Solutions Manual For
Explorations Introduction to
Astronomy 10th Edition By
Thomas Arny, Stephen
Schneider (All Chapters 1-18,
100% Original Verified A+ Grade)
Part 1: Chapter 10-18
Part 2: Chapter 1-9
All Chapters Arranged Reverse.
, Solutions Manual Part 1: Chapter 10-18
Chapter 10 Small Bodies Orbiting the Sun
CHAPTER 10 Small Bodies Orbiting the Sun
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.
1
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
,Chapter 10 Small Bodies Orbiting the Sun
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.
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.
2
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, Chapter 10 Small Bodies Orbiting the Sun
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.
Answers to Problems
1. Earth travels the circumference of its orbit in a year, so
V = 2π(1 AU) / 365 days = 2π AU (1.5 108 km / AU ) /(365 d 24 h / d 3600s / h)
= 9.42 108 km / 3.15 107 s 10 km / s = 30 km / s
(9..15 = 2.99)
2. Calculate the surface gravity and escape velocity for Ceres, assuming a radius of 487 km and
a mass of 9.43 1020 kg .
( )
2
gsurface = GM / R 2 = 6.67 10−11 m3kg −1s −2 9.43 10 20 kg / 487 km 103 m / km
= 0.265 m / s 2
Vesc = (2GM / R)1/ 2
( 2 6.67 10 ( ))
1/ 2
−11
= m3kg −1s −2 9.43 1020 kg / 487 km 103 m / km
= 508 m / s = 0.508 km / s
(On Earth it’s close to 11, 000 m / s or 11 km / s )
3. The semi-major axis is half the sum of the nearest and farthest points,
a = (0.19 AU + 1.97 AU) / 2 = 2.16 AU / 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 = a 3/ 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 = 1.5 .
5. Use the modified form of Kepler’s Third Law to find the mass of Pluto and Charon.
m + M = 4 2 d 3 / GP 2
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
3
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
Explorations Introduction to
Astronomy 10th Edition By
Thomas Arny, Stephen
Schneider (All Chapters 1-18,
100% Original Verified A+ Grade)
Part 1: Chapter 10-18
Part 2: Chapter 1-9
All Chapters Arranged Reverse.
, Solutions Manual Part 1: Chapter 10-18
Chapter 10 Small Bodies Orbiting the Sun
CHAPTER 10 Small Bodies Orbiting the Sun
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.
1
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
,Chapter 10 Small Bodies Orbiting the Sun
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.
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.
2
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, Chapter 10 Small Bodies Orbiting the Sun
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.
Answers to Problems
1. Earth travels the circumference of its orbit in a year, so
V = 2π(1 AU) / 365 days = 2π AU (1.5 108 km / AU ) /(365 d 24 h / d 3600s / h)
= 9.42 108 km / 3.15 107 s 10 km / s = 30 km / s
(9..15 = 2.99)
2. Calculate the surface gravity and escape velocity for Ceres, assuming a radius of 487 km and
a mass of 9.43 1020 kg .
( )
2
gsurface = GM / R 2 = 6.67 10−11 m3kg −1s −2 9.43 10 20 kg / 487 km 103 m / km
= 0.265 m / s 2
Vesc = (2GM / R)1/ 2
( 2 6.67 10 ( ))
1/ 2
−11
= m3kg −1s −2 9.43 1020 kg / 487 km 103 m / km
= 508 m / s = 0.508 km / s
(On Earth it’s close to 11, 000 m / s or 11 km / s )
3. The semi-major axis is half the sum of the nearest and farthest points,
a = (0.19 AU + 1.97 AU) / 2 = 2.16 AU / 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 = a 3/ 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 = 1.5 .
5. Use the modified form of Kepler’s Third Law to find the mass of Pluto and Charon.
m + M = 4 2 d 3 / GP 2
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
3
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
