Obtain an expression of acceleration produced by gravity of earth.

Imagined a earth to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface.

A point outside the earth is outside all the shells.

Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre.

For a point inside the earth, the situation is different. This is illustrated in figure.

Consider the earth to be made up of concentric shells.

A point mass $m$ is situated at a distance $r\left(r<\mathrm{R}_{\mathrm{E}}\right)$ from the centre.

The point $P$ lies outside the sphere of radius $r$.

For the shells of radius greater than $r$, the point $P$ lies inside. Hence, no gravitational force exert on mass $m$ kept at $\mathrm{P}$.

If mass of particle is $m$ at $\mathrm{P}$, mass of sphere $m_{r}$ with radius $r$, then the force on the mass $m$ at $P$ has a magnitude of force,

$\mathrm{F}=\frac{\mathrm{G} m \mathrm{M}_{r}}{r^{2}}$

We assume that the entire earth is of uniform density.

Hence its mass is $\mathrm{M}_{\mathrm{E}}=\left(\frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}\right) \rho$

$\therefore \frac{4}{3} \pi \rho=\frac{\mathrm{M}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^{3}}$

$\ldots(2)$

But $\mathrm{M}_{r}=\frac{4}{3} \pi r^{3} \rho$

$\ldots(3)$