The height at which the weight of a body beco | vedclass

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Question

$\frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}}$

$\frac{\mathrm{g} / 16}{\mathrm{g}}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$

$\mathrm{h}=3

Vedclass · Accepted Answer

$3R$

Vedclass · Answer

$\frac{g_{h}}{g}=\frac{R^{2}}{(R+h)^{2}}$

$\frac{\mathrm{g} / 16}{\mathrm{g}}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$

$\mathrm{h}=3 \mathrm{R}$

The height at which the weight of a body becomes $1/16^{th}$, its weight on the surface of earth (radius $R$), is

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