Question

Value of $g$ on the surface of earth is $9.8m/s^2$. Find acceleration due to gravity at depth $h=\frac{R}{2}$ from the surface ($R=$ radius of earth).

Answer 1

Gravitaitonal field inside the earth varies as:

$g^{\prime }=\frac{GM}{r^2}\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=\frac{GMr}{R^3}=g\frac{r}{R}$

At depth $\frac{R}{2}$,

$r=R-\frac{R}{2}$

$g^{\prime }=g\frac{\frac{R}{2}}{R}=\frac{g}{2}=4.9m/s^2$

Answer is $4.9m/s^2$.