Four identical particles of mass M are located at the corners of a square of side 'a' . What should be their speed if each of them revolves under the influence of others' gravitational field in a circular orbit circumscribing the square?

Engineering

Physics

Gravitational Force and Field

Question

$1 .16 \sqrt{\frac{GM}{a}}$

$1 .35 \sqrt{\frac{GM}{a}}$

$1 .41 \sqrt{\frac{GM}{a}}$

$1 .21 \sqrt{\frac{GM}{a}}$

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Net force on particle towards centre of circle is

$F_{C} = \frac{{GM}^{2}}{2 a^{2}} + \frac{{GM}^{2}}{a^{2}} \sqrt{2}$

$= \frac{{GM}^{2}}{a^{2}} (\frac{1}{2} + \sqrt{2})$

This force will act as centripetal force.

Distance of particle from centre of circle is $\frac{a}{\sqrt{2}} .$

$r = \frac{a}{\sqrt{2}}, F_{C} = \frac{{mv}^{2}}{r}$

$\frac{{mv}^{2}}{\frac{a}{\sqrt{2}}} = \frac{{GM}^{2}}{a^{2}} (\frac{1}{2} + \sqrt{2})$

$v^{2} = \frac{GM}{a} (\frac{1}{2 \sqrt{2}} + 1)$

$v^{2} = \frac{GM}{a} (1 .35); v = 1 .16 \sqrt{\frac{GM}{a}}$