Question

Particles having equal mass move in a circular path and attracted mutually gravitation and placed at equilibrium, find the velocity of particles?​

Answer 1

Answer:

1/2 × √ (G × (m/r) )

Explanation:

path : circular

masses : m & m

distance : d = 2r

Force of gravity : F = G × m × m / (4r^2)

i,e,...F = G/4 × (m/r)^2

centripetal Force : F' = m v^2 /r

since ...F = F'

G/4 × (m/r)^2 = m v^2 / r

hence .....v = 1/2 × √ (G × (m/r) )