A body of mass m is raised to aheight h from the surface of the earth where the acceleration due to gravity is g. Prove that the loss in eight due to variation in g is approximately 2 mgh`//`R, where R is the radius of the earth.

A body of mass m is raised to a height h from a point on the surface of the earth of the earth where the acceleration due to gravity is g. Prove that the loss of weight due to variation of g is (2mgh)/R , where R is the radius of the earth.

If R is the radius of the earth and .g. the acceleration due to gravity, the mass of the earth is

A body of mass m rises to a height h=R/5 from the surface of earth. If g is the acceleration due to gravity at the surface of earth, the increase in potential energy is (R = radius of earth)

The height at which the acceleration due to gravity becomes g/9 . (Where g = acceleration due to gravity) in terms of R. Where R is the radius of earth

A body of mass m rises to a height h=R//5 from the earth's surface where R is earth's radius. If g is acceleration due to gravity at the earth's surface, the increase in potential energy is

A body of mass m rises to a height h=R//5 from the earth's surface where R is earth's radius. If g is acceleration due to gravity at the earth's surface, the increase in potential energy is

A body of mass m rises to height h = R/5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is

An object of mass m is raised up to a small height h above the earth's surface. If the acceleration due to gravity on the earth's surface is g, prove that the weight of the object will decreases by (2mgh)/R with respect to that on the earth's surface . R =radius of the earth.