An object of mass m is raised from the surface of the earth to a certain height. The gain in its gravitational potential energy is 5 6 mgR , where g is the acceleration due to gravity at the surface of the earth and R is the radius of the earth. The height to which the object is raised is

Step-by-step solution

Let the height to which the object is raised be h . Initial gravitational potential energy at the surface of the earth: U_i = - GMm R Final gravitational potential energy at height h : U_f = - GMm R+h Gain in gravitational potential energy: U = U_f - U_i = - GMm R+h - (- GMm R ) U = GMm ( 1 R - 1 R+h ) = GMmh R(R+h) Using the relation GM = gR^2 , we can write: U = mgR^2 h R(R+h) = mgh 1 + h R Given that U = 5 6 mgR , we equate the two expressions: mgh 1 + h R = 5 6 mgR h R+h = 5 6 6h = 5R + 5h h = 5R Answer: 5R