Two particles of equal mass m go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Answer

A, B, D

Explanation

force on M at C due to gravitatioN/Al attraction
`vecF_(CB)=(Gm^2)/(2R^2)vecj`
`vecF_(CD)=-(Gm^2)/(4R^2)vecj`
`vecF_(CA)=-(Gm^2)/(4R^2) cos 45veci+(Gm^2)/(4R^2)sin 45vecj`
`:. vecF_C=vecF_(CA)+vecF_(CB)+vecF_(CD)`
`=(-Gm^2)/(4R^2)(2+1/2)veci+(Gm^2)/(4R^2)(2+1/2)vecj`
so resultant fore on C,
`vecF_C=(Gm^2)/(4R^2)(2/2+1)`
For moving along the circle
`F=(mv^2)/R`
`=V=(GM)/R ((2/2+1)/4 `