JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2018
- Hard
- MCQ
A body of mass \(m\) is moving in a circular orbit of radius \(R\) about a planet of mass \(M\). At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius \(\frac{R}{2}\) , and the other mass, in a circular orbit of radius \(\frac{3R}{2}\). The difference between the final and initial total energies is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \( - \frac{{GMm}}{{6R}}\)
Step-by-step Solution
Detailed explanation
Initial gravitational potential energy,
\({E_i} = - \frac{{GMm}}{{2R}}\)
Final gravitational potential energy,
\({E_f} = - \frac{{GMm/2}}{{2\left( {\frac{R}{2}} \right)}}\frac{{GMm/2}}{{2\left( {\frac{{3R}}{2}} \right)}}\)
\( = - \frac{{GMm}}{{2R}} - \frac{{GMm}}{{6R}}\)
\( = - \frac{{4GMm}}{{6R}} = - \frac{{2GMm}}{{3R}}\)
\(\therefore \) Difference between initial and final energy,
\({E_f} - {E_i} = \frac{{GMm}}{R}\left( { - \frac{2}{3} + \frac{1}{2}} \right) = - \frac{{GMm}}{{6R}}\)
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