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(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.Initial p

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$\frac{3}{4}mgR$

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(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.Initial potential energy at the surface of the earth $(r = R)$: $U_i = -\frac{GMm}{R}$.Final potential energy at a height $h = 3R$ from the surface,the distance from the center is $r = R + 3R = 4R$. Thus,$U_f = -\frac{GMm}{4R}$.The change in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{4R} - (-\frac{GMm}{R})$.$\Delta U = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the expression: $\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

The radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth,then the change in potential energy will be:

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