For particles of equal masses M that move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Explanation

Resultant gravitational force `M` at `A`

`F_(0)=(GM^(2))/((2Rcos45^(@))^(2))=(GM^(2))/((sqrt(2)R)^(2)) =(GM^(2))/(2R^(2))`
`F_(1)=(GM^(2))/((2R)^(2))=(GM^(2))/(4R^(2))`
Resultant force on `M`
`F=sqrt(2)F_(0)+F_(1)=(GM^(2))/(R^(2))(1/(sqrt(2))+1/4)`, along `AO`
This force `F` provides necessary centripetal force
`F=(Mv^(2))/R`
`(GM^(2))/(R^(2))(1/(sqrt(2))+1/4)=(Mv^(2))/R`
`v=[(4+sqrt(2))/(4sqrt(2)).(GM)/R]^(1/2)`
`=[(2sqrt(2)+1)/4.(GM)/R]^(1/2)`