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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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$\frac{5}{6} \mathrm{mgh}$
Given $\mathrm{h}=\frac{\mathrm{R}}{5}$
Potential Energy at surface of the earth, $\left(\mathrm{PE}_1\right)=-\frac{\mathrm{GMm}}{\mathrm{R}}$
Potential Energy at height $h=\frac{R}{5},\left(P_2\right)=-\frac{G M m}{R+\frac{R}{5}}=-\frac{5 G M m}{6 R}$
Increase in Potential energy $=\mathrm{PE}_2-\mathrm{PE}_1=-\frac{5 \mathrm{GMm}}{6 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMm}}{6 \mathrm{R}}$
(Using $g=\frac{G M}{R^2}$ and $h=\frac{R}{5}$ ) $\Rightarrow P_2-P_1=\frac{5}{6} \mathrm{mgh}$
Potential Energy at surface of the earth, $\left(\mathrm{PE}_1\right)=-\frac{\mathrm{GMm}}{\mathrm{R}}$
Potential Energy at height $h=\frac{R}{5},\left(P_2\right)=-\frac{G M m}{R+\frac{R}{5}}=-\frac{5 G M m}{6 R}$
Increase in Potential energy $=\mathrm{PE}_2-\mathrm{PE}_1=-\frac{5 \mathrm{GMm}}{6 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMm}}{6 \mathrm{R}}$
(Using $g=\frac{G M}{R^2}$ and $h=\frac{R}{5}$ ) $\Rightarrow P_2-P_1=\frac{5}{6} \mathrm{mgh}$
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