2020 Phase-2 GravitationVariation of g with depth Medium Question 22 of 66 in Gravitation

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)

A $\dfrac{R}{n}$
B $\dfrac{R}{n^2}$
C $\dfrac{R(n-1)}{n}$
D $\dfrac{Rn}{(n-1)}$

The answer is C.

Solution

At depth d, $g'=g\left(1-\dfrac{d}{R}\right)$. Setting $g'=\dfrac{g}{n}$: $\dfrac{1}{n}=1-\dfrac{d}{R}\Rightarrow\dfrac{d}{R}=1-\dfrac{1}{n}=\dfrac{n-1}{n}\Rightarrow d=\dfrac{R(n-1)}{n}$.

Concept Variation of g with depth

Gravitation in past papers

66 questions from this chapter have appeared across 26 exam years.