Derivative of Angular Component under Central Force

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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 $p$.


Let $\mathbf F$ be expressed as:

$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$


$\mathbf u_r$ 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.



\(\displaystyle F_\theta\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r^2 \dfrac {\d^2 \theta} {\d t^2} + 2 r \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}\) \(=\) \(\displaystyle 0\) 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.