Three particle each of mass `m`, are located at the vertices of an equilateral triangle of side a. At what speed must they move if they all revolve under the influence of their gravitational force of attraction in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ?

Explanation

Let `A,B` and `C` be the three masses and `O` the centre of the circumscribing circle The radius of this circle is
`R = (L)/(2) "sec" 30^(@) = (L)/(2) xx (2)/sqrt(3) = (L)/sqrt(3)`
let v be the speed of each mass `M` along the circle Let us consider the motion of the mass at A The force of gravitational attraction on it due to the masses at `B` and `C` are
`(GM^(2))/(L^(2))` along AB and `(GM^(2))/(L^(2))` along AC
The resultant force is therefore
`2 (GM^(2))/(L^(2)) cos 30^(@) = (sqrt3GM^(2))/(L^(2)) along AD`
This for preserving the triangle must be equal to the neceassary centripetal force That is
`(sqrt3GM^(2))/(L^(2)) = (Mv^(2))/(R) = (sqrt3Mv^(2))/(L) [because R = L//sqrt3]` or `v = sqrt(GM)/(L)`
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