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Universal Gravitation
CHAPTER OUTLINE 13.1 Newton’s Law of Universal Gravitation 13.2 Free-Fall Acceleration and the Gravitational Force 13.3 Kepler’s Laws and the Motion of Planets 13.4 The Gravitational Field 13.5 Gravitational Potential Energy 13.6 Energy Considerations in Planetary and Satellite Motion
ANSWERS TO QUESTIONS
*Q13.1 The force is proportional to the product of the masses and inversely proportional to the square of the separation distance, so we compute m 1 m 2 r^2 for each case: (a) 2 ⋅ 3 1 2 = 6 (b) 18 (c) 18 4 = 4.5 (d) 4.5 (e) 16 4 = 4. The ranking is then b > a > c = d > e.
*Q13.2 Answer (d). The International Space Station orbits just above the atmosphere, only a few hundred kilometers above the ground. This distance is small compared to the radius of the Earth, so the gravitational force on the astronaut is only slightly less than on the ground. We think of it as having a very different effect than it does on the ground, just because the normal force on the orbiting astronaut is zero.
*Q13.3 Answer (b). Switching off gravity would let the atmosphere evaporate away, but switching off the atmosphere has no effect on the planet’s gravitational field.
Q13.4 To a good fi rst 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.
*Q13.5 Having twice the mass would make the surface gravitational field two times larger. But the inverse square law says that having twice the radius would make the surface acceleration due to gravitation four times smaller. Altogether, g at the surface of B becomes (2 ms 2 )(2) 4 = 1 ms^2 , answer (e).
*Q13.6 (i) 42 = 16 times smaller: Answer (i), according to the inverse square law.
(ii) mv^2 r = GMm r^2 predicts that v is proportional to (1 r )^1 ^2 , so it becomes (14) 1 ^2 = 1 2 as large: Answer (f ).
(iii) (4^3 )^1 ^2 = 8 times larger: Answer (b), according to Kepler’s third law.
337
*Q13.7 Answer (b). The Earth is farthest from the sun around July 4 every year, when it is summer in the northern hemisphere and winter in the southern hemisphere. As described by Kepler’s second law, this is when the planet is moving slowest in its orbit. Thus it takes more time for the planet to plod around the 180° span containing the minimum-speed point.
Q13.8 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!
*Q13.9 Answer (c). Ten terms are needed in the potential energy:
U = U (^) 12 + U (^) 13 + U 14 (^) + U (^) 15 + U (^) 23 + U (^) 24 + U (^) 25 + U (^) 34 + U (^) 35 + U 45
Q13.10 The escape speed from the Earth is 11.2 kms and that from the Moon is 2.3 kms, smaller by a factor of 5. The energy required—and fuel—would be proportional to v^2 , or 25 times more fuel is required to leave the Earth versus leaving the Moon.
*Q13.11 The gravitational potential energy of the Earth-Sun system is negative and twice as large in mag- nitude as the kinetic energy of the Earth relative to the Sun. Then the total energy is negative and equal in absolute value to the kinetic energy. The ranking is a > b = c.
Q13.12 For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be located at the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it must be over the southern hemisphere for the other half. We could share with Easter Island a satellite that would look straight down on Arizona each morning and vertically down on Easter Island each evening.
Q13.13 Every point q on the sphere that does not lie along the axis connect- ing the center of the sphere and the particle will have companion point q ′ for which the components of the gravitational force perpendicular to the axis will cancel. Point q ′ can be found by rotating the sphere through 180° about the axis. The forces will not necessarily cancel if the mass is not uniformly distributed, unless the center of mass of the non-uniform sphere still lies along the axis.
Q13.14 Speed is maximum at closest approach. Speed is minimum at farthest distance. These two points, perihelion and aphelion respectively, are 180° apart, at opposite ends of the major axis of the orbit.
Q13.15 Set the universal description of the gravitational force, F GM^ m g R X X
= 2 , equal to the local description, Fg = ma gravitational , where M (^) X 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.
Q13.16 The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as one interpretation of Equation 13.1 would suggest. All the bits of matter that make up the Earth will pull in different outward directions on the extra particle.
Q13.17 Cavendish determined G. Then from g GM R
= 2 , one may determine the mass of the Earth.
338 Chapter 13
F pq
F pq′
p
q
q’ (behind the sphere)
FIG. Q13.
340 Chapter 13
P13.5 The force exerted on the 4.00-kg mass by the 2.00-kg mass is directed upward and given by F 24^4 2 j 24 2
6 67 10 11 4 00
= (^) ( × − ⋅ )
G
m m r
N m kg 2 2 kgg^ kg m N
( )( )
= × −
2 11
j
j The force exerted on the 4.00-kg mass by the 6.00-kg mass is directed to the left F 64^4 6 i 64 2 = G m m (− ) = −( 6 67 × 10 −^11 ⋅ )^4 r
ˆ (^). N m 2 kg^2.^000 6 00 4 00 10 0 10
2 11
kg kg m N
( )( )
= − × −
i
i
Therefore, the resultant force on the 4.00-kg mass is
F 4 (^) = F 24 (^) + F 64 = (^) (− 10 0. i ˆ^ +5 93. j ˆ^ ) × 10 −^11 N.
*P13.6 (a) The Sun-Earth distance is 1 496. × 1011 m and the Earth-Moon distance is 3 84. × 10 8 m, so the distance from the Sun to the Moon during a solar eclipse is
1 496. × 10 11 m − 3 84. × 10 8 m = 1 492. × 1011 m
The mass of the Sun, Earth, and Moon are M (^) S = 1 99. × 10 30 kg M (^) E = 5 98. × 10 24 kg and M (^) M = 7 36. × 10 22 kg
We have F Gm m SM (^) r
1 2 (^ × − )^ (^ × )^ × 2
11 2
20 1 492 10
( ) ( × )
= ×
. N
(b) FEM = ( 6 67. × 10 −^11 N m⋅ 2 kg^2 ) ( 5 98. × 10 24 ) 7 36. × 10222 8 2
20 3 84 10
( ) ( × )
= ×
. N
(c) FSE = ( 6 67.^ ×^10 −^11 N m⋅^2 kg^2 ) ( 1 99.^ ×^10 30 ) 5 98.^ ×^10244 11 2
22 1 496 10
( ) 3 55 10 ( × )
= ×
. N
(d) The force exerted by the Sun on the Moon is much stronger than the force of the Earth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. The Moon’s path is everywhere concave toward the Sun. Only by subtracting out the solar orbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass of this system.
P13.7 F GMm r
= = (^) ( × ⋅ ) − (^ )^ × 2 6 67. 10 11 N m 2 kg 2 1 50.^ kg 15 0.^100 4 50 10
3 2 2
10
− −
( ) − ( × )
= ×
kg m
N
FIG. P13.
Universal Gravitation 341
P13.8 Let θ represent the angle each cable makes with the vertical, L the cable length, x the distance each ball scrunches in, and d = 1 m the original distance between them. Then r = d − 2 x is the separation of the balls. We have
∑^ Fy =^0 :^ T^ cos^ θ −^ mg =^0
∑^ Fx =^0 :^ T^ Gmm r
sin θ − 2 = 0
Then tan θ = Gmm r mg^2
x L x
Gm (^2) − 2 g d 2 x^2
x d x Gm g
( − 2 ) 2 = L^2 − x^2
The factor Gm g
is numerically small. There are two possibilities: either x is small or else d − 2 x is small.
Possibility one : We can ignore x in comparison to d and L , obtaining
x 1
2 11 m
N m kg kg m s
2 2
(. ×^ − ⋅ )(^ ) ((. )
45 m (^) x = 3 06. × 10 − 8 m
The separation distance is r = 1 m − 2 3 06(. × 10 − 8 m ) = 1 000. m −61 3. nm. This equilibrium is stable.
Possibility two : If d − 2 x is small, x ≈ 0 5. m and the equation becomes
2
11 .
m
N m kg kg N
2 2
( ×^ − ⋅ )( ) r kkg
m m ( )
( 45 ) − (^2 0 5. )^2 r = 2 74. × 10 −^4 m
For this answer to apply, the spheres would have to be compressed to a density like that of the nucleus of atom. This equilibrium is unstable.
Section 13.2 Free-Fall Acceleration and the Gravitational Force
P13.9 a MG RE
( )
m s m s
2 (^2) toward the Earth.
*P13.10 (a) For the gravitational force on an object in the neighborhood of Miranda we have
m g
Gm m r
g Gm r
obj
obj Miranda Miranda 2
Miranda Mir
aanda 2
2 2
N m kg kg
× ⋅ (^) ( × ) ×
. 11.^19
m
m s 2 ( )
FIG. P13.
continued on next page
Universal Gravitation 343
P13.13 Applying Newton’s 2nd Law, (^) ∑ F = ma yields Fg = mac for each star: GMM r
M
2 2 r
2
v (^) or M r G
v^2
We can write r in terms of the period, T , by considering the time and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars’ common orbit, so 2 π r = vT. Therefore
M r G G
T
v^2^ v^2 v π
so,
M
T
G
2 2 220( ×^10 ) (^ 14 4^ )(^ 86 400 ) 6
v^3 3 π π
m s. d s d ..
11 1 26^10 32 63 3
( ×^ ⋅ )
− =^ ×^ =
N m kg 2 2 kg^ solarr masses
P13.14 By Kepler’s Third Law, T^2 = ka^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 00 0 570 2
(. ) 2 = (. ) ⎛. +^3
y
The farthest distance the comet gets from the Sun is
y = 2 75 6(. ) 2 3− 0 570. = 35 2. A.U. (out around the orbit of Pluto).
P13.15 T a GM
2
= π^ (Kepler’s third law with m << M )
M a GT
( × ) × − ⋅
4 4 4 22^10
2 3 2
2 8 3 11
π π^. .
m N m 2 kgg s 2 kg ( ) (^ × )^
= ×
2 1 90^1027
(approximately 316 Earth masses)
P13.16 (^) ∑ F = ma : Gm^ M r
m r
planet star planet 2
2
v
GM r
r
GM r r (^) x x ry y
y
star
star
v^2 2 3 3 3 2 3 2
ω
ω ω ω
ω ωω (^) x x ω y
y
r r
⎠⎟^
⎝⎜^
⎠⎟^
3 2 90 0 3 2 5 00
yr
5 00. yr
So planet Y has turned through 1.30 revolutionss.
FIG. P13.
FIG. P13.
FIG. P13.
344 Chapter 13
P13.17 GM
R d
R d T
J J
J ( + )
( + ) 2
2 2
4 π
GM TJ^2 2 R (^) J d^3 11
= ( + ) ( × − ⋅ ) ×
π
. N m 2 kg^2.^227 2 2 3 9 84 3 600 4 6 99 10 8 92
( kg) ( × ) = ( × + )
π d d ×× 10 7 m = 89 200km above the planet
P13.18 The gravitational force on a small parcel of material at the star’s equator supplies the necessary centripetal acceleration: GM m R
m R
s mR s s 2 s
2 =^ v = ω^2
so ω = = GM (^ ×^ − ⋅ )^ ( × R
s s 3
6 67. 10 11 N m 2 kg 2 ⎡2 1 99. 1030 kg)) ⎣ ⎤⎦ ( 10 0.^ ×^10 3 m)^3 ω = 1 63. × 10 4 rad s
P13.19 The speed of a planet in a circular orbit is given by
∑^ F^ =^ ma :^ GM^ m r
m r
sun 2
2 = v^ v^ =^
GM
r
sun
For Mercury the speed is vM = ( × ) ( × ) ( × )
11 30 10
m s
2 2 ==^ 4 79^ ×^10
. 4 m s
and for Pluto, vP = ( × ) ( × ) ( × )
11 30 12
m s
2 2 ==^ 4 74^ ×^10
. 3 m s
With greater speed, Mercury will eventually move farther from the Sun than Pluto. With original distances rP and rM perpendicular to their lines of motion, they will be equally far from the Sun after time t where
r t r t r r t
t
P P M M P M M P
2 2 2 2 2 2 2 2 2 2 2
5 9
− = (^) ( − )
v v v v
. 1 1 10 5 79 10 4 79 10 4
12 2 10 2 4 2
( × ) −^ ( × ) ( × ) −
m m m s
25 ( × ) =^9
×
×
m s
m m s
2 2 2 224 10 3 93 × 8 s =. yr
*P13.20 In T^2 = 4 π^2 a^3 GMcentral we take a = 3.84 × 108 m.
M (^) central = 4 π^2 a^3 GT^2 = 4 3 84^10 27 3
π 2 (.^8 / )(.
×
× ⋅ ×
m) (6.67 10 N m kg
3 − 11 2 2 886400 6 02^10
24 s) 2 =^. × kg This is a little larger than 5.98 × 1024 kg.
The Earth wobbles a bit as the Moon orbits it, so both objects move nearly in circles about their center of mass, staying on opposite sides of it. The radius of the Moon’s orbit is therefore a bit less than the Earth–Moon distance.
346 Chapter 13
Section 13.5 Gravitational Potential Energy
P13.24 (a) U GM m r = − E = − (^ ×^ ⋅ )^ ( ×
6 67. 10 −^11 N m 2 kg 2 5 98. 1024 kg))( (^) )
( + ) ×
= − ×
6 4 77^109
kg m
J
(b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward on the satellite and upward on the planet.
F GM m r
= E = (^ ×^ ⋅ )^ (^ × )
− 2
6 67. 10 11 N m 2 kg 2 5 98. 1024 kg 1100 2 569
kg 8.37 10 m
6 N
( ) ( × )
P13.25 (a)^ ρ^ π (^) π
( × ) ( × )
M =
r
S (^43) E 2
30 6 3
kg m
..84 × 10 9 kg m 3
(b) g GM r
S E
( × − ⋅ ) ( × ) 2
6 67. 10 11 N m 2 kg 2 1 99. 1030 kg 6 6 37 10
6 2 3 27^106
( × )
= ×
m
m s 2
(c) U GM m g r S E
( 6 67.^ ×^10 −^11 N m⋅^2 kg^2 ) 1 99.^ ×^1030 kgg^ kg m
J
( )(^ ) ×
= − ×
6 2 08^1013
P13.26 The height attained is not small compared to the radius of the Earth, so U = mgy does not apply;
U
GM M
r = − 1 2 does. From launch to apogee at height h ,
K (^) i + U (^) i + ∆ E (^) mch = K (^) f + Uf :
M^20
GM M
R
GM M
p i R h E p E
E p E
v − + = −
1 2 (10 0.^ ×^10 3 m s^ ) −^2 (6 67.^ ×^10 −^11 N m⋅^2 kg^2 ) 5 98.^ ×^100
24
11
kg 6.37 10 m
N m kg
6
2 2
×
⎝⎜^
= − ((. × − ⋅ )
×
× +
⎝⎜^
×
24
7
kg 6.37 10 m
m
(^6) h
( 22 s^2 ) −^ ( × m^2 s^2 ) = −^ × m^3 s^2 ×
. 7.^14
6 6 14 7
m
m m 3 s^2
× + =
×
×
h
. h
. m s m
m
2 2 =^ ×
= ×
7
7
h.
*P13.27 (a) U U U U U Gm m Tot (^) r
1 2 12
U Tot
N m^2 kg^2 kg = −
3 6 67( × 10 − ⋅ ) ( 5 00 × 10 − ) 0
. 11.^3
m
= − × − J
(b) Each particle feels a net force of attraction toward the midpoint between the other two. Each moves toward the center of the triangle with the same acceleration. They collide simultaneously at the center of the triangle.
P13.28 W U
Gm m r
∆ ⎛⎝^1 2 − 0 ⎞⎠
W =
( +^ 6 67.^ ×^10 −^11 N m⋅^2 kg^2 ) ( 7 36.^ ×^10 22 kg) 1 00.^ ×^100 1 74 10
3 6
kg 9 m
J
( ) ×
= ×
Universal Gravitation 347
P13.29 (a) Energy conservation of the object-Earth system from release to radius r :
K U K U GM m R h
g (^) h g (^) r E E
( + ) =^ ( + )
altitude radius 0 1 22
2 1 1
2
1 2
m GM m r
GM r R h
dr
E
E E
v
v
⎠⎟^
ddt
(b) dt^ dr dr i
f
i
f
f
i ∫ =^ ∫ −^ v =∫ v. The time of fall is
t GM r R h
dr
t
E R E
R h
E
E = −
1 2
×× × × × −
×
⎣⎢^
11 24
.. (^) r. (^6) m
−
×
× ∫
1 2
6 37 10 6
dr
. m
6.87 10 6 m
We can enter this expression directly into a mathematical calculation program. Alternatively, to save typing we can change variables to u^ = r 10 6
. Then
∆ t u = (^) ( × ) − ×
− − 7 977 10
1 2 6 6
1 2
. 6 . ddu
u
6 37
6 87
8 6 3 541 (^10 6) 1 2
.
.
∫
= ×
( )
− ∫
1 2
6 37
6 87 du .
.
A mathematics program returns the value 9.596 for this integral, giving for the time of fall ∆ t = 3 541. × 10 −^8 × 10 9 × 9 596. = 339 8. = 340 s
Section 13.6 Energy Considerations in Planetary and Satellite Motion
P13.30 (a) v solar escape Sun Sun
= = km s ⋅
M G
RE
(b) Let r = R (^) E S ⋅ x represent variable distance from the Sun, with x in astronomical units.
v = = ⋅
2 M G 42 1
R (^) E Sx x
Sun.
If v = 125 000 km 3 600 s
, then x = 1 47. A.U. = 2 20. × 1011 m (at or beyond the orbit of Mars, 125 000 km h is sufficient for escape).
P13.31^1 2
m^2 GM m^2 r r i E m f i
v + − vf
v^2^ v^2 i E E
or v (^) f v E E
GM
R
2 1
= 2 −^2
and v (^) f v E E
GM
R
2
1 2 2
v (^) f = ⎡⎣( 2 00 × (^10 4) ) − 1 25 × 10 ⎤⎦ = 1 66 × 10 2 8 1 2^4
... m s
Universal Gravitation 349
P13.35 (^) Fc = FG gives m r
GmM r
v E 2 = 2
which reduces to v = GM r
E
and period = 2 π^^ r^ = 2 π r r v GM (^) E (a) (^) r = RE + 200 km = 6 370 km + 200 km =6 570km
Thus,
period m m = (^) ( × ) ( × ) × −
6
6 π. (^11)
. N m kg kg s
( ⋅^2 2 ) ( × ) = × =
24 3
T.. mmin = 1 47. h
(b) v = = GM (^ ×^ − ⋅ )^ (^ × ) r
E 6 67^10 5 98^10
. 11.^24
N m 2 kg 2 kg 557 10
( × )
m
. km s
(c) K (^) f + U (^) f = K (^) i + Ui + energy input, gives
input (^) = − + −
m^2 m^2 GM m r
GM m f i r E f
E i
v v (1)
r R R
i E
i E
= = ×
= = ×
6
2
m
s v m s π
Substituting the appropriate values into (1) yields the minimum energy input (^) = 6 43. × 10 9 J
P13.36 The gravitational force supplies the needed centripetal acceleration.
Thus, GM m R h
m R h
E ( (^) E + ) (^) E
(^2) ( + )
v^2 or v^2 =
GM
R h
E E
(a) T r R (^) E h GM R (^) EEh
( + ) ( +)
2 π 2 π v T^
R h GM
E E
( + ) 2
3 π
(b) (^) v =
GM
R h
E E
continued on next page
350 Chapter 13
(c) Minimum energy input is ∆ E min = ( K f + U gf ) − ( K i − Ugi )
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^ i =^ m i
v^2
with
vi E^ E
R R
π π
. day s and U GM m gi R E E
Thus,
∆ E m
GM
R h
GM m R h m
E R
E
E E
min =^ E
π^2 ( 4400 s)^2
⎥ +^
GM m R
E E
or ∆ E GM m R^ h R R h
R m E E E E
E min =^
( + )
π^2 ( s)^2
P13.37 (a) Energy conservation for the object-Earth system from firing to apex:
K U K U
m GmM R
GmM R h
g (^) i g (^) f
i E E
E E
( + ) =^ ( + )
v^20
where 1 2
m GmM R
E E
v esc^2 =. Then
1 2
2
2
v v v
v v v
i E E
i
R
R h
esc 2 esc 2
esc 2 esc
2 RR
R h R h R
h
R
E E
i
E E E
v v^2 v v v
esc 2 esc 2
esc 2 esc
2 −− −^ =^
v
v v v i E v^ v
E E i E i E
R
R R R
h R
2
2 2 esc 2 esc 2 esc 2
vv v v
i i
2 2 esc
(b) h = ×^ (^ )
6 37 10 = ×
6 2 2 2
m 8.76 (^). (^77) m
(c) The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it is described by the same energy equation
v (^) i v E v E E
R
R h
h R h
esc esc = ..^
3 2 7 ( × ) (^6 )
×
× + ×
m s m m mm m s m s
2 2
⎝⎜^
= ×
= ×
8 4
vi.
(d) With v^ i <<^ v esc , h R^ R^ R GM
E i E i E E
≈^ v = v
(^2) v 2
2 esc^2
. But (^) g GM R
E E
= 2 , so^ h^ g
= vi
2 2
, in agreement with
0 2 = vi^2 + 2 (− g ) ( h − 0 )
352 Chapter 13
Additional Problems
*P13.40 (a) Let R represent the radius of the asteroid. Then its volume is
π R^3 and its mass is ρ 4 π 3
R^3.
For your orbital motion, (^) ∑ F = ma , Gm m R
m R
1 2 2 2 2 = v (^) , G R R R
ρ π 4 3
3 2
2
v
R
G
⎠⎟^
( ) × −
v^2 1 2^2 ρ π
m s kg 2 m^3 111
1 2 4 1100 4
N m kg 2 m ⋅ (^) ( )
⎠⎟^
= ×
π
(b) ρ π π
(^3 ) R = ( kg m (^3) ) (. × m) =. ×^66 kg
(c) v = 2 π R T
T
R
( × ) = × =
2 2 1 53^10
4 π π 4 v
m m s s h
(d) For an illustrative model, we take your mass as 90 kg and assume the asteroid is originally at rest. Angular momentum is conserved for the asteroid-you system: L L m R I
m R m R T
m
∑ i =∑ f = −
= −
2 2 1 2
2
v
v
v
ω π asteroid ==
1
1 2
π
π π
m R T
T m R m
asteroid
asteroid (^) v
(.^ ×× ) ( × ) ( )( )
(^16) kg 4 m kg m s
. 7 7 × 10 17 s =26 5. billion years
Thus your running does not produce significant rotation of the asteroid if it is originally stationary, and does not significantly affect any rotation it does have.
This problem is realistic. Many asteroids, such as Ida and Eros, are roughly 30 km in diameter. They are typically irregular in shape and not spherical. Satellites such as Phobos (of Mars), Adrastea (of Jupiter), Calypso (of Saturn), and Ophelia (of Uranus) would allow a visitor the same experience of easy orbital motion. So would many Kuiper-belt objects.
Universal Gravitation 353
P13.41 Let m represent the mass of the spacecraft, rE the radius of the Earth’s orbit, and x the distance from Earth to the spacecraft.
The Sun exerts on the spacecraft a radial inward force of F GM m s r x s E
( − )^2
while the Earth exerts on it a radial outward force of F GM m E x
= E
2 The net force on the spacecraft must produce the correct centripetal acceleration for it to have an orbital period of 1.000 year.
Thus,
F F GM m r x
GM m x
m r x
m S E r x S E
E E E
( − )
( − )
(^2 2) ( − )
v^2 2 ππ^ r^ x T
⎡ ( (^) E − ) ⎣
2
which reduces to GM r x
GM
x
r x T
S E
E E ( − )
( − ) (^2 )
2 2
4 π (^) (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. × 10 9 m and 1 48. × 10 9 m by substituting both of these as trial solutions, along with the following data: M (^) S = 1 991. × 10 30 kg, M (^) E = 5 983. × 10 24 kg, rE = 1 496. × 1011 m, and T = 1 000. yr = 3 156. × 10 7 s.
With x = 1 47. × 10 9 m substituted into equation (1), we obtain 6 052. × 10 −^3 m s 2 − 1 85. × 10 −^3 m s 2 ≈ 5 871. × 10 −^3 m s^2 or 5 868. × 10 −^3 m s 2 ≈ 5 871. × 10 −^3 m s^2 With x = 1 48. × 10 9 m substituted into the same equation, the result is 6 053. × 10 −^3 m s 2 − 1 82. × 10 −^3 m s 2 ≈ 5 870 8. × 10 −^3 m s^22 or 5 870 9. × 10 −^3 m s 2 ≈ 5 870 8. × 10 −^3 m s^2
Since the fi rst trial solution makes the left-hand side of equation (1) slightly less than the right hand side, and the second trial solution does the opposite, the true solution is determined as between the trial values. To three-digit precision, it is 1 48. × 10 9 m.
As an equation of fifth degree, equation (1) has five roots. The Sun-Earth system has five Lagrange points, all revolving around the Sun synchronously with the Earth. The SOHO and ACE satellites are 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 to gain 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.
Universal Gravitation 355
P13.45 (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 r
ring (^). When the object is at A,
this is − × ⋅ × ×
. 11 N m kg 2.36 10 kg kg
2 20 2 88 2 8 2
4 2 10
m m
J
( ) +^ ( × )
= −. ×
(b) When the object is at the center of the ring, the potential energy is
−
× ⋅ ×
×
. 11 N m 1 000 kg 2.36 10 kg kg
2 20 2 110 8 1 57^10
5 m
= −. × J
(c) Total energy of the object-ring system is conserved: K U (^) g (^) A K Ug (^) B
B
( + ) =^ ( + )
0 − 7 04 × 10 = −
. 4 J 1 000 kg v^2 1. 557 10
2 8 70 10 13
5
4 1 2
×
⎛ × ×
⎝⎜^
⎠⎟^
J
J
1 000 kg vB
..2 m s
P13.46 (a) The free-fall acceleration produced by the Earth is g
GM
r = 2 E = GM rE −^2 (directed downward) Its rate of change is dg dr
= GM E ( − 2 ) r −^3 = − 2 GM rE −^3
The minus sign indicates that g decreases with increasing height. At the Earth’s surface, dg dr
GM
R
E E
(b) For small differences, ∆ ∆
g ∆ r
g h
GM
R
E E
3 Thus,
∆ g GM h R
E E
3
(c) ∆ g =
2 6 67(. × 10 −^11 N m⋅ 2 kg^2 ) ( 5 98. × 10 24 kg)6 00. mm m
m s 2
( × )
= × −
6 3 1 85^105
P13.47 From the walk, 2 π r = 25 000m. Thus, the radius of the planet is r = = ×
m 2 m π
From the drop: ∆ y = 1 gt = g ( ) =
(^2) 29 2. s 2 1 40. m
so, g
MG
r
= × − =
2 3 28^1032
m s m s 2 ∴ M = 7 79. × 1014 kg
356 Chapter 13
P13.48 The distance between the orbiting stars is d = 2 r cos 30 ° = 3 r since
cos
° =. The net inward force on one orbiting star is
Gmm d
GMm r
Gmm d
m r Gm
2 2 2
2 30 30
2 30
cos cos
cos
2 2
2 2 2 2 3 2
2 2
r
GM
r
r rT G m^ M r T
T
⎝⎜^
π
π
π rr G M m
T
r G M m
3
3 1 2
π ⎟
P13.49 For a 6.00 km diameter cylinder, r = 3 000 m and to simulate 1 g = 9 80. m s 2
g r
r
g r
v^22
ω
ω. rad s
The required rotation rate of the cylinder is 1 rev 110 s (For a description of proposed cities in space, see Gerard K. O’Neill in Physics Today , Sept. 1974.)
P13.50 For both circular orbits,
∑^ F^ =^ ma :^ GM m r
m r
E 2
2
v
v =
GM
r
E
(a) The original speed is vi = ( × ⋅ ) ( × ) ×
. 11.^24
N m 2 kg 2 kg 00 2 10
6 5 7 79^103
m m
m s ( +^ × )
=. ×
(b) The final speed is vi = ( ×^ ⋅ ) ( × ) ×
. 11.^24
N m 2 kg 2 kg 00
6 7 85^103
m
m s ( )
=. ×
The energy of the satellite-Earth system is
K U m GM m r m
GM
r
GM
r
GM m g r
+ = 1 − E^ = E^ − E^ = − E
v^2
(c) Originally Ei = − ( 6 67. × 10 −^11 N m⋅ 2 kg 2 ) ( 5 98. × 10 24 kg ) 100 kgg m
J
( ) ( × )
= − ×
6 3 04^109
FIG. P13.
F g
v
FIG. P13.
r
d
r
F
F
F
continued on next page
