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 $\frac{3 R}{2}$ . The difference between the final and initial total energies is:

- \frac{G M m}{2 R}

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+ \frac{G M m}{6 R}

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\frac{G M m}{2 R}

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- \frac{G M m}{6 R}

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Solution

The correct option is D $- \frac{G M m}{6 R}$
$E_{i} = \frac{- {G M}_{m}}{2 R}$
$E_{f} = \frac{- G M (m / 2)}{2} (\frac{R}{2}) + ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{G M (\frac{m}{2})}{2 (\frac{3 R}{2})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = \frac{- 2 {G M}_{m}}{3 R}$
$E_{f} - E_{i} = \frac{- 2 {G M}_{m}}{3 R} - (\frac{{- G M}_{m}}{2 R})$
$= \frac{- {G M}_{m}}{R} (\frac{2}{3} - \frac{1}{2})$
$= \frac{{- G M}_{m}}{6 R}$

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The correct option is D −GMm6REi=−GMm2REf=−GM(m/2)2(R2)+⎛⎜ ⎜ ⎜ ⎜⎝GM(m2)2(3R2)⎞⎟ ⎟ ⎟ ⎟⎠=−2GMm3REf−Ei=−2GMm3R−(−GMm2R) =−GMmR(23−12) =−GMm6R