When a massive star dies, gravity causes its radius to contract quickly from the

Question

When a massive star dies, gravity causes its radius to contract quickly from the size of a giant star (R1 = 7.5 * 10^8m) to roughly the size of O'ahu (Rf= 4.0*10^4m), creating a very dense object known as a "neuron star." The initial giant star rotated with a period of 25 days, while the final neuron star rotates much faster.

For simplicity, assume that both before and after collapse, the start is a solid, uniform-density sphere, rotating like a rigid body about its center. Also, assume that the mass of the star (M = 6.0*10^30kg) remains constant during collapse. (No outside torques act on the star during the rapid contraction)

Question (show work): Find the FINAL ROTATION PERIOD of the neuron star in milliseconds and the INCREASE IN KINETIC ENERGY of the star, as a result of its contraction.

Explanation / Answer

The mass of the star is very small, less than that of the earth. Are you sure you have the right number?

Yes, that is actually the number I assumed.

We have for angular velocity: w = 2*pi/P where p is the period so:
p = 2*pi/w

For the star after collapse we can use this same equation (with capital letters):
P = 2*pi/W

Divide these to get: P = p*(w/W)

So to find the period P of the neutron star we need the ratio of the angular velocities.

Moment of inertia for a solid sphere = (2/5)*m*r^2
Angular momentum = I*w = [ (2/5)*m*r^2]*w = [(2/5)*m]*w*r^2

Since angular momentum must be conserved as the star contracts, we can equate the initial and final values of the momentm. (2/5)M is a constant for both so it can be taken out and we have (using capitol letters for the after):
w*r^2 = W*R^2
w/W = (R/r)^2

Using this in the equation for period we get:
P = p*(R/r)^2

And to evaluate we have:
p = 30 days
R/r = 4x10^4/8x10^8 = 5.0x10^(-5)
P = 30*[5.0x10^(-5)]^2 = 7.5x10^(-8) days = 6.48x10^(-3) seconds

So the final rotation period is 6.48 milliseconds

Rotational kinetic energy is given by: KE = (1/2)*I*w^2
Using I from above:
Increase in KE we get dKE = (1/5)*M*[R^2*W^2 - r^2*w^2]
Using w = 2*pi/p we get: dKE = (4*M*pi^2/5)*[R^2/P^2 - r^2/p^2] = 1.5x10^45 J

R = new radius = 4x10^4 m
r = Initial radius = 8x10^8 m
Difference in PE = (3/5)G*M[1/r - 1/R] = -2.5x10^46 J

So the loss in potential energy more than accounts for the gain in kinetic energy when the star shrinks

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