At a height of $10 \,km$ above the surface of the earth,the value of acceleration due to gravity is the same as that at a particular depth below the surface of the earth. Assuming uniform mass density for the earth,the depth is ............. $km$.

$A$ boy weighs $72 \ N$ on the surface of the earth. What will be the gravitational force on the boy at a height equal to half the radius of the earth (in $N$)?

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is [ $R$ = radius of earth,$g$ = acceleration due to gravity,$n=$ integer].

Give the value of acceleration due to gravity at height $12\, km$ from the surface of earth. (in $, m/s^2$)

At a height of $h$ $km$ from the surface of the Earth,the gravitational potential and the value of $g$ are $-5.4 \times 10^7\, J kg^{-1}$ and $6.0\, m s^{-2}$ respectively. Take the radius of the Earth as $6400\, km$.