A small ball of mass m is released at a heigh | vedclass

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(D) Let the initial position be at height $R$ above the surface (distance $2R$ from the center) and the final position be at depth $R/2$ inside the Ea

Vedclass · Accepted Answer

$\frac{7GMm}{R^3}$

Vedclass · Answer

(D) Let the initial position be at height $R$ above the surface (distance $2R$ from the center) and the final position be at depth $R/2$ inside the Earth (distance $R/2$ from the center).Initial potential energy $U_i = -\frac{GMm}{2R}$.Final potential energy at depth $d = R/2$ inside the Earth is given by $U_f = -\frac{GMm}{2R^3}(3R^2 - d^2) = -\frac{GMm}{2R^3}(3R^2 - (R/2)^2) = -\frac{GMm}{2R^3}(3R^2 - R^2/4) = -\frac{GMm}{2R^3}(\frac{11R^2}{4}) = -\frac{11GMm}{8R}$.Spring potential energy at compression $x = R/2$ is $U_s = \frac{1}{2}Kx^2 = \frac{1}{2}K(R/2)^2 = \frac{KR^2}{8}$.By the law of conservation of energy,$K_i + U_i = K_f + U_f + U_s$.Since the ball starts from rest and comes to rest momentarily,$K_i = 0$ and $K_f = 0$.$0 - \frac{GMm}{2R} = -\frac{11GMm}{8R} + \frac{KR^2}{8}$.$\frac{KR^2}{8} = \frac{11GMm}{8R} - \frac{GMm}{2R} = \frac{11GMm - 4GMm}{8R} = \frac{7GMm}{8R}$.$K = \frac{7GMm}{R^3}$.

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