If the acceleration due to gravity at a height 'h' from the surface of the Earth is 96% less than its value on the surface, then h = ____R where R is the radius of the Earth.

If R is the radius of the earth, then the value of acceleration due to gravity at a height h from the surface of the earth will become half its value on the surface of the earth if

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)

The value of acceleration due to gravity on the surface of earth is x . At an altitude of h from the surface of the earth, its value is y . If R is the radius of earth, then the value of h is

Considering the earth as a uniform sphere of radius R show that if the acceleration due to gravity at a height h from the surface of earth is equal to the acceleration due to gravity at the same depth, then h=1/2(sqrt(5)-1)R.

Considering the earth as a uniform sphere of radius R show that if the acceleration due to gravity at a height h from the surface of earth is equal to the acceleration due to gravity at the same depth, then h= 1/2(sqrt5-1) R

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 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 3 R above earth's surface to the acceleration due to gravity on the surface of earth is (R = radius of 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: