Three particles 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 ?
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Explanation

The resultant force on particle at a due to other two particles is
`F_(A)=sqrt(F_(AB)^(2)+F_(AC)^(2)+2F_(AB)F_(AC) cos 60^(@))=sqrt(3) (Gm^(2))/a^(2) ...(i)" "[ :' F_(AB)=F_(AC)=(Gm^(2))/a^(2)]`
Radius of the circle `r=a/sqrt(3)`
If each particle is given a tangential velocity v, so that the resultant force acts as the centripetal force, then `(mv^(2))/r=sqrt(3) (mv^(2))/a` ...(ii)
From (i) and (ii), `sqrt(3) (mv^(2))/a= (Gm^(2) sqrt(3))/a^(2) implies v=sqrt((Gm)/a)`
Time period `T=(2pir)/v=(2pia)/sqrt(3) sqrt(a/(Gm))=2pi sqrt(a^(3)/(3Gm))`