AkCalculators

Orbital period — Kepler's third law, derivations and applications

Orbital motion has been studied for centuries, with Johannes Kepler's empirical laws providing the first compact description of planetary motion. Kepler's third law — relating orbital period and semi-major axis — becomes a precise physical law when combined with Newton's gravitational theory. In Newtonian mechanics, the orbital period T for two bodies in a bound orbit is given by:

T = 2π \u221a(a^3 / (G (M + m)))

Here a is the orbit's semi-major axis (for circular orbits a is the orbital radius), G is the gravitational constant and M and m are the central and orbiting body masses. For most satellite problems m is negligible compared to M and the simplified relation T ≈ 2π √(a^3 / (G M)) is used.

Derivation (sketch)

Start from Newton's law for circular motion: centripetal acceleration v^2 / r must equal gravitational acceleration G M / r^2. For circular orbits v^2 / r = G M / r^2, hence v = √(G M / r). The period T = circumference / speed = 2π r / v, substitute v and simplify to get T = 2π √(r^3 / (G M)). Replacing r with semi-major axis a generalizes the formula to elliptical orbits when using orbital elements.

Units and practical inputs

Use SI units: meters for distance, seconds for time, kilograms for mass. The standard gravitational parameter μ = G M is often used in orbital mechanics because it combines constants and can be measured observationally. For Earth μ ≈ 3.986004418e14 m^3 s^-2 and for the Sun μ ≈ 1.32712440018e20 m^3 s^-2. This calculator accepts μ directly to simplify inputs.

Circular vs elliptical orbits

For a circular orbit a is simply the orbit radius. For elliptical orbits a is the semi-major axis — the average of periapsis and apoapsis distances from the focus. Kepler's third law in this form applies to both circular and elliptical orbits when a is used.

Applications and examples

Example 1 — Low Earth Orbit satellite: For a circular orbit at altitude 350 km above Earth's surface, a ≈ R_earth + 350 km ≈ 6.721e6 m, using Earth's μ ≈ 3.986004418e14 m^3 s^-2 gives T ≈ 2π √(a^3 / μ) ≈ 5.62e3 s ≈ 93.7 minutes.

Example 2 — Geostationary orbit: A geostationary satellite must have T = 24 hours = 86400 s. Rearranging gives a = (μ T^2 / (4 π^2))^(1/3) which yields a ≈ 4.216e7 m from Earth's center (≈ 35,786 km altitude).

Example 3 — Planetary orbit: Using the Sun's μ and Earth's semi-major axis a ≈ 1.496e11 m reproduces Earth's orbital period ≈ 365.25 days.

Limitations and corrections

Kepler/Newtonian formulas assume point masses (or spherically symmetric mass distributions) and ignore perturbations from other bodies, atmospheric drag, non-spherical gravity fields, and relativistic effects. For high-precision orbit determination use numerical integration with full perturbation models and referenced ephemerides.

Using this calculator effectively

Provide any two independent values among a, T, M or μ to compute the third. If you provide μ, you don't need to enter G or M. Use the precision selector for formatting and enable step-by-step derivations for lab reports. Remember to convert altitudes above a body's surface to orbital radius by adding the central body's mean radius.

Orbital period formulas are compact but underlie a wide range of spaceflight and astronomical calculations — from satellite mission design to estimating transit times in exoplanet studies. Use this tool for quick checks and educational purposes, and refer to specialist astrodynamics software for mission-critical planning.