Kepler's Third Law Calculator
Period equals the square root of 4 pi squared times radius cubed divided by G times mass
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 = 6.6743 × 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.
- Write Kepler's third law solved for period: T = √(4π² r³ / GM).
- Compute r³ = (6.771 × 10⁶)³ ≈ 3.103 × 10²⁰ m³.
- Compute the constant 4π² ≈ 39.48, then 4π² r³ ≈ 39.48 × 3.103 × 10²⁰ ≈ 1.225 × 10²² m³.
- Compute GM = (6.6743 × 10⁻¹¹) × (5.972 × 10²⁴) ≈ 3.986 × 10¹⁴ m³/s².
- 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 = 6.6743 × 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 (6.6743 × 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 = 6.6743 × 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.
References:
Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.
NASA, Orbits and Kepler's Laws. https://science.nasa.gov/resource/orbits-and-keplers-laws/
Circular Orbit Diagram
Kepler's third law applies to any closed orbit. This calculator uses the circular-orbit simplification: a body of mass m moves at constant speed along a circle of radius r centered on a much larger central mass M. The period T is the time to complete one revolution.
M — central mass (kg, dominant body) · m — orbiting body mass (cancels out of T) · r — orbital radius (center-to-center distance) · v — tangential orbital velocity · T — period (one full revolution) · G = 6.6743 × 10⁻¹¹ N·m²/kg².
Worked Examples
Satellite Navigation
What is the orbital period of a GPS satellite?
GPS satellites orbit Earth (M = 5.972 × 10²⁴ kg) at r ≈ 2.6561 × 10⁷ m from Earth's center (≈ 20,200 km altitude). The constellation is designed so each satellite repeats its ground track every sidereal day — verify by computing T.
- Knowns: r = 2.6561 × 10⁷ m, M = 5.972 × 10²⁴ kg
- G = 6.6743 × 10⁻¹¹ N·m²/kg²
- T = 2π × √(r³ / (G × M))
- T = 2π × √((2.6561 × 10⁷)³ / (6.6743 × 10⁻¹¹ × 5.972 × 10²⁴))
- T = 2π × √(1.873 × 10²² / 3.984 × 10¹⁴)
- T = 2π × √(4.702 × 10⁷)
T ≈ 43,089 s ≈ 11.97 h
Almost exactly half a sidereal day (11h 58m), so each satellite traces the same ground track twice per day. The constellation has 24+ satellites in 6 orbital planes, giving the four-satellite minimum needed for a position fix from almost anywhere on Earth's surface.
Kepler Verification
Does Kepler's law correctly predict Earth's year from its 1 AU orbit?
Earth orbits the Sun (M = 1.989 × 10³⁰ kg) at r = 1.496 × 10¹¹ m (one astronomical unit). Plug into Kepler's third law and confirm the calendar year matches.
- Knowns: r = 1.496 × 10¹¹ m, M = 1.989 × 10³⁰ kg
- G = 6.6743 × 10⁻¹¹ N·m²/kg²
- T = 2π × √(r³ / (G × M))
- T = 2π × √((1.496 × 10¹¹)³ / (6.6743 × 10⁻¹¹ × 1.989 × 10³⁰))
- T = 2π × √(3.348 × 10³³ / 1.327 × 10²⁰)
- T = 2π × √(2.523 × 10¹³)
T ≈ 3.156 × 10⁷ s ≈ 365.28 days
Within 0.02% of the sidereal year (365.256 days). The tiny gap comes from Earth's eccentricity (we used the mean a = 1 AU, but the law actually requires the semi-major axis) and rounding G and the Sun's mass. This is the same calculation Kepler did empirically — without G.
Exoplanet Astronomy
How short is the year on the "hot Jupiter" HD 209458 b?
HD 209458 b is a gas giant orbiting at r ≈ 7.103 × 10⁹ m (≈ 0.04747 AU) around its host star of mass ≈ 1.148 M_sun ≈ 2.283 × 10³⁰ kg. It was the first exoplanet observed to transit its star — calculate the expected transit period.
- Knowns: r = 7.103 × 10⁹ m, M = 2.283 × 10³⁰ kg
- G = 6.6743 × 10⁻¹¹ N·m²/kg²
- T = 2π × √(r³ / (G × M))
- T = 2π × √((7.103 × 10⁹)³ / (6.6743 × 10⁻¹¹ × 2.283 × 10³⁰))
- T = 2π × √(3.584 × 10²⁹ / 1.524 × 10²⁰)
- T = 2π × √(2.352 × 10⁹)
T ≈ 304,700 s ≈ 3.53 days
Matches the measured transit period of 3.524 days. Hot Jupiters orbiting that close are tidally locked, with day-side temperatures above 1,200 K. The radial-velocity wobble of HD 209458 was first reported in 1999, then confirmed by transit photometry months later.
Central Body Masses
These are the central-body masses M for the calculator — the mass of the dominant body the satellite orbits. Supply the semi-major axis (a) and/or the period (T) yourself; the orbiting body's own mass does not enter Kepler's third law. Click a row to load that body's mass into M while keeping your current orbital-radius and period inputs. For the full sortable list of bodies with radius, surface gravity, and escape velocity, see the planetary data reference.
| Body | Mass (kg) | Type | Source |
|---|---|---|---|
| Sun | 1.99×10³⁰ | Star | NASA Planetary Fact Sheet |
| Mercury | 3.3×10²³ | Rocky planet | NASA Planetary Fact Sheet |
| Venus | 4.87×10²⁴ | Rocky planet | NASA Planetary Fact Sheet |
| Earth | 5.97×10²⁴ | Rocky planet | NASA Planetary Fact Sheet |
| Moon | 7.34×10²² | Moon | NASA Planetary Fact Sheet |
| Mars | 6.42×10²³ | Rocky planet | NASA Planetary Fact Sheet |
| Jupiter | 1.9×10²⁷ | Gas giant | NASA Planetary Fact Sheet |
| Saturn | 5.68×10²⁶ | Gas giant | NASA Planetary Fact Sheet |
Masses are from the NASA Planetary Fact Sheet. Use the central body the satellite actually orbits — a low Earth orbit uses Earth's mass, a heliocentric orbit uses the Sun's.
Kepler's Third Law Formula
Kepler's third law relates the orbital period of a satellite to its semi-major axis and the mass of the central body it orbits. The Newtonian form below is derived by balancing gravity against centripetal acceleration for a circular orbit.
T = 2π × √(r³ / (G × M))
Where:
- T — orbital period — time to complete one revolution (s)
- r — orbital radius (m); for elliptical orbits, use the semi-major axis a in place of r
- G — Newton's universal gravitational constant, 6.6743 × 10⁻¹¹ N·m²/kg²
- M — mass of the central body the satellite orbits (kg)
The period grows as r^(3/2): quadrupling the radius gives roughly 8× the period. The mass of the orbiting satellite does not appear — a 1 kg sensor and a 400-ton space station at the same altitude share the same period. The law applies to any central-force orbit and is the cleanest way to get T from r and M, or M from observed T and r (the standard method for weighing planets and stars).
Related Calculators
- Gravitational Force Calculator — Newton's law F = Gm₁m₂/r² for any pair of masses
- Gravitational Acceleration Calculator — surface gravity from planet mass and radius using g = GM/r²
- Escape Velocity Calculator — minimum launch speed to leave a planet via v = √(2GM/R)
- Gravity Equations Hub — all four gravitational formulas in one comprehensive tool
- Circular Motion Calculator — analyze centripetal force and orbital motion parameters
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