Law 2

This one we can deduce. First of all, what is the torque on a planet exerted by the sun? It's zero, since the direction of the force is in the same direction as the vector displacement between the planet and the sun. So ${\bf\tau}~=~ {\bf r}\times {\bf F}~=~ 0$. In other words, the sun doesn't try to twist the planet.

So if the torque is zero, angular momentum is conserved. Let's figure out the angular momentum geometricaly. Now

$${\bf L}~=~ {\bf r}\times {\bf p}~=~ m ({\bf r}\times {d{\bf r}\over dt})$$ (1.6)

So what's ${\bf r}\times d{\bf r}$? This little picture should help

Here we have ${\bf r}$ at some point of time, and show how much area it sweeps out over a very short time $dt$. At the end of this time, ${\bf r}$ has moved by the vector $d{\bf r}$. So let's compute ${\bf r}\times d{\bf r}$. It points out of the page and has a magnitude

$$\vert{\bf r}\times d{\bf r}\vert ~=~ r dr \sin\theta ~=~ r dr_{\perp}$$ (1.7)

This is just twice the area of the blue region (for small $dr$), because the area of this triangle is

$$dA ~=~ {{1 \over 2}} r dr_{\perp}$$ (1.8)

So putting this all together

$$L ~=~ m \vert{\bf r}\times {d{\bf r}\over dt}\vert ~=~ m {2dA\over dt}$$ (1.9)

This says that the angular momentum is proportional to the rate of change of the area swept out. So the rate that area is swept out is constant. Ths is equivalent to to saying that the planet sweeps out equal areas in equal times.