Question

Three particles, each of the mass m are situated at the vertices of an equilateral triangle of side a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. (F=(Gm_(1)m_(2))/(r^(2)))

Answer 1

Hence the velocity of the particle is V = √ Gm /a

Explanation:

F 12= Gm^2 / a^2

F13 = Gm^2 / a^2

F12 + F13 =√ Gm^2 / a^2 + Gm^2 / a^2 + 2 Gm^2 / a^2 cos(60°)

F1 =  √3 Gm^2 / a^2

As we know that

F = m ω^2

F = m ω^2 x a /√3

m ω^2 x a /√3 = √3 Gm^2 / a^2

ω = 3G m / a^3

As we knwo that

mv^2/a√3  = √3 Gm^2 / a^2

V = √ Gm /a

Hence the velocity of the particle is V = √ Gm /a

Answer 2

Hey mate see the attachment for your answer