Kepler's Second Law (equal areas in equal times)

Define $\mathbf h = \mathbf r\times\dot{\mathbf r}$. Differentiating, $\dot{\mathbf h} = \dot{\mathbf r}\times\dot{\mathbf r} + \mathbf r\times\ddot{\mathbf r} = \mathbf 0 + \mathbf r\times\ddot{\mathbf r}$ (the first term vanishes since any vector crossed with itself is zero). For a central force, $\ddot{\mathbf r}$ is parallel to $\mathbf r$ — write $\ddot{\mathbf r} = f(r)\hat{\mathbf r}$ for some scalar function $f$ — so $\mathbf r\times\ddot{\mathbf r} = f(r)\,\mathbf r\times\hat{\mathbf r} = \mathbf 0$ as well. Hence $\dot{\mathbf h} = \mathbf r\times\ddot{\mathbf r} = -\dfrac{GM}{r^{3}}(\mathbf r\times\mathbf r) = \mathbf 0$: $\mathbf h$ is a fixed vector, constant in both magnitude and direction.

Because $\mathbf h$ has fixed direction and $\mathbf h = \mathbf r\times\dot{\mathbf r}$ is always perpendicular to $\mathbf r$, the position vector $\mathbf r$ is confined forever to the single fixed plane perpendicular to $\mathbf h$ — the motion is planar. Set up polar coordinates $(r,\theta)$ in that plane. Writing $\mathbf r = r\hat{\mathbf r}$ and $\dot{\mathbf r} = \dot r\hat{\mathbf r} + r\dot\theta\hat{\boldsymbol\theta}$, the cross product gives $\mathbf h = r\hat{\mathbf r}\times(\dot r\hat{\mathbf r}+r\dot\theta\hat{\boldsymbol\theta}) = r^{2}\dot\theta\,\hat{\mathbf z}$, so the scalar $h=r^{2}\dot\theta$ is itself constant.

In time $dt$, the radius vector sweeps a thin, nearly triangular sector of area $dA = \tfrac12 r\cdot(r\,d\theta) = \tfrac12 r^{2}\,d\theta$ (base $r\,d\theta$, height $r$, factor $\tfrac12$ for a triangle). Dividing by $dt$ gives exactly $\dfrac{dA}{dt} = \dfrac12 r^{2}\dot\theta = \dfrac{h}{2}$. Since $h$ is constant, the areal sweep rate $dA/dt$ is constant for all time — equal areas are swept in equal times, proving Kepler's second law. $\blacksquare$