A satellite is moving with a constant speed $v$ in circular orbit around 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 ejection, the kinetic energy of the object is

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

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

$2 m v^{2}$

$m v^{2}$

answer is D.

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Detailed Solution

At height r from center of earth orbital velocity = $\sqrt{\frac{G M}{r}}$
∴ By energy conservation

$K i n e t i c e n e r g y o f m + (- \frac{G M M}{R}) = 0 + 0$
(At infinity, PE=KE=0)
Kinetic energy of 'm' = $\frac{G M m}{r} = {(\sqrt{\frac{G M}{r}})}^{2} m = m v^{2}$

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