A particle is dropped on Earth from a height | vedclass

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(B) The gravitational potential energy at a distance $r$ from the center of the Earth is $U = -\frac{GMm}{r}$.Using the conservation of energy,the vel

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$\sqrt{\frac{2}{3}}$

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(B) The gravitational potential energy at a distance $r$ from the center of the Earth is $U = -\frac{GMm}{r}$.Using the conservation of energy,the velocity $v$ of the particle just before hitting the surface (at distance $R$ from the center) when dropped from height $h$ is given by $\frac{1}{2}mv^2 = GMm(\frac{1}{R} - \frac{1}{R+h})$.For the initial drop from height $h_1 = R$,the velocity of approach $v_1$ is $\frac{1}{2}mv_1^2 = GMm(\frac{1}{R} - \frac{1}{2R}) = \frac{GMm}{2R}$,so $v_1 = \sqrt{\frac{GM}{R}}$.For the rebound to height $h_2 = R/2$,the velocity of separation $v_2$ is $\frac{1}{2}mv_2^2 = GMm(\frac{1}{R} - \frac{1}{R + R/2}) = GMm(\frac{1}{R} - \frac{2}{3R}) = \frac{GMm}{3R}$,so $v_2 = \sqrt{\frac{2GM}{3R}}$.The coefficient of restitution $e$ is defined as $e = \frac{v_2}{v_1} = \frac{\sqrt{2GM/3R}}{\sqrt{GM/R}} = \sqrt{\frac{2}{3}}$.

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

$A$ particle is dropped on Earth from a height $R$ (radius of Earth) and it bounces back to a height $R/2$. The coefficient of restitution for the collision is (ignore air resistance and rotation of Earth).

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