Question

Derive an expression for acceleration due to gravity at depth '$h$' below earth's surface.

Answer 1

Let a body $P$ of mass $m$ be situated at a depth $h$ below the earth's surface.
The gravitational force of attraction on a body inside a spherical shell is always zero. Therefore, the body $P$ experiences gravitational attraction only due to the inner solid sphere. The mass of this sphere is
$M^{\prime }=volume\times density=\frac{4}{3}\pi {(R_e-h)}^3\rho$
where $\rho$ is mean density of the earth. Therefore, according to Newton's law of gravitation, the force of attraction on the body $P$ is
$\frac{G{M}^{\prime }m}{{(R_e-h)}^2}=G\frac{\frac{4}{3}\pi {(R_e-h)}^3\rho m}{{(R_e-h)}^2}$
$=\frac{4}{3}\pi G(R_e-h)\rho m$
This force must be equal to the weight of the body $m{g}^{\prime }$, where $g^{\prime }$ is the acceleration due to gravity at a depth $h$ below the surface of the earth. Thus,
$m{g}^{\prime }=\frac{4}{3}\pi G(R_e-h)\rho m....\left(i\right)$
Similarly, if a body be at the surface of the earth $(h=0$) where acceleration due to gravity is $g$, then
$mg=\frac{4}{3}\pi G{R}_e\rho m....\left(ii\right)$
Dividing eq.$\left(i\right)$ by $\left(ii\right)$ we have
$\Rightarrow \frac{g^{\prime }}{g}=\frac{R_e-h}{R_e}$
$\Rightarrow g^{\prime }=g(1-\frac{h}{R_e})$