A body of mass m is released from a height of | vedclass

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(B) According to the law of conservation of energy,the loss in gravitational potential energy is equal to the gain in kinetic energy.Initial height fr

Vedclass · Accepted Answer

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

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(B) According to the law of conservation of energy,the loss in gravitational potential energy is equal to the gain in kinetic energy.Initial height from the center of the Earth,$r_1 = R + 2R = 3R$.Final height from the center of the Earth,$r_2 = R + R = 2R$.Initial potential energy,$U_1 = -\frac{GMm}{r_1} = -\frac{GMm}{3R}$.Final potential energy,$U_2 = -\frac{GMm}{r_2} = -\frac{GMm}{2R}$.Change in potential energy,$\Delta U = U_1 - U_2 = -\frac{GMm}{3R} - (-\frac{GMm}{2R})$.$\Delta U = \frac{GMm}{2R} - \frac{GMm}{3R} = \frac{3GMm - 2GMm}{6R} = \frac{GMm}{6R}$.Since the body is released from rest,the gain in kinetic energy $K = \Delta U = \frac{1}{6} \frac{GMm}{R}$.

$A$ body of mass $m$ is released from a height of $2R$ above the Earth's surface. What will be its kinetic energy at a height of $R$ from the Earth's surface? (Where $R$ is the radius of the Earth)

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