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: As the star contracts, in order for its kinetic energy to increase, another form of energy must decrease: The "gravitational potential energy" of a self-gravitating sphere is given by: Ugr = -3GM^2/5R. Calculate deltaUgr for the star's contractiion, and show that this is more than enough energy to supply increase in kinetic energy of the star.
(The remaining excess energy creates a spectacular "supernova" explosion of light, heat, fusion etc.)

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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