PSE9e ISM chapter13 final tủ tài liệu training

13
Universal Gravitation
CHAPTER OUTLINE
13.1

Newton’s Law of Universal Gravitation

13.2

Free-Fall Acceleration and the Gravitational Force

13.3

Analysis Model: Particle in a Field (Gravitational)

13.4

Kepler’s Laws and the Motion of Planets

13.5

Gravitational Potential Energy

13.6

Energy Considerations in Planetary and Satellite Motion

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ13.1

Answer (c). Ten terms are needed in the potential energy:
U = U12 + U13 + U14 + U15 + U23 + U24 + U25 + U34 + U35 + U45

OQ13.2

The ranking is a > b = c. The gravitational potential energy of the
Earth-Sun system is negative and twice as large in magnitude 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.

OQ13.3

Answer (d). The satellite experiences a gravitational force, always
directed toward the center of its orbit, and supplying the centripetal
force required to hold it in its orbit. This force gives the satellite a
centripetal acceleration, even if it is moving with constant angular
speed. At each point on the circular orbit, the gravitational force is
directed along a radius line of the path, and is perpendicular to the
motion of the satellite, so this force does no work on the satellite.

OQ13.4

Answer (d). 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
684

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 13

685

becomes (2 m/s2)(2)/4 = 1 m/s2.
OQ13.5

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.

OQ13.6

Answer (b). The mass of a spherical body of radius R and
density ρ is M = ρV = ρ(4πR3/3). The escape velocity from the
surface of this body may then be written in either of the following
equivalent forms:

vesc =

2GM
R

and vesc =

2G ⎛ 4πρR 3 ⎞
=
R ⎜⎝ 3 ⎟⎠

8πρGR 2
3

We see that the escape velocity depends on the three properties
(mass, density, and radius) of the planet. Also, the weight of an
object on the surface of the planet is Fg = mg = GMm/R2, giving

G ⎡ ⎛ 4π R 3 ⎞ ⎤ 4
g = GM R = 2 ⎢ ρ ⎜
⎥ = πρGR
R ⎣ ⎝ 3 ⎟⎠ ⎦ 3
2

The free-fall acceleration at the planet’s surface then depends on the
same properties as does the escape velocity. Changing the value of g
would necessarily change the escape velocity. Of the listed
quantities, the only one that does not affect the escape velocity is the
mass of the object.
OQ13.7

(i)

Answer (e). According to the inverse square law, 1/42 = 16 times
smaller.

(ii)

Answer (c). mv2/r = GMm/r2 predicts that v is proportional to
(1/r)1/2, so it becomes (1/4)1/2 = 1/2 as large.

(iii) Answer (a). According to Kepler’s third law, (43)1/2 = 8 times
larger; also, the circumference is 4 times larger and the speed
1/2 as large: 4/(1/2) = 8.
OQ13.8

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.

OQ13.9

The ranking is b > a > c = d > e. The force is proportional to the
product of the masses and inversely proportional to the square of the
separation distance, so we compute m1m2/r2 for each case: (a) 2·3/12 =
6 (b) 18 (c) 18/4 = 4.5 (d) 4.5 (e) 16/4 = 4.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


686

Universal Gravitation

OQ13.10

Answer (c). 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 might think the gravitational force is zero or nearly zero,
because the orbiting astronauts appear to be weightless. They and the
space station are in free fall, so the normal force of the space station’s
wall/floor/ceiling on the astronauts is zero; they float freely around
the cabin.

OQ13.11

Answer (e). We assume that the elliptical orbit is so elongated that
the Sun, at one focus, is almost at one end of the major axis. If the
period, T, is expressed in years and the semimajor axis, a, in
astronomical units (AU), Kepler’s third law states that T 2 = a 3 .
Thus, for Halley’s comet, with a period of T = 76 y, the semimajor
axis of its orbit is

a=

3

(76)2

= 18 AU

The length of the major axis, and the approximate maximum distance
from the Sun, is 2a = 36 AU.

ANSWERS TO CONCEPTUAL QUESTIONS

CQ13.1

CQ13.2

CQ13.3

(a) The gravitational force is conservative. (b) Yes. An encounter with
a stationary mass cannot permanently speed up a spacecraft. But
Jupiter is moving. A spacecraft flying across its orbit just behind the
planet will gain kinetic energy because of the change in potential
energy of the spacecraft-planet system. This is a collision because the
spacecraft and planet exert forces on each other while they are
isolated from outside forces. It is an elastic collision because only
conservative forces are involved. (c) The planet loses kinetic energy
as the spacecraft gains it.
GM
Cavendish determined G. Then from g = 2 , one may determine
R
the mass of the Earth. The term “weighed” is better expressed as
“massed.”
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

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 13

687

evening.
CQ13.4

(a)

Every point q on the sphere that does not lie along the axis
connecting 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.

(b)

The forces will not necessarily cancel if the mass is not
uniformly distributed, unless the center of mass of the
nonuniform sphere still lies along the axis.

ANS. FIG. CQ13.4
CQ13.5

The angular momentum of a planet going around a sun is
conserved. (a) The speed of the planet is maximum at closest
approach. (b) The speed is a minimum at farthest distance. These two
points, perihelion and aphelion respectively, are 180° apart, at
opposite ends of the major axis of the orbit.

CQ13.6

Set the universal description of the gravitational force, Fg =

CQ13.7

(a)

In one sense, ‘no’. If the object is at the very center of the Earth
there is no other mass located there for comparison and the
formula does not apply in the same way it was being applied
while the object was some distance from the center. In another
sense, ‘yes’. One would have to compare, though, the distance
between the object with mass m to the other individual masses
that make up the Earth.

(b)

The gravitational force of the Earth on an object at its center
must be zero, not infinite as one interpretation of Equation 11.1
would suggest. All the bits of matter that make up the Earth
pull in different outward directions on the object, causing the
net force on it to be zero.

CQ13.8

GMX m
,
RX2
equal to the local description, Fg = magravitational, 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.

The escape speed from the Earth is 11.2 km/s and that from the

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


688

Universal Gravitation
Moon is 2.3 km/s, smaller by a factor of 5. The energy required—and
fuel—would be proportional to v2, or 25 times more fuel is required
to leave the Earth versus leaving the Moon.

CQ13.9

Air resistance causes a decrease in the energy of the satellite-Earth
system. This reduces the radius 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!

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 13.1
P13.1

Newton’s Law of Universal Gravitation

This is a direct application of the equation expressing Newton’s law of

gravitation:

(

(1.50 kg ) 15.0 × 10−3 kg
GMm
−11
2
2
F=
= 6.67 × 10 N ⋅ m /kg
2
r2
4.50 × 10−2 m

)

(

(

)

)

= 7.41 × 10−10 N
P13.2

For two 70-kg persons, modeled as spheres,

Fg =

Gm1m2 ( 6.67 × 10
=
r2

−11

N ⋅ m 2/kg 2 ) ( 70 kg ) ( 70 kg )

( 2 m )2

~ 10−7 N
P13.3

(a)

At the midpoint between the two objects, the forces exerted by the
200-kg and 500-kg objects are oppositely directed,
Gm1m2
r2

and from

Fg =

we have

∑F =

G ( 50.0 kg ) ( 500 kg − 200 kg )

( 2.00 m )

2

= 2.50 × 10−7 N

toward the 500-kg object.
(b) At a point between the two objects at a distance d from the
500-kg object, the net force on the 50.0-kg object will be zero
when
G ( 50.0 kg ) ( 200 kg )

( 4.00 m − d )

2

=

G ( 50.0 kg ) ( 500 kg )
d2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13

689

To solve, cross-multiply to clear of fractions and take the square
root of both sides. The result is
d = 2.45 m from the 500-kg object toward the smaller object .
P13.4

(a)

The Sun-Earth distance is 1.496 × 1011 m and the Earth-Moon
distance is 3.84 × 108 m, so the distance from the Sun to the Moon
during a solar eclipse is
1.496 × 1011 m − 3.84 × 108 m = 1.492 × 1011 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

FSM =
=

Gm1m2
r2
(6.67 × 10−11 N ⋅ m2 /kg 2 )(1.99 × 1030 kg )(7.36 × 1022 kg )

(1.492 × 10

11

m)

2

= 4.39 × 1020 N
(b)

FEM

(6.67 × 10
=

−11

N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg ) ( 7.36 × 1022 kg )

( 3.84 × 10

8

m)

2

= 1.99 × 1020 N

(c)

FSE =

(6.67 × 10

−11

N ⋅ m 2/kg 2 ) ( 1.99 × 1030 kg ) ( 5.98 × 1024 kg )

(1.496 × 10

11

m)

2

= 3.55 × 1022 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.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


690
P13.5

Universal Gravitation
With one metric ton = 1 000 kg,
F = m1 g =

g=

Gm1m2
r2

Gm2 ( 6.67 × 10
=
r2

−11

N ⋅ m 2/kg 2 ) ( 4.00 × 107 kg )

(100 m )2

= 2.67 × 10−7 m/s 2
P13.6

The force exerted on the 4.00-kg mass by the
2.00-kg mass is directed upward and given
by

mm
F12 = G 22 1 ˆj
r12

= ( 6.67 × 10−11 N ⋅ m 2/kg 2 )

( 4.00 kg )( 2.00 kg ) ˆj
( 3.00 m )2

= 5.93 × 10

−11

ANS. FIG. P13.6

ˆj N

The force exerted on the 4.00-kg mass by the 6.00-kg mass is directed to
the left:

mm
F32 = G 2 2 3 − ˆi
r32

( )

= ( −6.67 × 10−11 N ⋅ m 2/kg 2 )

( 4.00 kg )(6.00 kg ) ˆi
( 4.00 m )2

= −10.0 × 10−11 ˆi N
Therefore, the resultant force on the 4.00-kg mass is




F4 = F24 + F64 = −10.0ˆi + 5.93ˆj × 10−11 N 

(

*P13.7

)

The magnitude of the gravitational force is given by

Gm1m2 ( 6.672 × 10−11 N · m 2 /kg 2 ) ( 2.00 kg ) ( 2.00 kg )
F=
=
( 0.300 m )2
r2
= 2.97 × 10−9 N
P13.8

Gm1m2
, we
r2
5
would find that the mass of a sphere is 1.22 × 10 kg! If the spheres
have at most a radius of 0.500 m, the density of spheres would be at

Assume the masses of the sphere are the same. Using Fg =

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 13

691

least 2.34 × 105 kg/m3, which is ten times the density of the most dense
element, osmium.

The situation is impossible because no known element could compose
the spheres.
P13.9

We are given m1 + m2 = 5.00 kg, which means that m2 = 5.00 kg − m1.
Newton’s law of universal gravitation then becomes
m1m2
r2
⇒ 1.00 × 10−8 N

F=G

= ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )

( 5.00 kg ) m1 − m

2
1

(1.00 × 10
=

−8

6.67 × 10

Thus,

m12 − ( 5.00 kg ) m1 + 6.00 kg = 0

or

( m1 − 3.00 kg )( m1 − 2.00 kg ) = 0

m1 ( 5.00 kg − m1 )
( 0.200 m )2

N ) ( 0.040 0 m 2 )

−11

N ⋅ m /kg
2

2

= 6.00 kg 2

giving m1 = 3.00 kg, so m2 = 2.00 kg . The answer m1 = 2.00 kg and
m2 = 3.00 kg is physically equivalent.
P13.10

Let θ represent the angle each cable makes with the
vertical, L the cable length, x the distance each ball is
displaced by the gravitational force, and d = 1 m the
original distance between them. Then r = d − 2x is the
separation of the balls. We have

∑ Fy = 0: T cosθ − mg = 0
∑ Fx = 0: T sin θ −
Then

tan θ =

L −x

ANS. FIG. P13.10

Gmm
r 2 mg

x
2

Gmm
=0
r2

2

=

Gm
2
g ( d − 2x )

→

Gm
2
x ( d − 2x ) = g L2 − x 2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


692

Universal Gravitation

Gm
is numerically small. We expect that x is very small
g
compared to both L and d, so we can treat the term (d − 2x) as d, and
(L2 − x2) as L2. We then have
The factor

x (1 m )

2

(6.67 × 10

=

−11

N ⋅ m 2/kg 2 ) ( 100 kg )

( 9.80 m/s )
2

( 45.00 m )

x = 3.06 × 10−8 m

Section 13.2
P13.11

Free-Fall Acceleration and the Gravitational Force

The distance of the meteor from the center of Earth is R + 3R = 4R.
Calculate the acceleration of gravity at this distance.
g=

GM (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
r2
[4(6.37 × 106 m)]2

= 0.614 m/s 2 , toward Earth

P13.12

The gravitational field at the surface of the Earth or Moon is given by
GM
g= 2 .
R
The expression for density is

so

and

ρ =

M
M
=
,
V
4
π R3
3

4
M = π ρR 3
3

4
G ⎛ πρ R 3 ⎞
⎝3
⎠ 4

g=
= Gπρ R
2
R
3

Noting that this equation applies to both the Moon and the Earth, and
dividing the two equations,

4
Gπρ M RM
gM
ρ R
3
=
= M M
gE
4
ρE RE
GπρE RE
3

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13

693

Substituting for the fractions,

()

1 ρM 1
=
6
ρE 4
P13.13

(a)

ρM 4
2
=
=
ρE
6
3

and

For the gravitational force on an object in the neighborhood of
Miranda, we have

mobj g =

GmobjmMiranda
2
rMiranda

(6.67 × 10
Gm
g = 2 Miranda =
rMiranda

−11

N ⋅ m 2 / kg 2 ) ( 6.68 × 1019 kg )

( 242 × 10

3

m)

2

= 0.076 1 m/s 2
(b)

We ignore the difference (of about 4%) in g between the lip and
the base of the cliff. For the vertical motion of the athlete, we have

1 2
ay t
2
1
−5 000 m = 0 + 0 + ( −0.076 1 m/s 2 ) t 2
2
y f = y i + vyi +

⎛ 2 ( 5 000 m ) s 2 ⎞
t=⎜
⎟
⎝ 0.076 1 m ⎠
(c)

1/2

= 363 s

1
x f = xi + vxit + axt 2 = 0 + ( 8.50 m/s )( 363 s ) + 0 = 3.08 × 103 m
2

We ignore the curvature of the surface (of about 0.7°) over the
athlete’s trajectory.
(d)

vxf = vxi = 8.50 m/s

vyf = vyi + ay t = 0 − ( 0.076 1 m/s 2 ) ( 363 s ) = −27.6 m/s

(

)


Thus v f = 8.50ˆi − 27.6ˆj m/s = 8.502 + 27.62 m/s at

⎛ 27.6 m/s ⎞
tan −1 ⎜
= 72.9° below the x axis.
⎝ 8.50 m/s ⎟⎠

v f = 28.9 m/s at 72.9° below the horizontal

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


694

Universal Gravitation

Section 13.3
P13.14

(a)

Analysis Model: Particle in a Field (Gravitational)
g1 = g 2 =

MG
r + a2
2

g1y = − g 2 y
g y = g1y + g 2 y = 0

g1x = g 2 x = g 2 cos θ

cos θ =

(a

r

2

+ r2 )

1/2

ANS. FIG. P13.14

( )


g = 2g 2 x − ˆi
or


g=

(r

2MGr
2

+ a2 )

32

toward the center of mass

(b)

At r = 0, the fields of the two objects are equal in magnitude and
opposite in direction, to add to zero.

(c)

As r → 0, 2MGr(r 2 + a 2 )−3/2 approaches 2MG(0)/a 3 = 0.

(d)

When r is much greater than a, the angles the field vectors make
with the x axis become smaller. At very great distances, the field
vectors are almost parallel to the axis; therefore, they begin to
look like the field vector from a single object of mass 2M.

(e)

As r becomes much larger than a, the expression approaches
2MGr(r 2 + 02 )−3/2 = 2MGr/r 3 = 2MG/r 2 as required.

P13.15

The vector gravitational field at point O is given by

(

Gm
Gm
 Gm
g = 2 ˆi + 2 ˆj + 2 cos 45.0°ˆi + sin 45.0ˆj
l
l
2l

so

)

( )

1 ⎞ ˆ ˆ
 Gm ⎛
g = 2 ⎜1+
⎟ i + j or
l ⎝
2 2⎠

1⎞
 Gm ⎛
g = 2 ⎜ 2 + ⎟ toward the opposite corner.
l ⎝
2⎠

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 13

695

ANS. FIG. P13.15
P13.16

(a)

−11
2
2
30
3
GMm ( 6.67 × 10 N ⋅ m /kg ) ⎡⎣100 ( 1.99 × 10 kg ) ( 10 kg ) ⎤⎦
F=
=
2
r2
(1.00 × 104 m + 50.0 m )

= 1.31 × 1017 N

(b)

ΔF =

GMm GMm
− 2

2
rfront
rback

2
2
ΔF GM ( rback − rfront )
Δg =
=
2
2
m
rfront
rback

ANS. FIG. P13.16

Δg = ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )
2
2
⎡⎣100 ( 1.99 × 1030 kg ) ⎤⎦ ⎡( 1.01 × 10 4 m ) − ( 1.00 × 10 4 m ) ⎤
⎣
⎦
2
2
4
4
(1.00 × 10 m ) (1.01 × 10 m )

Δg = 2.62 × 1012 N/kg

Section 13.4
P13.17

Kepler’s Laws and the Motion of Planets

The gravitational force on mass located at distance r from the center of
the Earth is Fg = mg = GME m/r 2 . Thus, the acceleration of gravity at
this location is g = GME /r 2 . If g = 9.00 m/s2 at the location of the
satellite, the radius of its orbit must be

GME
r=
=
g

(6.67 × 10

−11

N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg )
9.00 m/s 2

= 6.66 × 106 m
From Kepler’s third law for Earth satellites, T 2 = 4π 2 r 3GMES, the
period is found to be
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


696

Universal Gravitation
r3
T = 2π
= 2π
GME

(6.67 × 10

−11

(6.66 × 10

6

m)

3

N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg )

= 5.41 × 103 s

or
⎛ 1h ⎞
T = ( 5.41 × 103 s ) ⎜
= 1.50 h = 90.0 min
⎝ 3 600 s ⎟⎠

P13.18

The gravitational force exerted by Jupiter on Io causes the centripetal
acceleration of Io. A force diagram of the satellite would show one
downward arrow.

∑ Fon Io = MIo a:

GMJ MIo MIo v 2 MIo ⎛ 2π r ⎞ 2 4π 2 rMIo
=
=
⎜
⎟ =
r2
r
r ⎝ T ⎠
T2

Thus the mass of Io divides out and we have Kepler’s third law with
m << M,
4π 2 r 3
4π 2 (4.22 × 108 m)3
⎛ 1d ⎞
MJ =
=
⎟
2
−11
2
2
2 ⎜

GT
(6.67 × 10 N ⋅ m /kg )(1.77 d) ⎝ 86 400 s ⎠

P13.19

2

and

M J = 1.90 × 1027 kg (approximately 316 Earth masses)

(a)

The desired path is an elliptical trajectory with the Sun at one of
the foci, the departure planet at the perihelion, and the target
planet at the aphelion. The perihelion distance rD is the radius of
the departure planet’s orbit, while the aphelion distance rT is the
radius of the target planet’s orbit. The semimajor axis of the
desired trajectory is then a = ( rD + rT )/2.

ANS. FIG. P13.19
If Earth is the departure planet, rD = 1.496 × 1011 m = 1.00 AU

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13

697

With Mars as the target planet,
⎛
⎞
1 AU
rT = 2.28 × 1011 m ⎜
= 1.52 AU
11
⎝ 1.496 × 10 m ⎟⎠

Thus, the semimajor axis of the minimum energy trajectory is
a=

rD + rT 1.00 AU + 1.52 AU
=
= 1.26 AU
2
2

Kepler’s third law, T2 = a3, then gives the time for a full trip
around this path as

T = a3 =

(1.26 AU )3

= 1.41 yr

so the time for a one-way trip from Earth to Mars is
Δt =

P13.20

1.41 yr
1
T=
= 0.71 yr
2
2

(b)

This trip cannot be taken at just any time. The departure must
be timed so that the spacecraft arrives at the aphelion when the
target planet is located there.

(a)

The particle does possess angular momentum, because it is
not headed straight for the origin.

(b)

Its angular momentum is constant. There are no identified
outside influences acting on the object.

(c)

Since speed is constant, the distance traveled between tA and
tB is equal to the distance traveled between tC and tD . The area
of a triangle is equal to one-half its (base) width across one side

times its (height) dimension perpendicular to that side.
So

1
1
bv0 ( tB − tA ) = bv0 ( tD − tC )
2
2

states that the particle’s radius vector sweeps out equal areas in
equal times.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


698
P13.21

Universal Gravitation
Applying Newton’s second law, ∑ F = ma yields
Fg = mac for each star:
GMM Mv 2
=
r
( 2r )2

or

M=

4v 2 r
G

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
2π r = vT. Therefore,

M=

ANS. FIG. P13.21

4v 2 r 4v 2 ⎛ vT ⎞
=
⎜
⎟
G
G ⎝ 2π ⎠

3
2v 3T 2 ( 220 × 10 m/s ) ( 14.4 d ) ( 86 400 s/d )
M=
=
πG
π ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )
3

so,

= 1.26 × 1032 kg = 63.3 solar masses

P13.22

To find the angular displacement of planet Y, we
apply Newton’s second law:

∑ F = ma:

Gmplanet Mstar
r2

=

mplanet v 2
r

Then, using v = rω ,

GMstar
= v 2 = r 2ω 2
r
GMstar = r 3ω 2 = rx3ω x2 = ry3ω y2
solving for the angular velocity of planet Y gives
⎛r ⎞
ωy = ωx ⎜ x ⎟
⎝ ry ⎠

32

⎛ 90.0° ⎞ 3 2
468°
=⎜
3 =
⎟
5.00 yr
⎝ 5.00 yr ⎠

ANS. FIG. P13.22

So, given that there are 360° in one revolution we
convert 468° to find that planet Y has turned through
1.30 revolutions .

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13
P13.23

699

By Kepler’s third law, T2 = ka3 (a = semimajor
axis). For any object orbiting the Sun, with T in
years and a in AU, k = 1.00. Therefore, for
Comet Halley, and suppressing units,

⎛ 0.570 + y ⎞
(75.6) = (1.00) ⎜
⎟⎠

⎝
2

3

2

The farthest distance the comet gets from the
Sun is
y = 2 ( 75.6 )

23

ANS. FIG. P13.23

− 0.570 = 35.2 AU

(out around the orbit of Pluto).
*P13.24

By conservation of angular momentum for the satellite, rp v p = ra va , or
vp
va

=

ra 2 289 km + 6.37 × 103 km 8 659 km
=
=
= 1.27

rp
459 km + 6.37 × 103 km
6 829 km

We do not need to know the period.
P13.25

For an object in orbit about Earth, Kepler’s third law gives the relation
between the orbital period T and the average radius of the orbit
(“semimajor axis”) as
⎛ 4π 2 ⎞ 3
T2 = ⎜
r
⎝ GME ⎟⎠

We assume that the two given distances in the problem statements are
the perigee and apogee, respectively.
Thus, if the average radius is

rmin + rmax 6 670 km + 385 000 km
=
2
2
5
= 1.96 × 10 km = 1.96 × 108 m

r=

The period (time for a round trip from Earth to the Moon) would be

r3
T = 2π
GME
= 2π

(6.67 × 10

−11

(1.96 × 10

8

m)

3

N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg )

= 8.63 × 105 s

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


700

Universal Gravitation
The time for a one-way trip from Earth to the Moon is then
⎛ 8.63 × 105 s ⎞ ⎛ 1 day ⎞
1

Δt = T = ⎜
⎟⎠ ⎜⎝ 8.64 × 10 4 s ⎟⎠ = 4.99 d
⎝
2
2

P13.26

The gravitational force on a small parcel of material at the star’s
equator supplies the necessary centripetal acceleration:

GMs m mv 2
=
= mRsω 2
Rs2
Rs
so

(6.67 × 10

GMs
ω=
=
Rs3

−11

N ⋅ m 2 /kg 2 ) ⎡⎣ 2 ( 1.99 × 1030 kg ) ⎤⎦

(10.0 × 10 m )

3

3

ω = 1.63 × 10 4 rad/s
P13.27

We find the satellite’s altitude from
GM J

(R

J

+d

)

=

2

(

4π 2 RJ + d
T

2

)

where d is the altitude of the satellite above Jupiter’s cloud tops. Then,

(

GM JT 2 = 4π 2 RJ + d

(6.67 × 10

−11

)

3

N ⋅ m 2 /kg 2 ) ( 1.90 × 1027 kg ) ( 9.84 × 3 600 )

2

= 4π 2 ( 6.99 × 107 + d )

3

which gives

d = 8.92 × 107 m = 89 200 km above the planet
P13.28

(a)

In T2 = 4 π2a3/GMcentral we take a = 3.84 × 108 m.

Mcentral =
=

4π 2 a 3
GT 2
4π 2 (3.84 × 108 m)3
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(27.3 × 86 400 s)2

= 6.02 × 1024 kg
This is a little larger than 5.98 × 1024 kg.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13
(b)

701

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.

P13.29

The speed of a planet in a circular orbit is given by

∑ F = ma:
(a)

GMsun m mv 2
=
r2
r

→ v=

GMsun
r

For Mercury, the speed is

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
vM =
5.79 × 1010 m
= 4.79 × 10 4 m/s
and for Pluto,

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
5.91 × 1012 m
= 4.74 × 103 m/s

vp =

With greater speed, Mercury will eventually move farther from
the Sun than Pluto.
(b)

With original distances rP and rM perpendicular to their lines of
motion, they will be equally far from the Sun at time t, where
2 2
rP2 + vP2 t 2 = rM2 + v M
t

2
rP2 − rM2 = ( v M
− vP2 ) t 2

t=
=

( 5.91 × 10

( 4.79 × 10

4

12

m ) − ( 5.79 × 1010 m )
2

2

m/s ) − ( 4.74 × 103 m/s )
2

2

3.49 × 1025 m 2
= 1.24 × 108 s = 3.93 yr
2.27 × 109 m 2/s 2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


702

Universal Gravitation

Section 13.5
P13.30

(a)

Gravitational Potential Energy
We compute the gravitational potential energy of the satelliteEarth system from

U=−
=−

GME m
r
(6.67 × 10−11 N ⋅ m2 /kg 2 )( 5.98 × 1024 kg )(100 kg )

( 6.37 + 2.00 ) × 106 m

= −4.77 × 109 J
(b), (c) The satellite and Earth exert forces of equal magnitude on each
other, directed downward on the satellite and upward on Earth.
The magnitude of this force is

F=
=

GME m
r2
(6.67 × 10−11 N ⋅ m2 /kg 2 )( 5.98 × 1024 kg )(100 kg )

( 8.37 × 10

6

m)

2

= 569 N
P13.31

The work done by the Moon’s gravitational field is equal to the
negative of the change of potential energy of the meteor-Moon system:

⎛ −Gm1m2
⎞
Wint = −ΔU = − ⎜
− 0⎟

⎝
⎠
r
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(1.00 × 103 kg)
Wint =
1.74 × 106 m
= 2.82 × 109 J
*P13.32

The enery required is equal to the change in gravitational potential
energy of the object-Earth system:

U = −G

GME
Mm
and g =
so that
RE2
r

1⎞ 2
⎛ 1
ΔU = −GMm ⎜
− ⎟ = mgRE
⎝ 3RE RE ⎠ 3
2
ΔU = ( 1 000 kg ) ( 9.80 m s 2 ) ( 6.37 × 106 m ) = 4.17 × 1010 J
3

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13
P13.33

(a)

703

The definition of density gives
3 ( 1.99 × 1030 kg )
MS
ρ=
=
= 1.84 × 109 kg/m 3
4 2 4π ( 6.37 × 106 m )3
πr
3 E

(b)

For an object of mass m on its surface, mg = GMS m/RE2. Thus,
−11
2
2
30
GMS ( 6.67 × 10 N ⋅ m /kg ) ( 1.99 × 10 kg )
g= 2 =
2

rE
(6.37 × 106 m )

= 3.27 × 106 m/s 2

(c)

Relative to Ug = 0 at infinity, the potential energy of the object-star
system at the surface of the white dwarf is

GMSm
rE

Ug = −

(6.67 × 10
=−

−11

N ⋅ m 2 /kg 2 ) ( 1.99 × 1030 kg ) ( 1.00 kg )
6.37 × 106 m

= − 2.08 × 1013 J
P13.34

(a)

Energy conservation of the object-Earth system from release to
radius r:

(K + U )
g

altitude h

(

= K + Ug

)

radius r

GME m 1
GME m
0−
= mv 2 −
RE + h 2
r
⎡
⎛1
1 ⎞⎤
v = ⎢ 2GME ⎜ −
⎥
⎝ r RE + h ⎟⎠ ⎦
⎣
(b)

f

f

i

i

∫ dt = ∫ −

1/2

=−

dr
dt

i

dr
dr
= ∫ . The time of fall is, suppressing units,
v
v
f

Δt =

RE +h

∫

RE

⎡
⎛1
1 ⎞⎤
⎢ 2GME ⎜ −
⎥
⎝ r RE + h ⎟⎠ ⎦
⎣

−1/2

dr

Δt = ( 2 × 6.67 × 10−11 × 5.98 × 1024 )
6.87×106 m

−1/2

1
⎡⎛ 1
⎞⎤
×
∫ 6 ⎢⎣⎜⎝ r − 6.87 × 106 m ⎟⎠ ⎥⎦
6.37×10 m

−1/2

dr

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


704

Universal Gravitation
We can enter this expression directly into a mathematical
calculation program.
Alternatively, to save typing we can change variables to u =

r
.
106

Then
Δt = ( 7.977 × 10

)

14 −1/2

6.87

1
⎛ 1
⎞
∫6.37 ⎜⎝ 106 u − 6.87 × 106 ⎟⎠

106

6.87

1 ⎞
⎛1
= 3.541 × 10
⎜ −
⎟
6 −1/2 ∫ ⎝ u
(10 ) 6.37 6.87 ⎠
−8

−1/2

106 du

−1/2

du

A mathematics program returns the value 9.596 for this integral,
giving for the time of fall

Δt = 3.541 × 10−8 × 109 × 9.596 = 339.8 = 340 s
P13.35

(a)

Since the particles are located at the corners of an equilateral
triangle, the distances between all particle pairs is equal to

0.300 m. The gravitational potential energy of the system is then
⎛ Gm1m2 ⎞
U Tot = U 12 + U 13 + U 23 = 3U 12 = 3 ⎜ −
r12 ⎟⎠
⎝

U Tot = −

3 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.00 × 10−3 kg )

2

0.300 m

= −1.67 × 10−14 J

(b)

Section 13.6
P13.36

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.

Energy Considerations in
Planetary and Satellite Motion

We use the isolated system model for energy:

Ki + Ui = Kf + Uf

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13

705

⎛ 1 1⎞ 1
1
mvi2 + GME m ⎜ − ⎟ = mv 2f
2
⎝ rf ri ⎠ 2
which becomes
⎛
1 2
1⎞ 1
vi + GME ⎜ 0 − ⎟ = v 2f
2
RE ⎠ 2
⎝

2GME
RE

or

v 2f = v12 −

and

⎛
2GME ⎞
v f = ⎜ v12 −
RE ⎟⎠
⎝

1/2

⎧⎪
⎡ 2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ⎤ ⎫⎪
2
4
v f = ⎨( 2.00 × 10 m/s ) − ⎢
⎥⎬
6.37 × 106 m
⎢⎣
⎥⎦ ⎪⎭
⎩⎪

1/2

= 1.66 × 10 4 m/s
*P13.37

To determine the energy transformed to internal energy, we begin by
calculating the change in kinetic energy of the satellite. To find the
initial kinetic energy, we use

vi2
GME
=
RE + h ( RE + h )2
which gives
Ki =
=

1
1 ⎛ GME m ⎞
mvi2 = ⎜
2
2 ⎝ RE + h ⎟⎠

−11
2
2
24
1 ⎡ ( 6.67 × 10 N ⋅ m kg ) ( 5.98 × 10 kg ) ( 500 kg ) ⎤
⎥
2 ⎢⎣
6.37 × 106 m + 0.500 × 106 m
⎦

= 1.45 × 1010 J

Also,

Kf =

1
1
2
mv 2f = ( 500 kg ) ( 2.00 × 103 m s ) = 1.00 × 109 J.
2
2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


706

Universal Gravitation
The change in gravitational potential energy of the satellite-Earth
system is

⎛ 1
GME m GME m
1 ⎞
−
= GME m ⎜ −
⎟
Ri
Rf
⎝ Ri R f ⎠

ΔU =

= ( 6.67 × 10−11 N ⋅ m 2 kg 2 ) ( 5.98 × 1024 kg ) ( 500 kg )

× ( −1.14 × 10−8 m −1 )

= −2.27 × 109 J
The energy transformed into internal energy due to friction is then

ΔEint = K i − K f − ΔU = ( 14.5 − 1.00 + 2.27 ) × 109 J
= 1.58 × 1010 J
P13.38

To obtain the orbital velocity, we use

or

∑F =

mMG mv 2
=
R2
R

v=

MG
R

We can obtain the escape velocity from
1
mMG
2
mvesc

=
2
R

or
P13.39

(a)

2MG
=
R

vesc =

2v

The total energy of the satellite-Earth system at a given orbital
altitude is given by
Etot = −

GMm
2r

The energy needed to increase the satellite’s orbit is then,
suppressing units,

ΔE =

GMm ⎛ 1 1 ⎞

−
2 ⎜⎝ ri rf ⎟⎠

(6.67 × 10 )( 5.98 × 10 ) 10
=
−11

24

2

⎞
kg ⎛
1
1
−
⎜
10 m ⎝ 6 370 + 100 6 370 + 200 ⎟⎠
3

3

ΔE = 4.69 × 108 J = 469 MJ

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 13
(b)

707

Both in the original orbit and in the final orbit, the total energy
is negative, with an absolute value equal to the positive kinetic
energy. The potential energy is negative and twice as large as
the total energy. As the satellite is lifted from the lower to the
higher orbit, the gravitational energy increases, the kinetic energy
decreases, and the total energy increases. The value of each
becomes closer to zero. Numerically, the gravitational energy
increases by 938 MJ, the kinetic energy decreases by 469 MJ,
and the total energy increases by 469 MJ.

P13.40

(a)

The major axis of the orbit is 2a = 50.5 AU so a = 25.25 AU.
Further, in the textbook’s diagram of an ellipse, a + c = 50 AU,
so c = 24.75 AU. Then
e=

(b)

c 24.75
=
= 0.980
a 25.25

In T2 = Ks a3 for objects in solar orbit, the Earth gives us

(1 yr )

2

= K s ( 1 AU ) K s =
3

(1 yr )2

(1 AU )3

Then

T =
2

(c)

U=−
=−

(1 yr )2 ( 25.25 AU )3
3

(1 AU )

→ T = 127 yr

GMm
r

(6.67 × 10−11 N ⋅ m2 / kg 2 )(1.99 × 1030 kg )(1.20 × 1010 kg )
50 ( 1.496 × 1011 m )

= −2.13 × 1017 J

*P13.41

For her jump on Earth,

1
mvi2 = mgy f
2

[1]

which gives

vi = 2gy f = 2 ( 9.80 m/s ) ( 0.500 m ) = 3.13 m/s

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


708

Universal Gravitation
We assume that she has the same takeoff speed on the asteroid. Here

1
GMA m
mvi2 −

= 0+0
2
RA

[2]

The equality of densities between planet and asteroid,

ρ=

ME
MA
=
4
4
π RE3
π RA3
3
3

implies
3

⎛R ⎞
MA = ⎜ A ⎟ ME
⎝ RE ⎠

[3]

Note also at Earth’s surface

g=

GME
RE2

[4]

Combining the equations [2], [1], [3], and [4] by substitution gives

1 2 GMA
v =
2 i
RA
GME
GME RA2
y
=
RE2 f
RE3
RA2 = y f RE = ( 0.500 m ) ( 6.37 × 106 m )

RA = 1.78 × 103 m
*P13.42

For a satellite in an orbit of radius r around the Earth, the total energy
GME
of the satellite-Earth system is E = −
. Thus, in changing from a
2r

circular orbit of radius r = 2RE to one of radius r = 3RE, the required
work is

W = ΔE = −

GME m GME m
1 ⎤
GME m
⎡ 1
+
= GME m ⎢
−
=
⎥
2rf
2ri
12RE
⎣ 4RE 6RE ⎦

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.