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JEE Main 201815 April
Exam
JEE Main
Year
2018
Date
15 April
Subject
Physics
Chapter
Gravitation

JEE Main 2018 · 15 April · Physics · Gravitation

A body of mass m is moving in a circular orbit of radius R about a...

A body of mass mm is moving in a circular orbit of radius RR about a planet of mass MM. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R2\dfrac{R}{2} . And the other mass, in a circular orbit of radius 3R2.\dfrac{3R}{2} . The difference between the final and the initial total energies is

Options

  1. A+Gm6R+ \dfrac{Gm}{6R}
  2. B−Gm2R- \dfrac{Gm}{2R}
  3. C−Gm6R- \dfrac{Gm}{6R}
  4. DGm2R\dfrac{Gm}{2R}

Correct answer

C. −Gm6R- \dfrac{Gm}{6R}

Step-by-step solution

Ei=−GMm2R,{E}_{i} = - \dfrac{GMm}{2R},

Ef=−GMm/22(R2)−GMm/22(3R2)=−GMm2R−GMm6R=−4GMm6R=−2GMm3R{E}_{f} = - \dfrac{GMm / 2}{2\left(\dfrac{R}{2}\right)} - \dfrac{GMm / 2}{2\left(\dfrac{3R}{2}\right)} = - \dfrac{GMm}{2R} - \dfrac{GMm}{6R} = - \dfrac{4GMm}{6R} = - \dfrac{2GMm}{3R}

Ef−Ei=GMm2R(−23+12)=−GMm6R.{E}_{f} - {E}_{i} = \dfrac{GMm}{2R}\left(- \dfrac{2}{3} + \dfrac{1}{2}\right) = - \dfrac{GMm}{6R} .
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