The modern way to derive Kepler's area law is to break the force into its
radial and transverse components, rather than the tangential and normal
components used by Hermann, and use polar coordinates whose origin is the
center of force. Assume then that the center of force is at the origin of a
polar coordinate system. Using vector notation, \(\operatorname{set}
\mathbf{u}_{r}=\mathbf{i} \cos \theta+\mathbf{j} \sin \theta\) and
\(\mathbf{u}_{\theta}=-\mathbf{i} \sin \theta+j \cos \theta .\) Show that \(d
\mathbf{u}_{r} / d \theta=\mathbf{u}_{\theta}\) and \(d \mathbf{u}_{\theta} / d
\theta=-\mathbf{u}_{r}\). Then show that if \(\mathbf{r}=r \mathbf{u}_{r}\), then
the velocity \(\mathbf{v}\) is given by \(r(d \theta / d t) \mathbf{u}_{\theta}+\)
\((d r / d t) \mathbf{u}_{r}\). Show next that the radial component \(a_{r}\) and
the transverse component \(a_{\theta}\) of the acceleration are given by
$$
a_{r}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2} \quad
\text { and } \quad a_{\theta}=r \frac{d^{2} \theta}{d t^{2}}+2 \frac{d r}{d
t} \frac{d \theta}{d t}
$$
Since the force is central, \(a_{\theta}=0 .\) Multiply the differential
equation expressing that fact by \(r\) and integrate to get \(r^{2} \frac{d
\theta}{d t}=k\), where \(k\) is a constant. Show finally that \(r^{2} \frac{d
\theta}{d t}=\) \(\frac{d A}{d t}\), where \(A\) is the area swept out by the
radius vector. This proves Kepler's law of areas.