The value of acceleration due to gravity at a depth of $ 1600 \,km $ is equal to: (Radius of Earth $ = 6400 \,km $) (in $\,ms^{-2}$)

Different planets have masses $M_1, M_2, M_3$ and radii $R_1, R_2, R_3$ respectively. The acceleration due to gravity on their surfaces are $g_1, g_2, g_3$ respectively. Based on the given graph,arrange their masses in descending order.

If the earth suddenly shrinks (without changing mass) to half of its present radius,the acceleration due to gravity will be

The figure shows the variation of acceleration due to gravity with distance from the center of a uniform spherical planet of radius $R$. What is $r_2 - r_1$?

The value of gravitational acceleration $g'$ at a height $h$ above the earth's surface is $\frac{g}{4}$. Then,what is the value of $h$ in terms of the earth's radius $R$?

Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is: