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IIT JEE Advanced 2011 Paper 2
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© examsnet.com
Question : 6
Total: 60
A satellite is moving with a constant speed V in a circular orbit about the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is
1
2
m
V
2
m
V
2
3
2
m
V
2
2
m
V
2
Validate
Solution:
The escape velocity and orbital velocity is given by
V
escape
=
√
2
V
orbit
Therefore, the velocity of the mass m is
V' =
√
2
V
Thus, the kinetic energy is
1
2
m
V
′
2
=
1
2
m
(
√
2
V
)
2
=
m
V
2
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