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 $\polar {r, \theta}$.

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:

 $\displaystyle F_\theta$ $=$ $\displaystyle m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} }$ $\displaystyle F_r$ $=$ $\displaystyle m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}$

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

Thus:

 $\displaystyle F_\theta$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle r^2 \dfrac {\d^2 \theta} {\d t^2} + 2 r \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}$ $=$ $\displaystyle 0$ multiplying through by $r$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac \d {\d t} \paren {r^2 \dfrac {\d \theta} {\d t} }$ $=$ $\displaystyle 0$ Product Rule for Derivatives $\displaystyle \leadsto \ \$ $\displaystyle r^2 \dfrac {\d \theta} {\d t}$ $=$ $\displaystyle h$ Derivative of Constant

for some constant $h$.

That is:

$\dfrac {\d \theta} {\d t} = \dfrac h {r^2}$

Hence the result, by definition of inverse proportion.

$\blacksquare$