First slide

Universal law of gravitation

Question

Three particles, each of mass M are moving in a circle under their mutual gravitational forces such that they always form an equilateral triangle of side a while rotating. Speed of each particle is:

Moderate

A

$\sqrt{\frac{2 GM}{a}}$
B

$\sqrt{\frac{3 GM}{a}}$
C

$\sqrt{\frac{GM}{2 a}}$
D

$\sqrt{\frac{GM}{a}}$

Solution

$2 Fcos \frac{60^{\circ}}{2} = \frac{{Mv}^{2}}{a / \sqrt{3}}$
where $F = \frac{{GM}^{2}}{a^{2}}$
So, $v = \sqrt{\frac{GM}{a}}$

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