The gravitational potential energy of a body of mass 'm' at the earth's surface $- m g R_{e}$ . Its gravitational potential energy at a height $R_{e}$ from the earth's surface will be (Here $R_{e}$ is the radius of the earth)

$- 2 m g R_{e}$

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$2 m g R_{e}$

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$\frac{1}{2} m g R_{e}$

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$- \frac{1}{2} m g R_{e}$

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Solution

The correct option is D
$- \frac{1}{2} m g R_{e}$

$△ U = U_{2} - U_{1} = \frac{m g h}{1 + \frac{h}{R_{e}}} = \frac{m g R_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{m g R_{e}}{2}$
$\Rightarrow U_{2} - (- m g R_{e}) = \frac{m g R_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2} m g R_{e}$

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