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 1.12. 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

$${\bf e}_r =(\cos\theta,\, \sin\theta),$$ (1.283)

$${\bf e}_\theta = (-\sin\theta,\, \cos\theta),$$ (1.284)

respectively.

We can write the displacement of our planet as

$${\bf r} = r\,{\bf e}_r.$$ (1.285)

Thus, the planet's velocity becomes

$${\bf v} = \frac{d{\bf r}}{dt} = \skew{3}\dot{r}\,{\bf e}_r + r\,\dot{\bf e}_r,$$ (1.286)

where $\dot{\phantom{t}}$ 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, from differentiating Equation (1.283) with respect to time,

$$\dot{\bf e}_r = \skew{5}\dot{\theta}\,(-\sin\theta,\,\cos\theta) = \skew{5}\dot{\theta}\,\,{\bf e}_\theta.$$ (1.287)

Thus,

$${\bf v} = \skew{3}\dot{r}\,\,{\bf e}_r + r\,\skew{5}\dot{\theta}\,\,{\bf e}_\theta.$$ (1.288)

The planet's acceleration is written

$${\bf a} = \frac{d{\bf v}}{dt} = \frac{d^{ 2}{\bf r}}{dt^{2}}= \sk... ...\ddot{\theta})\,{\bf e}_\theta + r\,\skew{5}\dot{\theta}\,\,\dot{\bf e}_\theta.$$ (1.289)

Again, ${\bf e}_\theta$ has a non-zero time-derivative because its direction changes as the planet moves around. Differentiation of Equation (1.284) with respect to time yields

$$\dot{\bf e}_\theta = \skew{5}\dot{\theta}\,(-\cos\theta,\,-\sin\theta) = - \skew{5}\dot{\theta}\,{\bf e}_r.$$ (1.290)

Hence,

$${\bf a} = (\skew{3}\ddot{r}-r\,\skew{5}\dot{\theta}^{\,2})\,{\bf ... ...ew{5}\ddot{\theta} + 2\,\skew{3}\dot{r}\,\skew{5}\dot{\theta})\,{\bf e}_\theta.$$ (1.291)

It follows that the equation of motion of our planet, (1.277), can be written

$${\bf a} = (\skew{3}\ddot{r}-r\,\skew{5}\dot{\theta}^{\,2})\,{\bf ... ...ot{r}\,\skew{5}\dot{\theta})\,{\bf e}_\theta = - \frac{G\,M}{r^{2}}\,{\bf e}_r.$$ (1.292)

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

$$\skew{3}\ddot{r}-r\,\skew{5}\dot{\theta}^{\,2} = - \frac{G\,M}{r^{2}},$$ (1.293)

and a tangential equation of motion,

$$r\,\skew{5}\ddot{\theta} + 2\,\skew{3}\dot{r}\,\skew{5}\dot{\theta} = 0.$$ (1.294)