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

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 ............. $km$.

The acceleration due to gravity at an altitude $h$ above the Earth's surface is half of its value on the surface of the Earth. If the radius of the Earth $R = 4000 \ mile$,the altitude $h$ is approximately ......... $mile$.

$A$ spherical planet far out in space has a mass $M_0$ and diameter $D_0$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

The radii of two planets $A$ and $B$ are $R$ and $4R$ and their densities are $\rho$ and $\rho/3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_A : g_B)$ will be

If $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the Earth's surface,then the mean density of the Earth is: