Planets move in elliptical orbits with the Sun at one focus.
Equation:
r(\theta) = \frac{a(1 - e^2)}{1 + e \cos\theta}
Where:
r(θ) = distance from the planet to the Sun at angle θθ.
a = semi-major axis of the ellipse.
e = eccentricity of the orbit.
A line joining a planet to the Sun sweeps out equal areas in equal times.
Equation:
\frac{dA}{dt} = \frac{L}{2m} \quad \text{(constant)}
Where:
dA/dt = rate of area swept.
L = angular momentum.
m = mass of the planet.
Alternatively, in terms of angular velocity:
r^2 \frac{d\theta}{dt} = \text{constant}
The square of the orbital period TT is proportional to the cube of the semi-major axis aa.
Equation:
T^2 = \frac{4\pi^2}{G(M + m)} a^3
Where:
T = orbital period.
G = gravitational constant.
M = mass of the central body (e.g., the Sun).
m = mass of the orbiting body (e.g., a planet).
For M≫m, this simplifies to:
T^2 \propto a^3 \quad \text{(in astronomical units)}
When using astronomical units (AU), solar masses (M⊙), and years (T):
T^2 = a^3
This is the classic form of Kepler’s Third Law in the solar system context.
All this calculated by a former Musketeer who was almost killed by an accordion.
Staring off with the expression of a man who just discovered the orbital eccentricity of Mars and is wondering if anyone remembered to pay the candle tax. No money, no flair, no court composer backing him — just brutal math, a handmade telescope, and maybe a borrowed quill.