Satellite Launching — 20-Question Quiz | EEEN 567 Satellite Engineering

Instructions

Attempt all 20 questions before checking answers. Each question carries 1 mark. Choose the single best option. Questions are organised in four blocks of five, one block per cognitive level of Bloom's taxonomy:

Level 1 — Remember (Q1–Q5) Level 2 — Understand (Q6–Q10) Level 3 — Apply (Q11–Q15) Level 4 — Analyze (Q16–Q20)

How to use this quiz Work through the blocks in order. On submission you will get an overall score plus a per-level breakdown so you can see whether gaps are in recall, comprehension, computation, or reasoning. Then study the answer key — every explanation teaches the underlying principle, not just the letter.

Your Score

0 / 20 (0%)

Block A — Level 1: Remember

Q1.Remember

The ideal rocket equation Δv = I_sp·g₀·ln(m₀/m_f), which underpins all launch vehicle sizing, is named after:

a) Wernher von Braun b) Robert Goddard c) Konstantin Tsiolkovsky d) Hermann Oberth

Answer: c. Konstantin Tsiolkovsky derived the rocket equation in 1903, linking achievable velocity change to exhaust velocity and mass ratio. Von Braun (V-2, Saturn), Goddard (first liquid-fuel rocket flight, 1926) and Oberth (theory) were pioneers too, but the equation carries Tsiolkovsky's name.

Q2.Remember

Specific impulse (I_sp) of a rocket engine is defined as thrust divided by:

a) Propellant volume flow rate b) Propellant weight flow rate (ṁ·g₀) c) Total propellant mass on board d) Chamber pressure

Answer: b. I_sp = F / (ṁ·g₀), with units of seconds. It measures how many seconds one unit weight of propellant can produce one unit of thrust — the standard figure of merit for propellant efficiency (LOX/LH₂ ≈ 420–460 s; solids ≈ 260–290 s).

Q3.Remember

The event "Max-Q" during launch denotes the point of maximum:

a) Engine thrust b) Altitude c) Axial acceleration d) Dynamic pressure, q = ½ρv²

Answer: d. Dynamic pressure peaks where air density is still appreciable while speed is already high (typically ~60–90 s after liftoff). Vehicles often throttle down through Max-Q to limit aerodynamic and structural loads — and the acoustic noise on the payload.

Q4.Remember

A Geostationary Transfer Orbit (GTO) is an elliptical orbit whose apogee is at approximately:

a) 35 786 km altitude (GEO radius 42 164 km) b) 400 km altitude c) 2 000 km altitude d) 100 000 km altitude

Answer: a. GTO has perigee in LEO (~200–400 km) and apogee at the geostationary altitude of 35 786 km (42 164 km from Earth's centre). An apogee burn then circularises at GEO — or a cryogenic upper stage can insert the satellite directly, skipping the transfer burn on the spacecraft.

Q5.Remember

In a typical launch vehicle telemetry chain, the subsystem whose sole function is to destroy the vehicle if it leaves the approved flight corridor is the:

a) Guidance computer (GNC) b) PCM frame encoder c) Flight Termination System (FTS) d) Power conditioning unit

Answer: c. The FTS is an independent radio-command destruct chain kept deliberately separate from flight avionics, so that a flight-computer failure cannot disable public-safety capability. It is armed on the pad and operated by the range safety officer per standards such as RCC 319.

Block B — Level 2: Understand

Q6.Understand

Launch vehicles discard (stage off) empty booster structures primarily because:

a) It improves the aerodynamics of the remaining vehicle b) Dead structural mass worsens the mass ratio of everything still burning c) It reduces the risk of lightning striking the vehicle d) International law requires debris to fall in designated zones

Answer: b. From Δv = v_e ln(m₀/m_f): once tanks are empty, their structure is mass carried to no benefit. Staging removes that dead mass, instantly improving the mass ratio of the remaining stages — the logarithmic equation makes this worth the complexity of separation events.

Q7.Understand

Equatorial launch sites such as Kourou (5.2° N) are strongly favoured for geostationary missions mainly because launching eastward from low latitude:

a) Provides the largest free rotational velocity bonus (≈465·cos φ m/s) and direct access to low inclinations b) Guarantees better weather than mid-latitude sites c) Allows the vehicle to climb above all weather in the first seconds d) Reduces the need for a second stage

Answer: a. Earth's surface speed at the equator is ≈465 m/s eastward; the useful component is scaled by cos(latitude). Low latitude also lets the vehicle reach the near-zero inclination of GEO/GTO with minimal plane-change cost — Kourou's location is a key reason Ariane captured the commercial GEO market.

Q8.Understand

The protective fairing is jettisoned roughly three minutes into flight (above ~100 km) because by then:

a) The satellite's batteries are fully charged b) The flight computer switches to its backup c) The second stage has already separated d) Aerodynamic heating and pressure have fallen to negligible levels, so the mass of the fairing is no longer justified

Answer: d. The fairing exists only to shield the payload from ascent aero-thermodynamic loads. Once the atmosphere is essentially vacuum, every kilogram of fairing is dead mass on the way to orbit — so it is jettisoned by pyrotechnic separation and often recovered on modern vehicles (e.g., fairing halves caught by ships or recovered from water).

Q9.Understand

A GEO satellite is usually injected into GTO rather than directly into GEO because:

a) GTO gives a free velocity increase from Earth's rotation b) GEO altitude cannot be reached by chemical rockets c) Reaching GEO directly needs ~4 km/s more from the launcher, whereas the satellite's own apogee motor (or electric propulsion) can supply the circularisation at lower launch cost d) GTO avoids the Van Allen radiation belts entirely

Answer: c. LEO→GEO total Δv is ≈ 3.9 km/s from a 400 km orbit. Asking the launcher to supply it all costs payload mass and money, so a compromise evolved: launcher delivers GTO (perigee burn only, ~2.45 km/s); the spacecraft raises apogee/circularises with its own propulsion. Electric-propulsion satellites take months but save enormous launch mass.

Q10.Understand

In a gravity-turn ascent, the vehicle is allowed to pitch over gradually under gravity and aerodynamics rather than being steered hard onto a pre-set path. The chief benefit is:

a) It keeps the trajectory vertical for easier range tracking b) Thrust is spent mainly on gaining orbital speed instead of fighting gravity, minimising steering and gravity losses c) It guarantees the payload reaches a polar orbit d) It maximises time spent at Max-Q for structural proof

Answer: b. Orbit is mostly kinetic energy (7.7 km/s horizontal), not altitude. Flying a near-zero-lift, slowly pitching gravity turn aligns thrust with the velocity vector so minimal force is wasted on steering loads — this is why ideal Δv to LEO (~9.3–10 km/s) exceeds the orbital speed by only ~1.5–2 km/s of losses.

Block C — Level 3: Apply

Q11.Apply

A stage has I_sp = 300 s, initial mass 500 000 kg and final mass 100 000 kg. Using the rocket equation, its ideal Δv is closest to:

a) 1.61 km/s b) 4.74 km/s c) 2.94 km/s d) 9.42 km/s

Answer: b. Δv = I_sp·g₀·ln(m₀/m_f) = 300 × 9.81 × ln(5) = 300 × 9.81 × 1.609 ≈ 4 736 m/s ≈ 4.74 km/s. Note the logarithm: even a 5:1 mass ratio yields under 5 km/s — hence multi-stage vehicles.

Q12.Apply

Using μ = 3.986×10¹⁴ m³/s² and R_E = 6 371 km, the circular orbit velocity at 400 km altitude is approximately:

a) 7.67 km/s b) 3.07 km/s c) 10.85 km/s d) 1.99 km/s

Answer: a. v_c = √(μ/r), r = 6 771 km = 6.771×10⁶ m → v_c = √(3.986×10¹⁴ / 6.771×10⁶) ≈ 7 673 m/s ≈ 7.67 km/s. Option (c) is the escape velocity there (√2 × 7.67 ≈ 10.85 km/s) — a common distractor.

Q13.Apply

A vehicle launches due east from Cape Canaveral (latitude 28.5° N). Earth's rotational surface speed is ≈465 m/s at the equator. The eastward velocity bonus is approximately:

a) 465 m/s b) 233 m/s c) 409 m/s d) 792 m/s

Answer: c. v_rot = 465·cos(28.5°) = 465 × 0.879 ≈ 409 m/s. This is free Δv that the vehicle does not need to supply — one reason performance figures quote "to 28.5° inclination": the same rocket achieves noticeably less to a polar orbit from the same pad.

Q14.Apply

At a certain radius the circular-orbit velocity works out to 7.0 km/s. The escape velocity at that same radius is:

a) 3.5 km/s b) 7.0 km/s c) 14.0 km/s d) 9.9 km/s

Answer: d. v_esc = √2 · v_c = 1.414 × 7.0 ≈ 9.9 km/s. The √2 factor comes directly from energy: escape needs kinetic energy equal in magnitude to the gravitational potential energy at that radius.

Q15.Apply

A Hohmann transfer is planned from a 400 km circular LEO (r₁ = 6 771 km) to GEO altitude (r₂ = 42 164 km). The perigee burn Δv₁ is approximately (μ = 3.986×10¹⁴ m³/s²):

a) 2.45 km/s b) 3.20 km/s c) 1.47 km/s d) 0.77 km/s

Answer: a. Transfer semi-major axis a_t = (r₁+r₂)/2 = 24 467 km. v_p = √[μ(2/r₁ − 1/a_t)] ≈ 10.15 km/s; circular speed at r₁ is 7.67 km/s; so Δv₁ ≈ 2.45 km/s. (The 1.47 km/s figure is the apogee circularisation burn Δv₂ — the classic trap in this question.)

Block D — Level 4: Analyze

Q16.Analyze

A launcher uses LOX/kerosene on its first stage (I_sp ≈ 311 s sea level) and LOX/hydrogen on its upper stage (I_sp ≈ 450 s). The best engineering justification for placing hydrogen on the upper stage — rather than the first — is:

a) Hydrogen's low density makes tanks lighter, and tank mass matters most at lift-off b) Hydrogen engines cannot ignite at sea level, so they must be used in vacuum c) Upper-stage dry mass is carried through all remaining Δv, so its higher I_sp compounds in the rocket equation; the first stage instead prioritises thrust density and sea-level performance d) Cryogenic handling is easier at altitude where ambient temperature is lower

Answer: c. Every kilogram saved on the upper stage avoids being multiplied through every subsequent burn — so high I_sp (even with bulky, heavy hydrogen tanks) pays off most there. The first stage must produce enormous thrust within a compact vehicle and fight sea-level nozzle losses, favouring dense kerosene. (Option b is false — hydrolox engines like the RS-25 ignite at sea level.)

Q17.Analyze

A commercial GEO comsat can fly on either of two launchers: Launcher X from Kourou (5.2° N) or Launcher Y from Plesetsk (62.8° N), with identical quoted performance. The most defensible analysis for choosing X is:

a) From Kourou the ~463 m/s rotational bonus and direct low-inclination access cut the plane-change and energy cost of GTO, raising usable payload; from Plesetsk the achievable GTO mass collapses after paying for high-latitude plane change b) Plesetsk's higher latitude gives a stronger rotational bonus, so it should be preferred c) Latitude affects only weather, not performance, so the choice should be made on price alone d) GTO inclination is set by the satellite, not the launch site, so both sites are equivalent

Answer: a. Rotational bonus scales as cos(latitude): 463 m/s at Kourou vs ~213 m/s at Plesetsk. Worse, from 62.8° N a direct ascent cannot reach GTO's near-equatorial plane — the plane change alone costs km/s, slashing delivered mass. Identical brochure performance is quoted for each site's "native" orbit; the fine print matters.

Q18.Analyze

Two identical vehicles launch the same day, one carrying a 2 000 kg payload, the other a 3 000 kg payload, to the same target orbit. Analysing via the rocket equation, the correct conclusion is:

a) Both achieve the same orbit; payload mass does not enter the rocket equation b) The heavier payload lowers the mass ratio m₀/m_f, reducing available Δv — so the 3 000 kg payload reaches a lower-energy orbit unless the vehicle compensates (more propellant, shorter burn, or larger rocket) c) The heavier payload increases thrust-to-weight ratio and thus reaches a higher orbit d) The 3 000 kg payload reaches orbit later but at the same energy

Answer: b. Payload adds to m_f of the last stage, so R falls and ln R falls — Δv falls. This is precisely the launcher's "performance curve": payload mass is traded against orbital energy, and why launch contracts quote kg-to-orbit.

Q19.Analyze

Coupled-loads analysis requires the satellite's first axial natural frequency (on its adapter) to be above ~100 Hz. The best analysis of the reason is:

a) Frequencies below 100 Hz interfere with S-band telemetry and must be avoided for EMC compliance b) The launcher cannot generate vibration above 100 Hz, so the requirement is a formality c) Frequencies above 100 Hz reduce aerodynamic drag on the fairing d) Low-frequency launcher bending and engine-dominant modes live below ~100 Hz; a stiff payload keeps dynamic coupling small so loads can be predicted conservatively and shock/acoustic environments remain bounded

Answer: d. Launcher primary bending modes and major forcing (POGO, engine pressure oscillation) sit below ~100 Hz. If the payload is stiff (typically ≥100–120 Hz requirement), it behaves quasi-statically to those modes, so loads analysis becomes manageable and test levels stay standardised. It is a dynamics/decoupling argument, not EMC or drag.

Q20.Analyze

A launch provider must decide between an expendable vehicle (ELV) and a reusable first stage (RLV) for a new constellation needing ~20 flights/year. Which analysis best captures the real trade?

a) Reuse saves manufacturing cost per flight but demands propellant/reserve mass and landing systems (reducing payload or orbit), plus refurbishment and fleet operations — economically favourable mainly at high, steady flight rates like this constellation's b) Reuse is always cheaper per flight because no hardware is discarded c) Expendables are always cheaper because they avoid recovery operations d) The choice affects only the launch provider's accounting, never the customer's payload mass

Answer: a. Recovery needs margin: boost-back/re-entry burns, landing legs, grid fins — mass that comes straight out of payload. Refurbishment (engines, thermal protection) and fleet management add recurring cost. The crossover depends on flight rate and payload penalty; at ~20 flights/year with steady demand, amortising development over many reuses usually wins — the SpaceX business model in a nutshell. Options (b), (c) are absolutes; (d) is false since the performance penalty directly reduces customer mass.

Post-Quiz Answer Key & Explanations

Complete answer table followed by a short explanation per question. Cross-reference with the Satellite Launching study guide sections indicated.

Q	Level	Answer	Core concept
1	Remember	c	Tsiolkovsky rocket equation
2	Remember	b	I_sp = F / (ṁ·g₀)
3	Remember	d	Max-Q = peak dynamic pressure
4	Remember	a	GTO apogee = GEO, 35 786 km
5	Remember	c	Flight Termination System
6	Understand	b	Staging improves mass ratio
7	Understand	a	Rotational bonus ∝ cos(latitude)
8	Understand	d	Fairing jettison rationale
9	Understand	c	GTO vs direct-GEO economics
10	Understand	b	Gravity turn minimises losses
11	Apply	b	Δv = 300·9.81·ln5 ≈ 4.74 km/s
12	Apply	a	v_c(400 km) ≈ 7.67 km/s
13	Apply	c	465·cos28.5° ≈ 409 m/s
14	Apply	d	v_esc = √2·v_c ≈ 9.9 km/s
15	Apply	a	Hohmann perigee burn ≈ 2.45 km/s
16	Analyze	c	I_sp value compounds on upper stage
17	Analyze	a	Site latitude vs GTO energy cost
18	Analyze	b	Payload ↑ ⇒ mass ratio ↓ ⇒ Δv ↓
19	Analyze	d	Frequency decoupling in coupled loads
20	Analyze	a	Reusability economics at flight rate