Question #108074
Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process and winding up with a radius 1.0% of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun’s new rotation be? (Take the Sun’s current period to be about 30 days) What would be its final KE in terms of its initial KE of today?
1
Expert's answer
2020-04-07T09:06:11-0400

For a single rotating body, the law of conservation of momentum must be fulfilled. We will assume (which is a rough approximation) that the mass of the star is distributed evenly over its volume. Then the moment of inertia of the ball is written as I=25mR2I=\frac{2}{5}mR^2, where m, R - its mass and radius [1]. the net angular momentum of a ball around a principal axis is L=IwL=Iw. When our Sun is normal star

(1) Lo=25MR2woL_o=\frac{2}{5}M_{\bigodot} R^2 \cdot w_o

After it collapses into a white dwarf, losing about half its mass in the process and winding up with a radius 1.0% (=0.01) of its existing radius the angular momentum will be

(2) L1=25M2(0.01R)2w1L_1=\frac{2}{5}\frac{M_{\bigodot}}{2} (0.01\cdot R)^2 \cdot w_1

According the law of conservation angular momentum (the lost mass carries away no angular momentum) we have L1=LoL_1=L_o . Thus for new angular velocity w1w_1 we have

(3) 25M2(0.01R)2w1=25MR2wow1=21002wo=2104wo\frac{2}{5}\frac{M_{\bigodot}}{2} (0.01\cdot R)^2 \cdot w_1=\frac{2}{5}M_{\bigodot} R^2 \cdot w_o\\ w_1=2 \cdot 100^2 w_o=2\cdot 10^4 w_o

wo=2π30day=2π30246060s=2π3.9107s1w_o=\frac{2\pi}{30 day}=\frac{2\pi}{30\cdot 24\cdot 60\cdot 60 s}=2\pi \cdot 3.9\cdot 10^{-7} s^{-1} Finally for new rotation we get

w1=2π7.8103s1w_1=2\pi \cdot 7.8\cdot 10^{-3} s^{-1} Thus our Sun's white dwarf will rotate with period T=17.8103s=128sT=\frac{1}{7.8\cdot 10^{-3}}s=128s

The KE we can find by the equation

Ek=Iw22=Lw2E_k=\frac{Iw^2}{2}=\frac{Lw}{2} Initial KE is Eo=Lowo2E_o=\frac {L_ow_o}{2}. The final KE is E1=L1w12E_1=\frac {L_1w_1}{2}. We can find a relationship

(4) E1Eo=L1w1Lowo=w1wo=2104\frac {E_1}{E_o}=\frac {L_1 w_1}{L_ow_o}=\frac {w_1}{w_o}=2\cdot 10^4

Where we taken into account the law of conservation angular momentum and (3). In our assumptions, the gravitational potential energy of the Sun's mass is completely converted into the kinetic energy of rotation of its remainder.

Answer: The Sun’s new rotation period will be about 128 s. The  KE of white dwarf will exceed the initial value for the Sun by 21042\cdot 10^4 times.

[1] https://en.wikipedia.org/wiki/Moment_of_inertia


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