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 R/2, and the other mass, in a circular orbit of radius 3R/2. The difference between the final and initial total energies is: (a) -GMm/2R (b) +GMm/6R (c) GMm/2R (d) -GMm/6R

Answer: (d). Orbital energy: E = -GMm/2r. Initial: Ei = -GMm/2R. Mass m/2 at R/2: -(GM(m/2))/(2(R/2)) = -GMm/2R. Mass m/2 at 3R/2: -(GM(m/2))/(2(3R/2)) = -GMm/6R. Ef - Ei = -GMm/2R - GMm/6R + GMm/2R = -GMm/6R. So the answer is (d) -GMm/6R.

A body of mass m is moving in a circular orbit of radius R about a… | JEE Main 2018 (15 Apr, Shift 1)

Q30JEE Main 2018 · 15 APR · SHIFT 1Difficult

Satellite Energy & Weightlessness

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:

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Answer: (d)

Orbital energy: $E = -\frac{GMm}{2r}$ . Initial: $E_i = -\frac{GMm}{2R}$ .
Mass $\frac{m}{2}$ at $\frac{R}{2}$ : $-\frac{GM(m/2)}{2(R/2)} = -\frac{GMm}{2R}$ .
Mass $\frac{m}{2}$ at $\frac{3R}{2}$ : $-\frac{GM(m/2)}{2(3R/2)} = -\frac{GMm}{6R}$ .
$E_f - E_i = -\frac{GMm}{2R} - \frac{GMm}{6R} + \frac{GMm}{2R} = -\frac{GMm}{6R}$ .
So the answer is (d) $-\frac{GMm}{6R}$ .

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Satellite Energy & Weightlessness question from JEE Main 2018 (15 Apr, Shift 1)

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