MEO 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, 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)

Medium Earth Orbit (MEO) is conventionally defined as:Remember

(a) 180 – 2 000 km altitude (b) Exactly 35 786 km altitude (c) Approximately 2 000 – 35 786 km altitude (d) 40 000 – 45 000 km apogee altitude

Explanation: MEO spans 2 000 km to the geostationary altitude. Option (a) is LEO; (b) is GEO; (d) describes HEO apogees.

The orbital period of a GPS satellite at 20 180 km is:Remember

(a) 24 hours (b) 11 h 58 min — exactly half a sidereal day (c) 95 minutes (d) 4 hours

Explanation: T = 2π√(a³/μ) at a = 26 551 km gives 43 082 s = 11 h 58 min, exactly half a sidereal day — the deliberate "repeating ground track" design. Option (c) is LEO; (a) is GEO/geosynchronous; (d) is O3b's ~4.8 h.

A GNSS receiver needs a minimum of how many satellites visible simultaneously to compute a full three-dimensional position fix?Remember

(a) 2 (b) 3 (c) 6 (d) 4

Explanation: Four unknowns: x, y, z and the receiver clock bias δt — hence four independent pseudorange equations. Three satellites would suffice only with a perfect receiver clock (a 1 µs clock error = 300 m position error).

The GNSS altitude band (19 000–23 000 km) is environmentally difficult because it lies:Remember

(a) Near the peak of the outer Van Allen radiation belt, requiring radiation-hardened electronics (b) Inside the densest orbital debris shell (c) Within the ionospheric F-layer, blocking all radio signals (d) In the region of maximum atmospheric drag

Explanation: The outer electron belt peaks near 20 000 km — right at GNSS altitude. Total dose and deep-dielectric charging force rad-hard parts (3–10× COST cost). Debris (b) is a LEO problem; drag (d) is a sub-600 km problem; the ionosphere (c) is a signal error source, not a blocker, and is below 1 000 km.

The nominal GPS constellation consists of:Remember

(a) 24 satellites in 3 planes of 8, inclination 56° (b) 24 satellites in 6 planes of 4, inclination 55° (c) 66 satellites in 6 polar planes (d) 3 satellites at 120° spacing

Explanation: GPS: Walker 55°:24/6 — 24 slots in 6 planes (4 per plane) at 55° inclination, guaranteeing ≥4 satellites above 5° elevation worldwide (typically 6–12). Option (a) describes Galileo's 3×8 geometry; (c) is Iridium (LEO); (d) is the GEO broadcast triad.

Level 2 — Understand (Q6–Q10)

GNSS designers chose the half-sidereal-day period primarily because it:Understand

(a) Minimizes atmospheric drag on the satellites (b) Maximizes Doppler shift for better receiver sensitivity (c) Makes each satellite retrace the same ground track every day, giving predictable, repeatable coverage geometry and symmetrically averaged perturbations (d) Doubles the satellites' apparent speed across the sky

Explanation: In one half-sidereal-day orbit Earth rotates exactly 180°, so after two orbits the satellite retraces its ground track — coverage geometry repeats daily (simple almanacs, predictable constellations), and J₂ resonant perturbations average symmetrically. Drag (a) is negligible at 20 000 km; Doppler (b) is a nuisance, not a feature; (d) is false.

Why do dual-frequency GNSS receivers (L1 + L2/L5) achieve far better accuracy than single-frequency receivers?Understand

(a) The ionosphere is dispersive: its delay differs at the two frequencies, so differencing the pseudoranges removes ~99% of the largest natural ranging error (b) Two frequencies double the satellite transmit power (c) The second frequency is a backup in case the first is jammed (d) Dual frequency halves the speed of light, halving all delays

Explanation: Ionospheric delay ∝ 1/f² (dispersive medium). Measuring the same path at two frequencies separates the ionospheric term from the geometric range — the classic dual-frequency correction, worth 1–100 m of accuracy. Options (b) and (d) are physically false; (c) describes a minor secondary benefit only.

The fundamental reason navigation constellations live in MEO rather than GEO is:Understand

(a) GEO satellites cannot carry atomic clocks (b) MEO has no radiation belts (c) GEO signals would be too strong for receivers (d) Geometry: 24 MEO satellites spread across the whole sky give strong 3-D dilution-of-precision, whereas visible GEO satellites cluster along one arc at similar elevation — no vertical geometry, no fix

Explanation: Positioning needs satellites at widely spread elevations and azimuths. MEO sats sweep from 5° to near-zenith with full-azimuth coverage; GEO sats (all near the equatorial plane) sit at one elevation band in one azimuth sector — the geometry matrix is singular for 3-D positioning. Atomic clocks (a) fly in GEO fine; MEO has the worst belts (b is false); signal strength (c) is a link-budget matter, not a geometry matter.

MEO spacecraft must be designed for solar-maximum radiation conditions even though most of the mission may occur in quieter years, because:Understand

(a) Components are cheaper when designed for worst case (b) Radiation damage is cumulative and unrepairable — a single violent solar-maximum year can deliver most of the mission's total dose, and failed parts cannot be swapped in orbit (c) Solar maximum improves GNSS accuracy, so satellites must be present (d) Ground stations only operate during solar maximum

Explanation: Total ionizing dose integrates over the mission and is irreversible; the outer-belt electron flux varies ~10× with the solar cycle. A 15-year life will certainly contain at least one solar-maximum period, so components must survive it. Option (a) is backwards — rad-hard parts cost more precisely because of this requirement.

GLONASS flies at 64.8° inclination while GPS uses 55°. The best explanation is:Understand

(a) 64.8° requires fewer satellites (b) 64.8° avoids the Van Allen belts (c) Russian territory extends to very high latitudes; the steeper inclination keeps satellites higher in the sky there, improving geometry, availability and signal path at 60°+ N (d) 64.8° is the critical inclination, so perigee stays fixed

Explanation: Inclination sets the latitude band of best service. GLONASS optimized for its national territory (Moscow 55.8°N, Murmansk 69°N); GPS optimized for global-average geometry. The critical inclination is 63.4° — and GLONASS's 64.8° is circular (e ≈ 0), so perigee precession (d) is irrelevant; the belt position (b) is set by altitude, not inclination.

Level 3 — Apply (Q11–Q15)

The orbital period of a Galileo satellite at 23 222 km altitude is approximately:Apply

(a) 14.1 hours (b) 11 h 58 min (c) 24 hours (d) 4.8 hours

Explanation: a = 29 593 km; T = 2π√(a³/μ) ≈ 50 664 s ≈ 14.1 h. Option (b) is GPS's half-sidereal-day; (d) is O3b's ~4.8 h at 8 062 km. Kepler's third law scales as a^3/2: higher MEO, longer period.

The service coverage half-angle of a GPS satellite (20 180 km) with a 5° elevation mask is approximately:Apply

(a) 23° (b) 71° (c) 81° (d) 45°

Explanation: ρ = arccos((6371/26 551)·cos5°) − 5° = arccos(0.2391) − 5° ≈ 76.2° − 5° = 71.2° — each satellite covers ~38% of the globe. Option (a) is a LEO figure; (c) is GEO's geometric maximum (0° mask).

The free-space path loss at GPS L1 (1.575 GHz) for a 20 180 km path is approximately:Apply

(a) 196 dB (b) 168 dB (c) 182 dB (d) 154 dB

Explanation: L_FS = 92.45 + 20log(1.575) + 20log(20 180) = 92.45 + 3.9 + 86.1 = 182.5 dB. Option (b) ≈ LEO at Ku; (a) ≈ GEO at C-band. The GPS signal arrives ~20 dB below thermal noise — recovered by ~43 dB of spread-spectrum processing gain.

The maximum Doppler shift at L1 from a GPS satellite (orbital velocity 4.19 km/s, taken as a worst-case radial bound) is approximately:Apply

(a) ±4 Hz (b) ±400 Hz (c) ±4 kHz (d) ±22 kHz

Explanation: Δf = (v/c)f₀ = (4.19/3×10⁵) × 1.575×10⁹ ≈ ±22 kHz as the absolute bound; realistic geometry (rise/set, cos-elevation, Earth-rotation factor R_E/(R_E+h) ≈ 0.24) reduces this to ±4–5 kHz — which is why option (c) is what a receiver actually sees, while the question asks for the v-as-radial bound. LEO at the same frequency gives ±40 kHz; GEO is negligible.

GPS satellite clocks are rate-offset by −4.45×10⁻¹⁰ before launch to compensate relativistic effects. If this correction were omitted, the ranging error would accumulate at approximately:Apply

(a) 11.5 km per day (b) 11.5 m per day (c) 1.2 km per day (d) 115 km per day

Explanation: ΔR per day = 4.45×10⁻¹⁰ × 86 400 s × 3×10⁸ m/s ≈ 11 540 m ≈ 11.5 km/day. The correction blends general relativity (gravitational blue shift, +45.7 µs/day) and special relativity (velocity, −7.2 µs/day) to net +38.5 µs/day, neutralized by the clock offset. Option (b) slips three orders of magnitude.

Level 4 — Analyse (Q16–Q20)

During an equatorial geomagnetic storm, single-frequency GNSS receivers across East Africa report metre-level jumps while dual-frequency receivers are unaffected. The correct diagnosis:Analyse

(a) The storm knocked satellites out of orbit (b) Solar radiation overloaded the receivers' processors (c) Equatorial ionospheric scintillation and TEC gradients ballooned the ionospheric delay — the dominant single-frequency error; dual-frequency receivers subtract it via the dispersive (1/f²) two-frequency correction (d) The storm changed the speed of light in the troposphere

Explanation: Storm-time equatorial plasma bubbles cause rapid total-electron-content (TEC) variations — the classic single-frequency failure mode after sunset. The differential measurement at two frequencies is immune because the ionosphere's refractive index depends on f² while the geometric delay does not. Satellites remain in orbit (a); processor overload (b) is not how scintillation manifests; the troposphere (d) is non-dispersive and storm-independent at GNSS bands.

A national agency proposes replacing its MEO GNSS usage with three GEO satellites broadcasting navigation signals over East Africa. Technical review concludes the system cannot deliver 3-D positioning. The decisive analysis:Analyse

(a) GEO satellites move too fast across the sky for ranging (b) All visible GEO satellites lie in the equatorial plane: their line-of-sight unit vectors span only a 2-D subspace (one elevation band, one azimuth sector) — the positioning geometry matrix is rank-deficient in the vertical, so vertical DOP diverges regardless of signal quality (c) GEO is too far away for the signals to arrive (d) Atomic clocks cannot operate above the Van Allen belts

Explanation: Dilution of Precision is a geometry property: with all transmitters on one arc near one elevation (≈ the station's latitude complement), the vertical component of the design matrix collapses — VDOP → ∞. MEO's mid-orbit altitude is the sweet spot precisely because satellites sweep 5°–90° elevations across all azimuths. Distance (c) only weakens the link margin (recoverable), clocks (d) work anywhere, and GEO's slow motion (a) would if anything help.

An operator of a MEO constellation at 20 000 km notices that spares are kept in dedicated orbital slots and launches are rare, whereas an LEO operator replaces failed satellites routinely. The structural reason:Analyse

(a) MEO satellites never fail, so spares are ceremonial (b) Launch vehicles cannot reach MEO, so broken satellites are repaired in orbit (c) LEO satellites are more reliable than MEO satellites (d) MEO replacement is slow (months of transfer) and expensive, so reliability must be insured in orbit: pre-positioned on-orbit spares re-phase within days, whereas LEO's low per-unit cost and short decay times favor launch-and-replace economics

Explanation: The economics follow the physics of the altitude: reaching 20 000 km costs months and tens of millions per replacement, so MEO systems buy insurance in advance (spares in plane, redundant payloads). LEO failures decay in years and units cost a fraction — churn is the business model. Options (a), (b), (c) are factually wrong; repairs in orbit (b) are not routine practice.

Rain-fade planning for an O3b-class MEO trunk link (8 062 km, Ku-band) versus an equivalent GEO link concludes the MEO link needs roughly 6 dB less rain margin. The quantified physical basis:Analyse

(a) Rain attenuation is path-integrated through the rain cell: the MEO's higher elevation geometry shortens the slant path through the troposphere, and the ~13 dB lower free-space path loss (175 vs ~205 dB at Ku) supplies margin that absorbs the fade — GEO must spend power or ACM to buy the same protection (b) It rains less at 8 062 km altitude (c) MEO satellites fly below the weather (d) The conclusion is wrong: rain fade depends only on frequency, so margins must be identical

Explanation: Specific attenuation (dB/km in rain) is altitude-independent — but the link margin is not. MEO's shorter slant path through rain plus its ~13 dB FSPL headroom means a given fade dents C/N far less. Option (d) confuses the fade itself with the budget's ability to absorb it; (b) and (c) are category errors (satellites are all above weather; rain lives in the troposphere below ~15 km).

A surveyor using RTK corrections achieves centimetre accuracy, while a phone 5 metres away using the same constellation achieves metre-level accuracy. The engineering analysis:Analyse

(a) The surveyor's receiver is closer to the satellites (b) Phones use a different, weaker constellation (c) Both use code pseudorange (~1–5 m precision); the surveyor adds carrier-phase measurement (2 mm wavelength → cm precision) plus differential/RTK corrections that cancel common errors (ionosphere, ephemeris, clocks) using a nearby reference — the satellites are identical, the measurement technique and error-modeling differ (d) Centimetre accuracy violates the GPS signal specification and must be a processing artefact

Explanation: Accuracy is a measurement-stack property, not a satellite property: carrier phase (2.4 mm at L1) + differencing against a base station removes shared biases and integer-ambiguity resolution delivers cm–mm. Option (a) is geometrically negligible (5 m vs 20 000 km); (b) is false (same signals); (d) is false (RTK is a mature, spec-consistent technique).

Post-Test Answer Key & Summary

Q	Key	Bloom Level	Concept Tested
1	c	Remember	MEO altitude regime
2	b	Remember	Half-sidereal-day period
3	d	Remember	Four satellites for a fix
4	a	Remember	Outer Van Allen belt at GNSS altitude
5	b	Remember	GPS Walker 55°:24/6
6	c	Understand	Repeating ground track design
7	a	Understand	Dual-frequency ionosphere correction
8	d	Understand	MEO geometry vs GEO for PNT
9	b	Understand	Cumulative dose → design for solar max
10	c	Understand	Inclination matched to service latitudes
11	a	Apply	Galileo period 14.1 h
12	b	Apply	Coverage half-angle 71.2°
13	c	Apply	FSPL 182.5 dB at L1
14	d	Apply	Doppler bound ±22 kHz (≈±5 kHz realized)
15	a	Apply	Relativistic drift 11.5 km/day
16	c	Analyse	Equatorial scintillation diagnosis
17	b	Analyse	GEO geometry rank-deficiency for PNT
18	d	Analyse	On-orbit spares vs launch-and-replace economics
19	a	Analyse	Rain margin from path geometry + FSPL
20	c	Analyse	Carrier-phase RTK vs code pseudorange

Grading guide: 18–20 = Excellent (GNSS-ready mastery); 14–17 = Good; 10–13 = Fair — revisit coverage geometry and link budgets; <10 = Review the MEO Satellite Systems study guide before re-attempting.