The mass of a planet and its diameter are three times those of the Earth. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$.

What is the value of acceleration due to gravity at a height equal to half the radius of the Earth,from its surface?

If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of that planet is (in $\,g$)

$A$ 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

Obtain the general equation of gravitational force at a distance $r$ from the centre of the Earth and derive the equation for acceleration due to gravity on the surface of the Earth.

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):