Orbital Period Calculator

Calculate orbital period using T = 2π·√(r³/GM) (Kepler's Third Law).

🛸 Orbits📐 T = 2π√(r³/GM)🌍 Kepler
Orbital radius (r) m
Central body mass (M) kg

Presets:

⚠️ Enter valid positive numbers.

What Is Orbital Period?

The orbital period formula T = 2π·√(r³/GM) comes from Kepler's Third Law, derived using Newton's law of gravitation and circular orbit centripetal force. Here T is the period (s), r is the orbital radius (m), G = 6.674×10⁻¹¹ N·m²/kg² is the gravitational constant, and M is the central body's mass (kg). This applies to any circular orbit.

Kepler's Third Law states T² ∝ r³ — period squared is proportional to semi-major axis cubed. The proportionality constant is 4π²/(GM). For Earth's solar orbit: T²/r³ = 4π²/(GM_sun) = constant for all planets. This allows the solar system's scale to be measured with just relative period measurements (done by Kepler from Earth in 1619).

Orbital period varies enormously: LEO satellite ≈ 90 minutes; GEO satellite = exactly 24 hours; Moon = 27.3 days; Earth around Sun = 365.25 days; Pluto = 248 years; Voyager 1 escaped the solar system in 1977 after a 12-year journey. Period increases as r increases — higher orbits are slower.

Geostationary orbit (GEO) is special: T = 24 hours exactly, so the satellite appears stationary above a fixed point on the equator. From T = 86,400 s and M_Earth, r_GEO = (GM_Earth × T²/(4π²))^(1/3) = 42,164 km from Earth's center = 35,786 km altitude.

Formula Reference Table

Solve ForFormulaNotes
Orbital periodT = 2π√(r³/GM)G = 6.674×10⁻¹¹ N·m²/kg²
Orbital radiusr = (GMT²/(4π²))^(1/3)Given T and M
Central massM = 4π²r³/(GT²)From orbit data
Kepler's 3rd lawT² = 4π²r³/(GM)T² ∝ r³
GEO altituder_GEO = 42,164 kmT = 86,400 s, M = M_Earth
Orbital velocityv = 2πr/T = √(GM/r)m/s

3 Worked Examples

Example 1
ISS Orbit Period

ISS orbits at r = 6,771 km = 6.771×10⁶ m, M_Earth = 5.972×10²⁴ kg.

  • T = 2π√(r³/GM) = 2π√((6.771×10⁶)³ / (6.674×10⁻¹¹ × 5.972×10²⁴))
  • T = 2π√(3.097×10²⁰ / 3.985×10¹⁴) = 2π√(7.77×10⁵)
  • T = 2π × 881.5 = 5,539 s = 92.3 min
✓ T = 5,539 s = 92.3 minutes
Example 2
Find GEO Radius

What orbital radius gives a 24-hour period?

  • r = (GM_E × T²/(4π²))^(1/3)
  • r = (6.674×10⁻¹¹ × 5.972×10²⁴ × (86400)²/(4π²))^(1/3)
  • r = (3.985×10¹⁴ × 7.465×10⁹ / 39.48)^(1/3) = (7.532×10²²)^(1/3)
  • r = 4.216×10⁷ m = 42,160 km; altitude = 35,789 km
✓ r = 42,160 km; altitude = 35,789 km (GEO)
Example 3
Find Star Mass from Exoplanet Orbit

Exoplanet orbits a star at r = 1.5×10¹¹ m with T = 1.5 years = 4.73×10⁷ s.

  • M = 4π²r³/(GT²)
  • M = 4π² × (1.5×10¹¹)³ / (6.674×10⁻¹¹ × (4.73×10⁷)²)
  • M = 4π² × 3.375×10³³ / (6.674×10⁻¹¹ × 2.237×10¹⁵)
  • M = 1.334×10³⁵ / 1.492×10⁵ = 8.94×10²⁹ kg ≈ 0.45 M_sun
✓ Star mass ≈ 0.45 solar masses (a cool red dwarf)

Real-World Applications

🛰️
Satellite Communications
GEO satellites at 35,786 km appear stationary, ideal for TV broadcasting and weather monitoring. MEO (GPS at 20,200 km, T≈12 hr) and LEO (Starlink at 550 km, T≈95 min) serve other purposes.
🌙
Lunar Missions
Apollo missions: T_lunar_orbit ≈ 2 hours (r≈1,800 km). Transfer orbit from Earth to Moon took 3 days. Precise T and r calculations were essential for trajectory design.
🌌
Exoplanet Detection
Transit timing: if an exoplanet periodically blocks starlight, T = time between transits. Combined with r (from stellar type), Kepler's law gives M_star. Radial velocity method: orbital velocity v = 2πr/T from Doppler shift.
⚗️
Binary Stars
Binary star systems: both stars orbit their center of mass. T and r measured from apparent motion; Kepler's 3rd law gives M₁ + M₂. Essential for stellar mass determination — the only direct mass measurements available.
☀️
Solar System Mapping
Kepler measured planetary periods from Earth by comparing angular speeds. T²/r³ = constant allowed relative distances in AU (Earth=1 AU). Combining with T gives the absolute solar system scale once 1 AU is measured by parallax.

Common Mistakes to Avoid

⚠️
Using altitude instead of orbital radius

r = R_central_body + altitude. For LEO at 400 km altitude: r = 6,371 + 400 = 6,771 km, not 400 km.

⚠️
Forgetting G constant units

G = 6.674×10⁻¹¹ N·m²/kg² = 6.674×10⁻¹¹ m³/(kg·s²). Use M in kg and r in meters for T in seconds.

⚠️
Using Kepler's 3rd Law without knowing M

T² = 4π²r³/(GM) — you need M to find T. Two planets orbiting the same star can use the ratio form T₁²/T₂² = r₁³/r₂³ without knowing M.

⚠️
Confusing orbital period with rotation period

Orbital period T is the time to complete one orbit. Rotation period (day) is the time the body takes to spin once. Earth's orbital period is 365.25 days; rotation period is 24 hours — very different!

⚠️
Wrong formula for elliptical orbits

T = 2π√(a³/GM) where a = semi-major axis (not radius) for elliptical orbits. For circular orbits, a = r.

Frequently Asked Questions

What is Kepler's Third Law? ▾
T² = 4π²a³/(GM): the square of the orbital period is proportional to the cube of the semi-major axis. Kepler discovered this empirically in 1619 from Tycho Brahe's planetary observations. Newton derived it theoretically from his law of gravitation.
Why do higher orbits take longer? ▾
Two reasons combine: (1) orbital velocity v = √(GM/r) decreases with r (slower speed). (2) circumference 2πr increases with r (longer path). Both work against faster periods: T ∝ r^(3/2) = r × r^(1/2). Doubling orbital radius increases period by 2^(3/2) = 2.83×.
What makes GEO special? ▾
T = 24 hours = 86,400 s matches Earth's rotation. The satellite stays over the same spot on the equator, enabling continuous coverage. At 35,786 km altitude, about 3 GEO satellites cover 95% of Earth's surface (except polar regions). Signal round-trip time ≈ 0.24 s — noticeable latency for video calls.
How is exoplanet mass determined from Kepler's law? ▾
Radial velocity method: measure Doppler shift of star due to planet's gravitational pull. From this, derive v_star and period T. The star and planet orbit the center of mass: M_planet × r_planet = M_star × r_star. Combined with r_total = r_planet + r_star from v and T, both masses can be found if M_star is known from stellar type.
What is a resonance orbit? ▾
Orbital resonances occur when two bodies have period ratios of small integers. Jupiter's moons Io, Europa, and Ganymede are in 1:2:4 resonance (T_Io:T_Europa:T_Ganymede = 1:2:4). Tidal forces maintain the resonance and pump orbital energy into tidal heating — making Io the most volcanically active body in the solar system.
How do scientists discover satellites' mass from orbits? ▾
By observing the orbital parameters of a moon (or probe flying past), scientists use M = 4π²r³/(GT²) to determine the body's mass without landing on it. This technique determined masses for most solar system bodies, including distant Pluto's mass from Charon's orbit.
What is the Hill sphere? ▾
The Hill sphere (radius r_H ≈ r_planet(M_planet/(3M_star))^(1/3)) is the region around a planet where a satellite can orbit stably without being captured by the star. Earth's Hill sphere ≈ 1.5 million km — the Moon orbits well within it. Beyond r_H, the satellite would eventually be pulled away by the Sun.
Can satellites orbit below LEO? ▾
Technically yes (very briefly), but atmospheric drag rapidly decays orbits below ≈160 km. The ISS at 400 km altitude loses ≈2 km/month to drag and requires regular reboosts. Extremely brief 'grazing orbit' rockets at 100 km altitude last minutes before re-entry.

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Formula Explorer connections

Interpretation: This relationship connects mass, distance, orbit or spacetime behavior through gravitation and astrophysical scaling. Assumption: Many calculations assume spherical bodies, point masses, circular orbits, weak fields or Newtonian gravity; relativistic regimes require the stated correction.

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