There are three identical point mass bodies each of mass `m` locted at the vertices of an equilateral triangle with side `r`. They are exerting gravitational force of attraction on each other, which can be given by Newton's law of gravitaiton. Each mass body produces its gravitational field in the surrounding region. the magnitude of gravitational field at a point due to a point mass body is the measure of gravitaitonal intensity at that point. The gravitational potential at a point in a gravitational field is the amount of workdone in bringing a unit mass body infinity to the given point without acceleration.

Answer the following questions :
At what speed must they move if they all revolve under the influence of one another's gravitation in a circular orbit circumsribing the triangle still preserving the equilateral triangle

Explanation

Refer to Fig. `(APC). 1`, force of attracting on body at `C` due to body at `A` is

`F_(1) = (Gm xx m)/(r^(2))`along `CA`
Force of attracting on body at `C` due to body at `B` is
`F_(2) = (Gm xx m)/(r^(2))` along `CB`
These force `vec(F_(1))` and `vec(F_(2))` are inclined at an angle `60^(@)`, so the resultant force on body at `C` is
`F = sqrt(F_(1)^(2) + F_(2)^(2) + 2F_(1)F_(2) cos 60^(@))`
`= sqrt(F_(1)^(2) + F_(1)^(2) + 2F_(1)F_(2) xx 1//2)`
`= sqrt(3)F_(1) = sqrt(3) (Gm^(2))/(r^(2))` along `CD`
Here, `CO = (2)/(3) CD = (2)/(3) xx AC sin 60^(@)`
`= (2)/(3) xx r (sqrt(3))/(2) = (r )/(sqrt(3))`.
When each body is decribing a circular orbit with centre of orbit at `O`, the force `F` provides the required centripetel force. The radius of the circular orbit is `OC (= r//sqrt(3))`. If ` upsilon` is the speed of the body in circular orbit, then
`(m upsilon^(2))/(r//sqrt(3)) = (sqrt(3) Gm^(2))/(r^(2))` or `upsilon = sqrt((Gm)/(r ))`
`:. upsilon = sqrt(((6.67 xx 10^(-11)) xx2)/(1)) = 1.155 xx 10^(-5) ms^(-1)`