A body of mass m kg. starts falling from a point 2R above the Earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the Earth’s surface [R-Radius of Earth, M-Mass of Earth, G-Gravitational Constant]

\frac{1}{2} \frac{G M m}{R}

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\frac{1}{6} \frac{G M m}{R}

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\frac{2}{3} \frac{G M m}{R}

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\frac{1}{3} \frac{G M m}{R}

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Solution

The correct option is B $\frac{1}{6} \frac{G M m}{R}$
Potential energy $U = \frac{- G M m}{r} = - \frac{G M m}{R + h} U_{i n i t i a l} = - \frac{G M m}{3 R} a n d U_{f i n a l} = - \frac{- G M m}{2 R} L o s s i n P E = g a i n i n K E = \frac{G M m}{2 R} - \frac{G M m}{3 R} = \frac{G M m}{6 R}$

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A body of mass m kg. starts falling from a point 2R above the Earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the Earth’s surface [R-Radius of Earth, M-Mass of Earth, G-Gravitational Constant]

The correct option is B 16GMmRPotential energy U=−GMmr=−GMmR+hUinitial=−GMm3R and Ufinal=−−GMm2RLoss in PE=gain in KE=GMm2R−GMm3R=GMm6R

The correct option is B $\frac{1}{6} \frac{G M m}{R}$
Potential energy $U = \frac{- G M m}{r} = - \frac{G M m}{R + h} U_{i n i t i a l} = - \frac{G M m}{3 R} a n d U_{f i n a l} = - \frac{- G M m}{2 R} L o s s i n P E = g a i n i n K E = \frac{G M m}{2 R} - \frac{G M m}{3 R} = \frac{G M m}{6 R}$