Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each particle

Explanation

The net gravitational force = centripetal force
The force between A & B is `F_(1)`
`F_(1) = (GMM)/((sqrt(2)R)^(2))`
The force between A & D is `F_(2)`
`F_(2) = (GMM)/((sqrt(2)R)^(2))`
The force between A & C is `F_(3)`
`F_(3) = (GMM)/((2R)^(2))`
The components of `F_(1) & F_(2)` along the radius `F_(1) cos 45^(@) & F_(2) cos 45^(@) [F_(1) = F_(2) = F]`
Net force = `2F cos 45^(@) + F_(3)`
`=(2GM^2)/((sqrt(2)R)^(2)) xx 1/(sqrt2) + (GM^2)/(4R^2)`
Net force = centripetal force
`(GM^2)/(4R^2) [2sqrt(2) + 1] = (MV^2)/(R)`
`V = 1/2[(GM)/R (2sqrt(2) + 1)]`.