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How is it possible to find the shortest possible rotation period of a pulsar from a mass and a radius?

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One simple method is to consider a particle of the star at the surface on the equator, this particle will feel two principal forces: a centrifugal force, $F_c$ generally acting to pull the particle off the surface and a gravitational $F_g$ (and strong nuclear?) force holding the particle to the surface.

If $F_c$ > $F_g$ then the particle will tend to leave the surface, if not then the particle will stay put.

As I suspect this is a homework question I'll leave the detailed mathematics out of it for now!

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Thanks, however, can you help me find the period of rotation of the particle on the equator of a rotating pulsar with only the mass and radius given? –  Joseph Flynn Aug 18 '11 at 17:23
    
well if you start with the inequality i have given you and sub in some appropriate formulas some things might start to become more obvious... –  Nic Aug 18 '11 at 17:26
    
In reality, as it spins up the geopotential will deform, outward at the equator, and outward at the poles, so the pulsar will become oblate. This increases (for fixed rotaton rate) the equatorial velocity, and decreases the gravitational field at the equator, so the star will adjust some more. How this plays out depends upon the equation of state. –  Omega Centauri Aug 18 '11 at 18:28
    
Ok, thanks very much! –  Joseph Flynn Aug 18 '11 at 22:56

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