Question

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 in figures below

The correct figure is

• A. $(i)$
• B. $(ii)$
• C. $(iii)$
• D. $(iv)$

Answer 1

Answer: (D)

The acceleration due to gravity at a depth $d$ below surface of earth is
$g^{\prime }=\frac{GM}{R^2}\left(1-\frac{d}{R}\right)=g\left(1-\frac{d}{R}\right)$
$g^{\prime }=0$ 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 $r<R,g^{\prime }=g\left(1-\frac{d}{R}\right)=\frac{g^r}{R}$
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.