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  • Calculating a Planet’s Orbital Period Using Kepler’s Third Law

    By Chris Deziel | Updated Aug 30, 2022

    The pioneering work of German astronomer Johannes Kepler (1571‑1630) and Danish astronomer Tycho Brahe (1546‑1601) yielded the first rigorous mathematical description of planetary motion. Their collaboration produced Kepler’s three laws of planetary motion, which later enabled Sir Isaac Newton (1643‑1727) to formulate the universal law of gravitation.

    Kepler’s Third Law Explained

    Kepler’s third law states that the square of a planet’s orbital period (P) is proportional to the cube of the semi‑major axis (d) of its orbit:

    P² = k·d³

    Here k is a proportionality constant equal to 4π²/(G M), where G is the gravitational constant and M is the mass of the Sun (the planet’s mass is negligible in comparison). Because the Sun’s mass dominates, we can safely treat M as the solar mass.

    Using Astronomical Units for Simplification

    When distance is expressed in astronomical units (AU)—the mean Earth‑Sun distance (~93 million miles)—and the period is measured in Earth years, the constant k reduces to 1. The law then simplifies to:

    P² = d³

    or, solving for the period:

    P = √(d³)

    To find a planet’s year in Earth years, substitute its average distance from the Sun in AU. For example, Jupiter’s orbital radius is 5.2 AU:

    P = √(5.2³) ≈ 11.86 Earth years.

    Determining Orbital Eccentricity

    The eccentricity (E) quantifies how much a planet’s orbit deviates from a perfect circle. It ranges from 0 (circular) to 1 (extremely elongated). For an elliptical orbit with aphelion distance a and perihelion distance p, the eccentricity is calculated as:

    E = (a − p)/(a + p)

    Venus has the most circular orbit (E ≈ 0.007), while Mercury’s is more elongated (E ≈ 0.21). Earth’s orbit sits in between with E ≈ 0.017.

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