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Gravitation question

2013 · Q150
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Gravitation question

2013 · Q150

NEETPhysicsGravitationMCQ+4 / −1
A body of mass 'm' is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be
  1. A
    3mgR
  2. B
    13{1 \over 3}31​mgR
  3. C
    mg2R
  4. D
    23{2 \over 3}32​ mgR
View written solutionFree

Correct answer: D

Gravitational potential energy at any point at a distance r from the centre of the earth is

U=−GMmrU = - {{GMm} \over r}U=−rGMm​

where M and m be masses of the earth and the body respectively.

At the surface of the earth, r = R

Ui=−GMmR{U_i} = - {{GMm} \over R}Ui​=−RGMm​

At a height h from the surface,

r = R + h = R + 2R = 3R    ( h = 2R (Given))

∴\therefore∴ Uf=−GMm3R{U_f} = - {{GMm} \over {3R}}Uf​=−3RGMm​

Change in potential energy,

ΔU=Uf−Ui\Delta U = {U_f} - {U_i}ΔU=Uf​−Ui​

=−GMm3R−(−GMmR)=GMmR(1−13) = - {{GMm} \over {3R}} - \left( { - {{GMm} \over R}} \right) = {{GMm} \over R}\left( {1 - {1 \over 3}} \right)=−3RGMm​−(−RGMm​)=RGMm​(1−31​)

=23GMmR=23mgR = {2 \over 3}{{GMm} \over R} = {2 \over 3}mgR=32​RGMm​=32​mgR  (∵g=GMR2)\left( \because{g = {{GM} \over {{R^2}}}} \right)(∵g=R2GM​)

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