The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit: \( T^2 \propto a^3 \).
Setup — Circular Orbit Approximation
We first prove the law exactly for circular orbits (it extends to ellipses via the area law). Consider a planet of mass \(m\) orbiting the Sun of mass \(M\) in a circle of radius \(r\) at speed \(v\).
Newton's law of gravitation supplies the centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
Cancel \(m\) and solve for \(v^2\):
\[ v^2 = \frac{GM}{r} \]
Express v in terms of Period T
The orbital circumference is \(2\pi r\), so:
\[ v = \frac{2\pi r}{T} \implies v^2 = \frac{4\pi^2 r^2}{T^2} \]
Equate and Solve
\[ \frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \]
\[ T^2 = \frac{4\pi^2}{GM} r^3 \]
Kepler's 3rd Law — Proved
\[ \boxed{T^2 = \frac{4\pi^2}{GM}\, r^3} \]
The constant of proportionality \(\dfrac{4\pi^2}{GM}\) depends only on the central mass, explaining why the same law holds for all planets. For ellipses, \(r \to a\) (semi-major axis) — proved once we have the 1st law.
A line joining a planet to the Sun sweeps out equal areas in equal times.
Setup — Polar Coordinates
Work in polar coordinates \((r, \theta)\) centred on the Sun. The position vector is \(\mathbf{r}\). The area swept in time \(dt\) is the area of the thin triangle:
\[ dA = \tfrac{1}{2}|\mathbf{r} \times d\mathbf{r}| = \tfrac{1}{2}|\mathbf{r} \times \mathbf{v}|\,dt \]
Define the specific angular momentum vector:
\[ \mathbf{L} = \mathbf{r} \times \mathbf{v} \implies \frac{dA}{dt} = \frac{|\mathbf{L}|}{2} \]
Key Step — Show L is Constant
Differentiate \(\mathbf{L}\) with respect to time:
\[ \frac{d\mathbf{L}}{dt} = \frac{d}{dt}(\mathbf{r} \times \mathbf{v}) = \mathbf{v}\times\mathbf{v} + \mathbf{r}\times\dot{\mathbf{v}} = \mathbf{0} + \mathbf{r}\times\mathbf{a} \]
Newton's gravity is a central force — it acts along \(\mathbf{r}\):
\[ \mathbf{F} = m\mathbf{a} = -\frac{GMm}{r^2}\hat{\mathbf{r}} \implies \mathbf{a} = -\frac{GM}{r^2}\hat{\mathbf{r}} \]
Therefore:
\[ \mathbf{r} \times \mathbf{a} = \mathbf{r} \times \left(-\frac{GM}{r^2}\hat{\mathbf{r}}\right) = -\frac{GM}{r^2}(\mathbf{r}\times\hat{\mathbf{r}}) = \mathbf{0} \]
Because \(\mathbf{r} = r\hat{\mathbf{r}}\), the cross product \(\mathbf{r}\times\hat{\mathbf{r}} = r(\hat{\mathbf{r}}\times\hat{\mathbf{r}}) = \mathbf{0}\).
Thus \(\dfrac{d\mathbf{L}}{dt} = \mathbf{0}\), so \(\mathbf{L} = \text{const}\). Therefore:
Kepler's 2nd Law — Proved
\[ \boxed{\frac{dA}{dt} = \frac{L}{2} = \text{constant}} \]
Equal areas in equal times is simply the conservation of angular momentum for any central force — it doesn't even require gravity to be inverse-square!
The orbit of a planet is an ellipse with the Sun at one focus.
Setup — Equations of Motion in Polar Form
In polar coordinates, Newton's 2nd law gives two components. Using dots for time derivatives:
\[ \ddot{r} - r\dot{\theta}^2 = -\frac{GM}{r^2} \qquad \text{(radial)} \]
\[ r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0 \qquad \text{(tangential)} \]
The tangential equation is just \(\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta}) = 0\), confirming \(L = r^2\dot{\theta} = \text{const}\).
Change of Variable — The Binet Substitution
Let \(u = 1/r\). Express \(\dot{r}\) and \(\ddot{r}\) in terms of \(u' = du/d\theta\):
\[ \dot{r} = \frac{dr}{dt} = -\frac{1}{u^2}\frac{du}{d\theta}\dot{\theta} = -Lu' \]
\[ \ddot{r} = -L\dot{\theta}\,u'' = -L^2u^2\,u'' \]
Using \(\dot\theta = Lu^2\) from the angular momentum relation \(L = r^2\dot\theta\).
Substitute into the Radial Equation
\[ -L^2u^2\,u'' - \frac{1}{u}\cdot L^2u^4 = -GMu^2 \]
Divide through by \(-L^2u^2\):
\[ u'' + u = \frac{GM}{L^2} \]
This is a simple harmonic oscillator equation in \(u(\theta)\) with a constant forcing term!
Solve the ODE
The general solution is:
\[ u(\theta) = \frac{GM}{L^2} + A\cos(\theta - \theta_0) \]
Choose \(\theta_0 = 0\) (orienting the axes to the perihelion). Converting back to \(r = 1/u\):
\[ r = \frac{1}{u} = \frac{L^2/GM}{1 + e\cos\theta} \]
where the
eccentricity is:
\[ e = \frac{AL^2}{GM} \]
Identify as a Conic Section
The equation \(r = \dfrac{\ell}{1 + e\cos\theta}\) with \(\ell = L^2/GM\) is the polar equation of a conic section with the origin at one focus:
\[
\begin{cases}
e < 1 & \implies \textbf{Ellipse} \\
e = 1 & \implies \text{Parabola} \\
e > 1 & \implies \text{Hyperbola}
\end{cases}
\]
For a bound orbit (planet with negative total energy), we must have \(e < 1\), so the orbit is an ellipse. The Sun sits at one focus by construction.
The semi-major axis follows from the perihelion \(r_{\min}\) and aphelion \(r_{\max}\):
\[ a = \frac{r_{\min}+r_{\max}}{2} = \frac{\ell}{1-e^2} = \frac{L^2/GM}{1-e^2} \]
Kepler's 1st Law — Proved
\[ \boxed{r(\theta) = \frac{L^2/GM}{1 + e\cos\theta}, \quad e < 1 \implies \text{Ellipse}} \]
Closing the Loop — 3rd Law for Ellipses
Now that we have the ellipse, we can confirm the 3rd law properly. The total area of the ellipse is \(\mathcal{A} = \pi ab\). From the 2nd law, \(dA/dt = L/2\), so:
\[ T = \frac{\pi ab}{L/2} = \frac{2\pi ab}{L} \]
Using \(b^2 = a\ell = aL^2/GM\) and squaring:
\[ T^2 = \frac{4\pi^2 a^2 b^2}{L^2} = \frac{4\pi^2 a^2 \cdot aL^2/GM}{L^2} = \frac{4\pi^2}{GM}\,a^3 \]
Which closes the full proof — the 3rd law holds exactly for elliptical orbits with \(r \to a\). ∎