The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R = \text{radius of the earth})$

Two satellites $P$ and $Q$ are moving in different circular orbits around the Earth (radius $R$). The heights of $P$ and $Q$ from the Earth's surface are $h_p$ and $h_Q$,respectively,where $h_p = R / 3$. The accelerations of $P$ and $Q$ due to Earth's gravity are $g_p$ and $g_Q$,respectively. If $g_p / g_Q = 36 / 25$,what is the value of $h_Q$?

Find the magnitude of acceleration due to gravity at a height of $10 \, km$ from the surface of the earth.

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

Obtain an expression for the acceleration due to gravity of the Earth at a depth $d$ below its surface.