JEE Main20265 April 2026Evening ShiftPhysicsGravitationActual
A body of mass m is taken from the surface of earth to a height equal to twice the radius of earth ( R_e ). The increase in potential energy will be _______. ( g is acceleration due to gravity at the surface of earth)
Options
- A1 2 mgR_e
- B3 4 mgR_e
- C1 4 mgR_e
- D2 3 mgR_e
Correct answer
D. 2 3 mgR_e
Step-by-step solution
Initial potential energy of the body at the surface of the earth is given by: U_i = - GMm R_e Final potential energy of the body at a height h = 2R_e from the surface of the earth is: U_f = - GMm R_e + h = - GMm R_e + 2R_e = - GMm 3R_e The increase in potential energy is: U = U_f - U_i = - GMm 3R_e - (- GMm R_e ) U = GMm R_e (1 - 1 3 ) = 2GMm 3R_e Acceleration due to gravity at the surface of the earth is g = GM R_e^2 , which gives GM = gR_e^2 . Substituting this value, we get: U = 2(gR_e^2)m 3R_e = 2 3 mgR_e Answe