At what depth $d$ below the Earth's surface does acceleration due to gravity become half of its value at the surface ($g/2$)?
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Question

At what depth $d$ below the Earth's surface does acceleration due to gravity become half of its value at the surface ($g/2$)?

Accepted Answer

Variation of $g$ with depth is $g' = g\left(1 - \frac{d}{R}\right)$. Setting $g' = \frac{g}{2}$, we get $1 - \frac{d}{R} = \frac{1}{2} \implies d = \frac{R}{2}$.

Gravitation

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