When the value of acceleration due to gravity ' $\mathrm{g}$ ' become $\frac{\mathrm{g}}{3}$ above surface of earth at height ' $h$ ' then relation between ' $h$ ' and ' $\mathrm{R}$ ' is $(\mathrm{R}= ext { radius of earth) }$

Question

When the value of acceleration due to gravity ' $\mathrm{g}$ ' become $\frac{\mathrm{g}}{3}$ above surface of earth at height ' $h$ ' then relation between ' $h$ ' and ' $\mathrm{R}$ ' is $(\mathrm{R}=	ext { radius of earth) }$

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g} \frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{g}}{3} \ & \frac{1}{3}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \ & \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \ & \mathrm{h}=(\sqrt{3}-1) \mathrm{R} \end{aligned}$ .

When the value of acceleration due to gravity ' $\mathrm{g}$ ' become $\frac{\mathrm{g}}{3}$ above surface…

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