Gravitation
127
E
XECUTE
:
11
2
2
22
2
2
6
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
⎛
⎞⎛
⎞
=
=
=
=
⎜
⎟⎜
⎟
⎝
⎠⎝
⎠
.
E
VALUATE
:
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
ω
we
calculated,
rad
a
g
=
at a point that is precisely 4.25 m from the rotation axis.
12.23.
I
DENTIFY
and
S
ET
U
P
:
Example 12.5 gives the escape speed as
1
2
/
,
v
GM R
=
where
M
and
R
are the mass and
radius of the astronomical object.
E
XECUTE
:
11
2
2
12
1
2(6.673
10
N
m
/kg
)(3.6
10
kg)/700 m
0.83 m/s.
v
−
=
×
⋅
×
=
E
VALUATE
:
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
,
v
GR
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.
I
DENTIFY
:
In part (a) use the expression for the escape speed that is derived in Example 12.5.
In part (b) apply
conservation of energy.
S
ET
U
P
:
3
4.5
10
m
R
=
×
. In part (b) let point 1 be at the surface of the comet.
E
XECUTE
:
(a)
The escape speed is
2
GM
v
R
=
so
2
3
2
13
11
2
2
(4.5
10
m)(1.0 m/s)
3.37
10
kg
2
2(6.67
10
N m /kg )
Rv
M
G
−
×
=
=
=
×
×
⋅
.
(b)
(i)
2
1
1
1
2
K
mv
=
.
2
1
0.100
K
K
=
.
1
GMm
U
R
= −
;
2
GMm
U
r
= −
.
1
1
2
2
K
U
K
U
+
=
+
gives
2
2
1
1
1
1
2
2
(0.100)(
)
GMm
GMm
mv
mv
R
r
−
=
−
.
Solving for
r
gives
2
2
1
3
11
2
2
13
1
1
0.450
1
0.450(1.0 m/s)
4.5
10
m
(6.67
10
N
m
/kg
)(3.37
10
kg)
v
r
R
GM
−
=
−
=
−
×
×
⋅
×
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.
E
VALUATE
:
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.
I
DENTIFY
:
The escape speed, from the results of Example 12.5, is
2
/
.
GM R
S
ET
U
P
:
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
=
×
.
E
XECUTE
:
(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.
E
VALUATE
:
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.
I
DENTIFY
:
The kinetic energy is
2
1
2
K
mv
=
and the potential energy is
GMm
U
r
= −
S
ET
U
P
:
The mass of the earth is
24
E
5.97
10
kg
M
=
×
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Energy, Kinetic Energy, Mass, Potential Energy, kg, Celestial mechanics

Click to edit the document details