Four identical particles of mass M are located at the corners of a square of side a. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

Correct option is C. 1.16GMa
Step 1: Draw diagram and find radius of circular orbit (Hypotenuse)2 = a2 + a2
Hypotenuse = √2a
Diameter = √2a
Radius = Diameter/2 = √2a/2 = a/√2
Step 2: Net force on particle towards center of circle
Both F = GM2/a2 are at 90°, so their resultant is √2GM2/a2 and act at angle bisector.
Net force on particle towards center of circle is
Fc = √(GM2/(2a2))2 + (GM2/a2)2 = GM2/a2(1/2 + 1)
Step 3: Calculation of speed
This resultant force will act as centripetal force. Distance of particle from center of circle is a/√2.
r = a/√2, Fc = mv2/r
mv2/(a/√2) = GM2/a2(1/2 + 1)
v2 = GMa(1/(√2)2 + 1)
v2 = GMa(1/2 + 1)
v2 = (3/2)GMa
v = √(3GMa/2) = 1.16GMa.