Kepler's Third Law Calculator

Johannes Kepler originally formulated that the square of a planet's orbital period (T) is directly proportional to the cube of the semimajor axis of its orbit (a). In modern Newtonian mechanics, this relationship is expressed as a^3 / T^2 = G * (M + m) / (4 * pi^2), where G is the gravitational constant, M is the mass of the central star, and m is the mass of the orbiting planet. When the orbiting body is much smaller than the host star, the planet's mass (m) becomes negligible, simplifying the calculation to purely reflect the central star's mass and the orbital radius.

Let us calculate the orbital period of an exoplanet orbiting a star with a mass equal to the Sun (M = 1.989 x 10^30 kg) at a semimajor axis of 4 astronomical units (approx. 5.98 x 10^11 meters). First, cube the semimajor axis distance: (5.98 x 10^11)^3 = 2.138 x 10^35 m^3. Next, divide by the gravitational constant modified term and multiply by 4 * pi^2 factoring in the stellar mass. Solving for the period squared yields approximately 7.15 x 10^15 seconds squared. Taking the square root gives an orbital period of roughly 8.46 x 10^7 seconds, which translates to about 2.68 Earth years.

Ensure that all input values use consistent SI units (meters, kilograms, and seconds) unless the calculator automatically handles unit conversions like astronomical units or solar masses. Keep in mind that Kepler's original harmonic law assumes a perfectly circular orbit; for highly elliptical orbits, always use the true semimajor axis rather than the periapsis or apoapsis distance alone.