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Gravitation
Variation of g with depth
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2020 Phase-2 1 question
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What is the depth at which the value of acceleration due to gravity becomes $\dfrac{1}{n}$ times the value that at the surface of earth? (radius of earth = R)At depth d, $g'=g\left(1-\dfrac{d}{R}\right)$. Setting $g'=\dfrac{g}{n}$: $\dfrac{1}{n}=1-\dfrac{d}{R}\Rightarrow\dfrac{d}{R}=1-\dfrac{1}{n}=\dfrac{n-1}{n}\Rightarrow d=\dfrac{R(n-1)}{n}$.