The height ' $h$ ' above the earth's surface at which the value of acceleration due to gravity ( g ) becomes $\left(\frac{\mathrm{g}}{3}\right)$ is ( $\mathrm{R}=$ radius of the earth)

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Aicharya · Accepted Answer

We know that, $\begin{aligned} & \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}ight)^2 \ & 	ext { For } g_h=\frac{g}{3} \ & \frac{g}{3}=g\left(\frac{R}{R+h}ight)^2 \ & \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \ & \sqrt{3} \mathrm{R}=\mathrm{R}+\mathrm{h} \ & 	herefore \quad h=(\sqrt{3}-1) \mathrm{R} \end{aligned}$

The height ' $h$ ' above the earth's surface at which the value of acceleration due to gravity ( g ) becomes…

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