The value of the acceleration due to gravity is g1 at a height h=(R)/(2)(R= radius of the earth) from the surface of the earth.It is again equal to (g1/2) at a depth d below the surface of the earth.The ratio d/R equals:

The value of the acceleration due to gravity is g_(1) at a height h=(R)/(2) (R=radius of the earth) from the surface of the earth.It is equal to 2g_(1) at a depth d below the surface of the earth.The ratio ((d)/(R)) is equal to:

The acceleration due to gravity at a depth R//2 below the surface of the earth is

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth

The acceleration due to gravity at a height 1km above the earth is the same as at a depth d below the surface of earth. Then :

If accelerations due to gravity at a height h and at a depth d below the surface of the earth are equal, how are h and d related ?

What is the value of acceleration due to gravity at a height equal to helf the radius of earth, from surface of earth ? [take g = 10 m//s^(2) on earth's surface]

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R form the surface of the earth is (g=acceleration due to gravity at the surface of the earth)

The depth at which the value of acceleration due to gravity is 1/n times the value at the surface, is (R=radius of the earth)

At what depth from the surface of the earth, the acceleration due to gravity will be half the value of g on the surface of the 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).