Geostationary Orbit
A geostationary satellite stays above a fixed longitude on the equator because its orbital angular velocity matches Earth's rotation. In the two-body point-mass model, a circular equatorial orbit at radius r obeys ω²r = GM/r², so r = (GM/ω²)^(1/3). The angular rate ω must match how fast Earth turns relative to inertial space: astronomers use the sidereal day (~23 h 56 min), not the 24 h solar day defined by the Sun's apparent return. Using the sidereal period reproduces the classical geostationary altitude near 35 786 km above Earth's mean equatorial radius. The simulator shows an Earth-fixed diagram: meridians rotate, while the spacecraft and its nadir line remain aligned. A checkbox swaps the period in the formula to illustrate why broadcast or meteorology communities sometimes quote a slightly different radius when they loosely say "24 h."
Who it's for: Introductory orbital mechanics after circular velocity and Kepler's third law; complements Hohmann transfer and space-elevator pages.
Key terms
- Geostationary orbit
- Sidereal day
- Angular velocity
- Two-body problem
- Sub-satellite point
- Orbital radius
- Earth rotation
How it works
**Geostationary** orbit: a **circular** equatorial orbit whose **angular speed** matches **Earth’s rotation**, so a satellite stays above **one longitude** (idealized). With **GM** for Earth and central gravity **ω²r = GM/r²**, hence **r = (GM/ω²)^{1/3}**. **Sidereal** day **T ≈ 23 h 56 min** gives the **true** GEO radius (~42164 km from the center). A **24 h solar** day is **longer**, so **ω = 2π/T** is **smaller** and the same formula yields a **slightly larger** **r** (not smaller)—the checkbox swaps **T** so you can see that mix-up. The canvas is **Earth-fixed** (meridians rotate); the **nadir** line links the sub-satellite point to the craft.
Key equations
Frequently asked questions
- Why sidereal and not exactly 24 hours?
- Earth completes 360° relative to distant stars in a sidereal day. The solar day is longer because Earth must rotate slightly more than 360° for the Sun to return to the same meridian as we orbit the Sun. Geostationary locking uses the sidereal rate.
- If I mistakenly use 24 h solar for ω = 2π/T, does the GEO radius increase or decrease?
- It increases. A longer day means a smaller angular velocity ω, and r = (GM/ω²)^{1/3} grows when ω shrinks. True geostationary altitude comes from the sidereal day.
- Is the orbit exactly a circle in reality?
- Operational satellites experience lunisolar perturbations, Earth's non-spherical gravity (J₂), and solar radiation pressure; station-keeping burns keep them near their assigned slots. The page uses a perfect circle and point mass.
- How does this relate to GPS or Starlink?
- GPS satellites are in medium Earth orbits with ~12 h periods—not geostationary. Starlink uses low Earth orbits. Only the special combination of equatorial plane, zero inclination, and sidereal-matched ω gives the "hovering" property.
More from Gravity & Orbits
Other simulators in this category — or see all 22.
Orbital Decay (Atmosphere)
Toy perigee drag: shrinking, circularizing ellipse; ISS lifetime intuition.
Mercury Perihelion Precession
GR Δω per orbit vs Newton; ~43″/century readout; amplified animation.
Earth–Moon Barycenter Wobble
Heliocentric path of Earth’s center: barycentric ellipse plus lunar epicycle (exaggerated).
Oberth Effect
Same prograde Δv at peri vs apo on one ellipse; higher ε when burning deep.
Three-Body Figure-Eight
Equal masses: Chenciner–Montgomery choreography in 2D (RK4, periodic orbit).
Restricted 3-Body (map)
CRTBP: escape vs collision vs chaos proxy; μ slider.