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

The gravitational force on each mass due to other `3` masses provides the necessary centripetal force . Consider the particle `A`, Net force on particle `A` is given by
`vecF_(A)=vecF_(AB)+vecF_(AD)+vecF_(AC)`
`impliesF_(AY)=(GM^(2))/((AB)^(2))cos45^(@)+(GM^(2))/((AD)^(2))cos45^(@)+(GM^(2))/((AC)^(2))`
`=(GM^(2))/((sqrt(2)R)^(2))(1)/(sqrt(2))+(GM^(2))/((sqrt(2)R)^(2))(1)/(sqrt(2))+(GM^(2))/((2R)^(2))=(2GM^(2))/(2sqrt(2)R^(2))+(GM^(2))/(4R^(2))=(GM^(2))/(R^(2))[(1)/(sqrt(2))+(1)/(4)]` along `AO`
This must provide the necessary centripetal force
Hence, `(GM^(2))/(R^(2))[(1)/(sqrt(2))+(1)/(4)]=(Mv^(2))/(R )`
`impliesv=sqrt((GM)/(R )[(2sqrt(2)+1)/(4)])`