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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Physics
Chapter
Gravitation

JEE Main 2018 · 15 April · Morning Shift · Online · 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\frac{R}{2}, and the other mass, in a circular orbit of radius 3R2\frac{3 R}{2}. The difference between the final and initial total energies is:

Options

  1. A−GMm2R-\frac{G M m}{2 R}
  2. B+GMm6R+\frac{G M m}{6 R}
  3. C−GMm6R-\frac{G M m}{6 R}
  4. DGMm2R\frac{G M m}{2 R}

Correct answer

C. −GMm6R-\frac{G M m}{6 R}

Step-by-step solution

Initial gravitational potential energy, E_i=-GMm2R\frac{G M m}{2 R}Final gravitational potential energy, Ef=−GMm/22(R2)−GMm/22(3R2)=−GMm2R−GMm6R=−4GMm6R=−2GMm3R\begin{aligned}E_f &=-\frac{G M m / 2}{2\left(\frac{R}{2}\right)}-\frac{G M m / 2}{2\left(\frac{3 R}{2}\right)}\\ &=-\frac{G M m}{2 R}-\frac{G M m}{6 R}\\ &=-\frac{4 G M m}{6 R}=-\frac{2 G M m}{3 R} \end{aligned}\thereforeDifferencebetweeninitialandfinalenergy,Ef−Ei=Difference between initial and final energy, E_f-E_i=\frac{G M m}{R}\left(-\frac{2}{3}+\frac{1}{2}\right)=-\frac{G M m}{6 R}$
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