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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Answer

(i)1(ii)

Explanation

In the figure three particles located at vertices A, B and C of equilateral triangle of side AB = BC = CA = a.

These particle move in a circle with O as the centre and radius
r=OA=OB=OC where
`r=(BD)/(cos30^@)=(a//2)/(sqrt3//2)=a/sqrt3`
The gravitational attraction force acting on a particle, say due to particle at B is `F_1=(Gmm)/a^2` and due to particle at C is `F_2` .
Here `F_1=F_2=F` (Say) =`(Gm^2)/a^2`
Therefore, the resultant force on the particles at A is
`=2Fcos30^@=(2Gm^2)/a^2 sqrt3/2 = sqrt3 (Gm^2)/a^2`
`F_r` is directed along AO..... Thus the net force on particle at A is radial. Similarly, the net force on particle at B and C at C is `F_r`, each directed towards centre O. This force provides the necessary centripetal force. If v is the required initial velocity of each particle , then `(mv^2)/r = sqrt3 (Gm^2)/a^2 ` or `v^2=sqrt3 (Gmr)/a^2`
Since `r=a/sqrt3`, we have `v^2=sqrt3(Gm)(sqrt3a)=(Gm)/a rArr v=sqrt((Gm)/a)`
Time period T = `(2pir)/v =(2pixxa//sqrt3)/sqrt((Gm)/a) =2pi(a^3/(3Gm))^(1//2)`