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Gravitation
127
EXECUTE:
11
2
2
22
2
26
2
(6.67 10
N m /kg )(4.8 10 kg)
1.30 m/s
(1.569 10 m)
Gm
g
R
−
×⋅
×
==
=
×
.
rad
g
a
=
gives
2
1.30 m/s
60 s
1 rev
(0.553 rad/s)
5.28 rpm
4.25 m
1 min
2 rad
g
r
ω
π
⎛⎞
⎛
⎞
=
=
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
.
EVALUATE:
The radius of Europa is about onefourth that of the earth and its mass is about onehundredth that of
earth, so
g
on Europa is much less than
g
on earth.
The lander would have some spatial extent so different points
on it would be different distances from the rotation axis and
rad
a
would have different values.
For the
we
calculated,
rad
ag
=
at a point that is precisely 4.25 m from the rotation axis.
12.23.
IDENTIFY
and
SET UP:
Example 12.5 gives the escape speed as
1
2/
,
vG
M
R
=
where
M
and
R
are the mass and
radius of the astronomical object.
EXECUTE:
11
2
2
12
1
2(6.673 10
N m /kg )(3.6 10 kg)/700 m
0.83 m/s.
v
−
=×
⋅
×
=
EVALUATE:
At this speed a person can walk 100 m in 120 s; easily achieved for the average person. We can
write the escape speed as
2
4
1
3
,
R
πρ
=
where
ρ
is the average
density of Dactyl. Its radius is much smaller
than earth&s and its density is about the same, so the escape speed is much less on Dactyl than on earth.
12.24.
IDENTIFY:
In part (a) use the expression for the escape speed that is derived in Example 12.5.
In part (b) apply
conservation of energy.
SET UP:
3
4.5 10 m
R
. In part (b) let point 1 be at the surface of the comet.
EXECUTE:
(a)
The escape speed is
2
GM
v
R
=
so
232
13
11
2
2
(4.5 10 m)(1.0 m/s)
3.37 10 kg
22
(
6
.
6
7
1
0
N
m
/
k
g
)
Rv
M
G
−
×
.
(b)
(i)
2
1
11
2
K
mv
=
.
21
0.100
K
K
=
.
1
GMm
U
R
=−
;
2
GMm
U
r
.
2 2
K
UKU
+
=+
gives
(0.100)(
)
GMm
GMm
mv
mv
R
r
−=
−
.
Solving for
r
gives
1
31
1
2
2
1
3
1
1
0.450
1
0.450(1.0 m/s)
(6.67 10
N m /kg )(3.37 10 kg)
v
rR G
M
−
=
−
××
⋅
×
and
45 km
r
=
.
(ii) The debris never
loses all of its initial kinetic energy, but
2
0
K
→
as
r
→∞
.
The farther the debris are from the comet&s center, the
smaller is their kinetic energy.
EVALUATE:
The debris will have lost 90.0% of their initial kinetic energy when they are at a distance from the
comet&s center of about ten times the radius of the comet.
12.25.
IDENTIFY:
The escape speed, from the results of Example 12.5, is
.
GM R
SET UP:
For Mars,
23
6.42 10 kg
M
and
6
3.40 10 m
R
.
For Jupiter,
27
1.90 10 kg
M
and
7
6.91 10 m
R
.
EXECUTE:
(a)
11
2
2
23
6
3
2(6.673 10
N m /kg )(6.42 10 kg)/(3.40 10 m)
5.02 10 m/s.
v
−
⋅
×
×
=
×
(b)
11
2
2
27
7
4
2(6.673 10
N m /kg (1.90 10
kg)/(6.91 10 m)
6.06 10 m/s.
v
−
⋅
×
×
=
×
(c)
Both the kinetic energy and the gravitational potential energy are proportional to the mass of the spacecraft.
EVALUATE:
Example 12.5 calculates the escape speed for earth to be
4
1.12 10 m/s
×
.
This is larger than our
result for Mars and less than our result for Jupiter.
12.26.
IDENTIFY:
The kinetic energy is
2
1
2
K
mv
=
and the potential energy is
GMm
U
r
SET UP:
The mass of the earth is
24
E
5.97 10 kg
M
.
EXECUTE:
(a)
32
9
1
2
(629 kg)(3.33 10 m/s)
3.49 10 J
K
=
×
(b)
11
2
2
24
7
E
9
(6.673 10
N m /kg )(5.97 10 kg)(629 kg)
8.73 10 J
2.87 10 m
GM m
U
r
−
×
×
×
.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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