about how many galaxies would we find? If we assume each galaxy has 100 billion stars (the same as the Milky Way), how many stars are there?

14. (We would find roughly 90 billion galaxies, and 9 × 1021 stars).

In degrees, the circumference of a sphere is 2R, and that circumference will span 360°. Equating this two different ways to measure the sky reduces to R = 57.3°. The surface area of a sphere is 4 R2. Substitute 57.3° for R and you find that there are approximately 41259 “square” degrees on a sphere. There are 60 arc minutes per degree, so there are 3600 square arc minutes per square degree. Therefore there are 41259 × 3600 = 1.48 × 108 square arcminutes on the sky. Our sample area is 5 square arcminutes, so there are: 1.48 × 108 square arcminutes / (5 square arcminutes/area) = 3 × 107 areas equal to the size of the Hubble Deep Field on the sky.

If each has 3000 galaxies, the total number of galaxies would be: 3 × 107 areas × 3000 galaxies/area = 9 × 1010 galaxies = 90 billion galaxies.

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If each galaxy has a 100 billion = 100 × 109 stars, then there are: Nstars = (90 × 109 galaxies) × (100 × 109 stars/galaxy) = 9 × 1021 stars!

15. The radial velocity of M31 toward the Milky Way is about 100 km/sec. If it maintains that speed, how long will it take to travel the 700-kpc distance between the two galaxies?

15. (About 6.8 billion years). Relate the time to distance and speed, t = d/v. d = 700 kpc = 700 × 103 pc × 3.09 × 1013 km/pc = 2.16 × 1019 km. so: t = d/v = 2.16 × 1019 km / (100 km/s) = 2.16 × 1017 s. and ( 2.16 × 1017 s) / (3.16 × 107 s/y) = 6.8 × 109 y = 6.8 billion years.

16. Assume that the disk of a galaxy like the Milky Way has a diameter of 40 kpc and has a surface area πR2. Likewise, each star covers an area determined by πR2. a. If there are 100 billion stars each the size of the Sun, what fraction of the galaxy’s disk is actually “covered” by a star? b. The fraction found in (a) expresses the approximate chance that when a single star passes through the galaxy’s disk, it will strike a star. If an identical galaxy passed through, how many stars do you estimate would undergo a collision?

16. (a) (1.3 × 10−13, or about one ten-trillionth). (b) no stars would collide. (a) In the problem, we are directed to consider each object to have a surface area as R2; we might expect this to be πR2, but since all we need is the fraction the π’s would cancel out anyway.

Fraction = (area of stars) / (area of disk) = [100 × 109 stars × (RSun)2] / [Rgalaxy]2 The Sun has a radius of 6.96 × 105 km. The galaxy has a diameter of 40 kpc, and therefore a radius of 20 kpc = 20 × 1000 pc × 3.09 × 1013 km/pc = 6.18 × 1017 km.

Fraction = [100 × 109 stars × (6.96 × 105 km)2] / [6.18 × 1017 km]2 = 4.8 × 1022 / (3.8 × 1035) = 1.3 × 10−13 16. (cont.) (b) No stars would undergo a direct collision. The chance of a collision would be about (1.3 × 10−13) × 1011 = 0.013, even with 100 billion stars in both galaxies. This means we might expect a single stellar collision to occur in one out of 100 galaxy collisions!

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Unit 77 13. Use Wien’s law (Unit 23) to estimate the temperature of gas in clusters if the peak emission wavelength is 0.1 nm. Compare this to the Sun’s core temperature of 15 million kelvin.

13. (29 × 106 K). Wien’s law states that T = 2.9 × 106 K nm/λmax. For a peak wavelength of 0.1 nm the temperature works out to be T = 2.9 × 10 6 K nm/0.1 nm = 2.9 × 107 K = 29 million Kelvin. This is comparable to the Sun’s core temperature.

14. The Great Attractor, located some 50 Mpc away, is suspected to have caused the Local Group to accelerate to a speed of 500 km/sec over the last 13 billion years. a. What is the average acceleration over that time, in m/sec2? b. Based on Newton’s law of gravitation, how big a mass is needed at that distance to produce the observed acceleration?

14. (a) About 1.6 × 10–12 m s–2. (b) About 5.7 × 1046 kg. a) Acceleration = change in velocity / change in time = ∆v / ∆t.

Assuming that it started at 0 velocity, a = (500 km/s – 0 km/s)/(13 billion years) = (500 × 103 m/s)/(13 × 109 y × 3.16 × 107 s/y) = 1.2 × 10−12 m/s2. (b) The acceleration is caused by the force of gravity between the Local Group and the Great Attractor. The force of gravity is F = GMGAMLG/R2. We can equate this to Newton’s first law, F = MLG × a.

MLG × a = GMGAMLG/R2 14. (cont.) The mass of the local group cancels out. The distance R is 50 Mpc = 1.545 × 10 24 m.

Solving for the mass of the Great Attractor, MGA = a × R2 / G. = (1.2 × 10−12 m/s2)× (1.545 × 1024 m)2 / (6.67 × 10−11 m3 kg−1 s−1) = 4.36 × 1046 kg = 2.2 × 1016 solar masses.

15. Compare the gravitational attraction between the Milky Way (M = 10 12 M⊙) and the Andromeda galaxy (M = 1012 M⊙, d = 2.5 Mly), to the gravitational attraction between the Local Group (M = 5 × 1012 M⊙) and the Virgo Cluster (M = 1015 M⊙, d = 54 Mly).

15. (4.7 × 1029 N, 5.1 × 1030 N). The gravitational attraction between the Milky Way and the Andromeda galaxy is:

180

F= =

G M M W M A nd d2 2 (6.67´ 10- 11N m 2kg- 2 )(1012 ´ 2´ 1030 kg) 2

15 é æ öù ê(2.5´ 106 ly)çç9.5´ 10 m ÷ ú ÷ ÷ ê ú ÷ ç l y è øú êë û

= 4.7´ 1029 N

Likewise, the gravitational attraction between the Local Group and the Virgo Cluster is 5.1 × 1030 N. 16. Suppose that the typical galaxy has a mass of 1011 M⊙. If the total mass of a cluster of 100 galaxies is 1015 M⊙, how much dark matter is contained in the cluster by the percentage of the total mass (ignoring the hot X-ray gas)?

16. (99% is dark matter).

The typical galaxy has a mass (of stars and gas) of 1011 solar masses. The total mass of 100 galaxies in the cluster is therefore 1013 solar masses. The total mass of the cluster is 1015 solar masses. So only 1013/ 1015 = 0.01 = 1% of the mass of the cluster is in the galaxies, and 99% of the mass must be dark matter.

17. Same as problem 16, but now the hot X-ray gas has a mass 10 times greater than the galaxies. How much dark matter is contained in the cluster by the percentage of the total mass?

17. (89% is dark matter). If the hot X-ray gas has a mass 10 times greater than the (normal) mass of the galaxies in the cluster, then there is 10 × 100 × 1011 solar masses = 1014 solar masses of material in hot gas. The total non-dark matter mass is 1014 solar masses of hot gas plus 1013 solar masses of galaxies, which is 0.11 × 1015 solar masses. The fraction of normal matter is (0.11 × 1015)/1015 = 0.11.

So it’s about 11% normal matter and 89% dark matter.

18. A spectral line of hydrogen has an absorption laboratory wavelength of 122 nm.

It is observed in the spectrum of a quasar at 130 nm, 180 nm, and 330 nm. What are the redshifts (z) of these clouds? What would be their distances, assuming Hubble’s law (and using V = c × z)?

18. (a) Redshifts of about 0.0656, 0.475, and 1.70. (b) distances of about 290, 2000, & 7300 Mpc. (a) The rest wavelength of the line is 122 nm. It is observed at 130, 180, and 330 nm.

Red-shift = z =  λ/ λ so z = (130 nm – 122 nm)/(122 nm) = 8/122 = 0.0656 181

z = (180 nm – 122 nm)/(122 nm) = 58/122 = 0.475 z = (330 nm – 122 nm)/(122 nm) = 208/122 = 1.70 (b) Using v = c × z and d = v/H = c × z / H, d = c × z / H = (3 × 105 km/s) × (0.0656) / (70 km/(s Mpc)) = 281 Mpc d = c × z / H = (3 × 105 km/s) × (0.475) / (70 km/(s Mpc)) = 2036 Mpc d = c × z / H = (3 × 105 km/s) × (1.7) / (70 km/(s Mpc)) = 7286 Mpc

Unit 78 12. If the radio galaxy NGC 383 has lobes that are 300 kpc in length, how long does it take material traveling at 90% of the speed of light to travel from the base to the tip of a single lobe?

12. (1.1 million years). Used the fact that the speed of light can be expressed as the distance light travels in one year divided by one year, c = ly/yr. So at 90% of the speed of light, material will travel 90% of a ly in one year. Using d = v t, solve for time: t = d/v = (3 × 105 pc × 3.26 ly/pc)/(0.9 × ly/yr) = 1.1 × 106 years = 1.1 million years.

13. The radius of a black hole is approximately 3 km times the mass of the black hole in solar masses (Unit 69). a. What is the diameter of a 109 M⊙ black hole in AU? b. For an accretion disk 10 times the radius of the black hole in (a), how many light-hours across would it be?

13. (a) About 40 AU. (b) About 55.6 hours (a) RSchwarz = 3 km × (M/MSun) = 3 km × 109 = 3 × 109 km.

Convert to AU: 3 × 109 km / (1.5 × 108 km/AU) = 3/1.5 × 109/108 AU = 20 AU. The diameter, then, is 40 AU. (b) Ten times the radius is 10 × 3 × 109 km = 3 × 1010 km, but we need the diameter rather than radius, so it is 6 × 1010 km. Light travels at about 3 × 105 km/s, so the light travel time would be: t = d/v = 6 × 1010 km / (3 × 105 km/s) = 2 × 105 s.

Divide by 3600s in an hour to get 55.6 h.

14. Hydrogen has a spectral line at 122 nm in the ultraviolet portion of the spectrum. What are the largest and smallest red-shifts a quasar could have for this line to be seen in the visible portion of the spectrum (400 to 700 nm)?

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14. (minimum z ~ 2.28 for the line to be visible in blue (400 nm); maximum z = 4.74 for the line to be visible in red light (700 nm). Red-shift is defined as: z =  λ/λ = (λrs − λo)/ λo, where λ rs is the red-shifted wavelength we observe, and λo is the original wavelength emitted in the rest frame of the quasar. The minimum shift to get into the visible would be to 400 nm, so: zmin = (400 nm – 122 nm)/122 nm = 2.28; Similarly, the maximum shift would be to 700 nm, so: zmax = (700 nm – 122 nm)/122 nm = 4.74.

15. An AGN is observed to double in brightness over 7 days and then to grow dim again in the next 7 days. What is an upper limit to the probable size of the AGN? Give your result in AU. 15. (About 1200 AU, or roughly 10 times the diameter of the solar system).

We have Diameter < (c × t). Here  t = 7 days = 7 d × 24 h/d × 3600 s/h = 604800 s.

So diameter < 3 × 105 km/s × 604800 s = 1.81 × 1011 km = 1213 AU. 16. Spectral broadening observed in a particular Seyfert galaxy suggests that the material is orbiting at 1200 km/sec at a radius of 0.5 parsecs from its center. Calculate the amount of mass within this radius.

16. (3.32 × 1038 kg). Following the procedure outlined in Unit 74 the mass within the given radius is found from M<R = V2 × R/G. For the given values this works out to be: M<R = (1.2 × 106 m/s)2 × (0.5 pc × 3.08 × 1016 m/pc) / (6.67 × 10−11 m3 / kg ∙ s2) = 3.32 × 1038 kg = 1.7× 108 M⊙.

17. A hydrogen line of gas orbiting in the nucleus of an active galaxy is Doppler shifted on the right side of the AGN to 123 nm, and on the left side to 121 nm. How fast is the gas rotating?

17. (About 2460 km/s). The hydrogen line is redshifted to 123 nm on the side of the galaxy that is rotating away from us, and to 121 nm (blue shifted) on the side that is rotating towards us; the rest wavelength should therefore be 122 nm (assuming everything is symmetric). The Doppler shift is: v/c =  λ/λ = (123 nm – 122 nm) / 122 nm = 0.00819, and the velocity must be (0.00819) × c = (0.00819) × 3 × 105 km/s = 2460 km/s.

183

Unit 79 11. Use Figure 79.3 to calculate the mass within 30 kpc for the overall rotation curve and separately for the contribution just from stars. By what factor is the implied mass from the total rotation curve greater than the curve from the luminous stars?

11. From figure 79.3 we can estimate that at 30 kpc the rotation speeds are approximately 210 km/s for the full rotation curve and 40 km/s for the contribution from stars. For the full rotation curve, 2 R (3.0´ 104 pc´ 3.08´ 1016 m /pc 5 M <R = V ´ = (2.10´ 10 m /s) ´ G 6.67´ 10- 11m 3kg- 1s- 2 2

= 6.1 × 1041 kg = 3.1 × 1011 M⊙.

For the rotation curve contribution from just the luminous stars we observe, 2 R (3.0´ 104 pc´ 3.08´ 1016 m /pc 4 M <R = V ´ = (4.0´ 10 m /s) ´ G 6.67´ 10- 11m 3kg- 1s- 2 = 2.2 × 1040 kg = 1.1 × 1010 M⊙.

2

This is more-or-less in agreement with the idea that there is about ten times more dark matter than visible matter in spiral galaxies.

12. What is the mass-to-light ratio of a galaxy with 1011 M⊙ in stars and gas and 5 × 1011 M⊙of dark matter?

12. (mass to light ratio ~ 6). The mass to light ratio is just as it sounds; the 1011 solar masses of stars and gas can be estimated to produce 1011 solar luminosities of light. The total mass of the galaxy would be the dark matter plus the stars and gas, MTot = 5 × 1011 MSun + 1011 MSun, so the ratio is (6 × 1011 MSun) / (1011 LSun) for a mass-to-light ratio of 6.

13. A spiral galaxy is observed in which the rotation speed remains 250 km/sec from 1 kpc out to the largest distances at which we can detect anything. How much mass exists out to 10 kpc? 11 kpc? 12 kpc? Can you generalize your results as to how much additional matter there is for each kiloparsec farther out you look?

13. (10 kpc: 2.9 × 1041 kg, 11 kpc: 3.2 × 1041 kg, 12 kpc: 3.5 × 1041 kg; there is an increase of about 3 × 1040 kg per Kpc). We want to use M<R = (V2 × R)/G, with values for R of 10, 11, 12… kiloparsecs. 1 kpc = 3.09 × 1019 m. G = 6.67 × 10–11 m3 kg–1 s2. V is constant at 250 km/s. As an example, M<R = (V2 × R)/G = ((250 × 103 m/s)2 × 10 × 3.09 × 1019 m / (6.67 × 10–11 m3 kg–1 s2)

184

= 2.9 × 1041 kg.

For 11 kpc, M<R = 3.2 × 1041 kg, and for 12 kpc, M<R = 3.5 × 1041 kg.

We can also see that since V is constant, for each kpc of distance, Mkpc = (V2 × R)/G = ((250 × 103 m/s)2 × 3.09 × 1019 m /(6.67 × 10–11 m3 kg–1 s2) = 2.89 × 1040 kg about 3 × 1040 kg is added (which is also evident from the pattern of answers).

(Divide results in kg by 2 × 1030 kg/solar mass to find the answers in solar masses if desired.)

14. In the Virgo Cluster there are galaxies measured to be traveling at about 1000 km/sec at a distance of 500 kpc from the center of the cluster. If you apply the mass formula to this velocity and radius, what interior mass do you find?

14. (About 2.3 × 1044 kg, or 1.2 × 1014 kg solar masses).

We want to use M<R = (V2 × R)/G, with values of 1000 km/s for V and 500 kpc for R. Convert 500 kpc = 500 × 103 pc × 3.09 × 1016 m/pc = 1.545 × 1022 m.

M<R = (V2 × R)/G = ((1000 × 103 m/s)2 × 1.545 × 1022 m /(6.67 × 10–11 m3 kg–1 s2 ) = 2.3 × 1044 kg = 2.3 × 1044 kg/(2 × 1030 kg/solar mass) = 1.2 × 1014 M⊙.

15. Suppose the Milky Way contains 10 billion M⊙ of dark matter at radii between 8 and 9 kpc from its center. a. What is the density of matter in the “shell” between the radii of 8 and 9 kpc?

(Hint: Subtract the volume of a sphere of radius 8 kpc from the volume of a sphere of radius 9 kpc.) b. How much mass of dark matter would there be in a volume equal to Earth?

What object has a similar mass?

15. (a) About 7.5 × 10–25 kg/L. (b) About 0.8 kg within the Earth’s radius; which is about 4 × 10−31 solar masses. a) Density = M/V, and we need the volume of the shell, which is the difference of the volumes of the spheres with a 9 kpc and 8 kpc radii.

9 kpc = 9 × 103 pc × 3.09 × 1016 m/pc = 2.781 × 1020 m.

8 kpc = 8 × 103 pc × 3.09 × 1016 m/pc = 2.472 × 1020 m. And 1000 L = 1m3 so: Vshell = (4/3 × πR93) – (4/3 × πR83) = 4/3 × π × (R93 – R83) = 4/3 × π × ((2.781 × 1020 m)3 – (2.472 × 1020 m)3) = 2.68 × 1061 m3 =2.68 × 1064 L Density = M/Vshell = (1010 × 2 × 1030 kg)/(2.68 × 1064 L) = 7.5 × 10−25 kg/L. Or, = M/Vshell = (1010 × 2 × 1030 kg)/(2.68 × 1061 m3) = 7.5 × 10−22 kg/m3.

185

15. (cont.) b) The amount of dark matter in a sphere equal in radius to the Earth would be M = density × volume = (7.5 × 10−22 kg/m3) × (4/3 × π × (6378 × 103m)3) = 7.5 × 10−22 kg/m3 × 1.1 × 1021 m3 = 0.815 kg.

This is between the mass of a 20 oz soda and a small pineapple.

In astronomical terms, this is 0.815 kg/(2 × 1030 kg/solar masses) = 4 × 10−31 solar masses.

(Even compared to the volume of space within the Earth’s orbital radius around the Sun, the result is only: M = density × volume = (7.5 × 10−22 kg/m3) × (4/3 × π × (1.5 × 1011m)3) = 7.5 × 10−22 kg/m3 × 1.4 × 1034 m3 = 1013 kg. still a tiny fraction of the Sun’s mass).

16. Most spiral galaxies show a decline in the amount of light they produce that drops by a factor of 2 every 2 kpc farther out you look. For example, 3 × 1010 L⊙ are produced within 2 kpc of the center; 1.5 × 1010 L⊙ are produced between 2 and 4 kpc of the center; half that amount between 4 and 6 kpc; and so on. Compare the amount of luminous matter found this way to the total amount of matter contained within these same regions if the galaxy has a flat rotation curve with a rotation speed of 250 km/sec. Find the ratio of dark matter to luminous matter within 2 kpc of the center; between 2 and 4 kpc from the center; and between 10 and 12 kpc from the center.

16. For this problem we assume that each solar luminosity of light is produced by 1 solar mass, which should be provide a reasonable estimate of the mass of luminous matter. To find the total amount of matter in a shell, we will use M<R = (V2 × R)/G, and subtract the amount from the smaller shell inside. Since the velocity is constant, this is quite easy if we are only interested in 2 kpc sections: MR4 to R2 = MR4 – MR2 = (V2 × R4)/G – (V2 × R2)/G = (V2)/G × (R4 – R2) = (V2)/G × (2 kpc) = V2 × 2 kpc/G Mover 2kpc = (2.5 × 105 m/s)2 × 2 × 103 pc × 3.09 × 1016 m/pc /(6.67 × 10−11 m3 kg−1 s−2) = 5.79 × 1040 kg = 2.9 × 1010 solar masses.

16. (cont.) There is always 2.9 × 1010 solar masses in each 2 kpc section. The amount of dark matter in each section of the galaxy should be the difference between the total amount of mass and the luminous mass. The chart below shows the amounts of luminous matter, the total matter, the inferred dark matter, and the ratio of dark matter to luminous matter.

Shell range (kpc)

Luminous M in shell (solar M)

Total Mass in shell (solar M)

Total Mass / Lum. Mass (ratio)

Dark Matter (Total – LM) (solar M)

Dark Mass / Lum. Mass (ratio)

0 to 2 2 to 4

3.00E + 10 1.50E + 10

2.90E + 10 2.90E + 10

0.97 1.93

−1.00E + 09 1.40E + 10

−0.03 0.93

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4 to 6 6 to 8 8 to 10 10 to 12

7.50E + 09 3.75E + 09 1.88E + 09 9.38E + 08

2.90E + 10 2.90E + 10 2.90E + 10 2.90E + 10

3.87 7.73 15.47 30.93

2.15E + 10 2.53E + 10 2.71E + 10 2.81E + 10

2.87 6.73 14.47 29.93

The ratio of dark matter to luminous matter is almost 0 in the center part of the galaxy—although there is some approximation in the model. By 2-4, there is almost as much dark matter as luminous matter, with the ratio of 0.93. At 10-12 kpc, there is 30 times as much dark matter as luminous matter in the shell.

Unit 80 11. When Hubble first estimated the Hubble constant, galaxy distances were still very uncertain, and he got a value for H of about 600 km/sec per Mpc. What would this have implied about the age of the universe? What problems would this have presented for cosmologists?

11. (The age would be ~ 1.6 billion years; see below).

Inverting the constant to find the Hubble time, we get: tH = 1/H = 1 Mpc/(600 km/s) = (106 × 3.09 × 1013 km) / (600 km/s) = 3.09 × 1019 s / 600 = 5.15 × 1016 s.

11. (cont.) This is 5.15 × 1016 s / (3.16 × 107 s/y) ~ 1.6 × 109 y, or only about 1.6 billion years—there are rocks on Earth much older than this, and Sun is also estimated to be much older than this. This of course was a source of great concern for cosmologists, since it makes no sense for the Earth and Sun to be older than the universe itself! This much higher Hubble constant would have implied that the universe was expanding much faster than current estimates. The greater expansion rate would mean that the universe would have taken less time to expand to its current size, and hence that the age of the universe is much less than the currently accepted value of 13 to 14 billion years.

12. A galaxy at a distance of 25 Mpc has a recession velocity of 1875 km/sec. What is Hubble’s constant based on this one galaxy? Why is this not a good way to determine Hubble’s constant?

12. (H = 75 km s−1 Mpc−1).

H = v/d = (1875 km/s) /(25 Mpc) = 75 km / (s Mpc). One reason this is not a good way to determine Hubble’s constant because there may be additional factors affecting the galaxy’s velocity which we have not considered (local or “peculiar” motion from interactions with other galaxies).

13. For many years, opposing opinions on the value of Hubble’s constant were debated.

Some astronomers argued that H = 50 km/sec per Mpc, while others claimed a value close to 90 km/sec per Mpc. Calculate tH for these two values of H. 187

13. (About 2 × 1010 years for H = 50 km/sec per Mpc; about 1 × 1010 years for the larger value H = 90 km/sec per Mpc) For the value of H = 50 km/sec per Mpc: tH = 1/Ho = 1/(50 km/s/Mpc) = 1 Mpc / (50 km/s) = 1 s × 1 Mpc / 50 km = 1 s × 106 pc × 3.0856 × 1016 m × (10−3 km/m)/ 50 km = 6.17 × 1017 s = (6.17 × 1017 s) / (3.16 × 107 s/yr) = 1.95 × 1010 yr For the value of H = 90 km/sec per Mpc tH = 1/Ho = 1/(90 km/s/Mpc) = (3.0856 × 1019 km s)/(90 km) = 3.43 × 1017 s = 1.09 × 1010 yr 14. What wavelength would visible light (400 to 700 nm) have if it were seen from a galaxy formed just half a billion years after the Big Bang? What if it came from matter when the universe was just 400,000 years old? (Refer to Table 80.1.)

14. (At 500 million years after the Big Bang, the range would be 4400 to 7700 nm; and for 400,000 years after the Big Bang, the range would be 400400 nm to 700700 nm (4.004 × 105 nm to 7.007 × 105 nm.)

If we see the galaxy as it was just a half billion years after the Big Bang, this implies a red-shift of 10 from Table 80.1. Red-shift z =  λ/λ = (λobserved – λemitted)/ λemitted. So taking z = 10 and the original emitted light λemitted = 400 nm, we get:  λ = 10 × 400 nm = 4000 nm, and λobserved = λe+λ = 400 nm + 4000 nm = 4400 nm.

Likewise, we get 7700 nm for the 700 nm light. Use the same reasoning for the galaxy if it formed just 400,000 years after the Big Bang, for which z = 1000, so for 400 nm:  λ = 1000 × 400 nm = 400,000 nm = 4 × 105 nm and λobserved = λe + λ = 400,400 nm.

This is a wavelength of 400 m or 0.4 mm, in the infrared portion of the EM spectrum.

15. Plot the light travel time (y axis) versus the redshift (x axis) based on the data from Table 80.1. Estimate the redshift for galaxies that are seen as they were 1, 2, and 3 billion years ago.

15. To graph the whole data set, it makes sense to use a log scale for the redshift. However, only the first few data points are needed to answer the question. Also, note that beyond a redshift of 10, the change in light travel time is quite modest, so a linear graph through z = 10 covers most of the rapid change in travel time vs. z.

1 billion years, z = 0.07; 2 billion years, z = 0.17; 3 billion years, z = 0.28.

188

Light Travel Time vs.

Light Travel Time vs.

Redshift

Redshift

Light Travel (Billions of years)

Light Travel (Billions of years)

15. (cont.)

Redshift

Redshift

16. Plot the distance when light was emitted (y axis) versus the redshift (x axis) based on the data from Table 80.1. Estimate the redshift at which a galaxy of a fixed physical size would have the smallest angular size. Estimate the distance at which the galaxy would look the same size as it does when at a redshift of 0.1.

16. Angular size varies inversely proportional with the distance. A galaxy of a fixed specific size would have the smallest angular size at the farthest distance—the peak of the graph at z = 2. At a redshift of 0.1, the galaxy was 1.2 billion light years away; note that the same galaxy would appear the same angular size if it was at a redshift of ~ z = 140. This is because the actual distance of the galaxy when the light was emitted would also be 1.2 billion light years for a redshift of ~ 140.

Distance vs. Redshift

Distance When Emitted (Billions of ly)

Distance When Emitted (Billion ly)

Distance vs. Redshift

Redshift

Redshift

189

Unit 81 11. The temperature of the universe needs to be less than 100 K before stars can form.

At what redshift does this occur?

11. (A redshift of about 33).

Recall from the text that the scaling of the temperature of the universe goes with redshift. The larger the redshift, the farther back into time we are observing, and the hotter the universe would have been. The microwave background today at 3 K implied a temperature of 3000 K at z = 1000. So we get the universe reaching a cooler temperature of ~ 100 K at a later time, with a smaller redshift of z = 33.

12. When our Solar System formed about 4.5 billion years ago, approximately what would the temperature of the CMB radiation have been? Would the structure in the CMB have looked the same as today? Explain your answer.

12. (About 7 K). When the Solar System was formed, the universe was about 2/3 rd its current age, and looking the data in the table in Unit 80 we can conclude that the redshift z for ~ 13.7 billion – 4.5 billion = 9.2 billion years ago is about 1.5. The redshift is smaller 7 billion years ago (z = 1) and larger 10.2 billion years ago (z = 2). The universe is therefore about z + 1 = 2.5 times larger now than it was then (or, it was about 0.4 the current size then, see Table values).

Therefore, the wavelength of the CMB would be 2.5 times shorter, and the temperature correspondingly 2.5 times larger—about 7 K or so (2.7 K × 2.5 ~ 7 K).

At that time we would have been seeing CMB light coming from regions that were nearer to us than what we see today (today, we would see these same regions as they are 4.5 billion years after the epoch of last scattering that the CMB represents). We would expect the kinds of structure to have been similar, because the structure of the CMB was set at last scattering, but the specific patterns would have been different and the angular sizes larger (since the CMB would have been “closer”).

13. Imagine that there were so many galaxies that 30-kelvin dust inside their interstellar clouds blocked our view in every direction. What would their redshift have to be for the dust to have a temperature like the CMB? What problems are there with this model for explaining the observed properties of the CMB? 13. (The red-shift z would be about 10.) Using Wien’s Law, 30 K corresponds to a wavelength of about 96667 nm (as λ), and the currently observed temperature of 2.726 K for the CBR corresponds to about 1063830 nm (as the red-shifted λ). The difference, λ, is 967163 nm.

Red-shift is defined by: z =  λ/λ = 967163 nm/96667 nm = 10.0005, or ~ 10.

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This scenario seems a difficult explanation for the CMB, however, because is seems to require a phenomenally large number of galaxies all with the same redshifts and dust temperatures. We would expect galaxies to have different red-shifts, or at least, slightly different temperatures.

Red-shifted dust emission from galaxies would not exhibit the phenomenal smoothness of the CMB. Why the galaxies no longer so completely fill all of space today, and why closer galaxies show such variation of temperature, composition, size, and structure, would also be questions requiring an answer from this model.

14. At the time of the last scattering epoch (at a redshift z ≈ 1000), a. what was the average density of normal matter, if today it is about 3 × 10 −31 kg/L? b. how large a sphere would have contained 1011 M⊙, about as much normal matter as in the Milky Way Galaxy? Express your answer in parsecs.

14. (a) About 3 × 10−22 kg/L; (b) The radius of the sphere is about 1750 pc.

At the time of the last scattering epoch (at a red-shift of z = 1000), (a) By table 79.1, at z = 1000 then the universe was about 0.001 times the size it is now. Thus, the diameter or radius was one thousandth what it is today. But density varies by the inverse of volume, and therefore the inverse cube of radius. Thus, if radius decreases, density increases greatly.   1/R3  1/(1/.001)3) = 109. So at that early era, the density was a billion times what it is today: Ρ = 109 × 3 × 10−31 kg/L = 3 × 10−22 kg/L. 14. (cont.) (b) The sphere’s radius would be about 1750 pc. V = M/D, so V = (1011 × 2 × 1030 kg)/(3 × 10−22 kg/L) = 6.67 × 1062 L = 6.67 × 1059 m3.

Since V = 4/3 ×  × R3, then R = (3 × 6.67 × 1059m3/(4))1/3 = 5.4 × 1019 m = 1750 pc.

15. Table 80.1 shows that material we see at a redshift of 1000 was emitted when the material was 45 million ly away from us. This distance determines the angular size of regions at this redshift, as given by the angular size formula in Unit 10.4. a. Show that a region that is 1° across at this redshift has a diameter of approximately 800,000 ly. b. How big is that region today?

15. (a) The relation from Unit 10 is L = d × A/57.3°, so the diameter is given by: L = (45 × 106 ly) × 1°/57.3° = 785,340 ly, or approximately 800,000 ly. (b) At a redshift of 1000, the universe was 0.001 = 1/1000 times the size it is today. The region would have expanded by a factor of 1000, so the region would be: 8 × 105 × 103 ly = 8 × 108 ly = 800 million light years across today.

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Unit 82 11. On a flat surface, the angles of a triangle always add up to 180°, but on a spherical surface they may add up to a larger number. a. Find a triangle on a spherical surface that contains three 90° angles. Illustrate your example. b. What is the largest sum of the angles you can find for a triangle on a spherical surface?

11. (a) One example of a triangle on a sphere with three 90-degree angles can be shown on an Earth globe by the equator and lines of constant longitude (meeting at a pole) at 0 and 90 degree longitudes. (See figure below) 11. (cont.) (b) Each angle of a triangle on a spherical surface can reach a maximum of 180 degrees, so the maximum sum would be 540 degrees. Align all three “sides” of the triangle along the equator to form a spherical triangle that covers an entire hemisphere.

12. For a surface with negative curvature, what is the smallest sum of angles you can find for a triangle? Illustrate your example.

12. For a surface with negative curvature, if the curvature is extreme enough or the triangle big enough, you can approach zero for the sum of the angles. To illustrate this, consider/draw a sequence of concentric triangles drawn on a saddle. Each larger triangle has smaller angles in the corners, eventually approaching zero degrees in each corner. The corners do not have to open very wide, because the curvature of the space spreads out the diverging sides of the triangle. (Drawing is approximate).

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13. If the average density of matter in the universe is 7 × 10 –31 kg/L, what is the radius of the volume that would form a sphere containing the equivalent mass of the Sun? 13. (930 ly). The volume of a sphere is given by V = 4 πR3/3. From the definition of mass density, ρ = M/V = M/ (4 πR3/3), we can arrive at an expression and a value for the radius, R = (3 M/4πρ)1/3 = (3 × 1 M⊙/ (4 π (7 × 10−31 kg/L)))1/3 = (3 × 2 × 1030 kg / (4 π (7 × 10−28 kg/m3)))1/3 = 8.8 × 1018 m = 930 ly.

14. If our Galaxy contains 1011 M⊙ within a radius of 100,000 parsecs, find its density.

How does this density compare to the critical density?

14. (About 1.6 × 10−27 kg/L, ~ 170 times the critical density).

Treating the galaxy as a sphere, we have ρ = M/V = 1011 × 1.99 × 1030 kg / (4/3 (100,000 pc × 3.09 × 1016 m)3) = 1.6 × 10−24 kg/m3 = 1.6 × 10−24 kg/m3 × 10−3 m3/1 L = 1.6 × 10−27 kg/L The critical density in the text is 9.3 × 10−30 kg/L; the galaxy is 173 times more dense.

15. The critical density depends on the value of Hubble’s constant, H. The value of H is still not precisely known. If H turns out to be 73 km/sec per Mpc, calculate how much the critical density changes compared to H = 70 km/sec per Mpc. 15. (The critical density would be about 9% higher, or 1.01 × 10−29 kg/L). From the text, c = 3 H2/(8 G), so c goes as the square of the Hubble constant. If the Hubble constant was 73 and not 70, c,73 = c × (73/70)2 = c × 1.09. The critical density would be 1.09 times as much – about 9% higher. The actual value would be = 9.3 × 10−30 kg/L × 1.09 = 1.01 × 10−29 kg/L. 16. With no cosmological constant, there are basically three distinct patterns of expansion possible: open and positively curved; open and flat; and closed and negatively curved. How many different kinds of universes can you identify if a cosmological constant is permitted?

Sketch what the expansion diagram looks like for each.

16. With no constant the choices are open, positive; open, flat; closed, negative. With a cosmological constant, the situation is much more complicated. If  > M, the universe will expand forever; if they are equal, it will stay the same size; if  is small enough, it won’t affect things very much. The tricky part is that  stays a constant value during the history of the universe, but M changes—as the universe gets larger, M gets smaller.

16. (cont.) The diagrams are simply representative, as the exact inflection points and curvature in the middle stages of a universe’s existence depend on  /M over time. However, even if the behavior of a universe is complicated in the middle stages, the behavior at the end is only one of three cases. Either the universe is still able to collapse (M > , always, and so positive curvature as well) or it expands forever—once  > M, eventually the expansion will accelerate as M will just keep dropping. If  = M at the maximum size of the universe, the 193

universe would stop—and remain a constant size—but as pointed out in the book this is an unstable equilibrium likely to revert to one of the other two cases.

In any universe which without dark energy would be open with positive or flat geometry, if we add dark matter, at some point  > M, and the expansion will start to accelerate. For a universe which would have been closed (positive curvature) without dark energy, the critical point is when the universe reaches maximum size. While a universe without dark energy will always collapse if it has M > 1 and positive curvature, with dark energy, if  > M at the point of maximum size, then the universe will not collapse but instead expand forever (and still have positive curvature, but it will look very flat!). If  = M at that point, the universe may be static for a period of time, but this is not an especially stable situation and it will probably eventually tip to  < M and collapse or  > M and expand forever. If  < M at the maximum expansion, we can imagine the universe has gotten larger than it might have if  = 0, but it is still able to overcome dark energy and collapse.

16. (cont.) It is also worth noting that cosmologists also consider universes where there is a negative cosmological constant—just as the positive one forces almost all the possible cases to eventually expand in an inflationary way, the negative cosmological constant causes all (or nearly all) the universes to collapse. Finally, a student might be tempted to take the three initial  = 0 situations and apply the possible ratios of /M. In that case, the results still simplify to the above, but we do get a chance to look at some of the intermediate behavior in the universes. Those cases might break down something like this (note that since M changes with time, one case can lead to another).

Letters refer to lines in this diagram:

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Let’s start with a M > 1 universe that would normally collapse and would have positive curvature. With M > 1, it still has positive curvature.

(1) If/while M >  > 0, the universe will expand a bit bigger than it would have with no cosmological constant, but it will still have positive curvature and collapse. (A) This is the only situation which still results in a collapse.

(2) If/while M =  > 0, the universe is static; if it started from a big bang, it could reach a maximum size and stop (B). This is unstable and it will probably tip to the above or below state if the size of the universe is perturbed and M changes.

(3) If/while  > M > 1, the universe will expand forever instead of collapsing, but it will still be closed with positive curvature (C); eventually the expansion will accelerate like (F) and then (G), making the universe look very flat.

16. (cont.) If M = 1, the universe would normally expand forever, but the rate would slow down, and the geometry would be flat. With  > 0, the universe will instead be positively curved, but as it expands it will look increasingly flat, and: (1) If/while M >  > 0, the universe will expand forever, at a faster rate than without  (D), until M drops such that  > M, when the expansion will accelerate as (G).

(2) If/while M =  > 0, the universe could potentially stop at the maximum size, but this would occur at the end of time (E); this state isn’t stable and may collapse to the above or below situation if the size of the universe is perturbed at all (changing M).

(3) If/while  > M > 1, the universe will expand forever at a much faster rate than without dark energy (like C or F, then G once  >> M).

If M < 1, then  could result in M +  < 1 (negative curvature), = 1 (flat), or > 1 (positive curvature). For each choice of geometry: 195

(1) If M >  > 0, the universe will expand (like D) until M drops so  > M, and then the expansion will accelerate (as G).

(2) If  > M > 1, the universe will expand forever at a much faster rate. (like F, and then G, once  >> M).

Unit 83 12. One second after the Big Bang, the density of the universe was about 0.1 kg/L. What was the radius of a spherical volume of the universe that contained as much mass as the Milky Way Galaxy?

12. (About 1.7´ 1013m ).The mass of the Galaxy is approximately 1012 M⊙= 2 × 1042 kg. If the density of the universe was 0.1 kg/L, then the volume occupied would be (2 × 1042 kg)/(0.1 kg/L), which is 2 × 1043 liters or 2 × 1040 cubic meters. The radius of this volume is:

R= 3

3V 4p

=3

3(2´ 1040 m 3) 4p

= 1.68´ 1013m

13. What wavelength of radiation is required to produce a proton–antiproton pair?

Take the mass of proton to be 1.67 × 10−27 kg.

13. (6.61 × 10−16 m). The total energy equivalent of the proton-antiproton’s mass is E = m c2 = (2 × 1.67 × 10−27 kg) × (3.0 × 108 m/s)2 = 3.01 × 10−10 J. Starting with the energy of a photon, E = h f, and using c = λ f, we arrive at: λ = h c/E = (6.62 × 10−34 J s) × (3.0 × 108 m/s) / (3.01 × 10−10 J) = 6.61 × 10−16 m = 6.61 × 10−7 nm.

14. As inflation began, the size of the currently visible universe expanded from about 10−30 m to twice that size in 10−35 sec. Calculate Hubble’s constant at that time, converting it to the same units used today. In the next 10−35 sec, the size doubled again. Show that Hubble’s constant remains the same.

14. (About 3 × 1054 km/(s Mpc); see below).

Consider the motion of the “edge” of the universe. For the sake of simplicity, assume you are in the middle and the edge is 10−30 m away (the “size” can be the radius or the diameter since the factor of ½ will drop out anyway).

The edge moves away 10−30 m = 10−33 km in 10−35 s, so it has a velocity of 10−33 km/10−35 s = 100 km/s. The distance of the edge starts out at:

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10−30 m = 10−30 m/(3.09 × 1016 m/pc) = 3.24 × 10−47 pc = 3.24 × 10−53 Mpc. So if v = H × d, then H = v/d = (100 km/s) / (3.24 × 10 −53 Mpc) = 3.09 × 1054 km/(s Mpc).

We can see how the distance drops out of this calculation: H = v/d = d/t × 1/d = 1/t.

It’s just the conversion factor and the time that are important. Putting in 2 × 10 −30 m for the distance would make no difference. If the distance is traveled in 10 −35 s we get: H = 1/t = (1/10−35 s) × (3.09 × 1013 km/pc) × (106 pc/Mpc) = 3 × 1054 km/(s Mpc) 15. The map in Figure 83.9 B is 360° wide and 180° tall. The radiation was produced at a redshift of about 1000. An angle of 1° on this map corresponds to about 250 kpc. a. Approximately how big, in degrees, are the largest high-density regions you see? How large is this in kiloparsecs? b. How big would the region be today? (Hint: Use the redshift.) c. Relate this size to components of the large-scale structure we see today (Unit 77).

15. (a) Looking at the CMB map, there are individual lumps perhaps 1 degree across and connected regions more than 10 degrees across that remain either higher or lower than the average over the whole map. The largest areas (drawn circles in the figure) are perhaps 30 degrees across but these are not continuous regions (maybe 1/6 of the height of the map).

If one degree is about 250 kpc, then these are regions of perhaps 250 kpc, 2.5 Mpc, and 7.5 Mpc at z = 1000. (b) Scaling these to today, they would be 1000 times bigger, and the corresponding regions are about 250 Mpc, 2.5 Gpc, or 7.5 Gpc (1 Gpc = billion pc = 3 billion ly) in size. (c) The individual lumps then correspond to the sizes of the largest superclusters today, which according to Unit 77 can range up to hundreds of millions of light years across. The larger regions would imply slight variations in the density of matter in widely separated portions of the universe, which would give rise to the spongy structure seen on billions of light year sizes in, e.g., Figure 77.8.

16. If the entire universe was 1 meter in diameter before inflation began, and if our visible universe was located at a random place in this volume, what is the probability we would be close enough to the “edge” of the universe to see it?

16. (Something like 3 × 10−31). We can proceed making the questionable assumption that the universe is a spherical 3-dimensional volume, and making a few very rough estimates based on Figure 83.7. Estimating very roughly from Figure 83.7, at the start of inflation the visible universe was about 10−30 m in radius. To be close enough to see the edge (then or now), at that time we would have needed to be within one radius, that is, 10 −30 m, of the edge of the 1m radius sphere that makes up the entire universe. We can estimate the probability of being located close enough to the edge by comparing the volume of the universe, and the volume of 197

a shell at the edge that’s 10−30 m thick. The easiest way to do this is probably to approximate the shell as the surface area of the sphere times the height—10−30 m. We’ll see most of the terms cancel out: Vshell/Vsphere = (4 R2 × h)/(4 R3/3) = h/(3 R) = 10−30 m / 3 m = 3 × 10−31.

16. (cont.) Likewise if we chose the best-case possible estimate from the figure, the size would be 10−20 m, and this would come out to be 3 × 10−21. Either way, an inconceivably small probability.

Unit 84 11. In Figure 84.4, the straight-line extrapolation of the current expansion rate (red line) gives the same age for the universe as the much more complex dark energy model (blue curve). This is why the simple calculation of the age of the universe in Unit 80 gave nearly the correct value. Use a sketch of these curves to estimate and discuss whether a straight-line extrapolation would have given a good estimate of the universe’s age 10 billion years ago, or 10 billion years in the future.

11. Draw lines tangent to the expansion.

If we lived 10 billion years in the past, the line intersects about −16 Gyrs, so we would think the universe was 6 billion years old. The actual age of the universe was 3.7 billion years at that time. If we lived 10 billion years in the future, the extrapolation intersects −5 Gyr. The straight line extrapolation would make us think that the universe was 15 billion years old, but in fact it would really be 23.7 billion years old at that time.

12. The following calculations can help us begin to understand why the Local Group is predicted to survive the effects of dark energy, but more distant galaxies are expected to be pulled away: a. Using the values of mass and size in the text, what is ΩM for the Local Group? Note that in the text, this is calculated out to the distance of neighboring groups. b. Instead, calculate the density inside of the central 1-Mpc radius region of the Local Group (where nearly all of the group’s mass resides). How does this density compare to critical density?

12. (a) M ~ 0.075 for the local group. (b) 1.2 × 10−28 kg/L. This is 14 times the critical density. (a) From the text, Unit 82 section 3, the mass of the local group is M = 2 × 1042 kg and the density = 7 × 10–31 kg/L = 7 × 10–28 kg/m3. The critical density c = 3H2/(8πG), which for H = 70 km/(s Mpc) = 9.3 × 10–30 kg/L (Unit 83.3).

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M = /c = (7 × 10–31 kg/L)/(9.3 × 10–30 kg/L) = 0.075. (b) The density inside the central 1 Mpc radius region of the Local Group can be calculated from the mass/volume. The radius is 0.5 × 106 pc × 3.09 × 1016 m/pc = 1.545 × 1022 m.

The volume is (4π/3 × R3) = 4π/3 × (1.545 × 1022 m)3 = 1.54 × 1067 m3 = 1.54 × 1070 L. And the mass is 2 × 1042 kg, if most of it is in this region. With these values:  = (2 × 1042 kg)/(1.54 × 1070 L) = 1.3 × 10−28 kg/L. and M = /c = (1.3 × 10−28 kg/L)/(9.3 × 10–30 kg/L) = 14 (If you treat 1 Mpc as the radius instead, you get M ~ 14/8 = 1.75).

13. Dark energy retains a constant density even as the universe expands. Today it has a value of ΩΛ = 0.7. Matter, by contrast, grows more dilute as space expands, so its density would have been larger in the past by a factor that depends on the redshift, as (z + 1)3.

At what redshift would dark energy and matter (currently ΩM = 0.3) have had the same density?

13. (At a redshift z = 0.3264). For M to have a density of 0.7, it would need to be 0.7/0.3 = 7/3 = 2.33 times the current value. The density goes as 1/volume cubed, and length in the universe goes as (z + 1), so the density in the past would be great by a factor of (z + 1)3 as noted in the problem. For 0.3 × (z + 1)3 = 0.7, z = (0.7/.3)1/3 – 1 = 0.3264. At z = 0.3264, M = DE. 14. The Super-Kamiokande experiment searches for proton decays by monitoring 20 million kilograms of water for decay events. Given that a single proton has a mass of 1.67 × 10−27 kg, how many decays would you expect on average in 1 year of observations if it takes 1040 years for a single proton to decay?

14. To arrive at a rough estimate for the number of protons in the water we can make the assumption that half the mass comes from protons with the other half due to the neutrons (and ignoring the mass of the electron). The number of protons is then: Np = 106 kg/(1.67 × 10−27 kg/proton) = 6 × 1032 protons.

With only 10−40 predicted decays per year, there are far too few protons in the water to expect to see a decay event in one year. In reality the Super-Kamiokande experiment places upper bounds on the number of expected event rates per year.

15. If the universe now begins doubling in size every 10 billion years, regions that are currently moving away from us at half the speed of light will start receding at c in 10 billion years. Using Hubble’s law, how far away from us are such regions currently? About how many doublings and how many years will it take a region that is currently receding at 3000 km/sec to be receding at the speed of light?

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15. (2143 Mpc now; between 6 and 7 doublings, over 66.4 billion years). Using Hubble’s Law, a region currently receding at ½ the speed of light would be a distance: d = v/Ho = (0.5 × 3 × 105 km/s)/(70 km/s Mpc) = 2143 Mpc away now.

Each doubling, the speed of recession doubles. A region currently receding at 3000 km/s is currently at 1/100 the speed of light. Solving 2n = 100 for n is the number of doublings needed.

After 6 doublings, the speed will have increased 64 times; after 7 doublings the speed will have increased 128 times, so it will take in between 6 and 7. We can find the exact number: n log 2 = log 100 → n = log (100) / log(2) = 6.64. This is a small enough number one can also look at it by brute force: 3000 → 6000 → 12000 → 24000 → 48000 → 96000 → 192000 → 384000.

Unit 85 11. For what fraction of Earth’s lifetime has Homo sapiens existed?

11. (About 0.01%). According to the unit, homo sapiens evolved about half a million years ago, so compared to the age of the Earth: 100% × 500,000 y/(4.5 × 109 y) = 100% × 5/4.5 × 105/109 = 100% × 1.1 × 10−4 = 1.1 × 10−2 % = 0.011%.

12. A bacterium has a mass of about 10−15 kg. It can replicate itself in 1 hour. If it had an unlimited food supply, how long would it take the bacterium, doubling in number every hour, to match the mass of Earth? (Hint: Solve for the number of hours N, where 2 N gives the number of bacteria you estimate.)

12. (The mass of bacteria would equal the mass of the Earth in just over 132 hours).

The mass of the Earth is 5.97 × 1024 kg. Divide by the mass of the bacterium, 10–15 kg, and we can see we will need: 5.97 × 1024 kg/(10–15 kg/bacterium) = 5.97 × 1039 bacteria.

Approximate 5.97 as 6. Let P represent the size of our bacteria population, and P = 2N, where N is the number of hours. Therefore we need 6 × 1039 = 2N. One can work this out by guessing N’s, by calculating the sequence of doublings, or by using logs. log (P) = N log (2) so N = log(P)/log(2) = log(6 × 1039)/log(2) = [log(6) + log(1039)]/log(2) = [log(6) + 39]/log(2) = [39.778]/0.301 = 132.15 Note that 2132 = 5.4 × 1039 which is just a bit under the necessary number of hours, and 2133 = 1.1 × 1040, which is over the necessary number.

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13. If a DNA chain is ionized, it can lead to mutations in the cell’s structure, which can ultimately lead to cancers in humans. The ionization energy for DNA is roughly 8.0 × 10–19 J. What wavelength of light does this correspond to? Where in the electromagnetic spectrum does this radiation fall?

13. (250 nm; UV). The energy of a photon is given by E = h × f = h × c/λ, which gives the wavelength: λ = h × c/E = (6.6 × 10−34 J ×s) × (3.0 × 108 m/s) / (8.0 × 10−19 J) = 2.5 × 10−7 m = 250 nm.

From Unit 22 we find that this radiation is in the ultraviolet region of the electromagnetic spectrum. This makes sense since skin cancer is caused by UV light.

14. From 1950 to 2000, the world population increased from 2.5 billion to 6 billion. If this trend continues, what will be the world population in 2050? In 2100?

14. (About 14 and 35 billion). The answer to the question depends on whether we consider the growth to be exponential or linear, and exponential is a better estimate of population growth.

Over fifty years, the population increased 6 B/2.5 B = 2.4 times. Therefore, we might expect in the next 50 years from 2000 to 2050 for the population to grow by the same factor, to 6 B × 2.4 = 14.4 billion, and in the succeeding 50 years to 14.4 B × 2.4 = 35 billion. (If the growth was linear, which is not a good approximation, we would add 3.5 billion every 50 years, for 9.5 billion and 12 billion).

Unit 86 13. How long would it take to get to the nearest star at the space shuttle’s speed of 30,000 kph?

13. (About 150,000 y). It’s 4.2 ly to the nearest star, so: t = d/v = 4.2 × 9.5 × 1012 km/ly / (3 × 104 km/h) = 1.33 × 109 h.

1.33 × 109 h / (365.25 × 24 h/y) = 1.5 × 105 y.

14. Use a method similar to the Drake equation to estimate the number of people getting a haircut at some given moment. That is, identify the different factors involved in deciding to get a haircut, estimate the probabilities of each, then multiply the probabilities times the number of people in, for example, the United States (approximately 260 million).

14. The Haircut Equation can be represented as: HC = Np × Fn × Fd × Fw where  

Np is the total population of a given area in question; Fn is the fraction of people in that population who need a haircut at any given time; 201

 

Fd is the fractional delay time in “inverse” days representing the length of time on average a person delays getting a haircut; and, Fw represents the fraction of the time that weather factors present an excuse for not getting a haircut.

(Other complicating factors could be considered, but let’s stop here.). Given 260 million people in the United States as Np, estimate Fn as ¼; Fd as 1/7 (that is, imagine that the average person delays getting a haircut for 7 days; and estimate Fw as 1/21, meaning that bad weather might prevent someone from getting a haircut every 3 weeks or so. Plugging in the numbers we get: HC = 300 million × ¼ × 1/7 × 1/21 = 0.51 million or about 510,000 haircuts every day in the United States, or during a 12-hour work day (coast to coast) about 708 any particular minute. (Remember, this is just a wild guess!) Note: some latitude should be given in what factors are considered. A result within a factor of 2 of around a million haircuts a day is pretty reasonable.

15. Appendix Table 10 lists the nearest stars to Earth. What fraction of these stars have a spectral type that may be suitable for life based on the criterion stated in the text? If we use this fraction instead of the 20% used in the example application of the Drake equation, how would the estimated number of intelligent civilizations change?

15. (29%; the estimated number would increase a bit). According to the text, life is more probable around F-, G-, and K-type stars. Of the 28 stars listed in Appendix Table 11, 8 are of these spectral types. This makes the fraction of suitable stars 8/28 = 2/7 = 29%. This fraction is larger than the 20% stated in the text, so the estimated number of intelligent civilizations would slightly increase.

16. Make your own estimate of the number of habitable planets in the Milky Way. What is the average distance between them?

16. Students should make their own guesses for the number of civilizations using the Drake equation, but the estimate can be used to find the distance as in Section 86.4. The formula determining the average areas in terms of the size of the Milky Way disk is 2

ædö 2 N I ´ p çç ÷ = p (RM W ) ÷ ÷ èç 2ø

which when solved will produce the distance between civilizations, d=

2RM W NI

.

So using the estimation of NI as 10000, we get 800 ly, as in the text.

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17. Voyager 1 is currently leaving the Solar System at a speed of 17 km/sec. Assuming a constant speed, how long would it take Voyager 1 to travel 4.24 ly, the distance to the nearest star?

17. (74,500 years) Speed is found from v = d/t, so the travel time would be: t = d/v = (4.22 ly × 9.46 × 1012 km/ly) / (17 km/s) = 2.35 × 1012 s = 74,500 years.

18. Suppose that we are the only civilization in the Galaxy, and all the other civilizations have died out. For a possible 10,000 civilizations calculated by the Drake equation and a Galaxy with an age of 10 billion years, what is the typical life span of a civilization? How old is our civilization?

18. (About a million years). The galaxy is 10 billion years old = 1010 y. Here we are basically assuming the civilizations live one after the over, so dividing the age of the galaxy by the number of civilizations we get 1010/104 = 106 years. A million years as the average age of a civilization. Our civilization is ~ 5000 years old.

19. One solution suggested for our growing population in the far future is to construct a sphere surrounding the Sun with a radius of 1 AU (Earth’s orbital distance). The amount of light received per square meter on the inside of this sphere would be similar to what we receive on Earth now. What is the surface area of such a sphere compared to the surface of Earth?

19. (It would be about 550 million times the size). The surface area of a sphere is 4 R2.

Comparing the Earth’s surface area, and a (Dyson) sphere at the Earth’s orbital distance: 4 Rorbit2/4 REarth2 = (Rorbit/REarth)2 = ((1.5 × 108 km)/(6378 km))2 = (2.35 × 104)2 = 5.5 × 108 The sphere would have a surface area 550 million times that of the Earth.

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