homework 07 – FRENNEA, KYLE – Due: Mar 7 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
On the way from a planet to a moon as
tronauts reach a point where that moon’s
gravitational pull is stronger than that of the
planet.
The masses of the planet and the
moon are, respectively, 5
.
38
×
10
24
kg and
7
.
36
×
10
22
kg.
The distance from the cen
ter of the planet to the center of the moon is
3
.
29
×
10
8
m.
Determine the distance of this point from
the center of the planet.
Correct answer: 2
.
94549
×
10
8
m (tolerance
±
1 %).
Explanation:
If
r
p
is the distance from this point to the
center of the planet and
r
m
is the distance
from this point to the center of the moon,
then from the formula
G m M
p
r
2
p
=
G m M
m
r
2
m
,
we obtain
q
=
r
m
r
p
=
radicalBigg
M
m
M
p
=
radicalBigg
7
.
36
×
10
22
kg
5
.
38
×
10
24
kg
= 0
.
116963
.
On the other hand,
r
p
+
r
m
=
R .
Eliminating
r
m
from the last two equalities,
we obtain
r
p
=
R
q
+ 1
=
3
.
29
×
10
8
m
(0
.
116963) + 1
= 2
.
94549
×
10
8
m
.
Question 2
part 1 of 1
10 points
A satellite moves in a circular orbit around
the Earth at a speed of 6
.
7 km
/
s.
Determine the satellite’s altitude above the
surface of the Earth.
Assume the Earth
is a homogeneous sphere of radius
R
earth
=
6370 km and mass
M
earth
= 5
.
98
×
10
24
kg .
You will need
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
Correct answer: 2518
.
86
km (tolerance
±
1
%).
Explanation:
The gravitational force provides the cen
tripetal acceleration.
G m M
r
2
=
m v
2
r
,
where
M
is the mass of the Earth and
m
is
the mass of the satellite. Solving for
r
yields
r
=
G M
v
2
= 8
.
88886
×
10
6
m
Then the height of the satellite above the
Earth’s surface is
h
=
r

R
earth
= 2518
.
86 km
Question 3
part 1 of 2
10 points
Given:
G
= 6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
.
A 1
.
9 kg mass weighs 18
.
62 N on the surface
of a planet similar to Earth.
The radius of
this planet is roughly 5
.
6
×
10
6
m.
Calculate the mass of of this planet.
Correct answer: 4
.
60583
×
10
24
kg (tolerance
±
1 %).
Explanation:
By Newton’s Law of Universal Gravitation,
W
=
G
m M
planet
r
2
,
so
M
planet
=
W r
2
G m
=
(5
.
6
×
10
6
m)
2
(6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
)
×
18
.
62 N
1
.
9 kg
=
4
.
60583
×
10
24
kg
.
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homework 07 – FRENNEA, KYLE – Due: Mar 7 2007, 4:00 am
2
Question 4
part 2 of 2
10 points
Calculate the average density of this planet.
Correct answer: 6261
.
16 kg
/
m
3
(tolerance
±
1 %).
Explanation:
The volume of the planet is
V
=
4
3
π r
3
so its average density is
ρ
=
M
planet
V
=
3
M
planet
4
π r
3
=
3 (4
.
60583
×
10
24
kg)
4
π
(5
.
6
×
10
6
m)
3
=
6261
.
16 kg
/
m
3
.
Question 5
part 1 of 1
10 points
Given:
G
= 6
.
673
×
10
−
11
N
·
m
2
kg
2
Mars has a mass of about 6
.
40
×
10
23
kg,
and its moon Phobos has a mass of about
9
.
11
×
10
15
kg.
If the magnitude of the gravitational force
between the two bodies is 4.33
×
10
15
N, how
far apart are Mars and Phobos?
Correct answer: 9
.
47907
×
10
6
m (tolerance
±
1 %).
Explanation:
Basic Concept:
F
g
=
G
m
1
m
2
r
2
Given:
m
1
= 6
.
40
×
10
23
kg
m
2
= 9
.
11
×
10
15
kg
F
g
= 4
.
33
×
10
15
N
G
= 6
.
673
×
10
−
11
N
·
m
2
kg
2
Solution:
r
2
=
G m
1
m
2
F
g
=
parenleftbigg
6
.
673
×
10
−
11
N
·
m
2
kg
2
parenrightbigg
·
(
6
.
4
×
10
23
kg
) (
9
.
11
×
10
15
kg
)
4
.
33
×
10
15
N
= 8
.
98528
×
10
13
m
2
Thus
r
=
radicalbig
8
.
98528
×
10
13
m
2
= 9
.
47907
×
10
6
m
Question 6
part 1 of 1
10 points
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 Spring '08
 Turner
 Mass, Work, General Relativity, Correct Answer, KYLE

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