A mine is located at depth $\frac{\mathrm{R}}{3}$ below earth's surface. The acceleration due to gravity at…
A mine is located at depth $\frac{\mathrm{R}}{3}$ below earth's surface. The acceleration due to gravity at that depth in mine is ( $\mathrm{R}=$ radius of earth, $\mathrm{g}$ = acceleration due to gravity)
$\mathrm{g}$
$3 g$
$\frac{2 \mathrm{~g}}{3}$
$\frac{\mathrm{g}}{3}$
Solution
The acceleration due to gravity at depth $\mathrm{d}$ is $\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$
given $\mathrm{d}=\mathrm{R} / 3$
$\therefore \quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{(\mathrm{R} / 3)}{\mathrm{R}}\right)=\frac{2}{3} \mathrm{~g}$
.