Geostationary Orbit Radius Calculator
Find the orbital radius, period or gravitational parameter from r = (μ · T² / 4π²)⅓ — Kepler's third law rearranged.
Inputs
Choose what to solve for, then fill in the other quantities. Scientific notation such as 3.986e14 is accepted. Units convert automatically.
Results
Enter values and press Calculate. Your answer will appear here.
Orbit geometry
Schematic view of the orbit plane. r is measured from the centre of the central body; h is the altitude above its surface. The drawing is scaled to your numbers but clamped for legibility, so it is not strictly to scale.
Radius vs. period (log–log)
Kepler's third law puts every circular orbit of a given body on one straight line: log r = ⅔ log T + constant. The slope is exactly ⅔. A 100× longer period needs a 104/3 ≈ 21.5× larger radius.
What is being calculated
Synchronous orbits of the Solar System
| Body | Sidereal rotation | μ (km³/s²) | Stationary radius | Altitude |
|---|
Computed by the same engine. The Moon's synchronous radius (~88 400 km) lies far outside its Hill sphere of ~60 000 km, so no stable lunar stationary orbit exists — a satellite there would be pulled away by the Earth.
Typical circular orbits
| Orbit | Period | Radius | Altitude | Speed |
|---|
Illustrative values for Earth (μ = 398 600.4418 km³/s², R = 6378.137 km).
Key relations and unit conversions
| Quantity | Relation |
|---|---|
| Kepler's third law | T = 2π √(r³ / μ) · T² = 4π² r³ / μ |
| Orbit radius | r = (μ T² / 4π²)1/3 |
| Gravitational parameter | μ = 4π² r³ / T² = G·M |
| Altitude | h = r − Rbody |
| Orbital speed | v = 2πr / T = √(μ / r) |
| Angular velocity | ω = 2π / T = √(μ / r³) |
| Gravity at the orbit | gr = μ / r² = v² / r = ω² r |
| Escape speed at r | vesc = √(2μ / r) = √2 · v |
| Geostationary condition | circular · equatorial · prograde · T = sidereal rotation period |
| Gravitational constant | G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² |
| Length | 1 km = 10³ m · 1 Mm = 10⁶ m · 1 mi = 1609.344 m · 1 nmi = 1852 m |
| Time | 1 min = 60 s · 1 h = 3600 s · 1 d = 86 400 s |
| Sidereal days | Earth 86 164.0905 s (23 h 56 m 4.1 s) · Mars 88 642.663 s · Jupiter 35 729.7 s |
| Gravitational parameter | 1 km³/s² = 10⁹ m³/s² · Earth 3.986 004 418 × 10¹⁴ m³/s² |
Formula
- r: orbital radius from the centre of the central body (m)
- μ: standard gravitational parameter of the central body, μ = G·M (m³/s²)
- T: orbital period (s)
- G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻², M = mass of the central body (kg)
Example — Earth geostationary orbit. The Earth's sidereal rotation period is T = 86 164.0905 s and μ = 3.986004418 × 10¹⁴ m³/s². Then
r = ( 3.986004418×10¹⁴ × 86 164.0905² / 4π² )1/3 ≈ 4.2164 × 10⁷ m ≈ 42 164 km
With an equatorial radius of 6378.137 km the altitude is h ≈ 35 786 km, and the orbital speed is v = 2πr/T ≈ 3075 m/s. Because T exactly equals the sidereal rotation period, the satellite hovers over one point on the equator — that is why satellite dishes can be bolted in place.
The same relation is Kepler's third law: T = 2π √(r³/μ), or equivalently T² = 4π² r³ / μ. Solving it for μ gives the gravitational parameter of any body once one orbit is measured — this is how μ is determined for planets and moons from spacecraft tracking data.
Three conditions must hold for a true geostationary orbit: the orbit must be circular, equatorial (inclination 0°), and prograde (moving with the rotation), with T equal to the body's sidereal rotation period — not the solar day. An equatorial orbit with the right period but the wrong eccentricity or inclination is geosynchronous: it repeats the same ground track each day but traces a figure-eight rather than standing still.
Advanced mode adds the central body's radius and rotation period so the altitude and the body's own stationary radius are reported, plus orbital speed, angular velocity, gravity at the orbit, revolutions per day, and the option to enter the body's mass M instead of μ (using μ = G·M).
References
- Wikipedia, Geostationary orbit: derivation of r ≈ 42 164 km for Earth
- Wikipedia, Kepler's laws of planetary motion: the third law T² ∝ r³
- Wikipedia, Standard gravitational parameter: μ = G·M for the Sun, planets and moons
- Wikipedia, Sidereal time: why the sidereal day is 86 164.09 s, not 86 400 s
- NIST SP 811, Guide for the Use of the SI: unit conversion factors
How to Use This Tool
- Choose what to solve for: r, T or μ. The matching input box is hidden.
- Pick a preset, or type the two known values and choose their units.
- In Advanced, add the body's radius for the altitude and its rotation period for the body's own stationary orbit.
- Click Calculate.
- Read the highlighted answer, then check the orbit diagram and the log–log chart.
What This Tool Offers
- Solves r = (μ T² / 4π²)1/3 for r, T or μ
- Gravitational parameter entered directly or derived from the body's mass (μ = G·M)
- Automatic unit conversion for radius, period and μ
- Altitude, orbital speed, angular velocity and gravity at the orbit
- Comparison with the body's own stationary radius from its rotation period
- Scaled orbit diagram and log–log r vs T chart with slope ⅔
- Ready-made presets for Earth, Mars, Jupiter, Saturn, Venus and the Moon
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| r | Orbital radius | m |
| μ | Earth's gravitational parameter | m3/s2 |
| T | Rotational period | s |
What it means
The geostationary orbit is a circular orbit in the equatorial plane with an orbital period equal to Earth’s sidereal rotation period. This makes the satellite appear stationary from the ground. The radius is derived from Kepler’s third law. Geostationary orbit is used for communication, weather, and navigation satellites because it provides constant coverage. The altitude is about 35,786 km above sea level. This orbit is unique and has limited slots, making it a valuable resource. The formula is essential for launching and station‑keeping of geostationary satellites. Understanding this orbit is critical for telecommunication engineers and space operators.
Worked example
Geostationary Orbit Radius – Two Examples
Real‑World| Parameter | Value |
|---|---|
| μ | 3.986×10¹⁴ |
| T | 86,164 s |
| Parameter | Value |
|---|---|
| T | 86,400 s |
Common mistakes
- Geostationary orbit radius: r = (μ·T²/(4π²))^(1/3).
- T: Orbital period (for geostationary, T = 86,164 s).
- μ: Earth’s μ.
- Resulting radius ~42,164 km (altitude ~35,786 km).
- Equatorial orbit required – inclination zero.
Applications
Geostationary orbit radius, r = (μ·T²/(4π²))^(1/3), determines the altitude at which a satellite's orbital period matches Earth's rotation (24 hours). This orbit is unique and ideal for communication and weather satellites that need to remain fixed over a point on the equator. Engineers use this formula to design satellite constellations, to size launch vehicles for GEO insertion, and to plan station‑keeping. The altitude is approximately 35,786 km above the equator. By understanding geostationary orbit, aerospace engineers can design missions that rely on constant coverage, enabling global telecommunications, broadcasting, and meteorological monitoring.
- Design of communication satellites (TV, internet, telephone)
- Meteorological satellites for continuous weather monitoring
- Navigation augmentation (e.g., SBAS)
- Launch vehicle design for GEO insertion (Δv requirements)
- Station‑keeping and orbital debris mitigation
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