Deriving Kepler's third law T² ∝ r³
From GMm/r² = mrω² and ω = 2π/T, we get T² = (4π²/GM) r³, so T² ∝ r³ for all bodies orbiting the same central mass. A graph of T² against r³ is a straight line through the origin with gradient 4π²/GM.
GMm/r² = mrω² ⇒ ω² = GM/r³ ⇒ T² = (4π²/GM) r³What this actually means
Standard four-line derivation. Gravity equals centripetal force in the ω form, cancel m to get ω² = GM/r³, substitute ω = 2π/T to get 4π²/T² = GM/r³, rearrange. Using the ω version from the start saves you a substitution.
The proportionality only holds for bodies orbiting the SAME central mass, since the constant contains M. Earth and Jupiter both orbit the Sun, so you can compare them. An Earth satellite and a Jupiter moon you cannot.
In practice the ratio form does everything: (r₂/r₁)³ = (T₂/T₁)². Jupiter takes 11.9 years, so r_J = 1.50 × 10¹¹ × 11.9^(2/3) = 7.82 × 10¹¹ m. No G, no M, no solar mass needed.
That two-thirds power is where calculator errors happen. r ∝ T^(2/3) and T ∝ r^(3/2). Write down which one you need before touching the calculator, since swapping them gives a plausible-looking but wrong number.
For the practical-skills angle: plot T² on the y-axis against r³ on the x-axis and you get a straight line through the origin, gradient 4π²/GM. Measure the gradient and you have found the mass of the central body. That is a favourite planning-question answer.
Watch out for the trap version. 'A 24-hour satellite is replaced by one of twice the mass, also 24-hour. Ratio of radii?' Mass cancelled in the derivation, so same period means same radius, ratio 1 : 1.
Applying T² ∝ r³ across two different central bodies, or mixing up T ∝ r^(3/2) with r ∝ T^(2/3).
Prove it — watch it be true
- Open the T² vs r³ builder and add several orbits at different radii.
- Confirm the plotted points fall on a straight line passing through the origin.
- Read the gradient and compare it with the displayed 4π²/GM value for that planet.