Gravitation

JEE Main level, single correct, medium

103 of 223

Question 103 of 223

A body of mass mm is taken from the surface of earth to a height equal to twice the radius of earth (ReR_e). The increase in potential energy will be _______.
(gg is acceleration due to gravity at the surface of earth)

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Answer: (D) 23mgRe\dfrac{2}{3}mgR_e

Initial potential energy of the body at the surface of the earth is given by:

Ui=−GMmReU_i = -\dfrac{GMm}{R_e}

Final potential energy of the body at a height h=2Reh = 2R_e from the surface of the earth is:

Uf=−GMmRe+h=−GMmRe+2Re=−GMm3ReU_f = -\dfrac{GMm}{R_e + h} = -\dfrac{GMm}{R_e + 2R_e} = -\dfrac{GMm}{3R_e}

The increase in potential energy is:

ΔU=Uf−Ui=−GMm3Re−(−GMmRe)\Delta U = U_f - U_i = -\dfrac{GMm}{3R_e} - \left(-\dfrac{GMm}{R_e}\right)

ΔU=GMmRe(1−13)=2GMm3Re\Delta U = \dfrac{GMm}{R_e} \left(1 - \dfrac{1}{3}\right) = \dfrac{2GMm}{3R_e}

Acceleration due to gravity at the surface of the earth is g=GMRe2g = \dfrac{GM}{R_e^2}, which gives GM=gRe2GM = gR_e^2.

Substituting this value, we get:

ΔU=2(gRe2)m3Re=23mgRe\Delta U = \dfrac{2(gR_e^2)m}{3R_e} = \dfrac{2}{3}mgR_e

Answer: 23mgRe\dfrac{2}{3}mgR_e

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