Gravitation

JEE Main level, single correct, medium

127 of 223

Question 127 of 223

An object of mass 1kg1kg is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be
[If, g=10ms−2g=10m{s}^{-2} and radius of earth =6400km=6400km]

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Answer: (A) 48MJ48MJ

Gravitational potential energy on the earth surface Ui=−GMmR{U}_{i}=\dfrac{-GMm}{R}
Gravitational potential energy at a height hh above the earth's surface Uf=−GMmR+h=−GMmR+3R=−GMm4R{U}_{f}=-\dfrac{GMm}{R+h}=-\dfrac{GMm}{R+3R}=-\dfrac{GMm}{4R}
Gain in gravitational potential energy
ΔU=Uf−Ui=−GMm4R+GMmR=GMm+4GMm4R=3GMm4R\Delta U={U}_{f}-{U}_{i} =-\dfrac{GMm}{4R}+\dfrac{GMm}{R}=\dfrac{GMm+4GMm}{4R}=\dfrac{3GMm}{4R}
Gravitational potential energy in terms of acceleration due to gravity is ∆U=34mgR∆U=\dfrac{3}{4}mgR
=34×1×10×64×105=\dfrac{3}{4}\times 1\times 10\times 64\times {10}^{5}
=48MJ=48MJ

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