Derivative of Angular Component under Central Force

Theorem
Let a point mass $p$ of mass $m$ be under the influence of a central force $\mathbf F$.

Let the position of $p$ at time $t$ be given in polar coordinates as $\left\langle{r, \theta}\right\rangle$.

Let $\mathbf r$ be the radius vector from the origin to $p$.

Then the rate of change of the angular coordinate of $p$ is inversely proportional to the square of the radial coordinate of $p$.

Proof
Let $\mathbf F$ be expressed as:
 * $\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$

where:
 * $\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$
 * $\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$
 * $F_r$ and $F_\theta$ are the magnitudes of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively.

From Motion of Particle in Polar Coordinates, the second order ordinary differential equations governing the motion of $m$ under the force $\mathbf F$ are:

However, we are given that $\mathbf F$ is a central force.


 * CentralForce.png

Thus:

for some constant $h$.

That is:
 * $\dfrac {\mathrm d \theta} {\mathrm d t} = \dfrac h {r^2}$

Hence the result, by definition of inverse proportion.