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

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$\frac{1}{2} m v^{2}$

$2 m v^{2}$

$\frac{3}{2} m v^{2}$

$m v^{2}$

answer is B.

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Detailed Solution

In circular orbit of satellite , potential energy
$= 2 \times (k i n e t i c e n e r g y) = - 2 \times \frac{1}{2} m v^{2} = - m v^{2} (∵ K E = \frac{G m M}{2 r}, P E = \frac{- G m M}{r})$
Just to escape from the gravitational pull, its total mechanical energy should be zero.

Therefore, its kinetic energy should be $+ m v^{2}$ .

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