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).

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period is

If the radius of the earth is $6347 \, km$,then what will be the difference between the acceleration of free fall and the acceleration due to gravity near the earth's surface?

At what distance from the centre of the earth,the value of acceleration due to gravity $g$ will be half that on the surface? ($R =$ radius of earth)

The height of any point $P$ above the surface of the Earth is equal to the diameter of the Earth. The value of acceleration due to gravity at point $P$ will be: (Given $g = $ acceleration due to gravity at the surface of the Earth)