When our sun eventually runs out of fuel, its core will collapse leaving a conde
ID: 585466 • Letter: W
Question
When our sun eventually runs out of fuel, its core will collapse leaving a condensed white dwarf star about the size of the Earth. Assume that 60% of the solar mass will be retained in the white dwarf.
1. Use the current solar rotation rate (about 1 rotation every 30 days) and look up the relevant physical parameters (mass, radius, etc) in order to calculate the rotational speed of our sun when it collapses to the size of a white dwarf.
2.Some more massive stars collapse even further, becoming neutron stars, which are approximately the size of a large city (say, a radius of 10 km). What would be the rotational speed of our sun if it collapsed to the size of a neutron star (again assume 60% mass is retained)?
Explanation / Answer
I guess first of all the question is incomplete, as we also need to know the new radius of the white dwarf, however, taking an assumption that the radius gets wind up to 5.0 percent of its existing radius.
Angular momentum is conserved. However, since it is smaller in radius, the rotation rate will increase.
From above,
L = 2/5 * M * R^2 * w
Using the current mass, radius and rotation rate:
M = 1.9891e30 kg
R=6.955e8 m
w = 2*pi/(30*24*60*60) rad/s
L = 9.329e41 kg m^2 / s
If the new radius is 5% of R, r=0.05R,
L = 2/5 * M * r^2 * w'
where w' is the new rotation rate.
Since L = 2/5 * M * R^2 * w,
w' = w / 0.05^2
w' = 400w
w' = 9.696 rad/s
This is one rotation every 1.8hr (108min).
Rotational KE is defined as:
KE = 1/2 * I *w^2
Since I' = 0.05^2 * I and w' = 400*w,
KE' = 1/2 * 0.05^2 * I * 400^2 * w^2
KE' = 400 * 1/2 * I *w^2
KE' = 400KE
It would have 400 times the rotational kinetic energy that it has today. This extra energy comes from converting gravitational potential energy into kinetic energy as it collapses.
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