Elliptic Orbits

In this section we will prove that the solutions to the equation of motion are conic sections in general, and ellipses in particular. For the sake of simplicity, we consider only the simplified version for an earth orbiting satellite (13). This equation gives ${\bf r}$ as a function of the time $t$. From the previous section we know that the motion is restricted to a plane. To continue we introduce polar coordinates in this plane and write:

$${\bf r}(r,\theta)=r{\bf\hat{r}},$$

and hence

$$\frac{d{\bf r}}{dt}=\frac{dr}{dt}{\bf\hat{r}}+r\frac{d{\bf\... ...rac{dr}{dt}{\bf\hat{r}}+r{\bf\hat{\theta}}\frac{d\theta}{dt},$$

where ${\bf\hat{\theta}}$ is a unit vector perpendicular to ${\bf\hat{r}}$, which forms a right hand system together with ${\bf\hat{r}}$. Using this we can rewrite the angular momentum as

$${\bf h}=m r{\bf\hat{r}}\times\left(\frac{dr}{dt}{\bf\hat{r}... ...theta}{dt} \left({\bf\hat{r}}\times{\bf\hat{\theta}}\right).$$

In particular, we have

$$h=\vert{\bf h}\vert= mr^2\frac{d\theta}{dt}= 2m\frac{dA}{dt},$$

where $A$ is the area swept over by ${\bf r}$. The conservation of angular momentum therefore directly implies Kepler's second law.

To continue observe that

$$0=\frac{d{{\bf\hat{\theta}}\cdot{\bf\hat{\theta}}}}{dt}=2{\bf\hat{\theta}}\cdot\frac{d{\bf\hat{\theta}}}{dt},$$

and hence $\frac{d{\bf\hat{\theta}}}{dt}$ is perpendicular to ${\bf\hat{\theta}}$ itself and therefore parallel to ${\bf\hat{r}}$. It turns out that

$$\frac{d{\bf\hat{\theta}}}{dt}= -{\bf\hat{r}}\frac{d\theta}{dt}.$$

Combining all these properties we can write:

$$\frac{d^2{\bf r}}{dt^2}={\bf\hat{r}}\left(\frac{d^2r}{dt^... ...rac{d^2\theta}{dt^2}+2\frac{dr}{dt}\frac{d\theta}{dt}\right),$$

and the entire equation of motion becomes

$${\bf\hat{r}}\left(m\frac{d^2r}{dt^2}-mr\left(\frac{d\theta}... ...{d^2\theta}{dt^2}+2m\frac{dr}{dt}\frac{d\theta}{dt}\right)=0.$$

Using the expression for $L$ above we get

$${\bf\hat{r}}\left(m\frac{d^2r}{dt^2}-mr\left(\frac{d\theta}... ...{M_E m}{r^2}\right)+{\bf\hat{\theta}}\frac1r \frac{dL}{dt}=0$$

Since the angular momentum is conserved $\frac{dh}{dt}=0$ we get

$$m\frac{d^2r}{dt^2}-mr\left(\frac{d\theta}{dt}\right)^2+G\frac{M_Em}{r^2}=0.$$ (16)

We can eliminate $\theta$ by using $\frac{d\theta}{dt}=\frac{h}{mr^2}$. Doing this reduces the equation of motion to a single non-linear second order ODE. Observe that we started out with a system of three second order non-linear ODE.

What follows is a number of little tricks to simplify (16) even further. First observe that

$$\frac{dr}{dt}=\frac{dr}{d\theta}\frac{d\theta}{t}=\frac{h}{mr^2}\frac{dr}{d\theta}.$$

Differentiating this again with respect to $t$ we get

$$\frac{d^2 r}{dt^2}=-\frac{2h}{mr^3}\frac{dr}{dt}\frac{dr}{... ...d\theta}\right)^2+ \frac{h^2}{m^2r^4}\frac{d^2r}{d\theta^2}.$$

Plugging this into (16) we get after some algebra

$$\frac1{r^2}\frac{d^2 r}{d\theta^2}-\frac{2}{r^3}\left(\frac{dr}{d\theta}\right)^2-\frac1r+\frac{Gm^2M_E}{h^2}=0.$$ (17)

All this did so far is change the independent variable from $t$ to $\theta$. (16) was a non-linear second order equation and as such difficult to solve.. To continue we make a change of dependent variable, namely let

$$r=\frac1w$$

Then

$$\frac{dr}{d\theta}=-\frac1{w^2}\frac{dw}{d\theta},$$

and

$$\frac{d^2r}{d\theta^2}=\frac2{w^3}\left(\frac{dw}{d\theta}\right)^2-\frac1{w^2}\frac{d^2w}{d\theta^2}.$$

We observe that this introduces another non-linear term of the form

$$\left(\frac{dw}{d\theta}\right)^2.$$

However, after substituting this we see that this new nonlinear term cancels with the existing nonlinear term and we get

$$\frac{d^2 w}{d\theta^2}+w=\frac{Gm^2M_E}{L^2}.$$ (18)

This new equation is now a linear second order equation with constant coefficients, the best of all possible worlds. This equation is easy to solve and its solutions are of the form

$$w=A\cos\left(\theta-\theta_0\right)+\frac{Gm^2M_E}{L^2},$$ (19)

where the constants $A$ and $\theta_0$ depend on the initial conditions. After re-introducing $r$ we get:

$$r=\frac1{A\cos\left(\theta-\theta_0\right)+\frac{Gm^2M_E}{L^2}}=\frac{p}{e\cos\left(\theta- \theta_0\right)+1},$$ (20)

where

$$p=\frac{L^2}{Gm^2M_E},$$

and

$$e=Ap.$$

(20) is of course the well known equation of an ellipse in polar coordinates with eccentricity $e$ and focus at the origin. We have thus established Kepler's first law, namely that the orbits are elliptical.