Question

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

Answer 1

Acceleration due to gravity at a depth $d$

Consider earth to be a homogeneous sphere of radius $R$ and mass $M$ with centre at $O$ .

Let $g$ be the value of acceleration due to gravity at a point $A$ on the surface of earth then

$g=\frac{GM}{r^2}$ ........$\left(1\right)$

If $\rho$ is uniform density of material of the earth, then

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

Putting value of $M$ in equation $\left(1\right)$ and equating it, we get:

$\therefore g=\frac{G\times \frac{4}{3}\pi R^3\rho }{R^2}=\frac{4}{3}\pi GR\rho$ .......$\left(2\right)$

Let $g^{\prime }$ be the acceleration due to gravity at the point $B$ at a depth $d$ below the surface of earth. The distance of the point $B$ from the centre of the earth is $(R–d)$. The earth can be supposed to be made of a smaller sphere of radius $(R–d)$ and a spherical shell of thickness $d$.

The body at $B$ is inside the spherical shell of thickness $d$. The force on body of mass $m$ at $B$ due to spherical shell is zero.

The body at $B$ is outside the surface of smaller sphere of radius $(R–d)$. The force on the body of mass $m$ at $B$ is due to smaller sphere of earth of radius $(R–d)$ is just as if the entire mass $M^{\prime }$ of the smaller sphere of earth is concentrated at the centre $O$.

$\therefore g^{\prime }=\frac{G{M}^{\prime }}{{(R-d)}^2}$

and $M^{\prime }=\frac{4}{3}\pi {(R-d)}^3\rho$

$\therefore g^{\prime }=\frac{G\times \frac{4}{3}\pi {(R-d)}^3\rho }{{(R-d)}^2}=\frac{4}{3}\pi G(R-d)\rho$ ........$\left(3\right)$

Dividing $\left(3\right)$ by $\left(2\right)$, we get

$\frac{g^{\prime }}{g}=\frac{\frac{4}{3}\pi G(R-d)\rho }{\frac{4}{3}\pi GR\rho }=\frac{R-d}{R}=\frac{R}{R}-\frac{d}{R}$

$\therefore g^{\prime }=g(1-\frac{d}{R})$

Hence, as depth increases, value of acceleration due to gravity decreases.