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 polar coordinates, ($r$, $\theta$), as illustrated in Figure 13. Here, $r=\sqrt{x^2+y^2}$ 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

$${\bf e}_r = (\cos\theta,\, \sin\theta),$$ (218)

$${\bf e}_\theta = (-\sin\theta,\, \cos\theta),$$ (219)

respectively.

Figure 14: An ellipse.

We can write the position vector of our planet as

$${\bf r} = r\,{\bf e}_r.$$ (220)

Thus, the planet's velocity becomes

$${\bf v} = \frac{d{\bf r}}{dt} = \dot{r}\,{\bf e}_r + r\,\dot{\bf e}_r,$$ (221)

where $\dot{~}$ 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 (218) with respect to time,

$$\dot{\bf e}_r = \dot{\theta}\,(-\sin\theta,\,\cos\theta) = \dot{\theta}\,\,{\bf e}_\theta.$$ (222)

Thus,

$${\bf v} = \dot{r}\,\,{\bf e}_r + r\,\dot{\theta}\,\,{\bf e}_\theta.$$ (223)

Now, the planet's acceleration is written

$${\bf a} = \frac{d{\bf v}}{dt} = \frac{d^2{\bf r}}{dt^2}= \dd... ...eta})\,{\bf e}_\theta + r\,\dot{\theta}\,\,\dot{\bf e}_\theta.$$ (224)

Again, ${\bf e}_\theta$ has a non-zero time-derivative because its direction changes as the planet moves around. Differentiation of Equation (219) with respect to time yields

$$\dot{\bf e}_\theta = \dot{\theta}\,(-\cos\theta,\,-\sin\theta) = - \dot{\theta}\,{\bf e}_r.$$ (225)

Hence,

$${\bf a} = (\ddot{r}-r\,\dot{\theta}^{\,2})\,{\bf e}_r + (r\,\ddot{\theta} + 2\,\dot{r}\,\dot{\theta})\,{\bf e}_\theta.$$ (226)

It follows that the equation of motion of our planet, (212), can be written

$${\bf a} = (\ddot{r}-r\,\dot{\theta}^{\,2})\,{\bf e}_r + (r\,... ...\dot{\theta})\,{\bf e}_\theta = - \frac{G\,M}{r^2}\,{\bf e}_r.$$ (227)

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

$$\ddot{r}-r\,\dot{\theta}^{\,2} = - \frac{G\,M}{r^2},$$ (228)

and a tangential equation of motion,

$$r\,\ddot{\theta} + 2\,\dot{r}\,\dot{\theta} = 0.$$ (229)