12.  Four particles of equal mass are moving around a circle of radius r due to their mutual gravitational attraction.  Find angular velocity.  Answer in the picture.

$w=\sqrt{\frac{GM}{r^3}\left[\frac{4+\sqrt{2}}{4\sqrt{2}}\right]}$

Dear Student,


Kindly refer to the following link where an answer to a similar query has been provided
https://www.meritnation.com/ask-answer/question/four-particles-of-equal-mass-m-move-along-a-circle-of-rad/physics/9762295

You should use $\omega =\frac{v}{R}$  ('R' is taken as the radius of circle )  to obtain the value of angular velocity after referring to the solution provided  and you will get 
the angular velocity as 


$\omega = \sqrt{\frac{GM}{2R^3}\left(\frac{1}{2}+\sqrt{2}\right)}$   , which can be modified into the result as shown in the query



Regards