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Gravitation question

2003 · Q164
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Gravitation question

2003 · Q164

NEETPhysicsGravitationMCQ+4 / −1
A body of mass m is placed on earth's surface which is taken from earth surface to a height of h = 3R, then change in gravitational potential energy is
  1. A
    mgR4{{mgR} \over 4}4mgR​
  2. B
    23mgR{2 \over 3}mgR32​mgR
  3. C
    34mgR{3 \over 4}mgR43​mgR
  4. D
    mgR2{{mgR} \over 2}2mgR​
View written solutionFree

Correct answer: C

Gravitational potential energy on earth’s surface = −GMmR - {{GMm} \over R}−RGMm​, where M and R are the mass and radius of the earth respectively, m is the mass of the body and G is the universal gravitational constant.

Gravitational potential energy at a height h = 3R

=−GMmR+h=−GMmR+3R=−GMm4R = - {{GMm} \over {R + h}} = - {{GMm} \over {R + 3R}} = - {{GMm} \over {4R}}=−R+hGMm​=−R+3RGMm​=−4RGMm​

∴\therefore∴ Change in potential energy

=−GMm4R−(−GMmR) = - {{GMm} \over {4R}} - \left( { - {{GMm} \over R}} \right)=−4RGMm​−(−RGMm​)

=−GMm4R+GMmR = - {{GMm} \over {4R}} + {{GMm} \over R}=−4RGMm​+RGMm​

=34GMmR = {3 \over 4}{{GMm} \over R}=43​RGMm​ =34mgR = {3 \over 4}mgR=43​mgR

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