Centrifugal Radius

In a diffuse cloud, the centrifugal force does not play an important role. Consider a gas parcel whose specific angular momentum is equal to $j$. When the gas element contracts till the distance $r$ from the center, the centrifugal force per mass of

$$\frac{v_\phi^2}{r}=\frac{j^2}{r^3},$$ (4.91)

works. Contraction stops when the gravitational acceleration $GM/r^2$ is balanced by the centrifugal force. Its radius (centrifugal radius) is expressed as

$$r_{c}=\frac{j^2}{GM}.$$ (4.92)

Assuming that the disk is near the hydrostatic balance in $z$-direction, total column density is related to the volume density on the $z=0$ plane as

$$\sigma=\left(\frac{2c_s^2\rho_c}{\pi G}\right)^{1/2}.$$ (4.93)

We can show that a nondimensional ratio of specific angular momentum to the mass is equal to the ratio of the free-fall time to the rotation period as follow:

$$\frac{c_sj}{GM}=\frac{c_s\omega_cr^2}{G\sigma\pi r^2}=\frac{\omega_c}{(2\pi G \rho_c)^{1/2}}.$$ (4.94)

Many numerical simulations confirm that the contracting disk in the runaway isothermal contraction phase has some `universality.' The ratio of the free-fall time to the rotation period is approximately equal to

$$\frac{\omega_c}{(2\pi G \rho_c)^{1/2}}\simeq 0.3,$$ (4.95)

regardless of the initial conditions (Matsumoto, Nakamura, & Hanawa 1997). Thus, $c_s j/GM \simeq 0.3$. Finally, we can see the centrifugal radius as

$$r_{c}=\frac{GM}{c_s^2}\left(\frac{c_sj}{GM}\right)^2\simeq 0.3\frac{GM}{c_s^2},$$ (4.96)

increases with time in proportional to the mass $M$, because gas element with large $j$ contracts later. As the centrifugal radius increases with time, the launching point of the outflow also expands with time.