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(A) The gravitational potential energy at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.At the surface of the earth,the

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$\frac{2}{3}mgR$

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(A) The gravitational potential energy at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.At the surface of the earth,the distance from the center is $r_i = R$.Therefore,the initial potential energy is $U_i = -\frac{GMm}{R}$.At a height $h = 2R$ from the surface,the distance from the center is $r_f = R + h = R + 2R = 3R$.Therefore,the final potential energy is $U_f = -\frac{GMm}{3R}$.The change in potential energy is $\Delta U = U_f - U_i$.$\Delta U = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R}(1 - \frac{1}{3}) = \frac{2}{3}\frac{GMm}{R}$.Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this,$\Delta U = \frac{2}{3} \frac{(gR^2)m}{R} = \frac{2}{3}mgR$.

$A$ body of mass $m$ is taken from the earth's surface to a height equal to twice the radius $(R)$ of the earth. The change in potential energy of the body will be

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