Three particle each of mass `m`, are located at the vertices of an equilateral triangle of side a. At what speed must they move if they all revolve under the influence of their gravitational force of attraction in a circular orbit circumscribing the triangle while still preserving the equilateral triangle ?

Options

A. `sqrt((Gm)/(a))`
B. `sqrt((2Gm)/(a))`
C. `sqrt((3Gm)/(a))`
D. `sqrt((4Gm)/(a))`

Answer

A

Explanation

From the figure,
`F_(A)=F_(AB)+F_(AC)=2[(GM^(2))/(a^(2))]cos 30^(@) =[(Gm^(2))/(a^(2)),sqrt(3)]`
Also, `r=(a)/(sqrt(3))`,
Now, `(mv^(2))/(r)=F|F_(A)|or (mv^(2)sqrt(3))/(a)=(Gm)/(a)=(Gm)/(a^(2))sqrt(3)`
`v=sqrt((Gm)/(a))`.