Value chain · CLM-1437
In plain terms
A satellite in geostationary orbit, about 36,000 km up, goes round the Earth once in the time the Earth takes to turn once, which is just under 24 hours. So it stays over the same spot on the ground.
The claim as recorded
ESA's space debris standard defines a geostationary Earth orbit as an "Earth orbit having zero inclination, zero eccentricity, and an orbital period equal to the Earth's sidereal rotation period", and gives the geostationary altitude as 35786 km. By Kepler's third law, a circular orbit of radius 6378 + 35786 = 42,164 km has a period of 86,163.6 s (23 h 56 min 3.6 s). This is 0.4 s less than one sidereal day (86,164 s), well inside the 3 s that rounding the two inputs to the kilometre allows, and about 3 min 56 s shorter than 24 hours (CALC-136).
Evidence behind this: T0: the orbital period is computed in CALC-136 from SRC-080's altitude and Earth radius, plus one physical constant that is not taken from a held source. The definition is SRC-080's: an ESA standard quoted for its own text (T1). SRC-416 (T3) says the same thing in plain words.
Scope and limits
The period is computed, not read from a source. No held source states GM, the length of a sidereal day or a GEO period in hours. The 86,164 s used for the comparison is also an unsourced constant, rounded to the second; the unrounded value, about 86,164.09 s, makes the difference 0.5 s. The two-body result ignores the Earth's uneven shape, the pull of the Sun and Moon and the pressure of sunlight; a real satellite must correct for their effects with station-keeping (CLM-1075). SRC-005 uses a GEO semi-major axis of 42,173.30429 km (PDF p.4, Table 2), which is 9.3 km more than 42,164 km. At that radius the same formula gives 86,192.1 s, 28 s longer than a sidereal day, and that page does not say why its value differs. SRC-416 is an interim release ("Interim Assessment Release 0.1"), and its sentence joins a period and a "speed" loosely. It is cited for the plain statement only, not as a final assessment. "One day" here means one sidereal day, one turn of the Earth relative to the stars, not the 24-hour solar day.
What it assumes
That the 35786 km altitude is measured above the 6378 km equatorial radius that the same standard gives for its LEO region. The standard does not say this for GEO, but the sum gives a period that matches a sidereal day to within the rounding of the inputs, which is consistent with it. A point-mass Earth and a circular two-body orbit. GM = 398,600.4418 km3/s2 and a sidereal day of 86,164 s, both physical constants that are not taken from a held source.
Where it comes from
This is the project's own working, shown in full:
printed p.12/80, definition 3.2.22, verbatim: "geostationary Earth orbit / Earth orbit having zero inclination, zero eccentricity, and an orbital period equal to the Earth's sidereal rotation period [ISO 24113:2023]". SRC-080 printed p.17/80, section 3.4, in the definition of the GEO protected region, verbatim: "where the geostationary altitude is 35786 km."; and in the definition of the LEO protected region: "a spherical Earth with an equatorial radius of 6378 km". SRC-416 PDF p.47 (printed A-2), Appendix A, verbatim: "A special case occurs at Geostationary (GEO) orbit where the orbital period of the satellite corresponds to the speed of Earth's rotation and the satellite appears stationary over the ground." CALC-136: T = 2 pi sqrt(a^3/GM) = 86,163.57 s at a = 42,164 km.
Basis: orbital period in seconds; distances in km
Cite this
energyaperture.com/c/clm-1437
CLM-1437, Energy Aperture, as of 2025-10.
Connected records
Other published records linked to this one. A list, not a judgement.