A body of mass m kg starts falling from a point 2R above the Earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the Earth’s surface [R Radius of Earth, M-Mass of Earth, G-Gravitational Constant]

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$\frac{1}{3} \frac{GMm}{R}$

$\frac{1}{2} \frac{GMm}{R}$

$\frac{1}{6} \frac{GMm}{R}$

$\frac{2}{3} \frac{GMm}{R}$

answer is C.

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Detailed Solution

Gravitational Potential energy $U = \frac{- GMm}{r} = \frac{- GMm}{R + h}$
$U_{initial} = - \frac{GMm}{3 R} and U_{final} = \frac{- GMm}{2 R} Loss in PE = gain inKE = \frac{GMm}{2 R} - \frac{GMm}{3 R} = \frac{GMm}{6 R}$

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