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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 $U = -\frac{GMm}{r}$.At the su

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$\frac{4 W}{3}$

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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 $U = -\frac{GMm}{r}$.At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.For a height $h = R$,the distance from the center is $r = R + R = 2R$. The potential energy is $U_f = -\frac{GMm}{2R}$.The energy required $W$ is the change in potential energy:$W = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{2R}$.For a height $h = 2R$,the distance from the center is $r = R + 2R = 3R$. The potential energy is $U_f' = -\frac{GMm}{3R}$.The energy required $W'$ is:$W' = U_f' - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{2GMm}{3R}$.Since $W = \frac{GMm}{2R}$,we have $\frac{GMm}{R} = 2W$.Substituting this into the expression for $W'$:$W' = \frac{2}{3} 	imes (2W) = \frac{4W}{3}$.

The energy required to take a body from the surface of the earth to a height equal to the radius of the earth is $W$. The energy required to take this body from the surface of the earth to a height equal to twice the radius of the earth is:

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