Explorations: Introduction to Astronomy 9th Edition by Thomas Arny & Stephen Schneider. ISBN 9781260

CHAPTER 1 T H E CYCLES OF THE SKY Lecture Suggestions Planetarium software may be helpful for this chapter. An orrery (mechanical model of the Sun and Earth) or even just an approximation of one can also help illustrate various motions and that the constellations change with the seasons. Set it on a table in the front of the room and then mark off constellations on the walls with chalk or paper decorations. Moving the model Earth around the model Sun then allows students to see how the stars visibly change and how the Sun “moves” through the Zodiac. In effect, this turns the room into a simple planetarium.

A flashlight with a fat beam shows the importance of angle to seasonal heating.

When directed directly at the wall the energy is concentrated in a small area. When shined obliquely at the wall, the beam covers a larger area, implying less concentration of heat and therefore a lower temperature. It’s also important to include the idea that the day is longer in the summer, which is sometimes overlooked.

A bright light source and tennis balls, basketballs, volleyballs or even golf balls can demonstrate features of eclipses and phases of the moon.

Answers to Thought Questions 1. If you were standing on Earth’s equator, looking due north you would see the north celestial pole on the horizon (and the south celestial pole on the horizon, looking due south).You cannot see the north celestial pole from Australia (it’s below the horizon), only the south celestial pole.

2. SKETCH FOR STUDENTS 3. The main astronomical reason why there are 12 zodiacal signs is that the Sun appears to move about 30 degrees per month across the background stars (360 / 30 = 12).

4. SKETCH FOR STUDENTS. At the horizon, setting or rising stars move perpendicularly to the horizon (so they are useful for East-West navigation). At the north pole, stars more or less do not set, they just circle in the sky. At a mid-latitude, the stars make an angle with the horizon. In the Northern Hemisphere, as they set they also move more toward the north. Extremely schematically: setting stars looking west: at the equator, | | |; at a mid-latitude, Northern Hemisphere: \ \ \; at the North Pole, ---.

5. When it is winter in New York, the Northern Hemisphere is tilted away from the Sun; therefore at that time the Southern Hemisphere is tilted towards the Sun and it’s summer in Australia. Although Paris is partway around the world from New York, it’s at about the same latitude and it is also winter there. The part of the sky that you see at night is the part of the sky away from the Sun, so everywhere on Earth sees the same “half” of the sky at night during a 24 hour period, as Earth rotates viewers into nighttime: all three locations should be able to see Orion, which straddles the celestial equator.

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6. If Earth’s orbit had no tilt, there would be no variation in the angle of sunlight over the year, nor would there be variation in the length of day—conditions would be somewhat like the equinox all the time and a 12 hour day everywhere, every day. There would be a slight variation in temperature over the year with higher temperatures in January based on the small change in Earth’s distance from the Sun, but this effect would be very small (clearly it does not affect the seasons induced by the tilt very much). Northern and Southern hemispheres would experience these weak seasons at the same time.

7. The position of sunrise along the eastern horizon changes during the year because Earth’s axis (and correspondingly, the celestial equator) is tilted at 23.5 degrees to the plane of its orbit (the ecliptic) and Earth maintains this same tilt throughout the year. At the equinoxes (March 21 and Sept. 23), the Sun lies on the celestial equator. Because the celestial equator cuts the horizon at the east and west points, the Sun will rise and set due east and due west, respectively. In winter, the tilt of Earth results in the Sun rising north of east and setting north of west, and in winter, the Sun rising south of east and setting south of west. At the winter solstice (Dec. 21), the Sun lies 23.5 degrees south of the celestial equator on the sky. It will therefore rise the most to the south of the East on the horizon and set the most to the south of West that year. At the summer solstice (June 21), the Sun lies 23.5 degrees north of the celestial equator. It will therefore rise the most north of East and set the most North of West.

8. We have time zones to keep our local time in approximate alignment with solar time, and to standardize time between different parts of countries and the world—using exact local solar time everywhere would be just as confusing as using one set of hours for the whole world. The sketch can show how it’s solar noon on one part of earth (the Sun is highest in the sky) and a very different time elsewhere (the Sun would be high or low in the sky), or just show a close-up of the difference between the local time in one time zone (say noon) and an adjacent time zone (say 1 pm).

9. Some possible ideas: (One or two of these or related ideas should be sufficient).

-You can see some phases during the day, so the geometry is incorrect for the phase to be a shadow.

-You can determine the Sun-Earth-Moon angle is not 180 degrees during most phases.

-Lunar eclipses (shadow on the Moon) do occur, and only during the 180 degree/full moon alignment.

-The curvature of the terminator during Moon phases is not consistent from phase to phase—if it was Earth’s shadow, it would always be the same shape. The changing terminator shape is consistent with a partially illuminated sphere.

-The radius of curvature of the terminator for phases does not match the radius of the shadow of Earth during an eclipse.

-Eclipses happen over a period of hours, while the phases change slowly over weeks, suggesting they are not caused by the same thing.

10. In this case, the sidereal month would remain 27.3 days as the periodic alignment with the stars would not change, but the solar month would be shorter because the Moon will reach new moon before re-aligning with the stars instead of after. The redrawn figure

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of 1.14 should indicate that this is the case (the right part of the diagram will happen in opposite order).

Answers to Problems 1. 360 degrees / 24 hours = 15 degrees/hr. A simple problem but a good number for students to know.

2. The equinox (Sept. 22) latitude of the Sun will be 90° - 55° = 35°. The highest angle = 35° + 23.5° = 58.5°, lowest = 35° - 23.5° = 11.5°.

3. MAKE SKETCH FOR STUDENTS. Use the phases diagram and label the time on different parts of Earth; then draw horizon lines tangent to Earth to the waxing crescent moon to determine the time it is 1st visible rising the east is at about 9 AM and it sets in the west at about 9 PM. 4. The Moon’s sidereal period is 27.3 days. Therefore, it moves…

360° / 27.3 days = 13.2°/day.

Earth must rotate an extra 13.2° to “catch up” to the Moon. Earth rotates about 15°/hr (Problem 1), so the Moon rises about 13.2°/(15°/hr) = 0.88 hr ≈ 53 minutes later each day.

5. Moon’s draconic orbital period is 27.2122 days = P. Moon’s synodic period is 29.5306 days = S. 242  P = 6,585.3524 days 223  S = 6,585.3238 days 242  P = 6585.3524 days /(365.24 days/yr)= 18.03 years.

The result gives a pattern of solar and lunar eclipses which repeat over about 18 years, but shift in location across Earth because there is not an even number of solar days in a year.

6. Circumference of moon’s orbit = C = 2  384,400 km = 2.42  106 km V = C / Period = 2.42  106 km / (27.322 d  24 h/d) = 3680 km/h Earth’s diameter = 6400 km  2 = 12,800 km Time = D / V = 12,800 km / (3680 km/h) = 3.5 hours (approx) Since Earth only moves a small distance around its orbit in a few hours of time, the time for the Moon to travel through Earth’s shadow is the dominant effect in determining the duration of the lunar eclipse. (See also next question).

7. We would need to decide exactly when to start and when to stop counting—the above estimate considers the leading edge of the moon going from one side of Earth to the other, but we should probably add the Moon’s diameter to the distance traveled so we

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move all the way out of Earth’s shadow. This would increase the time (by about diameter of Moon / diameter of Earth ~ 30%). However, Earth’s shadow is not exactly the size of Earth’s diameter—it is a bit smaller, and that would decrease the time (a noticeable correction). As the whole system moves around the Sun, there is also a slight shift in the shadows; since Earth and Moon are moving in the same direction this will lengthen the shadow passage very, very slightly.

8. The approach to the estimation is still fine: the Moon is still moving fast enough that any effect on the shadows from the Earth-Moon system moving around the Sun is negligible for our purposes. We do need to treat the radius as a ~ 200 km shadow passing overhead; 200 km / (3680 km/hr) = 0.05 h = 3.26 minutes.

Rmoon

Rsun d Sun

Moon

Answers to Test Yourself 1. (d) The north celestial pole is directly overhead.

2. (b) Altitude = latitude = 30°.

3. (a), (c), (d) Sun, Moon, stars would rise in West, rotate around Polaris clockwise.

4. (d) longer day and more concentrated Sun 5. (d) Above the arctic circle, or below the Antarctic circle, the daylight varies from 0 to 24 hours over the year.

6. (b) Waning gibbous. Must be same moon phase seen all over Earth.

7. (c) Full moon rises at 6pm.

8. (b) between first quarter (rises at noon) and full (rises at 6pm) 9. (c) 2 weeks 10. (c) Moon covering sun 11. (b) bigger moon, more eclipses, but not every month.

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CHAPTER 2 THE RISE OF ASTRONOMY Answers to Thought Questions 1. Despite the Moon illusion, you are actually closest to the Moon when it is at its highest point in the sky. As seen from over the North Pole,

2. If the stars were much closer than they really are, Aristarchus would have been able to demonstrate the stellar parallax caused by Earth’s orbital motion around the Sun. 3. The phases of Venus are caused by Venus orbiting the Sun, so keeping all distances and periods constant, it wouldn’t matter whether Earth and Venus were orbiting the Sun or Venus was orbiting the Sun and the Sun (and Venus) were orbiting Earth.

4. The Sun has a slightly larger angular diameter in January than in July because Earth’s orbit is an ellipse. The idea that orbits are ellipses is Kepler’s First Law. 5. The apparent motion of Jupiter is primarily a result of Earth’s motion around the Sun, not Jupiter’s motion. A sketch like the one below can help show this—consider the view from Earth over the year, with Jupiter moving only a little along its orbit. At position A, Jupiter is at opposition, and rises when the Sun sets. By the time Earth is at B, Jupiter has not moved very far (so its motion is neglected in this sketch), but now it is only 90 degrees away from the Sun—already high in the sky at Sunset. By position C, Jupiter would be “up” in the daytime. Moving to position D, Jupiter is moving past the Sun on the opposite side as before.

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6. If the same comet is only visible every 50 years or more (or much, much more!) then the comet must have a much longer p than Earth’s orbit. Consequently by Kepler’s third law it must have a corresponding larger a. If the comet needs to be close to Earth and Sun to be seen, then at least some of the time it must be at only 1 or 2 AU from the Sun; for this to fit with a large a and p, it must have an elliptical orbit with high eccentricity.

7. This should be less about technology and more about philosophy; and about getting students to look up contemporary work. One key difference is less personal, government and patron-sponsored religious/mystic motivation—astronomy is no longer in the game of predicting planetary positions as portents of war and peace. Also, astronomy has become increasingly based on fact and less influenced by philosophy (Ptolemy’s and Kepler’s motivations looking for perfection in the heavens; Kepler was conflicted but notably broke with his philosophical position (circular orbits) when confronted with scientific evidence). Increased technology has revealed entire branches of astronomy previously invisible to scientists (radio, x-ray, infrared, gravitational waves, etc.).

8. (Students must research the astronomers in question).

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Answers to Problems 1. This problem is a modern version of the method Eratosthenes used to measure the size of Earth. Given that the shadow length is 15 degrees, the distance in latitude between the two points on the asteroid must be 15 degrees, or 15°/360° = 1/24th the circumference of the asteroid. If the 15 degrees corresponds to 10 km, then the total distance around the asteroid must be 10 km × 24 = 240 km. The radius, R, of the asteroid is related to its circumference, C, by C = 2R. Thus R = C/2 = 240/2 = 38 km 2. This can be directly calculated, but it is easier to just look at the proportionalities involved. Since C = 360°/angle × distance, the circumference is directly proportional to the distance between Alexandria and Syene. If the distance were three times as much, and the angle the same, Eratosthenes would have calculated that the circumference of Earth was three times larger: 250,000 stadia × 3 = 750,000 stadia or 25,000 miles × 3 = 75,000 miles. (“Three times larger” should be an acceptable answer.)

3. From Earth, the Sun has an angular diameter of 0.5°. Angular size varies as the inverse of the distance, so if Mercury is 0.387 times as far, the Sun is 0.5°/0.387 = 1.29°. Pluto is 39.53 times Earth’s distance, so the Sun is 0.5°/39.53 = 0.0126°.

4. The Andromeda galaxy has an angular diameter of 5 degrees at a distance of 6 2.2 × 10 ly. We can find its true size by using the angular diameter formula L = 2dA/360°, where d = distance, A = angle subtended, and L = linear diameter. Thus, 6 5 L = 2 × 2.2 × 10 ly × 5°/360° = 1.92 × 10 ly.

5. If P = 64 years, you can use Kepler’s 3rd Law to estimate its distance from the Sun. 2 3 P = a , where P = period in years and a = average orbital radius in AU. We can find a by 2/3 2/3 2/3 1/3 taking the cube root of both sides to get P = a, or a = P = (64) = (64 × 64) 1/3

= (8 × 8 × 8 × 8) circular.

= 2 × 2 × 2 × 2 = 16 AU. This is the radius of the orbit if the orbit is

6. a3 = P2. (526)3 = 1.4553 × 108 = P2, so P = 12,063 years.

7. This is an application of Kepler’s third law, P2 = a3, where a is in AU and P is in years.

If P = 125 yrs, then a3 = 1252. Solving for a, we take the cube root of both sides to get a = (1252)1/3, where we have used the fact that the cube root of a number is the number to the 1/3 power. This can be solved with a calculator or by noticing that

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(125 x 125)1/3 = (25 x 5 x 5 x 25)1/3 = (25 x 25 x 25)1/3 = 25, so a = 25 AU. If the planet’s orbit is circular, then that is also the planet’s orbital radius.

8. This problem is another application of Kepler’s third law, P2 = a3, where a is in AU and P is in years. In this case, we are given a, and are asked to find P. Thus P2 = a3 = 163.

Solving for P by taking the square root and recalling that the square root is the number to the 1/2 power, we find that P = (163)1/2 = 64 yrs. (Note: in solving this problem, you can simplify the math by reversing the order of the power and the square root. That is, (163)1/2 = (161/2)3 = 43 = 64.)

Answers to Test Yourself 1. (d) Angular size is inversely proportional to distance so LM/LS= 1/400.

2. (b) Retrograde motion causes planets to stop their regular eastward motion with respect to the stars and move westward for a time.

3. (b) simplicity of models 4. (d) P2 = a3. 43 = 4 × 4 × 4 = 4 × 2 × 2 × 4 = 82 5. (a) Kepler’s 3rd Law relates a planet’s orbital period to the size of its orbit.

6. (e) Venus orbits the Sun. 7. (c) Parallax was conjectured but impossible to measure without high quality telescopes.

CHAPTER 3 GRAVITY AND MOTION Answers to Thought Questions 1. You would not want to kick a cinder block in space with your bare foot (ignoring the problems of having your foot bare in space) because it still has the same mass and inertia as on Earth. Kicking the block would hurt your foot as badly as if you’d kicked it on Earth’s surface.

2. Students should draw a sketch something like the ones below (either style or both).

It may feel as if the “imaginary” centrifugal force tries to push you out, but really you are trying to travel out at a tangent and the wall is stopping you. Keep in mind that the direction of acceleration in a circular orbit is always towards the center, and the net force must be in the same direction as net acceleration. The wall is constantly pushing you in to keep you going in a circle.

3. Yes, there is a force of gravity between the ISS and Earth; in fact it is the force keeping it in orbit, so yes, the ISS is affected by it. The astronauts also experience a force of gravity and are accelerated in orbit around Earth. They and the space station are accelerated together, and move together in orbit—all are continuously falling towards Earth together and so the astronauts “feel” weightless. There is no (appreciable) net force between the astronauts and the space station.

4. F = ma, so larger masses require larger forces for the same accelerations. The force to accelerate a car or to keep it moving at constant speed against friction comes ultimately from the power produced by burning gasoline in the engine. All other things being equal, you will need to use more energy in the engine to apply a larger force to move a larger vehicle the same distance as a smaller one.

5. Newton’s 3rd law will still apply. The Moon’s gravitational pull on Earth has exactly the same magnitude (but in the opposite direction of) Earth’s gravitational pull on the Moon.

6. When you walk, you accelerate forwards, so the force on you must be in the forward direction. You can’t push yourself forwards, so the ground must push you forwards. The ground pushing on you is a reaction to your foot pushing on the ground, so you must push backwards against the ground. So you push on the ground and the ground pushes on you, and the ground pushing on you makes you go forwards. Since the ground moves very little when you push on it, you can push hard, and it can push you back hard, and by F = ma you have a big force to move forwards. If you are walking on sand, it moves easily when you push back on it with your foot, and it’s much harder to make forward progress because you are more easily able to move the ground (so you can’t push so hard against it). Likewise, it moves you with less force.

7. Yes, because there is a force of gravity between any pair of objects with mass. The force decreases with the square of the distance between the objects. The book falls to the ground even though it is much further from the center of Earth than the center of you because Earth is much, much, much more massive than you are.

8. F = ma means F must have units of [kg] [m] / [s]2. Therefore F = GMm/d2 must also have those units, so [kg] [m] / [s]2 = [units of G] [kg][kg]/[m]2 simplifying, [units of G] = [kg]-1 [m]3 [s]-2 ; the units of G must be m3/(kg s2).

(Alternatively, [Newtons] = [units of G][kg]2[m]-2 and [units of G] = [N][m]2[kg]-2, but the intent is to remove the Newton unit, hence including F = ma in the problem setup).

9. The Sun’s gravitational force on Earth equals Earth’s gravitational force on the Sun, as there is only one value for the force of gravity between two objects, although the directions are opposite. F = GMm/d2 = GmM/d2. Earth pulls the Sun towards Earth just as strongly as the Sun pulls Earth towards the Sun. This is also an expression of Newton’s 3rd law of action and reaction. (How much each object actually accelerates depends on the relative masses, of course).

10. The two stars pull on each other with equal force, so the lower mass star will have a greater acceleration. Therefore, it will move more, and have the wider orbit (with a larger circumference). (See also the reasoning in TQ9 above.)

11. Surface gravity is given by g = GM/R2, so it depends on the ratio of the mass to the radius of the planet squared. A planet with a large radius, but made of low density materials such as ice, or light gases such as hydrogen, could have a lower value of g than a planet made of very dense material (iron) that was quite small.

Answers to Problems

1. F = ma so a = F/m. (a) 2F on m: a’ = F’/m’ = 2F/m = 2 (F/m) = 2a. (b) 2F on 2m: a’ = F’/m’ = 2F/(2m) = (2/2) (F/m) = (1)(a) = a. (c) 10F to m: a’ = F’/m’ = 10F/m = 10(F/m) = 10a. (d) 10F to 3m: a’ = F’/m’ = 10F/3m = (10/3)(F/m) = 10/3 a = 3.33 a 2. This is just F = ma again, but we have to figure out a. a = change in velocity / time = 2 m/s / 25 s = 0.08 m/s2.

F = ma = (2500 kg)(0.08 m/s2) = 200 Newtons 3. STUDENTS SHOULD DO THIS FOR THEMSELVES. The weight of a body is just the gravitational force exerted on it. You can therefore find your weight on Earth or on the Moon from their surface gravities.

2 2 • Earth’s surface gravity is 9.8 m/s while the Moon’s is 1.7 m/s .

• The Moon’s is thus 9.8/1.7 = about 5.8 times smaller than Earth’s.

• Thus your weight on the Moon is your weight on Earth divided by 5.8.

Alternatively, students can calculate the surface gravity of the Moon directly using g = GM/R2 and then their weight with F = mg; but they will have to come up with their mass in kilograms to calculate the force in Newtons and/or convert that back to pounds or compare to their Earth weight in Newtons (depending on instructor preference for the format of the answer).

4. From the problem, Neptune’s distance from the Sun is about 30 AU (more exactly it’s 30.05 AU). The orbital velocity, V, of a small mass around a much larger one can be 1/2 found from the formula in the chapter, namely, V = (GM/d) , where M is the Sun’s mass and d is Jupiter’s distance from the Sun. Substituting: -11

30

11

1/2

3

VJupiter = ( (6.7 × 10 m3 kg-1s-2 × 2 × 10 kg)/ (30 × 1.5 × 10 m) ) = 5.44 × 10 m/s.

The orbital period is the time it takes a body to complete an orbit.

Thus, orbital velocity = circumference/orbital period, V = C/P. Evaluating this we obtain 11 3 9 P = 2d/V = 2 × 30 × 1.5 × 10 m/ (5.44 × 10 m/s)= 5.2 × 10 seconds.

7

Since there are approximately 3.16 × 10 seconds in 1 year, P = about 165 years. Students should be encouraged to check a result like this against the data in the appendix or Kepler’s 3rd law.

11

20

5. The Milky Way Galaxy has a mass of about 10 MSun, and the Sun is 2.6 × 10 m from the center of the galaxy.

1/2 -11 30 11 20 V = (GM/R) = (6.7 × 10 m3 kg-1s-2 × 2 × 10 × 10 kg /2.6 × 10 m)1/2 5

V = 2.3 × 10 km/s 20 5 15 P = 2R/v = 2  × 2.6 × 10 /2.3 × 10 = 7.1 × 10 s.

One year is about 3.16 × 107 sec, so the period of the Sun’s orbit around the Milky Way 15 8 in years is 7.1 × 10 s /(3.16 × 107 s/yr) = 2.25 × 10 years = 225 million years.

6. The modified form of Kepler’s third law states that M = 4d3/GP2, where d is the orbital radius (assuming it is circular) and P is the orbital period. In this problem d = 5 × 1010 meters and P = 124 days. Before we can solve the problem, however, we need to convert P in days to P in seconds.

7 124 d × 24 hr/1d × 60 min/1hr × 60 sec/1min = 1.07 × 10 sec.

Inserting the values for a and P, gives M = 4 (5 × 1010 m)3/[6.67 × 10-11 m3 kg-1 s-2 (1.07 × 107 s )2] = 4 × 125 × 1030-(-11)/[6.67 × (1.072 × 107x2 )] kg = 4935/7.6 × 1041-14 kg = 6.5 × 1029 kg Since the Sun’s mass is 2 × 1030 kg, the mass of Gliese 581e is about 1/3 of a solar mass: 0.65 × 1030kg / (2 × 1030 kg/solar mass) = 0.33 solar masses.

7. In Section 3.7, Earth’s surface gravity is compared with the Moon’s. We will need the mass and radius of Jupiter and Pluto, which conveniently are given in terms of Earth’s radius and mass in the Appendix.

M Earth GM Earth 1 2 2 M Jupiter g Earth REarth 317.9 2 = 11.19 = 125 = 0.4 2 = = = g Jupiter GM Jupiter  R  317.9 317.9  1  11.19  Earth  2 R   Jupiter    R Jupiter  The surface gravity of Earth is about 4 tenths of the surface gravity on Jupiter; or Jupiter’s surface gravity is about 2.5 times Earth’s.

For Pluto,

M 1 g M Earth 0.1882 g Earth ==  Pluto2 =  0.0021 1 2 = 0.0021 = 16.8 R Jupiter    Earth   0.188 R  Pluto  The surface gravity of Earth is about 16.8 times as strong as the surface gravity of Pluto, which is about 1/16.8 = 0.059 that of Earth.

8. The escape velocity for Earth is found from:

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Vesc = (2GM/R)

Inserting the given values in the equation, we find that 1/2 -11 24 6 1/2 4 Vesc = (2GM/R) = (2 × 7 × 10 m3 kg-1 s-2 × 6 × 10 kg / (6 × 10 m)) = 1.18 × 10 m/s or Vesc = approx. 12 km/s 9. The conversion to mph helps put this speed into context for students.

11.8 km/s × (1 mile/1.609 km) × (3600 s/ 1h) = 26,400 miles/h = 26,400 mph 1/2

10. The escape velocity for the Sun may be found using the formula Vesc = (2GM/R) with the Sun’s mass and radius inserted. This gives -11 30 8 1/2 Vesc = (2 × 6.7 × 10 m3 kg-1 s-2 × 1.989 × 10 kg/(6.95 × 10 m)) = 6.2 × 105 m/s = approx. 620 km/s Or with the approximate values, -11 30 8 1/2 5 Vesc = (2 × 7 × 10 × 2 × 10 /(7 × 10 )) = 6.3 × 10 m/s = approx. 630 km/s 11. To compare the escape velocity of Mars and Saturn, the method is similar to the previous problem. The ratio simplifies to the expression, VSaturn    =  M Saturn  RMars  VMars  M Mars  RSaturn  We then evaluate each expression using the appropriate mass and radius. From the

appendix, MMars = 0.1 MEarth, MSaturn = 95 MEarth, RMars = 0.5 REarth and RSaturn = 9.5 REarth.

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Vesc(Saturn)/Vesc(Mars) = ( (95/0.1) x (0.5/9.5))

= ( (95/9.5) x (0.5/0.1) )1/2 = (10 x 5)1/2

= 7.1.

This is a good example of a question where the numbers can be easily estimated by writing out the values and shifting terms around.

12. To calculate the ratio of the escape velocities from the Moon and Earth, start with the formula for escape velocity: 1/2 1/2 Vesc-Moon = [2GMMoon/RMoon] and V esc-Earth = [2GMEarth/REarth] .

Next, divide the Moon’s escape velocity by Earth’s,  2GM Moon  2  R  V Moon  = Moon = 1 VEarth  2GM 2 Earth    REarth  1

M Moon M Earth = RMoon REarth

 M Moon  REarth       M Earth  RMoon 

We then evaluate each expression using the appropriate mass and radius; from the chapter, 81 MMoon = MEarth and REarth/RMoon = 3.8, so VMoon  1  =   (3.8) = 0.22 VEarth  81  Vesc-Moon/Vesc-Earth = 0.22.

In other words, the escape velocity of the Moon is about 1/5 of the escape velocity of Earth.

13. To find out if the pitcher can throw a ball fast enough to escape from Sinope, we need to find the escape velocity of Sinope. To do that, we use the escape velocity formula, V = (2GM/R)1/2.

Putting in the values for the mass and radius of Sinope, we find that V = [2 × 6 × 1016 kg × 6.7 × 10-11 m3kg-1s-2/(1.8 × 104 m)]1/2 = [2 × 6 × 6.7 × 1016-11-4/1.8 (m/s)2]1/2 = [44.67 × 10]1/2 m/s = 21 m/s which is smaller than the speed of the pitch. Thus the ball can escape.

Answers to Test Yourself 1. (a),(b),(c),(d) All show a mass tending to remain at rest or in uniform motion.

2. (d) A body moving along a curved path is not in uniform motion. The speed is constant but the direction changes so the velocity is not constant—it must be being accelerated.

The acceleration must be produced by a force.

3. (c) Newton’s work explained why Kepler’s laws worked with a physical reason: gravity.

4. (e) mass remains the same in both places.

5. (c) Newton’s 3rd law, the propellant’s action down has the net results of an action up for the rocket.

6. (a) True. Earth exerts the same gravitational force on you that you exert on Earth.

7. (d) If the distance between the two bodies is increased by a factor of 4, since gravity is an inverse square law the force is decreased by a factor of 4 × 4=16.

8. (a) Orbital velocity goes as one over the square root of the radius. 161/2 = 4, so the orbit would be 4 times slower.

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9. (b) 6 times larger. Vesc = (2GM/R) . The radii are the same but one planet is 36 times the mass of the other. Thus, the escape velocity from the more massive planet is 6 times greater than that from the less massive body.

Chapter 4

Light and Atoms

CHAPTER 4 LIGHT AND ATOMS Notes: This material is crucial for understanding much of the latter parts of the text.

However, some material can also be covered later, introduced with particular topics. For example, the sections on spectra and the Doppler shift can be given only a brief mention at this point and then treated more fully when discussing stars. The material on radiation in the atmosphere can be discussed with Earth.

The subject matter is tough for most students but lends itself to many demos that can enhance understanding. Use a small electromagnet made from nail, wire, and a battery to illustrate the link between electricity and magnetism. Use coil attached to milliammeter and a bar magnet to illustrate the link of magnetism to electricity. Use a comb charged by rubbing on a sweater and bits of paper to illustrate electrostatic attraction that holds electrons to the nucleus. A rheostat on a lamp shows Wien’s law as the filament changes color.

Answers to Thought Questions 1. We associate red with hot and blue with cold most likely because humans learned that red and yellow flames from fires were hot, and hot objects like coals get hot enough (a few thousand Kelvin) to glow red or orange. Objects hot enough to glow blue would probably be vaporized and are encountered much less often. On the other hand, lakes and oceans are blue and more likely to be cold, and frozen objects and ice often have blue hues. Perhaps also there is the association because people’s faces turn blue when they are cold, but flush red when they are hot.

2. Many night-vision cameras use IR detectors because all objects, particularly living things, radiate energy at IR wavelengths (they glow) regardless of whether there is visible light around. Warm-blooded animals are warmer than their surroundings and do actually glow in the infrared.

3. Just seeing a red object through a telescope would not be enough information to apply Wien’s law and deduce its temperature. The red object might be only be reflecting light (like the planet Mars), or it might be an ionized hydrogen cloud producing red line emission. (Also, even if the object is glowing—say a red giant or red dwarf star—to accurately deduce the temperature with Wien’s law requires obtaining at least a few different wavelength measurements to find the peak wavelength of the emission.)

4. Atoms do not emit a continuous spectrum because they can only emit or absorb energy (light) in discrete amounts corresponding to the differences between their energy levels.

5. Microwaves are absorbed by water molecules, causing them to vibrate and heat up.

The filling of a Pop-Tart contains more water molecules than the crust. The microwaves are absorbed by the water in the filling, heating it up more than the crust.

Chapter 4

Light and Atoms

6. Each element or ion produces a distinct pattern of spectral lines corresponding to its electron’s orbital energy levels.

7. Taking a spectrum of Venus would show absorption lines and bands corresponding to the atoms and molecules in its atmosphere.

8. The three main types of spectra are emission, absorption, and continuum. A special type of continuum spectrum is a blackbody spectrum. Emission spectra consist of specific, bright spectral lines at particular wavelengths. A continuous spectrum is emission at all wavelengths. A blackbody spectrum is the continuous spectrum radiated by a hot, dense object. An absorption spectrum is essentially a continuous spectrum which is “missing” photons at specific wavelengths where absorption lines are seen.

Usually this is the result of a continuous spectrum passing through a cool gas, which absorbs energy at the lines corresponding to the type of elements or molecules in the cool gas. An incandescent bulb has a hot, glowing filament which emits a continuous blackbody spectrum with a peak at a wavelength depending on the filament’s temperature. A compact fluorescent bulb contains an energized thin gas which emits an emission spectrum (typically, the line spectrum contains the lines of mercury, which is a common component of fluorescent bulbs; inside the bulb, ultraviolet radiation is often produced by the gas, but this is absorbed by the coating on the inside of the bulb which re-radiates visible light wavelengths).

A 23W fluorescent bulb can light up a room as effectively as a 100W incandescent bulb because a high fraction of the emitted light from the fluorescent bulb is lines at visible wavelengths. That is, much of the electrical energy used is converted to visible light. In the case of an incandescent, the peak wavelength is determined by the temperature of the filament, which is usually not high enough to be at visible wavelengths (and if it were, the bulb would likely also emit significant ultraviolet radiation). Consequently, considerable energy is emitted at infrared wavelengths by an incandescent bulb.

Incandescent bulbs are much hotter than fluorescent bulbs for this reason. Much of the efficiency of a fluorescent bulb is that much less electrical energy is wasted as heat (infrared radiation).

9. Adding water or carbon dioxide to our atmosphere reduces the heat loss from the planet. These molecules have absorption bands in the infrared with wavelengths ranging from several microns to several centimeters. This range in the infrared is the same range of wavelengths where most of the energy from incoming sunlight is re-radiated from the (warmed) ground. Thus water and carbon dioxide absorb this energy and re-radiate some of it back towards Earth instead of allowing it to escape to space. Earth becomes warmer in what is usually called the greenhouse effect. Potentially, if one increases the amount of water vapor to form more clouds, more sunlight would be reflected from the atmosphere.

However, enough radiation penetrates the clouds that usually the temperature would still increase. (Venus is covered in light-covered clouds, and although they are not water clouds, it still has the highest surface temperatures in the Solar System because of a greenhouse gas effect.)

Chapter 4

Light and Atoms

10. Oxygen and ozone molecules absorb UV radiation from the Sun (more so than the layers above and below). That extra energy heats the gas, raising the temperature of the ozone layer in comparison to the layers above and below it. When absorbed UV splits oxygen or ozone molecules, the freed atoms and partial molecules have increased kinetic energy.

Answers to Problems 1. To find how long it takes light to travel a given distance, divide the distance by the 6

11

8

speed of light. Since 150 × 10 km = 1.5 × 10 m, and the speed of light c = 3 × 10 m/s, 11 8 10-8 time = distance/velocity = 1.5 × 10 m/(3 × 10 m/s) = 15/3 × 10 s 2

= 5 × 10 s = 500s, or 8.3 minutes.

2. Signals travel to the spacecraft on Mars at the speed of light. If we send a command, the lander will receive the signal after a time equal to the light-travel time. However, the response from the spacecraft will take that time again to get back to Earth. The average orbital radius of Mars is 1.52 AU, so the closest approach to Earth is when Mars is 0.52 AU away (at opposition). The light travel time between Earth and Mars is t = D/c.

Substituting in numbers, 11 10 D = 0.52 AU × (1.5 × 10 m/AU) = 7.8 × 10 m 10

8

Thus, t = 7.8 × 10 m/(3 × 10 m/s) = 260s or 4.2 min. on average. The total time from signal being sent to receiving notification back will be twice that.

9

3.  = c. Since  =  GHz = 2.4 × 10 /s, 8 9  = (3 × 10 m/s) / (2.4 × 10 /s) = 0.125 m.

The wavelength of the cell phone or wireless signals is about 12.5 cm.

4. X-rays and radio waves travel at the same speed, the speed of light, so both reach Earth at the same time. However, the frequency of light rays varies inversely with their wavelength. Since  = c, after converting both wavelengths into meters we have: c X −ray X −ray  radio 6 10−2 m 8 = = = c   0.2510−9 m = 2.4  10 radio

radio

X −ray

The frequency of the X-rays is 240 million times larger than the frequency of the radio waves.

Chapter 4

Light and Atoms

5. Body temperature is about 300K. Using Wien’s Law as given in the book, max= 2.9 × 106 K nm / T = 2.9 × 106 K nm / 300 K = 9.67 × 103 nm ≈ 104 nm.

-3

-3

4

Note: Because 1 nm = 10 micrometers,  max = 10 × 10 micrometers = 10 micrometers.

This wavelength is usually said to be in the mid-infrared. Rattlesnakes have organs that detect heat/IR radiation so they can detect the presence of prey at night.

6

6. Using our version of Wien’s law, we have that  max = 2.9 × 10 K nm / T. Given that 6

max = 2000 nm, then T = 2.9 × 10 K nm/2000 nm = 1450 K ≈   6

6

7. We have max = 2.9 × 10 K nm / T. If T = 300 K, then max = 2.9 × 10 K nm / 300 K = 3

9.67 × 10 nm ≈ 104 nanometers = 10 micrometers, an infrared wavelength. Because we can’t see infrared wavelengths, we can’t see Earth’s glow (as is obvious given that the ground doesn’t glow in the dark).

8. STUDENTS SHOULD CREATE SKETCHES THAT SHOW SOMETHING SIMILAR TO FIGURE 4.15 AND 4.16. When emitting light, the electron drops to a lower orbital; it rises when the atom absorbs light.

9. Calculate the Doppler shift for blue light at  = 500 nm reflecting off a car traveling at V = 150 km/hr. It will help to convert the V to meters/second first, 150 km/hr × 1000 m/km × 1 hr / 3600 s = 41.7 m/s.

Now, V = c () so () = V/c and  = V/c × . Then, 8

-5

 = V/c ×  = 41.7 m/s / (3 × 10 m/s) × 500 nm = 6.95 × 10 nm.

-5 The shift in the wavelength is very small—only 6.95 × 10 nm. The reflected light would -5

have a wavelength of  −  = 500 nm − 6.95 × 10 nm ≈ 500 nm. The car would still look the same color blue to your eyes (imagine if it didn’t : you would see cars change color as they sped up or slowed down). However, a radar gun can detect the change in a radio wave bounced off the car (note that laser speed guns (LIDAR) employ a different technique).

10.  =  cm, and  = 21.010 cm. Since the observed wavelength is longer than the rest wavelength, the light has been redshifted, so the outer part of the galaxy you are observing must be moving away from you.  = 21.010 cm −  cm = 0.010 cm, and

Chapter 4

Light and Atoms

8

V = c () = 3 × 10 m/s × (0.010 cm / 21 cm ) = 142,900 m/s = 143 km/s.

Note that the typical rotation speed of outer parts of spiral galaxies is a little more than this, around 200 km/s.

Answers to Test Yourself 1. (e) They all travel at the same speed (in vacuum, anyway).

2. (b) Gamma rays have the shortest wavelength.

3. (e) Of the choices listed, radio photons have the lowest energy.

4. (c) Only Kelvin temperatures are directly proportional to thermal energy, so only the Kelvin temperature will double (to 560 K).

5. (b) 656 nm. The wavelength of line emission is not affected by temperature.

6. (d) Bright emission lines come from hot, rarefied gases.

7. (c) Hot interior shines through cooler low density gas.

8. (a) If the lines are shifted to longer wavelengths (redshifted), the object is moving away from us.

Chapter 5

Telescopes

CHAPTER 5 Telescopes Notes: This section introduces the different kinds of astronomical telescopes and detectors and emphasizes that astronomers study the heavens at many wavelength regions other than just visible light. The entire section may be omitted with no loss of continuity.

However, given the ubiquity of the CCD in modern scientific and consumer cameras it might be worth at least mentioning that the same basic technology in consumer electronics is used in most optical, infrared, and X-ray telescopes. Also, since the detection of exoplanets is presently an area of considerable scientific and public interest, if/when later discussing exoplanets it may be worth describing interferometers (which are now being used to directly image other planets and stars).

Answers to Thought Questions 1. Binoculars have larger lenses than your eyes and hence they collect more light and allow you to see dimmer objects. Binoculars with 50-mm lenses will collect four times as much light as those with 25-mm lenses, because the light collecting area scales with the square of the diameter. Furthermore, resolving power scales with the diameter, so the 50mm lenses should have about twice the resolving power (not considering atmospheric seeing effects).

2. When the pencil is perpendicular to the water’s surface, it looks straight. As you tilt the pencil from the perpendicular, the pencil looks more and more bent. Similarly, light rays entering our atmosphere are more bent (refracted) when they enter it at angles far from the perpendicular.

3. Even though part of the primary mirror is blocked by the secondary mirror, and the light reflected off the secondary mirror may even be reflected through a hole in the primary mirror, the light from distance planets or stars is coming into the telescope in parallel rays and is essentially incident everywhere on the primary mirror; this is how the primary can act as a light-gathering device. The incident light from the field is then reflected by the primary and concentrated (in a smaller physical area) on the secondary to be focused or fed into additional optics. The secondary and the hole do reduce the collecting area (compared to  times the radius of the mirror squared) by a small amount—this slightly reduces the brightness of the image, not what part of the field of view is visible.

4. High resolving power is much more desirable, as it permits you to distinguish objects at a finer angular scale. High magnification can be achieved with a combination of lenses on any telescope, but magnifying a low-resolution image will only make a bigger, unresolved, blurry image—just as once you can see the pixels of a digital photograph, zooming in further doesn’t add any detail, it just makes the pixels appear larger.

5. It would be incredibly expensive and impractical to build a single 36 km diameter telescope. The structure would be so huge that there would be no chance to steer it, objects could only be observed for a few seconds, and the field of view would be limited

Chapter 5

Telescopes

to what was directly above the telescope. In contrast, the VLA achieves the same spatial resolution for a fraction of the cost and materials, and can observe and track targets over the whole sky for periods of hours at a time. However, the sensitivity is much less than a single huge telescope. The total collecting area of the VLA is that of a 130 m diameter telescope (see Problem 6), a factor of (36,000 m / 130 m)2 = 76,700 times smaller than the collecting area of a single 36 km dish.

6. A 2-m telescope on Earth never reaches the limit of its resolving power of  = 0.02  (nm)/ D(cm) = 0.02 × 500 / 200 = 0.05 arcsec because we are observing through the turbulent atmosphere. The best “seeing” conditions only allow resolutions of a few tenths of an arcsecond.

7. All of these wavelengths are blocked by the atmosphere. 1 nanometer X-rays are absorbed by the ozone layer and ordinary oxygen in the atmosphere. 1 millimeter infrared waves are absorbed by water and carbon dioxide molecules in the atmosphere. 100 m radio waves are blocked by the electrically charged ions in the upper atmosphere.

8. Looking down a beach on a hot day, distant objects appear shimmering (sometimes even without binoculars) as hot air currents rise and refract the light trying to reach our eyes. In the same way, turbulent atmospheric blobs drifting across our line of sight cause star images to be blurred as the light is refracted and takes different paths through the atmosphere on its was to the aperture of the telescope. The “seeing” is a measure of how distorted the image appears.

Answers to Problems 2 1. Light gathering power goes as the collecting area. Area = D The pupil of the eye has a diameter Dpupil = 0.5 cm = 5 mm.

The small reflector has an aperture Dreflector = 10 cm = 100 mm.

2 2 Area of telescope / Area of eye = (Dreflector/Dpupil) = (100 mm / 5 mm) = 400 The collecting power of a 10 cm (4") reflector is 400 times greater than that of your eye.

2. EXPERIMENT FOR STUDENTS Theoretically,  = 0.02 /D, with  in nm and D in cm, so for 500 nm light and a 5 mm = 0.5 cm pupil,  = 0.02 × 500/0.5 = 20 arcsec. This is probably a bit higher resolution than the experimental results.

3. Theoretically,  = 0.02 /D = 0.02 × 500/0.5 = 20 arcsec. (see above) So, if the diameter of the crater is 2 arcminutes, you should be able to resolve it.

Chapter 5

Telescopes

4. Theoretically,  = 0.02 /D, with  in nm and D in cm. D = 25 m = 2500 cm and 2

9

8

 = 10 cm = 10 cm × 1 m / 10 cm × 10 nm / m = 10 nm. So, 8  = 0.02  / D = 0.02 × 10 / 2500 = 800 arcsec = 13.3 arcminutes.

9 For 1 m waves (10 nm), which are 10 times larger than 10 cm waves, we get a result 10 times larger:  = 8000 arcseconds or 133 arcminutes.

5. (1) How big an eye to see as well in the infrared at 12 micrometers as in visible light:  = 0.02 /D, and set visible = infrared. The eye has a pupil of D = 0.5 cm; visible light is -6 9 4 at 500 nm, and 12 micrometers is = 12 × 10 m × 10 nm/m = 1.2 × 10 nm (0.02  / D)visible = (0.02  / D)infrared so visible / Dvisible = infrared / Dinfrared 4 Dinfrared = infrared (Dvisible / visible) = (1.2 × 10 nm) × (0.5 cm / 500 nm) = 12 cm.

Your pupil would need to have a diameter of 12 cm to see 12 micrometer radiation at the same resolution as you can see 500 nm light.

8 (2) Radio light at wavelength of 10 cm = 10 nm. Solving with proportionalities: /D is 8

constant. The wavelength of the light is 10 nm / 500 nm = 200,000 times longer, so the diameter of the pupil would need to be 200,000 times larger, or 0.5 cm × 200,000 = 100,000 cm = 1000 m = 1 km!

6. If each of the VLA telescopes has a diameter of 25 meters the collecting area is  × (25m/2)2 = 491 m2 per telescope and 27 × 491 m2 = 13,257 m2 total. This is the same as one telescope of…

13,257 m2 =  × r2 so r = (13,257 / ) m = 65 m radius 65 meters, or diameter 130 m.

7. The CCD can record 0.80 of the photons striking it, and the photograph 0.04. To take an equal quality image in the same amount of time, we’d need to collect 0.80/0.04 = 20 times as much light for the photograph as for the CCD. The collecting area goes as the square of the diameter, so the diameter of the telescope taking the photograph would need to be (20)1/2 = 4.47 times larger than the diameter of the telescope using the CCD. 8. Hubble’s altitude is 569 km above Earth’s surface, and from the Appendix Earth’s radius is 6378 km. The radius of the orbit is therefore 6947 km, and the circumference C = 2 ×  × 6947 km = 43,649 km. Using the formula from the Chapter 3, the orbital velocity is…

Chapter 5

V=

GM = d

Telescopes

2 m 6.67 10−11 m3kg −1s−2  5.97 1024 kg 7 m = 7.57 103 = 5.7 10 s2 6947 103 m s

3

3

The orbital period is therefore 43649 × 10 m / (7.572 × 10 m/s) = 5765 s.

With 60 seconds per minute, that’s 96 minutes.

Since Hubble orbits Earth about every hour and a half, this clearly sets a limit on how long it can continuously observe a specific object to something like 45 minutes or half an hour when Earth would not be significantly in the way—but this time must be further reduced since some time must be taken to point the telescope at an object, and to allow the telescope to settle down from the vibrations induced from slewing to the target object.

Answers to Test Yourself 1. (c) Collecting area goes as diameter squared, so four times the diameter means sixteen times the light-gathering power.

2. (c) Higher resolution corresponds to finer detail.

3. (a) Since  = 0.02/D, you would have to make the mirror larger.

4. (d) Interferometers greatly increase the resolving power.

5. (b,e) CCDs do not need to be changed out and can record more photons in the same time. (Note that for (a), while CCDs can collect for a long time, so can film).

6. (a) The time pulses arrive would be different for different wavelengths. Less energetic photons interact and are delayed by the electrons in the ionized gas.

7. (b) adaptive optics to correct for the seeing in the atmosphere 8. (d) refraction makes the Sun appear higher than it actually is.

9. (d) Adaptive optics adjust the image compensating for atmospheric distortions.

Chapter 6

The Earth

CHAPTER 6 EARTH Notes: Demos can be helpful with this material. Some relatively easy ones: To illustrate mountain building at a subduction zone, slide a sheet of paper at a blackboard eraser or other solid object. The paper can be made to buckle upward much as Earth’s crust buckles when it forms folded mountain belts.

Bring in a gyroscope or top to demonstrate precession. You can also show precession with a thumbtack held by its point, spun, and dropped onto a table. The spinning tack will act like a tiny top. Encourage students to try this on their own.

Use an analogy to a blanket to illustrate that greenhouse effect works NOT by creating heat but by reducing heat loss.

Use a Crooke’s tube and magnet to illustrate bending of electron beam by magnetism. Relate to Earth’s field bending particles that create aurora.

Historical Note: A paper in Nature (Vol. 367, February 3, 1994) credits Abraham Ortelius, a 16th century geographer, with being the first to suggest (in 1596) that South America and Africa split apart to form two separate continents.

Answers to Thought Questions 1. When you heft fruit for weight, you are estimating the mass. Since fruit are approximately the same size (same diameter means same volume), you are also comparing the density, which will be higher for fruit with more water inside because water is fairly dense compared to dry plant matter. You are taking the average density of the fruit to find its composition in the same way that we measure the mean density of Earth to find its composition, knowing that iron is denser than silicate rocks.

2. Although the submarine’s walls, decks, engine, etc. may be mostly built out of steel or other metals, which have a density higher than water, remember that the average density of the submarine depends on its total mass divided by the total volume—much of the interior of the sub is filled with air, which has a much lower density than water. For the submarine to be able to rise and travel at the surface, the average density of the sub when the ballast tanks are mostly empty must be less than the density of water (so it rises).

However, if you fill the ballast tanks with water, the mass of the water in the tanks is much greater than the air it displaces. Since the volume of the submarine is the same, the average density of the submarine increases. For the sub to be able to submerge, the average density when the ballast tanks are full should be pretty close to the density of water—allowing the sub to be “neutrally buoyant” and with its engines and planes move itself to any depth in the water. If the average density of the sub is more than water, it will sink—and could get stuck on the sea floor.

3. The summit of Chimborazo is farther from the center of Earth than the summit of Mt. Everest is because Chimborazo is much closer to the equator than Mt. Everest. The rotation of Earth causes Earth to bulge at its equator by about 13 miles—that extra distance puts Chimborazo’s peak farther from the center, even though it is not as far above sea level (which is also affected by the rotation) as Everest.

Chapter 6

The Earth

4. Your cheek is the flesh around your hollow mouth cavity. It is soft tissue (unless you hit the cheekbone). Your forehead is the bone of your skull containing your brain. They make different sounds because sound waves (and seismic waves) travel differently through material of different densities and states (solid, liquid). The vibration of a sound wave is very similar to the vibration of seismic waves. Scientists study the interior of Earth by observing how seismic waves travel through it.

5. Although from about 1600 on, several people had suggested that the continents had drifted apart, they had only a little evidence in the form of the shape of coastlines and the match of fossils to support the hypothesis. In the absence of weightier evidence a hypothesis remains conjecture. In 1912, Wegener put forward the hypothesis that sea floor spreading had caused the continents to drift apart, supporting his ideas with additional fossil and geological evidence. The theory of plate tectonics did not receive wide acceptance until geologists were able to date the sea floor at different distances from oceanic ridges (1960s), thus testing predictions of the theory. Today, the motion of the plates can be measured directly with satellite imaging and the theory has become generally accepted with this highly accurate data. Through the development of a model, predictions, a variety of different, independent of tests and evidence, and increasingly accurate test results confirming the validity of the model, the scientific method has been use to validate the theory of plate tectonics.

6. Earth’s magnetic field is driven by rotational motions in the liquid (and solid) core of Earth, which in turn are related to the motion of Earth as a whole. If Earth rotated more slowly, we would expect the core to also rotate more slowly, and the field would not be as strong. We will see evidence of this correlation when we look at the other planets and see that in general slowly rotating planets have weaker fields and quickly rotating ones have stronger fields.

7. We can compare the mass of an atom of O2, H2O, and N2 easily by using atomic mass units. The mass of a molecule of O2 is about 16 + 16 = 32. N2 is a little lighter, 14 + 14 = 28. H2O is the lightest: 2 × 1 + 16 = 18. Dry air contains mostly oxygen and nitrogen molecules. In an equal volume of humid air, some of these would be replaced by lighter water molecules. The overall mass in the volume would be less, therefore the density of that volume of air will be less, and so the humid air would rise.

8. DIAGRAM FOR THE STUDENT—draw a diagram similar to Figure 6.26 showing how far the apparent target would move normal and faster rotation. The effect is larger with faster rotation.

9. If you threw a rock from the N pole towards a target on the equator, since Earth is rotating counterclockwise (W → E) the rotation carries the target westward and the rock misses it, appearing to be deflected to your right. (This can be illustrated with figures similar to Figure 6.26.)

Chapter 6

The Earth

Answers to Problems 1. If Earth’s radius was half of its real value, Earth’s volume would decrease to 1/8 of Earth’s real volume because volume of a sphere depends on the cube of the radius. Since the density is mass/volume, if the mass remained the same the density would increase (1/1/8 = 8) to 8 times the real density of Earth. If Earth’s mass were twice its real value, but the radius and volume the same, the density would be twice the real density (2/1 = 2).

2. The most common elements are O, Si, Al, Fe, Ca, Mg, Na, K (2% or more). All of these elements are produced in stellar fusion, the stable energy source powering the cores of stars. (Most of these elements are made during the red giant phase of stars’ lives.) All of the “precious” metals like gold (Au), silver (Ag) and platinum (Pt) are produced in supernovae explosions—a violent event that occurs when a massive star explodes.

3. The average density of the alloy would be the sum of the fraction of each element times its density, using the values provided: density = [(0.4 × 4.5) + (0.3 × 7.9) + (0.1 × 1.5) + (0.1 × 1.7) + (0.1 × 2.7)] g/cm3 = 4.76 g/cm3 = 4.8 g/cm3 4. If the seismic wave takes 700 seconds to travel down to the boundary and bounce back, the travel time to the boundary is half of 700 seconds, or 350 seconds. The distance to the boundary is found using D = vt. Thus, D = (8 km/sec) × 350 sec = 2800 km.

Figure 6.7 shows that the mantle/liquid core boundary is 6357 km – 3,500 km = 2857 km below the surface, in good agreement with our answer.

5. Every 5700 years, the percentage of the original sample that is still C-14 drops by half.

The graph should look something like the graph on the next page (which carries through seven half-lives), although some students might think to make a semi-log graph. Some leeway should be given on the answers depending on how students estimate them. Based on a good graph, less than 13% would be an age of about 17,000 years; or students can notice that you need about 3 half-lives: 100% to 50%; 50% to 25%; 25% to 12.5%; for about 17100 years. For less than 2%, the graph shows about 32,000 years, or a student might estimate it’s just less than another 3 half-lives (6 total): 12.5% to 6.25%, 6.25% to 3.125%, 3.125% to 1.56%, for 34200 years. (Although it is not the point of the problem, as it is beyond the scope of the book, you can also use logarithms to find the exact answers of 16777 and 32169 years via n = log (0.13) / log (0.5), etc.).

Chapter 6

The Earth

Perce nt re mainin g

6. 80 × 106 yr × 4 cm /yr = 3.2 × 108 cm = 3200 km. This is about half the current distance between some parts of Africa and South America. This is roughly consistent with what you can see in Figure 6.15.

7. The total mass of Earth’s atmosphere must be equal to the total mass of oxygen molecules plus the total mass of nitrogen molecules. The composition of the atmosphere by number is given as 1% argon atoms, 21% oxygen molecules, and 78% nitrogen molecules. Then there are 78/21 nitrogen molecules for each oxygen molecule, and 1/21 argon atoms for each oxygen molecule.

Let X = number of oxygen molecules.

The atomic mass of an atom of nitrogen is 14, an atom of oxygen, 16, and an atom of argon, 18, so the masses of single molecules of nitrogen and oxygen gas are Mnitrogen = 2 × 14 × 1.66 × 10-27 kg Moxygen = 2 × 16 × 1.66 × 10-27 kg And argon, a monatomic gas, has Margon = 18 × 1.66 × 10-27 kg

Chapter 6

The Earth

Overall we can say: Mtotal = (78/21)(X)(Mnitrogen) + (1/21)(X)(Margon) + XMoxygen Mtotal = X((78/21)(Mnitrogen) + (1/21)(Margon) + Moxygen) Then, X = Mtotal / ((78/21)(Mnitrogen) + (1/21)(Margon) + Moxygen) Substituting in masses, we find: X = (5.1 × 1018 kg ) / ((78/21)(2 × 14 × 1.66 × 10-27 kg) + (1/21)(18 × 1.66 × 10-27 kg) + (2 × 16 × 1.66 × 10-27 kg)) X = (5.1 × 1018 kg ) / ((1.66 × 10-27 kg)((78/21)(2×14) + (1/21)(18) + (2 × 16))) so X = 3.07 × 1045 / 136.86 = 2.24 × 1043. This very large number is the number of oxygen molecules in the atmosphere. The total mass of those molecules is Moxygen = 2.24 × 1043 × (2 × 16 × 1.66 × 10-27 kg) = 2.24 × 1.66 × 16 × 2 × 1043-27 kg = 119 × 1016 kg = 1.2 × 1018 kg.

8. DRAW FOR THE STUDENT. Since the air is flowing out from the center, and is then deflected to the right, it will look much the opposite of the low pressure system shown in the figure in the chapter.

Answers to Self-Test 1. (e) The mean density and seismic evidence shows that the core is iron and is at least partially molten.

2. (d) S waves do not travel well through liquid and are not detected at all locations.

3. (c) From a comparison of the amount of a radioactive element to its decay product, scientists can determine the age of a rock or other sample.

4. (d) The convection in the mantle affects the crust.

5. (a,d,e) earthquakes, volcanic activity, mountains. Note that (c), convection currents, are the cause of plate motion, not a result caused by it.

6. (a) motion in a liquid core 7. (c) it absorbs infrared light.

8. (b) Weather occurs in the troposphere.

9. (a) above the North (rotational) pole, since the ground would rotate a full 360 degrees below it.

Chapter 7

The Moon

CHAPTER 7 THE MOON Answers to Thought Questions 1. A newly paved road has no potholes. A very old road that hasn’t been paved for a long time has many. Similarly, the very old lunar surface is heavily cratered, but regions that have been covered with lava flows more recently (a sort of cosmic “re-paving”) have many fewer craters.

2. Bergmann’s rule suggests that larger animals are better adapted to existing in the cold than smaller ones—we might expect this means they are less likely to freeze to death—so therefore the larger animals must stay warm more easily. Larger animals have a larger volume, and a larger ratio of volume to surface area, so they lose heat less quickly. This is the same idea as is used in the chapter to explain the difference in temperature between Earth’s interior and the Moon’s interior.

3. Since Earth and the Moon were both hot and melted at the time the material that makes up the Moon was splashed into orbit, if comets brought water to Earth it makes sense it would have to happen after the formation of the Moon. Therefore the Moon should also have been bombarded with comets and have had significant water deposited on it, which does not seem to be the case. A counter-argument: the lack of water on the Moon could also be explained by the fact that it would quickly evaporate in the low or zero pressure atmosphere, and then the Moon’s low gravity would be unable to hold onto the water molecules, which would be lost to space fairly quickly. Evidence of water in some deep, dark craters adds weight to this possibility.

4. On Earth, wind, rain, etc. quickly erode footprints. On the Moon, no such processes occur. Thus, footprints last until obliterated by meteor impacts - a very, very slow process.

5. If the Moon were not in synchronous rotation, we would still observe the same phases, but the particular parts of the lunar surface lit up or in shadow would vary (we’d see what is now the “far side” sometimes). If Earth and Moon were both in synchronous rotation, the phases would look much the same as they do today, but would only be visible from the half of Earth that could see the Moon. In synchronous rotation, the same half of Earth would always face the Moon. (The period of the phases would also be different because for synchronous rotation the distance to the Moon would not be the same as what it is now).

6. Students should make a reasoned argument for future missions. Among other results, missions to the moon have allowed us to (1) place (retro) reflectors on the surface, to make exact measurements of the distance; (2) place seismic detectors on the surface, to learn about Moonquakes and the Moon’s internal structure to a greater detail than remote observing permits; (3) orbiting craft provide detailed gravitational data to map the Moon’s internal density; (4) the return of lunar rock samples provided critical and specific evidence about how the Moon’s crust is similar and different to Earth’s, which strongly affected our theories of the origin of the Moon (sec. 7.4); (5) missions allowed us to photograph the far side of the Moon, revealing fewer maria and in combination with other results allowing us to determine the offset of the Moon’s center from the crust (fig. 7.8); (6) probes in 2009

Chapter 7

The Moon

proved the presence of ice in shadowed craters near the Moon’s poles. This data has significantly constrained models of the Moon’s origin, composition, and structure.

7. If the day were 12 hours long, there would still be 2 tidal bulges and so the time between the high and low tides would be halved. Currently, as shown in Figure 7.19, if there is a low tide at 6 a.m., there is a high tide at noon. With the rotation period of Earth halved, low and high tides would be approximately 3 hours apart instead of approximately 6 hours apart.

8. As the Moon recedes from Earth, its gravitational impact is lessened and the tides will be shorter (think about the solar tides—the Sun is much more massive but so far away the tides are smaller than the lunar ones). If the Moon were twice as far from Earth, there would still be two tides each day (assuming Earth’s rotational period is about the same).

9. The tides occur about an hour later each day for the same reason that the Moon rises about an hour later each day: the Moon is moving in its orbit. It takes Earth about an hour to catch up to this motion. Since the what tide it is depends on the orientation of Earth and Moon (and where you are), it’s actually a little more than 6 hours between each tide, closer to 6 hours and 13 minutes, and closer to an hour more for the same tide the next day.

Answers to Problems 1. Moon’s mass to Earth’s mass: just more than 1% mMoon 734.9 1020 kg 73.49 1021 73 = = = 10−3  12 10−3 = 0.012 0.012 = 24 24 5.97 10 kg 6 mEarth 5.97 10 Moon’s radius to Earth’s radius: a bit more than a 25%: r 1.738 103 km 174 101 174 −1 −1 = 0.27 = Moon = = 10  2.7 10 = 0.27 64 102 64 rEarth 6.378 103 km 2. If the Mare Serentitatis has an angular diameter of 5 arc minutes, its linear diameter can be calculated according to the formula D/2d = A/360°, which relates angular diameter (A), true diameter (D), and distance (d) . Thus, D = 2dA/360°.

To use the formula, we must express A in degrees, not arc minutes. We do this by recalling that 1 degree contains 60 arc minutes. Inserting this value and the Moon’s distance in the formula, we get: D = 2dA/360° = 2(384,000 km)(5' )(1°/60')/360° = 558 km.

Note: the angular diameter has been rounded slightly, and the mare is not exactly circular, so the value may differ slightly from literature values.

3. The crater Tycho is 88 km wide.

D = 2dA/360°, so

Chapter 7

The Moon

A = (360°)(D)/(2d) = (360°) (88 km)/( 2×384,000 km) = 0.01313°. and 0.01313° × 60'/1° = 0.78’, or 0.78 arc minutes, or 47 arc seconds.

Theoretically, the human eye can manage a resolution of about 20 arc seconds, or 1/3 of an arc minute (see Chapter 5 problems). At least some people should be able to see this crater with the naked eye, under ideal conditions, and possibly with a filter if the Moon is very bright.

4. To calculate the Moon’s density (), divide its mass, M = 7.349 × 1020 kg = 7.349 × 1023 g, by its volume. Assuming it is a sphere, its volume is 4r3/3. Thus,  = (4r3/3). Now insert the values of M and r = 1738 km = 1738 × 105 cm = 1.7 × 108 cm to find,  = 7.3 × 1025 g/[4(1.7 × 108 cm)3/3] = {7.3/[(4) × (1.7)3]} × 1025 g / (108 cm)3 = 0.35 × 1025-(8×3) g/cm3 = 0.35 × 10 g/cm3 = 3.5 g/cm3.

The density of iron is 7.9 g/cm3. The very low value of the Moon’s density suggests there is a much smaller fraction of iron in the Moon than in Earth (average density of 5.5 g/cm3).

5. If the Moon were made of incompressible Swiss, its average density would equal the density of Swiss cheese, or 1.1 g/cm3. Since  = /V, then M = V. So, M = V = 4r3/3 = (1.1 g/cm3) × 4(1.7 × 108 cm)3/3 = 2.4 × 1025 g = 2.4 × 1022 kg This is of course about one-third of the Moon’s actual mass (7.3 × 1025 g). An equally valid and simpler solution to this problem is to notice that the ratio of the density of Swiss cheese to the actual density must be the same as the ratio of the Moon’s mass if made of cheese to the Moon’s actual mass and calculate accordingly: Mcheese = (1.1 g/cm3/ 3.5 g/cm3) Mreal = (1.1/3.5) 7.3 × 1025 g = 2.3 × 1025 g.

(The slight difference here is a result of rounding).

6. The Lunar Reconnaissance Orbiter orbits the Moon 50 km above the surface with an orbital period of 113 minutes. Assuming a circular orbit, the modified form of Kepler’s 3 2 Third Law (see Chapter 3) gives us m + M = 42d /GP , with M = mass of Moon, m = mass -11 3 2 of spacecraft, d = radius of orbit, and G = 6.67 × 10 m /(kg sec ).

Expressing d in meters and P in seconds so that units will cancel, P = 113 × 60 = 6780 sec d = 1738 + 50 km = 1788 km Inserting these values in the law gives

Chapter 7

The Moon

3

3

2

-11

3

2

m + M = 42 × (1788 × 10 m) /( (6780s) × 6.67 × 10 m /(kg s ) ) 22

= 7.36 × 10 kg. However, the mass of the spacecraft, m, is so tiny compared to the Moon’s mass that we can ignore it. Thus, the measurement of the Moon’s 22 mass is 7.36 × 10 kg (very close to the value in the Appendix).

7. Round trip light-travel time to the Moon and back is 2.56 seconds. The speed of light, c, 5

is 3 × 10 km/s. Using d = vt, with t = 2.56 sec / 2 for the time to the Moon, 5 D = ct = 3 × 10 km/s × 2.56/2 s = 384,000 km.

Conversations will be clumsy because of the extra 2.56s between when you finish asking a question and when you hear the reply.

8. The surface area of a sphere is 4r2, and the volume is 4/3r3. The ratio of the surface area to the volume is therefore 4r 2 3 = .

4 3  r r 3 So for the Moon, 3/r = 3/1738 km = 0.0017 km-1.

For Earth, 3/r = 3/6378 km = 0.00047 km-1.

The Moon’s SA to V ratio is 3.6 times larger than Earth’s – naively we might estimate the Moon would cool to the same temperature 3.6 times faster than Earth.

9. The rate at which the length of the day increases is 0.002 s/century.

To find out when Earth’s day was 5 hours long, we would need to consider how long it would take for the day to lengthen by 19 hours, or 19 × 60 min/hr × 60 s/min = 68,400s.

68,400s = 0.002s/century × time passed, so time passed = 68,400s / (0.002 s/century) = 3.42 × 107 centuries = 3.42 × 109 years = 3.4 billion years Answers to Self-Test 1. (d) The maria were formed after the highlands (and after the bombardment of the highlands).

2. (b) The Moon’s weak gravity makes it hard to hold onto an atmosphere (atoms and molecules are sufficiently heated to move fast enough to escape).

3. (a) We would observe both sides.

4. (c) Its mantle is cold and rigid.

5. (d) High tide to low tide is about 6 hours (high to high is 12 hours, half a day).

Chapter 7

6. (a) When it is high tide locally, the Moon is pulling you “up” the most.

The Moon

Chapter 8

Survey of Solar Systems

CHAPTER 8 SURVEY OF SOLAR SYSTEMS Answers to Thought Questions 1. Students should be encouraged to start with the actual IAU resolution. The importance of clearing an orbit is likely an arguable point. The importance of the object being spherical is also arguable, as the mass required for an icy body to be spherical is significantly less than the mass required for a rocky object. Students may wish to consider that some small bodies on their own would be classified as dwarf planets, but if those same bodies were in orbit around a planet, would be satellites.

2. In a biased sample, the numbers of different kinds of things you find in your survey of a population are off from the real values because the search method is better at finding certain kinds of objects than others. The Doppler method works best when there is a large Doppler shift. This happens when the planet is very large (or to a lesser extent, when the star is small) or when the planet is close to the star, or both. It is no surprise that many of the planets found with the Doppler method are massive and close to the star. STUDENTS SHOULD SPECIFY AN ATTEMPT TO SURVEY SOMETHING, AND HOW THE SEARCH METHOD COULD BIAS THE RESULT. Simple example: determining the population of each state by counting license plates on your local highway. Your result would be highly biased toward a high count for your state and neighboring states, even if those states had fewer people. On the other hand, if you were trying to figure out what cars sell the best, counting how many of different kinds of cars on the highway would probably not be as biased. You would, however, be biased if certain kinds of cars sell better in your area than in other areas. Another example: surveying how many people are in different age groups in the population. This will not work well if you are only counting people in an elementary school.

3. Some exoplanets differ from what we expect based on our solar system: (1) gas giants have been found much closer to their stars in our solar system, (2) more exoplanets have masses similar to or larger than Jupiter’s . This seems to suggest that the solar nebula theory—that gas giants would form at a distance from the central star where ices could condense and where helium and hydrogen could be accreted before being cleared by the stellar wind—could be incorrect in some ways. However, the evidence so far does not prove the solar nebula theory is wrong, or at least not entirely wrong. It’s even worth remembering that the inference that these exoplanets are gas giants is itself partly based on the solar nebula theory. We have only observed a few hundred systems, and the method of detection favors finding systems like the ones that have been found. These systems may or may not be typical of planetary systems, and may or may not have formed in a way similar to ours. Finally, the existing theory was created without any data on the original location of planets. Recent modeling suggests that under some conditions planets may migrate. Several large planets could migrate inwards and even collide with the central star. We can not even be sure this did not happen in our solar system. Therefore, while the current evidence does challenge some aspects of the solar nebula theory, it seems just as likely that the theory is simply incomplete than that it is flat out wrong.

Chapter 8

Survey of Solar Systems

4. STUDENTS SHOULD SUPPLY SOME IDEAS AND BRIEFLY JUSTIFY THEM. Possibilities include: gravity, Kepler’s laws, the energy output of the Sun, pressure, viscosity between rotating material, the composition of the solar nebula, thermodynamics / heat, magnetic fields, electric charge of material condensing to form grains, composition of material condensing to form grains, the temperature at various radii from the Sun, the initial momentum and energy of the collapsing cloud, the momentum and angular momentum of the cloud at different distances, the temperature at which various kinds of materials can condense or at which they evaporate, etc.

Answers to Problems 1. (a) 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. (b) For simplicity, values in the table are given in km(real) and meters(model), and arranged in order of size (and mass) of the planet. 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.

Body Sun Jupiter Neptune Earth Mercury Pluto Ceres

Radius (km)

Diameter(km)

6.96E+05 71492 24764 6378 2440 1160 487

(a) Model Diam (m)

1392000 142984 49528 12756 4880 2320 974

1.00000 0.10272 0.03558 0.00916 0.00351 0.00167 0.00070

Density kg/L 1.33 1.64 5.52 5.43 2.00 2.08

Volume (L) 0.567470448 0.02358486 0.000402925 2.25601E-05 2.42407E-06 1.79373E-07

(b) Model Masses (kg) 0.754735696 0.03867917 0.002224147 0.000122501 4.84814E-06 3.73096E-07

2. To calculate the densities of Venus and Jupiter, use density  = M/V. 3 You can calculate the volumes from the formula for the volume of a sphere, V = 4/3πR .

24 27 MVenus = 4.87 × 10 kg = 4.87 × 10 g 8 RVenus = 6,051 km = 6.051 × 10 cm 8 3 26 3 VVenus = V = 4/3π(6.051 × 10 cm) = 9.28 × 10 cm 27 26 3 3 Venus = M/V = 4.87 × 10 g / 9.28 × 10 cm = 5.25 g/cm 27

30

MJupiter = 1.9 × 10 kg = 1.9 × 10 g 9 RJupiter = 71,492 km = 7.15 × 10 cm 30 3 VJupiter = 1.53 × 10 cm 30 30 3 3 Jupiter = M/V = 1.9 × 10 g / 1.53 × 10 cm = 1.24 g/cm Venus has a density that is larger than that of rock, suggesting that it has an iron core. Jupiter is only slightly denser than water or ice, suggesting that it is largely made of lighter materials that are rich in hydrogen.

Chapter 8

Survey of Solar Systems

3. To calculate the escape velocities of Mercury and Jupiter, use the formula for escape 1/2 velocity from Chapter 3, Vesc = (2GM/R) .

27 MJupiter = 1.90 × 10 kg 7 RJupiter = 71,492 km = 7.1492 × 10 m 1/2

Vesc(Jupiter) = (2GM/R)

-11

27

7

1/2

= [(2 × 6.67 × 10 m3kg-1sec-2 × 1.9 × 10 kg) / (7.1492 × 10 m)] 9 1/2 4 = [3.5453 × 10 m2sec-2] = 5.95 × 10 m/sec 8 4 ≈ [36 × 10 m2sec-2] = 6 × 10 m/sec 23

MMercury) = 3.30 × 10 kg 6 RMercury = 2,440 km = 2.44 × 10 m Vesc(Mercury) = -11 23 6 1/2 = [2 × 6.67 × 10 m3 kg-1 sec-2 × 3.30 × 10 kg / (2.440 × 10 m) ] 7 1/2 3 = [1.80 × 10 m2sec-2] = 4.25 × 10 m/sec The escape velocity from Jupiter is about 60 km/sec, while that from Mercury is only about 4 km/sec. At a given temperature, light gases move faster than heavier gases. At higher temperatures, the molecules have higher velocities. Jupiter can hold on to its original atmosphere because it has a high escape velocity and because it is cold. Mercury is a small planet with a low escape velocity and it is very hot from being close to the Sun, so it has lost any atmosphere it might have had early in its history.

4. Calculate the escape velocities of Neptune and Mars, and compare the values to Earth’s escape velocity.

26 MNeptune = 1.02 × 10 kg, 3 7 RNeptune = 24,764 km × 10 m/km = 2.4764 × 10 m 1/2

Vesc(Neptune) = (2GM/R)

-11

26

7

1/2

= [2 × 6.67 × 10 m3 kg-1 sec-2 × 1.02 × 10 kg / (2.4764 km × 10 m)] 4 = 2.34 × 10 m/sec 23

MMars = 6.42 × 10 kg, 6 RMars = 3,396 km = 3.396 × 10 m 1/2

Vesc(Mars) = (2GM/R)

-11

23

= [2 × 6.67 × 10 m3 kg-1 sec-2 × 6.42 × 10 3 = 5.02 × 10 m/sec

3

6

1/2

kg / (3.396 × 10 m) ]

The escape velocity from Earth is about 11 km/s = 11 × 10 m/s, right in the middle, and Earth’s atmosphere is in the middle, too. Earth retains much more gas than Mars, which has a very thin atmosphere. However, Earth is unable to retain lighter gases like hydrogen and helium which Neptune can retain.

Chapter 8

Survey of Solar Systems

5. The star has about the same mass as the Sun, so Kepler’s third law can be used unaltered.

Given that the period is 29 days, you can solve for the semi-major axis of the orbit. To use the law, it is easiest to express the period in years rather than days. 29 days is 29/365.25 of a year. Given that P2 = a3, a = (P2)1/3 = ( (29/365.25)2 )1/3 = ( 0.0063 )1/3 = 0.18 AU. 6. The Doppler shift formula is V = c (); the shift is () = V/c. We need the velocity of the planet. Assuming the orbit is circular, that’s just the circumference divided by the period of 29 days.

11 11 C = 2r = 2 (0.18 AU × 1.5 × 10 m/AU) = 1.6965 × 10 m.

11

V = C/P = 1.6965 × 10 m / (29d × 24 hr/d × 60 m/hr × 60 sec/hr) 4 = 6.77 × 10 m/sec.

4

8

-4

So…    = V/c = 6.77 × 10 m/sec / (3 × 10 m/sec) = 2.26 × 10 = 0.000226.

7. To make calculations for the planet Gliese 851d, we need to use the modified version of Kepler’s third law because the star Gliese 851 has a mass of 0.31 solar masses. The planet’s orbit has a semi-major axis of 0.22 AU, which we will convert and use for d.

M = 4d3/GP2 with M in kg, P in seconds and d in meters.

11 10 d = 0.22 AU × 1.5 × 10 m/AU = 3.3 × 10 m 30

29

M = 0.31× 1.989 × 10 kg = 6.17 × 10 kg P2 = 4d3/(GM) 11 -11 29 = 4 (3.3 × 10 m/AU)3 / (6.67 × 10 m3 kg-1 sec-2 × 6.17 × 10 kg) 10 19 = 4 (×10 m)3/(4.115 × 10 m3 sec-2) 13 = 3.45 × 10 sec2 13 6 so P = (3.45 × 10 sec2)1/2 = 5.88 × 10 sec = 68 days. (This differs slightly from the known value (about 67 days) because of rounding.)

8. The areas of Earth and Sun’s disks are just the area of a circle, with the radius of Earth or Sun. The ratio will be: Earth/Sun = (REarth2)/(RSun2) = (REarth / RSun)2.

The radius of Earth is about 1/100 the radius of the Sun, so Earth will block a fraction of about (1/100)2 = (10-2)2 = 10-4 = 0.0001 of the light, or 0.01%. The exact figure would be: 5 2 -5 (REarth / RSun)2 = (6378 km/6.96 × 10 km) = 8.4 × 10 , or 0.0084%. Jupiter’s radius is about 1/10 the Sun’s, so it would block a fraction of 0.01, or 1% of the Sun’s light. The exact figure for Jupiter is: 5 2 (RJupiter / RSun)2 = (71,492 km / 6.96 × 10 km) = 0.0106, or 1.06%.

Chapter 8

Survey of Solar Systems

Answers to Test Yourself 1. (c) Mercury, Venus, and Earth are rocky with iron cores.

2. (d) Most of the planets spin counterclockwise.

3. (a) The Doppler-shift method is effective at finding planets that are massive and near the parent star.

4. (d) The Transit method works best for edge-on systems, so the planet will eclipse some light from the star.

5. (e) The solar nebula theory has to explain all these properties.

6. (d) There was a period of heavy bombardment.

7. (b) The small, cool planet will retain the most atmosphere if they all have the same mass.

Chapter 9

The Terrestrial Planets

CHAPTER 9 THE TERRESTRIAL PLANETS Answers to Thought Questions 1. One example of a resonance is the use of a tuning fork: when the fork is struck, if its pitch matches that of the object being tuned, the sound is amplified. Another example is that soldiers do not march in step across bridges in case they might set up a resonance in the structure. A famous destructive example is an opera singer breaking a glass with her voice, and a less apocryphal one the Tacoma Narrows bridge collapse (caused by the wind).

2. Venus is an example of a runaway greenhouse effect. Carbon dioxide traps the reradiated sunlight (infrared) within the atmosphere and raises the surface temperature.

Early in its history, any surface water it possessed was evaporated and broken up by sunlight in the upper atmosphere. Thus, Venus has permanently lost its water. The surface of Venus is hotter than that of Mercury (hot enough to melt lead), and the atmospheric pressure is 100 times greater than that at sea level on Earth. If we were to increase the carbon dioxide content of Earth’s atmosphere (by an unknown amount), less infrared radiation could escape, causing global warming. This would start to melt the polar ice caps and raise the sea level; the water would evaporate, increasing the cloud cover and make the effect even stronger. Many scientists believe this is already occurring.

3. STUDENT OPINIONS. 4. If Olympus Mons were on Earth, it would not be as large, unless more energy were available to create it. The surface gravity on Earth is considerably stronger than on Mars, so more energy would be required to create a volcano so tall.

5. You would need to find areas that had the highest atmospheric pressure, values higher than the average. The places on Mars that would have the highest atmospheric pressures are those with the lowest altitudes, such as Hellas Planitia (which has the global lowest altitude), Utopia Planitia near the Viking 2 landing site, or the bottom of Valles Marineris.

6. STUDENTS SHOULD FORM AN ARGUMENT. Key points here will revolve around the two major issues: Venus has too much atmosphere and heat, and Mars has far too little. Can students make an argument for why it would be easier to reverse Venus’ greenhouse, in a reasonable time, or to overcome the problems of Mars’ low gravity and therefore permanent difficulty in retaining an atmosphere. Keep in mind that Venus also has no water, and adding water to Venus’ atmosphere—while it may remove CO2—will complicate greenhouse effect since water is also a greenhouse gas.

7. Venus, Earth, and Mars all have roughly the same internal structure and composition.

The key difference is that Mars is much smaller than Earth and Venus. Consequently, the pressure is much less in the interior of Mars, and the material there is not compressed the way it is inside Earth and Venus. Mars’s mass is only about 1/10 the mass of Earth, but it

Chapter 9

The Terrestrial Planets

has a volume which is closer to one-sixth or one-seventh (and it compares similarly to Venus). Since it is made of the same basic stuff, we can conclude it is not as compressed.

Also, the core of Mars is a bit smaller in comparison to the rest of the planet. (In fact the uncompressed density of Mars is 3.3 g/cm3, but the compressed density is 3.9; for Earth the values are 4.2 and 5.5, and Venus 5.25 and 4.2 g/cm3, but knowing these values requiring looking outside the text.)

8. Evidence for the solar nebula hypothesis: The terrestrial planets all orbit in the same plane and in the same direction. The decrease of temperature with increasing distance from the Sun determines where different materials can condense and the makeup of the planets and asteroids matches this well. The view that planets were assembled from smaller planetesimals is supported by the many craters seen on their surfaces—craters formed by the last stages of collisions. It is also supported by the compositions of asteroids. In accordance with the processes discussed in Chapter 8, any original atmospheres that small planets had were lost; the atmospheres on Venus, Earth, and Mars formed later from volcanic eruptions or the evaporation of cometary debris.

9. Planetesimals may have affected the terrestrial planets: the Caloris Basin appears to be the result of a major impact on Mercury, and affected the shape of the terrain all over that world; another impact may have essentially removed much of Mercury’s crust. A major impact on Venus may have flipped the planet to result in its retrograde rotation. The origin of Earth’s Moon is most likely from a major impact, which mixed the impactors’ material with Earth, enriched Earth’s iron content, and ejected material from Earth’s crust that was incorporated into the Moon.

Answers to Problems 1. To calculate the escape velocity to launch a rocket from the surface of Mercury, use the escape velocity formula: 1/2 Vesc = (2GM/R) .

Inserting the mass and radius of Mercury, we find that 1/2 Vesc = (2GM/R) -11

= (2 × 6.67 × 10

23

3

1/2

m3 kg-1 s-2 × 3.30 × 10 kg /(2440 km × 10 m/km))

3

= 4.25 × 10 m/s.

Or, scaling from Earth, Vesc,Mercury = Vesc,Earth(MMercury/MEarth × REarth/RMecury)1/2 = 11.2 km/s (0.055/1 × 1/0.382)1/2 = 4.24 km/s.

(Note that the values differ because of various intermediate rounding.)

2. CREATE DIAGRAM FOR STUDENTS. This is easier to do if you don’t focus on Venus’ very long rotational period and just draw the situation for a generic retrograde planet, hence the suggestion of using a rotation period of about half the orbital period.

Since the direction of rotation is opposite the direction of revolution, starting out from

Chapter 9

The Terrestrial Planets

noon, the planet will rotate into a noon alignment again before rotating a full 360° around. (This is a lot like Thought Question 10 in Chapter 1).

3. When Venus is at inferior conjunction, it is directly between Earth and the Sun. The distance to Venus is the distance of Earth to the Sun minus the distance of Venus to the Sun: d = 1.000 AU – 0.723 AU = 0.277 AU. 11 10 = 0.277 AU × 1.50 × 10 m/AU = 4.16 × 10 m.

Once we know the distance, it’s just a matter of applying t = D/V, with V = c since radio signals travel at the speed of light.

10 8 t = (4.16 × 10 m) / (3 × 10 m/sec) = 139 sec.

The signal will take 139 sec, or about 2.3 minutes (2:19) to reach the spacecraft.

Pressure (% of surface value)

4. For the graphical approach, this is the result for the last few values:

Height (km)

The intersection of 0.6% and the curve happens at about 37 km. The full table of values is on the next page. The result of 37 km can also be estimated from looking at the table.

For the approach using logarithms, 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, corresponding to an altitude of 7.4 × 5 km = 37 km.

Chapter 9

The Terrestrial Planets

Height (km) 0 5 10 15 20 25 30 35 40 45

Pressure (% at surface) 100 50 25 12.5 6.25 3.125 1.5625 0.78125 0.390625 0.1953125

5. From Chapter 3, surface gravity is given by g = GM/R2. gsurface

gtop of Mons

= GM/R2 -11 23 3 = 6.67 × 10 m3 kg-1 s-2 × 6.42 × 10 kg / (3396 km × 10 m/km)2 = 3.71 m/s2 = GM/R2 -11 23 3 = 6.67×10 m3 kg-1 s-2 × 6.42×10 kg /[(3396 km+ 26 km) × 10 m/km]2

= 3.66 m/s2 This is a small but easily measurable difference. The difference is about 1.3% of the surface value: (3.71 – 3.66) / 3.71 × 100% = 0.05 / 3.71 × 100% = 1.35%.

6. Brightness decreases with the square of the distance from the source. Venus is at 0.723 AU, Earth is at 1 AU. The Sun will be brighter at Venus by (1/0.723)2 = 1.91 times, or almost twice as bright. Mars is at 1.524 AU; so compared to the brightness at Earth the Sun will be (1/1.524)2 = 1 / 2.32 = 0.431 times as bright as at Earth, or in other words, a little less than half as bright.

Answers to Test Yourself 1. (d) Mercury’s surface was not “active” long enough to erase the impact craters; Earth’s surface is still active with erosion and tectonic activity.

2. (a) Mercury has a relatively large iron core.

3. (c) Venus has a runaway greenhouse effect because of the CO2 in its atmosphere.

4. (d) Spectra have revealed that Venus’s clouds are droplets of sulfuric acid.

5. (b) Time can be estimated by counting craters.

6. (a, c, e) Evidence of liquid water once being present on Mars is provided by branching channels that appear to be riverbeds, teardrop islands, and compounds that form in reactions that occur in the presence of water.

Chapter 9

The Terrestrial Planets

7. (b) Venus has an atmospheric pressure about 100 times greater than that of Earth.

8. (a,b) All the terrestrial planets have an iron core and a silicate mantle.

Chapter 10

The Outer Planets

CHAPTER 10 THE OUTER PLANETS Answers to Thought Questions 1. If Jupiter were closer to the Sun it would become hotter. The hydrogen and helium in the atmosphere would move faster, causing some of it to be lost. With enough heating, it might evaporate down to a rocky/iron body similar to, but larger than Earth. [Consider the implications of this idea, in the context of migrating planets].

2. Europa’s and Enceladus’s relatively crater-free surfaces imply that some process has resurfaced them and destroyed old craters. One possibility is that liquid water has erupted from their interiors and flooded parts of their surfaces, covering old craters, and then freezing into a new solid surface.

3. Overall it would be a lot like having a comet in orbit. The average densities of Callisto and Ganymede are 1.94 and 1.85 g/cm3, and Tethys and Dione, which appear in Figure 10.23 to strongly resemble our Moon, have an average density even less (0.98 and 1.49 g/cm3). The Moon’s average density is 3.34 g/cm3. The Moon is made of rock; the gas giant moons are made largely of ices. If one were in orbit around Earth, some of these ices would melt from the combination of increased sunlight and also perhaps from tidal heating from interaction with Earth. Exactly how much melting would occur would depend in part on how effectively the surfaces absorbed sunlight (the average temperature on the Moon is still less than freezing) but once the process started, it would releasing methane, ammonia, and water.

Evaporated water, methane, and/or carbon dioxide if any formed or was present could potentially form a thin atmosphere that could create a greenhouse effect to further melt the surface and replenish the atmosphere. The surface features would disappear as the surface melted, and some of the gases would be lost to space, much like a comet tail, which would be quite a sight. The moons would show phases and cause tides similar to our Moon. With a mass about twice that of the Moon, Ganymede would cause tides twice as strong at the same distance. Callisto is only about 40% more massive than the Moon so its tides would be a little stronger. Tethys and Dione each have about 1% as much mass as our Moon, so their tides would be very weak (weaker, in fact, than the Sun’s). Over time, the objects would shrink as their icy layers evaporated, leaving behind a much smaller rocky core.

4. The Roche limits are 2.44 times the planet’s radii: Planet Limit Interior Moons in Table A.5 Saturn – 147053 km Pan, Atlas, Prometheus, Pandora Uranus – 62364 km Cordelia, Ophelia, Bianca, Cressida Objects can survive inside the Roche limit if they are held together by a force other than their own gravity. These moons would be expected to 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.

Chapter 10

The Outer Planets

5. Titan’s features are similar in shape to those found on Earth. More or less, material in the atmosphere appears to “rain” out of clouds and drain across the surface, reshaping rocks and forming rivers and lakes. Winds push the dunes around. The “volcanoes” erupt gases and material which resurfaces the moon. However, all of this occurs at much lower temperatures than for the terrestrial analogs. The rocks are made of very cold ice. At such temperatures, ice is as hard as steel or rock on Earth. The liquid in the lakes is made of hydrocarbons— ethane—not water (oxygen and hydrogen). Most different of all, the volcanoes do not erupt hot, molten rock, but are eject more or less cold molten hydrocarbons and water (though they are hotter than the surrounding surface and atmosphere). The “volcanoes” are more like Earth’s geysers than its volcanoes.

6. Earth’s sky is blue because small particles in the atmosphere scatter short wavelengths of light (blue light) more effectively than they scatter longer wavelengths (red light). The effect is most pronounced at sunset: the Sun appears reddish, because blue light is scattered out of the line of sight, but the opposite side of the sky looks blue (scattered sunlight). The situation is similar on Uranus, but the scattering particles are methane crystals, not dust, and that causes an important difference. As seen from outside, ice crystals of methane in the atmosphere scatter and reflect blue sunlight, making the atmosphere appear blue. However, the ice crystals also strongly absorb the red photons, unlike the scattering particles in Earth’s atmosphere.

Whereas the Sun appears redder from Earth’s surface because blue light is scattered out of the line of sight, from deep in Uranus’ atmosphere, the Sun should actually appear bluer than it would otherwise because the red photons have been absorbed. However, it seems unlikely that one would be able to see the Sun itself through the methane-ice clouds and haze. The scattering of blue light by the ice crystals suggests that at least in the upper atmosphere, the sky would appear blue from inside as well—but again, hazy and cloud-filled.

7. Uranus’s moons orbit its equator. Though tilted the system is extremely regular. The moons are made of more or less the same kinds of material as other gas giant moons (and gas giants, minus the hydrogen and helium)—frozen volatiles and rock. Together these details imply that, similar to how Earth’s moon was formed, material that splashed out of Uranus during a major collision coalesced into the orbiting moons. If the moons had formed prior to the collision, tidal forces and the angular momentum of the Uranus-moon systems would have opposed the forces in the collision trying to “tip” the planet (think of tilting a gyroscope).

Some of the moons might have been ejected and others could have irregular orbits. [However, if Uranus was tipped slowly, through gravitational interaction with Saturn, the same tidal forces might have helped also slowly tip the moons’ orbits.]

8. Uranus and Neptune are less massive than Jupiter and Saturn, and have lower escape velocities. Even though these planets are colder, the lighter gases can still escape their weaker gravitational fields.

9. Wind systems on the gas giants are driven by heat escaping from the interior; the bands result from the planets’ rapid rotation and the Coriolis effect. Neptune, being more massive, should have more internal heat than Uranus, but it is also farther from the Sun, so it is colder at the surface. This large temperature difference drives the circulation more strongly than it

Chapter 10

The Outer Planets

would be driven on Uranus. Uranus, being tipped, has disruptions to this kind of heat-flow driven circulation because of its extreme seasons. It seems possible that Uranus might exhibit atmospheric patterns more like the other gas giants during the parts of its orbit when one hemisphere is not perpetually pointed at the Sun. 10. STUDENTS’ OPINIONS WITH EXPLANATION. Answers to Problems 1. The time, t, it takes light to travel a distance, D, is t = D/c, where c is the speed of light, 3 5 × 10 km/s. Using the average distances from the Sun from the Appendix, Jupiter: Uranus:

6

5

3

t = D/c = (778.3 × 10 km) / 3 × 10 km/s = 2.5 × 10 s = 42 minutes 6 5 3 t = D/c = (2870 × 10 km) / 3 × 10 km/s = 9.6 × 10 s = 160 minutes

Since it would take twice this time to know if a command sent to a satellite was received and acknowledged, clearly equipment sent to the outer solar system needs to be highly automated— satellites could move pretty far off course just during the time to send and receive one command! Likewise, manned missions would have to make time-sensitive decisions on their own.

2. Again, we need to use t = D/v. This time the distance traveled is Jupiter’s circumference at the equator; the time is the period of rotation.

Distance = circumference = 2πR = 2π × 71492km = 449197 km. v = D/t = 449197 km / 9.9 hr = 45373 km/hr.

3. Let’s use Europa to calculate Jupiter’s mass using the moon’s orbital data. The modified form of Kepler’s third law is M = 4d3/GP2 with M in kg, P in seconds and d in meters. For 3 Europa, d = 671 × 10 km and P = 3.55 days × 24 hr/day × 3600s/hr = 306,720 s.

M = 4d3/GP2 3 3 -11 = 4 (671 × 10 km × 10 m/km)3 / (6.67×10 m3kg-1s-2 × 3067202s2) 26 = 4 (×10 )/(6.27 kg-1) 27 = 1.9 × 10 kg This is the same value as given in the Appendix for the mass of Jupiter.

4. There are several ways to solve this problem. First, through direct calculation: 3

26

29

Jupiter’s density, , is 1.33 g/cm . Saturn has a total mass of 5.68 × 10 kg = 5.68 × 10 g. If Saturn was a sphere with Jupiter’s average density, the volume of that sphere would be 29 3 29 3 V = M/ = 5.68 × 10 g / (1.33 g/cm ) = 4.27 × 10 cm .

The radius of a sphere of this volume is: 3 29 3 3 9 r = [(3V/(4)] = [(3 × 4.27 × 10 cm /(4)] = 4.67 × 10 cm = 46,700 km.

Jupiter’s radius is 71,492 km, so Saturn’s radius would be about 65% as large. (Saturn’s actual radius is 60,268 km, so it would be 77% the size it currently is.)

Chapter 10

The Outer Planets

The answer can also be found by using proportional reasoning. If Saturn and Jupiter have the same density, then the volume will depend only on the mass. The volume of a sphere grows as radius cubed. The ratio of Saturn’s mass to Jupiter’s mass is 26 27 26 26 5.68 × 10 kg / (1.90 × 10 kg) = 5.68 × 10 kg / (19.0 × 10 kg) = 5.68 / 19 = 0.2989.

Therefore, if Saturn had Jupiter’s density, it’s volume would be 0.2989 times the volume of 1/3 Jupiter, so the radius would be (0.2989) = 0.669 times the current radius of Jupiter, about 71,492 km × 0.669 = 47800 km.

Alternatively, one can start with the ratio of densities, with the ratio of Jupiter’s density to 3 3 Saturn’s being (1.33 g/cm )/(0.69 g/cm ) = 1.927. For Saturn to have the same density as Jupiter, it would be 1.927 times the old density. Since Saturn’s mass is assumed to stay the same, Vnew = Vorginal/1.927. Since V= 4/3r3, and volume depends on the cube of the radius, rnew = roriginal/(1.927)1/3 = 0.804 roriginal. Thus, the new radius would be about 80 percent of Saturn's current radius, or about 60,268 km × 0.804 = 48,500 km. This is 68% of Jupiter’s radius.

The slightly different answers are the result of rounding and the way that the assumption that Jupiter and Saturn are perfect spheres (which is not true) affects the individual calculations.

5. Saturn is 8.5 AU from Earth.

11 9 d = 8.5 AU × 1.5 × 10 m/AU × 1km/1000m= 1.275 × 10 km The rings are 270,000 km in diameter. The angular size formula is 9 -2 A = (360°)(D)/(2d) = (360°) (270000 km)/( 2 × 1.275 × 10 km) = 1.21 × 10 ° At 60 arc minutes per degree, this is 0.67 arc minutes, or 40 arc seconds.

6. Calculate the period of Saturn’s inner and outer rings: a = 90,000 km and 136,000 km.

26 Saturn’s mass is 5.68 × 10 kg.

P2 = 4d3/(GM) -11 = 4 (90000 × 1000 m)3 / (6.67×10 m3kg-1s-2 × 5.68×1026 kg) 8 = 2.418 × 10 s2 P = 15,550 s = 4.3 hours P2 = 4d3/(GM) -11 = 4 (136,000 × 1000 m)3 / (6.67×10 m3kg-1s-2 × 5.68×1026 kg) 9 = 2.62 × 10 s2 P = 51,197 s = 14.2 hours The rings can’t be solid: all parts of a solid object have to have the same period or it would shear apart.

7. Assume Enceladus is a sphere.

V = 4/3πR3 = 4/3π(499 km/2)3 = 4/3π(499 × 105 cm/2)3 = 6.51 × 1022 cm3 D = M/V = (1.08 × 1023 g) / (6.51 × 1022 cm3) = 1.66 g/cm3.

Chapter 10

The Outer Planets

The density is 1.66 g/cm3. This is a higher density than ice, indicating that in addition to icy materials, Enceladus must have some silicates and possibly even some iron. The rock and/or iron in Enceladus’ apparent composition may contain some radioactive isotopes that provide some internal heat in the form of radioactive decay. This could provide the energy to create tectonic activity on the surface, explaining areas of smooth terrain that apparently have been (relatively) recently renewed.

8. Use the equation from Chapter 3 to find the surface gravity of Neptune. Insert Neptune’s mass and radius from the appendix: 26 M = 1.02 × 10 kg. R = 24,764 km gsurface = GM/R2 -11 26 3 = 6.67 × 10 m3 kg-1 s-2 × 1.02 × 10 kg / (24,764 km × 10 m/km)2 = 11.1 m/s2 This is pretty close to the surface gravity on the surface of Earth (9.81 m/s2).

Answers to Test Yourself 1. (c) They are made of hydrogen, helium, and hydrogen-rich compounds.

2. (d) Rock and iron. It is likely to have a “small” core, similar to a terrestrial planet, but under huge pressure and high temperatures.

3. (d) Inside the Roche limit, tidal stresses will break apart a body that is only held together by gravity.

4. (e) The source is gravitational energy released by sinking material.

5. (b) The rotation axis has a large tilt.

6. (a) Miranda has a wide variety of difficult-to-explain surface features.

7. (a) A day. All the gas giants have rotation periods less than Earth’s.

Chapter 11

Meteors, Asteroids, and Comets

CHAPTER 11 METEORS, ASTEROIDS, AND COMETS Answers to Thought Questions 1. Both are debris, and both are heated debris, but the tails are not the same material or under the same conditions. The tail of a meteor is the material of the meteor heated by friction and vaporized as it falls into Earth’s atmosphere—it glows like a blackbody.

Comets have two tails, and neither are in Earth’s atmosphere. The dust tail is made of dust particles separated from the comet when the comet’s gravitational force is less than the force they experience from photons hitting them (light has energy and momentum even though it does not have mass). It reflects sunlight, but is not actually glowing. The gas/ion tail is pushed out by pressure (friction/collisions) from the solar wind. This is not the same as what happens to the meteor. The ion tail glows primarily because of fluorescence with ultraviolet light from the Sun, not because it is hot.

2. Construct a right triangle as below:

Neglecting the effect of the meteor moving horizontally, when you see it, if the meteor is 45° above the horizon, and 100 km up, you must be 100 km away from being directly below it. Someone closer will see it higher in the sky, and someone farther will see it lower in the sky. The distance of the observer is 100 km/tan(altitude), so there is some latitude in this question based on the estimate of what altitude the meteor can have and still be seen. If we estimate it would be hard to see the meteor if it was only 10° up, an observer 567 km away could see it, making the farthest distance between observers about 1134 km if two observers were on opposite sides.

Other factors students might consider: the speed of a meteor might be within a factor of two of 30 km/s (see Problems), so a meteor visible for five seconds could move a quarter (or more) of the distance to the 10° observer, affecting the estimate. We also have not considered the brightness of the meteor. A small meteor might be hard to distinguish from the sky brightness at a moderate distance.

3. Even with more mass in the asteroid belt, a planet could not form because of the gravitational interactions with Jupiter. Jupiter has already ejected considerable amounts of material from the asteroid belt.

Chapter 11

Meteors, Asteroids, and Comets

4. In the satellite picture, if the boats appear as points, the only way to estimate the size of the boat would be to assume that a larger boat would reflect more light and measure the brightness or size of the dot. However, different color boats reflect different amounts of light—a very large boat painted black might reflect less light than a small boat that was very white. If the coast guard used infrared images, however, the light would not be reflected. The boats are actually glowing—producing their own light—in the infrared. A bigger boat would produce more light than a smaller boat, even if they are the same temperature. The situation is essentially identical to studying asteroids—when looking with optical light, the reflectivity (albedo), as well as the size, affects the amount of light reflected. However, for two asteroids at the same temperature, the number of infrared photons will depend only on the sizes. (And even if they are at different temperatures, if you can get a spectrum or the intensity at a few wavelengths you could still determine the sizes as well as the temperatures!).

5. The composition of asteroids varies more or less smoothly from the inner edge of the belt to the outer edge of the belt, transitioning from rocky bodies with iron and silicates near Mars to objects with more carbon and volatiles and lower densities near Jupiter. The composition of the volatile-rich asteroids is more similar to Jupiter’s moons than Jupiter itself, since of course asteroids are not large enough to accumulate hydrogen and helium.

The composition trend traces what you would expect from condensation at different distances according to the solar nebula hypothesis. The existence of differentiated and undifferentiated bodies also supports the hypothesis by providing examples of planetesimals of various sizes, as does the presence of such objects at a radius where a planet would be expected (e.g., Bode’s Rule) but was not able to form because of Jupiter.

6. STUDENTS SHOULD MAKE UP CRITERIA. Some astronomers think that the planet classification is not well defined and that Pluto was reclassified mainly because of its size. Pluto has moons. Pluto has a spherical shape. Eris and Sedna are also spherical, and in fact Eris is larger than Pluto so it would also have to be a planet if size were the only criterion.

7. STUDENTS SHOULD JUSTIFY ANSWER. Certainly some of the craters in the pictures look round and resemble normal craters, and comets could no doubt occasionally collide with small bits of other comets, asteroids, or bits of themselves that have broken off. However, comets are also subject to extremes of heating and cooling and many of the features are probably a result of disruptions to the surface from explosive loss of material, expansion and contraction.

8. STUDENTS SHOULD PROVIDE REASONED OPINIONS. Answers should consider the relative energy of a bomb and the kinetic energy of the asteroid (in qualitative or quantitative terms). Students might compare the situation to mounting a rocket on the asteroid, or trying to break it in to pieces.

Chapter 11

Meteors, Asteroids, and Comets

Answers to Problems 1. Earth travels the circumference of its orbit in a year, so 8 V = 2 AU / 365 days = 2 AU × (1.5 × 10 km/AU) / (365 d × 24 h/d × 3600s/h) 8

7

= 9.42 × 10 km / 3.15 × 10 s ≈ 9/3 × 10 km/s = 30 km/s.

(9.42 / 3.15 = 2.99) 2. Calculate the surface gravity and escape velocity for Ceres, assuming a radius of 20 487 km and a mass of 9.43 × 10 kg.

-11 20 3 gsurface = GM/R2 = 6.67 × 10 m3 kg-1 s-2 × 9.43 × 10 kg / (487 km × 10 m/km)2 = 0.265 m/s2 1/2 Vesc = (2GM/R) -11

20

3

1/2

= (2 × 6.67 × 10 m3 kg-1 s-2 × 9.43 × 10 kg / (487 km × 10 m/km)) = 508 m/s = 0.508 km/s (On Earth it’s close to 11,000 m/s.)

3. The semi-major axis is half the sum of the nearest and farthest points, a = (0.19 AU + 1.97 AU)/2 = 2.16AU/2 = 1.08 AU. Since Icarus’ orbit extends farther than Earth’s orbit, it must cross the radius of Earth’s orbit (although not necessarily the orbit, depending on the inclinations) twice per orbit; once heading sunward and once heading away. Use Kepler’s 3rd Law to find the period, P = a3/2 = (1.08)3/2 = 1.12 yr, about 410 days. The average time between crossings is therefore is 205 days. Since Icarus’ aphelion point is about twice as far from the Sun as Earth’s, the crossing points are near the middle of Icarus’ orbit geometrically. However, since the asteroid travels faster when it travels the half of the orbit closer to the Sun, the time between the crossing inside Earth’s orbit to crossing outside will be a little shorter than 205 days, and the time spent outside Earth’s orbit a little longer than 205 days.

4. Pluto’s orbital period is 247.68 yrs. Neptune’s is 164.79 yrs. Dividing one by the other gives 247.68/164.79 = 1.5030 . . . which is very close to 3/2 = 1.5.

5. Use the modified form of Kepler’s Third Law to find the mass of Pluto and Charon. m + M = 4d3/GP2 Let M be Pluto’s mass and m be Charon’s. The period of the orbit is given as 6.387 days (which is 551,837 seconds), and the semimajor axis as 19591 km 4 3 -11 = 4 (1.9591 × 10 km × 10 m/km)3 / (6.67×10 m3kg-1s-2 × 551,8372 s2) 7 = 4 (×10 )3/(20.31 kg-1) 22 = m + M = 1.46 × 10 kg 22 20 In the Appendix, Pluto has a mass of 1.31 × 10 kg and Charon has a mass of 15.2 × 10 22 22 22 kg, so this agrees extremely well: 1.31 × 10 kg + 0.152 × 10 kg = 1.462 × 10 kg.

Chapter 11

Meteors, Asteroids, and Comets

6. To find the density of Charon we use the definition that density = mass/volume. We can find the volume by assuming Charon is a sphere and using the expression that the 3 volume of sphere is 4/3R . The radius of Charon is 600 km. Its volume is 3 7 3 23 3 4/3R = 4/3 (6.00 × 10 cm) = 9.05 × 10 cm 24

The mass of Charon is 1.5 × 10 g. Dividing this number by the volume gives the density, or, Density = M/V 24 23 3 = 1.5 × 10 g / (9.05 × 10 cm ) 24 24 3 3 = 1.5 × 10 g / (.905 × 10 cm ) = 1.5/0.905 = 1.66 g/cm .

This density suggests that Charon is made of ice (density ~ 1 g/cm3) and rock (density ~ 3 g/cm3). Since iron has a much higher density, there cannot be much of the metal present and Charon is unlikely to have an iron core (at least not one of much size).

2

3

7. Using Kepler’s Third Law, P = a , with P in years and a in AU, the semi-major axis of Swift-Tuttle must be a = P2/3 = 1332/3 = 26 AU. For the comet to cause the Perseid meteor shower, the orbit must intersect Earth’s.

Therefore, the perihelion passage must be at most 1 AU from the Sun, probably less. We can estimate the aphelion distance as approximately twice the semimajor axis, then, at 51 AU. (a = (perihelion + aphelion)/2, so aphelion = 2a – perihelion. If perihelion is small, then aphelion ~ 2a. ) The Kuiper belt, the source of short term comets, begins at about 50 AU. 2

3

8. Using Kepler’s Third Law, P = a , with P in years and a in AU, the semi-major axis of the comet must be 7 P = a3/2 = 500002/3 = 1.12 × 10 yr.

This is about 11 million years.

9. (The exact answer will depend on what distance the student chooses for the Oort cloud. A distance of 100,000 AU is shown in Figure 11.23) The temperature of a body decreases as the square root of its distance from the Sun. The temperature at 1 AU is 300 K. A comet that is in the Oort cloud might be at a distance of about 100,000 AU. Therefore, the change in the radius is a factor of ROort/R1AU = 100,000 /1 = 100,000 = 105.

The square root of this factor is (105)1/2 = 316.

Chapter 11

Meteors, Asteroids, and Comets

Since the temperature decreases as the radius increases, the temperature must drop by a factor of 316 times. If the temperature at 1 AU is 300 K, this means the temperature at the Oort cloud is about 1 K (300 K/316 = 0.94 K).

10. At 65 miles per hour, the SUV is going 29 m/s: 65 miles/hr × 1hr/3600s × 1.609 km/1mi × 1000m/km = 29 m/s.

2

6

The kinetic energy of an SUV is E = ½ mV = ½ (2700 kg) (29m/s)2 = 1.14 × 10 J The kinetic energy of the meteoroid is 2 6 E = ½ mV = ½ (0.010 kg) (30 km/s × 1000m/km)2 = 4.50 × 10 J. The meteoroid has about four times the kinetic energy of the SUV on the highway!

11. Calculate the energy of an impacting body from the expression for kinetic energy, 2 E = ½ mV .

4 m = 1000 kg and V = 30 km/s = 3 × 10 m/s (and 1 kg m2/s2 is a Joule).

4 2 11 E = 1/2 × 1000 kg × (3 × 10 m/s) = 4.50 × 10 J. To express this value in kilotons, divide it by the number of joules per kiloton: 11 12 4.50 × 10 J × 1 kiloton / (4 × 10 J) = 0.1125 kilotons The 1000 kg object impacts with an energy of just over a tenth of a kiloton.

Answers to Test Yourself 1. (d) Frictional heating. The meteoroid burns up on entry into Earth’s atmosphere forming a “shooting star” or meteor.

2. (b) Meteor showers are caused by Earth passing through cometary debris.

3. (a,c) Most meteorites are from comet and asteroid debris. Relatively few are pieces of Mars or the Moon.

4. (d) The asteroid belt is at 2–4 AU, between the orbits of Mars and Jupiter.

5. (d) Asteroid sizes are estimated based on the light they reflect.

6. (e) All of the above are possible depending on where the asteroid was formed.

7. (c) The mass of Pluto was determined by measuring the orbit of Charon.

8. (e) Both (b) tail always points away from the Sun and (d) composed of gas and dust evaporated from the nucleus are correct.

9. (a) Short period comets have orbits of decades and come from the Kuiper belt.

10. (c) Kinetic energy depends on velocity and mass.

11. (a,d,e) Evidence includes mass extinctions 65 million years ago, a layer of a rare element (iridium) and a layer of soot in the rock record.

Chapter 12

The Sun, Our Star

CHAPTER 12 THE SUN, OUR STAR Lecture Suggestions • Hydrostatic equilibrium and energy generation via fusion are concepts that are important in later chapters as well.

• Show images of solar features, videomagnetograms, and time-lapse movies of prominences, flares, and/or transits. Solar observatories such as the Big Bear Solar Observatory, Kitt Peak, and others have libraries as well as daily images on the web.

• You can use balloons and liquid nitrogen to illustrate perfect gas law.

• Magnets with Velcro taped to them can be used to “demonstrate” how the strong force overcomes the electrostatic repulsion when nuclei get close enough (nuclear fusion).

• Point out when describing sunspots that students can see the effect of a cooler temperature making a small region darker by sprinkling a few tiny drops of water on a glowing electric-stove burner. As the water evaporates, it cools the burner briefly and creates a short-lived dark spot.

• You can demonstrate electromagnetic force by using demos of the type where you generate a current by moving a magnet through a coil.

Answers to Thought Questions 1. Although the Sun’s corona is extremely hot, it is very tenuous. This hot, rarefied gas produces emission lines, not continuum like the photosphere. Even though the gas is very hot and energetic, there is relatively little matter in it and so it contains relatively little heat energy overall. It is also very far away (8 light minutes) and is not in direct contact with Earth.

2. Hydroelectric power is generated by the motion of falling water spinning turbines connected to generators. The Sun’s heat evaporates water, lifting it into the atmosphere. The lifted water can then fall as rain to supply rivers and lakes above a dam. When the water trapped behind a dam flows through the turbines, it generates electricity. The Sun’s energy lifted the water and therefore the Sun’s energy indirectly powers the turbines and generates the electricity.

3. Sailing uses wind as a power source, and the wind results from temperature differences in our atmosphere created by solar heating.

4. Scientists have shown that chemical processes cannot produce sufficient energy to support the Sun against gravitational collapse for long enough (4.5 billion years) to explain the age of the Solar System.

5. In other differentiated bodies in the solar system, dense material is able to sink past lighter material and the centers of the bodies are solid. In the Sun, a few things are different. First, the internal pressures and temperatures are far, far greater than in planets. The material at the center of the Sun, even though it is hydrogen and helium, is at a density of 160 g/cm3, many times denser than iron or lead, and it is still a gas because the temperature is tens of millions of K. The pressure of this gas holds up the Sun. The temperatures throughout the Sun are high enough to prevent the formation of molecules. In fact, not only would molecules that formed be immediately destroyed by collisions, but the temperatures are high enough that the

Chapter 12

The Sun, Our Star

atoms themselves are ionized. So large pieces of iron do not accumulate. Additionally, material in the radiative zone above the core does not move (of course, material does circulate in the convective layer above). So even though the Sun is hot and gaseous, material is unable to simply sink to the center. Differentiation does not occur in the same way as it would in a planet. (We will see later that in much larger stars than the Sun, iron can collect in the core, but this is the result of fusion creating it there, not differentiation).

6. Students should try to include relevant physics (appropriate to their level of knowledge).

To determine the structure of a star, you need to include gravity, the ideal gas law, the generation of energy from mass at the core, transport of energy outward, and the loss of energy at the surface (luminosity equation). More complex factors to include would be the opacity and composition of the material. To actually model the core you need to include the nuclear reactions that convert mass to energy, the rates of those equations, and the amounts of matter available. A typical stellar structure model includes differential equations to determine the pressure, density, temperature, and luminosity in successive spherical shells, combined with boundary conditions at the surface and core.

7. Wien’s law only works on objects emitting a thermal continuum spectrum (a blackbody).

Prominences and the corona are made of optically thin gases and produce line emission and non-thermal emission (from electrons spiraling in the magnetic field).

8. Consulting figure 12.12 or the equations in section 2, we see that the photons produced directly by the p-p chain (and subsequent positron annihilations) are all gamma rays. The spectrum would be a line spectrum of gamma rays, perhaps with some Doppler broadening because of the high temperatures in the core. The spectrum would not be continuous. (The question is artificial because the core must be extremely hot for fusion to occur, and therefore the particles have a high kinetic energy; realistically, the core would also produce a blackbody spectrum for a very high temperature but this is not a direct product of the p-p chain. We are imagining that the released energy does not interact with the matter in the core.)

9. If the energy blocked by sunspots was not released elsewhere on the Sun, it would build up inside the Sun. This would change the hydrostatic equilibrium, at least near the surface.

The trapped energy would heat the upper layers, and the heat energy would be used to push the outer layers out against gravity, making the Sun a little bigger. Similar changes happen as stars age (red giants) or in variable stars, which change their size in a regular pattern.

Answers to Problems 5 1. Exact and approximate answers will be shown. The radius of the Sun is 6.96 × 10 km, and the radius of Earth is 6378 km.

6.96 105km 6.96 R = 102 = 1.09 102 = 109.

Exactly, Sun = 3 6.378 10 km 6.378 REarth Rather approximately,

RSun 6.96 105km 6.96 = = 102  1102 = 100 .

3 REarth 6.378 10 km 6.378

Chapter 12

The Sun, Our Star

Since V = (4/3)R3, the exact calculation is 4 R3  R 3 V 3 6 Sun Sun 3 = 4 R3 =  R Sun  = 109 = 1.30 10 , V Earth

3

Earth

Earth

or, the Sun’s volume is approximately a million times (1003 = 106) that of Earth’s.

2. We can find the Sun’s linear diameter, D, using the relation between angular diameter,  true diameter, D, and distance, d, from Chapter 1: /360° = D/2d.

Solve for D by cross multiplying to get D = 2d  360°.

5 The Sun’s angular diameter, , is about 0.5 degrees and its distance d = 1.5 × 0 km.

8 6 D = 2 × 1.5 × 10 km (0.5°) / 360° = 1.31 × 10 km.

5

5

(This diameter makes the radius 6.55 × 10 km. The actual value is 6.96 × 10 km. The difference occurs because we used a rounded off value for the Sun’s angular diameter.)

3. We can calculate the mass of the Sun, M, using the modified form of Kepler’s third law, 3 2 m + M = 42d /(GP ), where m is Neptune’s mass, and d and P are Neptune’s distance from the Sun (30.05 AU) and orbital period (164.8 yr), respectively. First, we need to change the units to meters and seconds, 11

12

-11

9

a = 30.05 AU = 30.05 AU × 1.50 × 10 m/AU = 4.508 × 10 m 9 P = 164.8 yr = 164.8 yr × 365 d/yr × 24 h/d × 3600 s/h = 5.20 × 10 sec 2

m + M = 42 × (4.508 × 1012 m)3/(6.67 × 10 m3kg-1s-2 × (5.20 × 10 sec) ) 30 = 2.00 × 10 kg.

(The difference from the exact value is a result of rounding numbers that are cubed and squared and approximating Neptune’s mass as negligible.)

4. Show that the Sun’s surface gravity is about 30 times Earth’s (9.81 m/s2). Surface gravity is given by gsurface = GM/R2 . Create a ratio. An exact calculation is not required here.

2

R   1.989 1030 kg   1  g GM / R 2 M Sun Sun Sun Sun Earth = GM   R  =  5.974 1024 kg    100  g /R 2 = M Earth Earth  Sun     20Earth 29Earth   10 −24 10−4  3105−4 = 310 = 30    6 

2

5. Using the expression for escape velocity and the mass and radius of the Sun from the Appendix, we have 1/2 -11 30 8 1/2 Vesc = (2GM/R) = (2 × 6.67 × 10 m3kg-1s-2 × 2 × 10 kg / (6.96 × 10 m))

Chapter 12

The Sun, Our Star

5

= 6.17 × 10 m/s = 617 km/s.

The rising material in the granulation is moving at only about 1 km/s, so the surface of the Sun isn’t close to moving fast enough to fly off (although some individual particles probably are). If the escape velocity was the same speed as the rising material, much of the material coming up in the center of the granulation cells would be ejected from the surface.

6. Note: this problem is an approximation—it uses an approximate form of the ideal gas law, such that pressure is given by P = 8300 T, where  is the mass density of the gas. This requires quantities be in specific units as given in the problem, so units are left off in what follows.

The definition of density tells us that 3  = mass/volume = M/(4/3R ).

Pressure forces from interior are given by 2 Fpressure = P × Area = P × R , where we take the area to be the cross-sectional area of the Sun. Substituting the expression for the pressure into this expression gives 2 Fpressure = 8300 T × R .

Now substituting the definition of density into this expression gives 2 3 Fpressure = 8300 MTR /((4/3)R ) = 8300 × (3/4) × MT/R = 6225 MT/R. We next use the condition of hydrostatic equilibrium, which says that we must equate the pressure force and gravitational force. The question says to assume the Sun is split into 2 equal halves, a distance 1 R apart. The gravitational force between any two masses is 2 GMm/R . For the two halves of the Sun, this gives a gravitational force, 2 Fgrav = G (M/2)(M/2)/R For the Sun to be in hydrostatic equilibrium, Fpressure = Fgrav 2 2 6225 MT/R = GM /(4R ).

Solve this expression for T, and substitute: -11 30 8 6 T = GM/(4R × 6225) = 6.67 × 10 × 2 × 10 /(4 × 6.96 × 10 × 6225) = 8 × 10 K This is about half the 15 million K value expected. Given the approximations in this simple model, we shouldn’t be surprised that we don’t get exactly the right answer.

7. To calculate the Sun’s density (), divide its mass, M, by its volume. Assuming it is a sphere, its volume is 4R3/3. Thus  = (4R3/3). Now insert the values of M and R to find  = 2 × 1033 g/[4(7 × 1010 cm)3/3] = {2/[(4) × (7)3]} × 1033 g/1010 cm)3 = 0.0014 × 1033-10×3 g/cm3 = 1.4 g/cm3.

Jupiter’s density is 1.33 g/cm3, so the Sun’s average density is very close to Jupiter’s.

8. To find the energy that would be released by converting a 60 kg person entirely into energy, we just use E = mc2.

Chapter 12

The Sun, Our Star

E = (60 kg) (3 × 108 m/sec)2 = 5.4 × 1018 J. (This is an enormous amount of energy, about a million kilotons! Looked at another way, this is a little less than half the annual energy consumption of the US in 2009!)

9. The Sun’s total energy output is 4 × 1026 W = 4 × 1026 J/s. According to the chapter, 4.3 × 10-12 J are released from each p-p chain of reactions. To produce 4 × 1026 J each second requires (4 × 1026 J/sec) / (4.3 × 10-12 J/p-p chain) = 9.3 × 1037 p-p chains / sec.

10. Estimate the lifetime of the Sun. 10% of the Sun’s mass is available for fuel, 0.1 × 1.989 × 1030 kg = 1.989 × 1029 kg.

In each p-p chain, 6.6793 × 10-27 kg of this fuel is processed. So, Number of p-p chains = 1.989 × 1029 kg/(6.6793 × 10-27 kg) = 3 × 1055 p-p chains.

Each p-p chain releases 4.3 × 10-12 J, so the total energy produced over the lifetime of the Sun is 3 × 1055 p-p chains × 4.3 × 10-12 J/p-p chain = 1.29 × 1044 J. If the Sun radiates 4 × 1026 J/sec, then that total energy will last for 1.29 × 1044 J / 4 × 1026 J/sec = 3.225 × 1017 sec.

With 3.15 × 107 sec/yr, that’s 3.225 × 1017 sec / 3.15 × 107 s/yr ≈ 1017-7 = 1010 years.

The Sun should last about 10 billion years.

11. The exact answer depends on the method used to determine the effect of the sunspots on the temperature. In any case, it should be higher than the value of about +0.4° C that is shown for the year 2000. One argument is as follows: About 1900, the sunspot number was 30 and the temperature deviation was -0.4° C. About 1863, the sunspot number was 50 and the temperature deviation was -0.3° C About 1933, the sunspot number was 60 and the temperature deviation was -0.2° C. About 1963, the sunspot number was 60 and the temperature deviation was just a bit higher than -0.2° C. About 1978, the sunspot number was 80 and the temperature deviation was -0.1° C. From these values, we might roughly say that a change in sunspot number of about 30 is correlated with a change in temperature of about 0.2° C. From about 1982 to 2000, the sunspot number decreased from 90 to 60. We might have expected the temperature to return to the typical value for 60 sunspots, around -0.2° C. In 1982, the temperature deviation was roughly 0° C. Instead of dropping to -0.2° C by 2000, it rose to almost +0.4° C. The sunspot pattern suggests because the sunspots were dropping, the +0.4° C number is 0.2° C lower than we might expect. Had the sunspots remained steady, the temperature in 2000 would have been +0.6° C. From 1962 to 1982, the sunspot number increased from 60 to 90. If it had continued to increase from 90 to 120 by 2002, we might expect to see most of another 0.2° C added to the temperature, for a possible temperature deviation in 2000 of nearly 0.8° C! Answers to Test Yourself

Chapter 12

The Sun, Our Star

1. (d) Within the Sun, gas pressure balances gravitational force—the condition of hydrostatic equilibrium.

2. (d) The Sun is powered by the fusion of hydrogen into helium. Energy is released in this process because the final mass is somewhat less than the initial mass, the difference 2 being converted to energy via the relation that E = mc .

3. (a) Pressure increases by 4 times and density stays the same is the only combination that makes sense.

4. (b) The same. Neutrinos mostly just pass through everything.

5. (c) Doppler shifts are used to measure oscillations.

6. (a) Sunspots are cool—about 4500 K compared with the 6000 K photosphere.

7. (b) Differential rotation winds up the magnetic field, which is frozen into the plasma.

8. (c) The time between maximum numbers of sunspots is approximately 11 years, half of the 22 year solar cycle.

Chapter 13

Measuring the Properties of Stars

CHAPTER 13 MEASURING THE PROPERTIES OF STARS Lecture Suggestions Students may have no idea what “triangulation” means, so both concepts should be explained if describing parallax as an example of triangulation. Some students may also have trouble understanding that the H-R diagram is not related to locations of stars in space. An analogy of plotting height against weight for people may help. Having a contest for students to come up with their own mnemonics for the stellar types is always amusing.

Answers to Thought Questions 1. One could disprove that Proxima Centauri was the closest star to Earth if you could find another star that had a larger parallax angle.

2. It would be easier to measure parallax from Pluto because its orbit has a longer baseline (about 40 AU) than Earth’s orbit. This longer baseline would make it possible to measure parallaxes for stars 40 times more distant than from Earth. Also, Pluto’s lack of atmosphere gives seeing conditions comparable to a satellite around Earth. This would allow even more distant stars to be measured. (Or course, it would take a long time to get the measurements!)

3. There are two effects to consider here. The first is that the optics of the camera or telescope limit the resolving power of the camera or telescope (Chapter 5). Stars have very tiny angular sizes from Earth. If the diameter of the telescope is not large enough compared to the wavelength of light, and the limit of the resolution is bigger than the star, a bright star will be smeared out into a point the size of the resolution limit of the optics.

Imperfect optics and atmospheric distortion (if using a ground telescope) will smear the image into a larger blob. It will look like a bright central point that fuzzes out to dark in all directions. A brighter star will appear wider than a dim one, since more light is scattered. (This shape is called the point-spread function of a telescope).

Even if the optics are good enough, the recording device also needs to be able to capture an image at the angular size of the star—that is, the pixels on the imaging device have to be smaller than the imaged light. An object smaller than a pixel that is bright will just light up a whole pixel. Something similar happens with film, which is limited by the size and arrangement of the chemicals in the emulsion. (When building astronomical instruments, usually some effort is made to optimize the size of the pixels with the pointspread function).

4. Luminosity is the total energy output of a light source per unit of time. Brightness is how much of that energy strikes a specific perpendicular area, at a specific distance, per unit of time. Brightness is a bit of luminosity spread over an area. Brightness drops off with the square of the distance—the luminosity is spread over the surface of a sphere surrounding the light source, and the surface area of that sphere increases with the square of the distance. This is a bit similar to the difference between force and pressure, in that pressure is force spread over an area perpendicular to the force. In some cases, pressure also decreases with the square of distance from the source. The loudness (pressure) of a

Chapter 13

Measuring the Properties of Stars

sound wave at some distance from a speaker is somewhat similar to the brightness of light at some distance from a light bulb.

5. Primarily temperature. The chemical elements present in the B and M stars’ surfaces are the same, and assuming both are from the same population of stars, are in essentially the same ratios. The difference in the absorption lines from elements is an effect of the temperatures at the surfaces. At different temperatures, the atoms are in different ionization states and therefore their absorption lines have different strengths. Some additional sets of lines are found in the M-star spectra, because the surfaces of M stars are cool enough to permit some molecules to form. These would be broken up by collisions at the higher temperatures near the B star’s surface, so although the elements would be found on both stars, the molecules might only exist near the M star’s surface.

6. If we view a binary star face-on, with its orbit perpendicular to our line of sight (we see the orbit as a circle or ellipse), then the orbital motion is also perpendicular to our line of sight and therefore produces no Doppler shift. Thus, we cannot observe any radial velocities. Edge-on (we see the stars move back and forth along a line), the system will exhibit the maximum radial velocities and have the maximum Doppler shift. In between, the amount of shift will vary with from nothing at face-on to maximum at edge-on depending on the part of the motion that is radial to us (trigonometry depending on the angle is involved to the exact results for in-between orientations).

7. STUDENT SHOULD REDRAW DIAGRAM. Since the larger star is now brighter (yellow), the combined light should be brighter. The pattern during the eclipses is reversed: from 0-2, the light is from both stars and is at maximum; from 2 to 3 and during the dip beyond, the light will be dimmest while the cool star blocks light from the bright one; out of eclipse (from before 4 to 5) the light is from both stars and is at maximum; from 5 to 6 and the dip beyond, the light will be dimmed while the larger yellow star eclipses the red one, but not as severely as it was from 2 to 3; and after coming out of eclipse after 6 the light goes back to maximum.

8. In the visual binary, both of the stars in the binary pair should move compared to the background stars. If the visual binary is face-on, over the year the stars will move in a circle or ellipse and the motion will clearly not be from parallax. Only an edge-on binary might be confused. Even in an edge-on binary, one star will move in the opposite direction to what would be expected from parallax. If only one star is moving, the observation is probably of parallax. It should also be possible to compare the exact direction of the shift to the expected direction due to parallax—only a very small range of positions would be consistent. If observed over several years, any slight differences between the period of the binary and Earth's year would shift the motion out of agreement with a parallax shift.

Chapter 13

Measuring the Properties of Stars

9. A plot of distances vs. temperatures would look very different from an H-R diagram. If observational effects were ignored, we would see stars at all temperatures at all distances.

We would see more stars at lower temperatures (at all distances) because there are more low mass stars (which have lower temperatures) than high mass stars, and because nearly all stars spend some time as a red giant. Generally, there should be similar numbers at each distance for the same temperature. Realistic observational effects would bias the data to favor bright stars at larger distances, so as the distance increases there would be fewer stars at any temperature, and the number of low temperature stars would drop off quickly (although some will remain because red giants are very luminous).

10. To plot the star on an H-R diagram, we need to know the color (for the temperature) and the luminosity. Apparent magnitude is not enough information; we would need to determine the distance to the star somehow (perhaps through parallax or a binary orbit study) to convert the apparent magnitude of 3—essentially relative brightness—into an absolute magnitude (luminosity).

Answers to Problems 1. The equation for parallax is: d(pc) = 1/p(arc seconds). If for Sirius p = 0.377 arc seconds, d(pc) = 1/0.377 = 2.65 pc.

2. 61 Proxima Centauri has a parallax of 0.763 arc seconds. Its distance is therefore d(pc) = 1/p = 1/0.763 = 1.31 pc.

3. The parallax of Betelgeuse is 0.005 arc seconds. Therefore its distance is d = 1/p = 1/0.005 = 200 pc.

The error (uncertainty) in this measurement is +/–0.003", so we have to add and subtract this figure from the parallax angle to give the lower and upper limits to the distance estimate. dmin(pc) = 1/0.008 = 125 pc. The larger angle corresponds to the closer distance. dmax(pc) = 1/0.002 = 500 pc. The smaller angle corresponds to the farther distance.

4. STUDENTS SHOULD DO THE TRIANGULATION. Done carefully, this can work relatively well—if the error in the measured angle is down to a couple of degrees. This isn’t too hard with angles in the 50-80 degree ranges and objects that are 5 to 20 meters away.

5. Since  Ceti and  Centauri B have nearly the same luminosity, they are both the same kind of “standard candle” and the difference in apparent magnitudes is a brightness difference that results from one being farther away than the other. The difference in apparent magnitude for the two stars is 2.16, so the ratio of brightness is 2.5122.16 =7.3.

This means  Centauri B appears 7.3 times brighter than  Ceti, it must be closer. Using the method of standard candles from section 13.2, Bnear /Bfar = (dfar/dnear)2

Chapter 13

so

Measuring the Properties of Stars

7.3 = (dfar/dnear)2 2.7 = dfar/dnear

This means  Ceti is 2.7 times farther away than  Centauri B. 6. The wavelength of peak emission for Rigel is m = 200 nm. Use Wien’s Law to find the temperature: 6 6 T = 2.9 × 10 K nm / m = 2.9 × 10 K nm / 200 nm = 14,500 K. This is much hotter than the surface of the Sun, which is about 5800 K. (2.5 times hotter).

7. Alpha Centauri’s emission peaks at 500 nm. Applying Wien’s Law, 6 T = 2.9 × 10 K nm / 500 nm = 5800 K. The temperature of this star is approximately the same as our Sun’s. (Alpha Centauri is a similar spectral type to the Sun (G2)).

8. The problem is most easily solved using proportional reasoning, although it can also be solved via direct calculation. In both cases we start with the Stefan-Boltzmann law, 2 4 L = 4R  T , which can be solved for R to yield: Symbolically, .

Or, Arcturus has a radius that is around 40 times the radius of the Sun.

9. The problem is most easily solved using proportional reasoning, although it can also be solved via direct calculation. In both cases we start with the Stefan-Boltzmann law, 2 4 L = 4R  T , which can be solved for R to yield: L ) R= ( 4T 4 If the temperature of Sirius’s companion (Sirius B) is 27000 K, and the Sun’s is 5800 K, then Sirius B has a temperature (27000/5800) = 4.655 times that of the Sun (extra digits are being kept here to aid in the accuracy of the final calculation). If its luminosity is × -2 -2 4 1/2 10 times that of the Sun, the radius must be ( × 10 / 4.655 ) = that of the Sun. Symbolically,

Chapter 13

Measuring the Properties of Stars

( RSirius B = R Sun

LSB 4T 4SB

)

L ( ) 4T 4Sun

4

LSB  TSun  110−2  5800  110−2  =   = 4.614 10−3  = 4 LSun  TSB  1 4.655  27000 

=

4

.

Or, Sirius B has a radius that is a little more than 1/250th the radius of the Sun. Since Earth’s radius is about 1/100th the radius of the Sun, we should expect this to be close to the size of Earth. Multiplying by the size of the Sun to get the size of Procyon B in kilometers, 5

0.127 × 6.96 × 10 km = 3211 km.

This is a about 0.5 times the size of Earth (6400 km): RSirius B / REarth = 3211 km / 6400 km = 0.5.

10. Faint Balmer lines indicate either very high or very low surface temperatures, so the next step would be to locate and analyze other lines indicative of those particular states.

For example, helium lines indicate it’s a hotter star, but ionized calcium indicates it is a cooler star. Finding max would determine the surface temperature, and fine distinctions between the lines would lead to the specific spectral type.

11. The wavelength we measure, , of 400 nm, is longer than the “rest” wavelength, o, of 400.2 nm measured in the lab. The star’s light is blueshifted, which means that it is moving toward us. We can find out how fast by using the Doppler shift formula, 5

V = ( ) × c, where V = star’s velocity, and c = speed of light = 3 × 10 km/s.  =  − o = 400.0 nm – 400.2 nm = -0.2 nm 5 V = (-0.2 nm / 400.0 nm) × 3 × 10 = -150 km/s.

(The velocity comes out negative because the star is moving away ( − o < 0).)

12. To find the mass of a binary star pair we use the modified form of Kepler’s Third 3 2 Law, m + M = a /P , where P is measured in years, a in AU, and mass in solar masses.

P = 5 yr and a = 10 AU 3 2 m + M = 10 /5 = 1000/25 = 40 M ⊙.

The two stars’ combined mass is 40 solar masses.

13. To find the mass of a binary star pair we use the modified form of Kepler’s Third 3 2 Law, m + M = a /P , where P is measured in years, a in AU, and mass in solar masses.

Chapter 13

Measuring the Properties of Stars

3

2

3

2

m + M = a /P = 4 /2 = 64/4 = 16 M ⊙.

The two stars combined mass is 16 M⊙ solar masses.

14. In this problem we want to know the separation between the stars. We again use the modified form of Kepler’s Third Law, 3 2 m + M = a /P . m+M is 8 solar masses, and P is 1 years, 3 2 8=a /1 3 a =8 1/3 a = 8 = 2 AU. The separation in the binary is 2 AU. 15. Using Figure 13.23, Rigel: L = 105 L⊙, T = 10,000 K Barnard's Star: L = 0.0009 L⊙, T = 3,000 K For the Sun, L⊙ = 4R⊙2  T⊙4 and T⊙ = 5800 K. R = (L⊙ / (4 T⊙4))(1/2) R/R⊙ = (L/L⊙ )(1/2) (T⊙4 /T4)(1/2) = (L/L⊙ )(1/2) (T⊙/T)(2) So RRigel / R⊙ = (105 )(1/2) (5800 K / 10000 K)(2) = 106 RBarnard's Star / R⊙ = (0.0009 )(1/2) (5800 K / 3000 K)(2) = 0.11 Figure 13.23 shows that Rigel is very close to 100 solar radii, and Barnard's Star is very close to 0.1 solar radii.

16. From the chapter, L ≈ M 3 when the values are in solar units. If L = 5000, M = (L)(1/3) = (5000)(1/3) = 17.1 solar masses.

17. STUDENTS MUST DRAW AN H-R DIAGRAM USING THE DATA. Betelgeuse is a red giant, 40 Eridani B is a white dwarf.

Answers to Test Yourself 1. (d) d = 1/p = 1/0.04 = 25 pc.

2. (d) Luminosities vary over about 10 powers of 10, but masses only vary from 0.1 to 60 or so, so choice d is the false one.

3. (a) ¼ as bright means it’s twice as far away.

6 4 4 4. (e) T = 2.9 × 10 /400 = (2.9/4) × 10 ≈ 0.75 × 10 = 7500 K 5. (a) It must have a large radius (like a red giant; from Stefan-Boltzman law).

Chapter 13

Measuring the Properties of Stars

6. (e) O stars are the hottest in the list given (OBAFGKM).

7. (d) About 3 solar masses. Use M = a3 / P2 = 303/1002 =27,000 / 10,000 = 2.7.

8. (b) White dwarfs have small radii and are in the lower central part of the H-R diagram 9. (b) 90% of nearby stars are on the main sequence (90% of the stars in the disk are).

Chapter 14

CHAPTER 14

Stellar Evolution

STELLAR EVOLUTION

Answers to Thought Questions 1. If, late in their lives, stars became dimmer and dimmer (rather than evolve into giants), what would you expect of the relative numbers of blue, white, yellow and red stars visible in the night sky? Would any yellow stars be visible?

If we think only of the stars’ relative sizes, blue stars are always larger than white, which are always larger than yellow. Red stars are the smallest. If stars only get dimmer as they age, their luminosity never increases – their size must increase for this to happen.

The number of red and yellow stars would go down dramatically, as these are already dim to begin with. Variable yellow stars, however, may remain visible as they change size (due to non-evolutionary reasons.)

2. EXPERIMENT FOR STUDENTS. Gas shrinks and becomes more dense when cold.

3. (From the text:) The transformation from gas cloud to star proceeds through several stages. In the first stage, a dense clump within a cloud begins to collapse, its gas drawn inward by gravity, which compresses and heats it. In the second stage, any rotation of the gas clump makes it flatten into a disk, as we described in chapter 8 on the origin of the Solar System. In about a million years, the disk forms a small, hot, dense core at its center called a protostar, which marks the third stage.

4. EXPERIMENT FOR STUDENTS. 5. Only stars less massive than about 0.4 M☉ are completely convective, i.e., their material is completely mixed. Most stars can only use about 10% of their mass for fusion.

If a star could mix in the material from the outer layers, a star could sustain its mainsequence hydrogen-fusion phase longer. If it could use all the fuel, it could stay on the main sequence as much as 10 times longer! Normally, stars enter the red giant phase when the core contracts and hydrogen starts burning in a shell around the core. This would not happen effectively if most of the hydrogen in stars had already been fused into helium. Low mass stars would not become red giants, and would not eject planetary nebulae. Stars with high enough masses to fuse helium would still become red giants, but the hydrogen shell burning would not occur because there would be no hydrogen left.

The rest of the sequence of burning helium into carbon, carbon into oxygen, and so forth, could still occur, and stars could still explode as supernovae. There would be a lot more helium in the universe.

6. EXPERIMENT FOR STUDENTS. A video camera or overhead projector could be helpful if doing this in class.

7. EXPERIMENT FOR STUDENTS. 8. Look at the atomic numbers and mass of these elements: neon is 10/20, magnesium 12/24, sulfur 16/32, argon is 18/40, and calcium is 20/40. Fluorine is 9/19, sodium is 11/23, phosphorus is 15/31, and chlorine is 17/35. The nuclear reactions in the core of massive stars start when the hydrogen, which contains 1 proton, is depleted, and helium,

Chapter 14

Stellar Evolution

which has 2 protons and 2 neutrons, is abundant. All of the fusion reactions after helium are reactions in which even numbers of protons and neutrons are fused—it would be hard to produce a heavy element like fluorine or sodium, which has an odd number of protons and an odd number of nuclei, that would require splitting up a helium nucleus or finding available protons. Reactions which add even numbers are favored: add a helium nucleus to a neon nucleus and create an atom of magnesium. Combine 2 oxygen nuclei (8/16) and create sulfur, and so on.

9. Fusing two smaller nuclei into one larger nucleus either releases energy or requires energy, depending on the atomic number of the end product—elements with atomic numbers higher than iron (26) require energy to be input. Platinum and gold have atomic numbers 78 and 79 respectively, so these elements cannot be a power source in a stellar nucleus and are not created over long periods of time in the sizeable cores of massive stars. They can only be created by absorbing some of the energy of a very brief but extremely powerful supernova explosion and require that nuclei of the right sizes collide and fuse during the brief explosion. Iron (26) and silicon (14) are created in large quantities as their creation powers the sizeable stellar cores of massive red giants over a relatively much longer period of time.

10. In very old clusters, the hot blue stars should have long ago evolved off the main sequence because they consume their hydrogen faster than the low-mass stars and thus have shorter lifetimes. Blue stragglers arise either when low-mass stars collide, or a very close binary companion has recently dumped a large amount of material onto the other star, making it much more massive than before. The star’s interior re-organizes into a new hydrostatic equilibrium, it becomes a massive blue star, and its position changes on the H-R diagram.

Answers to Problems 1. DO SKETCHES. SHOULD RESEMBLE 14.23 and 14.26.

-20

2. The density of the cold cloud is 1.67 × 10 kg/m3. The mass of the Sun is 30 1.989 × 10 kg, so 30 -20 50 1.989 × 10 kg / (1.67 × 10 kg/m3) = 1.19 × 10 m3 of cloud would be required to have enough mass to make the Sun. The volume of the Sun, since it is a sphere, is 8 27 (4/3) ×  × R3 = (4/3) ×  × (6.96 × 10 m)3 = 1.41 × 10 m3 The ratio of the volume of the cloud to the volume of the Sun is 50 27 22 1.19 × 10 m3 / (1.41 × 10 m3) = 8.43 × 10 .

The volume of the cloud in cubic light years: 50 15 50 45 = 1.19 × 10 m3 × (1 ly / 9.46 × 10 m)3 = 1.19 × 10 m3 × 1 ly3 / 8.46 × 10 m3 = 141 ly3.

Chapter 14

Stellar Evolution

This would be a cube of about 5.2 light years per side—all that volume, at cloud density, would be needed to form the Sun. 3. This is a straight-forward velocity calculation. We just need to convert the light years into kilometers first: 2 ly × 9.46 × 1012 km = 1.89 × 1013 km Then, t = d/V = 1.89 × 1013 km/ (300 km/sec) = 6.31 × 1010 sec.

At 3.16 × 107 s/y this is about 2000 years.

4. Estimate the lifetime of the Sun. Core mass: 10% of the Sun’s mass is available for fuel, 0.1 × 1.989 × 1030 kg = 1.989 × 1029 kg.

In each p-p chain, 6.6793 × 10-27 kg of this fuel is processed. So, Number of p-p chains = 1.989 × 1029 kg/(6.6793 × 10-27 kg) = 3 × 1055 p-p chains.

Each p-p chain releases 4.3 × 10-12 J, so the total energy produced over the lifetime of the Sun is 3 × 1055 p-p chains × 4.3 × 10-12 J/p-p chain = 1.29 × 1044 J. If the Sun radiates 4 × 1026 J/sec, then that total energy will last for 1.29 × 1044 J / 4 × 1026 J/sec = 3.225 × 1017 sec.

With 3.2 × 107 sec/yr, that’s 3.2 × 1017 sec / 3.15 × 107 s/yr ≈ 1017-7 = 1010 years.

The Sun should last about 10 billion years.

5. To find the escape velocity from the atmosphere of a red giant we use the expression 1/2

for escape velocity given in Chapter 2. That is, Vesc = (2GM/R) . For our problem, M = 1 M☉ and R = 100 R☉.

1/2

Vesc = (2GM/R)

-11

= (2 × 6.67 × 10

30

8

1/2

m3kg-1s-2 × 2 × 10 kg / (100 × 6.96 × 10 m) )

4

= 6.19 × 10 m/sec = 62 km/sec Most planetary nebulae appear to be expanding at about 20 km/s, which is close to the number we found above. If the radius of this star were a bit larger, the escape velocity could be as small as 20 km/s.

Chapter 14

Stellar Evolution

6. To calculate the average density of a red giant, (), divide its mass, M, by its volume.

Assuming it is a sphere, its volume is 4R3/3. Thus,  =  (4R3/3) 1033 2.0 1033g g g 2.0 = 4  3 = 0.018 10 33− 36  3 =  12 3 12 3 4 3 (10 ) cm cm 3  (3 10 cm) 3  (3) g g = 0.018 10−3 = 1.8 10−5 cm3 cm3 This is about 55 times smaller than the density of air.

7. This is a straight-forward velocity calculation. We just need to convert the light years into kilometers first: 0.25 ly × 9.46 × 1012 km = 2.37 × 1012 km Then, t = d/V = 2.37 × 1012 km / (20 km/sec) = 1.18 × 1011 sec ~ 1011 sec.

At 3.16 × 107 s/y this is about 3700 years.

8. Calculate the Doppler shift using the formula V = c (), where  is the shift between the observed wavelength  and the rest wavelength ,  =  − . Solving for the shift,  = V / c  = (5000 km/sec) / (300,000 km/sec) = 0.0167 That is, the line will be shifted almost 2% from its expected value, and the actual shift in the wavelength is  =  × V/c = 0.688 nm × 0.0167 = 0.011 nm (So from part of the remnant coming towards us, the line would appear at 0.677 nm).

9. We estimate the age of a star cluster from the age of the most massive star still on the main sequence. The figure shows this mass to be a little more than halfway between 2 and 3 solar masses, so let’s estimate this as 2.6 M☉. We can find the age of such a star using the expression for a star’s main-sequence lifetime, 10

t = 10 M/L .

However, we need L as well as M. We can read L from the figure (in principle) or calculate it using the mass-luminosity relation. Estimating from the figure is tricky

Chapter 14

Stellar Evolution

because of the log scale, but the value might be around 20 L☉. The mass-luminosity law 3

says L = M , so 3

L = (2.6) = 17.6 L☉.

The expression for this star’s main-sequence lifetime is therefore 10

10

9

t = 10 yrs × M/L = 10 × 2.6/17.6 = 1.5 × 10 yrs.

This is also the age of the cluster, about 1.5 billion years.

Answers to Test Yourself 1. (e) Protostar → main sequence → red giant → white dwarf 2. (b) Gravity and mass drive what happens to a star.

3. (e) A planetary nebula occurs at the end of a low-mass star’s life.

10 4. (e) A star’s main sequence lifetime is t = 10 M/L. If L is 100 times larger, for the same mass the star will live 1/100 times as long as the Sun; if the mass is 5 times larger, for the same L the star will live 5 times longer. Both together mean the star will have a lifetime 20 times shorter than the Sun. 5. (c) It can get hot over a large range quickly.

6. (a,b,c,d) All are reasons why variable stars are useful (large luminosities means you can see them far away).

7. (c) A planetary nebula is a shell of gas made of the ejected outer layers of a low-mass red giant.

8. (b) contracts and heats 9. (b) The core stops contracting and heating before it ever reaches the temperature needed to fuse carbon or heavier elements.

10. (a,c,d,e) The only wrong choice is (b), because not all stars explode.

11. (a,d) The age and distance of the cluster can be determined from fitting the main sequence.

Chapter 15

Stellar Remnants

CHAPTER 15 STELLAR REMNANTS: WHITE DWARFS, NEUTRON STARS, AND BLACK HOLES Lecture Suggestions A demo using a rotating stool with weights in hand can be good for showing how conservation of angular momentum speeds up rotation as the mass is centralized. Adding flashlights can make the demo an introduction to pulsars.

Before introducing black holes, it’s helpful to review or describe escape velocity.

Make a table of escape velocities for the Sun, a white dwarf, and a neutron star. Show how escape velocity rises toward c.

The waterbed analogy can be helpful to describe black hole as “hole” in space-time.

Water going down the drain and soap suds accumulating and spiraling around drain hole are reasonable analogies for accretion disks. Note how the materials spirals faster as it moves in toward drain.

Answers to Thought Questions 1. Stars that are near to the dividing line between high and low-mass stars can process helium into heavier elements in their cores. So, some white dwarf stars might form after the collapse of the carbon core of such stars. Diamond is a form of carbon in which the atoms are arranged in a tetrahedron (pyramid) shape. Once cooled, even if they were made of diamond, the extremely high density of the surface and very high escape velocity would make it difficult to mine these stars. (Not to mention the problem that the nearest known white dwarf is at least four light-years away).

2. A white dwarf is quite similar to the core of an old, low-mass main sequence star: both are degenerate gases, both are primarily composed of helium; the helium flash that can occur in a star is a similar process to detonation (explosion) of a white dwarf. A binary system with an 8 solar mass blue star and a 1 solar mass white dwarf is initially difficult to explain: only a low mass star should leave behind a white dwarf, but the how could a low mass star have evolved through its whole life before the more massive blue star? The answer is that the white dwarf is essentially the leftover core material of a more massive star (say 7 solar masses) that reached the red giant phase first. When it expanded, its companion (say a yellow, 2 solar mass star) gravitationally captured the expanding envelope of the first star, increasing its mass to the 8 solar masses by stripping away most of the older star, leaving behind insufficient mass for an explosive end. The exposed core just collapsed into a white dwarf. The newly massive companion restructured, mixed new material into the core, and became a massive blue star.

3. It is not particularly surprising that not every supernova appears to have a pulsar.

Pulsars, rapidly rotating neutron stars, are produced by the explosion of massive stars in a Type II supernova. When a white dwarf collapses and explodes as a type I supernova, it may leave no remnant. Thus not every supernova explosion leaves a remnant. For those that do, in order to detect the pulsar, the beams of radiation must cross our line of sight.

Since the orientation of the pulsar’s beams to us is essentially random, there must be some pulsars in supernova remnants that we do not see because their beams of radiation do not happen to point toward us.

Chapter 15

Stellar Remnants

4. The period is the circumference divided by the rotational velocity: P = C / V = R / V From the chapter, we know that angular momentum is conserved by a collapsing star, and so MVR is a constant. So if a body collapses to 1/X of its initial radius, MVbeforeRbefore = constant = M (Vafter)(Rbefore/X) = M (XVbefore) (Rbefore/X) The velocity afterwards must be X times the initial rotational velocity. Therefore, Pafter = C/V = 2Rafter / Vafter = 2(Rbefore/X) / (XVbefore) Re-arranging this, we can factor it into something × Pbefore, 2

Pafter = (2Rbefore/Vbefore) × (1/X) × (1/X) = Pbefore / X 2 If the radius decreases to 1/X of what it was before, the period decreases to 1/X times what it was before.

5. The pulsing of variable stars results from the star actually getting larger and smaller, and changing temperature and luminosity accordingly. The star is variable in radius, temperature, and luminosity. These changes take reasonably long periods of time, typically days or longer. A rotating neutron star is not variable in radius, temperature or luminosity. The pulsations do not result from a change in size. The pulsations result from the rotation of the star, and radiation emitted from specific places on the star. The frequency of pulsation can be quite fast, even thousands of times per second, because it is a result of the fast rotation of the star. Pulsars are not variable stars.

6. In binary systems with white dwarfs, neutron stars, or black holes with companions, matter is often transferred from the companion star to the compact object. In all three systems, accretion disks can form around the compact object. In the white dwarf binary, hydrogen falling on the white dwarf surface builds up until it burns explosively, sometimes producing enough visible light to make a “new” star appear in the sky—a “nova”. In the neutron star binary, hydrogen falling on the neutron star also burns explosively, but because of the strong gravity of the neutron star it is heated to millions of degrees and produces bursts of X-rays. In some neutron star systems, material is channeled to specific hot spots on the neutron star which radiate regular X-ray pulses. In a black hole system, X-rays are also observed. These originate from the accretion disk itself, parts of which are heated to millions of degrees as gravitationally energy is released as material moves toward the black hole. The accretion disk gets very close to the black hole, and material traveling inward releases a large amount of gravitational energy, some of which becomes kinetic energy and then heat when collisions occur in the accretion disk.

7. If the Sun were replaced with a 1 solar mass black hole, neutron star, or white dwarf, this would have no effect on Earth’s orbit. The mass of the “Sun” would remain the same, and Earth’s orbit, which is far away from where the Sun is, would remain the same. The conditions at the surface of each of these objects are different, the escape velocity steadily increases from Sun to white dwarf to neutron star to black hole. The escape

Chapter 15

Stellar Remnants

velocity is the speed of light at the edge of the black hole. Also, the tidal forces get stronger from white dwarf to neutron star to black hole—the gravity at the edge of the black hole would be much more likely to tear an object apart than near a white dwarf, although that would still be much more extreme than at the edge of the Sun or on Earth’s surface. Finally, the temperature on the surface of the white dwarf would be several times higher than at the Sun’s surface; the neutron star, even higher (up to ~ 50,000 K).

(Students might also include comments about the luminosity from a white dwarf, neutron star, or black hole, which would all be less, or the smaller appearance from Earth, or event that gravitational lensing of surrounding stars might be visible during the “day” if the Sun were replaced by a black hole, although these areas are more or less beyond what the question asks.)

8. You could explain how escape velocity works by talking about throwing a ball into the air. If you throw it up with a little speed, it slows down and comes back. Throw it a little faster at the start, it gets up a little higher. Imagine throwing it so fast it doesn’t come back—you’d have to throw it the escape velocity, 11 km/s, to break free of Earth’s gravitational field. You could then talk about how the escape velocity of the Sun is about 600 km/s, and if you could shrink the Sun down to a radius of 3 km, its escape velocity exceeds the speed of light. This makes it plausible that if the core of a very massive star collapsed to a small enough size it could become a black hole and would only be “seen” by its effect on other matter.

9. If you jumped into a black hole feet first, the differential gravitational force between your feet and your head would stretch you out. Tidal stretching would rip you apart.

10.There are numerous possible answers to this one. Some of them are: a) lack of sunlight b) stars blocked by the body of the star would be visible as light wrapped around the black hole c) Excessive x-ray exposure, assuming an accretion disk is present.

There is one thing that would not be different – the orbit of the planet around the body.

At 5 AU, the planet is well outside of the surface of the star, so there are no additional tidal effects to take into account.

Answers to Problems 1. To find the average density of a white dwarf, we divide its mass by its volume. The 30

33

mass of a white dwarf is about 1 M☉ = 2 × 10 kg = 2 × 10 g. We can find its volume 3

4

7

using the formula V = 4R /3. The radius of a white dwarf is about 10 km = 10 m = 9

10 cm.

3

9

3

27

3

V = 4R /3 = 4(10 cm) /3 = 4.2 × 10 cm 33 27 3 5 3 Density = M/V = 2 × 10 g / 4.2 × 10 cm = 4.8 × 10 g/cm .

Chapter 15

Stellar Remnants

2. The formula for the escape velocity is vesc=√

2GM R

A 1MSun white dwarf with a radius of 104 km would have an escape velocity of

vesc=√

2(6.67×10−11 N∙𝑚2/𝑘𝑔2)(2×1030kg) 1×107𝑚

=5.2 x 106 m/s

A 1 MSun neutron star with a radius of 10 km would have an escape velocity of

vesc=√

2(6.67×10−11 N∙𝑚2/𝑘𝑔2)(2×1030kg) 1×104𝑚

=1.6 x 108 m/s.

3. The circumference of the neutron star would be 62.8 km: C = 2R = 2(10km) = 62.8 km.

Since the circumference is all 360° around the star, it’s 62.8km/360° = 0.174 km/°, so to travel 7° would be 7°  0.174 km/° = 1.22 km. The circumference of Earth is 40,212 km (radius = 6400 km), so on Earth it’s 40212 km/360° = 112 km/°, and it would take 7°  112 km/° = 784 km to travel 7° along the surface.

4. If an object has a mass M and is made up of N particles of mass m, then the number of particles it contains is N = M/m. Inserting the values of the star’s mass and that of the neutron, we find that N = 3.4 × 1030 kg/(1.7 × 10-27 kg) = 2 × 1030-(-27) = 2 × 1057 particles.

5. We find the total volume of the neutron star by multiplying the volume per neutron times the number of neutrons, or 10-45 m3 × 2 × 1057 = 2 × 1057-45 m3 = 2 × 1012 m3. Given that the volume of a cube is the length of one its sides (L) cubed (L3), we can find L from the volume by taking the cube root of the volume. Recalling that taking a cube root is the same as raising a number to the 1/3 power, we find that L = (2 × 1012 m3)1/3 = (2)1/3

Chapter 15

Stellar Remnants

1012×(1/3) m = 1.26 × 104 m = 12.6 km. This is comparable to the size of a neutron star, so we can conclude that in such a star the neutrons are nearly “touching” one another.

6. We imagine the core of a massive star collapsing and becoming a neutron star. The core has an initial radius of 104 km and a period of 0.5 d = 12 h = 43,200 s. If the core collapses to a radius of 10 km, we can find the period by exploiting the requirement that MVR = constant because of conservation of angular momentum. Using the results of TQ5, which are based on that idea, if the radius drops by a factor of X, the period drops by a factor of X squared.

Rbefore/Rafter = 104 km / 10 km = 103 So Pbefore/Pafter = (103)2 = 103×2 = 106 So Pafter = Pbefore / 106 = 43,200s / 106 = 0.0432 s This is similar to some of the pulsars described in the chapter; the Crab rotates 30 times a second, or with P = 1/30s = 0.033s; the pulsars in Figure 15.8 are a little slower; and the millisecond pulsars described in the chapter are a bit faster.

7. The Schwarzschild radius (Rs) is the distance at which Vesc = c. Solving the escape 2 velocity equation for Rs gives the equation shown in the chapter, Rs = 2GM/c .

Substituting in values for the Sun, -11 30 8 2 Rs = 2 × 6.67 × 10 m3kg-1s-2 × 1.99 × 10 kg / (3 × 10 m/s) 3

= 2.95 × 10 m ≈ 3 km.

Note: A simplified equation sometimes used is Rs = M/M☉ × 3km.

8. To find your own Schwarzschild radius, use your own mass in the expression 2 Rs = 2GM/c . For example, if you weigh 60 kg, then 2

-11

Rs = 2GM/c = 2 × 6.67 × 10 -26

8

2

m3kg-1s-2 × 60 kg/(3 × 10 m/s)

-25

= 8.89 × 10 m ≈ 10 m, a very tiny distance.

-15 To give you some idea as to how tiny this is, protons have a radius of about 10 m and -10

atoms have a radius of about 10 m. Thus, your Schwarzschild radius is at least ten orders of magnitude smaller than an atomic nucleus.

Chapter 15

Stellar Remnants

9. Use Wien’s law to find the wavelength of radiation from a 10 million K blackbody: 6 6 7 max= 2.9 × 10 K nm / T = 2.9 × 10 K nm / 10 K = 0.29 nm The wavelength of red light is about 650 nm and that of blue light is 450 nm; ultraviolet rays are a bit shorter, but it is actually X-ray photons that have wavelengths of this size (see the Table and Figure showing the electromagnetic spectrum in Chapter 4).

10. Find the mass of a system with a period of 34 days and a semi-major axis of 0.5 AU by using the modified form of Kepler’s third law (remembering to convert 34 days into years): 3

2

3

2

m + M = a / P = 0.5 /(34/365) = 14.4 solar masses Since the mass of a K0 star is about 0.79 solar masses (given), so the companion is 14.4 – 0.8 = 13.6 solar masses. If this large a mass were a star, it would be a bright blue O star.

However, the companion is not visible. Since this is too large a mass to be a neutron star or white dwarf, the logical explanation is that the companion is a black hole, and look for X-rays from the accretion disk or jets.

11. Find the mass of a system with a period of 8.4 years and a semi-major axis of 8 AU by using the modified form of Kepler’s third law: 3

2

3

2

m + M = a / P = 8 /(8.4) = 7.3 solar masses A B5 star is visible, but the companion is not visible. The B5 star has a mass of 5.9 solar masses (given), meaning the mass of the invisible companion is 7.3 – 5.9 = 1.4 solar masses. This mass is exactly the Chandrasekhar limit, and the companion could be a white dwarf. One could observe the system and look for nova bursts. However, it is also possible the companion is a neutron star, in which case one might look for X-ray pulsations or X-ray bursts.

12. Find the mass of a system with a period of 4 years and a semi-major axis of 4 AU by using the modified form of Kepler’s third law: 3 2 3 2 m+M = a / P = 4 /(4) = 4 solar masses An A0 type star is visible. A0 stars have a mass of 2.9 solar masses, so the invisible companion has a mass of 4 – 2.9 = 1.1 solar masses. This companion is most likely a white dwarf, and one should observe for visible light nova bursts.

Answers to Self-Test

Chapter 15

Stellar Remnants

1. (b,c,d,e) The radius decreases, density increases, may collapse if it exceeds 1.4 M☉, and may explosively fuse (burster) 2. (b) Some electrons must gain higher energy (change quantum states) to allow closer packing of the electrons.

3. (a) It was probably a white dwarf left behind after a star ejected its outer layers containing hydrogen.

4. (d) The neutron star’s radius is about 10 km, which is the size of a small city.

5. (b) Radio pulses are caused when the pulsar beams sweep past us.

6. (a) The acceleration of gravity is stronger near the neutron star.

7. (c) Eclipses of accretion disks around massive companions demonstrate they are massive bodies that are small and dark at visible wavelengths.

8. (e) The Schwarzschild radius is where the escape velocity reaches the speed of light.

9. (b) The photons are stretched and lose energy as they climb out of the strong gravitational field.

10. (c) The Sun will eject its outer layers into a planetary nebula and leave behind the core as a white dwarf.

Chapter 16

The Milky Way Galaxy

CHAPTER 16 THE MILKY WAY GALAXY Answers to Thought Questions 1. At the very edge of the galaxy, the Milky Way would look like a disk with a bulge in one direction on the sky. You would see very few stars if you looked in the direction opposite to the galactic center. At the nucleus, you would be centered in the bulge, with the globular clusters surrounding you in all direction. The disk would be a plane stretching around you in all directions, but it would be obscured by the dust in the same way that our view of the galactic center is obscured by the dust in the disk of the galaxy. Depending on whether we were actually right at the galactic center, we would either be in or close to the supermassive black hole with its accretion disk.

2. Interstellar reddening is a process by which a star looks redder than it should, because light of shorter wavelengths (especially blue) is scattered out of the line of sight by intervening interstellar dust. The scattering is strongest for blue wavelengths and the least for red wavelengths, varying smoothly in between. In the atmosphere at sunset, the Sun tends to look orange or red, and for more or less exactly the same reason. Particles in the atmosphere, especially dust, selectively scatter blue photons out of the line of sight to the Sun. This is the source of the blue light coming from the rest of the sky at sunset and all day long. The scattering does occur at any time of day (hence the sky is blue), but at sunset or sunrise, the path that light from the Sun takes through the atmosphere to the observer is much longer. As a result, more blue light is scattered, and the effect is stronger.

3. STUDENTS SHOULD PROVIDE SOME REASONED IDEAS. For the alien visitor, a shopping mall for an hour on a Wednesday afternoon in October will not give a very representative selection of the human population. You would probably only see a few (truant) children, and also very few adults from age 20 to 65 who have jobs, except for the ones working in the mall. You might see relatively more retired folks, and there could be an excess of adult women of all ages if there still more stay at home wives than stay at home husbands (this trend is evening out over time so it’s hard to tell). At an elementary school, your survey would be biased to have very large numbers of young children, but none younger than ~ 5 years old, relatively few adults, and would be missing the elderly entirely. A 9pm trip to the mall on the weekend would probably far over-represent the number of teenagers in the population, and there would be very few older people. Human clothing: you would get entirely different results in Florida or Chicago in December, because the temperature is a huge factor in what clothing people wear, and you would need to consider this to get an accurate answer—although an alien visitor might not know they needed to know it, just as astronomers sometimes do not know what effects they are not missing when surveying celestial objects.

4. STUDENTS SHOULD PROVIDE SOME REASONED IDEAS. Generally, you expect the candidate who appeals to older people to come out ahead on the survey of land-lines, the candidate who appeals to younger people on the survey of cell phone users, and an accurate result if you can call both groups, but only if you call them in proportion to the likeliness of cell phone or land line owners voting.

Chapter 16

The Milky Way Galaxy

5. STUDENTS SHOULD THINK OF EXAMPLES. Common ones include lines at amusement parks or concerts, traffic flow, etc.

6. EXPERIMENT FOR STUDENTS. This will produce a result that bears a resemblance to spiral arms.

7. The shape of a spiral arm is permanent, but the material in it is constantly changing, much like a wave at sea or even a wave around a ballpark. If the same stars stayed in the spiral arm, then for it to stay the same shape the stars at the outside end would have to have the same period as the ones at the end close to the center of the galaxy. Since those stars have to travel a much larger circumference to orbit, they would have to travel much faster. That is not the case—over much of the galaxy the stars travel at the same speed of a few hundred km/s (so a star twice as far out as another has twice as far to go; traveling at the same speed it will take twice as long). So the inner part of the spiral arm would lap the outer part, and the arms would get increasingly wrapped around the center as time went on.

8. We need to measure its distance from the Galactic center, probably using variable stars as standard candles. We would need to measure its velocity from Doppler shifts, and calculate its orbit. The modified form of Kepler’s Third Law could then be used in the same way as we used the orbits of the planets to calculate the mass of the Sun. 9. The best way to argue for the black hole interpretation of the dark mass at the center of the galaxy is to be able to set a limit on how much mass is within a specific volume at the center of the galaxy. Above a certain threshold, a collection of neutron stars would inevitably collide and coalesce under their own gravity, forming a black hole. The orbits of stars near the center might look different if there was a large, distributed mass of many neutron stars (or something else), or one large central point mass. A field of neutron stars would probably also include some pulsars, or X-ray bursters, which generate radiation different from the accretion disk of a black hole. Comparison of spectra from the center of the galaxy would help determine the nature of the object. Any other interpretation would have to be consistent with the existing infrared and X-ray observations of the center of the Galaxy, which have mapped the radiating objects in the center down to distances of a few hundred AU (see figure 16.27).

10. If there were no dust or gas in the galaxy, but the Milky Way were otherwise the same as it is, more stars would be visible in the sky, especially the massive, bright, blue stars and other giants. The Milky Way band would be brighter, and the center of the galaxy very bright, instead of partially blocked as seen in the pictures in the chapter. When the Milky Way is very old, and is no longer forming new stars, there will be no blue stars in the night sky—the only stars in the sky will be orange and red giants.

Chapter 16

The Milky Way Galaxy

Answers to Problems 1. STUDENTS SHOULD MAKE DIAGRAM. None of the individual stars visible in the night sky are from outside the Milky Way; very few of the stars we can see in the night sky are even very far across the Milky Way. 2. Use D = Vt. Re-arrange to form t = D/V. We’ll need to convert the distance in kiloparsecs into meters first: 16 20 D = 8 kpc × 1000 pc/kpc × 3.09 × 10 m/pc = 2.47 × 10 km.

20

8

11

t = D/V = (2.47 × 10 km) / (3 × 10 m/sec) = 8.24 × 10 sec 7

Divide this by 3.16 × 10 sec/yr to get 26,076 years (~ 26,100 years).

An alternative, and slightly easier, solution is to convert the 8 kpc into light-years, which is then the time: D = 8 kpc × 1000 pc/kpc × 3.26 ly/pc = 26,080 ly. t = D/V = 26,080 ly / (1ly/yr) = 26,080 years (~ 26,100 years).

(The slight difference in the answers is the result of rounding the speed of light, the number of ly in a pc, and the length of the year.)

3. We need to determine the number of atoms in the cloud, then multiply that number by the rate of emission: 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) = 1.19 × 1059 atoms.

The rate of emission is 1 photon per atom per 15 million years. It will be helpful to convert the time into seconds: 15 × 106 y × 3.16 × 107 sec/yr = 4.74 × 1014 sec Rate of emission = 1 photon/(15 million yr/atom) = 1photon/(4.7 × 1014 sec × atom) Finally then, Number of photons per second = Number of atoms × rate of emission = (1.19 × 1059 atoms) × 1 photon/(4.74 × 1014 sec × atom) = 2.51 × 1044 photons/sec 4. (a) This is similar to a half-life problem, with the appropriate formula being, Fraction remaining = (9/10)N, where N = the number of parsecs.

However, students might also solve this by brute force calculation.

For part a, after 10 pc, the fraction remaining is (9/10)10 = 0.35, or 35%. This is also easily enough directly calculated successively multiplying by 0.9 for each parsec: 0.9, 0.81, 0.73, 0.66, 0.59, 0.53, 0.48, 0.43, 0.38, 0.35.

Chapter 16

The Milky Way Galaxy

(b) There are several different ways to calculate this; one is to set up the formula ratio, ratio = (90/93)N. Then, (90/93)5 = 0.85 and (90/93)10 = 0.72.

Or, one can calculate the fractions of each directly, and then compute the ratio for the two steps: distance 1 2 3 4 5 6 7 8 9 10 blue: 0.90, 0.81, 0.73, 0.66, 0.59, 0.53, 0.48, 0.43, 0.38, 0.35. red: 0.93, 0.86, 0.80, 0.74, 0.70, 0.65, 0.60, 0.56, 0.52, 0.48. ratio at 5 pc = 0.59/0.70= .84 ratio at 10 pc = 0.35/0.48 = 0.73.

(These don’t match exactly the other answers exactly because of rounding at each step.

Without rounding so much the “brute force” ratios will match.)

5. To find the period of the Sun’s orbit around the center of the Milky Way, use D = Vt. The distance we want to use is the circumference of the orbit, which may be found from the radius of the orbit using C = 2R, where R is the distance of the Sun from the Galactic center: 13 17 R = 8 kpc = 8 kpc × 1000 pc/kpc × 3.09 × 10 km/pc = 2.47 × 10 km 17

18

so D = C = 2R = 2 × 2.47 × 10 km = 1.55 × 10 km 18 15 and P = t = D/V = 1.55 × 10 km / 220 km/s = 7.05 × 10 s.

15 7 8 finally, 7.05 × 10 s / (3.16 × 10 s/yr) = 2.23 × 10 years, or 223 million years.

6. To find the mass of the Milky Way based on the speed of objects at 16 kpc, we can use the equation M = V2 × R/G (one can also use Kepler’s third law, as done in the Astronomy by the numbers box). First, convert the 16 kpc into meters: 16 20 R = 16 kpc × 1000 pc/kpc × 3.09 × 10 m/pc = 4.94 × 10 m.

3

5

V = 230 km/sec = 230 × 10 m/sec = 2.30 × 10 m/sec.

5 2 20 -11 M = (2.30 × 10 m/sec) × 4.94 × 10 m / (6.67 × 10 m3 kg-1 sec-2) 31

3

2

-11

= 2.61 × 10 m /sec / (6.67 × 10 m3 kg-1 sec-2) 31+11 42 41 = 2.61/6.67 × 10 kg = 0.392 × 10 kg = 3.92 × 10 kg.

Finally, convert this value to solar masses: 41

30

11

3.92 × 10 kg × [1 solar mass / (1.99 × 10 kg)] = 1.97 × 10 solar masses.

11 ≈ 2 × 10 solar masses.

This result is a lower limit on the mass of the Milky Way, because it is only measuring the mass interior to 16 kpc. (Incidentally, this makes sense based on the results in the Astronomy by the numbers box in section 16.5, where the same velocity at a radius of 8 kpc yielded a result of 100 billion solar masses. The mass inside of R is proportional to R, after all.)

Chapter 16

The Milky Way Galaxy

The lower limit on the mass of the Andromeda Galaxy can be found by direct calculation 21 similar to the above, replacing 16 kpc with 35 kpc (1.07 × 10 m). However, it is simpler to notice that M is directly proportional to R, so, MAndromeda = (RAndromeda/RMW) MMW 11 = (35 kpc / 16 kpc) (1.97 × 10 solar masses) 11 = 4.31 × 10 solar masses.

The lower limit for the mass of Andromeda is a bit more than twice as much.

6

7. If the mass of the black hole at the center of the galaxy is 4 × 10 solar masses, we can apply the modified form of Kepler’s third law to find the period or semi-major axis of a star orbiting it: m + M = a3/P2, where M is in solar masses, a is in AU and P is in years. In this case, if m = mass of the orbiting star and M = mass of the galaxy, m will be tiny compared to M and we can ignore it. If the star’s closest approach is at 1200 AU, and its farthest is 2800 AU, then semi-major axis will be half of the sum of 2800 AU + 1200 AU = 4000 AU, so a = 2000 AU. 6

3

2

3

2

4 × 10 = a /P = 2000 /P 2

so and

3 3

6

9

6

9-3

P = (2 × 10 ) / 4 × 10 = 8 × 10 / 4 × 10 = 8/4 × 10 1/2 P = 2000 = 45 years.

= 2000

Star S0-20 has gone roughly 1/3 of the way around its orbit in 13 years in Figure 16.27B, and it was also traveling on the “fast” part of the orbit near the central black hole for part of this arc, so we can expect it to be rather slower on the other two-thirds.

13 yr × 3 + a little more ≈ 45 yr. The distances roughly match the figure as well.

(Students may have trouble remembering to apply Kepler’s second law when they compare to the figure, and consequently underestimate the period.)

8. Calculate the Schwarzchild radius for the black hole at the center of the galaxy using 6

4 × 10 solar masses for the mass. Substituting into the formula from Chapter 15, RSch = 2GM/c2 -11 6 30 8 2 = (2 × 6.67 × 10 m3kg-1s-2 × 4 × 10 × 1.99 × 10 kg)/(3 × 10 m/s) 10

7

= 1.18 × 10 m ≈ 1.2 × 10 km.

The shortcut way to do this problem is to realize that the Schwarzchild radius for 1 solar mass is 3 km, so for 4 million solar masses, it’s about 12 million km.

6

9. If the 4 × 10 solar mass black hole accumulated a million solar masses in a billion years, and the Milky Way is close to 13 billion years old (see “Age of the Milky Way” in the

Chapter 16

The Milky Way Galaxy

chapter), it must have accumulated the remaining 4 million solar masses over 12 billion years, at an average rate of: 6 9 6-9 -3 4 × 10 M☉ (12 × 10 yr) = 4/12 × 10 M ☉/yr = 0.33 × 10 M ☉/yr -4

= 3.3 × 10 M☉/yr That’s the same as the black hole absorbing 1 solar mass black hole every 3000 years.

-4 (1/(3.3333 × 10 M ☉/yr) = 3000 yr/M ☉).

Answers to Test Yourself 1. (b) The band of the Milky Way shows a concentration of stars.

2. (a,c) Most stars are not visible to the naked eye and have the same or less mass than the Sun. 3. (a) Pop I stars are young, blue, metal-rich, and in the disk.

4. (b) Pop II stars are old, red, metal poor, and make up most of the stars in globular clusters.

5. (a,b,d) The ISM can be seen as dark nebula and clouds show absorption spectra; matter creates absorption lines in stellar spectra; radio waves from the material are detected.

6. (d) Reflection nebula appear blue (they scatter/reflect blue photons).

7. (b,c,e) Evidence for spiral arms includes radio maps; gas clouds and HII regions within OB associations trace of spiral structure; other galaxies similar to the MW also have arms.

8. (a) Mass; in the same way we can use the orbits of the planets to determine the mass of the Sun. 9. (c,d) Evidence for black holes includes: other galaxies have them; orbits of stars near the center suggest a huge dark mass.

10. (d) The streams are material stripped out of smaller dwarf satellite galaxies.

Chapter 17

Galaxies

CHAPTER 17 GALAXIES Answers to Thought Questions 1. Hubble proposed that his tuning-fork diagram represented an evolutionary sequence of elliptical (E-type) galaxies evolving into spiral (S-type) galaxies. Since elliptical galaxies have little gas and dust for forming stars and they are made up of mostly old, Pop II stars with no hot, young blue stars, there is little or no evidence of ongoing star formation.

Elliptical galaxies appear to have used up or lost all their available gas earlier in their history, while spirals have abundant evidence of star formation. Thus, it is highly unlikely that E galaxies could evolve into S galaxies (at least without colliding with a gas-rich spiral or irregular). Today many astronomers believe that the opposite, an S galaxy losing its gas and becoming an E, is much more likely to happen (in clusters, for example).

2. DEMONSTRATE FOR STUDENTS. It would be impossible to make an error and classify a spherical galaxy as a highly elliptical one (record an E0 as an E7), but it would not be hard to mistake a highly elliptical galaxy for a less elliptical one if the long side was along the line of sight. There is not exactly a similar problem for spirals—spiral disks may appear in a variety of elliptical shapes depending on the angle at which they are seen, but it is known that the disk itself is circular and one can determine the angle. It can be difficult to assign the type of the spiral arms with certainty if the disk is highly inclined (edge-on). It can also be difficult to be certain if it is a barred spiral if the disk is edge-on.

3. If H were 100 km / sec / Mpc, which is a factor of 100/70 ~ 1.4 larger than the currently accepted value, the distances to galaxies and their inferred sizes would change. The measured recession velocities wouldn’t change. Since the distance = V / H, if H gets larger, then distance would get smaller by the same factor. Galaxies would be 1.4 times closer than we think now if H were really 100 instead of 70. The measured angular size would still be the same, but if a galaxy is closer the same angular size would mean that the galaxy must be smaller. Since L = 2d/360°, for H = 100, a galaxy would be (1.4 times) smaller than we now think (recall that the Moon and Sun are the same angular size, but the Moon is much closer and much smaller).

4. If you could travel instantly to any distance, you would need to travel out to a number of light years equal to your mother’s age. Light leaving Earth at the time she was born would then just be arriving there. Similarly, to look in on astronomers in ancient Egypt, you would need to be at a distance of about 5000 light-years. Since galaxies are millions of light-years away, we study them as they were millions of years ago. For astronomers to study truly young galaxies, they must study galaxies that have light travel time distances of billions of years. Instantaneous travel is of course impossible.

5. STUDENTS SHOULD REASON AN ANSWER. Some common things to think about: The sky is a very large place to search, and distant galaxies and small galaxies can be quite dim. Telescope time is limited. A telescope surveying a large area probably can’t point at any one part of it for long, so only the brightest or largest galaxies will be seen—especially at larger distances, the survey will miss a lot of smaller, dim galaxies. Surveying a smaller patch of sky for a longer time might seem to be the answer, but this approach runs the risk

Chapter 17

Galaxies

that the survey area might be mostly a void, where few galaxies would be found compared to the average, or a cluster, where too many would be found. Furthermore, if the survey covered a cluster, more ellipticals and more large ellipticals would be found compared to spirals; if the survey covered small groups, an excess of spirals or irregulars might be found. Keep in mind that until surveys were made, astronomers did not know that any of these trends existed—and there could be further trends that remain unidentified, because the trend can’t be observed (for example, what if the number of dim galaxies drastically increased at large distances, or drastically decreased—if you can’t see them anyway, you wouldn’t know!).

6. The jets coming out of active galaxies are much more powerful than those in bipolar outflows, but both have more or less the same structure. Material flowing in towards the central object accelerates and heats traveling in the accretion disk. Some of the material reaching the center falls onto the forming star or into the black hole. Some of the material is accelerated and flung away perpendicularly from the disk, most likely through a mechanism involving magnetic fields around the central object and in the disk. Highly collimated jets result. Astronomy has many examples of similar behavior happening at very different physical scales.

Answers to Problems 1. We can find the distance to the neighboring galaxy using the method of standard candles described in Chapter 13, by applying the inverse-square law for brightness to the Cepheid 2

variable. For two standard candles, Bo/B g = (d g/d o) .

Here, Bo = Brightness of Cepheid in Milky Way Bg = Brightness of other galaxy do = Distance to Cepheid in Milky Way dg = Distance to other galaxy 6

Since the Cepheid6appears 10 times fainter than one in our own galaxy, 2 B /B = 10 = (d /d ) . o

Therefore

g

g

o 3

dg/do = 10 .

The Cepheid in the other galaxy is 1000 times farther away than the one in our galaxy. If the Cepheid in our galaxy is 1000 pc away, the distance to the other galaxy is: 3 6 dg = 10 × do = 1000 × 1000 pc = 10 pc = 1 Mpc. 2. To find the angular diameter of the Milky Way as seen by an alien astronomer in the Virgo Cluster, solve the angular diameter formula for a: L/(2d) =  / 360°  = L/(2d) × 360°.

Chapter 17

Galaxies

Since the angular size of the Milky Way is about 30 kpc (Chapter 16),  = 30 kpc /(2 × 18 Mpc) × 360° = 30 kpc /(2 × 18,000 kpc) × 360° = 0.0955° = 0.0955° × 60 arc minutes / ° = 5.73 arc minutes.

The Milky Way would be about a tenth of a degree, or 6 arc minutes.

3. Using a Hubble constant of 70 km/ (sec Mpc) and Hubble’s law V = Hd, we get d = V/H = 19,250 km/s / [70 km/ (sec Mpc)] = 275 Mpc. 4. Use the Doppler shift formula to determine the recession velocity of a quasar showing a shift in a spectral line of  = 0.158, and then use Hubble’s law to find the distance.

5 4 V = c () = 3 × 10 km/s × 0.158 = 4.74 × 10 km/s.

4

d = V/H = 4.74 × 10 km/s / (70 km/s/Mpc) = 677 Mpc. (This is actually the redshift and approximate distance for 3C273, the first quasar for which the redshift was recognized to be cosmological, by Maarten Schmidt in 1963.)

5. Accelerated electrons traveling at nearly the speed of light traverse the 1.5 Mpc radio lobe from the center to the edge. This is just an application of t = D/V, with V = c, since the electrons are moving almost at the speed of light.

8 t = 1.5 Mpc / (3 × 10 m/s) 6 16 8 = (1.5 Mpc × 10 pc/Mpc × 3.09 × 10 m/pc) / (3 × 10 m/s) 14 14 7 6 = 1.55 × 10 s = 1.55 × 10 s / (3.15 × 10 s/yr) = 4.9 × 10 yr It takes the electrons a few million years. A simpler way to solve this exploits the idea that the electrons would travel 1 ly in 1 yr, and that 1 pc is 3.26 ly: 6 6 t = 1.5 × 10 pc × 3.26 ly/pc / (1 ly/yr) = 4.9 × 10 yr.

6. A radio galaxy 50 Mpc from Earth has a 1 billion solar mass black hole at the center which has a physical diameter of twice the Schwarzchild radius (event horizon to event horizon). To calculate the angular size, we first need to calculate the Schwarzchild radius, then use that and the distance in the angular size formula.

Rsch = 2GM/c2 -11 9 30 8 2 = (2 × 6.67 × 10 m3kg-1s-2 × 10 M☉ × 2 × 10 kg/M ☉) / (3 × 10 m/s) 12

= 3 × 10 m

Chapter 17

Galaxies

We can sanity check this with the fact that the Schwarzchild radius scales linearly with mass and a 1 solar mass black hole has a Schwarzchild radius of 3 km, so 1 billion solar masses 12 should have a radius of 3 billion km, or indeed, 3 × 10 m.

12 12 So, L = 3 × 10 m × 2 = 6 × 10 m, 6 16 24 and d = 50 Mpc × 10 pc/Mpc × 3.09 × 10 m/pc = 1.5 × 10 m 12 24 Then  = 360°  L/(2d) = 360°  (6 × 10 m) /(2 (1.5 × 10 m)) -10 = (2.3 × 10 )° Converting this tiny number of degrees to arc seconds by multiplying by 3600, as there are 60 arc minutes in a degree and 60 arc seconds in an arc minute: -10 -7 -6 a = (2.3 × 10 )° × 3600 arc seconds / ° = 8 × 10 arc seconds ≈ 10 arc seconds This is about 1 millionth of an arc second, and ten times smaller than the resolution limit in very long baseline array radio interferometry, which uses the entire Earth as the baseline of the telescope. To get an image, you need a few times higher resolution than the angular size of the object, or at least 100 times better than the current limit. It’s a hard business to image black holes! (This problem is loosely based on the galaxy NGC 3801.)

7. To find the minimum size of the quasar, we assume the time that it takes to brighten is equal to the difference in the time it takes the light from the closest and farthest parts of the quasar to reach us. That is, we will use D = V t, with D = diameter of the quasar, V = speed of light, and t = time to brighten. This assumes the entire object brightens at the same time, but we do not see it happen that way because it takes some time for the photons from the “back” of the object to reach us.

36 hours = 129,600 s 8 13 10 D = (3 × 10 m/s) × 129,600 s = 3.9 × 10 m = 3.9 × 10 km (= 260 AU).

This region is roughly 13 times larger than the Schwarzchild radius of the 1 billion solar mass black hole in problem 6, so it’s not unreasonable to think it could be the accretion disk and region surrounding a supermassive black hole in an active galactic nucleus.

8. STUDENTS SHOULD MAKE ESTIMATES. The Local Group is about 30 Milky Ways across (3,000,000 ly divided by 100,000 ly); the ratio is about 30.

LG/MW = 3,000,000 / 100,000 = 30 The Oort cloud is about 200,000 AU across (in diameter; see figure 11.23 in Chapter 11).

200,000 AU / (63,240 AU/ly) = 3.16 ly.

MW/SS = 100,000 ly / 3.16 ly = 32,000 The ratio of the size of the Milky Way to the size of the Local Group is about 1000 times smaller than the ratio of the solar system to the Milky Way (and that’s including the Oort

Chapter 17

Galaxies

Cloud, which is 100,000 times farther from the Sun than Earth is). Distances in the Local Group are almost comparable to the sizes of galaxies, and there are a few dozens of objects in the group. Distances in the Milky Way are much, much larger than the star systems, and there are hundreds of billions of star systems in the galaxy.

9. The total mass of the cluster is 1015 M☉. We can compute the amount of mass in the form of galaxies, and in the form of hot X-ray gas, and subtract these values from the total. The remainder is the dark matter.

100 galaxies with masses of 1011 M☉: 100 × 1011 M☉ = 1013 M☉ 10 times as much mass in X-ray gas as in galaxies: 10 × 1013 M☉ = 1014 M☉ This will be easy to calculate if we put everything to the same power of ten.

Total mass of everything

= 1015 M☉ = 100. × 1013 M☉

Total mass of hot gas

= 1014 M☉ = 10. × 1013 M☉

Total mass of galaxies

= 1013 M☉ = 1. × 1013 M☉

Total mass of dark matter

= (100 – 10 – 1) × 1013 M☉ = 89 × 1013 M☉.

The dark matter is 89% of the total mass.

[89 × 1013 M☉ / (100 × 1013 M☉)] × 100% = 89/100 × 100% = 89%.

10. To find the mass within each distance, the simplest way is to use the formula: 5 M = V2 × R/G. 100 km/sec = 10 m/sec.

16

20

R1 = 5 kpc × 1000 pc/kpc × 3.09 × 10 m/pc = 1.55 × 10 m.

16 21 R2= 50 kpc × 1000 pc/kpc × 3.09 × 10 m/pc = 1.55 × 10 m.

5

2

20

-11

5

2

21

-11

M1 = (2.00 × 10 m/sec) × 1.55 × 10 m / (6.67 × 10 m3 kg-1 sec-2) 40 = 9.30 × 10 kg 40 30 10 = 9.30 × 10 kg / (1.99 × 10 kg/M☉) = 4.67 × 10 M ☉ M2 = (2.10 × 10 m/sec) × 1.55 × 10 m / (6.67 × 10 m3 kg-1 sec-2) 42 = 1.03 × 10 kg 42 30 11 = 1.03 × 10 kg / (1.99 × 10 kg/M☉) = 5.15 × 10 M ☉ Students can alternatively use the formulas from Chapter 16 (page 438), making sure to convert R into au and P into years in order to obtain an answer in solar-masses.

Chapter 17

Galaxies

NGC 3200 has about 47 billion solar masses of material within a radius of 5 kpc, and about 500 billion solar masses within a radius of 50 kpc. Stars and gas make up most of the material in the 5 kpc radius; dark matter is the dominant component inside the 50 kpc radius.

Answers to Test Yourself 1. (d) Elliptical galaxies are populated by old, red stars and have little gas and dust.

2. (c) d = V/H = 14,000 km/sec / (70 km/s/Mpc) = 200 Mpc. 3. (b) The motion of M31 towards us is more significant than the universe’s expansion.

4. (d) Of the choices given, a ring galaxy is the only likely result.

5. (b) Seyfert galaxies have small, bright nuclei with rapid gas motions.

6. (a,b,c) The Local Group contains about forty galaxies, is a poor cluster, and is the group containing the Milky Way. 7. (a) Gas is stripped out of galaxies, turning them into ellipticals.

8. (b) Clusters may have 10 times as much mass in the form of hot gas as in the form of galaxies.

9. (d) Most of the mass of a galaxy is dark matter (found in the halo).

10. (b) In a galaxy, the ratio is about 10 parts dark matter to 1 part visible matter

Chapter 18

CHAPTER 18

Cosmology

COSMOLOGY

Answers to Thought Questions 1. STUDENTS SHOULD ANSWER FOR THEMSELVES. 2. In the (nonrelativistic) Doppler shift described in Chapter 4, the emitting object or the observer is moving through space, and the shift is observed as the waves are emitted or received. Imagine a stationary emitter. As you move through space away from it, you collect the waves more slowly than if you were standing still. So you observe a shift of the wavelength to a longer value because the peaks seem too far apart. In the case of the cosmological redshift, we can imagine we are moving away from the galaxy, but not through space; it is the expansion of space itself that stretches the light to a longer wavelength. The light is modified by the expansion of space, not motion through it. (Relativistic Doppler shifts get complicated). The cosmological redshift is more similar to the gravitational redshift that happens near a black hole or neutron star.

3. If the universe had not expanded since recombination, assuming for the moment we exist and could look at the sky, the cosmic microwave background radiation would not be microwave wavelength. Instead it would be at its original wavelength, 1000 times shorter, and also at a 1000 times higher temperature of about 3000 K. The sky would be bright—it would look like the surface of a red giant or other M-class star! The night and day sky would look the same (if the Sun was present, it might look like a bright yellow patch against the bright red).

4. STUDENT SHOULD PRESENT AN ARGUMENT. Several pieces of evidence which each require building up a little background would be to explain that the Big Bang resolves Olbers’s paradox, Hubble’s law, the existence and temperature of the CMBR, and ratios of hydrogen and helium in the universe. For example, introduce the idea of Olbers’s paradox by analogy to standing in a forest and looking at trees, expand to discussing stars in the sky and the universe going on forever. But, since the sky is dark, you can then resolve the problem by requiring a time when the universe began. Introduce the Doppler shift as the concept of a siren going past you; then explain that with telescopes we can “hear sirens” from distant galaxies and that indicates they are all moving away from us. This in turn suggests that in the past, they were all very close together. Explain that we can also detect light from everywhere we look (the CMBR) that is very much the same, but also very weak (in temperature), with an analogy to how smoke from a barbecue thins out from very concentrated and white to very thin and tenuous as it spreads out in the wind. Explain how the Big Bang resolves the existence and weakness (current energy) of these radio waves; that they are leftover heat from when the universe was very small and very hot (use an example of a hot air balloon about how hot gases expand). For nucleosynthesis, it will be necessary to explain that everything the world is made out of was created in stars, but when we look out into empty regions of space where no stars exist to have made the material, there are still some elements there. The Big Bang explains very neatly not only what we see there but how much of it there is.

Chapter 18

Cosmology

5. Stars formed shortly after the birth of the Universe will have the same percentage of helium as the Universe itself, namely about 24%. Stars that formed later will have more helium because they will contain some matter that was processed in other stars which exploded and distributed the processed material before the later stars formed. These stars therefore contain a higher percentage of helium than the oldest stars. For example, our Sun contains approximately 27% helium.

6. DEMONSTRATE FOR STUDENTS. It’s 2 triangles on a sphere—one inside and one “outside.” It’s only one on the flat or saddle shape.

7. During inflation, the universe expanded very quickly—the Hubble constant was very large. This expansion was driven by a large cosmological constant resulting from the nature of space itself when space is filled with large amounts of energy. After a tiny fraction of a second, the inflation period ended, and the expansion slowed. Today, it appears that the universe is expanding faster than at any time since inflation, and that the expansion is accelerating. This expansion is similar to inflation in that the expansion is driven by dark energy, which acts as a cosmological constant which results from the nature (energy) of space itself. Notice in Figure 18.27 that the blue line, the dark energy model, shows a similar slope at the very beginning (inflation) and the very end (dark energy driven expansion).

8. In Figure 18.26, the vertical axis is the measured redshift and the horizontal axis is essentially the measured distance based on the method of standard candles. Remember that the larger the distance, the further into the past we are looking: the present day is at the lower left corner of the graph. If the universe always expands at the same rate, then the relationship of the measured redshift to the actual distance would be a straight diagonal line—the green line. This represents the present day ratio of redshift to distance; the slope of the green line is the current value of Hubble’s constant. The Hubble law is V = Hd, or d = V/H. The age of the universe goes as 1/H. If Hubble’s constant is smaller, the line is flatter and there is a slower expansion; when Hubble’s constant is larger, the line is steeper, and there is faster expansion.

Therefore, if points in the past were ABOVE the green line, the line would be steeper in the past. This means Hubble’s constant would have been larger in the past, and the universe would have been expanding FASTER in the past than in the present. However, if the universe is expanding faster in the present than in the past, we would expect the line to be flatter in the past—so the most distant galaxies end up BELOW the green line. This indicates that the universe was expanding slower in the past than it is now. (Another way to think about this is that for each point on the graph you could calculate V = Hd. For the point to be above the green line, the H at that distance—and therefore at that time in the past—would have to be larger than H today, which would mean the universe was slowing down. The fact that the points are below the green line indicates the opposite is true, and H was smaller in the past). The universe appears to be accelerating in its expansion, which means it must have been expanding more slowly in the past, so the galaxies appear below the line.

9. If the universe continues to expand faster and faster, eventually the space between galaxies will be increasing so quickly that even at speed of light, light would never make it from one galaxy to the next. Space can expand faster than the speed of light. Our cosmic horizon will “shrink,” in the sense that distant galaxies we can currently “see” will disappear

Chapter 18

Cosmology

when their light can no longer reach us. In the very distant future, only what are now the nearest galaxies might be visible (depending on when the expansion overcomes the gravity holding the galaxies together). If the only objects in the night sky were in one’s own galaxy, cosmology would be very different, and very difficult, indeed!

Answers to Problems 1. Calculate the age of the universe for Hubble constants of 50 and 100 km/s/Mpc. The age of the universe can be estimated by eliminating the distance units from Hubble’s constant, 7 taking the inverse, and remembering that a year is about 3.16 × 10 s.

13 1 Mpc = 3.09 × 10 km so age =

1 H

= 50

1 km

=

1 50



sec• Mpc



km

106pc 1Mpc



3.09 1013km

=

1pc

1 50

19

 3.09 10 sec

s•Mpc 19

7

12

= 1/50 × 3.09 × 10 sec/(3.2 × 10 sec/yr) ≈ 1/50 × 10 yr 10 = 2 × 10 yr 10 (To 3 digits the answer is 1.96 × 10 yr).

If H were 50 km/s/Mpc, the implied age of the universe is 20 billion years. If H were 100, 12 10 age = 1/100 × 10 yr = 10 yr then the age comes out to only 10 billion years—a problem since some stars and the Milky Way are believed to be older!

2. The temperature of the universe at recombination was 3000 K. The temperature of the CMBR today is about 3K. Since recombination, the light of the CMBR has been stretched out to longer wavelengths by the expansion of the universe, so

now Trec 3000K = = = 1000 .

3K rec Tnow The light was stretched by a factor of 1000 – the universe must be about 1000 times wider today (and so a million times larger in volume) than it was at recombination.

9

3. Compared to when the average temperature was 3 × 10 K, the universe is a billion times larger:  now Tthen 3109 K = = = 10 9 then Tnow 3K If we shrank the Milky Way by a factor of a billion, it would have a diameter of

Chapter 18

5

Cosmology

9

-4

-4

4

10 ly / 10 = 10 ly = 10 ly × 6.3 × 10 AU/ly = 6.3 AU. At 6.3 AU, the entire Milky Way would have fit in the space between the Sun and Saturn (Jupiter’s orbit is 5.2 AU and Saturn’s orbit is 9.5 AU.) 9

4. If the increase in the universe’s diameter is by a factor of 10 , the change in the radius is also a factor of a billion, and the volume, which scales with the cube of the radius, would 9 3 27 increase (10 ) = 10 times its size when the universe was a few minutes old and it had a 9 temperature of 3 × 10 K. All of these are truly astronomical numbers.

5. The temperature of the universe at recombination was about 3000 K. Use Wien’s law to calculate the peak wavelength of the radiation at that temperature: 6 m = 2.9 × 10 K nm / T 6

= 2.9 × 10 K nm / 3000 K = 970 nm.

970 nm is in the infrared part of the spectrum. M-type stars have a surface temperature of 3000 K; the entire universe would have been much like the surface of a red dwarf or red giant (except there would only be H, He and a little Li and Be present).

6. The mass of the solar system is essentially the mass of the Sun, 2 × 1030 kg. If the density of the universe was 0.1 kg/L (= 0.1 g/cm3), then the volume occupied would be V = M/D = 2 × 1030 kg / (0.1 kg/L) = 2 × 1031 L. One liter is 1000 cm3, so this is also: 2 × 1031 L × 1000 cm3/L = 2 × 1034 cm3 = 2 × 1028 m3.

25

7. How many doublings add up to a factor of 10 ? We want to find x, in X 25 2 = 10 .

This can be solved exactly by using logarithms, X log 2 = 25 log 10 X = 25 log 10 / log 2 = 83.05 Or you can estimate it: 82 24 83 24 84 25 2 = 4.8 × 10 , 2 = 9.7 × 10 , 2 = 2 × 10 .

It may be surprising to think that the universe only needed to double in size 83 times during inflation, but that is the power of exponential growth. (You can demonstrate the speed of exponential growth by seeing how many times students can successively fold a piece of paper in half.)

Chapter 18

Cosmology

8. According to section 18.6, the critical density is 3 × 10–30 g/cm3. The dark energy is 73% of this, or 0.73 × 3 × 10–30 g/cm3 = 2.19 × 10–30 g/cm3.

We’ll have to make a few unit conversions. 1 L is 1000 cm3, and 1 g is 1000 kg, so this density is also 2.19 × 10–30 kg/L. (2.19 × 10–30 g/cm3 × 1000 cm3/L × 1kg/1000g = 2.19 × 10–30 kg/L.) So 1 L of space contains 2.19 × 10–30 kg of dark energy. Using the equivalence of matter and energy, E = mc2 = 2.19 × 10–30 kg × (3 × 108 m/sec)2 = 1.97 × 10–13 J and so there are about 2 × 10–13 J/L of dark energy in space.

Photons have energy E = hc/, so each 1 mm photon has and energy of E = hc/ = (6.63 × 10-34 J s × 3 × 108 m/s) / (10-3 m) = 1.99 × 10-22 J. So the dark energy in 1 L is equivalent to the energy of 2 × 10-13 J/(2 × 10-22 J/photon) = 10-13+22 = 109 photons ~ 1 billion photons!

(The exact number is 1.97/1.99 × 109 = 0.99 × 109, but it makes sense to round in this problem. A more exact calculation of the critical density using H = 70 km/sec/Mpc gives a value of approximately 9.2 × 10-30 g/cm3, which in turn produces a final result of 6.7 × 10-30 g/cm3 = 6.7 × 10-30 kg/L, an energy of 6.0 × 10-13 J, equivalent to about 3 billion photons.)

Answers to Test Yourself 1. (b) The evidence is cosmological redshift of spectra of galaxies.

2. (b) Time is inversely proportional to the value of H; the universe would be twice as old.

3. (d) It was radiation released at the recombination era, early in the history of the Universe.

4. (c) Significant amounts of hydrogen fused into helium in the first three minutes.

5. (b) Most of the time, it has not been very different – recombination happened about 13 billion years ago.

6. (a,c) Inflation explains the high uniformity of the background in different directions and the flatness of the universe.

7. (b) If the density exceeded the critical density, in the absence of dark energy the universe would contain enough matter for gravity to overcome the expansion and the universe would collapse.

Chapter 18

Cosmology

8. (a,c,d,e) Dark energy is represented by the cosmological constant, acts against gravity, causes space to expand faster, is evenly distributed through space. It is not associated with dark matter.

9. (c) The universe appears to be 72% dark energy, 23% dark matter, 4% regular matter (and light).

Essay 1

Backyard Astronomy

ESSAY 1 - Backyard Astronomy Answers to Thought Questions 1. If a planet is at opposition and you see it overhead it must be midnight because the planet is 180 degrees from the Sun and rises at sunset.

2. No. Objects in the western sky are setting; Mercury never gets very far from the Sun, so the only sky you could see it in near dawn would be if it rose just before the Sun, in the east.

Mercury is only visible in the western sky just after sunset (when it is on the correct side of the Sun to be seen then).

3. For Venus and Mercury to appear close in the sky, Mercury could be at almost any point in its orbit but Venus would need to be in the “middle parts” of the orbit as seen from Earth, within the 28° or so from the Sun that is the range that Mercury can reach. They might appear the closest when both are near inferior conjunction (see figure E1.8), between Earth and the Sun. Being so close to the Sun, Mercury can be difficult to spot even at greatest elongation, so they would not be easy to see when close together.

4. On a sunny day at sea, there would be a great contrast in the conditions on deck (especially with light reflected off the water, too) and below deck. An eye patch would allow a pirate to keep one eye dark adapted, so he could go below deck, switch the patch over, and quickly be able to see around the interior of the ship. Of course, navigating the ladder or stairs on a moving ship without depth perception offers its own perils.

Answers to Test Yourself 1. (d) Each hand is about 20°, each thumb is about 2°, and fingers are about 1°, so 42° would be two hands and a thumb.

2. (c, d) Altitude and azimuth, being measured in reference to the horizon, would both change.

Essay 1

Backyard Astronomy

3. (d) You are probably in the Southern hemisphere. The stationary star to the right is near a pole star, but if the star to the left has moved up, it is moving clockwise around the star to the right. This means the pole star near the star to the right is the south celestial pole, so north is in the opposite direction of the star to the right.

4. (b) Sunrise. If a planet is at inferior conjunction, it is seen against the disk of the Sun, so it must rise and set with the Sun. 5. (a) If Mercury is at greatest western elongation it is close to 28 degrees west of the Sun (imagine their relative positions at noon). It rises before the sun in the morning.

6. (d) Venus is an inferior planet (orbiting the Sun closer than Earth) and so can be at inferior conjunction, i.e., it can pass between Earth and the Sun. 7. (b, d) When your eye is dark adapted, its pupil is dilated and it is much more sensitive to light than during the day.

Essay 2

Special and General Relativity

ESSAY 2 – Special and General Relativity Answers to Thought Questions 1. STUDENTS SHOULD PROVIDE A REASONED RESPONSE. 2. LOOK FOR REASONED ANSWERS AS THESE QUESTIONS ARE OPEN TO SOME INTERPRETATION. Formally, at c, the Lorentz factor becomes infinite, suggesting time would be infinitely slowed down for a photon. Likewise, the time dilation effect somewhere where the escape velocity is equal to the speed of light—like at the event horizon of a black hole—would be as extreme as possible and hence to an outside observer, time would appear to slow to a stop at the event horizon (and any light emitting from such a region is redshifted to zero energy). However, the point of view of an observer at these locations is harder to imagine clearly: if a space traveler traveling at a high Lorentz factor essentially considers the universe moving past her at high speed, and does notice the time dilation, it is difficult to say what the photon traveling at c actually experiences, whether it is trapped at the event horizon or winging through space.

3. Students ideas should reflect the ideas of special relativity accurately. In particular, the times messages might be sent and received by a group that goes out, turns around, and comes back can be interesting.

Answers to Problems 1. To travel 100,000 light years in a time of 10 years, you would need to travel such that the 100,000 light years was contracted to an apparent distance of approximately 10 light years—a Lorentz factor of 10,000. Solving the equation for v/c, 2 1 2 1 v  = = 10,000 so 1 −  v  = = 10−4 and 1 −   = 10−8 .

2 10000 c c v 1−   c 2

v v So   = 1 − 10−8 and therefore   = 1 − 10−8 which is very, very, very close to 1— c c you have to be traveling at approximately the speed of light. Numerically, if you use a calculator with more than 8 places, you can solve this out as: -8 (1-10 )1/2 = (0.999999990)1/2 = 0.9999999949999999874999999375…

Alternatively, there is an algebraic solution: (1-A)1/2 = 1 – ½ A. So, -8 -8 (1-10 )1/2 = 1 – 0.5 × 10 = 0.999999995.

2. Mercury orbits the Sun at speeds from 59 km/s to 39 km/s at its nearest and farthest distances from the Sun. The fractions of the speed of light are: -4 59 km/s / 300,000 km/s = 1.967 × 10

Essay 2

Special and General Relativity

-4

39 km/s / 300,000 km/s = 1.3 × 10

The Lorentz factors are both approximately 1; the speeds are fairly slow (consult the Lorentz factor figure in the text). For the larger value:

=

1 2

v 1−   c

=

1 1 − (1.967 10−4 )

2

=

1

1 − (3.869110−8 )

=

1 0.999999961309

Using a calculator with more than 8 decimals (e.g., many of the moderately advanced pocket calculators or the calculators on computers), you can solve this out as =1.0000000193455005613725734750552 For the smaller value, 1 1 1 = = = = 2 2 −8 ) 4 − 1 − ( 1.69 10 v 1 − (1.3 10 ) 1−   c

1 0.9999999831

= 1.0000000084500001071037515083778 It is a small, but astrophysically important and measurable, difference between 1.00000001934 and 1.00000000845.

There would be a slightly increased Lorentz contraction of space, and time dilation, when nearer the Sun. Analytically, the ratio of the values would look something like this: 39 km 2

1 − c 2s near c 2 − 392 = = far c 2 − 592 59 km 2 1 − c 2s 3. The escape velocity from the Sun is 76 km/s at Mercury’s closest distance to the Sun and is 62 km/s at Mercury’s farthest distance. Using the gravitational time dilation formula, find how slow time runs in these two places.

The answers are again extremely close to 1 for both numbers, and require more than an 8 digit calculator (note that on some computers, the number of decimals can be set in the calculator application’s options).

Essay 2

Special and General Relativity

1 2

1

gives

76 km v 1 − esc2 1 − c 2s c dilation farther away.

2

1

for the larger dilation near the Sun, and 1−

62 km 2

for less

s

c2

Solving this out: -4 v/c = 76 km/s / 300,000 km/s = 2.533 × 10

=

1

1

=

1

=

1

=

1 − (6.416089 10−8 ) 1 − (2.533 10−4 ) v  1 −  esc   c  = 1.0000000320804465437325096364142 = 1.000000032 v 1 − cesc2

2

2

2

v/c = 62 km/s / 300,000 km/s = 2.067 × 10 =

1 v 1 − cesc2

2

=

1 v  1 −  esc   c 

2

=

1

=

0.99999993583911

-4

1 1 − (2.067  10−4 )

2

=

1

1 − (4.272489  10−8 )

=

1 0.99999995727511

= 1.0000000213624456845311089391345 = 1.000000021 This is again a small but clearly non-zero difference. Notice that in fact there is some time dilation happening at both the near and far points of the orbit—Mercury is close enough to the Sun we need to use General Relativity to understand even the farther part of the orbit. A few conclusions can be taken away from these numbers. First, if you compare the results here to problem 2, we notice that the General Relativistic effects we are looking at here are of the same approximate size as the Special Relativistic effects we calculated in problem 2.

It’s necessary to include both SR and GR to correctly predict Mercury’s orbit. Second, time is running at different speeds in the near and far parts of Mercury’s orbit—the difference in -8 the correction factors is 1.00000001934 -1.00000000845 = 1.089 × 10 from problem 2 and -8 1.1 × 10 from problem 3, together a few parts in 108. That means every for every 108 seconds that passes far away from Mercury, 2.2 seconds less pass in Mercury’s orbit. 108 seconds is only about 3.17 years, so Mercury’s timing would be off by a minute in only 86 years. Over astronomical time scales, Mercury’s timing would be quite out of sync with, say, Earth’s. Moreover, if Mercury is moving (from our perspective) at 59 km/s in its orbit, a difference in timing of two seconds would put the planet 120 km away from where you would expect it to be!

Additionally, the SR and GR corrections will affect the apparent mass of Mercury, it will be more massive closer to the Sun. Newton’s laws predict elliptical orbits, but Mercury itself would seem like a different object on a different path if you look at it close to the Sun or far from the Sun. Observationally, it has been known for centuries that the ellipse of Mercury’s orbit precesses around the Sun (the perihelion and aphelion points move); and that the measured precession of Mercury’s orbit cannot be explained by Newtonian mechanics; only since Einstein do we know it happens because of the relativistic effects.

Essay 2

Special and General Relativity

Answers to Test Yourself 1. (e) Everything seems normal to the astronaut (remember, motion is relative).

2. (c) All observers always measure the speed of light in a vacuum to be c.

3. (b) His sister will be older (it can be shown that Bob has accelerated compared to Earth when he turns around, and you can tell he is the one with the dilated time on the round-trip).

4. (c) The principle of equivalence states that gravity is the same as being in an accelerating reference frame.

5. (d,a,b,c) The time dilation is stronger near more massive bodies.

Essay 3

Keeping Time

ESSAY 3 - Keeping Time Answers to Thought Questions 1. The year is nearly 360 days long—and Earth completes one orbit around the Sun in that time. It’s a natural association, and rounding it off makes geometry a little easier.

2. The 8766 hours in a current year (365.25d × 24h/d) would be divided into 180 solar days of 48.7 hours (8766/170), or 48 hours 42 minutes. For each of these new solar days, Earth would move about 2/d (= 360/y / 180d/y) along its orbit. So Earth must be rotating an extra 2 (compared to the stars) for a solar day than a sidereal day; it’s rotation is 362/48.7h. The length of the sidereal day is then 360/(362/48.7h) = 48.4h.

3. At one point in history, February WAS the end of the year—for the ancient Roman calendar that is the basis of the modern calendar, which started in the spring, in March. This also helps explain the other subtractions made from February in Roman times.

4. At latitudes near the equator, the length of the day and the temperature change little with the passing of the seasons. Crops could be grown year-round. However, the amount of light at night changes with the full moon. It could make sense for an equatorial civilization to develop a lunar calendar. Even though the stars will be out of sync with the dates year-toyear, the full moon will be predictable. On the other hand, at high latitudes north or south of the equator, large changes in sunlight and temperature cause very noticeable seasons annually. Crops, building, and travel must be timed with the seasons. The Moon would be less important to a culture at a high latitude who would create a solar calendar.

5. STUDENT SHOULD SUGGEST CHANGES AND JUSTIFY THEM. Answers to Problems 1. Each degree of longitude corresponds to 24h / 360 hours of time = 1h / 15 = 60 min / 15 = 4 min/1.

Your friend’s clock will be 5 × 4 min/1 = 20 minutes ahead of yours.

2. At the equator, day and night are always about twelve hours each (neglecting effects of refraction in the atmosphere, etc.). The equator, being halfway from the North Pole to the South Pole, is the one place on Earth that does not move off of the terminator line between day and night because of Earth’s tilt. Only if Earth was tipped 90, so that at on that day from the equator the Sun would be half above and half below the horizon, would the equator not experience 12 hours of day and 12 hours of night. (See figure below, which shows the terminator at dawn.)

Essay 3

Keeping Time

3. The period of the phases is 29.5 days, so the 10 day “French Revolution week” would work pretty well for the whole cycle—three weeks (30 days) from full Moon to full Moon.

However, it’s probably more useful to see about a week pass between each phase, for which the 7 day week is better matched—four weeks between full Moons is 28 days. This is short of the whole cycle by 1.5 days, or 1.5/4 = 3/8ths of a day = 9 hours per phase. So, from one phase to the next it’s still closer to 7 days than 8. The 8 day week isn’t quite as good a match, 32 days over four weeks means it is long by 2.5 days per synodic month, or 0.625 = 5/8s = 15 hours too long phase to phase. The 7 day week is the best compromise.

4. The year is 365.2444 days long. Without leap days, the year would always be 365 days long, 0.2444 days short of the whole orbit each year. After one year, it would be 0.2444 days short. After 2 years, 0.4888. After 4 years, 0.2444 x 4 = 0.9776, the seasons and equinoxes would start a day late on the calendar. For the start of the seasons to shift all the way around 365 days back to the same date would take approximately 365 / .25 = 1460. or more exactly, 365 / 0.2444 = 1493.45 years (things line up again exactly in the middle of year).

Answers to Test Yourself 1. (b) The solar day is longer than the sidereal day, so the same stars rise earlier each night.

2. (c) If the axis were not tilted with respect to the ecliptic, night and day would be about equal during the year.

Essay 3

Keeping Time

3. (e) At the winter solstice (Dec. 21), the North Pole has 24 hours of night. The South Pole must have 24 hours of daylight.

4. (b) At present, we have a leap year every 4 years because 1/0.25=4. If the year was 365.2 days long, then we would divide the remainder (.2) into 1 to get 1/0.2=5, and we would have a leap year every 5 years.

Essay 4

Life in the Universe

ESSAY 4 - LIFE IN THE UNIVERSE Answers to Thought Questions 1. OPINION REQUIRED, but I personally would be depressed if we were the only life in the Universe.

2. SPECULATION PERMITTED. The other stars that formed along with the Sun would have been made up of the same stuff, and the Sun itself is reasonably typical, so we might infer that the nearby neighbors to the Sun are both similar stars and from the same original cloud. If there was anything special about the material or dynamics that made up our solar system (like the presence of certain compounds, or the height in the Galactic disk when the stars formed in the cloud, or something weirder) it’s reasonable to think the same things would have happened to the Sun’s relatives, and therefore it might improve the chances of finding life since life exists in our solar system. To the extent the cloud was perfectly typical, it would not affect the chances.

3. STUDENTS SHOULD PRESENT AN ARGUMENT. Jumping off points here would be the idea that life itself must have started many billions of years ago, but intelligent life required several billion years to develop, so the harder step might be developing intelligence. Of course, it might have been that life got started quickly on Earth accidentally, and it might be that once started, evolutionary pressures inevitably lead to intelligent life, so the harder step might be the first one.

4. STUDENT OPINIONS. Certainly if life is discovered elsewhere that is a similar biology to life on Earth (handedness of biologically produced chemicals, etc.) it will have been important that appropriate protection measures were taken, just to be able to prove that the discovered life was not accidentally transplanted by us.

5. OPINION. I suspect that the first reaction would be fairly chaotic, before any coherent response could be put forward. There would be a large faction worried about an alien invasion. I wonder how the major religions would take it?

6. By example on Earth, more advanced societies have destroyed, exploited and radically changed more primitive ones. Unless the aliens have some sort of "Prime Directive" of noninterference, we might have problems.

7. The middle annulus of the disk of the Milky Way is the best bet—the extreme edges have less star formation and colder temperatures, and might be slow to form stars and planets. They are also exposed to more extra-galactic radiation. Conversely, the central portion of the Milky Way and the disk are over-populated with young stars that explode and could destroy civilizations with gamma-ray bursts. Radiation from the center of the galaxy is intense, and tidal disruptions from the denser population of stars could create problems (comets, disrupted systems, etc.).

8. STUDENT’S MESSAGE. Answers to Problems

Essay 4

Life in the Universe

1. According to the essay, homo sapiens evolved about half a million years ago, so compared to the age of Earth, 100% × 500,000 yr / (4.5 × 109 y) = 100% × (5 / 4.5) × (105/109) = 100% × 1.1 × 10-4 = 1.1 × 10-2 % = 0.011%.

This is just more than 1 hundredth of a percent, a pretty small number!

2. The answer to the question depends on whether we consider the growth to be some kind of exponential or linear, and exponential is a much, much better estimate of population growth.

Over fifty years, the population increased 6B/2.5B = 2.4 times. Therefore, we might expect in 2050 to increase to 6 B × 2.4 = 14.4 billion, and by 2100 to increase to 14.4 B × 2.4 = 35 billion.

(If the growth were linear, which is not a good approximation, we would add 3.5 billion every 50 years, for 9.5 billion and 12 billion).

3. STUDENTS HAVE TO SELECT A STAR AND CALCULATE THE TIME FROM t = D/V. For the 2nd part, 50,000 light years  9.46  1012 km/ly / (10 km/s) = 4.73  1016 s.

There are 3.16  107 s/yr so that’s 1.50  109 yr – only a billion and a half years! Material really could get around. Recall that the Sun’s orbit around the Galaxy takes about 220 million years.

Students should give a reasonable argument that considers the needs of bacteria and the conditions faced.

4. Assume that the speed remains the average that it has been over the first 20 years. Pluto is about 39.5 AU from the Sun, so the Voyagers were at about 79 AU after 20 years. We can estimate the Voyagers’ speed: V = D/t = 79 AU / 20 yr = 3.95 AU / yr.

It’s convenient to express this speed in light-years per year: 3.95 AU/yr × 1 ly / (63240 AU) = 6.25 × 10-5 ly/yr.

So, 4.2 ly / (6.25 × 10-5 ly/y) = 6.7 × 104 yr = 67,000 yr.

It will take Voyager 67,000 years to travel as far as the distance to the nearest star.

5. 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 the “Arguments for Many Worlds” section. The formula determining the average areas in terms of the size of the Milky Way disk is  d 2 2R N    =  (R )2 which solves to d = MW .

I MW 2 NI If, for example, NI is 10,000, we get

Essay 4

Life in the Universe

d=

2RMW 2  50,000 ly 105 ly = = = 1000 ly , as in the text.

102 NI 104

6. For there to be a civilization within 30 light years of our own, 2 d = 2R / N  N = (2R / d )

Substituting the radius of the Milky Way (50,000 ly) and the distance to the nearest civilization,

N = (2(50,000 ly) / 30 ly) = (100000 / 30) = 1.1107 2

2

We would need 11 million civilizations.

Answers to Test Yourself 1. (a) There are fossil algae more than 3 billion years old.

2. (c) Conditions on the early Earth were suitable to make the building blocks of life.

3. (c) The biggest problem for a life orbiting a high-mass star is that high-mass stars only last a few tens of millions of years.

4. (d) The number of technically advanced civilizations.

5. (b,d,e) Otherwise we’d be colonized, we should receive signals, we might be an exception.

Note that civilizations keeping to themselves (a) still requires other civilizations to exist.

6. (b) Gaia - life changes its own environment.

7. (d) If the Universe were very different, we wouldn't be here to observe it.

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The Cosmic Landscape

PREVIEW: THE COSMIC LANDSCAPE Note: The video "Powers of Ten" is an excellent visual summary of the material. Some narration by the instructor may help students better understand the numbers that flash on the screen during the video.

The problems in this chapter are a good place to have students practice working numerical problems by hand to estimate solutions, rather than always reaching for the calculator.

Answers to Thought Questions 1. Earth, Solar System, (optional: galactic spiral arm), Milky Way Galaxy, Local Group, Virgo Supercluster, (optional: cluster of superclusters), the Universe.

2. Student responses could be examined based on (1) is the idea testable/answerable by science, (2) did they provide a clear hypothesis and if appropriate, discuss any obvious or existing data that provide a starting point, (3) did they provide good questions to try to answer and good tests to try; (4) did they discuss the implications of failed tests for the hypothesis. It may be interesting to let students present these ideas and discuss whether the testing of the hypothesis is in line with the ideas of the scientific method.

3. This is a somewhat open-ended question, but any theory of a new force would need to pass all existing experimental tests with equal or better agreement than the old theory, and provide explanations and pass experimental tests (including predictive tests) for phenomena at size scales, energy ranges, or in some other region that are not explained by existing theory. A new theory could replace a previous theory, if, for example, it extended the physical range of parameters to which the old theory applied. This has happened many times in physics. A good example of this would be how the general theory of relativity explains gravity over a broader range of circumstances than Newton’s universal law of gravity, and can, for example, explain the orbit of Mercury where Newton’s laws fail. Similarly, a combined theory of electromagnetism explains the phenomena of electricity and magnetism.

Therefore, any scientific explanation for ghosts or psychic powers must be consistent with, or provide a better explanation for, existing theories of electromagnetism, gravity, special relativity, and so on. One could only postulate a hypothesis that explained ghosts as manifestations of a new part of the electromagnetic spectrum if the postulate, in addition to passing the usual scientific tests for falsification, prediction, etc., also did not contradict existing results. Similarly, one could not postulate a mechanism for the exchange of ideas telepathically via some kind of radio transmission between organisms and simultaneously claim faster-than-light communication was possible, at least not without providing some credible and testable reason to believe that the special theory of relativity is incorrect! To consider a different issue, frequently, claims of ghost sightings refer to changes in temperature related to where the ghosts appear. To formulate a scientifically grounded theory, one would need to produce hypotheses that are consistent with other known rules of thermodynamics.

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The Cosmic Landscape

In short, there is plenty of room to propose new hypotheses, but they cannot contradict known science (unless they are able to convincingly supersede existing theories).

Answers to Problems 5 3 1. RSun = 7  10 km. REarth = 6.4  10 km. To show that the Sun is about 100 times the size of Earth, divide the Sun’s radius by Earths: 5

3

RSun/ REarth= 710 km/ (6.410 km).

We can use scientific notation to easily show that this is almost 100, because 7/6.4 is approximately 1. Therefore, 5

3

5

3

RSun/ REarth ≈ 10 km / 10 km = 10 / 10 = 105-3 = 102 = 100.

(The exact answer is 109 times.)

13

8

2. Using the given values, divide 1 ly = 1  10 km by 1 AU = 1.5  10 km to find the number of AU in one light-year.

13 8 13-8 5 4 4 10 /(1.510 ) = 10 /1.5 = 10 /1.5 = (10/1.5)  10 = 6.7  10 = about 67,000.

(The exact answer using unrounded values is about 63,240 AU.) 3. There are several ways to solve this problem. One easy way is to compare Earth’s radius or diameter to the Moon’s radius or diameter. The Moon’s diameter is about ¼ Earth’s. Since C = π × D, then the circumference is also ¼ Earth’s. So if a volleyball is about 68 cm in diameter, the Moon model would be 0.25 × 68 cm = 17 cm. The Moon’s diameter is C/π= 17 cm / π = 5.4 cm.

A slightly more accurate calculation using the values of Earth and Moon’s size from the Appendix (the Moon’s radius is 1738 km; Earth is 6378) shows the Moon’s radius to be 0.27 times Earth’s; therefore the diameter and circumference are also 0.27 times Earth’s; this gives a circumference for the model of the Moon of 18.4 cm and a diameter of 5.8 cm.

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

5

4. Use the formula time = distance / velocity. For light, V = c = 3  10 km/s. The distance 8

to Eris in the Appendix is 67.7 AU, and 1 AU =1.5  10 km, so…

8 10 d = 67.7 AU  1.5x10 km/AU ≈ 100  108 km = 10 km.

5

The speed of light, c, is 3  10 km/s. So, 10 5 4 t = 10 km / (3  10 km/s) = 3.33  10 s.

Divide by 60 s/min to get 564 minutes.

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The Cosmic Landscape

5. The Milky Way has a diameter of 100,000 ly, which we will represent as 2cm. According to Table P.1, the Local Group is about 50 times bigger than the Milky Way, with a radius of 2.5  106 ly and diameter of 5  106 ly (= 50  100,000 ly). Therefore, the Local Group would be about 100 cm in diameter.

For an exact calculation, Size (Local Group) / Size (Milky Way) = constant, so 5 106 ly 5 106 ly = x so x=  2 cm = 5 10 6−5  2 cm = 5 10  2 cm = 100cm 5 5 10 ly 2cm 10 ly The Virgo Supercluster is about 20 times the radius of the Local group, so it would be about 20  100 cm = 2000 cm or 20 meters in diameter at this scale.

The Visible Universe can be represented by a distance of 14 billion light years. Although technically this is not an accurate measurement of the physical size because of the expansion of the universe, it will do for the estimate. 14  109 ly / (50  106 ly) ~ 3  102, so the model of the Visible Universe would be another 300 times bigger than the Supercluster, or 6000 meters, or 6 km.

There are many ways to solve this problem using proportions and the data in Table P.1. It can be a good exercise for working out scientific notation by hand.

6. Given one object's speed and distance relative to another object, we can find how long it took the two to move apart by using the relation1 D = v t, where D is the distance, v is the speed, and t is the time. Solving for t gives, t = D/v. Inserting values for v and D, we find t = 3  108 ly / (6,000 km/s). We have two different distance units: ly on top and km below, and need to express ly in km. From the appendix we can find that 1 ly = 9.5  1012 km. Thus, km 3108 ly  9.5 1012 ly = 3  9.5 108+12−4  km = 4.8 1017 s t= km km 6 6000 s s To express this in years, we need to divide by the number of seconds in a year = about 3.2  107. Thus, t = 4.8 1017 seconds/3.2  107 seconds/yr = 1.5  1010 yrs, or 15 billion years.

This is one way that astronomers deduce the age of the Universe (soon after the Big Bang, the matter in those galaxies was close together).

1

This is the version of the formula used in the Appendix.

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The Cosmic Landscape

7. Key here is to help students understand the question being asked: if we want to know how many times smaller than a bacterium is a hydrogen atom, we need to divide the size of the bacterium by the size of the hydrogen atom. Since the hydrogen atom is smaller than the bacterium, the number we get by doing this will be larger than 1. (size of bacterium)/(size of hydrogen) = 10-6 m/10-10 m = 10-6-(-10) = 104 The hydrogen atom is 10,000 times smaller than the bacterium. It’s important for students to check that their answer makes sense—if they get 10-4, they should realize that would indicate the hydrogen was BIGGER than the bacterium—and since that doesn’t make sense that indicates they inverted the division. On the other hand, it is true to say that the size of the hydrogen atom is 10-4 of the size of the bacterium.

8. (105  (102)3 ) / (100  104  (108)1/2) = (105  102 x 3 ) / (102  104  108/2) = (105  106 ) / (102+4+4) = 1011 / 1010 = 10.

9. (8  106)2 / (2  10-3)3 = (82  106 2) / (23  10-3x3) = 64/8  1012 /10-9 = 8  1012 109 = 8  1021.

10. (3 x 105)2 / (4 x 104)1/2 = (32  105  2) / (2  104/2) = 9  1010 / (2  102) = 9/2  1010-2 = 4.5  108.

Answers to Test Yourself 1. (c) Jupiter's diameter is about 10 times larger than Earth's diameter.

2. (c) Over time they move 3. (b) The light year is a unit of distance.

4. (c) Your cosmic address is Earth, Solar System, Milky Way, Local Group.

5. (e) All of them are.

6. (a, c, d, e) Choice b (the Moon is uglier than Earth) is a matter of opinion.