A body of mass m rises to a height h = R/5 fr | vedclass

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(C) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.At height $h$,the potential energy is $U_f = -\frac{GMm}{

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$\frac{5}{6}mgh$

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(C) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.At height $h$,the potential energy is $U_f = -\frac{GMm}{R+h}$.The increase in potential energy is $\Delta U = U_f - U_i = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = GMm \left( \frac{h}{R(R+h)} \right)$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting $GM$,we get $\Delta U = \frac{gR^2 mh}{R(R+h)} = \frac{mgh}{1 + h/R}$.Given $h = R/5$,so $h/R = 1/5$.Therefore,$\Delta U = \frac{mgh}{1 + 1/5} = \frac{mgh}{6/5} = \frac{5}{6}mgh$.

$A$ body of mass $m$ rises to a height $h = R/5$ from the earth's surface,where $R$ is the earth's radius. If $g$ is the acceleration due to gravity at the earth's surface,the increase in potential energy is:

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