A body of mass m is taken from earth's surface to q a height equal to radius of earth . the change in potential will be

m g

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\frac{1}{2} m g R

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2 m g R

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\frac{1}{4} m g R

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Solution

The correct option is D $\frac{1}{2} m g R$
Given,
Mass $= m,$
Height $h = R$
Potential energy at surface $U_{s} = \frac{- G M m}{R} = - m g R$ $(g R^{2} = G M)$
Potential energy at height $h = R$ , $U_{h} = - \frac{G M m}{R + R} = \frac{- m g R}{2}$
Change in potential energy $Δ U = \frac{- m g R}{2} - (- m g R) = \frac{m g R}{2}$

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