If $g$ is the acceleration due to gravity on the earth's surface, the change in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth is

\frac{m g R}{2}

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2 m g R

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m g R

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- m g R

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Solution

The correct option is A $\frac{m g R}{2}$
$Gain in potential energy = △ U = U_{2} - U_{1}$
$△ U = (- \frac{G M m}{2 R}) - (- \frac{G M m}{R})$
$△ U = \frac{G M m}{R} - \frac{G M m}{2 R}$
$△ U = \frac{G M m}{2 R}$
$△ U = \frac{m g R^{2}}{2 R} A s, G M = g R^{2}$
$△ U = \frac{m g R}{2}$

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