At what depth does the acceleration due to gravity become $\frac{g}{4}$? ($R$ = radius of the Earth)

The diameters of two planets are in the ratio $4:1$. Their mean densities have the ratio $1:2$. The ratio of gravitational acceleration on the surface of the planets will be:

The ratio of the time periods of a simple pendulum at heights $2 R_E$ and $3 R_E$ from the surface of the earth is ($R_E$ is the radius of the earth).

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 $6400\, km$.

$A$ ball is launched from the top of Mt. Everest,which is at an elevation of $9000 \, m$. The ball moves in a circular orbit around the Earth. Acceleration due to gravity near the Earth's surface is $g$. The magnitude of the ball's acceleration while in orbit is

Write the difference between $G$ and $g$.