9.1 Periods

Definition 32 Configuration and orbital period

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A configuration period is a minimal positive time \(T {\gt} 0\) such that \(q(t + T) = q(t)\) for all \(t\). An orbital period adds the condition that the swept area over \([0, T]\) equals \(\pi ab\).

Proposition 33 One period traverses the ellipse once

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For a bounded orbit with configuration period \(T\), the orbit traces the ellipse exactly once over \([0, T]\).

Proof ▶

Lean proves that there are orthonormal vectors \(e_A\) and \(e_B\) in the orbital plane and a function \(\theta (t)\) such that

\[ q(t) = a(\cos \theta (t) - e)e_A + b \sin \theta (t)e_B. \]

So \(\theta (t)\) parametrizes the ellipse. The same theorem also gives a derivative formula for \(\theta \). Since \(x(t) = a(\cos \theta (t) - e)\) and \(y(t) = b\sin \theta (t)\) together determine \(q(t)\) in the orbital plane, the angle \(\theta (t)\) determines \(q(t)\). Note that for the eccentric anomaly, \(r(t) = a(1 - e\cos \theta (t))\).

The Lean theorem gives the areal velocity formula in terms of \(\theta \):

\[ \frac{d\mathcal A}{dt} = \frac{ab}{2}(1 - e\cos \theta (t))\, \dot\theta (t). \]

The second law says this equals the positive constant \(\| L\| /(2m)\). Since \(0 \le e {\lt} 1\), we have \(1 - e\cos \theta {\gt} 0\), so it follows that

\[ \dot\theta = \frac{\| L\| }{mab(1 - e\cos \theta )} {\gt} 0. \]

So \(\theta (t)\) is strictly increasing. The orbit is traversed in one direction.

The Lean theorem also states that \(\theta (T) = \theta (0) + 2\pi \). Therefore \(q\) traces the ellipse exactly once.

Theorem 34 Area of one full orbit

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If \(T\) is a configuration period, then \(\mathcal{A}(0, T) = \pi ab\).

Proof ▶

By Proposition 33, the orbit sweeps the interior of the ellipse exactly once over \([0, T]\). The area enclosed by an ellipse with semi-axes \(a\) and \(b\) is \(\pi ab\).

Theorem 35 Kepler’s third law

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For an orbital period \(T\),

\[ T^2 = \frac{4\pi ^2 m}{\mu }\, a^3. \]

Proof ▶

By the second law, the area swept over one period is:

\[ \mathcal{A}(0, T) = \frac{\| L\| }{2m} \cdot T. \]

By Theorem 34, this equals \(\pi ab\). So:

\[ T = \frac{2\pi m ab}{\| L\| }. \]

Squaring:

\[ T^2 = \frac{4\pi ^2 m^2 a^2 b^2}{\| L\| ^2}. \]

From Proposition 24(c), \(b^2 = a\ell = a\| L\| ^2/(m\mu )\), so \(\| L\| ^2 = mb^2\mu /a\). Substituting:

\[ T^2 = \frac{4\pi ^2 m^2 a^2 b^2}{mb^2\mu /a} = \frac{4\pi ^2 m}{\mu }\, a^3. \]