Blog #26 - Angular Momentum

Angular Momentum




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 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 sunlike 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 ∼300 km/s?

We begin by finding the equation for the angular momentum, L. We know we can relate the moment of inertia for a sphere and angular velocity to their linear equivalents to turn rotational movement into instantaneous linear movement. \[L = I \omega = M R v \] Using the information in the problem we can equate the rotational speed of the gas cloud and the speed of the rotating star to find the speed at the surface of the star. \[M_{\star} R_{\star} v_{\star} = M_{cloud} R_{cloud} v_{cloud} \] We know the cloud and the sun-like collapsed star have roughly the same mass so our equation for the speed of the star surface becomes \[v_{\star} = \frac{R_{cloud} v_{cloud}}{R_{\star}} \] Plugging in the given values we get \[v_{\star} = \frac{.05 pc*1 \frac{m}{s}}{2.3 * 10^{-8} pc} = 2170 \frac{km}{s} \] This value is much higher than the given \(300 \frac{km}{s}\) for the Sun. 

(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 = GM/r where M is the mass and r is the radius.)

We again begin by solving for the rotational inertia by equating the moment of inertia and velocity of a sphere (star) to the inertia of the disk. \[L_s = \frac{2}{5} M_s R_s v_s  = \frac{1}{2} M_d R_d v_d = L_d \] We also know that the speed of the disk can be described by Keplerian velocity so we substitute \[L_d = \frac{1}{2} M_d R_d * (\sqrt{\frac{G M_s}{R_d}}) \] Notice the mass of the velocity term is dictated by the central star as it is assumed to be much more massive than the surrounding disk material. Simplifying and solving we get \[M_d = \frac{4 (M_s)^{\frac{1}{2}} R_s v_s}{5 (R_d)^{\frac{1}{2}} G^{\frac{1}{2}}} \]Solving for our constants, we get \[M_d = 1.71*10^{29} \space kg \]

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

Comparing the angular momentums \[\frac{L_s}{L_j} = \frac{M_s R_s v_s}{M_j R_j v_j}\] which gives us a ratio of around 18 times the angular momentum for the Sun given the sun moves at around 2 km/s and Jupiter moves around 12 km/s.