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.

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

ω = 2π/T · r³ = GM/ω² · v = √(GM/r) · h = r − R⊕

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.

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