The value of g at depth h is two third the value that on the earth's surface. The value of h in terms of radius of earth R is

At what height h above the earth's surface, the value of g becomes g/2 (where R is the radius of the earth)

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

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

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

What is the depth at which the value of acceleration due to gravity become 1/n times the value that at the surface of earth ? (radius of earth = R)

The value of g at a certain height h above the free surface of the earth is x//4 where x is the value of g at the surface of the earth. The height h is

The value of g at a certain height h above the free surface of the earth is x//4 where x is the value of g at the surface of the earth. The height h is

Give the equation for the value of g at a depth 'd' from the surface of Earth. What is the value of 'g' at the centre 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