10 Kepler’s Laws: Combined Statement
Let \(\gamma \) be a non-collision solution with \(H {\lt} 0\) and \(L \neq 0\). Then:
First law. The orbit lies in the plane \(\Pi \) perpendicular to \(L\) and is an ellipse with the origin at one focus. Its eccentricity is \(e = \| A\| /(m\mu ) {\lt} 1\), its semi-major axis is \(a = -\mu /(2H)\), and its semi-latus rectum is \(\ell = \| L\| ^2/(m\mu )\).
Second law. The areal velocity \(d\mathcal{A}/dt = \| L\| /(2m)\) is constant, so equal areas are swept in equal times.
Third law. Every configuration period \(T\) satisfies \(T^2 = (4\pi ^2 m/\mu )\, a^3\).
For the same solution, the map \(u \mapsto L_u\) is an \(\mathfrak {so}(3)\)-comoment (Theorem 17), and on the shell \(H = h {\lt} 0\) the map \((u, v) \mapsto L_u + K_v\) is an \(\mathfrak {so}(4)\)-comoment (Theorem 21). Combining these statements with Theorem 36 gives the full comoment-map proof of Kepler’s laws.