# Gravitation, neutron starlooks simple

1. Jan 13, 2004

### 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. Lets 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

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

2. Jan 13, 2004

### 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.

3. Jan 14, 2004

### 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 ?

4. Jan 14, 2004

### jamesrc

Not quite. I was thinking you should set

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

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

5. Jan 14, 2004

### rdn98

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