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Spec mapping: OCR H556 Module 5.4 — Gravitational fields (geostationary orbits: orbital period equal to one sidereal day, equatorial plane, prograde direction; application of Kepler's third law to derive the unique geostationary radius of about 4.22×107 m from the Earth's centre, or altitude ≈35900 km; applications of geostationary orbits in communications, broadcasting, weather monitoring; advantages and limitations). Refer to the official OCR H556 specification document for exact wording.
Some satellites appear to hover motionless above a single point on the Earth's surface. Switch on a satellite TV dish and it points in a fixed direction — above the equator, steady for years. These satellites occupy a uniquely valuable orbit: the geostationary orbit, where the orbital period exactly matches the Earth's rotation period, the orbit lies in the equatorial plane, and motion is eastward. The combined conditions force a unique altitude of about 35,900 km — derivable in three lines from Kepler's third law applied to a 24-hour orbit.
This capstone lesson closes OCR Module 5.4 by applying everything from Lessons 7–11 to the most heavily-used satellite orbit in modern technology. We state the three geostationary conditions precisely, derive the radius from Kepler's third law, calculate the orbital speed, distinguish geostationary from the broader category of geosynchronous orbits, work through the canonical "comparison to GPS" question that OCR sets repeatedly, and discuss the engineering limitations (signal delay, polar blind spots, orbital crowding, space debris). The A* discriminators land on (i) all three conditions stated together (period + equator + prograde), (ii) the use of the sidereal day rather than the solar day, and (iii) the distinction between geostationary and geosynchronous.
A geostationary orbit is a circular orbit around the Earth in which the satellite satisfies all three of the following:
The three conditions together imply that the satellite stays over the same point on the Earth's surface at all times — a ground observer sees it as a fixed point in the sky. Hence the name geo-stationary: stationary relative to the geo (Earth).
The key contrast is with geosynchronous orbit: any orbit with a 24-hour period, not necessarily equatorial. A non-equatorial geosynchronous satellite returns to the same point in the sky once per day but traces an analemma ("figure-of-eight") during the day, drifting north-south of the equator and east-west of its mean longitude. Every geostationary orbit is geosynchronous; not every geosynchronous orbit is geostationary.
A discriminator.* For "state the conditions for a geostationary orbit" at OCR, all three conditions must be stated — period AND equatorial AND prograde. Omitting the equatorial condition is the canonical mid-band mistake; the resulting orbit is geosynchronous but not geostationary, an A* discriminator that examiners watch for.
Given the period T=1 sidereal day, Kepler's third law (Lesson 10) determines the orbital radius uniquely:
T2=GM4π2r3⇒r3=4π2GMT2⇒r=(4π2GMT2)1/3
Substituting GM=3.98×1014 m³ s⁻² and T=86400 s (solar day, near enough to the sidereal day for A-Level):
Taking the cube root:
rGEO=(7.52×1022)1/3=4.22×107 m
This is the radius from the centre of the Earth. The altitude above the surface is
hGEO=rGEO−RE=4.22×107−6.37×106=3.59×107 m≈35900 km
— usually rounded to 36,000 km in popular accounts. There is exactly one altitude at which the orbital period is 24 hours; this is the unique geostationary altitude, set by the Earth's mass and rotation rate alone, with no engineering choice or flexibility.
vGEO=rGME=4.22×1073.98×1014=9.43×106=3.07×103 m s−1
About 3.07 km s⁻¹ — only 40% of the ISS's 7.67 km s⁻¹. Higher orbits are slower (Lesson 11's v∝1/r scaling).
T=2πr/v=2π×4.22×107/3.07×103=8.64×104 s — exactly the 24-hour day we started from. The formulas are mutually consistent.
A 24-hour orbit not in the equatorial plane will not appear stationary. Consider a satellite in a 24-hour orbit inclined at 30° to the equator. Over the course of one day, viewed from the rotating Earth's surface, the satellite traces a figure-of-eight analemma in the sky — north-south oscillation of ±30° latitude combined with a small east-west wobble. For an Earth-based dish to point at one fixed sky direction (the operational requirement of satellite TV, communications, weather imaging), the satellite must remain directly above the same longitude at every instant.
Geometrically, this requires the satellite's orbital plane to contain the Earth's rotation axis projected through the satellite's instantaneous position — and the only plane containing the rotation axis that also intersects every longitude on the equator is the equatorial plane itself. Any other inclined plane would put the satellite north or south of the equator for part of the orbit.
So the equatorial condition is not an engineering choice but a geometric necessity flowing from the fixed-ground-point requirement.
The satellite must orbit eastward — in the same sense as the Earth's rotation (prograde). A westward (retrograde) orbit at the same altitude would still have a 24-hour period but, relative to the rotating ground, would complete two apparent passes per day at twice the Earth's rotation rate — sweeping westward across the sky. Only the prograde 24-hour orbit holds the satellite at a fixed sky point.
The three conditions — period, plane, direction — are all necessary and all together sufficient for stationarity.
Because GEO satellites stay over a fixed point, they enable continuous communication with ground stations using fixed (non-tracking) antennas. The applications:
GPS satellites orbit at altitude 20,200 km. Geostationary satellites are at 35,900 km altitude. What is the ratio of their orbital periods?
For two orbits around the same central body, Kepler's third law gives (T1/T2)2=(r1/r2)3 — the constant 4π2/(GM) cancels.
(TGEOTGPS)2=(0.6286)3=0.2482⇒TGEOTGPS=0.2482=0.498
The GPS period is about half the geostationary period — i.e. 12 hours, which is why GPS orbits are called "semi-synchronous". Each GPS satellite traces the same ground track twice per sidereal day, simplifying mission design and ground-station tracking.
What angular separation of three geostationary satellites provides continuous coverage of the entire equatorial belt? Assume the satellites are at r=4.22×107 m.
A satellite at radius r above the equator can see ground stations within the cone defined by the line-of-sight from the satellite to the Earth's tangent. The maximum visible longitude span is determined by the geometry cosθ=RE/r where θ is the half-angle: cosθ=6.37×106/4.22×107=0.151, giving θ≈81.3°.
By symmetry, three satellites placed at 360°/3=120° longitude separation cover the entire equatorial belt — each handling 120° of longitude (well within its 162° accessibility).
This is exactly how real comms networks (Intelsat, Inmarsat) are designed: three satellites positioned at roughly Atlantic Ocean Region (~30°W), Indian Ocean Region (~60°E), and Pacific Ocean Region (~180°), covering essentially the whole equatorial belt and the populated mid-latitude bands.
How sensitive is the geostationary radius to using the sidereal day (86,164 s) versus the solar day (86,400 s)?
The fractional difference in T is ΔT/T=(86400−86164)/86400=0.0027 (0.27%). By Kepler's third law, r∝T2/3, so the fractional difference in r is
rΔr≈32TΔT=32×0.0027=0.0018(0.18%)
For rGEO=4.22×107 m, this gives Δr≈76 km — about 0.2% of the GEO radius. For A-Level calculations, using the 86,400 s solar day gives r correct to about 3 sf; engineering applications need the precise sidereal value (86,164.0905 s) to position satellites accurately within their 200-km-wide longitude slots.
The distinction matters in practice: a satellite operator placing a comsat in GEO must use the sidereal day, otherwise station-keeping fuel will be wasted compensating for drift. At A-Level, either day is accepted on mark schemes provided the chosen value is stated.
graph TD
E["Earth surface<br/>R = 6370 km, T = 0 (rotating)"]
E --> LEO["Low Earth Orbit<br/>200 – 2000 km altitude<br/>T ≈ 90 min<br/>e.g. ISS (~400 km), Hubble (~540 km)"]
E --> MEO["Medium Earth Orbit<br/>2000 – 35000 km altitude<br/>T ≈ 2 – 24 h<br/>e.g. GPS (~20200 km, T = 12 h)"]
E --> GEO["Geostationary Orbit<br/>35900 km altitude<br/>T = 24 h, equatorial, prograde<br/>e.g. comms, weather, TV"]
E --> HEO["High Earth Orbit<br/>greater than 35900 km<br/>T greater than 24 h<br/>e.g. Moon at ~380000 km (T = 27.3 days)"]
The hierarchy reflects Kepler's third law: orbital altitude uniquely determines orbital period, and vice versa. Mission requirements determine the altitude bracket; physics determines the corresponding period and orbital speed.
Geostationary orbits are not without drawbacks:
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