Derivation of the day: velocity evolution of large bodies (two ways) \(u > v_{esc}\)

Pre-Collision

I. Inelastic Collisions

Post-Collision

We assume that collisions at different angles average out such that the head-on and the tail-on cases are the only ones that are interesting to derive.  Letting \(\vec{u}\) and \(\vec{v}\) represent velocities of magnitude \(u\) and \(v\) and \(\vec{v}' = \vec{v} + \Delta v\) we can write:
\[ M\vec{v} + m\vec{u} = M\vec{v}'+m\vec{v}' \\
M\vec{v} + m\vec{u} = (M+m)(\vec{v}+\Delta v) \\
m(\vec{u}-\vec{v}) = M\Delta v \]

II. Elastic Collisions

Pre-Collision

Once again, we let  \(\vec{u}\) and \(\vec{v}\) represent velocities of magnitude \(u\) and \(v\) and \(\vec{v}' = \vec{v} + \Delta v\).  We also write \(\vec{u}' = \vec{u} + \Delta u\).  Conserving both kinetic energy and momentum, we write:
\[M\vec{v}+m\vec{u} = M\vec{v}' + m\vec{u}' \\ M\vec{v}^2+m\vec{u}^2 = M\vec{v}'^2+m\vec{u}'^2\]
Rearranging and dividing,
\[\vec{v}+\vec{v}' = \vec{u} + \vec{u}' \\
2\vec{v} +\Delta v = 2\vec{u} + \Delta u\]
Going back to the equation for momentum, we have \(\Delta u = -\frac{M}{m}\Delta v\).
\[2\vec{v} + \Delta v = 2\vec{u} - \frac{M}{m}\Delta v \\
\left(1+\frac{M}{m}\right)\Delta v = 2(\vec{u}-\vec{v}) \approx \vec{u}-\vec{v} \\
\frac{M}{m} >> 1 \implies M\Delta v \approx m(\vec{u}-\vec{v})\]
Which is the same result as for inelastic collisions.

Rate of Change
We can write the rate of change of the momentum as
\[\frac{dp}{dt} = M\Delta v \times \frac{\text{collisions}}{\text{time}} = M\Delta v \times \frac{\Omega \sigma}{m u}R^2 \text{relative velocity}\]
We add the head-on and tail-on components.  Plugging in scalar quantities with signs the head-on case has \( M\Delta v = -m(u+v) \) and for the tail-on case, \( M\Delta v = m(u-v) \) with related relative velocities \( (u+v) \) and \( (u-v) \) respectively. So:
\[M\frac{dv}{dt}=\frac{\Omega \sigma}{m u}R^2(-m(u+v)^2 + m(u-v)^2)\\M\frac{dv}{dt}=-\frac{\Omega \sigma}{u}R^2(2uv)\]
Rearranging, we find:
\[\frac{1}{v}\frac{dv}{dt}=-\Omega \sigma \frac{R^2}{M} = -\frac{\Omega \sigma}{\rho R}\]