Angular momentum of the sun and a white dwarf

AI Thread Summary
The discussion focuses on calculating the angular momentum of the Sun and its future state as a white dwarf. The Sun's current angular momentum is calculated to be approximately 1.1E42 kgm²/s, using its mass, radius, and rotation period. As a white dwarf, the angular momentum is expected to decrease to about 7.5E37 kgm²/s, assuming the mass remains constant and the radius decreases. The conservation of angular momentum principle is highlighted, indicating that without external torque, the total angular momentum remains constant. Finally, the rotation period of the white dwarf is estimated to be around 150 seconds.
Fendergutt
Messages
6
Reaction score
0

Homework Statement



The Sun rotates about its own axis once every 26.0 days. Its mass is m_sun = 2.0E30 kg and radius is r_sun = 7.0E8 m. Assume that the Sun is a solid sphere with uniform density.

Astrophysicists predict that the Sun will collapse into a white dwarf in some billions of years. Its density will be high, and its radius 5.8E6 m. Assume the mass of the Sun to be unaltered (a good approximation for our calculation).

(i) What is the Sun's angular momentum now?
(ii) What is the Sun's angular momentum as a white dwarf?
(iii) How long will it take for the white dwarf Sun to rotate about its own axis?

Homework Equations



I = (2/5)*m*r^2
L = Iω = Iωk (k being the z-axis unit vector)

The Attempt at a Solution



(i)
ω_sun = (2pi)/(26 days) = 2.8E-6 rad/s

L = I_sun*ω_sun*k = (2/5)*m_sun*r_sun²*ω_sun*k = (2/5)*2.0E30 kg * (7.0E8 m)² * ω_sun * k = 1.1E42 kgm²/s

(ii)
I don't know the new spin angular velocity of the white dwarf, do I? If I leave ω_white = ω_sun the formula for L above with inserted new value for radius r_white gives appr. 7.5E37 kgm²/s k, ie. the angular momentum is lowered. Can this be correct?

(iii)
I don't know what to do here. I want to find the new period T, but I suspect there is something odd in (ii). Is there a relation I have overseen?



Thanks for any help.
 
Physics news on Phys.org
Fendergutt said:
(i)
ω_sun = (2pi)/(26 days) = 2.8E-6 rad/s

L = I_sun*ω_sun*k = (2/5)*m_sun*r_sun²*ω_sun*k = (2/5)*2.0E30 kg * (7.0E8 m)² * ω_sun * k = 1.1E42 kgm²/s
I didn't check the arithmetic, but the method is correct.

(ii)
I don't know the new spin angular velocity of the white dwarf, do I? If I leave ω_white = ω_sun the formula for L above with inserted new value for radius r_white gives appr. 7.5E37 kgm²/s k, ie. the angular momentum is lowered. Can this be correct?
What's required to change the angular momentum? Does that apply here?
 
Doc Al said:
I didn't check the arithmetic, but the method is correct.


What's required to change the angular momentum? Does that apply here?

Ah, so you are saying to me: Angular momentum is a conserved quantity: a system's angular momentum stays constant unless an external torque acts on it.

Then I set up T_white = ( I_white * 2pi * k ) / L which gives a rotation period of 1.5E2 s or 150 s.

(Mathematically: L features numbers and the unit vector k; In the calculation above I mentally ignored the unit vector because it was a feature of both nominator and denominator -- I'm I right to do this?)
 
All good.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top