HEO 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, J₂ = 1.0826×10⁻³, 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 period of a Molniya orbit is:Remember

(a) 24 hours (one solar day) (b) 12 hours (one solar day) (c) 11 h 58 min (one half sidereal day) (d) 95 minutes (like LEO)

Explanation: T = 2π√(a³/μ) with a = 26 560 km gives 43 082 s = ½ sidereal day exactly, so each satellite retraces the same ground track every day. Option (b) is 4 minutes too long — those 236 seconds matter; the orbit synchronizes with the stars, not the sun.

The critical inclination of a Molniya orbit — the angle at which perigee precession vanishes — is:Remember

(a) 55° (b) 63.4° (c) 90° (polar) (d) 98.6° (sun-synchronous)

Explanation: i_crit = arccos(1/√5) = 63.43°: the J₂ term (5cos²i − 1) vanishes and the ellipse freezes in inertial space. GPS uses 55° (a), which would let perigee precess ~30°/year in a Molniya orbit.

Apogee dwell — the satellite lingering near apogee — is a direct consequence of:Remember

(a) Kepler's second law: equal areas swept in equal times, so the satellite moves slowest at apogee (b) Atmospheric drag slowing the satellite at high altitude (c) The Earth's magnetic field braking the satellite (d) Solar radiation pressure pushing the satellite at apogee

Explanation: Angular momentum conservation (Kepler's second law) gives r²·dν/dt = const — the satellite moves slowest where it is farthest (v_a = 1.6 km/s vs v_p = 9.2 km/s for Molniya). Drag, magnetism and radiation pressure are perturbations, not the cause.

A Tundra orbit — used by Sirius XM satellite radio — has:Remember

(a) Period of 12 h and e ≈ 0.7 (b) Period of 24 h and e ≈ 0.7 (c) Period of 12 h and e ≈ 0.27 (d) Period of 24 h (one sidereal day), e ≈ 0.27, i = 63.4°

Explanation: Tundra: a = 42 164 km (24 h), e ≈ 0.27, critical inclination — each satellite dwells over its apogee region for ~2/3 of the day, so two phased satellites give continuous coverage. Option (a) describes Molniya's period with Tundra's wrong e.

Continuous high-latitude coverage from a Molniya orbit requires a minimum of ______ satellites per orbital plane:Remember

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

Explanation: Three satellites phased T/3 ≈ 8 h apart: each holds the ~8-hour "watch" centred on its ~10.8 h apogee dwell, handing over calmly to the next. Two would nominally tile the day but leave zero redundancy — one failure means a 6-hour blackout (see Q19).

Level 2 — Understand (Q6–Q10)

Why must a Molniya satellite's apogee be placed over the northern hemisphere while its perigee sits over the south?Understand

(a) The southern hemisphere has no ground stations (b) Atmospheric drag is stronger over the equator (c) The satellite spends ~90% of its period near apogee (Kepler's second law) — the dwell point must sit over the region being served; perigee's fast sweep is "wasted" over the unused hemisphere (d) Earth's magnetic field is weaker over the southern hemisphere

Explanation: The entire economic logic of HEO is time allocation: at e = 0.7 the apogee-side half of the sky consumes 90.6% of the period. Apogee over the service region converts orbital mechanics into coverage; perigee becomes a 68-minute non-event over the wrong hemisphere. Options (a), (b), (d) are irrelevant to orbital geometry.

Without the critical inclination, a Molniya constellation would fail because:Understand

(a) The orbital period would change with time (b) Atmospheric drag would circularize the orbit (c) The satellite would spiral into the Earth (d) J₂ perturbations would make the argument of perigee precess (~30°/year at i = 55°), walking the apogee out of the northern service hemisphere within about three years

Explanation: dω/dt ∝ (5cos²i − 1): at 63.4° this vanishes and the ellipse freezes; at 55° perigee precesses ~30°/yr, dragging apogee south — the dwell advantage evaporates. Option (a) is false (period depends only on a, which J₂ barely changes); (b) and (c) are not J₂ effects.

The argument of perigee of an operational Molniya satellite is ω = 270°. This places:Understand

(a) Perigee over the southern hemisphere and apogee over the northern hemisphere — the service geometry (b) Perigee over the northern hemisphere for faster coverage there (c) The orbit plane perpendicular to the equator (d) Apogee directly over the equator

Explanation: With ω = 270° (perigee at the southernmost point of the orbit, below the equator) the apogee sits at the northernmost point — over the service region. Option (b) would waste the fast perigee motion over the customers and park the dwell over Antarctica.

Why must the three Molniya satellites of a continuous-coverage constellation share the SAME orbital plane, rather than three different planes (RAANs)?Understand

(a) Different planes would violate the critical inclination condition (b) Apogees at different RAANs would sit over different longitudes, fragmenting coverage into three separated zones with gaps between them; phasing within one plane keeps a single seamless dwell region (c) Three planes would triple the launch cost (d) The satellites would collide if placed in different planes

Explanation: Coverage continuity demands that the dwell regions overlap; one plane with 8-h phasing gives a single unbroken high-elevation zone. Multiple planes give multiple apogee longitudes — three patchy coverages instead of one continuous one. The critical inclination applies per-plane regardless of RAAN (a is false); collision risk exists in any constellation (d is not the argument).

The "perigee gap" — the hours around perigee when the Molniya satellite cannot serve its northern users — exists because:Understand

(a) The satellite switches off its transmitter to save power (b) The high speed at perigee breaks the radio link (c) Geometry: near perigee the satellite is over the southern hemisphere, below the horizon or at unusably low elevation for northern stations (d) Rain attenuation peaks at perigee altitude

Explanation: At perigee the satellite is at 1 600 km altitude over the southern hemisphere — geometrically invisible from 60°N for most of the pass. The link is easy (short range) but unreachable: HEO's constraint is geometry, not power. Options (a), (b), (d) are not physical causes.

Level 3 — Apply (Q11–Q15)

The semi-major axis of a Molniya orbit (T = ½ sidereal day = 43 082 s) is:Apply

(a) 42 164 km (b) 26 560 km (c) 20 180 km (d) 6 921 km

Explanation: a = (μT²/4π²)^1/3 = (3.986×10⁵ × 43 082²/39.478)^1/3 ≈ 26 560 km. Option (a) is the 24-h (GEO/Tundra) value; (c) is GPS's MEO altitude; (d) is a 550 km LEO radius.

For a Molniya orbit with a = 26 560 km and e = 0.7, the perigee and apogee altitudes are approximately:Apply

(a) 500 km and 35 786 km (b) 1 597 km and 35 786 km (c) 500 km and 45 152 km (d) 1 597 km and 38 781 km

Explanation: r_p = a(1−e) = 7 968 km → altitude 1 597 km; r_a = a(1+e) = 45 152 km → altitude 38 781 km (above GEO!). Options confusing radius (c: 45 152 km is r_a, not the altitude) or GEO's altitude (b) are the designed traps.

Using the time equation t/T = (E − e sin E)/2π with E = 2·atan(√((1−e)/(1+e))·tan(ν/2)), the fraction of a Molniya period spent in the apogee-side half of the sky (ν from 90° to 270°) is:Apply

(a) 90.6% (≈ 10.8 hours) (b) 75.0% (9.0 hours) (c) 50.0% (6.0 hours) (d) 33.3% (4.0 hours)

Explanation: At ν = 90°: E = 0.7955 rad, sin E = 0.7140, so t/T = (0.7955 − 0.4998)/2π = 4.71%; the apogee half is 1 − 2×4.71% = 90.6% ≈ 10.8 h. Option (c) is the naive (wrong) angle-ratio answer — the exam classic.

The orbital velocities of a Molniya satellite at perigee and apogee are approximately:Apply

(a) 7.6 km/s and 3.1 km/s (b) 9.2 km/s and 3.1 km/s (c) 9.2 km/s and 1.6 km/s (d) 7.6 km/s and 7.6 km/s (constant)

Explanation: v = √(μ(2/r − 1/a)): v_p = √(3.986×10⁵ × (2/7 968 − 1/26 560)) ≈ 9.2 km/s; v_a ≈ 1.6 km/s. The ratio v_p/v_a = (1+e)/(1−e) = 5.67 ✓. Option (a) confuses GEO's 3.07 km/s.

During apogee dwell, a station at 60°N observes the Molniya satellite (apogee sub-point 63.4°N, same meridian) at an elevation of approximately:Apply

(a) 24° (b) 86° (c) 63° (d) 45°

Explanation: γ = 3.4° with k = R_E/r_a = 0.1411: sin el = (cos 3.4° − 0.1411)/√(1 + 0.1411² − 2×0.1411×cos 3.4°) = 0.8571/0.8590 ≈ 0.998 → el ≈ 86° — nearly overhead. Option (a) is what a GEO satellite on the same meridian would give at 60°N (21.9°). This elevation advantage is HEO's entire raison d'être.

Level 4 — Analyse (Q16–Q20)

Two years after launch, operators notice the Molniya constellation's apogee point drifting southward at roughly 2.5° per month; coverage of the Arctic is degrading. Records show the satellites were injected at i = 55° instead of the specified 63.4°. The correct analysis:Analyse

(a) The satellites are running low on propellant, causing orbital decay (b) Lunar-solar gravity is dragging the apogee south; it will return in six months (c) Atmospheric drag at perigee is lowering apogee (d) At i = 55°, 5cos²i − 1 ≠ 0, so J₂ makes the argument of perigee precess at ≈ 30°/year (~2.5°/month): the ellipse rotates in its plane, walking the apogee out of the service hemisphere — no amount of station-keeping fuel fixes a fundamentally wrong inclination cheaply

Explanation: The drift rate matches the J₂ prediction exactly (~30°/yr at 55° vs exactly zero at 63.43°), and its direction (out of the northern hemisphere) identifies ω-precession rather than decay or drag — those change altitudes, not the ellipse's orientation in the plane. The 2.5°/month signature + wrong-inclination records = geometric death sentence; the fix is re-injection at 63.4°, not more fuel. Options (a), (b), (c) produce different signatures (altitude loss, periodic libration, energy decay).

A link study finds that from 60°N, a Molniya satellite at apogee (38 781 km) has nearly the same FSPL as a GEO satellite (187.7 vs 188.0 dB at 1.5 GHz) — yet the Molniya system delivers far better availability in arctic weather. The correct analysis:Analyse

(a) The Molniya satellite moves slower, so rain cannot attenuate its signal (b) Path length is nearly identical, so the advantage is not path loss but ELEVATION: ~86° overhead means the signal crosses only ~2 km of troposphere vertically, while GEO's ~22° elevation forces a ~5× longer atmospheric path through rain and noise — the margin that GEO must buy with power or ACM, HEO gets from geometry (c) The Molniya apogee is outside the atmosphere entirely (d) FSPL is lower at 63.4° inclination

Explanation: Both links are ~40 000 km; FSPL differs by 0.3 dB — negligible. The availability gap comes from elevation: atmospheric path ∝ 1/sin(el) — 22° → ~2.7× the vertical column of rain, clouds and noise temperature vs 86°. Options (a), (c), (d) are physical nonsense (inclination does not enter FSPL).

A HEO spacecraft's radiation design must survive both the inner belt (proton-dominated) at perigee and the outer belt (electron-dominated) at apogee — twice per 12-hour orbit. The design consequence that follows from this specific environment is:Analyse

(a) No shielding is needed because the passes are brief (b) Only solar-array charging matters; electronics are unaffected (c) Charging regimes alternate (electron deep-dielectric charging at apogee dwell, proton-induced upsets at perigee), so both shielding AND charge-drain design are mandatory — and with ~1 460 belt transits per year, design must assume worst-case solar-maximum conditions (d) The radiation is identical to LEO, so COTS electronics suffice

Explanation: HEO's uniqueness is alternating regimes: hours of electron charging at apogee (inside the outer belt) bracket brief proton exposure at perigee (inner belt edge) — 2×730 = ~1 460 transits/year. Mitigations: rad-hard parts, conductive coatings, charge-drain resistors, and design for solar-maximum belt flux (damage is cumulative and irreducible). Option (d) fails: LEO at 550 km sits below the belts and uses COTS; HEO cannot.

An operator proposes replacing the classic three-satellite Molniya constellation with two satellites phased 6 h apart to save cost, arguing "the dwell window is 10.8 hours, so two windows cover the 24-hour day." The decisive analysis:Analyse

(a) The arithmetic barely closes (two 10.8 h windows centred 6 h apart do overlap), but the design is fragile: zero redundancy — a single satellite failure opens a 6-hour daily outage — and every handover occurs at the window edge (~5° elevation) with both satellites at their worst geometry; three satellites buy redundancy and calm, high-elevation handovers for one extra spacecraft (b) Two satellites cannot cover the day at all, since each covers only 10.8 h (c) Two satellites would collide at the phasing point (d) Cost is identical, so the question is moot

Explanation: Two windows of 10.77 h spaced 6 h apart overlap by ~5.8 h — coverage is geometrically continuous (so b is wrong), but: (1) no spare: one failure → 6 h/day blackouts, unacceptable for early-warning/utility comms; (2) handover happens at the window edges where elevation ≈ 5° — the exact low-elevation conditions HEO exists to avoid. Three sats phase 8 h apart put handovers at ~35–40° elevation with the third satellite as insurance. Option (c) is false (phasing in time, not co-location).

A mission must choose between (a) one tracking antenna following the dwell satellite and (b) three fixed antennas with make-before-break satellite switching for a Molniya communications service. The analysis supporting the classic Molniya choice (tracking) for continuous trunk traffic is:Analyse

(a) Fixed antennas cannot point above 45° elevation (b) Tracking antennas never lose lock, by definition (c) Switching causes the satellite's transmitter to overheat (d) The dwell satellite's apparent motion is slow (~0.05°/s) — trivial for a modern tracking mount — and continuous tracking avoids any switch interruption; switching suits broadcast/consumer traffic (Sirius XM) where a seconds-long make-before-break event every 8–12 h is acceptable, but trunk customers paying for uninterrupted service prefer the seamless tracked link

Explanation: The dwell rate (~12× GEO's apparent motion) is well within tracking-mount capability (1–5°/s standard), so seamless trunk service is achievable with one antenna; the 3-fixed + switch architecture trades a periodic interruption and triple antenna count for simpler drives. The choice is service-driven: continuous carrier-grade traffic → track; bursty consumer/broadcast traffic → switch. Options (a), (b), (c) are factually wrong.

Post-Test Answer Key & Summary

Q	Key	Bloom Level	Concept Tested
1	c	Remember	Molniya period = ½ sidereal day
2	b	Remember	Critical inclination 63.4°
3	a	Remember	Kepler's second law basis of dwell
4	d	Remember	Tundra orbit parameters
5	b	Remember	3 satellites per plane
6	c	Understand	Apogee placement logic
7	d	Understand	J₂ perigee precession
8	a	Understand	ω = 270° service geometry
9	b	Understand	Single-plane phasing rationale
10	c	Understand	Perigee gap geometry
11	b	Apply	a = 26 560 km from Kepler
12	d	Apply	r_p/r_a → 1 597 / 38 781 km
13	a	Apply	Dwell fraction 90.6% via time equation
14	c	Apply	Velocities 9.2 / 1.6 km/s
15	b	Apply	Elevation ≈ 86° at 60°N
16	d	Analyse	Wrong-inclination drift diagnosis
17	b	Analyse	Elevation (not path loss) advantage
18	c	Analyse	Alternating radiation regimes
19	a	Analyse	Two-satellite fragility analysis
20	d	Analyse	Track vs switch architecture trade

Grading guide: 18–20 = Excellent (HEO-mission ready); 14–17 = Good; 10–13 = Fair — revisit the time equation and critical inclination; <10 = Review the HEO Satellite Systems study guide before re-attempting.