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Question

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

A
12 mv2
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B
mv2
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C
32 mv2
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D
2mv2
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Solution

The correct option is B mv2

At height r from center of earth, orbital velocity of a satellite of mass m around the earth of mass M,
v=GMr...(1) Just to escape from the gravitational pull, its total mechanical energy should be zero.

K.E+P.E=0

K.E+(GMmr)=0

Substituting the value of GM from (1), we get

K.E+v2rmr=0

K.E=mv2

Hence, option (b) is correct answer.

Alternate Solution:

In circular orbit of a satellite,

Potential energy =2× kinetic energy

=2×12mv2

=mv2

Just to escape from the gravitational pull, its total mechanical energy should be zero.i.e.

K.E+P.E=0

K.E=P.E=+mv2

Therefore, its K.E should be +mv2

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