Satellite Engineering Quiz

Core fact. A geostationary orbit has a semi-major axis a = 42,164 km measured from the centre of the Earth. The altitude above the surface is h = a − R⊙ = 42,164 − 6,378 = 35,786 km.

Why the other options are wrong:

• A (20,200 km) is the GPS constellation altitude in MEO — a different orbit class entirely.
• C (42,164 km) is the classic exam trap: it is the orbital radius (distance from Earth's centre), not the altitude above the surface. Examiners deliberately include it to test whether you understand the distinction between r and h.
• D (1,000 km) is a typical Molniya perigee altitude / a mid-LEO imaging altitude — far below geostationary.

Memory anchor. Remember the pair as a couplet: “GEO: 42,164 from the centre, 35,786 from the ground.” If you can derive it from Kepler's third law (see Worked Example 3 in the study guide), you never need to memorise it.

Core fact. Kepler's three laws describe: (1) orbit shape — ellipse with the central body at one focus; (2) areal velocity is constant (satellite sweeps equal areas in equal times, moving fastest at perigee); (3) period–size relationship — T² = (4π²/μ)a³.

Why the other options are wrong:

• A states Kepler's first law.
• B states Kepler's second law.
• D is physically false: for circular orbits v = √(μ/r), so velocity decreases with orbital radius — LEO moves at ~7.8 km/s while GEO crawls at 3.07 km/s.

Exam tip. When a question says “Kepler's third law,” translate immediately to “the T²–a³ formula” — it is the only one of the three that contains a formula and is the one used in every numerical orbit problem.

Core fact. The Earth rotates once relative to the distant stars in a sidereal day = 86,164 s = 23 h 56 min 4 s. A geostationary satellite must match this period exactly to remain fixed over one longitude.

Why not 24 hours? A mean solar day (24 h) is measured relative to the Sun. Because Earth orbits the Sun while rotating, it must turn about 361° to bring the Sun back to the same meridian — roughly 1° more than one full rotation. That extra ~4 minutes means the solar day is ~236 s longer than the sidereal day. If a satellite used a 24 h period, it would drift ~0.986°/day westward along the equator — obviously unacceptable for a “fixed” satellite.

Why the other options are wrong:

• C (11 h 58 min) is the half-sidereal-day period used by GPS and Molniya satellites.
• D is a distractor with no orbital basis.

Practical note. The sidereal day also explains why a GPS satellite's ground track repeats exactly every (sidereal) day, which is essential for predictable navigation geometry.

Core fact. The GPS constellation flies in six orbital planes at an altitude of 20,200 km (a = 26,578 km), inclined 55°, with a period of 11 h 58 min — the classic MEO navigation architecture. A nominal constellation of 24 satellites guarantees at least 4 satellites in view anywhere on Earth at all times (the minimum for 3-D position fix plus receiver clock correction).

Why the other options are wrong:

• A (LEO ~550 km) describes Starlink, not navigation; LEO geometry changes far too quickly for stable ranging (and would need thousands of satellites for equivalent availability).
• C (GEO) gives poor, slowly changing geometry at high latitudes and cannot cover the poles at all — unsuitable for a global navigation system.
• D (Molniya HEO) is a Russian communications orbit, not a GNSS orbit.

Comparative anchor. The four GNSS systems all chose MEO: GPS 20,200 km • GLONASS 19,100 km • Galileo 23,222 km • BeiDou 21,528 km (plus GEO/IGSO augmentations for regional integrity over Asia). MEO is the navigation sweet spot between LEO instability and GEO geometry limits.

Core fact. Because the Earth is oblate (equatorial bulge, J2 term), an elliptical orbit's line of apsides (perigee–apogee line) slowly rotates in inertial space — apsidal precession. The precession rate depends on inclination i. Solving for the rate equal to zero gives the critical inclination:

At i = 63.4°, apogee stays permanently over the same (northern) hemisphere with essentially no station-keeping — the entire design rationale of the Molniya orbit. Russia, lacking geostationary visibility at its latitudes, used three Molniya satellites to keep a satellite high over the Northern Hemisphere 24/7.

Why the other options are wrong:

• A describes nodal precession (rotation of the orbital plane), exploited by sun-synchronous orbits at ~98° — a different phenomenon with a different critical angle.
• C confuses launch-site geometry with orbit mechanics; the inclination choice is driven by perturbation physics, not launch azimuth.
• D is wrong for the same reason as A — sun-synchronous requires a retrograde ~98° orbit.

Exam tip. Keep the two “critical angles” separate: 63.4° → frozen apsides (Molniya) • 98° → sun-synchronous nodal regression (Earth observation).

Concept. Three conditions must all hold for a satellite to appear stationary: (1) orbital period equals the Earth's sidereal rotation period (86,164 s); (2) the orbit lies in the equatorial plane (i = 0°); (3) the satellite travels west-to-east, the same sense as Earth's rotation. Option B captures all three. Under these conditions the satellite's angular rate exactly cancels Earth's, so its sub-satellite point (the point on Earth directly below it) never moves from one equatorial longitude.

Why the other options are wrong:

• A is physically false — the satellite is in free fall and moves at ~3.07 km/s in inertial space. What is zero is its velocity relative to the rotating ground, not its absolute velocity.
• C appeals to distance, but apparent motion depends on relative angular rate, not range; the Moon (384,400 km away) visibly moves across the sky precisely because its period (27.3 days) differs from Earth's rotation.
• D would be enormously wasteful and is not done — GEO satellites only fire thrusters for small station-keeping corrections (~m/s per year), not to hover.

Connected idea. If any of the three conditions fails — same period but non-zero inclination or eccentricity — the satellite is merely geosynchronous and traces a figure-8 (analemma) in the sky, requiring a tracking antenna. Only the full triple condition earns “geostationary.”

Concept. Two physical facts combine to force a LEO constellation:

1. Small footprint. From 800 km the horizon-limited coverage radius is roughly 2,750 km, i.e., each satellite sees only about 3–6% of Earth's surface (and much less at useful elevation angles).
2. Short pass duration. With a ~100-minute period, a satellite rises, transits and sets over a given station in roughly 10–20 minutes. Continuous service therefore requires another satellite to rise as one sets — a handover process that must be seamless and frequent.

Stitching continuous global (or even continuous regional) coverage out of these brief, small footprints demands overlapping satellites in multiple orbital planes — hence Iridium's 66, OneWeb's ~648, and Starlink's thousands.

Why the other options are wrong:

• A reverses causality; power limits link margin, not geometric coverage.
• C is fabricated — ITU rules govern spectrum and filings, not constellation size.
• D is backwards: LEO satellites move fastest (~7.5 km/s), which is the very reason passes are short.

Concept. Latency in a satellite link is dominated by propagation delay, t = d/c, where d is the slant range and c = 3×10⁸ m/s. The GEO range (~35,786 km) is roughly 70 times a typical LEO range (~500–1,500 km), so:

A ~250 ms round trip is perceptible in voice (overlap), awkward for video conferencing and online gaming, and harmful to TCP throughput because the congestion window grows slowly when every ACK takes a quarter-second. LEO's few-ms delay makes the satellite link behave like terrestrial fibre for interactive traffic.

Why the other options are wrong:

• A violates physics — all radio waves in free space travel at c regardless of the satellite's altitude.
• C confuses link budget with delay: transmit power raises SNR, not signal speed.
• D is false — LEO signals actually traverse the ionosphere twice, causing scintillation/fading that GEO links experience less severely.

Concept. Kepler's second law is a statement of conservation of angular momentum, h = r²ω = constant. Consequently the satellite's angular velocity ω is smallest where r is largest — at apogee. Integrated over the orbit, a satellite spends a disproportionate fraction of its period near apogee. For the Molniya orbit (e ≈ 0.72, T = 11 h 58 min) about 8 of the 12 hours elapse within the high-altitude half of the orbit over the Northern Hemisphere.

Design payoff. This dwell is not a curiosity — it is the entire mission concept. A satellite positioned so its apogee sits over northern latitudes delivers ~8 hours of high-elevation coverage per orbit, so only three satellites, phased 8 hours apart, maintain continuous service at latitudes where GEO is useless.

Why the other options are wrong:

• A — no apogee hold is used; thrust is reserved for periodic correction of perturbations.
• C — at ~39,700 km altitude the atmosphere is a hard vacuum; drag is utterly negligible.
• D — solar gravity perturbs the orbit secularly but is not what creates the dwell; the dwell exists even in a pure two-body orbit.

Concept. Earth's equatorial bulge (J2) exerts a torque on inclined orbits that makes the orbital plane rotate about Earth's axis — nodal regression. For a prograde orbit the node regresses westward; for a retrograde orbit (i > 90°) it advances eastward. Setting the regression rate to +0.9856°/day (matching the Sun's eastward drift along the ecliptic) yields an inclination of about 98°.

Operational consequence. Matching the Sun's apparent motion keeps the angle between the orbit plane and the Sun–Earth direction constant, so the satellite always crosses a given latitude at the same local solar time (e.g., 10:30 a.m. descending node for Landsat). Imagery therefore has consistent sun angle and shadowing season after season — indispensable for change detection in agriculture, forestry and water monitoring (directly relevant to Kenyan resource-survey missions).

Why the other options are wrong:

• A — SSO altitudes (~600–900 km) sit in the inner Van Allen region; shielding, not inclination, manages radiation.
• C describes GEO, an entirely different mechanism (matching rotation period, not solar angle).
• D — retrograde launches actually sacrifice eastward rotational boost (equatorial rotational velocity ~0.46 km/s helps only prograde launches).

Step-by-step solution.

Sanity checks. The result must lie in the LEO band (90–127 min) — it does. A useful rule of thumb: LEO period ≈ 84.5 min + 0.013 min per km of altitude, giving 84.5 + 800×0.013 = 94.9… use only as a coarse check; Kepler's law is exact. Options A (85 min) is the period at very low altitude; C and D belong to MEO/GEO regimes.

Marking guide. Full marks require: correct conversion of altitude to radius (a common error is forgetting to add R⊙ — using a = 800 km gives a nonsense 36-min period), correct substitution, and a unit-consistent answer.

Step-by-step solution.

Physical interpretation. This is the circular-orbit condition: gravity supplies exactly the centripetal force m v²/r, so v = √(GM/r). Note the trend — lower orbits need higher velocity (at the surface itself it would be ~7.9 km/s), while GEO needs only 3.07 km/s. This inverse relationship surprises many students but is central to mission design.

Why the other options are wrong:

• A (3.07 km/s) is the GEO velocity — substituting a = 42,164 km gives exactly this.
• C (9.81 km/s) misuses g's numerical value (9.81 is m/s², not km/s).
• D (11.2 km/s) is Earth's escape velocity at the surface — reached only by taking r = R⊙ and using v = √(2μ/r) (the vis-viva energy equation with the √2 factor), not the circular formula.

Step-by-step solution.

System significance. A ~196 dB loss dwarfs everything else in the link budget. It is compensated by transmitter EIRP (tens of dBW), antenna gains (tens of dBi each end) and an ultra-low-noise receiver (G/T). The formula's scaling laws are worth memorising: +6 dB per doubling of distance and +6 dB per doubling of frequency. This is why GEO operators moved to Ku/Ka band (more gain for the same dish) and why LEO broadband can use handheld terminals (its ~24–28 dB smaller loss).

Error watch. Mixing km and m inside the logarithm, or using log instead of 20log (the 20 — not 10 — comes from power being proportional to amplitude squared), are the two classic slips.

Step-by-step solution.

Scaling insight. Gain grows with the square of D/λ: doubling dish diameter gives +6 dB; doubling frequency (halving λ) also gives +6 dB. The efficiency η accounts for illumination taper, spillover and surface errors and typically lies between 0.5 and 0.75 for practical parabolic antennas.

Why the other options are wrong: 30 dBi (A) corresponds to a ~0.5 m dish; 35 dBi (B) to about 1 m; 44.5 dBi (D) exceeds what 2 m at 6 GHz can deliver (it would need ~4.5 m). Estimating antenna size from gain (or vice versa) is a valuable order-of-magnitude skill for the link-budget section of the course.

Step-by-step solution.

Interpretation. Each day the satellite circles the Earth nearly 14 times while the Earth rotates once beneath it, so the ground track shifts westward by about 360°/13.7 ≈ 26° of longitude per orbit. In ~13.7 days (one “repeat cycle” for some designs) the tracks re-align. This rapid revisit is the fundamental LEO advantage for imaging and low-latency communications.

Why the other options are wrong: A (10.5) results from using ~140 min (a GPS/MEO period, forgetting to convert); C (16.2) corresponds to a ~600 km orbit; D (24) would require a period of exactly 1 h — physically impossible in bound orbit (the 84.5-min surface-skimming orbit is the floor).

Cross-check. Compare with Q11: 800 km → ~101 min → 14.2 orbits/day; 1,000 km → ~105 min → 13.7 orbits/day. The gentle decline with altitude is a good sanity pattern to internalise.

Requirement analysis. Decompose the mission: (i) continuous 24/7 service — no gaps, no handovers acceptable for broadcast distribution; (ii) regional (East Africa) — one persistent footprint is enough; (iii) low-cost fixed dishes — mass-market receivers must be simple and must not track. The analysis maps each requirement to an orbit property:

GEO uniquely satisfies all three simultaneously — this is why essentially all direct-to-home TV in Africa (DStv via Eutelsat/SES craft) uses GEO. The ~240 ms round-trip latency, fatal for interactive broadband, is irrelevant for one-way broadcast.

Why the other options fail: A — LEO dishes would need motorised tracking and seamless handovers at ~7.5 km/s satellite motion, exploding receiver cost. C — HEO coverage of equatorial East Africa is brief per orbit, not continuous. D — three MEO satellites still move across the sky and would force every home dish to track.

Trade-off analysis. A rigorous comparison must score both architectures on every system dimension, not cherry-pick one:

Verdict logic. There is no universal winner: the correct engineering answer is service-dependent — interactive mass-market broadband favours LEO despite system complexity; trunk backhaul, maritime/aero connectivity and broadcast favour GEO. Option B is the only response that analyses both sides symmetrically, which is what “Analyze”-level marking rewards.

Why the others fail: A reverses the complexity/latency mapping; C is false (GEO cannot see the poles; LEO polar shells can); D contradicts the d/c physics of Q8.

Geometry analysis. GEO visibility from latitude φ is governed by the central angle ψ between the station and the sub-satellite point: cosψ = cosφcosΔλ. The elevation angle follows from cosε = sinψ · (a/R⊙) geometry, giving the textbook result that the maximum possible elevation to any GEO satellite from latitude φ is ε_max = 90° − φ − 8.7°. Hence:

• At 70°N: ε_max ≈ 11° — the satellite hugs the horizon, through the thickest troposphere, vulnerable to terrain blockage, multipath and rain/snow attenuation.
• At 81.3°: ε_max = 0° — the GEO arc is on the horizon; beyond this latitude GEO is not visible at all.

No amount of transmit power repairs geometry: the signal path at 5–10° elevation traverses several times the rain column of a 30°+ path and is easily blocked by terrain or buildings.

Solution analysis. The Molniya HEO (i = 63.4°, T = 11 h 58 min, e ≈ 0.72) places apogee (~39,700 km altitude, sub-satellite latitude up to 63°N) over the Northern Hemisphere, where the satellite dwells ~8 h per orbit (Q9). Three satellites, phased so one is always near apogee, deliver continuous, high-elevation coverage of latitudes 50–90°N — precisely the Russian/Scandinavian/Canadian problem. A polar-LEO constellation is the fallback but needs dozens of satellites and suffers short passes; HEO is the minimum-satellite solution.

Why the others fail: A and C invoke GEO where geometry forbids it; D's single LEO satellite gives minutes of coverage per pass — not “uninterrupted.”

Step-by-step analysis. The question isolates the range dependence of FSPL:

Interpretation. Frequency cancels in the ratio — the comparison is pure geometry. A 27.6 dB advantage means the LEO link delivers the same received power with ~575 times less transmit power, or equivalently supports a dish about 5.4 times smaller in diameter (gain ∝ D²). This single number explains the entire difference in terminal economics between GEO VSAT networks and LEO broadband/handheld services.

Error watch. Students who compute 10log instead of 20log get 13.8 dB (half the correct value) — recall the “20” comes from power ∝ field². Students using the altitude difference (35,786 − 1,500) instead of the ratio also err; only the ratio matters in dB.

Why the others fail: C inverts the physics (path loss depends on range, not satellite speed); D assumes frequency alone determines loss and ignores the d term entirely; A (24 dB) results from comparing slant ranges of 35,786 km and ~2,250 km — a defensible estimate if one used LEO apogee-like range, but the question fixes 1,500 km.

Physics analysis. Atmospheric density falls roughly exponentially with altitude: even at 300 km, residual air (number density ~10¹²–10¹³ m^-3, rising by an order of magnitude during solar-maximum heating) exerts drag on a satellite. Drag does negative work, removing orbital energy: the semi-major axis shrinks, the orbit decays, and the satellite re-enters. The lifetime scales very steeply with altitude because of the exponential atmosphere.

• 300 km: decay in weeks to a few years depending on ballistic coefficient and solar activity. (This is why the ISS at ~420 km re-boosts every few months using visiting vehicle thrusters.)
• 800 km: drag is negligible; natural lifetimes run from decades to centuries — objects placed there in the 1960s are still aloft.

Engineering consequence — two-sided: For mission longevity, choose higher altitude (800 km) or accept regular re-boost (300–400 km crewed stations). For orbital-debris mitigation, the logic reverses: deploy short-lived constellations (Starlink at ~550 km) so failed satellites decay and burn up within ~5 years, whereas objects abandoned at 800+ km persist as collision-generating debris unless actively de-orbited — a key theme in modern space-sustainability policy (e.g., the 25-year disposal guideline and FCC 5-year rule).

Why the others fail: A ignores drag entirely (false below ~600 km); C reverses cause and effect (800 km's higher velocity is more than offset by far lower density — drag ∝ ρv², and ρ wins); D states a physically impossible density ordering (the atmosphere thins with altitude, never thickens).