GEO Satellite Systems Quiz — ECE 521E Radar & Satellite Engineering

Instructions to Candidates

This quiz contains 20 multiple-choice questions covering the first four levels of Bloom's taxonomy (Remember, Understand, Apply, Analyse).
Select the best answer for each question, then press "Check Answers" for instant marking with explanations, or consult the Post-Test Answer Key at the end.
Constants: μ = 3.986×10⁵ km³/s², R_E = 6 371 km, sidereal day = 86 164 s, c = 3×10⁵ km/s, k = R_E/(R_E+h_GEO) = 0.1511, FSPL = 92.45 + 20log f(GHz) + 20log d(km).
Total time: 45 minutes (indicative).

Level 1
Remember (Q1–5)

Level 2
Understand (Q6–10)

Level 3
Apply (Q11–15)

Level 4
Analyse (Q16–20)

Level 1 — Remember (Q1–Q5)

The altitude of a geostationary satellite above the equator is:Remember

(a) 20 180 km (b) 35 786 km (c) 8 062 km (d) 780 km

Explanation: 35 786 km is the canonical figure (semi-major axis 42 164 km minus Earth's equatorial radius). Option (a) is GPS/MEO altitude; (c) is O3b (MEO); (d) is Iridium's LEO altitude.

The orbital period that makes a satellite geostationary equals:Remember

(a) One solar day: 24 h 00 min 00 s (b) One half sidereal day: 11 h 58 min (c) One sidereal day: 23 h 56 min 4 s (86 164 s) (d) One anomalistic year

Explanation: Synchronism must be with the stars — Earth's rotation against the inertial frame is 86 164 s. The solar day (option a) is ~4 min longer because Earth advances ~1°/day along its orbit. Option (b) is the GPS/Molniya design choice.

Three geostationary satellites spaced 120° apart provide continuous coverage of:Remember

(a) All latitudes to about ±70° with ≥10° elevation — essentially the whole globe except the polar regions (b) The entire globe including both poles (c) Only the equatorial belt between ±23.5° (d) One third of the Earth each, with gaps between footprints

Explanation: Each GEO footprint reaches an Earth-central half-angle of ~81.3° (geometric horizon); at 120° spacing the footprints overlap by ~2×71.4° − 120° ≈ 23° at the 10° mask — continuous coverage to ±70° latitude. The poles (γ = 90° > 81.3°) are geometrically unreachable — option (b) is the classic misconception.

The dominant long-term station-keeping burden on a GEO satellite is:Remember

(a) Atmospheric drag (b) Solar radiation pressure on the solar arrays (c) Earth's magnetic field (d) North–south correction against lunar–solar gravity, which grows orbital inclination at ~0.9°/year

Explanation: Lunar–solar torques tilt the orbital plane secularly (~0.9°/yr); without N–S burns the satellite develops a daily figure-of-eight. E–W drift (Earth's triaxiality) is a gentle oscillation, and drag (a) is negligible at 35 786 km. N–S keeping consumes most propellant and sets the ~15-year lifetime.

At end of life, a GEO satellite is disposed of by:Remember

(a) De-orbiting it into the atmosphere over the Pacific (b) Raising it ~300 km into the graveyard orbit above the operational arc (c) Leaving it in place and switching off the transponders (d) Lowering it to 20 000 km to join the GNSS constellation

Explanation: A dead GEO satellite drifting uncontrolled would cross every operational slot; disposal to ~300 km above GEO (with passivation) isolates it from the arc for millennia. Re-entry (a) is impossible with GEO-class propellant reserves — this is a key GEO/LEO contrast, since LEO sats at 550 km get free disposal from drag.

Level 2 — Understand (Q6–Q10)

A geostationary satellite's period must equal the sidereal (not the solar) day because:Understand

(a) The sun's gravity perturbs solar-synchronous orbits (b) The solar day varies with the seasons (c) After one orbit the satellite must re-align with the same Earth-fixed meridian, and Earth rotates 360° relative to the stars in 86 164 s — 4 minutes less than the sun-facing day (d) ITU regulations define the broadcast day in sidereal units

Explanation: "Fixed over a meridian" is defined against Earth's inertial rotation. In 86 400 s Earth turns 360° + 0.986° relative to the sun; a satellite with a solar-day period would slip ~1°/day westward (~110 km/day at the equator) — visibly wandering within a month. The mismatch is pure celestial geometry, not regulation or perturbation.

All three conditions — circular orbit, equatorial plane, sidereal-day period — are required for a truly geostationary satellite because:Understand

(a) Each defect produces residual apparent motion: nonzero e gives east–west libration, nonzero i gives north–south figure-of-eight drift — fixed dishes would need tracking margins (b) Only circular orbits are stable against meteorites (c) The ITU only licenses satellites with all three properties (d) Any one condition alone is sufficient for stationarity

Explanation: A geosynchronous satellite with i = 5° traces a daily analemma (±5° N–S); with e = 0.001 it librates east–west by ~0.2°. The fixed-dish economy — GEO's entire value proposition — survives only with all three. Option (d) is the classic error: "geosynchronous" (period only) is weaker than "geostationary".

GEO satellites cannot serve polar regions because:Understand

(a) The ionosphere blocks signals above 75° latitude (b) Solar radiation blinds satellite receivers near the poles (c) Polar ice reflects the signals back into space (d) Pure geometry: the orbit plane is the equator, so the central angle from any GEO satellite to a pole is 90°, exceeding the maximum visible angle of ~81.3°

Explanation: Visibility requires central angle γ ≤ arccos(R_E/(R_E+h)) = 81.3°; a pole is always at γ = 90° from every point of the equatorial ring. No antenna, power, or frequency fixes a line-of-sight that does not exist — only inclined/elliptical (Molniya) or polar LEO constellations can serve the poles.

A multi-carrier transponder's TWTA is operated "backed off" from saturation primarily to:Understand

(a) Reduce the satellite's power consumption (b) Preserve amplifier linearity: several carriers through a saturated TWTA generate intermodulation products that fall inside other transponders (c) Comply with ITU emission masks (d) Extend the TWTA's lifetime by running it cool

Explanation: A nonlinear amplifier is a mixer: n carriers → n(n−1) intermod products at sums/differences of carrier frequencies, splattering energy into neighbouring channels. Back-off (IBO/OBO, typically 3–10 dB) buys linearity with EIRP — which is why multi-carrier EIRP budgets quote the backed-off value, not the saturated rating.

TCP connections over GEO links (≈250 ms RTT) routinely stall without PEP acceleration because:Understand

(a) GEO satellites corrupt TCP checksums (b) TCP cannot physically reach 35 786 km (c) The bandwidth-delay product exceeds the default window: with ~250 ms RTT, a standard window caps throughput at a few hundred kbps regardless of raw C/N (d) Rain fade resets TCP sessions

Explanation: Throughput ≈ window/RTT: a 64 kB window over 0.25 s caps at ~2 Mbps, and loss-triggered slow-start collapses across the long pipe. PEPs (TCP spoofing at the gateway) acknowledge locally, keeping the satellite pipe full. The physics is in the delay-bandwidth product — not corruption (a) or rain (d).

Level 3 — Apply (Q11–Q15)

Using Kepler's third law, the semi-major axis of a geostationary orbit is:Apply

(a) 26 560 km (b) 29 593 km (c) 35 786 km (d) 42 164 km

Explanation: a = (μT²/4π²)^1/3 = (3.986×10⁵ × 86 164²/39.478)^1/3 ≈ 42 164 km. Option (c) is the altitude (a minus Earth's radius); (a) is Molniya; (b) is Galileo. Exam trap: semi-major axis ≠ altitude.

The elevation angle of a GEO satellite seen from 60°N on the same meridian is approximately:Apply

(a) 21.9° (b) 30.0° (c) 15.0° (d) 45.0°

Explanation: sin el = (cos60° − 0.1511)/√(1 + 0.1511² − 2×0.1511×cos60°) = 0.3489/0.9337 = 0.3737 → el ≈ 21.9°. Option (b) naively computes 90° − 60°; (c) misplaces the satellite's central angle. This is why high-latitude GEO users suffer rain/terrain penalties.

The free-space path loss at 12 GHz for a GEO slant range of 37 500 km is approximately:Apply

(a) 168 dB (b) 182 dB (c) 205.5 dB (d) 222 dB

Explanation: L_FS = 92.45 + 20log(12) + 20log(37 500) = 92.45 + 21.6 + 91.5 = 205.5 dB. Option (a) is LEO-at-Ku — the ~37 dB difference between (a) and (c) is precisely why consumer terminals work in LEO but GEO needs 0.6 m+ dishes with 50+ dBW satellite EIRP.

The maximum daily eclipse duration for a GEO satellite at the equinox, when the orbit plane passes through Earth's shadow centre, is approximately:Apply

(a) 12 minutes (b) 70 minutes (c) 6 hours (d) 36 minutes

Explanation: Shadow half-angle at the orbit: arcsin(R_E/a) = arcsin(0.1511) = 8.69°. Duration = (2×8.69/360) × 86 164 s ≈ 4 160 s ≈ 69–70 min. Option (d) halves the full shadow crossing; (a) is a LEO eclipse; the battery must carry the full payload through this every night of the eclipse seasons.

A DTH downlink: satellite EIRP 52 dBW, 12 GHz, slant 37 500 km, user terminal G/T = 12 dB/K, misc losses 2 dB. The C/N₀ and the C/N in a 27.5 MHz DVB-S2 channel are:Apply

(a) 74 dBHz and 0 dB (b) 85 dBHz and 20 dB (c) 95 dBHz and 15 dB (d) 85 dBHz and 10.7 dB

Explanation: C/N₀ = 52 + 12 − 205.5 − 2 + 228.6 = 85.1 dBHz; C/N = 85.1 − 10log(27.5×10⁶) = 85.1 − 74.4 = 10.7 dB. DVB-S2 8PSK 3/4 needs ≈ 10 dB — a realistic 0.7 dB clear-weather margin, with ACM (not static margin) covering rain. Option (b) adds 10log wrongly (b is a distractor for 10× bandwidth error).

Level 4 — Analyse (Q16–Q20)

A tropical DTH operator on Ku-band reports daily outages at 15:00–17:00 during the wet season, while a co-located C-band link on the same satellite is unaffected. The technically correct analysis:Analyse

(a) Convective rain cells at 12 GHz produce 10–30 dB fades that exhaust the Ku link's small clear-weather margin; C-band at 4 GHz suffers perhaps 1–2 dB from the same cells — the operator must add ACM, uplink power control, or migrate critical services to C-band (b) The satellite drifts out of its station-keeping box every afternoon (c) The C-band transponder is newer and therefore immune to weather (d) The Ku link is being jammed by a competitor every afternoon

Explanation: Rain attenuation scales steeply with frequency (≈ f² in this range): the same afternoon convective cell that costs Ku 15 dB costs C-band ~1–2 dB. The time-of-day correlation with convection and the C-band control channel together isolate propagation — not drift (which would be time-independent), hardware age (c), or jamming (no such evidence). This is the textbook argument for C-band's survival in tropical markets.

Every March and September, enterprise VSAT customers at a Nairobi teleport lose their Ku links for a few minutes around local noon — on a predictable calendar, clearing by itself. The analysis:Analyse

(a) The satellite's batteries are exhausted from the eclipse season (b) The teleport's air conditioning fails at solar noon (c) Sun outage: near the equinoxes the sun passes behind the satellite as seen from the station, and the sun's ~10 000 K noise temperature swamps the receiver within the antenna beamwidth for a few minutes — a predictable, unavoidable geometry, managed by scheduling (d) The satellite is eclipsed by the moon

Explanation: The signature is diagnostic: equinoctial + noon + self-clearing + minutes-long = the sun crossing within HPBW/2 + 0.27° of the satellite direction. For a 1.2 m Ku dish (HPBW ≈ 1.5°) outages run up to ~2 min/day over about two weeks around each equinox. Battery failure (a) would not be noon-punctual; lunar eclipse (d) is rare and night-time; mitigation is operational scheduling, not hardware.

After a rate increase, an operator squeezes 12 carriers into one transponder by running the TWTA at saturation. Customers on neighbouring transponders now report intermittent interference. The analysis:Analyse

(a) The transponder's channel filters have failed and must be replaced (b) At saturation the TWTA is strongly nonlinear: 12 carriers generate thousands of third-order intermodulation products whose frequencies fall into adjacent transponders — the fix is restoring output back-off (≈ several dB) and/or linearization, accepting the EIRP penalty (c) Neighbouring satellites are interfering because of station-keeping drift (d) The interference is sun outages returning early

Explanation: Intermodulation order: n carriers → C(n,2)×2 third-order products (2f_i−f_j, f_i+f_j−f_k) landing inside the same and adjacent channels — exactly the symptom (neighbours hurt, onset coincident with the operating change). Filter failure (a) would not correlate with the carrier-plan change; adjacent-satellite interference (c) is orbit-wide, not transponder-selective; sun outages (d) are calendrical, not configuration-correlated.

A Scandinavian broadcaster at 68°N can receive a GEO DTH satellite at only 5° elevation and suffers constant rain outages. Management asks engineering to "fix it with a bigger dish." The analysis:Analyse

(a) A 3 m dish will fully solve it: gain beats all geometry (b) The solution is to move the satellite closer in its orbital slot (c) The solution is a higher-power satellite on the same slot (d) At 5° elevation the problem is geometry, not gain: the signal traverses ~10× the vertical atmospheric path (rain column, gaseous absorption, high noise temperature from warm ground spillover) — no dish recovers that; the engineering answer is HEO (Molniya-class), polar-LEO, or terrestrial distribution

Explanation: Elevation at 68°N: sin el = (cos68° − 0.1511)/√(1.0228 − 2×0.1511×cos68°) → el ≈ 7°. At 5–7° elevation the rain path length through a cell is ~10× the zenith value, ground noise couples into the sidelobes, and terrain clearance fails — each ~1–3 dB of penalty that dish area cannot buy back. A Molniya apogee over the Arctic gives 60–90° elevation: the correct orbital fix. Options (b) and (c) misunderstand that the deficit is atmospheric path and noise geometry, not EIRP.

A regulator must choose between awarding an orbital slot at 21°E to (A) a GEO broadcast platform or (B) approving a 3-satellite Molniya-HEO system for the same European coverage. Comparing total system behaviour, the strongest analysis is:Analyse

(a) GEO delivers the European mid-latitudes with a single fixed-dish satellite — simplest ground segment, cheapest terminals, mature interference regime; HEO's 3-satellite, tracked/ switched ground segment only pays off poleward of ~55°N where GEO elevation collapses below ~20° — for pan-European coverage the GEO award is correct, with HEO reserved for the Arctic niche (b) HEO always dominates GEO because its satellites come closer (c) GEO always dominates HEO because three satellites beat one (d) The two systems are indistinguishable for coverage purposes

Explanation: The coverage geometry decides: at 50°N GEO gives ~33–38° elevation — perfectly healthy; HEO's advantages (60–90° elevation, dwell) are irrelevant there, while its costs (3 tracked sats, constellation ops, belt transits every orbit) are real. Poleward of ~55°N the comparison inverts (GEO 21.9° at 60°N vs Molniya ~86°). "Always dominates" claims (b, c) and equivalence (d) both fail the quantitative geometry test.

Post-Test Answer Key & Summary

Q	Key	Bloom Level	Concept Tested
1	b	Remember	GEO altitude 35 786 km
2	c	Remember	Sidereal day 86 164 s
3	a	Remember	Three-satellite coverage, polar limit
4	d	Remember	N–S station-keeping driver
5	b	Remember	Graveyard-orbit disposal
6	c	Understand	Sidereal vs solar day
7	a	Understand	The three GEO conditions
8	d	Understand	Polar invisibility geometry
9	b	Understand	Back-off vs intermodulation
10	c	Understand	Bandwidth-delay product / PEP
11	d	Apply	GEO semi-major axis 42 164 km
12	a	Apply	Elevation 21.9° at 60°N
13	c	Apply	FSPL 205.5 dB
14	b	Apply	Eclipse duration ≈ 70 min
15	d	Apply	C/N₀ 85 dBHz, C/N 10.7 dB
16	a	Analyse	Ku vs C-band rain diagnosis
17	c	Analyse	Sun-outage signature
18	b	Analyse	Intermodulation from saturated multi-carrier TWTA
19	d	Analyse	Low-elevation geometry at high latitude
20	a	Analyse	GEO vs HEO coverage-regime analysis

Grading guide: 18–20 = Excellent (GEO-systems mastery); 14–17 = Good; 10–13 = Fair — revisit look angles and link budgets; <10 = Review the GEO Satellite Systems study guide before re-attempting.