# Neutron star's angular velocity

1. Jun 3, 2005

### MAPgirl23

Under some circumstances, a star can collapse into an extremely dense
object made mostly of neutrons and called a neutron star. The density of a
neutron star is roughly 10^14 times as great as that of ordinary solid
matter. Suppose we represent the star as a uniform, solid, rigid sphere,
both before and after the collapse. The star`s initial radius was 7.0 x
10^5 km (comparable to our sun); its final radius is 16 km.

If the original star rotated once in 30 days, find the angular speed of
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** I think you need to make two assumptions to answer this question. The first assumption is that the mass of the star is the same after collapse as before. The second assumption is that the angular momentum of the star is the same after the collapse as before (angular momentum is conserved, but if the star had, for example, ejected matter during its collapse, the ejecta could have carried away some of its angular momentum).

If the moments of interia of the sphere before and after collapse are I_1 and I_2 respectively, and its angular speeds before and after colapse are w_1 and w_2 then from our second assumption we have:

(I_1)(w_1) = (I_2)(w_2)
=> w_2 = (I_1)(w_1)/(I_2)

You are given w_1 (you are are told that the star rotates once in 30 days before the collapse). And although you do not know I_1 and I_2 directly, you can find their ratio, because each one is proportional to the mass of the star and the square of its radius. From our first assumption the mass of the star is constant, so

(I_1)/(I_2) = (R_1)^2/(R_2)^2

where R_1 is the radius of the start before collapse, and R_2 is the radius after collapse. So

w_2 = (w_1)(R_1)^2/(R_2)^2

now using my logic I got
(R_1)^2/(R_2)^2 = (7.0 x 10^5)^2/(16)^2 = 1.914 x 10^9
w_1 = 1 rotation/2.592 x 10^6 sec = 3.85 x 10^-7

so: If the original star rotated once in 30 days, find the angular speed of the neutron star. 738.4 rad/s which was wrong.

What did I do wrong?

2. Jun 3, 2005

### Nylex

This bit is where you've gone wrong. It should be:

$$\omega_{1} = \frac{2\pi}{T}$$, where T is your orbital period, ie 2.592 x 10^6 s.

3. Jun 3, 2005

### siddharth

Shouldn't you use the above information below?

4. Jun 3, 2005

### apchemstudent

I see where you went wrong. You got the right answer, but not the right units. You have 1 revolution/sec... multiply your answer by 2pi...

Last edited: Jun 3, 2005
5. Jun 3, 2005

### MAPgirl23

Thanks, I see what I did.

6. Jun 3, 2005

### siddharth

Wait a minute. How can you say that the mass of the star is conserved? What happened to all the protons (and the electrons) in the star?

Initially you have M1=[4/3(pi)(r_1)^3] X (D1) and M2=[4/3(pi)(r_2)^3] X (D2) .
Since the ratio D1/D2 is given, by substituting the above values, you can get the ratio
I1/I2.