Why $r$ is a function of $\theta$? (polar coordinate)

Question

I am reading "Key Points of Mechanics" by Haruo Yoshida.

He wrote $r$ was a function of $\theta$. (please see below.)

But if $\theta(t_0)=\theta(t_1)$ and $r(t_0) \neq r(t_1)$ for $t_0 \neq t_1$, $r$ is not a function of $\theta$.

Please explain why $r$ is a function of $\theta$.

The equations of motion in a polar coordinate system are the following:

$m (\ddot{r}-r\dot{\theta}^2) = f_r$,
$m(r\ddot{\theta}+2\dot{r}\dot{\theta})=f_{\theta}$.

If $f_r := -\frac{GMm}{r^2}, f_{\theta}:=0$, then
$\ddot{r}-r\dot{\theta}^2 = -\frac{GM}{r^2}$,
$r\ddot{\theta}+2\dot{r}\dot{\theta}=0$.

$\frac{d}{dt}(r^2\dot{\theta}) = 2r\dot{r}\dot{\theta}+r^2\ddot{\theta} = r(r\ddot{\theta}+2\dot{r}\dot{\theta})=0$.

So, $r^2\dot{\theta}$ is constant.
Let $h := r^2\dot{\theta}$.

$\ddot{r}-r\dot{\theta}^2 = \ddot{r}-r(\frac{h}{r^2})^2 = \ddot{r}-\frac{h^2}{r^3} = -\frac{GM}{r^2}$.

$r$ is a function of $\theta$ and the following equation holds:
$\frac{d}{d\theta}(\frac{1}{r^2}\frac{dr}{d\theta})=\frac{1}{r}-\frac{GM}{h^2}$.

Answer 1

I think what you're concerned about is that, after rotating through $2\pi$ radians, $r$ can change. The general approach in orbital mechanics is to let $\theta$ change continuously, even though that means it has to eventually go outside $[0,\,2\pi)$. For example, suppose the object spirals exponentially into the origin: then constants $a,\,b$ exist for which $r=a\exp(-b\theta)$ describes the path, provided we see $\theta$ become arbitrarily large over time. (There's a discussion here of the case where the object spirals exponentially outward instead.)

It's been pointed out by @YvesDaoust that some other paths repeat old values of $\theta$ even without this looping effect. In this case, we'll have no choice but to take $r$ as a multivalued function of $\theta$.

Answer 2

This can arise with any parametric equation of a curve. In

$$\begin{cases}x=x(t)=t^2,\\y=y(t)=t\end{cases}$$

it is impossible to express $y$ as a true, univocal function of $x$,

$$y=f(x),$$ as $$y=\pm\sqrt x.$$

You can understand the author's statement as "$r$ is a multifunction of $\theta$", or just loosely "$r$ varies with $\theta$".

Answer 3

It turns out that the solution to these equations does have $r$ as a function of $\theta$, but this depends greatly on this particular potential function.

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Starting with $$ \ddot{r}-r\dot\theta^2=-\frac{GM}{r^2}\tag1 $$ and $$ r\ddot\theta+2\dot{r}\dot\theta=0\tag2 $$ equation $(2)$ says that $$ \dot\theta=\frac{k_1}{r^2}\tag3 $$ Plugging $(3)$ into $(1)$ gives $$ \ddot{r}=\frac{k_1^2}{r^3}-\frac{GM}{r^2}\tag4 $$ Apply $\frac{\mathrm{d}}{\mathrm{d}t}=\frac{k_1}{r^2}\frac{\mathrm{d}}{\mathrm{d}\theta}$: $$ \begin{align} \dot{r} &=\frac{k_1}{r^2}\frac{\mathrm{d}r}{\mathrm{d}\theta}\\ &=-k_1\frac{\mathrm{d}\frac1r}{\mathrm{d}\theta}\tag5 \end{align} $$ and again: $$ \begin{align} \ddot{r} &=-\frac{k_1^2}{r^2}\frac{\mathrm{d}^2\frac1r}{\mathrm{d}\theta^2}\\ &=\frac{k_1^2}{r^3}-\frac{GM}{r^2}\tag6 \end{align} $$ Therefore, $$ \begin{align} \frac{\mathrm{d}^2\frac1r}{\mathrm{d}\theta^2} &=\frac{GM}{k_1^2}-\frac1r\tag7 \end{align} $$ Thus, $$ \frac{GM}{k_1^2}-\frac1r=k_2\cos(\theta-\theta_0)\tag8 $$ Solving for $r$ yields $$ r=\frac1{\frac{GM}{k_1^2}-k_2\cos(\theta-\theta_0)}\tag9 $$ which is the polar equation for an ellipse, but the important part is that $r$ is a function of $\theta$.