Plane polar coordinates

Figure 4.1: Plane polar coordinates.

We can determine the instantaneous position of our planet in the $x$-$y$ plane in terms of standard Cartesian coordinates, $x$, $y$, or plane polar coordinates, $r$, $\theta $, as illustrated in Figure 4.1. Here, $r=(x^{\,2}+y^{\,2})^{1/2}$ and $\theta=\tan^{-1}(y/x)$. It is helpful to define two unit vectors, ${\bf e}_r\equiv {\bf r}/r$ and ${\bf e}_\theta\equiv {\bf e}_z\times {\bf e}_r$, at the instantaneous position of the planet. The first always points radially away from the origin, whereas the second is normal to the first, in the direction of increasing $\theta $. As is easily demonstrated, the Cartesian components of ${\bf e}_r$ and ${\bf e}_\theta$ are

$\displaystyle {\bf e}_r$ $\displaystyle = (\cos\theta,\, \sin\theta),$ (4.8)
$\displaystyle {\bf e}_\theta$ $\displaystyle = (-\sin\theta,\, \cos\theta),$ (4.9)


We can write the position vector of our planet as

$\displaystyle {\bf r} = r\,{\bf e}_r.$ (4.10)

Thus, the planet's velocity becomes

$\displaystyle {\bf v} = \frac{d{\bf r}}{dt} = \skew{3}\dot{r}\,{\bf e}_r + r\,\dot{\bf e}_r,$ (4.11)

where $\dot{\phantom r}$ is shorthand for $d/dt$. Note that ${\bf e}_r$ has a non-zero time derivative (unlike a Cartesian unit vector) because its direction changes as the planet moves around. As is easily demonstrated, by differentiating Equation (4.8) with respect to time,

$\displaystyle \dot{\bf e}_r = \skew{5}\dot{\theta}\,(-\sin\theta,\,\cos\theta) = \skew{5}\dot{\theta}\,\,{\bf e}_\theta.$ (4.12)


$\displaystyle {\bf v} = \skew{3}\dot{r}\,\,{\bf e}_r + r\,\skew{5}\dot{\theta}\,\,{\bf e}_\theta.$ (4.13)

The planet's acceleration is written

$\displaystyle {\bf a} = \frac{d{\bf v}}{dt} = \frac{d^{\,2}{\bf r}}{dt^{\,2}}= ...
...\ddot{\theta})\,{\bf e}_\theta + r\,\skew{5}\dot{\theta}\,\,\dot{\bf e}_\theta.$ (4.14)

Again, ${\bf e}_\theta$ has a nonzero time derivative because its direction changes as the planet moves around. Differentiation of Equation (4.9) with respect to time yields

$\displaystyle \dot{\bf e}_\theta = \skew{5}\dot{\theta}\,(-\cos\theta,\,-\sin\theta) = - \skew{5}\dot{\theta}\,{\bf e}_r.$ (4.15)


$\displaystyle {\bf a} = (\skew{3}\ddot{r}-r\,\skew{5}\dot{\theta}^{\,2})\,{\bf ...
...ew{5}\ddot{\theta} + 2\,\skew{3}\dot{r}\,\skew{5}\dot{\theta})\,{\bf e}_\theta.$ (4.16)

It follows that the equation of motion of our planet, Equation (4.2), can be written

$\displaystyle {\bf a} = (\skew{3}\ddot{r}-r\,\skew{5}\dot{\theta}^{\,2})\,{\bf ...
...{r}\,\skew{5}\dot{\theta})\,{\bf e}_\theta = - \frac{G\,M}{r^{\,2}}\,{\bf e}_r.$ (4.17)

Because ${\bf e}_r$ and ${\bf e}_\theta$ are mutually orthogonal, we can separately equate the coefficients of both, in the preceding equation, to give a radial equation of motion,

$\displaystyle \skew{3}\ddot{r}-r\,\skew{5}\dot{\theta}^{\,2} = - \frac{G\,M}{r^{\,2}},$ (4.18)

and a tangential equation of motion,

$\displaystyle r\,\skew{5}\ddot{\theta} + 2\,\skew{3}\dot{r}\,\skew{5}\dot{\theta} = 0.$ (4.19)