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 the figure. The correct figure is

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

Answer 1

Answer: (D)

$(iv)$

Answer 2

The acceleration due to gravity at a depth $d$ below the surface of earth is
$g' = g\left(1-\frac{d}{R}\right) = g\left(\frac{R-d}{R}\right) = g \frac{r}{R} \quad\ldots\left(i\right)$
where $R - d = r =$ distance of location from the centre of the earth. When $r = 0$, $g' = 0$
From $\left(i\right)$, $g \propto r$ till $R = r$, for which $g' =g$
For $r > R$, $g' = \frac{gR^{2}}{\left(R+h\right)^{2}} = \frac{gR^{2}}{r^{2}}$
or $g' \propto \frac{1}{r^{2}}$
Here, $R + h = r$
Therefore, the variation of $g$ with distance $r$ from centre of earth will be as shown in figure.