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8.1
GRAVITY
EXERCISES
Section 8.2 Universal Gravitation
12.
At rest on a uniform spherical planet, a body’s weight is proportional to the surface gravity,
gG
M
R
=
/.
2
Therefore,
(/)( / )
( /)
.
gg
MM RR
pE
p E
E p
==
2
2 Since
MM
R
R
p
E
/,
/
.
12
13.
INTERPRET
In this problem we want to use astrophysical data to find the Moon’s acceleration in its circular orbit
about the Earth.
DEVELOP
The gravitational force between two masses
mm
and
is given by Equation 8.1:
F
Gm m
r
=
2
, where
r
is
their distance of separation. The acceleration of the Moon in its orbit can be computed by considering the
gravitational force between the Moon and the Earth.
EVALUATE
Using Equation 8.1, the gravitational force between the Earth (mass
M
E
) and the Moon (mass
m
m
) is
F
GM m
r
Em
mE
=
2
where
r
mE
is the distance between the Moon and the Earth. By Newton’s second law,
Fm
a
=
, the acceleration of the
Moon is
a
F
m
GM
r
m
E
mE
=
××
−
2
11
2
2
6 67 10
5 97 10
(.
/
)
Nmk
g
⋅
24
82
32
385 10
269 10
kg
m
ms
)
)
./
×
=×
−
where we have used the astrophysical data given in Appendix E.
ASSESS
An alternative way to find the acceleration of the Moon as it orbits the Earth is to note that it completes a
nearly circular orbit of 385,000 km radius in 27 days. Based on this information, the (centripetal) acceleration of
the Moon is
a
v
r
r
T
mE
mE
=
×
×
2
2
2
28
4
4
3 85
10
27 3
86 400
π
)
,
m
s
m/s
)
.
2
273 10
−
Note that since the Moon’s orbit is actually elliptical (with 5.5% eccentricity), the values based on circular orbits are
reasonable.
14.
If the surface gravity of the Earth were three times its present value, with no change in mass, then
GM
R
E
/
2
=
3
2
GM
R
EE
where
R
E
is the present radius. Thus,
RR
E
//
.
%
1 3
57 7
gives the new, shrunken radius.
15.
INTERPRET
In this problem we are asked to use astrophysical data to find the gravitational acceleration near the
surface of
(a)
Mercury and
(b)
Saturn’s moon Titan.
DEVELOP
The gravitational force between two masses
and
is given by Equation 8.1:
F
Gm m
r
=
2
, where
r
is
their distance of separation. So for an object of mass
m
near a celestial body of mass
M
and radius
R
, the
gravitational force between them is
F
GMm
R
=
2
. By Newton’s second law,
a
=
, the gravitational acceleration near
the surface of the gravitating body is
a
GM
R
=
2
8
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View Full Document8.2
Chapter 8
EVALUATE
With reference to the first two columns in Appendix E, we find
(a)
g
GM
R
Mercury
Mercury
Mercury
2
Nm
==
×
−
(.
667 10
11
⋅
2
2
2
4
62
0 330
10
244 10
370
/)
(
.
)
)
./
kg
kg
m
m
×
×
=
s
2
(b)
g
GM
R
Titan
Titan
Titan
2
Nmk
g
×
−
/
)
11
2
2
⋅
)
)
0 135 10
258 10
135
24
2
×
×
=
kg
m
ms
ASSESS
The measured values are
g
Mercury
=
2
and
g
Titan
=
14
2
.
So our results are in good agreement
with the data.
16.
Newton’s law of the universal gravitation (Equation 8.1), with
mmm
12
, gives
m
Fr
G
×
×⋅
−
−
26
2
11
2
025 10
014
.(
.
)
.
/
.
/
kg
kg
2
857
=
(Newton’s law holds as stated, for two uniform spherical bodies, if the distance is taken between their centers.)
17.
INTERPRET
In this problem we want to find the gravitational force between the astronaut and the space shuttle.
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 Fall '10
 Pheong

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