Gravitation

JEE Main level, single correct, medium

148 of 223

Question 148 of 223

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:

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Answer: (C) −GMm6R-\frac{G M m}{6 R}

Initial gravitational potential energy,
Ei=−GMm2RE_i=-\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}
∴\therefore Difference between initial and final energy,
Ef−Ei=GMmR(−23+12)=−GMm6RE_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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