Gravitation

JEE Main level, single correct, medium

126 of 223

Question 126 of 223

A body of mass is taken from earth surface to the height hh equal to twice the radius of earth (Re)({R}_{e}), the increase in potential energy will be : (g=g= acceleration due to gravity on the surface of earth)

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Answer: (C) 23mgRe\dfrac{2}{3}mg{R}_{e}

The potential energy of a particle at a distance r(>Re)r(\gt {R}_{e}) from the centre of Earth is,
U=−GMemrU=-G\dfrac{{M}_{e}m}{r}
Initially, the particle is at a height Re{R}_{e} from the surface. Therefore, r=Rer={R}_{e}.
∴Ui=−GMemRe∴{U}_{i}=-G\dfrac{{M}_{e}m}{{R}_{e}}
When height is increased to 2Re2{R}_{e} from surface, r=2Re+Re=3Rer=2{R}_{e}+{R}_{e}=3{R}_{e}
∴Uf=−GMem3Re∴{U}_{f}=-G\dfrac{{M}_{e}m}{3{R}_{e}}
Thus, the change in potential energy is
∆U=Uf−Ui=−GMem3Re+GMemRe=23GMemRe=23GMeRe2mRe=23mgRe(asg=GMeRe2)∆U={U}_{f}-{U}_{i} =-G\dfrac{{M}_{e}m}{3{R}_{e}}+G\dfrac{{M}_{e}m}{{R}_{e}} =\dfrac{2}{3}G\dfrac{{M}_{e}m}{{R}_{e}} =\dfrac{2}{3}G\dfrac{{M}_{e}}{{{R}_{e}}^{2}}m{R}_{e} =\dfrac{2}{3}mg{R}_{e}\left(\text{as}g=G\dfrac{{M}_{e}}{{{R}_{e}}^{2}}\right)

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