Question

Arrive at the expression for acceleration due to gravity at a point below the surface of the earth

Answer 1

Let a point inside the earth at a depth $\sf{d}$ from earth's surface is considered, at which the acceleration due to gravity is going to be found out.

We know the expression for acceleration due to gravity on the surface of earth is,

$\longrightarrow\sf{g=\dfrac{GM}{R^2}\quad\quad\dots(1)}$

But as we go deeper the earth by a depth $\sf{d,}$ its mass and radius change from $\sf{M}$ and $\sf{R}$ to $\sf{M'}$ (say) and $\sf{R-d}$ respectively.

Similarly, since earth is spherical in shape, volume of earth changes from $\sf{\dfrac{4}{3}\,\pi R^3}$ to $\sf{\dfrac{4}{3}\,\pi(R-d)^3.}$

So the acceleration due to gravity at the depth will be,

$\longrightarrow\sf{g'=\dfrac{GM'}{(R-d)^2}\quad\quad\dots(2)}$

Since the density of earth is same,

$\longrightarrow\sf{\dfrac{M}{\left(\dfrac{4}{3}\,\pi R^3\right)}=\dfrac{M'}{\left(\dfrac{4}{3}\,\pi(R-d)^3\right)}}$

$\longrightarrow\sf{\dfrac{M}{R^3}=\dfrac{M'}{(R-d)^3}}$

$\longrightarrow\sf{M'=\dfrac{M\,(R-d)^3}{R^3}}$

Hence (2) becomes,

$\longrightarrow\sf{g'=\dfrac{GM\,(R-d)^3}{R^3\,(R-d)^2}}$

$\longrightarrow\sf{g'=\dfrac{GM}{R^2}\cdot\dfrac{R-d}{R}}$

From (1),

$\longrightarrow\sf{\underline{\underline{g'=g\left(1-\dfrac{d}{R}\right)}}}$

Hence we've arrived at the expression!