Case Study 1: The GPS Constellation — Why 20,200 km?

"The Global Positioning System is a dual-use satellite-based navigation system providing accurate, continuous, worldwide, all-weather position, velocity, and time information." — U.S. Government GPS description (paraphrased characterization; treat as illustrative)

Executive Summary

The Global Positioning System is the most-used space asset in daily life: billions of receivers, from phones to airliners to farm tractors, depend on it every second. Yet the single decision that made GPS possible was not a clever circuit or a better clock — it was the choice of orbit. GPS satellites fly in medium Earth orbit at an altitude of about $20{,}200\ \text{km}$, in six planes tilted at $55^\circ$, with a period of exactly half a sidereal day. In this case study we reverse-engineer that choice: we take the system's requirements and show that the MEO orbit falls out of them almost inevitably. We compute the period and confirm it is semi-synchronous, quantify the coverage that lets just $24$ satellites serve the whole planet, explain why four satellites (not three) must be visible at once, and total the price the altitude charges in radiation and deployment delta-v.

Skills applied

• Circular period from Kepler's third law and the semi-synchronous ground-track idea (§9.2, §9.3).
• Coverage geometry: the horizon half-angle and its consequence for constellation size (§9.1, §9.2).
• The coverage-versus-few-satellites trade that defines MEO (§9.2, §9.6).
• The radiation cost of the middle altitudes (§9.2) and reaching orbit as a delta-v line item (Chapter 3).

Background

Satellite navigation solves an old problem — where am I? — with a beautifully modern trick. Each GPS satellite carries an atomic clock and continuously broadcasts a timestamped signal. A receiver measures how long each signal took to arrive, multiplies by the speed of light to get a distance to each satellite, and solves for the one location consistent with all those distances. The method is called trilateration, and it has two hard requirements that pull in opposite directions:

1. From anywhere on Earth, at any time, the receiver must see enough satellites (at least four) spread across the sky to get a good geometric fix.
2. Each satellite is an expensive, atomic-clocked, radiation-hardened machine, so the constellation must be as small as possible.

"See many, everywhere, with few satellites" is the whole design tension, and altitude is the dial that resolves it. Let us watch the numbers force the answer.

Phase 1 — The period: a satellite that repeats its track

GPS orbits at $r \approx 26{,}560\ \text{km}$ (altitude $\approx 20{,}200\ \text{km}$ above the mean radius). First, what is its period?

import math
MU = 3.986e5              # km^3/s^2
r = 26560.0              # km, GPS orbital radius
T = 2 * math.pi * math.sqrt(r**3 / MU)
print("period (s):", round(T))
print("period (h,min):", int(T // 3600), "h", round((T % 3600) / 60))
print("fraction of sidereal day:", round(T / 86164, 3))
# Expected output:
# period (s): 43078
# period (h,min): 11 h 58
# fraction of sidereal day: 0.5

The period is half a sidereal day — $11\ \text{h}\ 58\ \text{min}$. A GPS satellite therefore circles the Earth exactly twice per day and, crucially, retraces the same ground track every day. This semi-synchronous repeat makes the constellation's geometry predictable: operators know precisely where each satellite will be over each ground antenna, day after day, which simplifies tracking, uploading fresh clock corrections, and constellation management. Sanity check: half of $86{,}164\ \text{s}$ is $43{,}082\ \text{s}$, and we computed $43{,}078\ \text{s}$ — agreement to within rounding of the radius. The altitude was chosen, in part, to land on this clean period.

Phase 2 — Coverage: how much of the Earth does one satellite see?

From altitude $h$, a satellite's signal reaches the ground out to a geometric horizon at Earth-central half-angle $\lambda = \arccos\!\big(R_\text{eq}/(R_\text{eq}+h)\big)$. Compare the three regimes:

import math
R = 6378.0
for name, h in [("LEO", 550), ("MEO-GPS", 20200), ("GEO", 35786)]:
    lam = math.degrees(math.acos(R / (R + h)))
    print(name, h, "km -> half-angle", round(lam, 1), "deg")
# Expected output:
# LEO 550 km -> half-angle 23.0 deg
# MEO-GPS 20200 km -> half-angle 76.1 deg
# GEO 35786 km -> half-angle 81.3 deg

A GPS satellite sees out to a $76^\circ$ half-angle — a cap covering roughly a third of the Earth's surface at once, nearly as much as a GEO satellite. That enormous footprint is the payoff of the high altitude: a receiver near the equator or at mid-latitude can see satellites from horizon to horizon in every direction. Contrast the LEO satellite's mere $23^\circ$: to guarantee four LEO satellites in view everywhere would take hundreds. MEO's wide footprint is exactly what lets a small constellation do the job.

Phase 3 — The constellation: 24 satellites, 6 planes, 55°

The baseline GPS constellation is 24 satellites in 6 orbital planes (four per plane), the planes spaced $60^\circ$ apart in longitude and each inclined at $55^\circ$ to the equator. Two design choices deserve a word:

• Six planes, four per plane. With each satellite seeing a third of the globe, spreading 24 of them evenly guarantees that at least four — usually eight to twelve — are above the horizon from any point on Earth at any moment. (The modern constellation flies ~31 operational satellites, adding robustness and better geometry, but the 24-slot baseline is what the coverage was designed around.)
• $55^\circ$ inclination, not $90^\circ$. A purely equatorial constellation would starve the poles; a purely polar one would over-serve them and thin out the equator. The $55^\circ$ compromise gives strong, even geometry across the populated mid-latitudes while still reaching high latitudes acceptably.

Why not put the satellites lower and cheaper? Because the coverage arithmetic explodes. Dropping to LEO shrinks each footprint by more than a factor of three in angle, and the number of satellites needed for continuous four-in-view coverage climbs into the hundreds — each still needing a precision clock. The altitude that minimizes total system cost is the one where a couple of dozen satellites suffice, and that is MEO.

Phase 4 — Why four satellites, not three

Trilateration in three-dimensional space would need only three distances — if the receiver's clock were perfect. It is not. A receiver uses a cheap quartz oscillator whose offset from true GPS time is unknown, and since every distance is (speed of light) × (measured time), a clock error $\delta t$ corrupts every range by $c\,\delta t$. A one-microsecond error is a $300\ \text{m}$ position error.

The fix is elegant: treat the clock offset as a fourth unknown and add a fourth satellite to solve for it. Four ranges, four unknowns ($x, y, z$, and the clock bias). This is why your phone "needs to see four satellites," and why the receiver ends up with a better clock than it paid for — it solves for time as a free by-product of solving for position.

# Unknowns and the observations needed to solve for them:
unknowns = ["x", "y", "z", "receiver clock bias"]
print("unknowns:", len(unknowns))
print("minimum satellites in view:", len(unknowns))
# Expected output:
# unknowns: 4
# minimum satellites in view: 4

This requirement feeds straight back into Phase 3: the constellation is not sized for one satellite in view but for four well-spread ones, everywhere, always — which is precisely why 24 MEO satellites, not twelve, form the baseline.

Phase 5 — The price of the altitude

No orbit is free. MEO charges GPS in two currencies.

• Radiation. At $20{,}200\ \text{km}$, GPS satellites live among the Van Allen radiation belts that LEO satellites sit safely beneath. Every navigation satellite must be built with radiation-hardened electronics and extra shielding — mass and cost that a low orbit would not demand. This is the hidden tax of the middle altitudes flagged in §9.2.
• Delta-v to deploy. Reaching a $20{,}200\ \text{km}$ orbit costs meaningfully more than reaching LEO. As a rough gauge, the delta-v from a LEO parking orbit up to GPS altitude is on the order of $3.5$–$4\ \text{km/s}$ (a two-burn transfer, in the spirit of the GTO climb of §9.3). That is delta-v the launch vehicle and upper stage must supply for every satellite — a real line item in the Chapter 3 sense.

The system accepts both costs because the alternative — a LEO navigation constellation of hundreds of radiation-soft but numerous satellites, or a GEO one with hopeless polar geometry — would cost far more or simply not work. MEO is not a compromise GPS settled for; it is the optimum the requirements pointed to.

Discussion Questions

1. The GPS ground track repeats every sidereal day. Name two distinct operational advantages of a repeating ground track for the people who run the constellation.
2. Modern GPS flies ~31 satellites, not the baseline 24. Given that 24 already guarantee four-in-view, what is bought by the extra seven? (Think about geometry quality and failures.)
3. Galileo orbits slightly higher than GPS (~$23{,}222\ \text{km}$) at ~$56^\circ$. Using the period you would compute, argue whether Galileo is also semi-synchronous, and what that implies about its ground-track repeat.
4. Why would a GEO-based navigation system give poor accuracy for a user in Norway, even though the GEO satellite is technically visible there?

Your Turn: Extensions

• Option A (analysis). Compute the exact GPS orbital radius that gives a period of precisely half a sidereal day ($43{,}082\ \text{s}$) by inverting Kepler's third law, and compare it to the $26{,}560\ \text{km}$ we used. How many kilometers apart are they, and why might the operational orbit differ slightly from the ideal semi-synchronous value?
• Option B (geometry). Estimate, using the $76^\circ$ coverage half-angle, what fraction of Earth's surface one GPS satellite covers, and use it to sanity-check that 24 evenly spread satellites can keep four in view everywhere. (A spherical-cap area is $2\pi R^2(1-\cos\lambda)$.)
• Option C (design). Sketch the orbit choice for a regional navigation system serving only one country near the equator (as India's NavIC does). Would you still choose MEO, or could a few geosynchronous and inclined-geosynchronous satellites suffice? Justify with coverage geometry.

Key Takeaways

• GPS's defining decision was its orbit: MEO at $\approx 20{,}200\ \text{km}$, six planes, $55^\circ$, half-sidereal-day period.
• That altitude is the sweet spot of the "see-four-everywhere-with-few-satellites" trade: high enough for a huge footprint (so $\sim 24$ satellites suffice), low enough to avoid GEO's polar blind spot.
• The four-satellite requirement comes from solving for the receiver's clock bias as a fourth unknown — a consequence that sizes the whole constellation.
• MEO's benefits are paid for in radiation (the Van Allen belts) and deployment delta-v — the universal lesson that every orbit is a bargain with the space environment and the rocket equation.
• Reverse-engineering a real system is the best proof of §9.6's thesis: the orbit is not chosen from a table; it is derived from the mission's requirements.