If, the earth's radius= 6.4 x 10^6 m earth's mass= 6 x 10^34 kg angular velocity of the earth and sattelite = 7.3 x 10^-5 rads-1 and the mass of the sattelite is 50 kg, what is the radius of the orbit of the satellite. This is a gravitational field question.
Answers
Answer:
Mass of the Earth, M=6.0×10
24
kg
m=200 kg
R
e
=6.4×10
6
m
G=6.67×10
−11
Nm
2
kg
−2
Height of the satellite,h=400 km=4×10
5
m
Total energy of the satellite at height h=(1/2)mv
2
+(−G
(R
e
+h)
M
e
m
)
Orbital velocity of the satellite, v=
R
e
+h
GM
e
Total energy at height h =
2
1
R
e
+h
GM
e
m
−
R
e
+h
GM
e
m
Total Energy=−
2
1
R
e
+h
GM
e
m
The negative sign indicates that the satellite is bound to the Earth.
Energy required to send the satellite out of its orbit = – (Bound energy)
=
2(R
e
+h)
GM
e
m
=
2(6.4×10
6
+4×10
5
)
6.67×10
−11
×6×10
24
×200
=5.9×10
9
J
If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply 5.9×10
9
J of energy to just escape it.
Explanation:
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