A man can jump 2 m high on the surface of the earth. How high he can jump on a planet where the acceleration due to gravity is `(g)/(12)`, where g is the acceleration due to gravity on the surface of the earth ?

The acceleration due to gravity becomes (g/2) where g = acceleration due to gravity on the surface of the earht at a height equal to

Acceleration due to gravity is __ at the surface of earth.

Let g be the acceleration due to gravity on the earth's surface.

What will be the acceleration due it gravity at a depth in Earth. where g is acceleration due to gravity on the earth ?

Find the height (in terms of R, the radius of the earth) at which the acceleration due to gravity becomes g/9(where g=the acceleration due to gravity on the surface of the earth) is

If g_(e) is acceleration due to gravity on earth and g_(m) is acceleration due to gravity on moon, then

A simple pendulum makes 20 oscillations in one second on the surface of the earth. Determine the time period of the simple pendulum on the surface of a planet where the acceleration due to gravity is one fourth of the acceleration due to gravity on the surface of the earth .

The height at which the acceleration due to gravity becomes (g)/(9) (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is

The height at which the acceleration due to gravity becomes (g)/(9) (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :