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Patel (sap785) – homework #6 – shubeita – (58645)
1
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beFore answering.
001
10.0 points
Two satellites have circular orbits about the
same planet. Their masses are
m
and 3
m
,
respectively, with orbits oF radii
r
and 2
r
,
respectively.
2
r
r
3
m
m
What is the ratio oF the orbital speeds oF
the two satellites?
1.
v
3
m
v
m
=
√
2
2.
v
3
m
v
m
=
1
√
2
correct
3.
v
3
m
v
m
=
1
2
4.
v
3
m
v
m
= 9
5.
v
3
m
v
m
= 3
6.
v
3
m
v
m
= 2
7.
v
3
m
v
m
=
1
9
8.
v
3
m
v
m
=
1
3
9.
v
3
m
v
m
=
√
3
10.
v
3
m
v
m
=
1
√
3
Explanation:
The only Force acting on an orbiting satel
lite is the Force oF gravity
F
=
GM m
r
2
,
where
M
is the planet’s mass and
m
is the mass oF
the satellite itselF. Consequently, the satel
lite is in Free Fall with acceleration
g
=
GM
r
2
directed towards the planet’s center.
±or a circular orbit, the FreeFall accel
eration equals the centripetal acceleration
a
c
=
v
2
r
,
so
GM
r
2
=
v
2
r
v
=
r
GM
r
∝
r
1
r
,
regardless oF the satellite’s mass
m
.
Consequently, For two satellites in circular
orbits around the same planet,
v
3
m
v
m
=
r
r
m
r
3
m
=
r
r
2
r
=
1
√
2
.
002
10.0 points
An apparatus like the one Cavendish used to
fnd
G
has a large lead ball that is 7
.
8 kg in
mass and a small one that is 0
.
03 kg. Their
centers are separated by 0
.
055 m.
±ind the Force oF attraction between them.
The value oF the universal gravitational con
stant is 6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
.
Correct answer: 5
.
16161
×
10
−
9
N.
Explanation:
Let :
d
= 0
.
055 m
,
G
= 6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
,
m
1
= 7
.
8 kg
,
and
m
2
= 0
.
03 kg
.
By Newton’s universal law oF gravitation,
F
=
G
m
1
m
2
d
2
= (6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
)
(7
.
8 kg) (0
.
03 kg)
0
.
055 m
2
=
5
.
16161
×
10
−
9
N
.
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View Full DocumentPatel (sap785) – homework #6 – shubeita – (58645)
2
003
10.0 points
Calculate the mass of the earth from the pe
riod of the moon (27
.
3 d) and its mean orbital
radius of 3
.
84
×
10
8
m.
The universal gravitational constant is
6
.
673
×
10
−
11
N
·
m
2
/
kg
2
.
Correct answer: 6
.
02115
×
10
24
kg.
Explanation:
Let :
T
m
= 27
.
3 d
,
r
m
= 3
.
84
×
10
8
m
,
and
G
= 6
.
673
×
10
−
11
N
·
m
2
/
kg
2
.
Using Kepler’s third law
T
M
2
=
4
π
2
GM
E
r
m
3
M
s
=
4
π
2
r
m
3
GT
m
2
=
4
π
2
(3
.
84
×
10
8
m)
3
6
.
673
×
10
−
11
N
·
m
2
/
kg
2
×
p
1
27
.
3 d
P
2
p
1 d
24 h
P
2
p
1 h
3600 s
P
2
=
6
.
02115
×
10
24
kg
.
004
10.0 points
Given:
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
Calculate the work required to move a
planet’s satellite of mass 1200 kg from a cir
cular orbit of radius 2
R
to one of radius 3
R
,
where 7
.
1
×
10
6
m is the radius of the planet.
The mass of the planet is 4
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