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A body of mass $m$ rises to height $h=R / 5$ from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is
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The correct answer is:
$\frac{5}{6} m g h$
$\Delta U=\frac{m g h}{1+h / R}$
Substituting $R=5 h$ we get
$\Delta U=\frac{m g h}{1+1 / 5}=\frac{5}{6} m g h$
Substituting $R=5 h$ we get
$\Delta U=\frac{m g h}{1+1 / 5}=\frac{5}{6} m g h$
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