Download Chapter 9: Gravity and Circular Motion

Chapter 9: Gravity and Circular Motion
Answers and Solutions
1.
Picture the Problem: Two bowling balls attract each other gravitationally.
Strategy: Solve the law of universal gravitation for the separation between the balls.
Solution: 1. Solve Newton’s law
of universal gravitation for r:
F =G
2. Substitute the numerical values:
r=
m1 m2
r2
⇒ r2 = G
(6.67 × 10
−11
m1 m2
F
N ⋅ m 2 kg 2
⇒r= G
)(
m1 m2
F
6.1 kg )(7.2 kg )
3.1 × 10 −9 N
= 0.94 m
Insight: A force of 3.1×10−9 N is equivalent to the weight of 320 nanograms, roughly 100 times larger than the mass of
a living cell. Gravitational forces are very weak!
2.
Picture the Problem: Two identical cars attract each other gravitationally.
Strategy: Solve the law of universal gravitation for the mass of the cars.
Solution: 1. Solve Newton’s law
of universal gravitation for m:
F =G
2. Substitute the numerical values:
m=
m2
r2
⇒
r2F
= m2
G
(4.7 m )2 (4.5 × 10− 6
6.67 × 10
−11
2
⇒m=
N
r2F
G
) = 1200 kg
N ⋅ m kg 2
Insight: A force of 4.5×10−6 N is equivalent to the weight of 460 micrograms, roughly 720 times smaller than the mass
of a 0.33-g steel BB (0.177 caliber).
3.
Picture the Problem: This is a follow up question to Guided Example
9.3. The Millennium Eagle is located along the x axis, at a distance of
2.00 km from point B, as indicated in the figure at right.
Strategy: We find the magnitude of the gravitational forces at location
A by using F = GmM r 2 , where m1 = m and m2 = M. The masses are
given in the problem statement, and the distance r can be found from
the distances in the sketch.
Solution: 1. Use the
Pythagorean theorem to find
the distance r from point A to
each asteroid:
r 2 = x 2 + ya2
2. Substitute the known values
to find the forces F1 and F2:
F1 = F2 = G
r=
2.00 km
(2000 m )2 + (1500 m)2
= 2500 m
mM
r2
(
= 6.67 × 10−11 N ⋅ m 2 kg 2
)
(2.50 × 10
7
)(
kg 3.50 × 1011 kg
(2500 m )
2
) = 93.4 N
Insight: The gravitational forces of the asteroids on the Millennium Eagle are higher than the 52.0 N found in the
example because the distance has decreased substantially.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–1
Chapter 9: Gravity and Circular Motion
4.
Pearson Physics by James S. Walker
Picture the Problem: The asteroid Ceres gravitationally attracts a nearby astronaut.
Strategy: Apply Newton’s law of universal gravitation to find the gravitational force on the astronaut.
Solution: Calculate the force:
F =G
(
)
8.7 × 10 20 kg (95 kg )
m1m2
−11
2
2
=
6.67
×
10
N
⋅
m
/kg
= 0.028 N
2
r2
14106 m
(
)
(
)
Insight: If the astronaut stood on the surface of Ceres (radius 487 km) the force would be 23 N (5.2 lb).
5.
If the separation between two masses is doubled, the gravitational force between them will decrease by a factor of 4.
This is because the force is inversely proportional to the square of the separation distance.
6.
Gravitational forces must be added as vectors.
7.
The gravitational force on a pair of masses is an attractive force that acts along the line that connects the centers of the
two masses.
8.
Newton’s law of universal gravitation shows that the force that attracts all masses to other masses is proportional to the
product of the masses and inversely proportional to the square of the distance between them. Therefore, the greater the
mass, the greater the force, and the greater the distance, the weaker the force of gravity.
9.
The force of Earth’s gravity is nearly as strong in orbit as it is on the surface of Earth, and the force of Earth’s gravity
extends past the Moon and out to infinity. The astronauts experience weightlessness because they and their spacecraft
are in constant free fall.
10. The gravitational forces on object B depend the masses of the objects and their separation distance. Because the
distances to objects A and C are the same, we need only note the masses to conclude that the gravitational force exerted
by object C (with mass 2m) is twice as great as the gravitational force exerted by object A (with mass m). Therefore, the
net gravitational force on object B will be toward the right.
11. Picture the Problem: The three objects in the figure at the right attract each
other gravitationally.
Strategy: Apply Newton’s law of universal gravitation to find the gravitational
force on each object due to the other two objects. Remember that each force
is attractive, and that forces add as vectors, so that we must keep track of the
directions of the forces. Let the positive direction be toward the right. Finally,
compare the magnitudes of the net forces on each object to determine their
ranking..
r
m (2 m )
m (2 m )
m2
+G
=G 2
2
2
r
r
(2 r )
Solution: 1. Calculate the forces on
object A, letting F = Gm 2 r 2 :
FA = FAB + FAC = G
2. Calculate the forces on object B:
FB = − FBA + FBC = − G
3. Calculate the forces on object C:
FC = − FCA − FCB = − G
2⎤
⎡
⎢⎣ 2 + 4 ⎥⎦ = 2.5 F
(2m) m + G (2m)(2m) = G m2
r2
r2
r2
[−2 + 4] = 2.0 F
(2m) m − G (2m)(2m) = G m2 ⎡ − 2 − 4⎤ = − 4.5F
⎥⎦
r2
r 2 ⎢⎣ 4
(2r )2
4. By comparing the magnitudes of the forces we arrive at the ranking FB < FA < FC .
Insight: Although object B is in the middle and is therefore closer to the other two objects, the forces act on it in
opposite directions, partially cancelling each other and leaving object B with the smallest net force of the three objects.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–2
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
12. Picture the Problem: A can of soup gravitationally attracts a jar of pickles.
Strategy: Apply Newton’s law of universal gravitation to find the gravitational force on the can and the jar.
Solution: 1. (a) Calculate the force:
Fcan on jar = G
m1m2
(0.18 kg )(0.34 kg )
= 6.67 × 10 −11 N ⋅ m 2 /kg 2
r2
(0.42 m)2
(
)
= 2.3 × 10 −11 N
2. (b) Because of the symmetry of
the situation the forces are equal:
Fjar on can = Fcan on jar = 2.3 × 10 −11 N
Insight: These forces are tiny, equivalent to the weight of 2.4 nanograms, roughly the mass of a living cell.
13. Picture the Problem: Two flower pots gravitationally attract each other.
Strategy: Apply Newton’s law of universal gravitation to find the gravitational force between the two flower pots.
Solution: 1. (a) Calculate the force:
F1 = G
m1m2
(1.6 kg )(1.6 kg ) = 1.7 × 10−10 N
= 6.67 × 10 −11 N ⋅ m 2 /kg 2
2
r
(1.0 m )2
2. (b) Calculate the force
for the new distance:
F3 = G
m1m2
(1.6 kg )(1.6 kg ) = 1.9 × 10−11 N
= 6.67 × 10 −11 N ⋅ m 2 /kg 2
2
r
(3.0 m )2
(
)
(
)
Insight: A force of 1.9×10−11 N is tiny, equivalent to the weight of 1.9 nanograms, roughly the mass of a living cell.
14. Picture the Problem: Two volleyball players attract each other gravitationally.
Strategy: Solve the law of universal gravitation for the separation between the players.
Solution: 1. Solve Newton’s law
of universal gravitation for r:
F =G
2. Substitute the numerical values:
r=
m1 m2
r2
⇒ r2 = G
(6.67 × 10
−11
m1 m2
F
N ⋅ m 2 kg 2
⇒r= G
)(
) ( 3.3 × 10
m1 m2
F
66 kg 72 kg )
−7
N
= 0.98 m
Insight: A force of 3.3×10−7 N is equivalent to the weight of 34 micrograms, roughly 10,000 times smaller than
the mass of a 0.33-g steel BB (0.177 caliber).
15. Picture the Problem: The mass of a baseball gravitationally attracts a batter.
Strategy: Solve the law of universal gravitation for the mass of the baseball batter.
m1m2
r2
r2F
= m1m2
G
r2F
m2 G
Solution: 1. Solve Newton’s law
of universal gravitation for m:
F =G
2. Substitute the numerical values:
(0.77 m )2 (1.1 × 10− 9 N )
m1 =
=
(0.15 kg ) (6.67 × 10−11 N ⋅ m 2 kg 2 )
⇒
⇒ m1 =
65 kg
Insight: A force of 1.1×10−9 N is equivalent to the weight of 112 nanograms, roughly 100 times larger than the mass of
a living cell.
16. Picture the Problem: This is a follow up question to Guided Example 9.4. The acceleration of gravity at the 370-km
altitude of the International Space Station is slightly smaller than the acceleration of gravity at Earth’s surface.
Strategy: First, use F = GmM E r 2 to find the force due to gravity on the space station. Then, set F = mgh to find the
acceleration due to gravity at that height.
Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–3
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
Solution: 1. Begin with F = GmM E r 2
and substitute RE + h for r. This gives the
force F due to gravity at a height h above
Earth’s surface:
F =G
2. Substitute mgh for the force F at the height h.
Next, cancel the mass m and solve for gh:
mg h = G
3. Substitute the numerical values to calculate gh:
g h = 6.67 × 10 −11 N ⋅ m 2 kg 2
mM E
mM
=G
2
r
(RE + h)2
mM E
(RE + h)
2
⇒ gh = G
(
)
ME
(RE + h)2
5.97 × 1024 kg
(6.37 × 10
6
+ 370 × 103 m
)
2
= 8.77 m/s 2
Insight: The force of Earth’s gravity is nearly as strong in orbit as it is on the surface of Earth, and the force of Earth’s
gravity extends past the Moon and out to infinity. The astronauts experience weightlessness because they and the space
station are in constant free fall.
17. Picture the Problem: The acceleration of gravity at an altitude h above Earth’s surface is reduced due to the increased
distance from the center of Earth.
Strategy: Set the acceleration of gravity at an altitude h equal to one-half the acceleration at h = 0 using the expression
in Guided Example 9.4, and solve for h.
Solution: Set g h = 12 g 0 and solve for h:
GM E
gh =
2
E
2R
(RE + h)
2
= ( RE + h )
2
2 RE = RE + h
h=
(
)
=
1 ⎛ GM E ⎞ 1
⎜
⎟ = g0
2 ⎝ RE2 ⎠ 2
2 − 1 RE =
(
)(
)
2 − 1 6.37 × 106 m = 2.64 × 106 m
Insight: This altitude is over seven times higher than the 370-km altitude of the International Space Station.
18. Picture the Problem: The lunar rover on the Apollo missions had the same mass on Earth’s surface and on the Moon’s
surface, but a different weight on each.
Strategy: Use the acceleration of gravity on Earth (9.81 m/s2) and the Moon (1.62 m/s2, from Quick Example 9.5)
to find the weight W = mg of the lunar rover.
(
)
Solution: 1. Find the rover’s weight on Earth:
Won Earth = mg = (225 kg ) 9.81 m/s 2 = 2210 N
2. Find the rover’s weight on the Moon:
Won Moon = mg m = (225 kg ) 1.62 m/s 2 = 365 N
(
)
Insight: On the Moon the rover weighs about one-sixth of what it does on Earth’s surface.
19. Picture the Problem: A 4.6-kg mass experiences a gravitational attraction to Earth that depends upon its distance from
the center of Earth.
Strategy: Use Newton’s law of universal gravitation to find the distance from Earth’s center that would produce
the given force (weight) for a 4.6-kg mass. Then use Newton’s second law to find the acceleration.
Solution: 1. (a) Solve Newton’s
law of universal gravitation for r:
r=
=
GM E m
F
(6.67 × 10
−11
)(
)
N ⋅ m 2 /kg 2 5.97 × 1024 kg (4.6 kg )
2.2 N
= 2.9 × 107 m
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–4
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
2. (b) Solve Newton’s second law for a:
a=
F 2.2 N
=
= 0.48 m/s 2
m 4.6 kg
Insight: The distance 2.9×107 m is about 4.6RE . If the mass were on Earth’s surface, it would weigh 45 N (10 lb).
20. Picture the Problem: Both Earth and the Moon exert a force on other masses in their vicinity according to Newton’s
law of universal gravitation.
Strategy: Use a ratio of gravitational forces to find the mass of the Moon in terms of the mass of Earth. Note that if you
solve Newton’s law of gravity for m2 you find m2 = F r 2 Gm1 .
Solution: Determine the ratio mM mE
by solving F = Gm1m2 r 2 for m2 :
mM FM rM2 Gm1 FM rM2
=
=
=
mE
FE rE2 Gm1
FE rE2
( 16 FE ) ( 14 rE )2
FE rE2
=
1
1
⇒ mM =
mE
96
96
Insight: Using a ratio can be a powerful tool for solving a question like this one, where few hard data are given, only
relationships between various quantities. The exact ratio of masses is 7.35 × 1022 kg 5.97 × 1024 kg = 1 81.2.
21. The methods of calculus can be used to support the assumption that the gravitational force due to a large spherical mass
(like the Sun or the Moon) is exactly the same as the force due to a point mass located at the center of the large spherical
mass. That is, the object gravitationally behaves as if all its mass were concentrated at its center.
22. A black object reflects no light, and black can be considered the absence of light. A black hole is an object with gravity
so strong that light can’t escape from it, so it is known as a “black” hole.
23. The expression g = GM R 2 indicates that the acceleration due to gravity at the surface of a planet is proportional to the
mass and inversely proportional to the square of the radius. Therefore, if both the mass and the radius of a planet were
doubled, the acceleration of gravity at its surface would decrease by a factor of 2.
24. The gravitational force of a large spherical star behaves as if all the mass of the star were concentrated at a point at its
center. Therefore, a spaceship that is orbiting the star will not be affected by changes in the star’s radius as long as the
mass of the star remained the same.
25. Picture the Problem: The mass of Mercury can be determined from its surface gravity and its radius.
Strategy: Solve the general expression for the acceleration of gravity, g = GM R 2 , for the mass of the Moon.
M
R2
R2 g
=M
G
Solution: 1. Solve g = GM R 2 for M:
g=G
2. Substitute the numerical values:
(2400 × 10 m) (3.7 m/s ) = 3.2 × 10
M =
(6.67 × 10 N ⋅ m kg )
⇒
3
−11
2
2
2
2
23
kg
Insight: The accepted value for the mass of Mercury is 3.3022×1023 kg.
26. Picture the Problem: The acceleration of gravity at large distances from Earth can be found from its general
expression.
Strategy: Use the general expression for the acceleration of gravity, g = GM r 2 , to calculate its magnitude.
Solution: Calculate the
acceleration of gravity:
gr = G
ME
ME
1 M
1
1
=G
= G 2E = g = 9.81 m/s 2 = 2.45 m/s 2
2
2
r
(2 RE ) 4 RE 4 4
(
)
Insight: The acceleration of gravity at a distance of three Earth radii from the center of Earth is one-ninth its value at
the surface of Earth, or 19 9.81 m/s 2 = 1.09 m/s 2 .
(
)
Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–5
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
27. Picture the Problem: The acceleration of gravity at the surface of Mars can be found from its general expression.
Strategy: Use the general expression for the acceleration of gravity, g = GM R 2 , to calculate its magnitude.
Solution: Calculate the
acceleration of gravity:
g=G
M Mars
6.4 × 1023 kg
= 6.67 × 10 −11 N ⋅ m 2 kg 2
= 3.69 m/s 2
2
2
3
RMars
3400 × 10 m
(
)
(
)
Insight: This is accurate to within rounding error of the reported value of 3.711 m/s2.
28. Picture the Problem: At a large distance from the Moon its acceleration of gravity is only 0.50 m/s2.
Strategy: Solve the general expression for the acceleration of gravity, g = GM r 2 , for the distance from the Moon.
Solution: 1. Solve g = GM r 2 for r:
g=G
2. Substitute the numerical values, using
7.35×1022 kg for the mass of the Moon:
r=
M
r2
⇒
GM
= r2
g
(6.67 × 10
−11
⇒r=
)(
GM
g
N ⋅ m 2 kg 2 7.35 × 10 22 kg
0.50 m/s 2
) = 3.1 × 10
6
m
Insight: This distance of 3130 km is about 1.8RM, where RM is the 1740-km radius of the Moon. At the surface of the
Moon the acceleration of gravity is 1.62 m/s2.
29. Picture the Problem: This is a follow up question to Guided Example
9.8. A 1200-kg car rounds a corner of radius r = 45 m with a speed
v = 12 m/s.
Strategy: In this system the force of static friction provides the
centripetal force required for the car to move in a circular path. That is
why the force of friction is directed toward the center of the circle—it
is keeping the car from skidding off the road. Set the static friction
force equal to the centripetal force to find the minimum coefficient of
static friction required to keep the car traveling in a circle.
Solution: 1. Set the static friction
force equal to the centripetal force,
and solve for μs :
f s, max = μs N = μs m g = m
μs =
v2
r
v2
gr
(12 m/s)
v2
=
= 0.33
gr
9.81 m/s 2 (45 m )
2
2. Substitute the numerical values:
μs =
(
)
Insight: The car could stay on the road, even if the coefficient of static friction were zero, if the road was banked such
that the force of gravity provides the necessary centripetal force to keep the car traveling in a circle.
30. Picture the Problem: Your car travels along a circular path at constant speed.
Strategy: Static friction between the tires and the road provides the centripetal force required to make your car travel
along a circular path. Set the static friction force equal to the centripetal force and calculate its value.
Solution: Set the static friction force
equal to the centripetal force:
mv 2 (1300 kg )(16 m/s )
= f cp = macp =
=
= 5.6 kN
59 m
r
2
f s, max
(
)
Insight: The maximum static friction force is μs mg = (0.88)(1300 kg ) 9.81 m/s 2 = 11.2 kN, which corresponds to a
maximum cornering speed (without skidding) of 23 m/s (50 mph).
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–6
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
31. A force that produces circular motion must always be directed toward the center of the circle, and must be equal to
f cp = mv 2 r , where m is the mass of the object, v is the constant speed of the object, and r is the radius of its path.
32. A centripetal force must be equal to f cp = mv 2 r , where m is the mass of the object, v is the constant speed of the
object, and r is the radius of its path. The centripetal force must always be directed toward the center of the circle,
perpendicular to the tangential velocity of the object that is in circular motion.
33. Centripetal acceleration is different from other accelerations because it produces a change in the direction of the
velocity, but not a change in the speed of the object. It is similar to other accelerations in that it is equal to the rate
of change of the velocity, and has the same units of m/s2.
34. Yes. The steering wheel changes the velocity of a car by changing the direction of its motion. Any change in velocity
requires an acceleration.
35. The centripetal acceleration of an object that is moving in a circle is given by acp = v 2 r . Therefore, if both the speed
and the radius of an object’s path are doubled, the centripetal acceleration increases by a factor of 2.
36. Picture the Problem: Four objects travel in circles, requiring varying amounts of centripetal force.
Strategy: Use the expression for centripetal force to calculate the amount of force required to keep each object moving
in a circle at constant speed. Use the calculated values to rank the centripetal forces.
mA vA2 (60 kg )(1 m/s )
=
= 12 N
rA
5m
2
f cp, A =
Solution: 1. Calculate the centripetal forces:
f cp, B
m v 2 (5 kg )(2 m/s )
= B B =
= 10 N
rB
2m
f cp, C
m v 2 (24 kg )(2 m/s )
= C C =
= 12 N
rC
8m
2
2
mD vD2 (2 kg )(4 m/s )
=
= 32 N
rD
1m
2
f cp, D =
2. By inspection of the force magnitudes we arrive at the ranking B < A = C < D.
Insight: The speed of an object has the largest effect upon the centripetal force required to keep it moving at constant
speed along a circular path.
37. Picture the Problem: Four objects travel in circles, requiring varying amounts of centripetal acceleration.
Strategy: Use the expression for centripetal acceleration to calculate the amount of acceleration required to keep each
object moving in a circle at constant speed. Use the calculated values to rank the centripetal accelerations.
Solution: 1. Calculate the
centripetal accelerations:
vA2 (1 m/s )
=
= 0.2 m/s 2
rA
5m
2
acp, A =
v 2 (2 m/s )
= C =
= 0.5 m/s 2
rC
8m
2
acp, C
vB2 (2 m/s )
=
= 2 m/s 2
rB
2m
2
acp, B =
v 2 (4 m/s )
= D =
= 16 m/s 2
rD
1m
2
acp, D
2. By inspection of the acceleration magnitudes we arrive at the ranking A < C < B < D.
Insight: The speed of an object has the largest effect upon the centripetal acceleration required to keep it moving
at constant speed along a circular path.
Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–7
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
38. Picture the Problem: A bucket swings along a vertical circular path at constant speed.
Strategy: The water must exert a force on the bucket to ensure it does not spill. By Newton’s third law, the bucket
must therefore exert a force on the water that is greater than zero. Write Newton’s second law in the vertical direction
for the water in the bucket when it is at the top of the circle. Set the normal force equal to zero and solve for the
minimum speed of the bucket. Let upward be the positive direction.
Solution: Write Newton’s
second law for the water, set
N to zero and solve for v:
∑F
y
= − N − mg = − m v 2 r
0 + mg = m v 2 r
v = rg =
(1.3 m) (9.81 m/s 2 ) =
3.6 m/s
Insight: The fact that mass cancels out on both sides of the equation above, and that the minimum speed is independent
of mass, is evidence for the equivalence of gravitational mass and inertial mass.
39. Picture the Problem: A car experiences a centripetal acceleration as it rounds a corner.
Strategy: Solve the expression for centripetal force for the radius of its path.
2
Solution: Solve acp = v r for r:
v2
acp =
r
v 2 (15 m/s )
⇒r=
=
= 51 m
acp
4.4 m/s 2
2
Insight: The speed of an object has the largest effect upon the centripetal acceleration required to keep it moving at
constant speed along a circular path.
40. All planets orbit along elliptical paths.
41. Kepler’s second law states that as a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount
of time. This does not change at any time throughout its orbit. This law is a consequence of the conservation of angular
momentum.
42. Kepler’s third law states that if the radius of a planet’s orbit is increased, its orbital period also increases. The period, T,
of a planet increases as its distance from the Sun, r, is raised to the 3/2 power.
43. The orbital period of a planet does not depend on the mass of the planet, but it does depend on the mass of the star that it
orbits. The larger the mass of the star, the shorter the orbital period of the planet, given that the orbital radius remains
the same.
44. A geosynchronous satellite has an orbital period that matches the 24 hr rotational period of Earth.
45`. No. A satellite must be moving relative to the center of Earth to maintain its orbit, but the North Pole is at rest relative
to the center of Earth. Therefore, a satellite cannot remain fixed above the North Pole. All geosynchronous orbits are
coplanar with the equatorial plane of Earth.
46. Picture the Problem: The orbital radius of a satellite is quadrupled.
Strategy: Use Kepler’s third law and the expression for the orbital speed of a satellite to answer the questions about
how the orbit period and orbital speed change when the orbital radius is quadrupled.
Solution: 1. (a) When the orbital radius is quadrupled the orbital period of a satellite will increase.
2. (b) Use Kepler’s third law to make a
ratio of the orbital periods:
(
(
2π
Tnew
=
Told
2π
)
)r
32
GM E rnew
GM E
32
old
⎛r ⎞
= ⎜ new ⎟
⎝r ⎠
old
32
⎛ 4r ⎞
= ⎜ old ⎟
⎝ r ⎠
32
= 43 2 = 8
old
3. From the ratio we can see that Tnew = 8 Told , or the period increases by a factor of 8.
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9–8
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
GM E rnew
vnew
=
vold
4. (c) Make a ratio of the orbital speeds:
rold
=
rnew
=
GM E rold
rold
1
1
=
=
4rold
4 2
5. From the ratio we can see that vnew = 12 vold , or the period decreases by a multiplicative factor of
1
2
.
Insight: The periods of the outer planets are larger, and the orbital speeds are smaller, than those of the inner planets.
47. Picture the Problem: Venus orbits the Sun with a period of 1.94×107 s.
Strategy: Solve Kepler’s third law for the radius of the orbital path of Venus.
2
⎛ 2π ⎞ 3 4π 2 3
T2 = ⎜
r
⎟ r =
GM s
⎝ GM s ⎠
Solution: 1. Solve Kepler’s third law for r:
GM s 2
T = r3
4π 2
r=
2. Substitute the numerical values:
3
(6.67 × 10
⇒r=
−11
3
GM s 2
T
4π 2
)(
N ⋅ m 2 kg 2 2.00 × 1030 kg
4π
2
) 1.94 × 10 s
(
)
7
2
11
= 1.08 × 10 m
Insight: This distance is 1.4 times smaller than the 1.50×1011 m radius of Earth’s orbit.
48. Picture the Problem: The Apollo capsule orbited the Moon at an altitude of 110 km above the Moon’s surface.
Strategy: Use Kepler’s third law to determine the period of orbit using the mass and radius of the Moon from those
given in Quick Example 9.5.
Solution: Apply Kepler’s third law:
⎛ 2π ⎞ 3/ 2
T =⎜
⎟r
⎝ GM M ⎠
⎡
=⎢
⎢
⎣
(6.67 × 10
2π
−11
2
N ⋅ m /kg
2
)(7.35 × 10
22
⎤
⎥ 1.74 × 106 + 110 × 103 m
kg ⎥
⎦
)
(
)
3/ 2
T = 7140 s = 1.98 h
Insight: This period turns out to be a bit larger than the 1.44-h orbit period of a satellite that is 110 km above Earth’s
surface.
49. Picture the Problem: Jupiter orbits the Sun with a radius of 7.784×108 km, or 7.784×1011 m.
Strategy: Use Kepler’s third law to determine the period of orbit using the mass of the Sun (2.00×1030 kg) and the orbit
radius of Jupiter (7.784×1011 m) from a table of astronomical data.
⎛ 2π ⎞ 3/ 2
Solution: 1. Apply Kepler’s third law: T = ⎜
⎟r
⎝ GM M ⎠
⎡
=⎢
⎢
⎣
(6.67 × 10
2π
−11
N ⋅ m 2 /kg 2
)(
⎤
⎥ 7.784 × 1011 m
30
2.00 × 10 kg ⎥
⎦
)
(
)
3/ 2
1y
= 11.8 y
3.16 × 107 s
2. The orbital period of a planet does not depend upon the planet’s mass.
T = 3.74 × 108 s ×
Insight: This calculated period is slightly smaller than the 11.86-y measured period due to rounding error.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9–9
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
50. Picture the Problem: A GPS satellite travels around Earth in a circular orbit.
Strategy: Use the expression for orbital speed to find its value for a GPS satellite. Recall that the orbit radius is
r = RE + h , where RE = 6.37×106 m.
Solution: Calculate the orbital speed:
v=
=
GM E
=
r
(6.67 × 10
GM E
RE + h
−11
)(
N ⋅ m 2 kg 2 5.97 × 1024 kg
6
3
6.37 × 10 + 20, 200 × 10 m
) = 3870 m/s
Insight: The orbits of these GPS satellites are designed so that at least three satellites are overhead every spot on Earth
at any one time. The orbit speed is considerably slower than the 7900 m/s orbital speed of low Earth orbits.
51. The gravitational force between two people passing on the street is on the order of 10−7 N, a force equivalent to the
weight of a few tens of micrograms of mass. Such forces are too small to notice.
52. The gravitational force between two objects is inversely proportional to the square of the separation distance, so that
cutting the separation distance in half will quadruple the gravitational force.
53. If the gravitational force between two objects depended upon the sum of their masses, the force would be nonzero even
if one of the objects had zero mass. That is, there would be a gravitational force between a mass and a point in empty
space, which is inconsistent with observations.
54. Picture the Problem: Four pairs of objects exert gravitational forces on each other.
Strategy: Use Newton’s law of universal gravitation to calculate the amount of force between each pair of objects. Use
the calculated values to rank the gravitational forces.
Solution: 1. Calculate the gravitational forces:
FA = G
FB = G
m1, A m2, A
rA2
m1, B m2, B
=G
=G
mm
m2
G
=
r2
r2
m (2 m )
=
1 m2
G
2 r2
(2r )2
m1, C m2, C
(2m)(3m) = 3 G m2
FC = G
G
=
2 r2
rC2
(2r )2
m1, D m2, D
(4m)(5m) = 20 G m2
FD = G
=G
2
9
rD
r2
(3r )2
rB2
2. By inspection of the force magnitudes we arrive at the ranking B < A < C < D.
Insight: The separation distance between two objects has the largest effect upon the gravitational force between them.
55. Picture the Problem: Two peaches attract each other gravitationally.
Strategy: Use Newton’s law of universal gravitation to find the force between the peaches.
Solution: 1. (a) Find the force:
F =G
m1m2
(0.16 kg )(0.16 kg ) = 2.7 × 10−11 N
= 6.6710 −11 N ⋅ m 2 /kg 2
r2
(0.25 m )2
2. (b) Repeat for the new distance:
F =G
m1m2
(0.16 kg )(0.16 kg ) = 6.8 × 10−12 N
= 6.6710 −11 N ⋅ m 2 /kg 2
r2
(0.50 m )2
(
(
)
)
Insight: Doubling the distance between the peaches cut the gravitational force by a factor of 4.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 10
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
56. Picture the Problem: Two bowling balls attract each other gravitationally.
Strategy: Use Newton’s law of universal gravitation to find the force between the bowling balls.
Solution: Find the force:
F =G
m1m2
(6.8 kg )(7.1 kg ) = 5.7 × 10−9 N
= 6.6710 −11 N ⋅ m 2 /kg 2
r2
(0.75 m )2
(
)
Insight: Increasing the distance between the balls from 0.75 m to 1.5 m will decrease the force from 5.7 nN to 1.4 nN.
57. Picture the Problem: An apple and an orange attract each other gravitationally.
Strategy: Use Newton’s law of universal gravitation to find the force between the two fruits.
Solution: 1. (a) Find the force:
F =G
m1m2
(0.11 kg )(0.24 kg ) = 2.4 × 10−12 N
= 6.6710 −11 N ⋅ m 2 /kg 2
2
r
(0.85 m)2
(
)
2. (b) The force the apple exerts on the orange is equal and opposite to the force the orange exerts on the apple, so its
magnitude must be 2.4 × 10 −12 N .
Insight: Halving the distance between the two fruits would quadruple the force between them, but it would still be tiny.
58. Picture the Problem: A spaceship is attracted gravitationally to both Earth and the Moon.
Strategy: Use Newton’s law of universal gravitation to relate the attractive forces from Earth and the Moon. Set the
force due to Earth equal to twice the force due to the Moon when the spaceship is at a distance r from the center of
Earth. Let R = 3.84 × 108 m, the distance between the centers of Earth and the Moon. Then solve the expression for
the distance r.
G
Solution: 1. (a) Set FE = 2 FM
using Newton’s law of gravity
and solve for r:
ms mE
r
2
ms mM
= 2× G
(R − r )2
mE ( R − r ) = 2mM r 2
2
r=
⇒ R − r = 2mM mE r ⇒ R = 2mM mE r + r
R
1 + 2mM mE
=
(
3.84 × 108 m
1 + 2 7.35 × 1022 kg
) (5.97 × 10
24
kg
)
= 3.32 × 108 m
2. (b) The answer to part (a) is independent of the mass of the spaceship because the spaceship’s mass is included
in the force between it and both the Moon and Earth, and so its value cancels out of the expression.
Insight: The distance in part (a) is the same for any mass, and corresponds to about 52 Earth radii or about 86% of the
distance R between Earth and the Moon.
59. Picture the Problem: Earth, the Moon, and the Sun attract each other gravitationally.
Strategy: Add the gravitational force on the Moon due to the other two masses using Newton’s law of gravity. The
forces are always attractive and add as vectors. We will take the force toward the left (toward the Sun) to be positive
and the force toward Earth to be negative.

⎛M
M M
M M
M ⎞
Solution: Add the forces: FM = G S2 M − G E2 M = GM M ⎜ 2 S − 2 E ⎟
rS-M
rE-M
⎝ rS-M rE-M ⎠
⎡
2.00 × 1030 kg
= 6.67 × 10−11 N ⋅ m 2 /kg 2 7.35 × 1022 kg ⎢
⎢ 1.50 × 1011 − 3.84 × 108 m
⎣
(
)(

FM = 2.40 × 1020 N toward the Sun
)
(
−
⎤
5.97 × 10 24 kg ⎥
2
3.84 × 108 m ⎥⎦
) (
2
)
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 11
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker

Insight: Note that if you compare the two terms that contribute to FM , you will see that the Sun exerts a force on the
Moon (4.38×1020 N) that is 2.2 times larger than the force Earth exerts on the Moon (1.98×1020 N).
60. Picture the Problem: Three equal masses are arranged at the vertices of an equilateral
triangle as shown in the figure at right.
Strategy: Each mass will be gravitationally attracted to the other two masses. The vector sum
of the forces will be along a line toward the midpoint of the other two masses, due to
symmetry. An examination of the geometry reveals that this force equals 2 F cos 30° , where F
is the gravitational force between any two masses. Use this relation together with Newton’s
law of gravity to find the force exerted on one of the masses.
m
F
r
60°
m
m
2
⎛
m2
−11 N ⋅ m ⎞ (6.75 kg )
cos
30
2
6.67
10
cos 30°
°
=
×
⎜⎝
r2
kg 2 ⎟⎠ (1.25 m )2
2
F = 2×G
Solution: Find 2 F cos 30° , where
F = Gm 2 r 2 from Newton’s law of gravity:
= 3.37 × 10 −9 N
Insight: Doubling the masses will quadruple the gravitational force, because in this case the gravitational force depends
upon the square of the mass of each object and the square of the separation distance.
61. Picture the Problem: A 2.0-kg mass is gravitationally attracted to three
other masses.
Strategy: Add the gravitational forces using the component method of
vector addition. Use Newton’s law of gravity and the geometry of the
problem to determine the magnitudes of the forces. Let m1 = 1.0 kg,
m2 = 2.0 kg, m3 = 3.0 kg, and m4 = 4.0 kg.
Solution: 1. Find the x component of the force on m2 . The
cosine of φis the adjacent side of
the triangle (r12) divided by the
hypotenuse (r24):
Fx = G
=G
m1m2
mm
+ G 2 2 4 cos φ
2
r12
r24
⎛m m r ⎞
m1m2
m m ⎛r ⎞
+ G 2 2 4 ⎜ 12 ⎟ = Gm2 ⎜ 21 + 43 12 ⎟
2
r12
r24 ⎝ r24 ⎠
r24 ⎠
⎝ r12
(
= 6.67 × 10
−11
⎧
⎫
4.0 kg )(0.20 m )
(
⎪ 1.0 kg
⎪
N ⋅ m /kg (2.0 kg ) ⎨
+
⎬
2
2
2 3/ 2
⎪ (0.20 m ) ⎡(0.20 m ) + (0.10 m ) ⎤ ⎪
⎣
⎦ ⎭
⎩
2
2
)
Fx = 1.3 × 10−8 N
2. Find the y component
of the force on m2 . The sine of φ
is the opposite side of the
triangle (r14) divided by the
hypotenuse (r24):
Fy = G
=G
m2 m3
mm
+ G 2 2 4 sin φ
2
r23
r24
⎛m m r ⎞
m2 m3
m m ⎛r ⎞
+ G 2 2 4 ⎜ 14 ⎟ = Gm2 ⎜ 23 + 43 14 ⎟
2
r23
r24 ⎝ r24 ⎠
r24 ⎠
⎝ r23
(
= 6.67 × 10

3. Use the components of F to
find its magnitude and direction.
−11
⎧
⎫
4.0 kg )(0.10 m )
(
⎪ 3.0 kg
⎪
+
N ⋅ m /kg (2.0 kg ) ⎨
⎬
2
2
2 3/ 2
0.10
m
(
)
⎡
⎤
⎪
⎪
0.20 m ) + (0.10 m )
(
⎣
⎦ ⎭
⎩
2
2
)
Fy = 4.5 × 10−8 N
F = Fx 2 + Fy 2 =
(1.3 × 10
−8
) (
2
N + 4.5 × 10 −8 N
)
2
= 4.7 × 10 −8 N
⎛ Fx ⎞
⎛ 1.3 × 10 −8 N ⎞
= tan −1 ⎜
= 16 to the left of downward
⎟
⎝ 4.5 × 10−8 N ⎟⎠
⎝ Fy ⎠
θ = tan −1 ⎜
Insight: Doubling all the masses will quadruple the gravitational force, because the gravitational force depends upon
the product of the masses of each object. Therefore, the force would stay exactly the same if we doubled all the masses
and doubled the length of the sides of the rectangle.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 12
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
62. Picture the Problem: Three objects are arranged in the manner shown at right.
Strategy: Use Newton’s law of universal gravitation to determine the relative
magnitudes of the forces experienced by each object.
Solution: 1. Find the force on object A:
FA =
GM (2 M ) GM (3M ) 11 GM 2
+
=
4 L2
L2
(2 L )2
2. Find the force on object B:
FB =
G (2 M )(3M ) GM (2 M )
GM 2 16 GM 2
−
=4 2 =
2
2
L
L
L
4 L2
3. Find the force on object C:
FC =
G (3M )(2 M ) GM (3M ) 27 GM 2
+
=
4 L2
L2
(2 L )2
4. By comparing the magnitudes of the forces we arrive at the ranking, A < B < C.
Insight: If object C were to switch places with object B it would experience a force of 62 GM 2 L2 , less than either
object A
(
7
2
)
GM 2 L2 or object B
(
13
2
)
GM 2 L2 , so the new ranking would be C < A < B.
63. The acceleration due to gravity at the surface of a planet is given by g = GM R 2 . If the mass of the planet is doubled,
while the radius remains the same, the acceleration of gravity at the surface will also double.
64. The acceleration due to gravity at the surface of a planet is given by g = GM R 2 . If the radius of the planet is doubled,
while the mass remains the same, the acceleration of gravity at the surface is reduced by a factor of four.
65. Picture the Problem: Three objects are arranged in the manner shown at right.
Strategy: Use Newton’s law of universal gravitation to determine the relative
magnitudes of the accelerations experienced by each object. In a previous problem
we found the forces: FA = 114 GM 2 L2 , FB = 4 GM 2 L2 , and FC = 274 GM 2 L2 .
Solution: 1. Find the acceleration of object A:
aA =
FA 11 GM 2 L2 11 GM 33 GM
=
=
=
mA
M
4
4 L2
12 L2
2. Find the acceleration of object B:
aB =
FB 4GM 2 L2
GM 24 GM
=
=2 2 =
mB
L
2M
12 L2
3. Find the force on object C:
aC =
FC 27 GM 2 L2 27 GM
=
=
mC
4
3M
12 L2
4. By comparing the magnitudes of the accelerations we arrive at the ranking, B < C < A.
Insight: If object C were to switch places with object B it would experience an acceleration of GM L2 , less than either
object A
(
14
4
)
GM L2 or object B
(
13
4
)
GM L2 , so the new ranking would be C < B < A.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 13
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
66. Picture the Problem: The acceleration of gravity at Titan’s surface is determined by its mass and radius.
Strategy: Calculate the acceleration of gravity g = GM R 2 on the surface of Titan. Use the mass and radius data given
in the problem.
Solution: Solve g T = G M T RT2 for Titan:
(
g T = 6.67 × 10 −11 N ⋅ m 2 /kg 2
)
(1.35 × 10 kg ) = 1.36 m/s
(2.57 × 10 m)
23
2
2
6
Insight: This acceleration of gravity isn’t much different from the 1.62 m/s2 found on the surface of Earth’s Moon.
Titan has more mass but also a larger radius than our Moon, and the two characteristics yield a similar surface gravity.
67. Picture the Problem: The acceleration of gravity at a planet’s surface is determined by its mass and radius.
Strategy: Calculate the acceleration of gravity g = GM R 2 on the surfaces of Mercury and Venus. Use the mass and
radius data from a table of astronomical values to find the surface gravities of these planets.
Solution: 1. (a) Solve g P = G M P RP2
for Mercury:
2
P
2. (b) Solve g P = G M P R for Venus:
(
g M = 6.67 × 10 −11 N ⋅ m 2 /kg 2
(0.0553 × 5.97 × 10 kg ) = 3.70 m/s
(2.440 × 10 m)
(0.816 × 5.97 × 10 kg ) = 8.87 m/s
)
(6.052 × 10 m)
24
)
2
2
6
24
(
g V = 6.67 × 10 −11 N ⋅ m 2 /kg 2
2
6
2
Insight: Although the mass of Mercury is smaller than that of Mars, its smaller radius and higher density results in a
nearly identical acceleration of gravity at the surface. The acceleration of gravity on Venus is similar to that on Earth.
68. Picture the Problem: The acceleration of gravity at an altitude h above Earth’s surface is reduced due to the increased
distance from the center of Earth.
Strategy: Set the acceleration of gravity at an altitude h equal to one-fourth the acceleration at h = 0 using the
expression in Guided Example 9.4, and solve for h.
Solution: Set g h = 14 g 0 and solve for h:
gh =
2
E
4R
GM E
(RE + h)
2
= ( RE + h )
2
=
1 ⎛ GM E ⎞ 1
⎜
⎟ = g0
4 ⎝ RE2 ⎠ 4
4 RE = RE + h
(
)
h = (2 − 1) RE = (1) 6.37 × 106 m = 6.37 × 106 m
Insight: When the altitude is equal to Earth’s radius, the distance to the center of Earth is twice its value at the surface.
However, the acceleration of gravity is inversely proportional to the square of the distance, so that the acceleration
drops by a factor of 4.
69. Picture the Problem: The acceleration of gravity due to Earth, at a distance equal to the Moon’s orbit radius, is
diminished because of the increased distance to the center of Earth.
Strategy: Use g = GM r 2 to find the acceleration of gravity, using r = 3.84 × 108 m, the distance between Earth and
the Moon.
Solution: Substitute the known
values into the expression for g:
g=G
ME
5.97 × 1024 kg
= 6.67 × 10 −11 N ⋅ m 2 /kg 2
= 0.00270 m/s 2
2
2
8
r
3.84 × 10 m
(
)
(
)
Insight: Another way to solve this question is to realize the Moon orbits Earth at a distance of about 60RE , so the
acceleration of gravity due to Earth at that location is
1
602
g=
1
3600
(9.81 m/s ) = 0.00273 m/s , almost correct!
2
2
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 14
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
70. Picture the Problem: Earth and a satellite attract each other gravitationally.
Strategy: Use Newton’s law of gravity to find the gravitational force on the satellite while it is in orbit.
Solution: Calculate the force:
F =G
(
)
(480 kg ) 5.971024 kg
mM E
2
2
−11
=
×
⋅
= 0.16 kN
6.67
10
N
m
/kg
2
r2
35106 m
(
)
(
)
Insight: The distance between the satellite and the center of Earth increases by a factor of 5.5 as it goes from the surface
to its orbit, and the gravitational force on the satellite decreases by a factor of 5.52 = 30 from 4.8 kN to 0.16 kN.
71. Picture the Problem: A 12-kg mass experiences a gravitational attraction to Earth that depends upon its distance from
the center of Earth.
Strategy: Use Newton’s law of universal gravitation to find the distance from Earth’s center that would produce
the given force (weight) for a 12-kg mass. Then use Newton’s second law to find the acceleration.
Solution: Solve Newton’s law of
universal gravitation for r:
GM E m
F
r=
(6.67 × 10
=
−11
)(
)
N ⋅ m 2 /kg 2 5.97 × 1024 kg (12 kg )
15 N
= 1.8 × 107 m
Insight: The distance 1.8×107 m is about 2.8RE . If the mass were on Earth’s surface, it would weigh 118 N (26 lb).
72. Picture the Problem: Both Earth and the Moon exert a gravitational force on a spaceship, but in opposite directions.
Strategy: Use Newton’s universal law of gravitation to find the attractive forces from Earth and the Moon. Set the
force due to Earth equal to the force due to the Moon when the spaceship is at a distance r from the center of Earth. Let
R = 3.84 × 108 m, the distance between the centers of Earth and the Moon. Then solve the expression for the distance r.
Solution: 1. (a) Set FE = FM
using Newton’s law of gravity
and solve for r:
G
ms mE
r
2
= G
ms mM
(R − r )2
mE ( R − r ) = mM r 2
2
r=
⇒ R − r = 2mM mE r ⇒ R = 2mM mE r + r
R
1 + mM mE
=
3.84 × 108 m
1+
(7.35 × 10
22
kg
) (5.97 × 10
24
kg
)
= 3.46 × 108 m
2. (b) The net gravitational force on the spaceship and the astronauts will steadily decrease, reaching zero at the location
found in part (a), and then gradually increase in the opposite direction (toward the Moon). However, the astronauts are
accelerated at the same rate as the spaceship and so they will appear to float, not walk.
Insight: The distance in part (a) is about 90% of the distance between Earth and the Moon! Earth’s gravity is the
dominant force through most of the journey to the Moon.
73. Picture the Problem: Both Earth and Mars accelerate other masses near their surfaces according to Newton’s law of
universal gravitation.
Strategy: Use a ratio of gravitational accelerations to find the mass of the Mars in terms of the mass of Earth. Note that
if you solve g = GM R 2 for M, you find M = gR 2 G .
Solution: Determine the ratio M M M E by
solving g = GM R 2 for M:
M M g M RM2 G g M RM2 (0.38 g E )(0.53RE )
=
=
=
= 0.11
ME
g E RE2 G
g E RE2
g E RE2
2
M M = 0.11 M E
Insight: Using a ratio can be a powerful tool for solving a question like this one, where only relationships between
various quantities are given. The exact ratio of masses is 6.4185 × 1023 kg 5.9736 × 1024 kg = 0.10745.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 15
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
74. The spinning basket exerts an inward force on the clothes, keeping them moving in a circle, but does not exert a force
on the water that is able to escape through the holes in the basket and can be drained away.
75. People on the outer rim of a rotating space station must experience a force directed toward the center of the station
in order to follow a circular path. This force is applied by the “floor” of the station, which is really its outermost wall.
Because people feel an upward force acting on them from the floor, just as they would on Earth, the sensation is like
an “artificial gravity.”
76. Picture the Problem: A car travels along an elliptical path at
constant speed.
Strategy: Static friction between the car’s tires and the road provides
the centripetal force required to make the car travel along a circular
path. The smaller the radius of the circle, the greater the required
force.
Solution: The radius of the car’s path is smallest at point A in the
diagram, so the required force is greatest there and so is the
likelihood of skidding. The radius of the car’s path is largest at point
C, so the likelihood of skidding is smallest there. The ranking is
therefore C < B < A.
Insight: A comet that travels along an elliptical path experiences the largest gravitational force when it is closest to the
Sun. In such a case the required centripetal force is largest both because v is greatest and r is least at that point. In orbit
terminology the point closest to the Sun is called perihelion.
77. Picture the Problem: A car travels along a circular path at constant speed.
Strategy: Static friction between the tires and the road provides the centripetal force required to make the car travel
along a circular path.
Solution: 1. (a) No, the velocity of the car is not constant because the direction changes (it always points tangent
to the path).
2. (b) Yes, the speed of the car is constant as stated in the question.
3. (c) Yes, the magnitude of the acceleration a = v 2 r remains constant because the speed and the radius of curvature
remain constant.
4. (d) No, the direction of the acceleration does not remain constant: it always points toward the center of the circle.
This requires the acceleration to point in a different direction in space, depending upon the position of the car.
Insight: If the car’s speed were to change, the answers to all four questions would be “no”!
78. Picture the Problem: Astronauts in a human centrifuge travel along a circular path at constant speed.
Strategy: Solve acp = v 2 r for the speed required to attain the desired acceleration.
Solution: Solve acp = v 2 r for v:
v = r acp = r (9.0 g ) =
(15 m )(9.0) (9.81 m/s 2 ) =
36 m/s
Insight: This speed corresponds to 23 revolutions per minute for the centrifuge, or 1 revolution every 2.6 s.
79. Picture the Problem: Clothes in a rotating drum travel along a circular path at constant speed.
Strategy: Solve acp = v 2 r for the speed required to attain the given acceleration.
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9 – 16
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
Solution: Solve acp = v 2 r for v:
v = r acp =
(0.31 m )(27 m/s 2 ) =
2.9 m/s
Insight: This speed corresponds to 89 revolutions per minute for the rotating drum, or 1 revolution every 0.67 s.
80. Picture the Problem: A test tube travels along a circular path at constant speed inside a centrifuge.
Strategy: Solve acp = v 2 r for the speed required to attain the desired acceleration.
v = racp = r (52, 000 g ) =
Solution: Solve acp = v 2 r for v:
(0.075 m )(52, 000)(9.81 m/s 2 )
= 200 m/s = 0.20 km/s
Insight: This speed corresponds to 25,000 revolutions per minute for the centrifuge, or 415 revolutions per second.
81. Picture the Problem: A car travels along a circular path at constant speed.
Strategy: Static friction between the tires and the road provides the centripetal force required to make the car travel
along a circular path. Set the static friction force equal to the centripetal force and calculate its value.
mv 2 (1400 kg )(18 m/s )
=
= 7.2 kN
63 m
r
2
Solution: Set the static friction force
equal to the centripetal force:
f s, max = f cp = macp =
(
)
Insight: The maximum static friction force is μs mg = (0.85)(1400 kg ) 9.81 m/s 2 = 11.7 kN, which corresponds to a
maximum cornering speed (without skidding) of 23 m/s (50 mph).
82. Picture the Problem: A car follows a circular path at constant speed
as it passes over a bump as shown in the figure at the right.
Strategy: The centripetal acceleration is downward, toward the center of
the circle, as the car passes over the bump. Write Newton’s second law in
the vertical direction and solve for the normal force N, which is also the
apparent weight of a passenger.
Solution: 1. Write Newton’s
second law for a passenger
and solve for N:
2. Insert the numerical values:
∑F
y
= N − mg = − macp = − m v 2 r
(
N = m g − v2 r
)
⎡
(12 m/s)2 ⎤ = 380 N = 0.38 kN
N = (67 kg ) ⎢9.81 m/s 2 −
⎥
35 m ⎦⎥
⎣⎢
Insight: This apparent weight is 42% less than the normal 0.66-kN weight of the passenger.
83. Picture the Problem: A car follows a circular path at constant speed as
it passes over a bump as shown in the figure at the right.
Strategy: The centripetal acceleration is downward, toward the center
of the circle, as the car passes over the bump. Write Newton’s second
law in the vertical direction and set the normal force N, which is also the
apparent weight of the passenger, equal to zero. Then solve for the speed
of the car.
Solution: 1. Write Newton’s
second law for a passenger, and
set N to zero:
∑F
2. Now solve for v:
v = rg =
y
= N − mg = − macp = − m v 2 r
0 − mg = − m v 2 r
(35 m )(9.81 m/s 2 ) = 19 m/s
Insight: The car has zero normal force on it as well, meaning it is now airborne!
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 17
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
84. Picture the Problem: The free-body diagram for Jill is shown at the right.
Strategy: The center of Jill’s circular motion is the pivot point of the vine. There are two
forces acting on Jill, the tension due to the vine upward and gravity downward. These
two forces add together to produce her centripetal acceleration in the upward direction.
Write Newton’s second law for Jill in the vertical direction and solve for T:
Solution: Write Newton’s second
law for Jill in the vertical direction
and solve for T:
∑F
y

v
= T − W = macp
r

T

W
2
T = W + macp = mg + m v r
2
⎡
2.4 m/s ) ⎤
(
2
= (63 kg ) ⎢9.81 m/s +
⎥
6.9 m ⎥⎦
⎢⎣
T = 670 N = 0.67 kN
Insight: The tension in the vine will be at its maximum at the bottom of her circular path because it is at that point that
the vine must both support her weight and provide the upward centripetal force. Her speed is a maximum there as well,
making the centripetal force the largest at that point.
85. Picture the Problem: Free-body diagrams for the top and bottom
positions on the Ferris wheel are shown at the right.
Top of Ferris Wheel
r

N
Strategy: Find the speed of the seat by dividing the circumference
of its circular path by the time it takes to complete a cycle. Write
Newton’s second law for the passenger at the top of the Ferris
wheel and a second Newton’s second law for a passenger at the
bottom, and solve each for the normal force.

v
Solution: 1. (a) The normal force on you by the Ferris wheel seat
is equal to your apparent weight. At the top of the Ferris wheel
your centripetal acceleration points downward and the normal
force is the smallest, but at the bottom of the Ferris wheel the
normal force must both support your weight and provide the
upward centripetal acceleration, so it is at its maximum there.

W

N

v

W
r
Bottom of Ferris Wheel
2. (b) Write Newton’s second law at the top of the
Ferris wheel and solve for your apparent weight N top :
∑F
3. Find the speed of the seat:
v=
4. Calculate the apparent weight N top :
⎛
(1.6 m/s )2 ⎞
N top = m g − v 2 r = (55 kg ) ⎜ 9.81 m/s 2 −
⎟
7.2 m ⎠
⎝
= N top − mg = − macp = − mv 2 r
y
(
N top = m g − v 2 r
)
2π r 2π (7.2 m )
=
= 1.6 m/s
T
28 s
(
)
= 520 N = 0.52 kN
5. Write Newton’s second law at the bottom of the
Ferris wheel and solve for your apparent weight N bottom :
∑F
y
= N bottom − mg = macp = mv 2 r
⎛
(1.6 m/s )2 ⎞
N bottom = m g + v 2 r = (55 kg ) ⎜ 9.81 m/s 2 +
⎟
7.2 m ⎠
⎝
(
)
= 560 N = 0.56 kN
Insight: The centripetal acceleration is 0.36 m/s2 so that the centripetal force is 3.6% of your 0.54 kN weight.
86. No. Kepler’s second law states that the radius of Mars’s orbit sweeps out the same amount of area per time throughout
its orbit, but the area it sweeps out differs from that of the other planets because its orbit radius and period are different.
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9 – 18
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
87. Kepler’s second law states that the radius of a planet’s orbit sweeps out the same amount of area per time throughout
its orbit. Therefore, if in 2 months a planet sweeps out the area A, in 1 month it sweeps out area 0.5A and in 3 months
it sweeps out area 1.5A.
88. The phrase “free fall” means that gravity is the only force that affects the satellite’s motion. It does not imply that the
satellite is falling straight down. Because gravity is the only force that affects the motion of a satellite in orbit, it is in
free fall, but its horizontal velocity is sufficiently large that it never gets any closer or farther from Earth’s surface.
Therefore, friend 1 is correct.
89. Once the orbit period and orbit radius of a satellite such as Charon can be measured, the mass of the planet it orbits
(Pluto in this case) can be calculated using Kepler’s third law. In reality the mass of Charon is large compared to the
mass of Pluto, so that Kepler’s third law in the form given in the text had to be adjusted to account for the motions of
the two bodies about their center of mass.
90. The Moon has no atmosphere to slow down a projectile, so that if you throw the rock sufficiently fast (1680 m/s) it will
orbit the Moon (in about 1.8 hours) and might hit you in the back!
91. It makes more sense to think of the Moon as orbiting the Sun, with Earth providing a smaller force that makes the Moon
“wobble” back and forth in its solar orbit.
92. Picture the Problem: The average distance from Earth to the Moon is increasing at the rate of 3.8 cm per year.
Strategy: Consider Kepler’s third law of orbits when answering the conceptual question.
Solution: 1. (a) Kepler’s third law states that T = (constant ) r 3 2 . Increasing the orbit radius r will therefore increase the
orbit period T. We conclude that the length of the month will increase.
2. (b) The best explanation is A. The greater the radius of an orbit, the greater the period, which implies a longer month.
Statements B and C are both false.
Insight: The angular momentum of Earth-Moon system is not conserved due to tidal friction, which converts some
of the mechanical energy of the system into thermal energy through the motion of ocean waters on Earth and the
deformation of the rocks of both Earth and the Moon.
93. Picture the Problem: A GPS satellite travels around Earth in a circular orbit.
Strategy: Use Kepler’s third law to find the orbit period of the GPS satellite.
Solution: Calculate the period:
⎛ 2π ⎞ 3/ 2
T =⎜
⎟r
⎝ GM E ⎠
⎡
=⎢
⎢
⎣
(6.67 × 10
2π
−11
2
N ⋅ m /kg
2
)(5.97 × 10
24
⎤
⎥ 2.02 × 107 + 6.37 × 106 m
kg ⎥
⎦
)
(
)
3/ 2
T = 43100 s = 12.0 h
Insight: The orbits of these GPS satellites are designed so that at least 3 satellites are overhead every spot on Earth
at any one time.
94. Picture the Problem: A planet orbits the star Iota Horologii according to Kepler’s third law.
Strategy: Use a ratio to find the orbit radius of the extrasolar planet by comparing its orbit with Earth’s orbit around
the Sun.
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9 – 19
Chapter 9: Gravity and Circular Motion
Pearson Physics by James S. Walker
⎛r⎞
⎜⎝ r ⎟⎠
E
Solution: Make a ratio using
Kepler’s third law and solve for r:
32
=
T GM s 2π
=
TE GM s 2π
⎛T⎞
r = rE ⎜ ⎟
⎝ TE ⎠
23
T
TE
⎛ 320 d ⎞
= 1.50 × 1011 m ⎜
⎝ 365 d ⎟⎠
(
)
23
= 1.4 × 1011 m
Insight: If we knew the period of the planet were exactly 320 days, we could keep an extra significant figure and report
1.37×1011 m as the orbit radius. Note that if we use Astronomical Units (1 AU = 1.50×1011 m) to describe the orbit
radius and years to describe the orbit period, Kepler’s third law can be written T 2 = r 3 .
95. Picture the Problem: The moon Phobos travels in a circular orbit around Mars.
Strategy: Use Kepler’s third law and the mass and radius of Mars to determine the orbit period.
Solution: Calculate the orbit period:
⎛ 2π ⎞ 3/ 2
T =⎜
⎟r
⎝ GM M ⎠
⎡
=⎢
⎢
⎣
(6.67 × 10
2π
−11
2
N ⋅ m /kg
2
)(0.108 × 5.97 × 10
24
⎤
⎥ 9378 × 103 m
kg ⎥
⎦
)
(
)
3/ 2
T = 27,500 s = 7.64 h
Insight: Mars rotates on its axis about every 24.62 hours, so the moon Phobos would cross the Martian sky about
3 times a day!
96. Picture the Problem: The moon Ganymede travels in a circular orbit around Jupiter.
Strategy: Solve Kepler’s third law for the mass of Jupiter.
Solution: Solve Kepler’s
third law for M J :
⎛ 2π ⎞ 3/ 2
T =⎜
⎟r
⎝ GM J ⎠
⇒T2 =
2
4π 2 3
r
GM J
3
2π
⎛ 2π ⎞ r
⎛
MJ = ⎜ ⎟
=
⎝ T ⎠ G ⎜⎝ 6.18 × 105
⇒ MJ =
4π 2 3
r
T 2G
(
)
3
2
1.07 × 109 m
⎞
= 1.90 × 1027 kg
⎟
s ⎠ 6.67 × 10−11 N ⋅ m 2 /kg 2
Insight: Jupiter is 318 times more massive than Earth!
97. Picture the Problem: A satellite travels in a circular orbit around Earth.
Strategy: Use Kepler’s third law to find the period of the satellite. Use the mass and radius of Earth given in a table of
astronomical values. When the altitude h = 2 RE , the orbit radius r = RE + h = 3RE .
Solution: 1. (a) Calculate T for r = 3RE :
⎛ 2π ⎞
3/ 2
T =⎜
⎟ (3RE )
⎝ GM E ⎠
⎡
=⎢
⎢
⎣
(6.67 × 10
2π
−11
2
N ⋅ m /kg
2
)(5.97 × 10
24
⎤
⎥ 3 × 6.37 × 106 m
kg ⎥
⎦
)
(
)
3/ 2
T = 26,300 s = 7.31 h
2. (b) The period does not depend on the mass of the satellite because in general, the period of an orbiting planet or
moon depends only on the mass being orbited.
Insight: The fact that the orbit period of an object is independent of its mass, as discussed in part (b), is ultimately due
to the equivalence of gravitational and inertial mass. This equivalence lays the foundation for Einstein’s General Theory
of Relativity.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 20
Pearson Physics by James S. Walker
Chapter 9: Gravity and Circular Motion
98. Picture the Problem: Deimos and Phobos travel in circular orbits about Mars.
Strategy: Solve Kepler’s third law for the orbit radius of Deimos. Use the mass of Mars given in a table of astronomical
values, M M = 0.108 × 5.97 × 1024 kg = 6.45 × 1023 kg.
Solution: 1. (a) Because T ∝ r 3/ 2 and Deimos has the greater period, Deimos is farther from Mars than is Phobos.
⎛ T GM M ⎞
r=⎜
⎟
⎝ 2π ⎠
2. (b) Solve Kepler’s third law for r:
(
2/3
⎡ 1.10 × 105 s
=⎢
⎢
⎣
) (6.67 × 10
)(
)
N ⋅ m 2 /kg 2 6.45 × 1023 kg ⎤
⎥
⎥
2π
⎦
−11
2/3
r = 2.36107 m
Insight: The orbit radius of Deimos, 23,600 km, is larger than the 9378-km orbit of Phobos, as predicted in part (a).
99. Picture the Problem: A geosynchronous satellite orbits Earth with a period of 24.0 h and an altitude of 35,800 km.
Strategy: Use the given expression for the orbital speed, v = GM E r , keeping in mind that the altitude h is given
and that r = RE + h.
v=
Solution: Calculate the orbital speed:
G ME
=
RE + h
(6.67 × 10
−11
)(
N ⋅ m 2 /kg 2 5.97 × 1024 kg
6.37 × 106 + 3.58 × 107 m
)
= 3070 m/s = 3.07 km/s
Insight: The speed of an object in low Earth orbit is about 7.91 km/s, but this isn’t a low orbit, because the altitude of
3.58×107 m is about 5.62 Earth radii above the surface, or a total orbit radius of 6.62 RE .
100. Picture the Problem: Earth travels around the Sun in a circular orbit.
Strategy: Use the given expression for the orbital speed, v = GM r , where M is the mass of the Sun (2.00×1030 kg)
and r is the orbit radius of Earth (1.50×1011 m).
v=
Solution: Calculate the orbital speed:
GM
=
r
(6.67 × 10
−11
)(
N ⋅ m 2 /kg 2 2.00 × 1030 kg
1.50 × 1011 m
)
= 29,800 m/s = 29.8 km/s
Insight: The orbital speed of Earth as it travels around the Sun is almost 87 times the speed of sound in air!
101. Picture the Problem: The tiny moon Dactyl travels around 243 Ida in an approximately circular orbit.
Strategy: Solve Kepler’s third law for the mass of 243 Ida, using the orbit distance and period given in the problem.
Solution: 1. (a) Solve Kepler’s third law for the mass of 243 Ida, using the orbit distance and period given in the
problem.
2. (b) Solve Kepler’s
third law for M 243 Ida :
M 243 Ida
(
)
3
2 3
2
89 × 103 m
2π
⎛ 2π ⎞ r
⎛
⎞
=⎜ ⎟
=
= 8.9 × 1016 kg
⎝ T ⎠ G ⎜⎝ 19 h × 3600 s/h ⎟⎠ 6.67 × 10 −11 N ⋅ m 2 /kg 2
Insight: This asteroid is a little bit bigger than Manhattan. The moon Dactyl is only about 1.4 km (less than 1 mi)
in diameter.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 21
Pearson Physics by James S. Walker
Chapter 9: Gravity and Circular Motion
102. Picture the Problem: The formula for orbital speed of a satellite can be derived from dimensional analysis.
Strategy: Use dimensional analysis, reducing G into its fundamental dimensions, and noting that ME has the dimension
of mass and r has the dimension of length.
G = 6.67 × 10 −11 N ⋅ m 2 kg 2
Solution: 1. Reduce G into its
fundamental dimensions
2
2
force )(length )
mass × acceleration ] (length )
(
[
=
=
(mass)2
(mass)2
⎡
length ⎤
2
length )
⎢(mass )
2 ⎥(
(time) ⎥⎦
⎢
(length )3
=⎣
=
(mass )2
(mass)(time)2
⎡ (length ) 3 2 ⎤
⎢
⎥ (mass )
2
⎢
⎥
mass
time
(
)
(
)
G ME ⎣
(length )2
⎦
=
=
r
(length )
(time)2
2. Combine the dimensions of G with
those of ME and r to get dimensions
of (length time) :
2
3. Take the square root of the
expression to find v:
v=
G ME
=
r
(length )2
(time)2
=
length
time
⇒ v=
G ME
r
Insight: The most reliable way to derive the expression is to set the centripetal force required to keep the satellite in a
circular orbit equal to the gravitational force between the satellite and Earth. Solve the resulting equation for the speed
of the satellite.
103. Picture the Problem: Two satellites travel in circular orbits about Earth.
Strategy: Use the given expression for the orbital speed of a satellite to answer the conceptual question. Then calculate
the speed, keeping in mind that the altitude h is given and that r = RE + h.
Solution: 1. (a) The given expression for the orbit speed is inversely proportional to the square root of the orbit radius;
that is, v ∝ 1 r . That means the orbit speed decreases as the radius increases, so satellite 2 will have the greater
orbital speed.
2. (b) Calculate the orbital speed:
v=
G ME
=
RE + h
G ME
=
RE + RE
(6.67 × 10
−11
)(
N ⋅ m 2 /kg 2 5.97 × 1024 kg
6
2 × 6.37 × 10 m
)
= 5590 m/s = 5.59 km/s
Insight: A satellite that orbits at an altitude of two Earth radii has an orbit speed of 4.56 km/s, slower than the satellite
at the lower orbit, as predicted.
104. Picture the Problem: You weigh yourself on a scale inside an airplane flying due east above the equator. The airplane
then turns around and heads due west with the same speed.
Strategy: Consider the speed of the aircraft relative to the center of Earth when answering the conceptual question.
Solution: If you fly to the east, which is the direction of Earth’s rotation, you have a greater speed relative to the center
of Earth than if you fly to the west. As a result, the centripetal force required to maintain your circular motion is greater,
and your apparent weight is less. We conclude that the reading on the scale will increase slightly when the airplane
turns around and heads due west with the same speed relative to Earth’s surface.
Insight: There would be no difference at all in the scale reading if the plane were flying due north and then turned
around and flew due south. In both such cases the airplane has the same speed relative to the center of Earth.
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9 – 22
Pearson Physics by James S. Walker
Chapter 9: Gravity and Circular Motion
105. The orbital speed of a satellite depends on the mass of Earth and the orbit distance r from the center of Earth. Because
the satellite and the International Space Station each orbit the same Earth at the same orbit radius, the orbital speed of
the satellite is equal to the orbital speed of the space station.
106. The orbital speed of a satellite is given by v = GM r , where M is the mass of the object that the planet or satellite is
orbiting, and r is the orbit radius. It can be seen from this expression that increasing the orbit distance r will decrease
the orbital speed.
107. Picture the Problem: A rotating clearview screen produces a large centripetal acceleration at its rim.
Strategy: Calculate the centripetal acceleration from the given linear speed and radius.
v 2 (35 m/s )
= 1
= 6300 m/s 2 = 6.3 km/s 2
r
0.39
m
)
2(
2
acp =
Solution: Calculate the centripetal acceleration:
Insight: This acceleration is 640 times larger than the acceleration of gravity, so a rain drop on the screen experiences a
“centrifugal” force 640 times greater than its own weight, causing it to be thrown off the rotating screen.
108. Picture the Problem: A dragonfly rotates at a high rate in order to remove water from its body.
Strategy: Solve the expression for centripetal acceleration for the radius.
v 2 (2.3 m/s )
=
= 0.021 m = 2.1 cm
acp
250 m/s 2
2
r=
Solution: Solve acp = v 2 r for r:
Insight: This acceleration is 25 times larger than the acceleration of gravity, so a water drop on the dragonfly
experiences a “centrifugal” force 25 times greater than its own weight, causing the water to be thrown off the
dragonfly’s body.
109. Picture the Problem: The Skylab spacecraft fell back to Earth out of orbit due to the friction it experienced in the upper
reaches of Earth’s atmosphere.
Strategy: Consider the mechanical energy of the orbiting satellite when answering the conceptual question.
Solution: As the radius of Skylab’s orbit decreased its speed did increase because the gain in its kinetic energy due to
the loss of gravitational potential energy was much larger than the loss of its kinetic energy due to friction. Another
approach is to recall that the orbit speed of a satellite is given by v = GM E r , so that the speed increases with
decreasing radius.
Insight: You might think that friction would slow Skylab—just like other objects are slowed by friction—but by
dropping Skylab to a lower orbit, and freeing up gravitational potential energy, friction was initially responsible for
an increase in speed. In the lower, denser portions of the atmosphere the friction did in fact slow down the spacecraft.
110. Picture the Problem: An astronaut jumps straight upward, momentarily comes to rest, and falls back to the surface of
the planet.
Strategy: The height of the astronaut’s jump is very small compared with the radius of the planet, so we assume that the
acceleration of gravity is constant during the jump. Use conservation of mechanical energy to find an expression for the
acceleration of gravity as a function of the initial speed and the maximum jump height, and set it equal to GM R 2 .
Then solve for the mass M of the planet.
Solution: 1. Set Ei = Ef and solve for g:
KEi + PEi = KEf + PEf
1
2
mvi2 + 0 = 0 + mgh
g = vi2 2h
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9 – 23
Pearson Physics by James S. Walker
Chapter 9: Gravity and Circular Motion
2. Now let g = GM R 2 and solve for M:
vi2 GM
= 2
2h
R
(
)
2
3860 × 103 m (3.10 m/s )
R 2 vi2
=
= 1.85 × 1024 kg
M =
2Gh 2 6.67 × 10 −11 N ⋅ m 2 /kg 2 (0.580 m )
(
2
)
Insight: As the height h increases for the same initial speed vi , the mass of the planet decreases. A less massive planet
will not accelerate the astronaut as much and she will rise to a greater height.
111. Picture the Problem: A child travels along a circular path at constant speed while sitting on a merry-go-round.
Strategy: The static friction force between the child and the surface of the merry-go-round provides the inward
centripetal force required to keep the child traveling in a circle. The normal force is equal to the weight of the child
because she is not accelerating in the vertical direction. Write Newton’s second law in the horizontal direction and
solve for the coefficient of static friction.
∑F
Solution: Write Newton’s second law in the
horizontal direction and solve for μs :
x
= μs mg = m v 2 r
(2.2 m/s )
v2
=
= 0.21
rg (2.3 m ) 9.81 m/s 2
2
μs =
(
)
Insight: The speed of 2.2 m/s corresponds to 9.1 revolutions per minute. If the coefficient could be doubled to 0.42 the
child could travel at 3.1 m/s or 13 revolutions per minute and still say, “Look Mom, no hands!”
112. Picture the Problem: The orbit period of a geosynchronous satellite matches the rotational period of the planet.
Strategy: This problem is similar to Active Example 9.12. Set the orbit period equal to the rotational period of the
planet and solve for distance r and then the altitude h. The rotational period of Mars is
24.6229 h × 3600 s/h = 8.86424 × 104 s. Use the given mass for Mars, and note its radius is 3.4×106 m.
Solution: 1. Square both sides of
Kepler’s third law and solve for r:
4π 2 r 3
T =
GM
2
⎛ GMT 2 ⎞
⇒r=⎜
⎝ 4π 2 ⎟⎠
(
1/3
)(
)(
)
1/3
⎡ 6.67 × 10 −11 N ⋅ m 2 /kg 2 0.108 × 5.9710 24 kg 8.86424 × 10 4 s 2 ⎤
⎥
r=⎢
⎢
⎥
4π 2
⎣
⎦
7
r = 2.0510 m
2. Solve r = R + h for h:
h = r − R = 2.05107 m − 0.34107 m = 1.71 ×107 m
Insight: This distance is less than half of the geosynchronous orbit radius of 4.22×107 m for Earth. Although the
rotation periods of the two planets are nearly equal, Mars has much less mass than does Earth.
113. Picture the Problem: The hockey puck travels in a circle with the string
tension providing the necessary centripetal force.
Strategy: The tension in the string will be the same everywhere as long as
the lip of the hole in the tabletop is frictionless. From Newton’s second law
for the hanging mass M, the tension in the string equals T = Mg . Set the
tension equal to the centripetal force required to keep the hockey puck
traveling in a circle of radius r at constant speed v, then solve for v.
Solution: Set the string tension equal to
the centripetal force and solve for v:
T = Mg = macp = m v 2 r
Mrg
m
Insight: Another way to look at the problem is to ask what mass M would
be required to keep m traveling in a circle of radius r at constant speed v,
and the answer is M = mv 2 rg .
v=
Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 24
Pearson Physics by James S. Walker
Chapter 9: Gravity and Circular Motion
114. Reports and specific data will vary. Reports should mention how timing signals from a minimum of three satellites are
used to triangulate the position of the GPS receiver on Earth’s surface. As of December 2012, the GPS system uses
32 satellites. Orbit altitude is approximately 20,200 km. Orbital period is 11 hrs 58 min. For civilain use, the location
accuracy is within 7.8 m with a 95% confidence level.
115. Because both the astronauts and the spacecraft are in free fall, they do not accelerate relative to one another. This
gives the impression of weightlessness relative to the spacecraft, but in reality gravity is fairly strong, providing
the centripetal force necessary to keep the astronauts and their spacecraft in circular orbit around Earth. You would
experience a similar effect as you went over the top of a hill on a high-speed roller coaster, or if you had the misfortune
to ride in an elevator that was plummeting in free fall.
116. Picture the Problem: The roughly spherical Comet Wild 2 has
a radius of 2.7 km, and the acceleration due to gravity on its
surface is 0.00010g. The two curves in the figure at right show
the surface acceleration as a function of radius for a spherical
comet with two different masses, one of which corresponds to
Comet Wild 2.
Strategy: Use the given radius and surface acceleration of the
comet to select the appropriate curve.
Solution: The point (2.7 km, 0.00010g) is a point on curve II,
not curve I. This corresponds to choice B.
Insight: The indicated point is also very close to the density of ice, so we can conclude the Comet Wild 2 is composed
primarily of ice.
117. Picture the Problem: Comet Wild 2 is a roughly spherical body of radius 2.7 km and a surface gravitational
acceleration of 1.0×10−4g.
Strategy: Use the equation g = G M R 2 to determine the mass of the comet.
(
)
9.8 × 10 − 4 m/s 2 (2700 m )
g RC2
=
= 1.07 × 1014 kg = 1.1 × 1014 kg
Solution: Solve g = G M R for M . The M =
G
6.67 × 10 −11 N ⋅ m 2 /kg 2
calculated value is choice C.
2
2
Insight: Although this seems like a very large number (the comet would weigh 110 billion tons on Earth!)
it is a very small mass compared to Earth. The comet’s mass is about 2.2% of the mass of all the water
in Lake Michigan (4.9×1015 kg).
118. Picture the Problem: Comet Wild 2 is a roughly spherical body of radius 2.7 km and a surface gravitational
acceleration of 1.0×10−4g. We imagine that it has a small satellite that orbits at 5.4 km from the center of the comet.
Strategy: Use the mass of the comet and Kepler’s third law to find the orbit period.
Solution: Apply Kepler’s third law. The
calculated value is choice D.
T=
2π
G MC
r3 2 =
(6.67 × 10
2π
−11
2
N ⋅ m /kg
2
)(1.07 × 10
14
kg
)
(5400 m)3 2
= 2.95 × 104 s = 8.2 h
Insight: This orbit period is much longer than the 4.0 h orbit period of a satellite that orbits at twice the radius of Earth.
The period of the International Space Station (in low Earth orbit) is 1.5 h.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9 – 25