A body of mass $m$ is raised to a height $10 R$ from the surface of the earth, where $R$ is the radius of the earth. The increase in potential energy is ( $G =$ universal constant of gravitation, $M =$ mass of the earth and $g =$ acceleration due to gravity)

\frac{G M m}{11 R}

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\frac{G M m}{10 R}

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\frac{m g R}{11 G}

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\frac{10 G M m}{11 R}

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Solution

The correct option is D $\frac{10 G M m}{11 R}$
PE at the earth surface $= \frac{- G M m}{R}$
PE at a height $h$ above the earth's surface $= \frac{- G M m}{(R + h)}$
$∴$ Change in PE $= \frac{- G M m}{(R + h)} - (- \frac{G M m}{R}) = \frac{G M m h}{R (R + h)}$
Substituting $h = 10 R$
$△ P E = \frac{G M m \times 10 R}{R (R + 10 R)} = \frac{10 G M m}{11 R}$

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