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Kepler's Third Law Calculator

Calculate the orbital period from orbital radius using Kepler's third law T² = (4π²/GM)r³. Relate the orbital period and semi-major axis for planets, moons, and satellites in gravitational systems.

Reviewed by Christopher FloiedPublished Updated

This free online kepler's third law calculator provides instant results with no signup required. All calculations run directly in your browser — your data is never sent to a server. Enter your values below and see results update in real time as you type. Perfect for everyday calculations, homework, or professional use.

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Orbital Period

3e+7 s

How to Use This Calculator

1

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2

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3

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4

Explore parameter sensitivity

Try adjusting individual input values to see how the output changes. This is a quick and effective way to develop intuition about how different parameters influence the result and to identify which inputs have the largest effect.

When to Use This Calculator

  • •Use the Kepler's Third Law Calculator when you need accurate results quickly without the risk of manual computation errors or unit conversion mistakes.
  • •Use it to verify calculations made by hand or in spreadsheets — an independent check can catch errors before they lead to costly decisions.
  • •Use it to explore how changing input parameters affects the output — a quick way to develop intuition and identify the most influential variables.
  • •Use it when collaborating with others to ensure everyone is working from the same numbers and applying the same assumptions.

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About Kepler's Third Law Calculator

The Kepler's Third Law Calculator determines the orbital period from the orbital radius and central body mass. Johannes Kepler discovered in 1619 that the square of a planet's orbital period is proportional to the cube of its semi-major axis: T² ∝ r³. Newton later showed this proportionality constant involves the central body's mass. This law enables calculating orbital periods for any satellite around any body — from the ISS around Earth to exoplanets around distant stars. It is the foundation of celestial mechanics and was historically used to weigh the Sun and planets.

The Math Behind It

Kepler's Third Law in its Newtonian form: T² = (4π²/GM)r³, or equivalently T = 2π√(r³/GM). **Historical context**: Kepler's original form (T² ∝ a³) was empirical — discovered by analyzing Tycho Brahe's planetary observations. Newton proved it theoretically using his law of gravitation, and importantly, identified the proportionality constant as 4π²/GM. **Key implications**: 1. **Period depends only on r and M**: Not on the orbiting body's mass 2. **T² ∝ r³**: Doubling orbital radius increases period by factor 2√2 ≈ 2.83 3. **Weighing celestial bodies**: Measuring T and r for a satellite yields M = 4π²r³/(GT²) **Solar system verification**: - Mercury: r = 0.387 AU, T = 0.241 yr → T²/r³ = 1.002 - Earth: r = 1 AU, T = 1 yr → T²/r³ = 1.000 - Jupiter: r = 5.20 AU, T = 11.86 yr → T²/r³ = 0.999 The ratio T²/r³ is constant for all objects orbiting the same central body. **For elliptical orbits**: Replace r with a (semi-major axis). The law works for any eccentricity. **Binary systems**: For two comparable masses: T² = 4π²a³/G(M₁ + M₂). This is used to measure stellar masses in binary star systems. **Exoplanet detection**: The transit method measures T; combining with radial velocity data yields orbital radius and planet mass. Kepler's third law is essential for characterizing the thousands of known exoplanets. **Tidal locking**: Over time, tidal forces can slow rotation until the orbital period equals the rotation period. Earth's Moon is tidally locked (same face always toward Earth). Eventually, Earth's rotation will slow to match the Moon's orbital period.

Formula Reference

Kepler's Third Law

T = 2π√(r³/GM)

Variables: r = orbital radius (m), G = 6.674×10⁻¹¹, M = central body mass (kg)

Worked Examples

Example 1: Earth's Orbit

r = 1.496×10¹¹ m, M = 1.989×10³⁰ kg (Sun)

Step 1:T = 2π√(r³/GM)
Step 2:= 2π√((1.496×10¹¹)³ / (6.674×10⁻¹¹ × 1.989×10³⁰))
Step 3:= 2π√(3.348×10³³ / 1.327×10²⁰)
Step 4:= 2π√(2.524×10¹³)
Step 5:= 2π × 5.024×10⁶ = 31,560,000 s

31.56 million seconds ≈ 365.3 days — one year.

Example 2: ISS Orbit

r = 6,771,000 m, M = 5.972×10²⁴ kg (Earth)

Step 1:T = 2π√((6.771×10⁶)³ / (6.674×10⁻¹¹ × 5.972×10²⁴))
Step 2:= 2π√(3.103×10²⁰ / 3.986×10¹⁴)
Step 3:= 2π√(7.785×10⁵) = 2π × 882.3 = 5543 s

5543 seconds ≈ 92.4 minutes per orbit.

Common Mistakes & Tips

  • !Using orbital altitude instead of orbital radius — add the central body's radius to the altitude.
  • !Forgetting to use the correct mass — the central body, not the orbiting object.
  • !Applying the simple form T² ∝ r³ when comparing orbits around different central bodies — the constant depends on M.
  • !Using the wrong units — SI units (meters, kilograms, seconds) are required for the formula with G.

Related Concepts

Orbital Velocity

Speed in orbit derived from the same gravitational balance.

Gravitational Force

The force that governs orbital mechanics.

Escape Velocity

Speed to leave orbit entirely.

Used in These Calculators

Calculators that build on or apply the concepts from this page:

Orbital Velocity Calculator

Frequently Asked Questions

How was Kepler's third law discovered?

Kepler spent years analyzing Tycho Brahe's precise observations of Mars. In 1619 he published the 'harmonic law' showing T² ∝ a³ for all planets. He considered this his greatest achievement, as it revealed a deep mathematical harmony in planetary motion.

Can we use this to weigh the Sun?

Yes! Rearranging: M = 4π²r³/(GT²). Using Earth's orbital data: M = 4π² × (1.496×10¹¹)³ / (6.674×10⁻¹¹ × (3.156×10⁷)²) = 1.989×10³⁰ kg. This is how the Sun's mass was first accurately determined.

Does this work for elliptical orbits?

Yes, with r replaced by the semi-major axis a (the average of closest and farthest distances from the central body). The law applies to any eccentricity from circular (e = 0) to highly elongated (e → 1).

Why are higher orbits slower?

From T = 2π√(r³/GM), period grows as r^(3/2). Both longer circumference and slower orbital speed contribute: v = √(GM/r) decreases with r, and the path 2πr increases. Together, T grows super-linearly with r.

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