13 CHAPTER OUTLINE 13.1 Newton’s Law of Universal Gravitation 13.2 Measuring the Gravitational Constant 13.3 Free-Fall Acceleration and the Gravitational Force 13.4 Kepler’s Laws and the Motion of Planets 13.5 The Gravitational Field 13.6 Gravitational Potential Energy 13.7 Energy Considerations in Motion Planetary and Satellite Universal Gravitation ANSWERS TO QUESTIONS Q13.1 Because g is the same for all objects near the Earth’s surface. The larger mass needs a larger force to give it just the same acceleration. Q13.2 To a good first approximation, your bathroom scale reading is unaffected because you, the Earth, and the scale are all in free fall in the Sun’s gravitational field, in orbit around the Sun. To a precise second approximation, you weigh slightly less at noon and at midnight than you do at sunrise or sunset. The Sun’s gravitational field is a little weaker at the center of the Earth than at the surface subsolar point, and a little weaker still on the far side of the planet. When the Sun is high in your sky, its gravity pulls up on you a little more strongly than on the Earth as a whole. At midnight the Sun pulls down on you a little less strongly than it does on the Earth below you. So you can have another doughnut with lunch, and your bedsprings will still last a little longer. Q13.3 Kepler’s second law states that the angular momentum of the Earth is constant as the Earth orbits the sun. Since L m r = ω , as the orbital radius decreases from June to December, then the orbital speed must increase accordingly. Q13.4 Because both the Earth and Moon are moving in orbit about the Sun. As described by F ma gravitational centripetal = , the gravitational force of the Sun merely keeps the Moon (and Earth) in a nearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in its orbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, at new moon or full moon. Q13.5 Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameter of the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit, however, must travel faster. Thus, the effect of air resistance is to speed up the satellite! Q13.6 Kepler’s third law, which applies to all planets, tells us that the period of a planet is proportional to r 32 . Because Saturn and Jupiter are farther from the Sun than Earth, they have longer periods. The Sun’s gravitational field is much weaker at a distant Jovian planet. Thus, an outer planet experiences much smaller centripetal acceleration than Earth and has a correspondingly longer period. 381
Transcript
13
CHAPTER OUTLINE
13.1 Newton’s Law of Universal Gravitation13.2 Measuring the Gravitational Constant13.3 Free-Fall Acceleration and the Gravitational Force13.4 Kepler’s Laws and the Motion of Planets13.5 The Gravitational Field13.6 Gravitational Potential Energy 13.7 Energy Considerations in
Motion
Planetary and Satellite
Universal Gravitation
ANSWERS TO QUESTIONS
Q13.1 Because g is the same for all objects near the Earth’s surface.The larger mass needs a larger force to give it just the sameacceleration.
Q13.2 To a good first approximation, your bathroom scale reading isunaffected because you, the Earth, and the scale are all in freefall in the Sun’s gravitational field, in orbit around the Sun. Toa precise second approximation, you weigh slightly less atnoon and at midnight than you do at sunrise or sunset. TheSun’s gravitational field is a little weaker at the center of theEarth than at the surface subsolar point, and a little weaker stillon the far side of the planet. When the Sun is high in your sky,its gravity pulls up on you a little more strongly than on theEarth as a whole. At midnight the Sun pulls down on you alittle less strongly than it does on the Earth below you. So youcan have another doughnut with lunch, and your bedspringswill still last a little longer.
Q13.3 Kepler’s second law states that the angular momentum of the Earth is constant as the Earth orbitsthe sun. Since L m r= ω , as the orbital radius decreases from June to December, then the orbital speedmust increase accordingly.
Q13.4 Because both the Earth and Moon are moving in orbit about the Sun. As described byF magravitational centripetal= , the gravitational force of the Sun merely keeps the Moon (and Earth) in a
nearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in itsorbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, atnew moon or full moon.
Q13.5 Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameterof the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit,however, must travel faster. Thus, the effect of air resistance is to speed up the satellite!
Q13.6 Kepler’s third law, which applies to all planets, tells us that the period of a planet is proportional tor 3 2 . Because Saturn and Jupiter are farther from the Sun than Earth, they have longer periods. TheSun’s gravitational field is much weaker at a distant Jovian planet. Thus, an outer planet experiencesmuch smaller centripetal acceleration than Earth and has a correspondingly longer period.
381
382 Universal Gravitation
Q13.7 Ten terms are needed in the potential energy:
U U U U U U U U U U U= + + + + + + + + +12 13 14 15 23 24 25 34 35 45 .
With N particles, you need iN N
i
N− =
−
=∑ 1
21
2
a f terms.
Q13.8 No, the escape speed does not depend on the mass of the rocket. If a rocket is launched at escapespeed, then the total energy of the rocket-Earth system will be zero. When the separation distance
becomes infinite U = 0a f the rocket will stop K = 0a f. In the expression 12
02mvGM m
rE− = , the mass
m of the rocket divides out.
Q13.9 It takes 100 times more energy for the 105 kg spacecraft to reach the moon than the 103 kgspacecraft. Ideally, each spacecraft can reach the moon with zero velocity, so the only term that needbe analyzed is the change in gravitational potential energy. U is proportional to the mass of thespacecraft.
Q13.10 The escape speed from the Earth is 11.2 km/s and that from the Moon is 2.3 km/s, smaller by a factorof 5. The energy required—and fuel—would be proportional to v2 , or 25 times more fuel is requiredto leave the Earth versus leaving the Moon.
Q13.11 The satellites used for TV broadcast are in geosynchronous orbits. The centers of their orbits are thecenter of the Earth, and their orbital planes are the Earth’s equatorial plane extended. This is theplane of the celestial equator. The communication satellites are so far away that they appear quiteclose to the celestial equator, from any location on the Earth’s surface.
Q13.12 For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be locatedat the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it mustbe over the southern hemisphere for the other half. We could share with Easter Island a satellite thatwould look straight down on Arizona each morning and vertically down on Easter Island eachevening.
Q13.13 The absolute value of the gravitational potential energy of the Earth-Moon system is twice thekinetic energy of the moon relative to the Earth.
Q13.14 In a circular orbit each increment of displacement is perpendicular to the force applied. The dotproduct of force and displacement is zero. The work done by the gravitational force on a planet in anelliptical orbit speeds up the planet at closest approach, but negative work is done by gravity andthe planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in onecomplete orbit is zero.
Q13.15 Every point q on the sphere that does not liealong the axis connecting the center of thesphere and the particle will have companionpoint q’ for which the components of thegravitational force perpendicular to the axiswill cancel. Point q’ can be found by rotatingthe sphere through 180° about the axis. Theforces will not necessarily cancel if the mass isnot uniformly distributed, unless the center ofmass of the non-uniform sphere still lies alongthe axis.
Fpq
Fpq
p
q
q’ (behind the sphere)
FIG. Q13.15
Chapter 13 383
Q13.16 Speed is maximum at closest approach. Speed is minimum at farthest distance.
Q13.17 Set the universal description of the gravitational force, FGM m
RgX
X
= 2 , equal to the local description,
F mag = gravitational , where MX and RX are the mass and radius of planet X, respectively, and m is the
mass of a “test particle.” Divide both sides by m.
Q13.18 The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as oneinterpretation of Equation 13.1 would suggest. All the bits of matter that make up the Earth will pullin different outward directions on the extra particle.
Q13.19 Cavendish determined G. Then from gGMR
= 2 , one may determine the mass of the Earth.
Q13.20 The gravitational force is conservative. An encounter with a stationary mass cannot permanentlyspeed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planetwill gain kinetic energy as the planet’s gravity does net positive work on it.
Q13.21 Method one: Take measurements from an old kinescope of Apollo astronauts on the moon. From themotion of a freely falling object or from the period of a swinging pendulum you can find theacceleration of gravity on the moon’s surface and calculate its mass. Method two: One coulddetermine the approximate mass of the moon using an object hanging from an extremely sensitivebalance, with knowledge of the position and distance of the moon and the radius of the Earth. Firstweigh the object when the moon is directly overhead. Then weigh of the object when the moon isjust rising or setting. The slight difference between the measured weights reveals the cause of tidesin the Earth’s oceans, which is a difference in the strength of the moon’s gravity between differentpoints on the Earth. Method three: Much more precisely, from the motion of a spacecraft in orbitaround the moon, its mass can be determined from Kepler’s third law.
Q13.22 The spacecraft did not have enough fuel to stop dead in its high-speed course for the Moon.
SOLUTIONS TO PROBLEMS
Section 13.1 Newton’s Law of Universal Gravitation
P13.1 For two 70-kg persons, modeled as spheres,
FGm m
rg = =× ⋅−
−1 22
11
27
6 67 10 70 70
210
.~
N m kg kg kg
m N
2 2e jb gb ga f .
P13.2 F m gGm m
r= =1
1 22
gGm
r= =
× ⋅ × ×= ×
−−2
2
11 4 3
27
6 67 10 4 00 10 10
1002 67 10
. ..
N m kg kg
m m s
2 22e je j
a f
384 Universal Gravitation
P13.3 (a) At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objectsare oppositely directed,
and from FGm m
rg =1 22
we have FG
∑ =−
= × −50 0 500 200
0 2002 50 102
5.
..
kg kg kg
m N
b gb ga f toward the 500-kg object.
(b) At a point between the two objects at a distance d from the 500-kg objects, the net force onthe 50.0-kg object will be zero when
G
d
G
d
50 0 200
0 400
50 0 5002 2
.
.
. kg kg
m
kg kgb gb ga f
b gb g−
=
or d = 0 245. m
P13.4 m m1 2 5 00+ = . kg m m2 15 00= −. kg
F Gm m
r
m m
m m
= ⇒ × = × ⋅−
− =×
× ⋅=
− −
−
−
1 22
8 11 1 12
1 12
8
11
1 00 10 6 67 105 00
0 200
5 001 00 10 0 040 0
6 67 106 00
. ..
.
.. .
..
N N m kg kg
m
kg N m
N m kg kg
2 2
2
2 22
e j b ga f
b g e je j
Thus, m m12
15 00 6 00 0− + =. . kg kgb gor m m1 13 00 2 00 0− − =. . kg kgb gb ggiving m m1 23 00 2 00= =. . kg, so kg . The answer m1 2 00= . kg and m2 3 00= . kg is physically
equivalent.
P13.5 The force exerted on the 4.00-kg mass by the 2.00-kg mass isdirected upward and given by
F j j
j
244 2
242
112
11
6 67 104 00 2 00
3 00
5 93 10
= = × ⋅
= ×
−
−
Gm m
r.
. .
.
.
N m kg kg kg
m
N
2 2e j b gb ga f
The force exerted on the 4.00-kg mass by the 6.00-kg mass isdirected to the left
F i i
i
644 6
642
112
11
6 67 104 00 6 00
4 00
10 0 10
= − = − × ⋅
= − ×
−
−
Gm m
r.
. .
.
.
e j e j b gb ga f N m kg kg kg
m
N
2 2 FIG. P13.5
Therefore, the resultant force on the 4.00-kg mass is F F F i j4 24 641110 0 5 93 10= + = − + × −. .e j N .
Chapter 13 385
P13.6 (a) The Sun-Earth distance is 1 496 1011. × m and the Earth-Moon distance is 3 84 108. × m , so thedistance from the Sun to the Moon during a solar eclipse is
1 496 10 3 84 10 1 492 1011 8 11. . .× − × = × m m m
The mass of the Sun, Earth, and Moon are MS = ×1 99 1030. kg
ME = ×5 98 1024. kg
and MM = ×7 36 1022. kg
We have FGm m
rSM = =× × ×
×= ×
−1 22
11 30 22
11 220
6 67 10 1 99 10 7 36 10
1 492 104 39 10
. . .
..
e je je je j
N
(b) FEM =× ⋅ × ×
×= ×
−6 67 10 5 98 10 7 36 10
3 84 101 99 10
11 24 22
8 220
. . .
..
N m kg N
2 2e je je je j
(c) FSE =× ⋅ × ×
×= ×
−6 67 10 1 99 10 5 98 10
1 496 103 55 10
11 30 24
11 222
. . .
..
N m kg N
2 2e je je je j
Note that the force exerted by the Sun on the Moon is much stronger than the force of theEarth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. TheMoon’s path is everywhere concave toward the Sun. Only by subtracting out the solarorbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass ofthis system.
Section 13.2 Measuring the Gravitational Constant
P13.7 FGMm
r= = × ⋅
×
×= ×−
−
−
−2
113
2 2106 67 10
1 50 15 0 10
4 50 107 41 10.
. .
.. N m kg
kg kg
m N2 2e j
b ge je j
386 Universal Gravitation
P13.8 Let θ represent the angle each cable makes with the vertical, L thecable length, x the distance each ball scrunches in, and d = 1 m theoriginal distance between them. Then r d x= − 2 is the separation ofthe balls. We have
Fy∑ = 0 : T mgcosθ − = 0
Fx∑ = 0 : TGmm
rsinθ − =2 0
FIG. P13.8
Then tanθ =Gmmr mg2
x
L x
Gm
g d x2 2 22−=
−a f x d xGm
gL x− = −2 2 2 2a f .
The factor Gm
gis numerically small. There are two possibilities: either x is small or else d x− 2 is
small.
Possibility one: We can ignore x in comparison to d and L, obtaining
x 16 67 10 100
9 8452
11
m N m kg kg
m s m
2 2
2a f e jb g
e j=
× ⋅−.
.x = × −3 06 10 8. m.
The separation distance is r = − × = −−1 2 3 06 10 1 000 61 38 m m m nm. . .e j .
Possibility two: If d x− 2 is small, x ≈ 0 5. m and the equation becomes
0 56 67 10 100
9 845 0 52
112 2.
.
.. m
N m kg kg
N kg m m
2 2
a f e jb gb g a f a fr =
× ⋅−
−
r = × −2 74 10 4. m .
For this answer to apply, the spheres would have to be compressed to a density like that of thenucleus of atom.
Section 13.3 Free-Fall Acceleration and the Gravitational Force
P13.9 aMG
RE
= = =4
9 8016 0
0 6132b g.
..
m s m s
22 toward the Earth.
P13.10 gGMR
G
RG R
R
= = =2
43
2
3
43
ρπ ρ
πe j
Ifgg
M
E
G R
G R
M M
E E= =
16
43
43
π ρ
π ρ
thenρρ
M
E
M
E
E
M
gg
RR
=FHGIKJFHGIKJ =FHGIKJ =
16
423
a f .
Chapter 13 387
P13.11 (a) At the zero-total field point, GmM
rGmM
rE
E
M
M2 2=
so r rMM
rr
M EM
EE
E= =××
=7 36 105 98 10 9 01
22
24.. .
r r rr
r
E M EE
E
+ = × = +
=×
= ×
3 84 109 01
3 84 103 46 10
8
88
..
..
m
m1.11
m
(b) At this distance the acceleration due to the Earth’s gravity is
gGM
r
g
EE
E
E
= =× ⋅ ×
×
= ×
−
−
2
11 24
8 2
3
6 67 10 5 98 10
3 46 10
3 34 10
. .
.
.
N m kg kg
m
m s directed toward the Earth
2 2
2
e je je j
Section 13.4 Kepler’s Laws and the Motion of Planets
P13.12 (a) vr
T= =
×
×= ×
2 2 384 400 101 02 10
33π π b g
b g m
27.3 86 400 s m s. .
(b) In one second, the Moon falls a distance
x atvr
t= = =×
×× = × =−1
212
12
1 02 10
3 844 101 00 1 35 10 1 352
22
3 2
82 3
.
.. . .
e je j
a f m mm .
The Moon only moves inward 1.35 mm for every 1020 meters it moves along a straight-linepath.
P13.13 Applying Newton’s 2nd Law, F ma∑ = yields F mag c= for each star:
GMM
r
Mvr2 2
2
a f = or Mv rG
=4 2
.
We can write r in terms of the period, T, by considering the time anddistance of one complete cycle. The distance traveled in one orbit is thecircumference of the stars’ common orbit, so 2πr vT= . Therefore
Mv rG
vG
vT= = F
HGIKJ
4 42
2 2
π FIG. P13.13
so, Mv TG
= =×
× ⋅= × =
−
2 2 220 10 14 4 86 400
6 67 101 26 10 63 3
3 3 3
1132
π π
m s d s d
N m kg kg solar masses
2 2
e j a fb ge j
.
.. .
388 Universal Gravitation
P13.14 Since speed is constant, the distance traveled between t1 and t2 is equal to the distance traveledbetween t3 and t4 . The area of a triangle is equal to one-half its (base) width across one side times its(height) dimension perpendicular to that side.
So12
122 1 4 3bv t t bv t t− = −b g b g
states that the particle’s radius vector sweeps out equal areas in equal times.
P13.15 Ta
GM2
2 34=
π(Kepler’s third law with m M<< )
Ma
GT= =
×
× ⋅ ×= ×
−
4 4 4 22 10
6 67 10 1 77 86 4001 90 10
2 3
2
2 8 3
11 227π π .
. ..
m
N m kg s kg
2 2
e je jb g
(Approximately 316 Earth masses)
P13.16 By conservation of angular momentum for the satellite,
r v r vp p a a=v
vrr
p
a
a
p= =
+ ×
+ ×= =
2 289 6 37 106 37 10
8 6591 27
3
3
km km459 km km
km6 829 km
..
. .
We do not need to know the period.
P13.17 By Kepler’s Third Law, T ka2 3= (a = semi-major axis)
For any object orbiting the Sun, with T in years and a in A.U.,k = 1 00. . Therefore, for Comet Halley
75 6 1 000 570
22
3
. ..a f a f=
+FHG
IKJ
y
The farthest distance the comet gets from the Sun is
y = − =2 75 6 0 570 35 22 3. . .a f A.U. (out around the orbit of Pluto)
FIG. P13.17
P13.18 F ma∑ = :Gm M
r
m v
rplanet star planet
2
2
=
GMr
v r
GM r r r
rr
x x y y
y xx
yy
star
star
yr yr
= =
= = =
=FHGIKJ =
°FHG
IKJ =
°
2 2 2
3 3 3 2 3 2
3 2
3 290 05 00
3468
5 00
ω
ω ω ω
ω ω ω.
. .
So planet has turned through 1.30 revolutionsY .
FIG. P13.18
Chapter 13 389
P13.19GM
R d
R d
TJ
J
J
+=
+
d id i
2
2
2
4π
GM T R d
d
d
J J2 2 3
11 27 2 2 7 3
7
4
6 67 10 1 90 10 9 84 3 600 4 6 99 10
8 92 10 89 200
= +
× ⋅ × × = × +
= × =
−
π
π
d ie je jb g e j. . . .
.
N m kg kg
m km above the planet
2 2
P13.20 The gravitational force on a small parcel of material at the star’s equator supplies the necessarycentripetal force:
GM mR
mvR
mRs
s ss2
22= = ω
so ω = =× ⋅ ×
×
−GMR
s
s3
11 30
3 3
6 67 10 2 1 99 10
10 0 10
. .
.
N m kg kg
m
2 2e j e je j
ω = ×1 63 104. rad s
*P13.21 The speed of a planet in a circular orbit is given by
F ma∑ = :GM m
rmv
rsun2
2
= vGM
r= sun .
For Mercury the speed is vM =× ×
×= ×
−6 67 10 1 99 10
5 79 104 79 10
11 30
104
. .
..
e je je j
m
s m s
2
2
and for Pluto, vP =× ×
×= ×
−6 67 10 1 99 10
5 91 104 74 10
11 30
123
. .
..
e je je j
m
s m s
2
2.
With greater speed, Mercury will eventually move farther from the Sun than Pluto. With originaldistances rP and rM perpendicular to their lines of motion, they will be equally far from the Sunafter time t where
r v t r v t
r r v v t
t
P P M M
P M M P
2 2 2 2 2 2
2 2 2 2 2
12 2 10 2
4 2 3 2
25
98
5 91 10 5 79 10
4 79 10 4 74 10
3 49 102 27 10
1 24 10 393
+ = +
− = −
=× − ×
× − ×=
××
= × =
e je j e je j e j
. .
. .
..
. . m m
m s m s
m m s
s yr2
2 2
390 Universal Gravitation
*P13.22 For the Earth, F ma∑ = :GM m
rmv
rmr
rT
s2
2 22= = FHG
IKJ
π.
Then GM T rs2 2 34= π .
Also the angular momentum L mvr mr
Tr= =
2π is a constant for the Earth.
We eliminate rLT
m=
2π between the equations:
GM TLT
ms2 2
3 2
42
= FHGIKJπ
πGM T
Lms
1 2 23 2
42
= FHGIKJπ
π.
Now the rate of change is described by
GM TdTdt
GdM
dtTs
s12
1 01 2 1 2−FHG
IKJ +FHG
IKJ =
dTdt
dMdt
TM
TT
s
s= −
FHGIKJ ≈2
∆
∆ ∆
∆
T tdM
dtT
M
T
s
s≈ −
FHGIKJ = −
×FHG
IKJ − ×
×
FHG
IKJ
= × −
2 5 0003 16 10
3 64 10 21
1 82 10
79
2
yr s
1 yr kg s
yr1.991 10 kg
s
30.
.
.
e j
Section 13.5 The Gravitational Field
P13.23 g i j i j= + + ° +Gml
Gml
Gml2 2 22
45 0 45 0cos . sin .e j
so g i j= +FHG
IKJ +
GMl2 1
12 2e j or
g = +FHGIKJ
Gml2 2
12
toward the opposite corner
y
m
O
m
xm
l
l
FIG. P13.23
P13.24 (a) FGMm
r= =
× ⋅ ×
× += ×
−
2
11 30 3
4 217
6 67 10 100 1 99 10 10
1 00 10 50 01 31 10
. .
. ..
N m kg kg kg
m m N
2 2e j e je je j
(b) ∆FGMmr
GMmr
= −front2
back2
∆∆
gF
m
GM r r
r r= =
−back2
front2
front2
back2
e jFIG. P13.24
∆
∆
g
g
=× × × − ×L
NMOQP
× ×
= ×
−6 67 10 100 1 99 10 1 01 10 1 00 10
1 00 10 1 01 10
2 62 10
11 30 4 2 4 2
4 2 4 2
12
. . . .
. .
.
e j e j e j e je j e j
m m
m m
N kg
Chapter 13 391
P13.25 g gMG
r a1 2 2 2= =+
g gy y1 2= − g g gy y y= +1 2
g g gx x1 2 2= = cosθ cosθ =+
r
a r2 2 1 2e jg i= −2 2g x e j
or g =+
22 2 3 2
MGr
r ae j toward the center of mass
FIG. P13.25
Section 13.6 Gravitational Potential Energy
P13.26 (a) UGM m
rE= − = −
× ⋅ ×
+ ×= − ×
−6 67 10 5 98 10 100
6 37 2 00 104 77 10
11 24
69
. .
. ..
N m kg kg kg
m J
2 2e je jb ga f .
(b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward onthe satellite and upward on the planet.
FGM m
rE= =
× ⋅ ×
×=
−
2
11 24
2
6 67 10 5 98 10 100569
. . N m kg kg kg
8.37 10 m N
2 2
6
e je jb ge j
P13.27 U GMm
r= − and g
GMR
E
E
= 2
so that ∆U GMmR R
mgRE E
E= − −FHG
IKJ =
13
1 23
∆U = × = ×23
1 000 9 80 6 37 10 4 17 106 10 kg m s m J2b ge je j. . . .
P13.28 The height attained is not small compared to the radius of the Earth, so U mgy= does not apply;
UGM M
r= − 1 2 does. From launch to apogee at height h,
K U E K Ui i f f+ + = +∆ mch :12
0 02M vGM M
R
GM M
R hp iE p
E
E p
E− + = −
+
12
10 0 10 6 67 105 98 10
6 67 105 98 10
5 00 10 6 26 103 99 106 37 10
6 37 103 99 101 26 10
3 16 10
3 2 1124
1124
7 714
6
614
77
. ..
..
. ...
...
.
× − × ⋅××
FHG
IKJ
= − × ⋅×× +
FHG
IKJ
× − × =− ×
× +
× + =×
×= ×
−
−
m s N m kg kg
6.37 10 m
N m kg kg
6.37 10 m
m s m s m s m
m m s m s
m
2 26
2 26
2 2 2 23 2
3 2
2 2
e j e j
e j
e j e j
h
h
h
h = ×2 52 107. m
392 Universal Gravitation
P13.29 (a) ρπ π
= =×
×= ×
MrS
E43
2
30
6 39
3 1 99 10
4 6 37 101 84 10
.
..
kg
m kg m3e j
e j
(b) gGM
rS
E
= =× ⋅ ×
×= ×
−
2
11 30
6 26
6 67 10 1 99 10
6 37 103 27 10
. .
..
N m kg kg
m m s
2 22e je j
e j
(c) UGM m
rgS
E= − = −
× ⋅ ×
×= − ×
−6 67 10 1 99 10 1 00
6 37 102 08 10
11 30
613
. . .
..
N m kg kg kg
m J
2 2e je jb g
P13.30 W UGm m
r= − = −
−−F
HGIKJ∆ 1 2 0
W =+ × ⋅ × ×
×= ×
−6 67 10 7 36 10 1 00 10
1 74 102 82 10
11 22 3
69
. . .
..
N m kg kg kg
m J
2 2e je je j
P13.31 (a) U U U U UGm m
rTot = + + = = −FHG
IKJ12 13 23 12
1 2
123 3
UTot
2 2 N m kg kg
m J= −
× ⋅ ×= − ×
− −−
3 6 67 10 5 00 10
0 3001 67 10
11 3 2
14. .
..
e je j
(b) At the center of the equilateral triangle
*P13.32 (a) Energy conservation of the object-Earth system from release to radius r:
K U K U
GM mR h
mvGM m
r
v GMr R h
drdt
g h g r
E
E
E
EE
+ = +
−+
= −
= −+
FHG
IKJ
FHG
IKJ = −
e j e jaltitude radius
012
21 1
2
1 2
(b) dtdrv
drvi
f
i
f
f
iz z z= − = . The time of fall is
∆
∆
t GMr R h
dr
tr
dr
EER
R h
E
E
= −+
FHG
IKJ
FHG
IKJ
= × × × × −×
FHG
IKJ
LNM
OQP
−+
−−
×
×
z
z
21 1
2 6 67 10 5 98 101 1
6 87 10
1 2
11 246
1 2
6 37 106
. ... m m
6.87 10 m6
We can enter this expression directly into a mathematical calculation program.
Alternatively, to save typing we can change variables to ur
=106 . Then
∆tu
duu
du= × −×
FHG
IKJ = × −FHG
IKJ
− −−
−
−
z z7 977 101
101
6 87 1010 3 541 10
10
10
1 16 87
14 1 2
6 6
1 26
6 37
6 878
6
6 1 2
1 2
6 37
6 87
..
...
.
.
.
e je j
A mathematics program returns the value 9.596 for this integral, giving for the time offall ∆t = × × × = =−3 541 10 10 9 596 339 8 3408 9. . . s .
Chapter 13 393
Section 13.7 Energy Considerations in Planetary and Satellite Motion
P13.3312
1 1 12
2 2mv GM mr r
mvi Ef i
f+ −FHG
IKJ =
12
01 1
22 2v GM
Rvi E
Ef+ −
FHG
IKJ =
or v vGMRf
E
E
212 2
= −
and v vGMRf
E
E= −FHG
IKJ1
21 2
2
v f = × − ×LNM
OQP = ×2 00 10 1 25 10 1 66 104 2 8
1 24. . .e j m s
P13.34 (a) vM GRE
solar escapeSun
Sun km s= =
⋅
242 1.
(b) Let r R xE S= ⋅ represent variable distance from the Sun, with x in astronomical units.
vM GR x xE S
= =⋅
2 42 1Sun .
If v =125 000 km
3 600 s, then x = = ×1 47 2 20 1011. . A.U. m
(at or beyond the orbit of Mars, 125 000 km/h is sufficient for escape).
P13.35 To obtain the orbital velocity, we use FmMG
Rmv
R∑ = =2
2
or vMG
R=
We can obtain the escape velocity from12
mvmMG
Resc2 =
or vMGR
vesc = =2
2
P13.36v
R hGM
R hi
E
E
E
2
2+=
+b g
K mvGM mR hi i
E
E= =
+FHG
IKJ =
× ⋅ ×
× + ×
L
NMM
O
QPP = ×
−12
12
12
6 67 10 5 98 10 500
6 37 10 0 500 101 45 102
11 24
6 610
. .
. ..
N m kg kg kg
m m J
2 2e je jb ge j e j
The change in gravitational potential energy of the satellite-Earth system is
∆UGM m
RGM m
RGM m
R RE
i
E
fE
i f= − = −
FHG
IKJ
= × ⋅ × − × = − ×− − −
1 1
6 67 10 5 98 10 500 1 14 10 2 27 1011 24 8 1 9. . . . N m kg kg kg m J2 2e je jb ge jAlso, K mvf f= = × = ×
Substituting the appropriate values into (1) yields the
minimum energy input = ×6 43 109. J
Chapter 13 395
P13.38 The gravitational force supplies the needed centripetal acceleration.
Thus,GM m
R h
mvR h
E
E E+=
+b g b g2
2
or vGMR h
E
E
2 =+
(a) Tr
vR hE
GMR h
E
E
= =+
+
2 2π π b gb g
TR hGME
E=
+2
3
πb g
(b) vGMR h
E
E=
+
(c) Minimum energy input is ∆E K U K Uf gf i gimin = + − −e j e j .It is simplest to
launch the satellite from a location on the equator, and launch it toward the east.
This choice has the object starting with energy K mvi i=12
2
with vR R
iE E= =
21 00
286 400
π π. day s
and UGM m
RgiE
E= − .
Thus, ∆E mGMR h
GM mR h
mR GM m
RE
E
E
E
E E
Emin =
+FHG
IKJ − +
−LNMM
OQPP +
12
12
4
86 400
2 2
2π
sb g
or ∆E GM mR h
R R hR m
EE
E E
Emin =
++
LNMM
OQPP−
22
2
86 400
2 2
2b g b gπ
s
P13.39 EGMm
rtot = − 2
∆
∆
EGMm
r r
E
i f= −
FHG
IKJ =
× ×
+−
+FHG
IKJ
= × =
−
21 1 6 67 10 5 98 10
210 1
6 370 1001
6 370 200
4 69 10 469
11 24 3
8
. .
.
e je j kg10 m
J MJ
3
P13.40 gGm
rEE
E
= 2 gGm
rUU
U
= 2
(a)gg
m rm r
U
E
U E
E U
= = FHGIKJ =
2
2
2
14 01
3 701 02.
.. gU = =1 02 9 80 10 0. . .a fe j m s m s2 2
(b) vGmresc E
E
E, =
2; v
Gmresc U
U
U, =
2:
vv
m rm r
esc E
esc U
U E
E U
,
,
..
.= = =14 03 70
1 95
For the Earth, from the text’s table of escape speeds, vesc E, .= 11 2 km s
∴ = =vesc U, . . .1 95 11 2 21 8a fb g km s km s
396 Universal Gravitation
P13.41 The rocket is in a potential well at Ganymede’s surface with energy
UGm m
r
m
U m
11 2
112
23
6
16
2
6 67 10 1 495 10
2 64 10
3 78 10
= − = −× ⋅ ×
×
= − ×
−. .
.
.
N m kg
kg m
m s
2
2
2 2
e je j
The potential well from Jupiter at the distance of Ganymede is
UGm m
r
m
U m
21 2
112
27
9
28
2
6 67 10 1 90 10
1 071 10
1 18 10
= − = −× ⋅ ×
×
= − ×
−. .
.
.
N m kg
kg m
m s
2
2
2 2
e je j
To escape from both requires
12
3 78 10 1 18 10
2 1 22 10 15 6
22 6 8
2
8
m v m
v
esc2 2
esc2 2
m s
m s km s
= + × + ×
= × × =
. .
. .
e j
P13.42 We interpret “lunar escape speed” to be the escape speed from the surface of a stationary moonalone in the Universe:
12
2
22
mvGM m
R
vGMR
vGMR
m
m
m
m
m
m
esc2
esc
launch
=
=
=
Now for the flight from moon to Earth
K U K U
mvGmM
RGmM
rmv
GmMr
GmMR
GMR
GMR
GMr
vGM
rGM
R
i f
m
m
E m
m
E
E
m
m
m
m
E m
m
E
E
+ = +
− − = − −
− − = − −
a f a f12
12
4 12
2
2
launch2
elimpact2
elimpact2
v GMR
Mr
MR
Mr
G
G
m
m
m
m
E
E
Eimpact
el
6 8 6 8
2 2
kg1.74 10 m
kg3.84 10 m
kg6.37 10 m
kg3.84 10 m
kg m
N m kg
= + + −FHG
IKJ
LNMM
OQPP
=× ×
×+
×
×+
×
×−
×
×
FHG
IKJ
LNMM
OQPP
= × + × + × − ×
= × ⋅ ×−
23
23 7 36 10 7 36 10 5 98 10 5 98 10
2 1 27 10 1 92 10 9 39 10 1 56 10
2 6 67 10 10 5 10
2
1 2
22 22 24 24 1 2
17 14 17 16 1 2
11 17
. . . .
. . . .
. .
e je j kg m km s
1 211 8= .
Chapter 13 397
*P13.43 (a) Energy conservation for the object-Earth system from firing to apex:
K U K U
mvGmM
RGmMR h
g i g f
iE
E
E
E
+ = +
− = −+
e j e j12
02
where 12
mvGmM
RE
Eesc2 = . Then
12
12
12
1
2
2
2
2
2
2
2
2
v v vR
R h
v vv RR h
v vR hv R
hv R
v vR
v R v R v Rv v
hR v
v v
iE
E
iE
E
i
E
E
E
iE
E E i E
i
E i
i
− = −+
− =+
−=
+
=−
− =− +
−
=−
esc2
esc2
esc2 esc
2
esc2
esc2
esc2
esc2
esc2
esc2
esc2
esc2
(b) h =×
−= ×
6 37 10
11 2 8 761 00 10
6 2
2 27.
. ..
m 8.76 m
a fa f a f
(c) The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it isdescribed by the same energy equation
v vR
R hv
hR h
v
iE
E E
i
2 2 2 3 2 7
6 7
8
4
1 11 2 102 51 10
6 37 10 2 51 10
1 00 10
1 00 10
= −+
FHG
IKJ = +FHG
IKJ = ×
×× + ×
FHG
IKJ
= ×
= ×
esc esc
2 2
m s m
m m
m s
m s
..
. .
.
.
e j
(d) With v vi << esc , hR vv
R v RGM
E i E i E
E≈ =
2
2
2
2esc
. But gGM
RE
E
= 2 , so hv
gi=2
2, in agreement with
0 2 02 2= + − −v g hi b ga f .
P13.44 For a satellite in an orbit of radius r around the Earth, the total energy of the satellite-Earth system is
EGM
rE= −
2. Thus, in changing from a circular orbit of radius r RE= 2 to one of radius r RE= 3 , the
required work is
W EGM m
rGM m
rGM m
R RGM m
RE
f
E
iE
E E
E
E= = − + = −
LNM
OQP=∆
2 21
41
6 12.
398 Universal Gravitation
*P13.45 (a) The major axis of the orbit is 2 50 5a = . AU so a = 25 25. AUFurther, in Figure 13.5, a c+ = 50 AU so c = 24 75. AU
Then eca
= = =24 7525 25
0 980..
.
(b) In T K as2 3= for objects in solar orbit, the Earth gives us
1 12 3 yr AUb g a f= Ks Ks =1
1
2
3
yr
AU
b ga f
Then T 22
331
125 25=
yr
AU AU
b ga f a f. T = 127 yr
(c) UGMm
r= − = −
× ⋅ × ×
×= − ×
−6 67 10 1 991 10 1 2 10
50 1 496 102 13 10
11 30 10
1117
. . .
..
N m kg kg kg
m J
2 2e je je je j
*P13.46 (a) For the satellite F ma∑ =GmM
rmv
rE
202
=
vGM
rE
0
1 2
= FHGIKJ
(b) Conservation of momentum in the forward direction for the exploding satellite:
mv mv
mv mv m
v vGM
r
i f
i
iE
∑ ∑=
= +
= = FHGIKJ
c h c h5 4 0
54
54
0
0
1 2
(c) With velocity perpendicular to radius, the orbiting fragment is at perigee. Its apogeedistance and speed are related to r and vi by 4 4mrv mr vi f f= and12
44 1
24
42 2mvGM m
rmv
GM mri
Ef
E
f− = − . Substituting v
v rrf
i
f= we have
12
12
22 2
2vGM
rv rr
GMri
E i
f
E
f− = − . Further, substituting v
GMri
E2 2516
= gives
2532
2532
732
2532
1
2
2
GMr
GMr
GM rr
GMr
rr
r r
E E E
f
E
f
f f
− = −
−= −
Clearing of fractions, − = −7 25 322 2r r rrf f or 7 32 25 02r
r
r
rf fFHGIKJ −FHGIKJ + = giving
r
rf =
+ ± −=
32 32 4 7 25
145014
2 a fa f or
1414
. The latter root describes the starting point. The outer
end of the orbit has r
rf =
257
; rr
f =257
.
Chapter 13 399
Additional Problems
P13.47 Let m represent the mass of the spacecraft, rE the radius of the Earth’s orbit, and x the distance fromEarth to the spacecraft.
The Sun exerts on the spacecraft a radial inward force of FGM m
r xs
s
E
=−b g2
while the Earth exerts on it a radial outward force of FGM m
xEE= 2
The net force on the spacecraft must produce the correct centripetal acceleration for it to have anorbital period of 1.000 year.
Thus, F FGM m
r x
GM mx
mvr x
mr x
r xTS E
S
E
E
E E
E− =−
− =−
=−
−LNMM
OQPPb g b g b g
b g2 2
2 22π
which reduces toGM
r x
GMx
r x
TS
E
E E
−− =
−
b gb g
2 2
2
2
4π. (1)
Cleared of fractions, this equation would contain powers of x ranging from the fifth to the zeroth.We do not solve it algebraically. We may test the assertion that x is between 1 47 109. × m and1 48 109. × m by substituting both of these as trial solutions, along with the following data:MS = ×1 991 1030. kg , ME = ×5 983 1024. kg , rE = ×1 496 1011. m, and T = = ×1 000 3 156 107. . yr s .
With x = ×1 47 109. m substituted into equation (1), we obtain
6 052 10 1 85 10 5 871 103 3 3. . .× − × ≈ ×− − − m s m s m s2 2 2
or 5 868 10 5 871 103 3. .× ≈ ×− − m s m s2 2
With x = ×1 48 109. m substituted into the same equation, the result is
6 053 10 1 82 10 5 870 8 103 3 3. . .× − × ≈ ×− − − m s m s m s2 2 2
or 5 870 9 10 5 870 8 103 3. .× ≈ ×− − m s m s2 2 .
Since the first trial solution makes the left-hand side of equation (1) slightly less than the right handside, and the second trial solution does the opposite, the true solution is determined as between thetrial values. To three-digit precision, it is 1 48 109. × m.
As an equation of fifth degree, equation (1) has five roots. The Sun-Earth system has five Lagrangepoints, all revolving around the Sun synchronously with the Earth. The SOHO and ACE satellitesare at one. Another is beyond the far side of the Sun. Another is beyond the night side of the Earth.Two more are on the Earth’s orbit, ahead of the planet and behind it by 60°. Plans are under way togain perspective on the Sun by placing a spacecraft at one of these two co-orbital Lagrange points.The Greek and Trojan asteroids are at the co-orbital Lagrange points of the Jupiter-Sun system.
400 Universal Gravitation
P13.48 The acceleration of an object at the center of the Earth dueto the gravitational force of the Moon is given by
a GM
d= Moon
2
At the point A nearest the Moon, a GM
d rM
+ =−a f2
At the point B farthest from the Moon, a GM
d rM
− =+a f2 FIG. P13.48
∆a a a GMd r dM= − =−
−LNMM
OQPP+
1 12 2a f
For d r>> , ∆aGM r
dM= = × −2
1 11 1036. m s2
Across the planet,∆ ∆gg
ag
= =×
= ×−
−2 2 22 109 80
2 26 106
7..
. m s
m s
2
2
*P13.49 Energy conservation for the two-sphere system from release to contact:
− = − + +
−FHGIKJ = = −LNM
OQP
FHG
IKJ
GmmR
Gmmr
mv mv
Gmr R
v v Gmr R
212
12
12
1 12
1
2 2
21 2
(a) The injected impulse is the final momentum of each sphere,
mv m Gmr R
Gmr R
= −LNMOQP
FHG
IKJ = −FHG
IKJ
LNM
OQP
2 21 2
31 2
12
1 12
1.
(b) If they now collide elastically each sphere reverses its velocity to receive impulse
mv mv mv Gmr R
− − = = −FHGIKJ
LNM
OQPa f 2 2
12
131 2
P13.50 Momentum is conserved:
m m m m
M Mi i f f
f f
f f
1 1 2 2 1 1 2 2
1 2
2 1
0 2
12
v v v v
v v
v v
+ = +
= +
= −
Energy is conserved:
K U E K U
Gm mr
m v m vGm m
r
GM MR
Mv M vGM M
R
vGM
Rv v
GMR
i f
if f
f
f f
f f f
+ + = +
− + = + −
− = + FHGIKJ −
= = =
a f a f
a f a f a f
∆
0 012
12
212
12
12
212
24
23
12
13
1 21 1
22 2
2 1 2
12
1
2
1 2 1
Chapter 13 401
P13.51 (a) avrc =2
ac =×
×=
1 25 10
1 53 1010 2
6 2
11
.
..
m s
m m s2e j
(b) diff m s2= − = =10 2 9 90 0 312 2. . .GMr
M =×
× ⋅= ×−
0 312 1 53 10
6 67 101 10 10
11 2
1132
. .
..
m s m
N m kg kg
2
2 2
e je jFIG. P13.51
P13.52 (a) The free-fall acceleration produced by the Earth is gGM
rGM rE
E= = −2
2 (directed downward)
Its rate of change is dgdr
GM r GM rE E= − = −− −2 23 3a f .
The minus sign indicates that g decreases with increasing height.
At the Earth’s surface, dgdr
GMR
E
E
= −2
3 .
(b) For small differences,
∆
∆
∆g
r
g
hGMR
E
E
= =2
3 Thus, ∆gGM hR
E
E
=2
3
(c) ∆g =× ⋅ ×
×= ×
−−
2 6 67 10 5 98 10 6 00
6 37 101 85 10
11 2 24
6 35
. . .
..
N m kg kg m
m m s
22e je ja f
e j*P13.53 (a) Each bit of mass dm in the ring is at the same distance from the object at A. The separate
contributions −Gmdm
r to the system energy add up to −
GmM
rring . When the object is at A,
this is
− × ⋅ ×
× + ×= − ×
−6 67 10 1 000
1 10 2 107 04 10
11
8 2 8 2
4..
N m kg 2.36 10 kg
kg m m J
2 20
2 e j e j.
(b) When the object is at the center of the ring, the potential energy is
−× ⋅ ×
×= − ×
−6 67 101 10
1 57 1011
85.
. N m 1 000 kg 2.36 10 kg
kg m J
2 20
2 .
(c) Total energy of the object-ring system is conserved:
K U K U
v
v
g A g B
B
B
+ = +
− × = − ×
=× ×FHG
IKJ =
e j e j0 7 04 10
12
1 000 1 57 10
2 8 70 1013 2
4 2 5
4 1 2
. .
..
J kg J
J1 000 kg
m s
402 Universal Gravitation
P13.54 To approximate the height of the sulfur, set
mvmg hIo
2
2= h = 70 000 m g
GMrIo = =2 1 79. m s2
v g hIo= 2 v = ≈2 1 79 70 000 500.a fb g b g m s over 1 000 mi h
A more precise answer is given by
12
2
1 2mv
GMmr
GMmr
− = −
12
6 67 10 8 90 101
1 82 101
1 89 102 11 22
6 6v = × ××
−×
FHG
IKJ
−. .. .
e je j v = 492 m s
P13.55 From the walk, 2 25 000πr = m. Thus, the radius of the planet is r = = ×25 000
3 98 103 m2
mπ
.
From the drop: ∆y gt g= = =12
12
29 2 1 402 2. . s ma f
so, gMGr
= = × =−2 1 40
29 23 28 102
32
.
..
m
s m s2a f
a f ∴ = ×M 7 79 1014. kg
*P13.56 The distance between the orbiting stars is d r r= °=2 30 3cos since
cos303
2°= . The net inward force on one orbiting star is
Gmmd
GMmr
Gmmd
mvr
Gmr
GMr
rrT
Gm
Mr
T
Tr
G M
Tr
G M
m
m
2 2 2
2
2 2
2 2
2
2 3
2
22 3
3
3
3
1 2
30 30
2 303
4
34
4
2
cos cos
cos
°+ + °=
°+ =
+FHG
IKJ =
=+
=+
F
HGG
I
KJJ
π
π
π
π
e j
e j
r
30°d
r
30°
60°
F
F
F
FIG. P13.56
P13.57 For a 6.00 km diameter cylinder, r = 3 000 m and to simulate 1 9 80g = . m s2
gvr
r
gr
= =
= =
22
0 057 2
ω
ω . rad s
The required rotation rate of the cylinder is 1 rev110 s
(For a description of proposed cities in space, see Gerard K. O’Neill in Physics Today, Sept. 1974.)
Chapter 13 403
P13.58 (a) G has units N m
kgkg m m
s kgm
s kg
2
2
2
2 2
3
2⋅
=⋅ ⋅
⋅=
⋅
and dimensions GL
T M
3
2=⋅
.
The speed of light has dimensions of c =LT
, and Planck’s constant has the same dimensions
as angular momentum or h =⋅M LT
2
.
We require G c hp q r = L , or L T M L T M L T L M T1 0 03 2 2p p p q q r r r− − − − = .
Thus, 3 2 1p q r+ + =
− − − =− + =
2 00
p q rp r
which reduces (using r p= ) to 3 2 1p q p+ + =
− − − =2 0p q p
These equations simplify to 5 1p q+ = and q p= −3 .
(e) Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy isso large that the total energy decreases by
E Ei f− = − × − − × = ×3 04 10 3 08 10 4 69 109 9 7. . . J J Je j .
(f) The only forces on the object are the backward force of air resistance R, comparatively verysmall in magnitude, and the force of gravity. Because the spiral path of the satellite is notperpendicular to the gravitational force, one component of the gravitational force pulls
forward on the satellite to do positive work and make its speed increase.
Chapter 13 405
P13.61 (a) At infinite separation U = 0 and at rest K = 0 . Since energy of the two-planet system isconserved we have,
012
121 1
22 2
2 1 2= + −m v m vGm m
d(1)
The initial momentum of the system is zero and momentum is conserved.
Therefore, 0 1 1 2 2= −m v m v (2)
Combine equations (1) and (2): v mG
d m m1 21 2
2=
+b g and v mG
d m m2 11 2
2=
+b g
Relative velocity v v vG m m
dr = − − =+
1 21 22b g b g
(b) Substitute given numerical values into the equation found for v1 and v2 in part (a) to find
v141 03 10= ×. m s and v2
32 58 10= ×. m s
Therefore, K m v1 1 12 321
21 07 10= = ×. J and K m v2 2 2
2 3112
2 67 10= = ×. J
P13.62 (a) The net torque exerted on the Earth is zero. Therefore, the angular momentum of the Earthis conserved;
mr v mr va a p p= and v vr
ra pp
a=FHGIKJ = × F
HGIKJ = ×3 027 10
1 4711 521
2 93 104 4...
. m s m se j
(b) K mvp p= = × × = ×12
12
5 98 10 3 027 10 2 74 102 24 4 2 33. . .e je j J
UGmM
rpp
= − = −× × ×
×= − ×
−6 673 10 5 98 10 1 99 10
1 471 105 40 10
11 24 30
1133
. . .
..
e je je j J
(c) Using the same form as in part (b), Ka = ×2 57 1033. J and Ua = − ×5 22 1033. J .
Compare to find that K Up p+ = − ×2 66 1033. J and K Ua a+ = − ×2 65 1033. J . They agree.
406 Universal Gravitation
P13.63 (a) The work must provide the increase in gravitational energy
W U U U
GM M
r
GM M
r
GM M
R y
GM M
R
GM MR R y
W
g gf gi
E p
f
E p
i
E p
E
E p
E
E pE E
= = −
= − +
= −+
+
= −+
FHG
IKJ
=× ⋅F
HGIKJ ×
×−
×FHG
IKJ
=
−
∆
1 1
6 67 105 98 10 100
16 37 10
17 37 10
850
1124
6 6.
.. .
N mkg
kg kg m m
MJ
2
2 e jb g
(b) In a circular orbit, gravity supplies the centripetal force:
GM M
R y
M v
R yE p
E
p
E+=
+b g b g2
2
Then, 12
12
2M vGM M
R ypE p
E=
+b gSo, additional work = kinetic energy required
=× ⋅ ×
×
= ×
−12
6 67 10 5 98 10 100
7 37 10
2 71 10
11 24
6
9
. .
.
.
N m kg kg
kg m
J
2
2
e je jb ge je j
∆W
P13.64 Centripetal acceleration comes from gravitational acceleration.
vr
M Gr
rT r
GM T r
r
r
c
c
2
2
2 2
2
2 2 3
11 30 3 2 2 3
4
4
6 67 10 20 1 99 10 5 00 10 4
119
= =
=
× × × =
=
− −
π
π
π. . .e ja fe je jorbit km
P13.65 (a) Tr
v= =
× ×
×= × = ×
2 2 30 000 9 46 10
2 50 107 10 2 10
15
515 8π π .
.
m
m s s yr
e j
(b) Ma
GT= =
× ×
× ⋅ ×= ×
−
4 4 30 000 9 46 10
6 67 10 7 13 102 66 10
2 3
2
2 15 3
11 15 241π π .
. ..
m
N m kg s kg
2 2
e je je j
M = ×1 34 10 1011 11. ~ solar masses solar masses
The number of stars is on the order of 1011 .
Chapter 13 407
P13.66 (a) From the data about perigee, the energy of the satellite-Earth system is
E mvGM m
rpE
p= − = × −
× ×
×
−12
12
1 60 8 23 106 67 10 5 98 10 1 60
7 02 102 3 2
11 24
6. .. . .
.a fe j e je ja f
or E = − ×3 67 107. J
(b) L mvr mv rp p= = °= × ×
= × ⋅
sin sin . . . .
.
θ 90 0 1 60 8 23 10 7 02 10
9 24 10
3 6
10
kg m s m
kg m s2
b ge je j
(c) Since both the energy of the satellite-Earth system and the angular momentum of the Earthare conserved,
at apogee we must have12
2mvGMm
rEa
a− =
and mv r La a sin .90 0°= .
Thus,12
1 606 67 10 5 98 10 1 60
3 67 10211 24
7.. . .
.a f e je ja fv
ras
−× ×
= − ×−
J
and 1 60 9 24 1010. . kg kg m s2b gv ra a = × ⋅ .
Solving simultaneously,12
1 606 67 10 5 98 10 1 60 1 60
9 24 103 67 102
11 24
107.
. . . .
..a f e je ja fa f
vv
aa
−× ×
×= − ×
−
which reduces to 0 800 11 046 3 672 3 10 02 7. .v va a− + × =
so va =± − ×11 046 11 046 4 0 800 3 672 3 10
2 0 800
2 7b g a fe ja f
. .
..
This gives va = 8 230 m s or 5 580 m s . The smaller answer refers to the velocity at the
apogee while the larger refers to perigee.
Thus, rL
mvaa
= =× ⋅
×= ×
9 24 10
1 60 5 58 101 04 10
10
37.
. ..
kg m s
kg m s m
2
b ge j.
(d) The major axis is 2a r rp a= + , so the semi-major axis is
a = × + × = ×12
7 02 10 1 04 10 8 69 106 7 6. . . m m me j
(e) Ta
GME= =
×
× ⋅ ×−
4 4 8 69 10
6 67 10 5 98 10
2 3 2 6 3
11 24
π π .
. .
m
N m kg kg2 2
e je je j
T = =8 060 134 s min
408 Universal Gravitation
*P13.67 Let m represent the mass of the meteoroid and vi its speed when far away.No torque acts on the meteoroid, so its angular momentum is conserved asit moves between the distant point and the point where it grazes the Earth,moving perpendicular to the radius:
FIG. P13.67L Li f= : m mi i f fr v r v× = ×
m R v mR v
v vE i E f
f i
3
3
b g ==
Now energy of the meteoroid-Earth system is also conserved:
K U K Ug i g f+ = +e j e j :
12
012
2 2mv mvGM m
Ri fE
E+ = −
12
12
92 2v vGM
Ri iE
E= −e j
GMR
vE
Ei= 4 2 : v
GMRi
E
E=
4
*P13.68 From Kepler’s third law, minimum period means minimum orbit size. The “treetop satellite” inFigure P13.35 has minimum period. The radius of the satellite’s circular orbit is essentially equal tothe radius R of the planet.
F ma∑ = :GMm
Rmv
RmR
RT2
2 22= = FHG
IKJ
π
G VR R
RT
G RR
T
ρπ
ρ ππ
=
FHGIKJ =
2 2 2
2
32 3
2
4
43
4
e j
The radius divides out: T G2 3ρ π= TG
=3πρ
P13.69 If we choose the coordinate of the center of mass at the origin, then
0 2 1=−
+
Mr mrM mb g
and Mr mr2 1=
(Note: this is equivalent to saying that the net torque must be zero andthe two experience no angular acceleration.) For each mass F ma= so
mrMGm
d1 12
2ω = and MrMGm
d2 22
2ω =FIG. P13.69
Combining these two equations and using d r r= +1 2 gives r rM m G
d1 22
2+ =+b g a f
ω
with ω ω ω1 2= =
and T =2πω
we find Td
G M m2
2 34=
+πa f .
Chapter 13 409
P13.70 (a) The gravitational force exerted on m2 by the Earth (mass m1 ) accelerates m2 according to:
m gGm m
r2 21 22= . The equal magnitude force exerted on the Earth by m2 produces negligible
acceleration of the Earth. The acceleration of relative approach is then
gGmr2
12
11 24
7 2
6 67 10 5 98 10
1 20 102 77= =
× ⋅ ×
×=
−. .
..
N m kg kg
m m s
2 22e je j
e j.
(b) Again, m2 accelerates toward the center of mass with g2 2 77= . m s2 . Now the Earthaccelerates toward m2 with an acceleration given as
m gGm m
r
gGm
r
1 11 22
12
2
11 24
7 2
6 67 10 2 00 10
1 20 100 926
=
= =× ⋅ ×
×=
−. .
..
N m kg kg
m m s
2 22e je j
e jThe distance between the masses closes with relative acceleration of
g g grel2 2 2 m s m s m s= + = + =1 2 0 926 2 77 3 70. . . .
P13.71 Initial Conditions and Constants:
Mass of planet: 5 98 1024. × kgRadius of planet: 6 37 106. × mInitial x: 0.0 planet radiiInitial y: 2.0 planet radiiInitial vx : +5 000 m/sInitial vy : 0.0 m/s
