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    Orbital mechanics

    Orbital Period Calculator

    How long one orbit takes for a given semi-major axis around any Solar System body.

    T = 2π √(a³ / μ)

    Inputs

    Results

    Orbital period
    23.9343 hours
    23 h 56 min
    Mean orbital speed (circular)
    3.07467 km/s
    Worked solution
    1. a³ = 7.4959 × 10^22 m³
    2. a³/μ = 188,060,000 s²
    3. T = 2π √(a³/μ) = 86,164 s (23 h 56 min)

    How the result changes

    Period (hours) plotted against Semi-major axis (km), from one quarter to four times your input, with your input marked.05010015020050,000100,000150,000Semi-major axis (km)Period (hours)your input
    Period vs semi-major axis

    How it works

    Kepler's third law, refined by Newton, says the square of the period grows with the cube of the orbit's size. Doubling the semi-major axis lengthens the period by a factor of 2^1.5 ≈ 2.83.

    At 42,164 km from Earth's centre, the period equals one sidereal day (23 h 56 min), which is why geostationary satellites appear fixed in the sky.

    Variables

    • T — orbital period
    • a — semi-major axis (orbital radius for a circle)
    • μ — gravitational parameter GM

    Assumptions and limits

    • Two-body problem with the orbiting mass negligible
    • Period depends only on the semi-major axis, so the result also holds for ellipses

    For education only. These models are simplified and are not suitable for spacecraft navigation or safety-critical engineering. See our scientific accuracy policy.