The ratio of acceleration due to gravity at a height 3 R above earth's surface to the acceleration due to gravity on the surface of earth is
(R = radius of earth)

The ratio of acceleration due to gravity at a height 3R above earth 's surface to the acceleration due to gravity on the surface of the earth is (where R=radius of earth)

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

At what height above the earth surface, the acceleration due to gravity will be half that on the surface of earth ? Suppose R is the radius of earth.

Calculate the acceleration due to gravity at a height of 1600 km from the surface of the Earth. (Given acceleration due to gravity on the surface of the Earth g_(0) = 9.8 ms^(-2) and radius of earth, R = 6400 km).

The height at which the acceleration due to gravity becomes g/9 (where g = 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 :

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 :

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 :