Question

Four particles abc each of mass m are kept at the corners of square each mass moves in a circle of radius r due to gravitational force find the velocity rs

Answer 1

If side of the square =a

and radius of the circle=r

then r^2=a^2/2

now on any particle the gravitational force of attraction due to the other 3 particles

F=1/a^2×Gm^2[(2)^1/2+1/2]

F=1/a^2×Gm^2×[2×(2)^1/2+1]×1/2

So this must be equals to the centrifugal force on the particle i.e= mv^2/r

So F= mv^2/r

or v = (Fr/m)^1/2