Where does all the momentum of the spinning protoplanetary disk go?

We are bounded in a nutshell of Infinite Space: Week 10: Worksheet #17: Problem #1: Where does all the momentum of the spinning protoplanetary disk go? 

1. Angular momentum. In this problem we will obtain some intuition on why a disk must form during star formation if angular momentum is to be preserved.

(a) Cloud angular momentum. Consider a typical interstellar cloud core that forms a single star. You can assume it has a mass of $1 \M_\odot$ and a diameter of 0.1 pc. A typical cloud rotational velocity is 1 m/s at the cloud edge. Calculate the angular momentum of the cloud assuming constant density. If the core collapses to form a Sun-like star, what would the velocity at the surface of the star be if angular momentum is conserved? How does this compare with the break-up velocity of the Sun which is $\sim$ 300 km/s?

(b) Disk angular momentum. Assume that all the angular momentum is instead transferred to a disk of size 10 AU and negligible height. How massive must the disk be? You can assume constant density. (Hint: You can assume that the disk rotates with a Keplerian velocity given by $v = \sqrt{GM/r}$ where M is the mass and r is the radius.)

c) Solar System. The Sun has a surface rotational velocity of $\sim$ 2 km/s at the equator. How do the angular momenta of the Sun and Jupiter compare?

a. First off, let us lay out all the base data and equations we’ll need: $$v = 1 m/s$$ $$\omega = \frac{v}{R}$$ $$R = 0.05 pc = 1.5 \times 10^{15} m$$ $$R_\odot = 7 \times 10^8 m$$ $$M_\odot = 2 \times 10^{30} kg ,$$ and the equation for angular momentum:   $$L = I \omega .$$

Now we expand this for a sphere, and set the angular momentums equal to one another in order to find the rotational velocity: $$L = \frac{2}{5} M R^2 \omega,$$ $$L_{cloud}= L_\star,$$ $$\frac{2}{5} M_\odot (0.05 pc)^2 \frac{1 \frac{m}{s}}{1.5 \times 10^{15} m } = \frac{2}{5} M_\odot (R_\odot)^2 \omega_\star.$$ Now it’s just a matter of plugging in the correct values we saw a bit ago, and separating out the value we wish to find: $$\frac{(1.5 \times 10^{15} m )^2}{(7 \times 10^8 m)^2} \cdot \frac{1}{1.5 \times 10^{15} m} = \omega_\star,$$  and simplified, we have that $$\omega_\star = 3 \times 10^{-3},$$ which by   $$\omega = \frac{v}{R},$$ can be turned into $$v_\star = 2.1 \times 10^6 m/s$$ $$v_\star = 2.1 \times 10^3 km/s ,$$ meaning this velocity is much greater than the break up velocity of the Sun, meaning the total angular momentum has been syphoned off in other ways so as to preserve the Sun.

b. Furthermore, the definitions of angular momentum allow us to find the mass of the disk in which the mass has been compressed. So using: $$L_{disk}= L_{cloud}$$ $$I_{disk} = \frac{1}{2} MR^2,$$ we can now write out this expression with some of the values we saw in the last part of the problem: $$L_{disk}= L_{cloud}$$   $$\frac{1}{2} M_{disk} R^2 \omega = \frac{2}{5} M_\odot (0.05 pc)^2 \frac{1 \frac{m}{s}}{1.5 \times 10^{15} m },$$ and understanding we can rewrite the angular velocity in terms of the mass, we see how: $$\omega = \frac{v}{R} =  \frac{\sqrt{\frac{GM}{R}}}{R},$$ and then inserted in the equation: $$\frac{1}{2} M_{disk} R^2 \frac{\sqrt{\frac{GM}{R}}}{R} = \frac{2}{5} M_\odot (0.05 pc)^2 \frac{1 \frac{m}{s}}{1.5 \times 10^{15} m },$$

Now we can begin simplifying the expression, solving for the mass of the disk:  

$$M_{disk} R_{disk} \sqrt{\frac{GM}{R_{disk}}} = \frac{4}{5} M_\odot (0.05 pc) (1 \frac{m}{s}),$$ $$M_{disk} = \frac{   \frac{4}{5} M_\odot (1.5 \times 10^{15} m) (1 \frac{m}{s})   } {    R_{disk} \sqrt{\frac{GM}{R_{disk}}}       },$$ beginning to expand the more difficult portions of equation and plug in the values:   $$M_{disk} =\frac{   \frac{4}{5} M_\odot (1.5 \times 10^{15} m) (1 \frac{m}{s})   } {    R_{disk}^{1/2} G^{1/2} M^{1/2}       },$$   $$M_{disk}^{3/2} = \frac{   \frac{8}{5} \times 10^{33} g (1.5 \times 10^{15} m) (1 \frac{m}{s})   } {   (10 AU)^{1/2} G^{1/2}      },$$ and now we further simplify and can thus solely solve for the mass:  $$M_{disk} = \left(   \frac{   \frac{8}{5} \times 10^{33} g (1.5 \times 10^{15} m) (1 \frac{m}{s})   } {   (10 AU)^{1/2} G^{1/2}      }  \right)^{2/3},$$ quickly establishing the values we shall need for the last bit: $$1 AU = 1.5 \times 10^{13} m,$$ we now finish off the equation: $$M_{disk} = \left(   \frac{   2.4 \times 10 ^{48} \frac{gm^2}{s} } {   (1.5 \times 10^{14} cm )^{1/2} (6.67 \times 10^{-8} cm^3 g^{-1} s^{-2} )^{1/2}      }    \right)^{2/3}$$ $$M_{disk} = \left(\frac{2.48 \times 10^{48} \frac{gm^2}{s}  }{ 3163 cm^2~g^{-1/2} s^{-1} }\right)^{2/3},$$ and applying a quick conversion for the numerator so as to get the right dimensions: $$M_{disk} = \left(\frac{2.48 \times 10^{52} \frac{g~cm^2}{s}  }{ 3163 cm^2~g^{-1/2} s^{-1} }\right)^{2/3},$$ we find that: $$M_{disk} = 3.9 \times 10 ^{32} g$$

c. Next, we can use our knowledge of angular momentum to express the comparison of Jupiter and the Sun’s angular momentum. We know that the angular momentum of a celestial body can be expressed by the orbital and rotational momentum: $$L_\odot = L_{\odot, rot} + L_{\odot, orb},$$ and $$L_{Jup} = L_{Jup, ~rot} + L_{Jup,~ orb},$$ and how each of these are defined as:   $$L_{rot} = \frac{2}{5} MRV = \frac{2}{5} MR^2 \omega$$ $$L_{orb} = MRV = MR^2 \omega.$$ We now apply these definition to the search for the angular momentum of the Sun:   $$L_\odot = \frac{2}{5} M_\odot R_\odot v_\odot + M_\odot R_\odot^2 \omega_\odot,$$ plugging in the values we have been using throughout the entire problem, along with defining the angular velocity in ters of the period of rotation, which is the same as the period of rotation of Jupiter, for which we use the relationship of $P^2 = a^3$ to find the period in years and convert it into seconds: $$L_\odot = \frac{2}{5} (2\times 10^{30} kg) (696,300 km ) (2 km/s) + (2 \times 10^{30} kg ) (696300 km)^2 \frac{2\pi}{\pi \times 10^7 \times (5.2 AU)^{3/2} } ,$$ and we are left with two components of the angular momentum, yet clearly one is the dominant force: $$L_\odot = 1.1 \times 10^{36} kg km^2 s^{-1} + 1.63 \times 10^{34}  kg km^2 s^{-1}$$ $$L_\odot =1.1 \times 10^{36} kg km^2 s^{-1}$$

Now for the angular momentum of Jupiter, we follow the same steps as we just did for the Sun:

$$L_{Jup} = L_{Jup, ~rot} + L_{Jup,~ orb},$$ and defining each of the components: $$L_{Jup}= \frac{2}{5} M_{Jup} R_{Jup}^2 \omega_{Jup} + M_{Jup} R_{\odot \to Jup}^2 \omega_{\odot \to Jup},$$ we start plugging in the values which we have obtained from texts $$L_{Jup} = \frac{2}{5} (2\times 10^{27} kg) (70,000 km)^2 \frac{2\pi}{10 h \cdot 3600 \frac{s}{h}} + (2\times 10^{27} kg) (5.2 AU \times 1.5 \times 10^8 \frac{km}{AU})^2\frac{2\pi }{\pi \times 10^7 \cdot (5.2)^{3/2} } ,$$ and after the final simplification we find that $$L_{Jup} = 5.77 \times 10^{31}  kg km^2 s^{-1} + 2.05 \times 10^{37} kg km^2 s^{-1},$$ and again we see one of the terms becomes insignificant because of the several orders of magnitude in difference: $$L_{Jup} =  2.05 \times 10^{37} kg km^2 s^{-1}$$

Finally, we can compare the angular momenta of Jupiter and the Sun and find that: $$\frac{L_{Jup}}{L_\odot} = \frac{2.05 \times 10^{37} }{1.1 \times 10^{36} } \approx 20,$$ meaning Jupiter takes up a lot of the Angular momentum, allowing the sun to spin at slower rates and maintain its ability.

References:

Carroll, B. W., & Ostlie, D. A. (2007). An Introduction to Modern Astrophysics. San Francisco: Pearson: Addison Wesley.