Question

Discuss the variation of acceleration due to gravity with depth. What happens to $g$ at the centre of the earth?

Answer 1

The acceleration due to gravity on the surface of the earth is

$g=\frac{GM}{R^2}$ $----------------\left(1\right)$

Consider $p$ be the density of the material of the earth

$mass=volume\times density$

$M=\frac{4}{3}\pi R^3\times p$

Now substituting in equation $\left(1\right)$

$\begin{array}{c}g=\frac{G}{R^2}\times \frac{4}{3}\pi R^3\times p\\ =\frac{4}{3}\pi GRp\end{array}$ $-----------\left(2\right)$

Now, Consider the body be taken to the depth $d$ below the surface of the earth.

Then, the acceleration due to gravity at the depth below the surface of the earth is

$g_d=\frac{4}{3}\pi G\left(R-d\right)p$ $---------\left(3\right)$

Now, divide equation $\left(3\right)$ by $\left(2\right)$

$\begin{array}{c}\frac{g_d}{g}=\frac{R-d}{R}=\left(1-\frac{d}{R}\right)\\ g_d=g\left(1-\frac{d}{R}\right)\end{array}$

This is an expression for the acceleration due to gravity at the depth below the surface of the earth and this expression shows the acceleration due to gravity decrease as we move down into the earth. and At the center of the earth $d=R$.

Hence, Acceleration due to gravity at the center of the earth is $0$