2nd Law: Radius vector sweeps out equal areas in equal times.

Proof) From the rotational analogue of Newton's 2nd law, namely
$\vec{\tau } = \displaystyle\frac{d\vec{L}}{dt}$ where $\vec{\tau}=\vec{r}\times\vec{F}$ (torque) and

$$\vec{L} = I\vec{\omega } = \vec{r}\times\vec{p}$$ (8)

we find that

$$L = mvr\sin\theta = \ \mbox{constant}$$ (9)

which implies $dA = \frac{L}{2m}dt = \ \mbox{constant}\ \times dt$ (see diagram above)