📂Classical Mechanics

Orbit Equation of a Particle under Central Force

Orbit Equation of a Particle Subject to Central Force¹

A particle of mass $m$ subject to a central force can be described by its motion equation in polar coordinates as follows.

$$\begin{equation} m\ddot{\mathbf{r}}=F(r)\hat{\mathbf{r}} \label{eq1} \end{equation}$$

$F(r)$ represents the central force acting on the particle. The acceleration in polar coordinates can be expressed as:

$$\ddot{\mathbf{r}}=\mathbf{a}=\left( \ddot{r}-r\dot{\theta}{}^{2} \right)\hat{\mathbf{r}} +\left(2\dot{r}\dot{\theta}+r\ddot{\theta} \right)\hat{\boldsymbol{\theta}}$$

Hence, separating the motion equation $\eqref{eq1}$ into components yields:

$$\begin{align} m\left( \ddot{r}-r\dot{\theta}{}^{2} \right) &= F(r) \label{eq2} \\ m\left( 2\dot{r} \dot{\theta} +r\ddot{\theta}\right) &= 0 \nonumber \end{align}$$

Given that $\frac{1}{r}\frac{ d }{ dt }(r^{2}\dot{\theta})=2\dot{r} \dot{\theta} +r\ddot{\theta}$, the second equation can be rewritten as:

$$\frac{ d }{ d t }(r^{2}\dot{\theta})=0$$

This implies that $r^{2}\dot{\theta}$ is a constant. Let’s define this constant as $l$.

$$\begin{equation} r^{2}\dot{\theta}=\text{constant}=l \label{eq3} \end{equation}$$

Meanwhile, the magnitude of the angular momentum of a particle moving under a central force is as follows.

$$\begin{equation} L=mr^{2}\dot{\theta} \tag{4} \label{eq4} \end{equation}$$

Therefore, due to $\eqref{eq3}$ and $\eqref{eq4}$, the following equation:

$$\left| l \right| =\frac{ L}{ m }=\left| \mathbf{r} \times \mathbf{v} \right|$$

holds true. Thus, $l$ can be interpreted as the angular momentum per unit mass of the particle. The fact that $l$ is constant indicates that the angular momentum of a particle moving under a central force is conserved, which is a known result. To derive the equation of the particle’s trajectory in space, which is independent of time, let’s introduce a new variable as below.

$$u=\frac{1}{r}$$

Then, by $\eqref{eq3}$, $\dot{\theta}=lu^{2}$ holds. Now, calculating $\dot{r}$ and $\ddot{r}$ results in:

$$\begin{align*} \dot{r} &= \frac{ d }{ d t}\left( \frac{1}{u} \right) \\ &= \dfrac{d}{du}\left(\frac{1}{u} \right) \frac{ d u}{ dt } = -\frac{ 1 }{ u^{2} }\frac{ d u}{ d t } \\ &= -\frac{ 1}{ u^{2} }\frac{ d u}{ d \theta }\frac{ d \theta}{ d t } =-\frac{1}{u^{2}}\dot{\theta} \frac{ d u}{ d\theta } \\ &= -l\frac{ d u}{ d\theta } \end{align*}$$

$$\begin{align*} \ddot{r} &= -l\frac{ d }{ dt }\left( \frac{ d u}{ d \theta } \right) \\ &= -l \frac{ d }{ d\theta }\left( \frac{ d u}{ d\theta } \right)\frac{ d \theta}{ d t } \\ &= -l\dot{\theta} \frac{ d ^{2}u }{ d \theta^{2} } \\ &= -l^{2}u^{2}\frac{ d ^{2}u}{ d \theta^{2} } \end{align*}$$

Substituting these into $\eqref{eq2}$ yields the following equation.

$$\begin{align*} && -ml^{2}u^{2}\frac{ d ^{2} u}{ d \theta^{2} }-ml^{2}u^{3}&=F({\textstyle \frac{1}{u}}) \\ \implies && \frac{ d^{2} u}{ d \theta^{2}}+u&=-\frac{1}{ml^{2}u^{2}}F({\textstyle \frac{1}{u}}) \end{align*}$$

Here, if the force $F$ is proportional to the inverse square of the distance, similar to gravity, the orbit becomes an ellipse. This application to the motion of planets is known as Kepler’s First Law.