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(B) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.The initi

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

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(B) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.The initial position is $2R$ above the Earth's surface,so the distance from the center is $r_i = R + 2R = 3R$.Initial potential energy: $U_i = -\frac{GMm}{3R}$.The final position is $R$ above the Earth's surface,so the distance from the center is $r_f = R + R = 2R$.Final potential energy: $U_f = -\frac{GMm}{2R}$.According to the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy $(KE)$:$KE = U_i - U_f = -\frac{GMm}{3R} - (-\frac{GMm}{2R})$$KE = \frac{GMm}{2R} - \frac{GMm}{3R} = \frac{3GMm - 2GMm}{6R} = \frac{GMm}{6R}$.

$A$ body of mass $m$ starts falling from a point $2R$ above the Earth's surface. What is 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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