CALC-136

Period of a circular orbit at geostationary radius, against one sidereal day

Calculation · Cited by CLM-1437

What this is

This is the project's own arithmetic, published in full so it can be checked rather than taken on trust. It is not a source. Every input it uses is cited to a record, and the conclusion it reaches only carries as far as those inputs do.

REGISTERED 2026-10-14. Drafted for the lessons records set (commit f8cfb1ce), checked by an independent reader (FIX, all applied, round 2 PASS), landed by SBSP-coord. verified: VER-2352 CONFIRMED 2026-10-14.

Feeds CLM-1437. Basis: seconds; km.

Quantity	Value	Source	Stated or assumed
Geostationary altitude	35786 km	SRC-080 printed p.17/80, section 3.4	STATED
Earth equatorial radius	6378 km	SRC-080 printed p.17/80, section 3.4 (LEO protected region)	STATED for the LEO region; ASSUMED to be the base of the GEO altitude
Semi-major axis a	6378 + 35786 = 42,164 km	this CALC	ASSUMED (circular orbit, zero eccentricity as SRC-080 defines GEO)
Earth gravitational parameter GM	398,600.4418 km3/s2	physical constant, not from a held source	ASSUMED (unsourced constant, as the inputs of CALC-099 are)
One sidereal day, for comparison	86,164 s	physical constant, not from a held source	ASSUMED (unsourced constant, rounded to the second)
Sensitivity case: SRC-005 semi-major axis	42,173.30429 km	SRC-005 PDF p.4 (printed 235), Table 2, GEO row	STATED

Working. Kepler's third law for a two-body circular orbit: T = 2 pi sqrt(a^3 / GM).

a      = 6378 + 35786                         = 42,164 km
a^3    = 42,164^3                             = 7.495928e13 km3
a^3/GM = 7.495928e13 / 398,600.4418           = 1.880562e8 s2
sqrt   =                                        13,713.3582 s
T      = 2 pi x 13,713.3582                   = 86,163.57 s = 23 h 56 min 03.6 s = 23.93433 h
T - 86,164 s (sidereal day)                   = -0.43 s (-0.0005 per cent)
T - 86,400 s (24 h)                           = -236.43 s (about 3 min 56 s)
radius whose period is exactly 86,164 s       = 42,164.14 km (altitude 35,786.14 km above 6378 km)
same T with GM rounded to 398,600 km3/s2     = 86,163.62 s

Sensitivity to SRC-005's semi-major axis.

a      = 42,173.30429 km  (9.30 km, 0.0221 per cent, above 42,164 km)
T      = 86,192.09 s = 23 h 56 min 32.1 s
T - 86,164 s                                  = +28.09 s (+0.0326 per cent)
dT between the two radii                      = +28.52 s (0.0331 per cent), as expected from dT/T = 1.5 da/a = 0.0331 per cent

Script, run with python -I under Python 3.14.0. Its output matches the figures above.

import math
GM = 398600.4418; R_EQ = 6378.0; H_GEO = 35786.0; A_SRC005 = 42173.30429; SIDEREAL_DAY = 86164.0
period = lambda a: 2 * math.pi * math.sqrt(a ** 3 / GM)
print(period(R_EQ + H_GEO), period(A_SRC005), period(A_SRC005) - period(R_EQ + H_GEO))

Result. A circular orbit at SRC-080's 35786 km altitude above a 6378 km equatorial radius has a period of 86,163.6 s, which is 23 h 56 min 3.6 s. This is 0.4 s less than one sidereal day of 86,164 s, inside the 3 s that rounding the inputs allows, and 3 min 56 s shorter than 24 hours. The calculation agrees with SRC-080's definition: a period equal to the Earth's sidereal rotation period. At SRC-005's semi-major axis of 42,173.3 km, the same formula gives a period 28 s longer than a sidereal day.

Where this fails. The inputs 35786 km and 6378 km are given to the nearest km. One km of radius moves the period by about 3 s, so the result is good to about plus or minus 3 s. GM and the sidereal day are physical constants that are not in any held source. The agreement shows that SRC-080's two figures are consistent with those constants to within the rounding of the inputs. The unrounded sidereal day, about 86,164.09 s, makes the difference 0.5 s. This does not check the constants against a source. SRC-080 gives 6378 km as the radius of the LEO protected region; that the GEO altitude is measured from the same radius is this CALC's assumption. The model is two-body and circular, with a point-mass Earth, so it ignores the Earth's oblateness and the pull of the Sun and Moon. SRC-005 PDF p.4 does not say why its semi-major axis is 9.3 km larger, so the 28 s gap is not explained here.

Cite this

energyaperture.com/calc/calc-136

CALC-136. The working above, not a source document.