Derivation of the day: collision rate part II

Background
Continuing from the previous derivations of this quantity, we now consider the regime \(\alpha^{1/2}v_H < u < v_H\).  In this case, the random velocity is smaller than the hill velocity and we must consider the gravitational field from the sun.

Quantities

\(\sigma\) = surface mass density
\(m\) = particle mass
\(u\) = random velocity
\(\Omega\) = angular speed
\(H\) = \(u/\Omega\) = scale height
\(R\) = radius of large body
\(R_H\) = Hill radius
\(v_H\) = Hill velocity

Derivation
\[ \frac{dN}{dt} = \text{Hill Entry Rate}\times\text{P} \\
\text{P} = \text{Probability of collision}\]
We can write the Hill entry rate as before \(\frac{\sigma}{m}\Omega R_H^2\) and need just find the probability of a collision.  This is given by the area of the disk in which the small body will collide with the large body divided by the area of intersection between the disk and the large body's Hill sphere which we approximate as \(\text{scale height}\times R_H\).  We find the radius of the first area (the impact parameter \(b\)) by conserving angular momentum:
\[mv_H b = m v_{esc} R \\
b = \frac{v_{esc}}{v_H}R\]
Which gives us (where \(R = \alpha R_H\) as found before)
\[P = \frac{\left(v_{esc}R/v_H\right)^2}{\left( u R_H/\Omega\right)} \\ \\
v \propto R^{-1/2} \implies v_{esc} = \alpha^{-1/2}v_H \\ \\
P = \frac{\alpha^{-1}v_H^2 \times \alpha^2 R_H^2 \times \Omega}{v_H^2 \times u R_H} \\
P = \frac{\alpha v_H}{u}\]
Putting the entire equation together, we find that the collision rate is given by:
\[\frac{dN}{dt} = \left(\frac{\sigma\Omega}{m} R_H^2\right) \: \alpha\frac{v_H}{u}\]