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Kepler's Third Law (Ratio Form)

T12T22=a13a23\frac{T_1^{2}}{T_2^{2}} = \frac{a_1^{3}}{a_2^{3}}

Worked example: Mars at 1.524 AU vs Earth → T1 = 1.8814 yr — press Try an example to run it live, then adjust anything.

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Kepler's Third Law (Ratio Form) explained

a1a2T1T2

Kepler's "harmonic law" of 1619 says T² ∝ a³ for everything circling the same central body, and the ratio form lets you compare two orbits without knowing G or the central mass at all. Using Earth as the reference body (T₂ = 1 yr, a₂ = 1 AU), Mars at a₁ = 1.524 AU must take T₁ = √(1.524³) ≈ 1.88 years to circle the Sun — precisely its observed year. Run the other way, an asteroid found with a 5.2-year period must orbit at (5.2)2/3(5.2)^{2/3} ≈ 3.0 AU, in the heart of the asteroid belt.

The law works for any shared centre: Jupiter's moons obey it among themselves, as do Earth's satellites. A geostationary satellite (T = 23.93 h) and the Moon (T = 27.32 d ≈ 655.7 h) give a ratio aMoon/ageo=(655.7/23.93)2/3≈9.1a_{\text{Moon}}/a_{\text{geo}} = (655.7/23.93)^{2/3} \approx 9.1 — and indeed the Moon's 384 400 km orbit is about nine times the 42 164 km geostationary radius. When the ratio form fails, something unseen is tugging: discrepancies in Uranus's motion led astronomers straight to Neptune in 1846.

Kepler's Third Law (Ratio Form) formula

T12T22=a13a23\frac{T_1^{2}}{T_2^{2}} = \frac{a_1^{3}}{a_2^{3}}
Where
  • T1T_1= Period of body 1 (s)
  • T2T_2= Period of body 2 (s)
  • a1a_1= Semi-major axis of body 1 (m)
  • a2a_2= Semi-major axis of body 2 (m)

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Part of this formula set

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