Kepler — Período Orbital

Kepler's third law: orbital period

Kepler's 3rd law states T² = (4π²/GM)·a³. T is the orbital period, a the semi-major axis, G the gravitational constant and M the central mass. Inside the solar system, if you measure a in astronomical units (AU) and T in years, the constant drops to 1 and you're left with T² = a³. Mars is a clean example: a = 1.524 AU gives T² = 3.54, so T ≈ 1.88 years, right where observation puts it. Closer to home, the ISS sits at r = 6,781 km from Earth's center and orbits in T ≈ 92 minutes, a figure you get with Earth's mass M = 5.972·10²⁴ kg.

Applications

Planning interplanetary missions and Hohmann transfer windows. Characterizing exoplanets picked up by TESS and JWST, where the period comes from the transits and the semi-major axis from Kepler's law plus the stellar mass. Laying out the GPS satellite constellation (T = 12 h, a = 26,560 km), working out geostationary orbits, and estimating masses in binary-star systems.

FAQ

Why does T² = a³ work without units in the solar system? With M set to one solar mass, picking AU and years makes 4π²/GM equal to 1. Step outside the solar system and you're back to the full equation.

Does it apply to satellites around Earth? It does. Just swap the Sun's mass for Earth's mass (5.972·10²⁴ kg). A satellite at a = 7,000 km comes out to T ≈ 97 min.

What if the orbit is highly elliptical? The 3rd law runs on the semi-major axis a, which is the average of the perihelion and aphelion radii. Because of that, eccentricity leaves T untouched for a given a.