Muraj, Hamza – Quiz 4 – Due: May 3 2006, 10:00 pm – Inst: Florin
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
Two iron spheres,
of mass
m
and 2
m
,
respectively,
and equally spaced points “
r
apart” are shown in the figure.
A
m
B
C
D
2
m
E
r
r
r
r
r
r
At which location would the net gravita
tional force on an object due to these two
spheres be a minimum?
1.
D
2.
C
correct
3.
B
4.
E
5.
A
Explanation:
Let
F
=
G
m
0
m
r
2
, the mass of an object
be
m
0
, and the distance between adjacent
locations and the centers of the spheres be
r
.
F
A
=
G
m
0
m
r
2
+
G
m
0
(2
m
)
(5
r
)
2
=
27
25
F
F
B
=
G
m
0
m
r
2

G
m
0
(2
m
)
(3
r
)
2
=
7
9
F
F
C
=
G
m
0
(2
m
)
(2
r
)
2

G
m
0
m
(2
r
)
2
=
1
4
F
F
D
=
G
m
0
(2
m
)
r
2

G
m
0
m
(3
r
)
2
=
17
9
F
F
E
=
G
m
0
(2
m
)
r
2
+
G
m
0
m
(5
r
)
2
=
51
25
F
Thus the gravitational force at
C
is mini
mum.
002
(part 1 of 1) 10 points
Two satellites have circular orbits about the
same planet.
The masses of the two satellites are respec
tively
m
m
=
m
and
m
3
m
= 3
m
. Both satel
lites have circular orbits with respective radii
r
m
=
r
and
r
3
m
= 2
r
.
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
3
3.
v
3
m
v
m
=
1
9
4.
v
3
m
v
m
=
1
2
5.
v
3
m
v
m
= 3
6.
v
3
m
v
m
=
√
2
7.
v
3
m
v
m
=
√
3
8.
v
3
m
v
m
= 9
9.
v
3
m
v
m
=
1
√
3
10.
v
3
m
v
m
=
1
√
2
correct
Explanation:
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Muraj, Hamza – Quiz 4 – Due: May 3 2006, 10:00 pm – Inst: Florin
2
The only force acting on an orbiting satel
lite is the force of gravity,
F
=
G M m
r
2
,
where
M
is the planet’s mass and
m
is the
mass of the satellite itself. Consequently, the
satellite is in free fall with acceleration
g
=
G M
r
2
directed towards the planet’s center.
Note:
This acceleration is independent on
the satellite’s mass
m
.
For a circular orbit, the freefall accelera
tion equals the centripetal acceleration,
g
=
a
c
=
v
2
r
,
and therefore
G M
r
2
=
v
2
r
=
⇒
v
=
r
G M
r
,
regardless of the satellite’s mass
m
.
Conse
quently, for two satellites in circular orbits
around the same planet,
v
3
m
v
m
=
r
G M
r
3
m
r
G M
r
m
=
r
r
m
r
3
m
=
r
r
2
r
=
1
√
2
.
003
(part 1 of 1) 10 points
A “synchronous” satellite, which always re
mains above the same point on a planet’s
equator, is put in orbit about a planet similar
to Jupiter.
This planet rotates once every 5
.
3 h, has a
mass of 1
.
9
×
10
27
kg and a radius of 6
.
99
×
10
7
m.
Given that
G
= 6
.
67
×
10

11
N m
2
/
kg
2
,
calculate how far above Jupiter’s surface the
satellite must be.
Correct answer: 3
.
54316
×
10
7
m.
Explanation:
Basic Concepts: Solution:
According to Kepler’s third law:
T
2
=
4
π
2
G M
r
3
where
r
is the radius of the satellite’s orbit.
Thus, solving for
r
r
=
‰
G
M T
2
4
π
2
1
3
=
{
6
.
67
×
10

11
N m
2
/
kg
2
}
1
3
×
‰
(1
.
9
×
10
27
kg) (19080 s)
2
4 (3
.
1415926536 )
2
1
3
= 1
.
05332
×
10
8
m
.
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 Spring '09
 Turner
 Mass, General Relativity, Correct Answer, Gravitational constant, Muraj

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