The acceleration due to gravity at a depth d below surface of earth is

$g' = {{GM} \over {{R^2}}}\left( {1 - {d \over R}} \right) = g\left( {1 - {d \over R}} \right)$

g' = 0 at d = R.

i.e., acceleration due to gravity is zero at the centre of earth.
Thus, the variation in value g with r is

For, r > R,

$g' = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}} = {{g{R^2}} \over {{r^2}}} \Rightarrow g' \propto {1 \over {{r^2}}}$

Here, R + h = r

For r < R, $g' = g\left( {1 - {d \over R}} \right) = {{gr} \over R}$

Here, $R - d = r \Rightarrow g' \propto r$

Therefore, the variation of g with distance from centre of the earth will be as shown in the figure.