The estimable Barney Maunder-Taylor asked at MathsJam:

How come the inverse square law leads to elliptical orbits and equal area swept in equal time?

I only know the answer to one of those questions. The differential equations for the inverse square laws work out to be:

$\diffn{2}{r}{t} - r \left (\diff{\theta}{t}\right) = - \frac k {r^2}$

$2 \diff{r}{t} \diff {\theta}{t} + r \diffn{2} {\theta}{t} = 0$

You can rewrite the second one as $\frac 1r \frac{\d}{\dt} \left( r^2 \diff{\theta}{t} \right) = 0$. Since $r$ isn't infinite, the derivative must be zero-- so $r^2 \diff{\theta}{t} = k$.

Now, in an infinitesimal instant $\delta t$, how much area is swept out by an orbit? It's the area of a sector, with radius $r$ and angle $\delta \theta$, so the rate at which area is swept out is $\frac 12 r^2 \diff{\theta}{t}$ -- which, from the previous paragraph, is constant.

As for showing it's an ellipse, I've filled half a notebook without luck. However, Barney had a follow-up question:

Where do you get those equations from?

Start with two orthogonal unit vectors, $\bb {\hat r}$ and $\bb {\hat \theta}$, defined as you'd expect. (They're both functions of $\theta$ but not of $r$; how far around you are changes their direction, but how far away you are doesn't).

If you increase $\theta$ infinitesimally, $\bb {\hat r}$ becomes something like $\bb {\hat r} \cos(\delta\theta) + \bb {\hat \theta} r \sin(\delta\theta)$. Loosely, the $\cos(\delta\theta)$ is 1, and $\sin(\delta\theta)$ is $\delta \theta$, so $\diff{\bb {\hat r}}{t} = r \bb {\hat \theta} \diff{\theta}{t}$. Similarly, $\diff{\bb {\hat \theta}}{t} = - \bb {\hat r} \diff{\theta}{t}$. (Apparently, this 'loosely' stuff is non-standard calculus, but it's totally legit.)

So, if we've got a position vector $\bb R= r \bb {\hat r}$ , its first time derivative is $\diff{R}{t} = \diff{r}{t} \bb {\hat r} + r \diff{\theta}{t} \bb {\hat \theta}$ (using the product rule).

Applying it again gives $\diffn{2}{R}{t} = \diffn{2}{r}{t} \bb {\hat r} + \diff{r}{t} \left(r \diff{\theta}{t} \bb {\hat \theta}\right) + \diff{r}{t} \diff{\theta}{t} \bb {\hat \theta} + r \diffn{2}{\theta}{t} \bb {\hat \theta} - r \left(\diff{\theta}{t}\right)^2 \bb {\hat r}$

Tidy tidy tidy: $\diffn{2}{R}{t} = \left(\diffn{2}{r}{t} - r\left(\diff{\theta}{t}\right)^2\right) \bb {\hat r} + \left(2 \diff{r}{t} \diff{\theta}{t} + r \diffn{2}{\theta}{t}\right) \bb {\hat \theta}$, as required.

If the only force is centripetal and inverse-square, then $\diffn{2}{r}{t} - r \left(\diff{\theta}{t}\right)^2 = - \frac{k}{r^2}$ and $2 \diff{r}{t}\diff{\theta}{t}+ r \diffn{2}{\theta}{t} = 0$.