Three particles each of mass m=1 kg, are located at the vertices of an equilateral triangle of side a =4m. The speed at which they must move to revolve in a circular orbit, under the inluence of mutual gravitational force, circumscribing and preserving the equilateral triangle is `v=nsqrt(G)`, then the value of n is

Answer

`0.50`

Explanation

The net gravitational force on any particle
due to the other two is
`F_(g)=(2Gm^(2))/(a^(2))cos(30^(@))=(sqrt(3)Gm^(2))/(a^(2))`
Nor `r=a/(sqrt(3))`
For circular motion
`F_(g)=(sqrt(3)Gm^(2))/(a^(2))=(mv^(2))/r`
solvig it we obtain
`v=0.50sqrt(G)`