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(B) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.At the earth's surface,$r_1

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$\frac{3}{4}mgR$

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(B) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.At the earth's surface,$r_1 = R$,so $U_1 = -\frac{GMm}{R}$.At a height $h = 3R$,the distance from the center is $r_2 = R + h = R + 3R = 4R$.So,$U_2 = -\frac{GMm}{4R}$.The change in gravitational potential energy is $\Delta U = U_2 - U_1 = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.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,$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

$A$ body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h = 3R$. The change in gravitational potential energy of the body is

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