The height above the surface of the earth where acceleration due to gravity becomes $\left(\frac{g}{9}\right)$ is ( $R$ is radius of the earth, $g$ is acceleration due to gravity)

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Aicharya · Accepted Answer

$\begin{aligned} & g^{\prime}=\frac{g}{\left(1+\frac{h}{R}ight)^2} \ & \frac{g}{9}=\frac{g}{\left(1+\frac{h}{R}ight)^2} \ & h=2 R\end{aligned}$

The height above the surface of the earth where acceleration due to gravity becomes…

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