Kepler's Third Law

By Theorem 1, the areal sweep rate $dA/dt=h/2$ is constant, so integrating over one full period $T$ gives the total enclosed area $A = \tfrac{h}{2}T$. For an ellipse with semi-major axis $a$ and semi-minor axis $b$, geometry gives $A=\pi ab$. Equating the two: $A = \pi a b = \dfrac{h}{2}T$.

Next, tie $h$ to the shape via Binet's equation: the semi-latus rectum of the orbit is $p=h^{2}/GM$, and for an ellipse the standard relations $b=a\sqrt{1-e^{2}}$ and $p=a(1-e^{2})=b^{2}/a$ hold. Combining $p=h^{2}/GM$ with $p=b^{2}/a$ gives $h^{2}=\dfrac{GMb^{2}}{a}$, so $h = \sqrt{\dfrac{GMb^{2}}{a}} = b\sqrt{\dfrac{GM}{a}}$.

Substitute this expression for $h$ into $\pi ab=\tfrac{h}{2}T$ and solve for $T$: $T=\dfrac{2\pi ab}{h}=\dfrac{2\pi ab}{b\sqrt{GM/a}}=2\pi a\sqrt{\dfrac{a}{GM}}=2\pi\sqrt{\dfrac{a^{3}}{GM}}$. Squaring both sides gives $T^{2} = \dfrac{4\pi^{2}}{GM}a^{3}$, which is Kepler's third law — and it holds for every ellipse regardless of $e$, since $e$ canceled out completely. $\blacksquare$