Kepler Third Law

There are two calculation routes. Within the solar system you use the pure ratio: in the units AU and years, T² = a³ holds for the Sun, because the Earth orbit sets both constants to 1. Mars with a = 1.524 AU therefore has T = √(1.524³) ≈ 1.88 years. For satellites or other central bodies you need the SI form T² = 4π²·a³/(G·M) with T in seconds, a in metres and M in kilograms. Choose the route by the question: ratios between planets work without G and M, absolute quantities such as a satellite altitude demand the full formula.

A geostationary satellite appears fixed above one point of the equator, so its period is one sidereal day T = 86,164 s. Rearrange the law for the semi-major axis: a³ = G·M·T²/(4π²). With G·M_Earth = 3.986×10¹⁴ m³/s² you get a³ = 3.986×10¹⁴ × 7.42×10⁹/39.48 ≈ 7.50×10²², so a ≈ 42,160 km from the Earth centre. Subtracting the Earth radius of 6,371 km leaves an orbital altitude of about 35,790 km above the equator. Television and weather satellites sit there. The most common mistakes are not subtracting the Earth radius or using the 24-hour solar day instead of the sidereal day.

Rearrange the SI form for the central mass: M = 4π²·a³/(G·T²). You only need the orbital data of one satellite. Example, Earth mass from the lunar orbit: a = 3.844×10⁸ m and T = 27.32 days = 2.36×10⁶ s give M = 39.48 × 5.68×10²⁵/(6.674×10⁻¹¹ × 5.57×10¹²) ≈ 6.0×10²⁴ kg, in good agreement with the literature value 5.97×10²⁴ kg. The solar mass (from the Earth orbit), the Jupiter mass (from the Galilean moons) and the masses of exoplanet systems were determined by the same pattern. Important: the mass of the small satellite cancels; you always obtain the mass of the central body.

Because the constant contains only properties of the central body. Newton derivation shows it: the gravitational force G·M·m/r² provides the centripetal force m·ω²·r of the orbit. The planet mass m cancels, and with ω = 2π/T what remains is T²/r³ = 4π²/(G·M). The right side holds only natural constants and the solar mass M, so Mercury through Neptune yield the same numerical value although their periods range from 88 days to 165 years. Exactly this told Newton that his theory of gravity was right: it reproduces Kepler empirically found law. Around another central body, such as Earth or Jupiter, the same structure holds with a different M and therefore a different constant.

According to Kepler first law, planetary orbits are ellipses with the Sun at one focus. The semi-major axis a is half the length of the longest axis of the ellipse and at the same time the mean of the smallest and largest solar distances: a = (r_perihelion + r_aphelion)/2. Exactly this a enters the third law, not the current distance, which changes constantly. For the nearly circular Earth orbit a is practically the orbital radius (1 AU = 1.496×10¹¹ m). Remarkably, the period depends only on a, not on the eccentricity. A comet with the same semi-major axis as a planet takes exactly as long per orbit, no matter how elongated its path is.