Q120 / 180 · NEET 2026 (Cancelled) · Physics
The amount of work done to raise a mass '$m$' from the surface of the Earth to a height equal to the radius of the Earth '$R$', will be :
- A$2\ mgR$
- B$mg\dfrac{R}{4}$
- C$mgR$
- Correct answerD$mg\dfrac{R}{2}$
Solution
Initial potential energy at the surface of the Earth is $U_i = -\dfrac{GMm}{R}$
Final potential energy at height $h = R$ is $U_f = -\dfrac{GMm}{R + R} = -\dfrac{GMm}{2R}$
Work done $W = \Delta U = U_f - U_i$
$W = -\dfrac{GMm}{2R} - \left(-\dfrac{GMm}{R}\right) = \dfrac{GMm}{2R}$
Using the relation $g = \dfrac{GM}{R^2}$, we get $GM = gR^2$
Substituting this in the expression for work done:
$W = \dfrac{gR^2 m}{2R} = mg\dfrac{R}{2}$
Answer: $mg\dfrac{R}{2}$
Final potential energy at height $h = R$ is $U_f = -\dfrac{GMm}{R + R} = -\dfrac{GMm}{2R}$
Work done $W = \Delta U = U_f - U_i$
$W = -\dfrac{GMm}{2R} - \left(-\dfrac{GMm}{R}\right) = \dfrac{GMm}{2R}$
Using the relation $g = \dfrac{GM}{R^2}$, we get $GM = gR^2$
Substituting this in the expression for work done:
$W = \dfrac{gR^2 m}{2R} = mg\dfrac{R}{2}$
Answer: $mg\dfrac{R}{2}$
Hence, the correct answer is option: 'D'.