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The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be a sphere of radius $R$ of uniform density is as shown figure below
The correct figure is
$(i)$
$(ii)$
$(iii)$
$(iv)$
Solution
The acceleration due to gravity at a depth d below surface of earth is
$g^{\prime}=\frac{G M}{R^{2}}\left(1-\frac{d}{R}\right)=g\left(1-\frac{d}{R}\right)$
$g^{\prime}=0$ 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, $r>R$
$g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}=\frac{g R^{2}}{r^{2}} \Rightarrow g^{\prime} \propto \frac{1}{r^{2}}$
Here, $R+h=r$
For $\quad r$
Here, $R-d=r \Rightarrow g^{\prime} \propto r$
Therefore, the variation of $g$ with distance from centre of the earth will be as shown in the figure.
