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The height at which the weight of the body becomes $\left(\frac{1}{9}\right)^{\text {th }}$ its weight on the surface of earth is ( $R=$ radius of earth)
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$2 \mathrm{R}$
$\begin{array}{ll} & \mathrm{W}_{\mathrm{h}}=\frac{\mathrm{W}}{9} \\ & \mathrm{mg}_{\mathrm{h}}=\frac{\mathrm{mg}}{9} \\ & \mathrm{~g}_{\mathrm{h}}=\frac{\mathrm{g}}{9} \\ \therefore \quad & \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{GM}}{9 \mathrm{R}^2} \\ \therefore \quad & \mathrm{R}+\mathrm{h}=3 \mathrm{R} \\ \therefore \quad & \mathrm{h}=2 \mathrm{R}\end{array}$
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