The dependence of acceleration due to gravity | vedclass

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Question

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}

Vedclass · Accepted Answer

$(iv)$

Vedclass · Answer

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.

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

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