In $g(r)=\frac{4}{3} \pi \mathrm{Grp}, \frac{4}{3} \pi \mathrm{Gp}$ is constant.

$\therefore g(r) \propto r$

Means, the gravitational acceleration $(g)$ at a point inside the earth is directly proportional to the distance of that point from the centre of the earth.

And $g(r)=\frac{\text { GM }}{r^{2}}$, where $r \gg>\mathrm{R}_{\mathrm{E}}$ so, $g(r) \propto \frac{1}{r^{2}}$ where $r \gg>\mathrm{R}_{\mathrm{E}} .$ Hence starting from the centre of the earth $g(r)$ increases in directly proportion as $r$ increases and then outside the surface $g(r)$ decreases as inverse square of distance.

The variations in gravitational acceleration with below the surface of earth and above the height from the surface is shown as in figure.