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. The speed of each particle is:

Options

A. `[(Gm)/r((1+2sqrt(2))/4)]^(1/2)`
B. `sqrt((GM)/r)`
C. `sqrt((GM)/r(1+2sqrt(2)))`
D. `[1/2(Gm)/r((1+2sqrt2)/2)]^(1/2)`

Answer

A

Explanation

`impliesmv_(0)^(2)=2Fcos45^@+F_(1)=(2Gm^(2))/((sqrt(2))^(2)) 1/(sqrt(2))+(Gm^(2))/(4r^(2))`
`(mv_(0)^(2))/r=(GM^(2))/(4r^(2))[2sqrt(2)+1]`
`impliesv_(0)=[(Gm(2sqrt(2)+1))/(4r)]^(1/2)`