Formula & Calculator

Kepler's Third Law

Relates a body's orbital period to the semi-major axis of its orbit.

AstronomyOrbital MechanicsFundamental

Kepler's Third Law Calculator T² = (4π² / GM) · a³

T = 2π · √(a³ / μ)
T = orbital period (s)  ·  a = semi‑major axis (m)  ·  μ = GM (m³/s²)
⟹ Solve T, a, μ
m
m³/s²
s
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T = 2π · √(a³/μ)  ·  μ = GM is the standard gravitational parameter. Valid for elliptical orbits (e > 0).

Interpretation

T² = (4π²/GM) a³. Relates orbital period (T) and semi‑major axis (a). For objects orbiting a central mass M. Used in celestial mechanics and exoplanet detection.

T² = (4π² / GM) a³
Kepler's Third Law

Variables

SymbolQuantityUnit
TOrbital periods
GGravitational constant
MCentral masskg
aSemi-major axism

What it means

Kepler’s third law states that the square of the orbital period (T) of a planet is proportional to the cube of the semi‑major axis (a) of its orbit. The constant of proportionality depends on the central mass (M) and the gravitational constant (G). This law was derived empirically by Johannes Kepler and later explained by Newtonian gravity. It is fundamental to celestial mechanics, used to determine masses of planets, stars, and even black holes. In astronomy, it enables the calculation of orbital distances from periods (or vice versa), and is essential for understanding the dynamics of binary star systems, exoplanets, and satellite orbits. The simplified form T² = a³ (years and AU) applies to the solar system. Understanding this law is crucial for astrophysicists and planetary scientists.

Worked example

Kepler's Third Law – Two Detailed Examples

Real‑World
Scenario: An astrophysicist is studying the orbit of a newly discovered exoplanet around a Sun‑like star. They measure the semi‑major axis of the planet's orbit as 1.496 × 10¹¹ m (1 AU). Using the known mass of the star (1.989 × 10³⁰ kg) and the gravitational constant G = 6.6743 × 10⁻¹¹ m³/kg·s², they want to calculate the orbital period T. This allows them to determine the year length on that planet and compare it to Earth's orbit.
ParameterValue
G (m³/kg·s²)6.67430e-11
M (kg)1.989e30
a (m)1.496e11
1T² = (4π² / (G M)) × a³
2Compute constant: 4π² / (6.67430e-11 × 1.989e30) ≈ 2.974e-19
3a³ = (1.496e11)³ ≈ 3.347e33, multiply gives T² ≈ 9.95e14, T ≈ 3.155e7 s ≈ 1 year
Result 3.156e7 s ✓ Approximately 1 year
Scenario: A space mission planner needs to calculate the orbital period of a satellite around Earth. The satellite is in a circular orbit at an altitude of 400 km, so its orbital radius (a) is Earth's radius (6.371×10⁶ m) plus 400,000 m = 6.771×10⁶ m. Using Earth's mass (5.972×10²⁴ kg) and G, they compute the period to schedule communication passes and propulsion manoeuvres.
ParameterValue
G6.67430e-11
M5.972e24
a6.771e6
1T² = (4π² / (6.67430e-11 × 5.972e24)) × (6.771e6)³
2Computed T ≈ 5.521e3 s ≈ 92 minutes
Result 5.52e3 s (≈92 min) ✓ Low Earth orbit period
Insight: Kepler's third law relates orbital period to semi‑major axis and central mass. The constant 4π²/(GM) is specific to the central body.

Common mistakes

  • Units: T in seconds, a in metres, M in kg, G in m³/(kg·s²).
  • Period squared: T² is proportional to a³ – not a².
  • Orbiting mass: M is the mass of the central body (e.g., Sun) – not the orbiting object.
  • Simplified form: For solar system (a in AU, T in years): T² = a³ (when M = 1 solar mass).
  • Elliptical orbits: a is the semi‑major axis, not the average radius.

Applications

Kepler's third law, T² = (4π²/GM) a³, relates the orbital period (T) of a planet or satellite to the semi‑major axis (a) of its orbit, with the constant depending on the mass of the central body (M). This law is the foundation of orbital mechanics, enabling astronomers to determine the masses of planets, stars, and even black holes from the orbits of their companions. Space mission planners use it to calculate the required orbital parameters for satellites and interplanetary probes. By measuring the orbital period and distance, scientists can infer the mass of the central object, which is crucial for understanding planetary systems and stellar evolution. This law also underpins the discovery of exoplanets via radial velocity and transit timing variations. Understanding Kepler's third law is essential for any work involving celestial dynamics.

  • Determination of planetary and stellar masses from orbital motion
  • Design of satellite orbits and interplanetary trajectories
  • Discovery and characterisation of exoplanets
  • Calculation of asteroid and comet orbital elements
  • Fundamental education in celestial mechanics

Frequently Asked Questions

Q01What is Kepler's Third Law formula and what does it describe?
A01

T² = (4π² / GM) a³. It describes the relationship between a body's orbital period (T) and the semi‑major axis (a) of its orbit. It applies to any two bodies orbiting under mutual gravity and is fundamental to celestial mechanics.

Q02What do T, a, G, and M represent in this equation?
A02

  • T – orbital period (time for one complete orbit).
  • a – semi‑major axis of the orbit (average distance from the central body).
  • G – gravitational constant (6.674×10⁻¹¹ N·m²/kg²).
  • M – mass of the central body (e.g., the Sun, Earth, or a planet).

Q03What is the simplified form of Kepler's Third Law for objects orbiting the Sun?
A03

When using years for T and astronomical units (AU) for a, the law simplifies to T² = a³ (since the constants cancel in these units). This form is extremely useful for quick calculations in the solar system.

Q04How does the law change for elliptical orbits compared to circular ones?
A04

Kepler's Third Law holds for any elliptical orbit, with a being the semi‑major axis. The period depends only on the semi‑major axis, not on the eccentricity. This is a remarkable result: two orbits with the same a have the same period, regardless of how elongated they are.

Q05How is Kepler's Third Law derived from Newton's laws?
A05

Starting from Newton's law of gravitation and centripetal force for a circular orbit: m v²/a = GMm/a². With v = 2πa/T, we get 4π²a/T² = GM/a², which rearranges to T² = (4π²/GM) a³. The derivation extends to ellipses via calculus.

Q06How is Kepler's Third Law applied to satellites around Earth?
A06

Using Earth's mass (M_E ≈ 5.972×10²⁴ kg), we can compute the orbital period for any given semi‑major axis. For a geostationary orbit, where T = 24 hours, solving gives a ≈ 42,164 km (about 35,786 km above Earth's surface).

Q07What is the significance of the exponent 3/2 in the relationship between T and a?
A07

Since T² ∝ a³, it follows that T ∝ a^(3/2). This means that orbital period increases faster than the size of the orbit. For instance, doubling the semi‑major axis increases the period by a factor of 2^(3/2) ≈ 2.83.

Q08How does Kepler's Third Law help in determining the mass of celestial bodies?
A08

By measuring the orbital period and semi‑major axis of a satellite or planet, we can compute the mass of the central body: M = (4π² a³) / (G T²). This is how we determine masses of planets, stars, and even supermassive black holes (using orbiting stars or gas).

Q09What is the difference between Kepler's Third Law and the vis‑viva equation?
A09

Kepler's Third Law relates the period to the semi‑major axis. The vis‑viva equation gives the speed at any point in an orbit: v² = GM (2/r − 1/a). They complement each other: one gives the time scale, the other gives the instantaneous velocity.

Q10How does Kepler's Third Law apply to binary star systems?
A10

For two stars orbiting their common centre of mass, the law is modified: T² = (4π² / (G(M₁+M₂))) a³, where a is the separation between the stars and M₁+M₂ is the total mass. This allows astronomers to measure stellar masses from binary orbits.