Gravitation, neutron starlooks simple

[rdn98]

A neutron star is formed when a star has burned all its nuclear fuel and begins to collapse in upon itself. It then packs roughly the mass of our Sun into a region with the same radius as that of a small city while continuing to spin at very rapid rate. Let's say you have a neutron star with a radius of 13 km and rotational velocity of 103 rotations per minute.

---------------------------------------------------------------------a) What is must be the minimum mass so that the material on its surface remains in place?

First thing I did was convert rotational velocity to translational velocity.
so (103 rev/min)(2pi/1rev)(1min/60secs)*13000m= A (lets just keep it simple for now)

Well, I want the minimum mass, so I looked into the gravitatin chapter, and the only thing that pops out at me is the escape speed formula

v=sqrt(2*G*M/R)
where G is the gravitation constant
M is my variable
and R is my radius.

So I plugged in my velocity, and solved for M, but its not working out right. Am I missing something here, or am I on the right track? *sigh* Too much time wasted on this problem..lol

 

[jamesrc]

You should be trying to set the gravitational attraction between a test mass on the surface of the planet and the planet equal to the centripetal force of that test mass given the rotational velocity of the planet.

 

[rdn98]

Sorry, you lost me for a second.

Are you saying setup the gravitation attaction equation equal to the centripetal force equation?

So (G*Me*m)/R^2=Me*v^2/R ?

 

[jamesrc]

Not quite. I was thinking you should set

\frac{GmM_n}{R^2} = \frac{mv^2}{R} = m\omega^2R

and solve for M_n, the mass of the neutron star. This applies to a mass on the equator of the star.

 

[rdn98]

Thank you so much man. I figured it out. Now I can rest easily.