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.

Options

A. `sqrt((GM)/(r ))`
B. `sqrt((2sqrt(2)GM)/(r ))`
C. `sqrt((GM)/(r ) 2(sqrt(2) + 1))`
D. `sqrt((GM)/(r )(2sqrt(2) + 1)/(4))`

Answer

D

Explanation

Refer to Fig. Resultant force on particle 1,

`F_(r) = sqrt(2) F + F' = sqrt(2) (Gm xx m)/((sqrt(2)r)^(2)) + (Gm xx m)/(4r^(2)) = (m upsilon^(2))/(r )`
or `upsilon = sqrt((Gm)/(r )((2sqrt(2) + 1)/(4)))`