Chapter 12 Solutions
8.
Picture the Problem: All three masses attract each other gravitationally.
Strategy: Add the gravitational force on each mass due to the other two masses using equation 12-1.
Solution: 1. (a) Add
the forces:
M
M E MS
M M
M 
 G E2 M  GM E  2 S  2 M 
2
rE-S
rE-M
 rE-S rE-M 


2.00 1030 kg
7.35 1022 kg 
  6.67 1011 N  m 2 /kg 2  5.97 1024 kg  

2
2
 1.50  1011 m   3.84 108 m  


FE  G
FE  3.56 1022 N toward the Sun
2. (b) Add the forces:
M
MSM M
M M
M 
 G E2 M  GM M  2 S  2 E 
2
rS-M
rE-M
 rS-M rE-M 


2.00 1030 kg
5.97 1024 kg 
  6.67  1011 N  m 2 /kg 2  7.35 1022 kg  

 1.50 1011  3.84 108 m 2  3.84 108 m 2 


FM  G
FM  2.40 1020 N toward the Sun
3. (c) Add the forces:
M
MSM M
M M
M 
 G S2 E  GM S  2 M  2 E 
2
rS-M
rS-E
r
 S-M rS-E 


7.35  1022 kg
5.97  1024 kg 
  6.67  1011 N  m 2 /kg 2  2.00 1030 kg  

 1.50  1011  3.84  108 m 2 1.50  1011 m 2 


FS  G
FS  3.58  1022 N toward the Earth-Moon system
12.
Picture the Problem: The 2.0-kg mass is gravitationally attracted to the other
three masses.
Strategy: Add the gravitational forces using the component method of vector
addition. Use equation 12-1 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. (a) Find the x component of the force on m2 :
m1m2
mm
 G 22 4
r122
r24
mm
mm
 G 1 2 2  G 22 4
r12
r24
Fx  G
  6.67  1011
Fx  1.3  108 N
cos 
 m1 m4 r12 
 r12 
   Gm2  2  3 
r
r24 
 24 
 r12


 4.0 kg  0.20 m 
 1.0 kg

2
2
N  m /kg   2.0 kg  


2
2
2 3/ 2
0.20
m






0.20 m    0.10 m 


 

m2 m3
mm
 G 2 2 4 sin 
r232
r24
m m r 
mm
m m r 
 G 2 2 3  G 2 2 4  14   Gm2  23  43 14 
r23
r24  r24 
r24 
 r23


 4.0 kg  0.10 m 
 3.0 kg

11
2
2
  6.67  10 N  m /kg   2.0 kg  


2
2
2 3/ 2
0.10
m
  0.20 m    0.10 m   





Fy  4.5 108 N
2. Find the y component
of the force on m2 :
Fy  G
3. Use the components of F to
find its magnitude and direction.
F  Fx 2  Fy 2 
 Fy
 Fx
  tan 1 
1.3 10
8
N    4.5  108 N   4.7  108 N
2
2
8

N
1  4.5  10
  tan 
  74 below horizontal and to the left
8
1.3

10
N



4. (b) If the sides of the rectangle are all doubled, all forces are reduced by a factor of 22  4; the directions of the forces are
unchanged.
18.
Picture the Problem: The acceleration of gravity at a planet’s surface is determined by its mass and radius.
Strategy: Apply a formula similar to equation 12-4 to find the acceleration of gravity on the surface of Titan. Use the mass and
radius data given in the problem.
Solution: Solve g T  G M T R for Titan:
2
T
g T   6.67 1011 N  m 2 /kg 2 
1.35 10 kg   1.36 m/s
 2.57 10 m 
23
6
2
2
43.
Picture the Problem: An object is located at the surface of the Earth and later at an altitude of 350 km.
Strategy: Use equation 12-8 to find the gravitational potential energy of the object as a function of its distance r  RE  h from
the center of the Earth. Then take the difference between the values at h = 0 and h = 350 km and compare it with the
approximate change in potential, U  mgh.
Solution: 1. (a) Calculate U   G
MEm
11
2
2
RE  h U 0   6.67 10 N  m /kg


5.97 10
24
kg   8.8 kg 
6.37 10 m
6
  5.5 108 J
at h = 0:
45.
5.97 10
kg  8.8 kg 
U h    6.67 10
3. (c) Take the difference U :
U  U h  U 0    5.50 108 J     5.21108 J   2.9 107 J
4. Compare with mgh:
U  mgh  8.8 kg   9.81 m/s 2  350 103 m   3.0 107 J
11
N  m /kg
2
2
 6.37 10
24
2. (b) Calculate U at h = 350 km:
6
 350 103 m
  5.2 108 J
Picture the Problem: The rocket is given enough kinetic energy to completely escape the Earth or the Moon.
Strategy: The rocket completely escapes the Earth when it is infinitely far away, which is when its gravitational potential energy
is zero. Set the kinetic energy equal to the magnitude of the initial (negative) gravitational potential energy in order to find the
energy needed to escape. Use the radius and mass data for the Earth and Moon given in the inside back cover of the book.
K  Ui  G
Solution: 1. (a) Calculate
K  U   U i  U i for the Moon:
MMm
RM
  6.67  1011 N  m 2 /kg 2 
K  U i  G
2. (b) Calculate K  Ui for the Earth:
MEm
RE
  6.67  10
52.
 7.35 10 kg   39, 000 kg   1.110
1.74 10 m 
22
11
6
J
 5.97 10 kg   39, 000 kg   2.4 10

 6.37 10 m 
24
11
N  m /kg
2
2
12
6
J
Picture the Problem: The projectile rises from one Moon radius at launch to 365 km above the surface at the highest point of its
travel.
Strategy: Use conservation of energy to determine the initial kinetic energy required to change the distance of the projectile
from RM to RM  h from the center of the Moon. Then use equation 7-6 to find the initial speed of the projectile.
Ki  U i  K f  U f
Solution: 1. Set Ei  Ef
and solve for Ki  12 mv :
2
i
2. Now multiply by 2 m
and solve for vi :
1
2
mvi2  G
mM M
mM M
 0G

RM
RM  h
1
2
 1
1 
mvi2  GmM M 


 RM RM  h 
 1
1 
vi  2GM M 


R
R
 M
M h


1
1
 2  6.67  1011 N  m 2 /kg 2  7.35  10 22 kg  


6
6
3
1.74

10
m
1.74

10

365

10
m


vi  988 m/s
56.
Picture the Problem: The projectile is launched at 11.2 km/s from the surface of the Earth, but slows down as it converts its
initial kinetic energy into gravitational potential energy.
Strategy: Set the mechanical energy at the launch equal to the mechanical energy when the speed is equal to the one-half the
escape speed. Then determine the gravitational potential energy and therefore the distance such a projectile would be from the
center of the Earth.
Solution: 1. Set Ei  Ef and simplify the
expression by multiplying both sides by 2 m :
Ki  U i  Kf  U f
M m
M m
1
mvi2  G E  12 mvf2  G E
2
RE
rf
2GM E 2GM E
2

 vf  vi2
rf
RE
2. Substitute vi  ve  2GM E RE and
2GM E 2GM E  1 2GM E   2GM E 

 4


rf
RE
RE   RE 

1
1

 rf  4 RE  4  6.37  106 m   2.55  107 m
rf 4 RE
vf  12 ve 
1
2
2GM E RE and solve for rf :
72. Picture the Problem: The 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 (equation 8-8) 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 according to equation 12-4. Then solve for the mass of the planet.
Solution: 1. Set Ei  Ef and solve for g:
Ki  U i  K f  U f
1
2
mvi2  0  0  mgh
g  vi2 2h
2. Now let g  GM R 2 and solve for M:
vi2 GM
 2
2h
R
 3860 103 m   3.10 m/s 
R 2 vi2
M 

 1.85  1024 kg
2Gh 2  6.67 1011 N  m 2 /kg 2   0.580 m 
2
2
74.
Picture the Problem: The three masses attract each other gravitationally and are
positioned as shown in the figure at right.
Strategy: Determine the magnitudes of the forces from the 2.00 kg ( F2 ) and
3.00 kg ( F3 ) masses using Newton’s Law of Universal Gravitation. Use the
magnitudes and directions of the two forces to add them using the component
method of vector addition. Combine the components of the vector sum to find
the magnitude and direction of the net force on the
1.00-kg mass. Let downward and to the left be the positive directions.
1.00 kg  2.00 kg 
m1m2
  6.67 1011 N  m2 /kg 2 
 1.33 1012 N
2
2
r
10.0
m


Solution: 1. Find the magnitude of F2 :
F2  G
2. Find the magnitude of F3 :
F3   6.67 1011 N  m2 /kg 2 
3. The horizontal components
subtract and the vertical
components add:
1.00 kg  3.00 kg 
 2.00 1012
2
10.0
m


N
F1x  F3 x  F2 x   2.00  1012 N  cos 60  1.33 10 12 N  cos 60
 3.35 1013 N (to the left)
F1 y  F3 y  F2 y   2.00  1012 N  sin 60  1.33 10 12 N  sin 60
 2.88 1012 N (downward)
4. Combine the components
to find the magnitude and
the direction:
F1  F1x 2  F1 y 2 
  tan 1
 3.35 10
13
N    2.88  1012 N   2.90  1012 N
2
2
 2.88  1012 N 
 tan 1 
  83.4 below horizontal, to the left
13
F1x
 3.35  10 N 
F1 y