Q111 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be ($g$ = acceleration due to gravity on earth's surface) (R - radius of earth)
- Correct answerA$\dfrac{2}{3}\text{mgR}$
- B$\dfrac{1}{3}\text{mgR}$
- C$\dfrac{1}{2}\text{mgR}$
- D$3\text{mgR}$
Solution
Initial potential energy of the body on the surface of the earth is $U_i = -\dfrac{GMm}{R}$
Final potential energy of the body at height $h = 2R$ is $U_f = -\dfrac{GMm}{R+h} = -\dfrac{GMm}{R+2R} = -\dfrac{GMm}{3R}$
Increase in potential energy is $\Delta U = U_f - U_i = -\dfrac{GMm}{3R} - \left(-\dfrac{GMm}{R}\right) = \dfrac{2GMm}{3R}$
Using the relation $g = \dfrac{GM}{R^2}$, we have $GM = gR^2$
Substituting this in the expression for $\Delta U$, we get $\Delta U = \dfrac{2(gR^2)m}{3R} = \dfrac{2}{3}mgR$
Answer: $\dfrac{2}{3}\text{mgR}$
Final potential energy of the body at height $h = 2R$ is $U_f = -\dfrac{GMm}{R+h} = -\dfrac{GMm}{R+2R} = -\dfrac{GMm}{3R}$
Increase in potential energy is $\Delta U = U_f - U_i = -\dfrac{GMm}{3R} - \left(-\dfrac{GMm}{R}\right) = \dfrac{2GMm}{3R}$
Using the relation $g = \dfrac{GM}{R^2}$, we have $GM = gR^2$
Substituting this in the expression for $\Delta U$, we get $\Delta U = \dfrac{2(gR^2)m}{3R} = \dfrac{2}{3}mgR$
Answer: $\dfrac{2}{3}\text{mgR}$
Hence, the correct answer is option: 'A'.