Minimum rotation period of pulsar, based on mass and radius?

I need to show that the minimum possible rotation period T of a star of mass m and radius r is contained by the inequality:





T 鈮?(4蟺2r3/GM)^(1/2) [G is gravitational constant]





I've never been good at "show that" questions. I don't want the answer, I would just like to be shown which formulas are involved.





Thanks in advance.|||You Imagine a "rock" or any piece of matter "m" on the Pulsar's surface. There are two forces on the rock; The force of gravity "GMm/R^2" pulling it in towards the center and the Normal force "N" pushing up or out that keeps the rock from falling into the center. These two opposite forces give a net force that equals the centripetal force;


GMm/R^2 - N = mv^2/R





N = GMm/R^2 - mv^2/R





The rotation has the effect of throwing the rock out, off the surface, so the faster the rotation speed the smaller "N" will be (force of gravity is constant), but it can't be negative because Normal forces can only push. So the rotation speed "v" has to be such that "N %26gt;/= 0" . Zero occurs when the rock is just about to leave the surface. Another words the surface matter is about to fly off and the Pulsar is essentially spinning off all its mass.


So


GMm/R^2 - mv^2/R %26gt;/= 0





The linear velocity "v" is the circumference divided by the period; v = 2piR/T


So now you can write; (m cancels);


GM/R^2 %26gt;/= 4(pi)^2R/T^2





Solve this for T^2 , and then T and you have your result.