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(D) The total mechanical energy is conserved. Let $r_1$ be the initial distance from the center of the Earth and $r_2$ be the final distance from the

Vedclass · Accepted Answer

$\frac{1}{4} \frac{GMm}{R}$

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(D) The total mechanical energy is conserved. Let $r_1$ be the initial distance from the center of the Earth and $r_2$ be the final distance from the center of the Earth.Initial distance from center: $r_1 = 3R + R = 4R$.Final distance from center: $r_2 = R + R = 2R$.Initial potential energy: $U_1 = -\frac{GMm}{r_1} = -\frac{GMm}{4R}$.Final potential energy: $U_2 = -\frac{GMm}{r_2} = -\frac{GMm}{2R}$.Since the body starts from rest,initial kinetic energy $K_1 = 0$.By the law of conservation of energy: $K_1 + U_1 = K_2 + U_2$.$0 + (-\frac{GMm}{4R}) = K_2 + (-\frac{GMm}{2R})$.$K_2 = \frac{GMm}{2R} - \frac{GMm}{4R} = \frac{GMm}{4R}$.

$A$ body of mass $m$ starts falling from a distance $3R$ above the Earth's surface. When it reaches a distance $R$ above the surface of the Earth (radius $R$,mass $M$),its kinetic energy is:

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