Satellite internet · Session 4 of 8
The geometry of being in view
How long is a satellite usable, how much ground does it cover, and how many do you need?
Go out just after sunset and find a satellite — a steady point of light crossing the sky, not blinking. Follow it. Within four or five minutes it is gone behind the trees, and nothing has broken. It simply went somewhere your line of sight cannot reach. Anything that was talking to it had, by then, already found another one; a Starlink terminal does this all day, and the only reason you never notice is that the handover is planned before the satellite goes.
By the end
- Derive the horizon condition cos γ = R / r and extend it to a minimum elevation angle.
- Calculate footprint radius, pass duration and slant range at a stated minimum elevation.
- Explain why a constellation's satellite count follows from geometry, and why visibility is necessary but not sufficient for service.
The idea
Whether a satellite is usable from a given town is a geometry problem with an exact answer. Let R be Earth's radius, r the orbital radius, and γ the central angle between the town and the satellite's ground position — the angle at Earth's centre. When the satellite sits exactly on the horizon, the line of sight is tangent to Earth's surface, the triangle has a right angle at the town, and cos γ = R / r. At 550 km that gives γ = 23.0°.
No real terminal works down to the horizon. Signals skimming the ground cross far more atmosphere, meet buildings and hills, and arrive weak, so a minimum elevation angle ε is specified — around 25° for a Starlink user terminal. Including it gives cos(ε + γ) = (R / r) · cos ε, which reduces to the horizon case when ε = 0. At 550 km with ε = 25°, γ collapses from 23.0° to just 8.46°.
That collapse is the whole design problem. The footprint — the circle of ground from which the satellite is usable — has radius Rγ (with γ in radians), so it shrinks from 2,557 km at the horizon to 940 km at 25°. In area, using the spherical cap 2πR²(1 − cos γ), that is a fall from 20.3 million km² to 2.77 million — a factor of more than seven, bought by one sensible engineering requirement.
Small footprints and fast satellites mean short passes. The satellite sweeps through angle at the rate v / r = 0.0628° per second, so crossing a 2γ = 16.9° window takes about 269 s — 4.5 minutes, and less for any pass that is not straight overhead. A connection therefore cannot belong to one satellite. It is handed from one to the next in a planned handoff, and the network must always have a successor ready.
Distance also varies enormously within a single pass. The slant range d — the actual line-of-sight distance — is 550 km when the satellite is overhead and 1,123 km at the 25° edge, a factor of 2.04. Received power falls as 1/d², so the edge of the footprint receives 4.2 times less, about 6.2 dB down. The link has to be designed for the worst case, not the best.
Motion that fast also shifts the frequency. The Doppler effect changes the received frequency by Δf = f · vᵣ / c, where vᵣ is the component of velocity along the line of sight. With f = 12 GHz and the full orbital speed of 7.59 km/s as a bound, Δf reaches about 304 kHz — far wider than the channel spacing, so terminals must track and correct it continuously. Geostationary systems, which barely move relative to the ground, never face this.
Finally, geometry is necessary but nowhere near sufficient. A visible satellite still needs working hardware, a free beam, spare capacity, and a path onward — either a ground station within its footprint, or an inter-satellite laser link to one that has it. A satellite over the mid-Atlantic may be perfectly visible and completely useless.
The elevation-angle triangle
Gather
Paper Earth circle, compass, protractor, counters, non-stretch string, calculator and a recording sheet.
Build and investigate
- Draw an Earth circle of radius 10 cm and a concentric orbit circle of radius 14 cm. Label it an exaggerated geometry model — at true scale the orbit would be 10.9 cm and indistinguishable from the surface. Mark a town on the circumference and 12 equally spaced positions, 30° apart, on the orbit.
- Place one counter on the orbit and stretch string straight from the town to it. Count the satellite as usable only if the string leaves the surface outward without crossing Earth's interior. Verify with the protractor that this is the same as cos γ ≤ R / r.
- Advance the satellite one position at a time and record visibility across all 12 steps. Count usable steps, not whether service happened at all. Repeat on the on-screen model, which uses the same 10 : 14 ratio and the same 30° steps, and confirm the two agree.
- Repeat with four equally spaced satellites across the same 12 steps. At each step, choose a visible satellite and record every switch and every gap. Then disable one satellite and run it again.
- Now do it exactly. For the real shell at h = 550 km, calculate γ from cos(ε + γ) = (R / r) cos ε at ε = 0°, 25° and 40°. Tabulate γ, footprint radius Rγ and pass duration 2γ ÷ 0.0628° per second for each.
- Calculate the slant range at the zenith and at the 25° edge, form the ratio, square it for the power ratio and convert to decibels with 10 log₁₀. State which end of the pass the link budget has to survive.
Run the numbers
One relation does the work, and every quantity a constellation designer argues about falls out of it.
| Relation | Expression | What it means |
|---|---|---|
| Horizon condition | cos γ = R / r | satellite exactly on the horizon, ε = 0 |
| With minimum elevation | cos(ε + γ) = (R / r) · cos ε | ε = minimum usable elevation angle |
| Footprint radius | s = Rγ | γ in radians; ground distance along the surface |
| Footprint area | A = 2πR²(1 − cos γ) | area of a spherical cap |
| Angular rate | ω = v / r | 0.0628°/s at 550 km; pass time ≈ 2γ / ω |
| Slant range | d = R · sin γ ÷ cos(ε + γ) | true line-of-sight distance |
| Doppler shift | Δf = f · vᵣ / c | vᵣ = velocity along the line of sight |
Core · Grade 7–8
- Using cos γ = R / r with R = 6371 km and r = 6921 km, calculate γ at the horizon. Convert to radians and find the footprint radius Rγ in km. How far from the town does usability end?
- Repeat at ε = 25° using cos(ε + γ) = (R / r) cos ε. Tabulate γ and Rγ at ε = 0°, 25° and 40°, then describe in one sentence what raising the minimum elevation does to coverage.
- At ω = 0.0628°/s, how long is an overhead pass usable at ε = 25°? Divide your answer into 24 hours to estimate how many handoffs a continuously connected terminal performs in a day.
Deeper · Grade 9–12
- Derive cos(ε + γ) = (R / r) cos ε from the sine rule. In triangle O–T–S with O at Earth's centre, the angle at T between the line of sight and the local horizontal is ε, so the interior angle OTS is 90° + ε. Write the sine rule, simplify, and confirm it reduces to cos γ = R / r when ε = 0.
- Calculate the slant range at zenith and at the ε = 25° edge. Form d_edge / d_zenith, square it for the free-space power ratio, and express it in decibels as 10 log₁₀. A link budget must close at the worst geometry — how many decibels of margin does that demand?
- A single 550 km satellite covers 2.77 million km² at ε = 25°. Earth's surface is 510 million km². Divide to get a lower bound on the number of satellites needed for instantaneous global coverage, then list three reasons the real constellation is several times larger than your bound.
Measure and record
For each of the 12 time steps, record the visible satellite IDs, the chosen ID, and whether a handoff, an outage or no change occurred. Coverage fraction = steps with a usable satellite ÷ 12. Record the exact calculations separately, with ε, γ, footprint radius, pass duration and slant range.
Worked example
At ε = 0°: cos γ = 6371 ÷ 6921 = 0.9205, so γ = 23.0° = 0.401 rad and the footprint radius is 6371 × 0.401 = 2,557 km. At ε = 25°: cos(25° + γ) = 0.9205 × cos 25° = 0.8343, so 25° + γ = 33.46° and γ = 8.46° = 0.1476 rad, giving a radius of 940 km. One requirement cut the usable radius by a factor of 2.7 and the area by more than seven. Pass duration follows: 2 × 8.46° ÷ 0.0628° s⁻¹ = 269 s, about 4.5 minutes. And the model result alongside it: if your 12-step run shows a usable satellite at 9 steps, the sampled coverage is 9 ÷ 12 = 75%. That is a statement about twelve instants. It says nothing about the gaps between them — which is exactly the error the exact calculation protects you from, since the geometry is continuous and your sampling is not.
Try the idea
A moving constellation
Advance time and find a satellite visible from the town at the top of Earth. This exaggerated 2D model counts straight-line visibility to the horizon only.
One step = 30° of model motion, so twelve steps complete a lap — the same twelve positions you marked on paper. Earth radius = 85 drawing units and orbit radius = 119, the same 10 : 14 ratio as the paper model, so the two should agree step for step. Here cos γ = 10/14 gives a usable arc of 88.8° while four satellites sit 90° apart: the gap is real, and 30° sampling steps straight over it. No real orbital time, altitude or capacity is implied.
Pause and explain
Two satellites are visible. One will drop below your minimum elevation at the next step. Must the connection fail? Then: your terminal can see a satellite over the mid-Atlantic with plenty of capacity, 4,000 km from any ground station, and the constellation has no inter-satellite links. Can it serve you?
Check your reasoning after your attempt
First: no, not necessarily — a handoff to the other visible satellite preserves it, which is the normal case and why constellations are sized for overlap rather than bare coverage. Second: no, it cannot. Without a laser crosslink, a satellite is a mirror that must see both the user and a gateway at the same time; with a 940 km footprint radius at ε = 25°, a ground station 4,000 km away is far outside it. The satellite is visible, powered, idle and useless. This is the difference between geometric visibility and service, and it is why mid-ocean and polar coverage waited for crosslinks.
Save in your mission notebook
The 12-step coverage table with handoffs and outages marked; the exact γ, footprint and pass-duration table at three elevation angles; the slant-range ratio in decibels; one sentence distinguishing sampled from continuous coverage.
Parent / teacher notes
Run the paper model and the on-screen model side by side in step 3; they now use the same 10 : 14 ratio and the same 30° steps, so if a learner's counts disagree, something in the method is wrong and it is worth finding. The sampled-versus-continuous trap in step 4 is the intellectual centre of the lesson — let learners claim 100% coverage, then have them calculate 2γ and the 90° spacing. Grade 7–8 can use the supplied γ values and do the footprint and pass-duration arithmetic. Grade 9–12 should derive the elevation relation from the sine rule and do the decibel conversion.
Suggested timing: 10 minutes predict, 15 read and explain, 30 build and investigate, 25 measure and calculate, 10 independent explanation. Adapt the activities rather than rush unfinished reasoning.
Take it further
Calculate γ and the footprint for a 1,200 km shell at ε = 25° and compare with 550 km. Higher satellites see more ground and need fewer of them — so list what gets worse: slant range and therefore path loss, one-way delay, and orbital lifetime after failure. Then state which way you would decide and on what evidence.