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(B) The gravitational potential energy on the surface of the earth is given by $U_1 = -\frac{GMm}{R}$,where $M$ is the mass of the earth and $R$ is th

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$\left(\frac{3}{4}\right) mgR$

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(B) The gravitational potential energy on the surface of the earth is given by $U_1 = -\frac{GMm}{R}$,where $M$ is the mass of the earth and $R$ is the radius of the earth.At a height $h = 3R$,the distance from the center of the earth is $r = R + h = R + 3R = 4R$.The gravitational potential energy at this height is $U_2 = -\frac{GMm}{4R}$.The change in gravitational potential energy is $\Delta U = U_2 - U_1$.$\Delta U = -\frac{GMm}{4R} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.Using the relation $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the expression for $\Delta U$:$\Delta U = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

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

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