Three identical stars, each of mass M, form an equilateral triangle (stars are positioned at the corners) that rotates around the centre of the triangle. The system is isolated and edge length of triangle is ‘L’. The minimum amount of work done that is required to dismantle the system is (Assume L is much greater than size of stars)

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$\frac{3 G M^{2}}{2 L}$

$\frac{3 G M^{2}}{4 L}$

$\frac{3 G M^{2}}{L}$

$\frac{G M^{2}}{2 L}$

answer is B.

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Detailed Solution

Given mass of each star= M
Length of triangle =L
Total energy of system initially =KE+PE
$P E = - 3 \frac{G M^{2}}{L}$
$K E = 3 (\frac{1}{2} M V^{2})$
Where $\frac{M V^{2}}{R} = 2 \frac{G M^{2}}{L^{2}} \cos 30^{0} = \frac{\sqrt{3} G M^{2}}{L^{2}} & R = \frac{L}{\sqrt{3}}$

$∴$ $K E = 3 (\frac{1}{2} M V^{2}) = \frac{3}{2} \times \frac{G M^{2}}{L}$
Hence, $T E_{i} = - \frac{3 G M^{2}}{L} + \frac{3 G M^{2}}{2 L}$ $= - \frac{3}{2} \frac{G M^{2}}{L}$
$T E_{f} = 0$
Work done = $T E_{f} - T E_{i}$ $= \frac{3}{2} \frac{G M^{2}}{L}$