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Kepler's Third Law Calculator

Period equals the square root of 4 pi squared times radius cubed divided by G times mass

Solution

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How It Works

Kepler's Third Law of Planetary Motion states that the square of an orbital period is proportional to the cube of the orbital radius: T² = (4π² / GM) × r³, where G = 6.6726 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant and M is the mass of the central body the satellite orbits. This calculator rearranges the equation to solve for the period T, the orbital radius r, or the central mass M given the other two quantities. It works for any two-body orbit — planets around the Sun, moons around a planet, or satellites around Earth.

Example Problem

Calculate the orbital period of a satellite in low Earth orbit at a radius of 6.771 × 10⁶ m (about 400 km altitude, the ISS orbit). Earth's mass is M = 5.972 × 10²⁴ kg.

  1. Write Kepler's third law solved for period: T = √(4π² r³ / GM).
  2. Compute r³ = (6.771 × 10⁶)³ ≈ 3.103 × 10²⁰ m³.
  3. Compute the constant 4π² ≈ 39.48, then 4π² r³ ≈ 39.48 × 3.103 × 10²⁰ ≈ 1.225 × 10²² m³.
  4. Compute GM = (6.6726 × 10⁻¹¹) × (5.972 × 10²⁴) ≈ 3.986 × 10¹⁴ m³/s².
  5. Divide and take the square root: T = √(1.225 × 10²² / 3.986 × 10¹⁴) = √(3.073 × 10⁷) ≈ 5,544 s ≈ 92.4 minutes — the textbook ISS orbital period.

Key Concepts

Kepler's Third Law is a direct consequence of Newton's gravitational law combined with the requirement that gravity supplies the centripetal force of a circular orbit. The proportionality constant 4π²/GM is fixed for any given central body, so all satellites of that body — regardless of mass — follow the same T² ∝ r³ curve. This is why every Sun-orbiting planet's period in years squared equals its semi-major axis in AU cubed: Earth's relation 1² = 1³ defines those units. The law applies in the limit where the orbiting body's mass is much smaller than the central mass; for comparable masses, the full two-body formula uses the sum M₁ + M₂.

Applications

  • Spacecraft mission planning: computing transfer orbit periods and timing rendezvous with target bodies.
  • Astronomy: estimating the mass of a planet, star, or galaxy from the orbital period and radius of a satellite, moon, or companion star.
  • Exoplanet discovery: deriving the orbital radius of a transiting planet from its observed period and the host star's mass.
  • Geostationary orbit design: solving for the radius (≈ 42,164 km) that gives Earth-orbiting satellites a 23h 56m (sidereal-day) period.
  • Educational physics: demonstrating that the same gravitational law governs falling apples, lunar orbits, and planetary motion.

Common Mistakes

  • Using altitude above a planet's surface instead of orbital radius from the center — radius r in the formula is measured from the center of the central body, not its surface.
  • Mixing up the central mass and the satellite mass — Kepler's third law uses M, the much larger central body, and ignores the orbiter's mass in the standard form.
  • Forgetting to use SI units (kg, m, s) with G = 6.6726 × 10⁻¹¹ — mixing AU, years, and solar masses requires the rearranged Sun-centric form, not the raw SI equation.
  • Applying the law to highly elliptical orbits as if r were a single radius — for non-circular orbits, r is the semi-major axis of the ellipse.
  • Squaring T but cubing r when remembering the relation backwards — the period is squared (T²), the radius is cubed (r³).

Frequently Asked Questions

How do you calculate orbital period using Kepler's third law?

Apply T = √(4π² r³ / GM), where r is the orbital radius from the center of the central body, M is the central body's mass, and G is the universal gravitational constant (6.6726 × 10⁻¹¹ N·m²/kg²). Make sure r is in meters and M is in kilograms before computing; the result T will be in seconds.

What is the formula for Kepler's third law?

T² = (4π² / GM) × r³, which rearranges to T = √(4π² r³ / GM) for period, r = ∛(GMT² / 4π²) for radius, and M = 4π² r³ / (GT²) for mass. T is the orbital period, r the orbital radius, M the central mass, and G = 6.6726 × 10⁻¹¹ N·m²/kg².

What does Kepler's third law tell us?

It says the square of any orbiting body's period is proportional to the cube of its orbital radius. For Sun-orbiting bodies measured in years and AU, this simplifies to T² = r³ exactly. The proportionality constant is fixed by the central body's mass — so any two satellites of the same body satisfy T₁²/r₁³ = T₂²/r₂³.

Can Kepler's third law calculate planet mass?

Yes. Rearranging gives M = 4π² r³ / (GT²) — measure a satellite's orbital period T and radius r, then solve for the central body's mass. This is how astronomers weigh distant stars, planets, and even galaxies using their companions' orbital data.

Does Kepler's third law work for elliptical orbits?

Yes — for elliptical orbits the same relation holds when r is replaced by the semi-major axis a (half the longest diameter of the ellipse). T² = (4π² / GM) × a³. The circular form is just the special case where eccentricity is zero.

Why is the radius cubed and the period squared?

It falls out of equating gravity (∝ 1/r²) to the centripetal force needed for circular motion (∝ r/T²). Setting GM/r² = 4π²r/T² and solving gives T² = (4π²/GM)r³ directly — the squares and cubes are forced by the gravitational inverse-square law.

Reference: Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.

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