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(B) Let $M$ be the mass of the earth and $R$ be its radius. The body of mass $m$ is initially at a height $R$ above the surface,so its distance from t

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$\sqrt{ gR }$

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(B) Let $M$ be the mass of the earth and $R$ be its radius. The body of mass $m$ is initially at a height $R$ above the surface,so its distance from the center of the earth is $r_A = R + R = 2R$.Using the Law of Conservation of Mechanical Energy between point $A$ (at height $R$) and point $B$ (at the surface):$E_A = E_B$$K_A + U_A = K_B + U_B$Since the body is initially at rest,$K_A = 0$.$0 - \frac{G M m}{2R} = \frac{1}{2} m v^2 - \frac{G M m}{R}$$\frac{1}{2} m v^2 = \frac{G M m}{R} - \frac{G M m}{2R}$$\frac{1}{2} m v^2 = \frac{G M m}{2R}$$v^2 = \frac{G M}{R}$Since $g = \frac{G M}{R^2}$,we have $G M = g R^2$.Substituting this into the equation for $v^2$:$v^2 = \frac{g R^2}{R} = g R$$v = \sqrt{g R}$

$A$ body which is initially at rest at a height $R$ above the surface of the earth of radius $R$,falls freely towards the earth. Its velocity on reaching the surface of the earth is:

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