Model a star as a rigid uniform spherical mass of radius R_0 spinning with a per

Question

Model a star as a rigid uniform spherical mass of radius R_0 spinning with a period T_0. (In fact, such a model of a star like the Sun as a rigid body is not very realistic, but we will go ahead with it anyway.) If the star collapses under gravitational forces with to a radius R and loses no mass nor angular momentum, what is the new period of rotation of the star? Consider a star with twice the mass of the Sun, yet with an initial radius that is equal to the radius of the Sun and with a initial period of rotation of 25.0 d. If this star collapses and becomes a neutron star with a radius of 10.0 km, what is the new rotational period of the star? Assume (unrealistically) the star satisfies the conditions of the model in part of this problem. Such small neutron stars are created by the gravitational collapse of massive stars when they exhaust their options for generating energy via nuclear fusion reactions in their cores. In fact, in the formation of neutron stars via supernova explosions, a significant percentage of the mass of the star is lost to the system, perhaps as much as 95%. This means, of course, that the initial star had to have a mass considerably greater than what we assumed in part As a result of the mass loss and the concomitant loss of the angular momentum associated with the Spelled material, neutron stars do not spin as rapidly as the calculation in part would indicate. Nonetheless, neutron stars (which always have masses between about 1.5 and 3.5 times times the Sun, and radii of only about 10 km) do rotate quite the neutron star in the middle of the Crab Nebula in the constellation Taurus (the bull) rotates 30 times each second!

Explanation / Answer

let new period is T

since angular momentum remain same

L1=L2

0.4*M1*Ro^2*wo=0.4*M1*R^2*w

(b)

we got w=wo*(Ro/R)^2

we know

To=2 *pi/wo=>wo=2*pi/To

so 2pi/T=(2*pi/To)*(Ro/R)^2

so T=To*(R/Ro)^2

(b)

let new period is T

since angular momentum remain same

L1=L2

0.4*(2*Msun)*(695700)^2*wo=0.4*(1.4*Msun)*10^2*w

we got w=(2*3.45*10^9 )wo

T=2*pi/w=2pi/((2*3.45*10^9 )wo)=3.61*10^(-9) day=0.312 ms

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