Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. They revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is

Question

Dr. Abubacker · Accepted Answer

(a) Given, $F_{1} = F_{2} = F\ and\ 	heta = 60{^\circ}$
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$Resultant\ force = \sqrt{3}\ F$
$	herefore Force\ on\ mass\ at\ A\ due\ to\ mass\ at\ B\ and\ C$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \sqrt{3}\left( \frac{{GM}^{2}}{L^{2}} ight)$
Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction.
$ie,\ \ \frac{{Mv}^{2}}{\left( L/\sqrt{3} ight)} = \sqrt{3}\left( \frac{{GM}^{2}}{L^{2}} ight)$
$or\ \ v = \sqrt{\frac{GM}{L}}$

Q. Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. They revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is

Explanation

Reviewed by Dr. Abubacker

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