Rocket Science Physics — 7th & 8th Grade
8th Grade Physics · Advanced Course Syllabus

Rocket Science
Physics III

The capstone course. Covers Hohmann transfers, mission architecture, advanced propulsion, reentry physics, orbital mechanics with calculus-level intuition, and full rocket system design.

5Units
20Topics
100+Examples
7thPrerequisite
Prerequisites: Vectors · I_sp · Rotational Motion · Atmosphere Model · Δv Budgets · Projectile Motion

Course Overview

8th grade Rocket Science Physics is where students work at near-professional level. Topics include orbital maneuver planning (Hohmann transfers, bi-elliptic transfers), advanced propulsion systems (ion drives, nuclear thermal, solar sails), thermal management during reentry, and the systems-engineering perspective that ties a complete mission together. Students develop full mission Δv budgets and design staged rocket systems from first principles.

Learning Objectives

Execute Hohmann transfer calculations between orbits
Analyze bi-elliptic transfers and when they beat Hohmann
Calculate reentry heating using kinetic energy analysis
Understand ion drive thrust and propellant mass flow
Apply the vis-viva equation to elliptical orbits
Design multi-stage rocket systems to meet Δv budgets
Analyze solar sail acceleration and mission profiles
Calculate Oberth effect maneuver advantages
Build complete end-to-end mission Δv budgets
Compare propulsion technologies using common metrics
8 · Unit 01 Orbital Maneuvers — Changing Orbits
1.1 The Vis-Viva Equation — Speed at Any Point in an Orbit

The vis-viva equation gives the orbital speed of a spacecraft at any point in an elliptical orbit, given its current distance r from the center and the semi-major axis a of the orbit. For a circular orbit, r = a and it reduces to the circular orbital velocity formula from 6th grade. GM = 3.986 × 10¹⁴ m³/s² for Earth.

Vis-Viva Equation v² = GM × (2/r − 1/a) | For circular: a = r → v = √(GM/r)
Worked Examples
EXAMPLE 1Speed at perigee of an elliptical orbit
Given
  • Perigee altitude: 300 km → r_p = 6,671 km = 6.671×10⁶ m
  • Apogee altitude: 36,000 km → r_a = 42,371 km
  • Semi-major axis a: (r_p + r_a)/2
Solution
  1. a = (6.671×10⁶ + 4.237×10⁷)/2 = (6.671 + 42.37)×10⁶/2 = 24.52×10⁶ m
  2. v² = GM × (2/r_p − 1/a)
  3. = 3.986×10¹⁴ × (2/6.671×10⁶ − 1/24.52×10⁶)
  4. = 3.986×10¹⁴ × (2.997×10⁻⁷ − 4.077×10⁻⁸)
  5. = 3.986×10¹⁴ × 2.589×10⁻⁷ = 1.032×10⁸ → v = 10,158 m/s
Answerv at perigee = 10,158 m/s — much faster than circular at same altitude (7,726 m/s).
EXAMPLE 2Speed at apogee of same orbit
Given
  • r_a: 4.237 × 10⁷ m
  • a: 2.452 × 10⁷ m (from Ex. 1)
Solution
  1. v² = 3.986×10¹⁴ × (2/4.237×10⁷ − 1/2.452×10⁷)
  2. = 3.986×10¹⁴ × (4.722×10⁻⁸ − 4.077×10⁻⁸)
  3. = 3.986×10¹⁴ × 6.45×10⁻⁹ = 2.571×10⁶ → v = 1,603 m/s
  4. Note: v_p × r_p = v_a × r_a (conservation of angular momentum check): 10158×6.671 = 1603×42.37 ≈ 67,763 ✓
Answerv at apogee = 1,603 m/s — very slow at the top of the orbit.
EXAMPLE 3Verify circular orbit as special case of vis-viva
Given
  • Circular orbit: r = a = 7×10⁶ m
Solution
  1. v² = GM(2/r − 1/a) = GM(2/r − 1/r) = GM/r
  2. v = √(GM/r) = √(3.986×10¹⁴ / 7×10⁶)
  3. v = √(5.694×10⁷) = 7,546 m/s
  4. Identical to circular velocity formula — vis-viva is a generalization! ✓
Answerv = 7,546 m/s — circular orbit is just a special case of vis-viva.
EXAMPLE 4Speed of Apollo CM at lunar distance from Earth
Given
  • r (at Moon's distance): 3.844×10⁸ m
  • Trans-lunar a: (r_perigee + r_Moon)/2 = (6.6×10⁶ + 3.844×10⁸)/2 = 1.955×10⁸ m
Solution
  1. v² = 3.986×10¹⁴(2/3.844×10⁸ − 1/1.955×10⁸)
  2. = 3.986×10¹⁴(5.202×10⁻⁹ − 5.115×10⁻⁹)
  3. = 3.986×10¹⁴ × 8.7×10⁻¹¹ = 34,678 → v = 186 m/s
  4. Apollo arrived at the Moon moving only ~186 m/s — barely faster than a jet plane!
AnswerApollo CM speed at Moon's distance ≈ 186 m/s — nearly coasted to a stop.
EXAMPLE 5Find semi-major axis needed for a target apogee speed
Given
  • Target v at apogee: 500 m/s
  • Apogee r: 4.2×10⁷ m
Solution
  1. From vis-viva: v² = GM(2/r − 1/a)
  2. (500)² = 3.986×10¹⁴(2/4.2×10⁷ − 1/a)
  3. 250000 / 3.986×10¹⁴ = 4.762×10⁻⁸ − 1/a → 6.27×10⁻¹⁰ = 4.762×10⁻⁸ − 1/a
  4. 1/a = 4.762×10⁻⁸ − 6.27×10⁻¹⁰ ≈ 4.699×10⁻⁸ → a = 2.128×10⁷ m
AnswerSemi-major axis a = 2.128×10⁷ m (21,280 km) for 500 m/s apogee speed.
1.2 Hohmann Transfer — The Most Efficient Orbit Change

A Hohmann transfer moves a spacecraft between two circular orbits using the minimum possible Δv. It requires exactly two engine burns: one to enter an elliptical transfer orbit at the lower orbit, and one at the higher orbit to circularize. It is used for satellite deployment, Moon missions, and interplanetary travel.

Hohmann Transfer Δv₁ = v_transfer_perigee − v_circular_low | Δv₂ = v_circular_high − v_transfer_apogee
Worked Examples
EXAMPLE 1Transfer from ISS orbit (400 km) to GPS orbit (20,200 km)
Given
  • r₁: 6.771×10⁶ m (ISS)
  • r₂: 2.657×10⁷ m (GPS)
  • a_transfer: (r₁+r₂)/2 = 1.667×10⁷ m
Solution
  1. v_circular at r₁: v₁ = √(GM/r₁) = √(3.986×10¹⁴/6.771×10⁶) = 7,673 m/s
  2. v_transfer at perigee (r₁): v_tp = √(GM(2/r₁−1/a)) = √(3.986×10¹⁴(1.481×10⁻⁷−6.0×10⁻⁸)) = 9,945 m/s
  3. Δv₁ = 9,945 − 7,673 = 2,272 m/s
  4. v_circular at r₂ = 3,873 m/s | v_transfer at apogee = 2,540 m/s
  5. Δv₂ = 3,873 − 2,540 = 1,333 m/s
AnswerΔv₁ = 2,272 m/s | Δv₂ = 1,333 m/s | Total Δv = 3,605 m/s
EXAMPLE 2Transfer time for Hohmann maneuver
Given
  • a_transfer: 1.667×10⁷ m (from Ex. 1)
  • Transfer is half an ellipse
Solution
  1. Full transfer orbit period via Kepler: T = 2π√(a³/GM)
  2. a³ = (1.667×10⁷)³ = 4.633×10²¹
  3. T_full = 2π√(4.633×10²¹ / 3.986×10¹⁴) = 2π√(1.162×10⁷) = 2π × 3,409 = 21,421 s
  4. Transfer time = T/2: 21,421/2 = 10,710 s ≈ 2.98 hours
AnswerTransfer time = 10,710 s ≈ 2.98 hours to reach GPS altitude.
EXAMPLE 3Earth-to-Mars Hohmann transfer Δv
Given
  • r_Earth orbit: 1.496×10¹¹ m (1 AU)
  • r_Mars orbit: 2.279×10¹¹ m (1.524 AU)
  • GM_Sun: 1.327×10²⁰ m³/s²
Solution
  1. a_transfer = (1.496+2.279)×10¹¹/2 = 1.888×10¹¹ m
  2. v_Earth orbit = √(1.327×10²⁰/1.496×10¹¹) = 29,785 m/s
  3. v_transfer at Earth: √(GM_Sun(2/r_E − 1/a)) = 32,726 m/s
  4. Δv₁ = 32,726 − 29,785 = 2,941 m/s (to leave Earth orbit)
  5. At Mars: v_Mars = 24,131 m/s; v_transfer = 21,480 m/s → Δv₂ = 2,651 m/s
AnswerTotal heliocentric Δv ≈ 5,592 m/s (plus Earth/Mars gravity well costs).
EXAMPLE 4Earth-Mars transfer time (Hohmann)
Given
  • a_transfer: 1.888×10¹¹ m
  • GM_Sun: 1.327×10²⁰ m³/s²
Solution
  1. T_full = 2π√(a³/GM_Sun)
  2. a³ = (1.888×10¹¹)³ = 6.726×10³³
  3. T = 2π√(6.726×10³³/1.327×10²⁰) = 2π√(5.07×10¹³) = 2π×7.12×10⁶ = 4.474×10⁷ s
  4. Transfer time = T/2 = 2.237×10⁷ s ÷ 86400 = 259 days
AnswerEarth-to-Mars Hohmann transfer ≈ 259 days (~8.5 months).
EXAMPLE 5Deorbit burn to reenter from ISS altitude
Given
  • Current orbit: 400 km circular (r = 6.771×10⁶ m)
  • Reentry interface: 120 km (r = 6.491×10⁶ m)
Solution
  1. Transfer orbit: a = (6.771+6.491)×10⁶/2 = 6.631×10⁶ m
  2. v_circular = 7,673 m/s at 400 km
  3. v_transfer at 400 km: √(GM(2/6.771×10⁶−1/6.631×10⁶)) = √(3.986×10¹⁴(2.954×10⁻⁷−1.508×10⁻⁷)) = 7,585 m/s
  4. Δv_deorbit = 7,673 − 7,585 = 88 m/s retrograde burn
AnswerDeorbit burn = 88 m/s retrograde — small burn, but it starts reentry!
8 · Unit 02 Reentry Physics & Thermal Management
2.1 Reentry Heating — Converting Orbital KE to Heat

When a spacecraft reenters the atmosphere, it must convert enormous kinetic energy into heat — about 62 GJ for a 1,000 kg craft returning from orbit. A fraction of this goes into the heat shield; the rest heats the shockwave in front of the vehicle. This is why reentry is one of the most dangerous phases of spaceflight.

Reentry Energy KE_entry = ½mv_entry² | Q_shield = f × KE_entry (f = 0.01–0.05 typically)
Worked Examples
EXAMPLE 1Total KE to be dissipated during Crew Dragon reentry
Given
  • m: 9,500 kg (Crew Dragon capsule)
  • v_entry: 7,800 m/s
Solution
  1. KE = ½ × m × v²
  2. KE = 0.5 × 9500 × (7800)²
  3. KE = 0.5 × 9500 × 60,840,000
  4. KE = 288,990,000,000 J = 289 GJ
  5. Equivalent to 69 tons of TNT exploding in ~6 minutes!
AnswerKE = 289 GJ — all of this must be dissipated safely during reentry.
EXAMPLE 2Heat shield energy absorption (f = 3%)
Given
  • KE_total: 289 GJ (from Ex. 1)
  • Heat shield fraction f: 0.03 (3%)
Solution
  1. Q_shield = f × KE = 0.03 × 289×10⁹ = 8.67×10⁹ J = 8.67 GJ
  2. The remaining 97% heats the shockwave and surrounding air (plasma).
  3. PICA-X ablator can absorb ~8 MJ/kg, so: shield mass = 8.67×10⁹/8×10⁶ = 1,084 kg
AnswerShield absorbs 8.67 GJ — 97% goes into the shock wave, not the vehicle.
EXAMPLE 3Reentry peak deceleration (g-force)
Given
  • v_entry: 7,800 m/s
  • v_final: 0 (parachutes deploy)
  • Reentry duration: ~400 s (simplified)
Solution
  1. Average deceleration: a = Δv/t = 7800/400 = 19.5 m/s²
  2. In g-units: 19.5 / 9.8 = 1.99 g average
  3. Peak deceleration for capsules is typically 3–8 g during peak heating.
  4. Shuttle had lower peak g (~1.5g) due to its lifting reentry trajectory.
AnswerAverage ~2 g; peak 3–8 g — humans can survive up to ~9 g briefly.
EXAMPLE 4Atmospheric entry angle — too steep vs too shallow
Given
  • Corridor width: ~1–3° for capsule
  • Too steep (>6°): max g > 12 g, fatal heating
  • Too shallow (<1°): skips off atmosphere
Solution
  1. Entry angle determines deceleration rate via: F_drag ∝ sin(γ) where γ is flight path angle.
  2. Optimal angle ~3°: distributes heating over ~400 s, keeps peak g under 8.
  3. Skip zone (γ < 1°): KE carries vehicle out of atmosphere like a stone skipping water.
  4. Burn zone (γ > 6°): deceleration so rapid the heat shield can't ablate fast enough.
AnswerSafe corridor is ~1–5° — miss it by even 2° and the mission fails.
EXAMPLE 5Lunar return reentry — compare to LEO reentry
Given
  • v_LEO reentry: 7,800 m/s
  • v_lunar reentry: 11,000 m/s
  • m: 5,000 kg (same capsule)
Solution
  1. LEO KE: ½×5000×7800² = 1.521×10¹¹ J = 152.1 GJ
  2. Lunar KE: ½×5000×11000² = 3.025×10¹¹ J = 302.5 GJ
  3. Ratio: 302.5/152.1 = 1.99× more energy
  4. Lunar reentry requires a shield rated for twice the energy — much harder mission!
AnswerLunar return has 2× the reentry energy — Apollo needed a much thicker heat shield.
8 · Unit 03 Advanced Propulsion Systems
3.1 Ion Drives — High I_sp, Low Thrust

Ion drives ionize a propellant (usually xenon) and accelerate ions electrostatically to exhaust velocities of 20–80 km/s, giving I_sp of 2,000–10,000 s. Thrust is tiny (millinewtons), but operated continuously for months or years, they achieve enormous Δv very efficiently. Used by Dawn, Hayabusa, and planned for Gateway.

Ion Thruster F = ṁ × v_e | v_e = I_sp × g₀ | Power = ½ṁv_e² = F×v_e/2
Worked Examples
EXAMPLE 1Thrust from an ion drive (NASA NSTAR)
Given
  • I_sp: 3,100 s
  • ṁ: 2.79×10⁻⁶ kg/s (xenon)
Solution
  1. v_e = I_sp × g₀ = 3100 × 9.807 = 30,402 m/s
  2. F = ṁ × v_e = 2.79×10⁻⁶ × 30,402
  3. F = 0.0848 N = 84.8 mN (millinewtons!)
  4. Roughly the weight of a large paperclip — but runs for years!
AnswerNSTAR thrust = 84.8 mN — tiny but massively efficient over long durations.
EXAMPLE 2Δv from ion drive running for 1 year
Given
  • F: 84.8 mN
  • m_spacecraft: 450 kg
  • Duration: 1 year = 3.156×10⁷ s
Solution
  1. Acceleration: a = F/m = 0.0848/450 = 1.884×10⁻⁴ m/s²
  2. Δv = a × t: 1.884×10⁻⁴ × 3.156×10⁷ = 5,946 m/s
  3. That's nearly 6 km/s — enough to transfer between inner planets!
  4. A chemical rocket to do the same would need enormous fuel mass.
Answer1 year of ion drive operation gives Δv ≈ 5,946 m/s — planet-transfer capable.
EXAMPLE 3Xenon propellant consumed in 1 year
Given
  • ṁ: 2.79×10⁻⁶ kg/s
  • Duration: 3.156×10⁷ s
Solution
  1. m_xenon = ṁ × t = 2.79×10⁻⁶ × 3.156×10⁷
  2. m_xenon = 88.1 kg
  3. Compare: chemical rocket for same Δv on 450 kg spacecraft needs
  4. Δv = v_e×ln(m₀/m_f): with v_e=3050 m/s → m_fuel = m_f(e^(5946/3050)−1) = 450×(e^1.95−1) = 450×6.03 = 2,713 kg!
  5. Ion drive uses 88 kg vs chemical 2,713 kg — 31× less propellant.
Answer88 kg xenon gives same Δv as 2,713 kg chemical propellant — 31× more efficient!
EXAMPLE 4Power required by ion drive
Given
  • F: 84.8 mN
  • v_e: 30,402 m/s
  • Efficiency η: 65%
Solution
  1. Jet power (KE output): P_jet = ½Fv_e = 0.5 × 0.0848 × 30402 = 1,289 W
  2. Input power needed: P_in = P_jet / η = 1289 / 0.65 = 1,983 W ≈ 2 kW
  3. A single NSTAR thruster needs only 2 kW — solar panels can supply this easily.
AnswerIon drive requires 2 kW — easily powered by spacecraft solar panels.
EXAMPLE 5Compare ion drive vs chemical for same Δv budget
Given
  • Target Δv: 2,000 m/s
  • Spacecraft dry mass: 500 kg
  • Chemical I_sp: 300 s; Ion I_sp: 3,000 s
Solution
  1. v_e (chemical) = 300×9.807 = 2,942 m/s | v_e (ion) = 3000×9.807 = 29,421 m/s
  2. Chemical: m₀/m_f = e^(2000/2942) = e^0.680 = 1.974 → m_fuel = 500×0.974 = 487 kg
  3. Ion: m₀/m_f = e^(2000/29421) = e^0.068 = 1.070 → m_fuel = 500×0.070 = 35 kg
  4. Ion uses 35 kg vs 487 kg — 13.9× less propellant for same Δv!
AnswerIon: 35 kg fuel | Chemical: 487 kg fuel — for the same 2,000 m/s Δv.
8 · Unit 04 Oberth Effect & Gravity Assists
4.1 The Oberth Effect — Burning Deep in a Gravity Well

The Oberth Effect states that a rocket burn is more effective — delivers more mechanical energy — when performed at high speed (deep in a gravity well) than at low speed. This is because rocket exhaust carries away less total energy when the rocket is moving fast. This is why all deep-space missions burn at periapsis, and it's the principle behind lunar slingshot maneuvers.

Oberth Effect ΔKE = Δv × v_burn + ½(Δv)² vs ΔKE_slow = ½(Δv)² (huge difference!)
Worked Examples
EXAMPLE 1Energy gain from same burn at high vs low speed
Given
  • Δv burn: 100 m/s (same for both)
  • m: 1,000 kg
  • v_fast: 10,000 m/s (periapsis)
  • v_slow: 500 m/s (apoapsis)
Solution
  1. Burn at fast: ΔKE = ½m[(v+Δv)²−v²] = ½×1000×[(10100)²−(10000)²]
  2. = 500×(102,010,000−100,000,000) = 500×2,010,000 = 1.005×10⁹ J
  3. Burn at slow: = 500×[(600)²−(500)²] = 500×(360,000−250,000) = 5.5×10⁷ J
  4. Ratio: 1.005×10⁹ / 5.5×10⁷ = 18.3× more energy from same burn!
AnswerFast burn: 1.005 GJ vs slow burn: 55 MJ — 18× more energy at periapsis!
EXAMPLE 2Lunar flyby to gain interplanetary speed
Given
  • Approach speed to Moon: 900 m/s
  • Periapsis lunar flyby speed: 2,400 m/s
  • Burn at periapsis: 500 m/s Δv
Solution
  1. Without Oberth: 500 m/s added to 900 m/s approach = 1,400 m/s departure
  2. With Oberth (burn at 2,400 m/s periapsis): departure speed at infinity
  3. v_∞² = (2400+500)² − v_esc_moon² − v_approach² (energy conservation)
  4. Simplified: departure velocity gain ≈ 500 + 1,500 = ~2,000 m/s effective!
  5. The Oberth effect multiplied the 500 m/s burn into ~2,000 m/s of departure speed gain.
AnswerOberth effect: 500 m/s burn at lunar periapsis ≈ 4× more effective than in free space.
EXAMPLE 3Why all rockets pitch over before the gravity turn completes
Given
  • Rocket speed at 100 km altitude: 2,000 m/s
  • Remaining burn Δv: 4,000 m/s
Solution
  1. Burn mostly horizontal at high speed is more efficient (Oberth).
  2. Energy from horizontal burn: ΔKE ≈ m×v×Δv = m×2000×4000 = 8,000m MJ
  3. Energy from vertical burn: thrust fights gravity, no Oberth benefit.
  4. This is why Falcon 9 is nearly horizontal by 100 km altitude — maximizing Oberth advantage.
AnswerHorizontal burn at high speed uses Oberth effect — maximizes energy from same propellant.
EXAMPLE 4Gravity assist from Jupiter — Voyager-style calculation
Given
  • Jupiter orbital speed: 13,070 m/s
  • Spacecraft approach v_∞: 5,640 m/s
  • Maximum v_∞ after gravity assist: approach v_∞ + 2×v_Jupiter
Solution
  1. Max velocity gain from Jupiter flyby (prograde): Δv = 2×v_∞×v_J/(v_∞+v_J)
  2. = 2×5640×13070/(5640+13070)
  3. = 147,448,800/18,710 = 7,881 m/s speed gain (simplified)
  4. Final v_∞ ≈ 5,640 + 7,881 = ~13,500 m/s (enough to escape solar system!)
AnswerJupiter gravity assist can add ~7,881 m/s — Voyager escaped the solar system this way!
EXAMPLE 5Bi-elliptic transfer — when does it beat Hohmann?
Given
  • r₁: 7×10⁶ m (LEO)
  • r₂: 1.05×10⁸ m (very high orbit, ~15× LEO)
  • Rule of thumb: bi-elliptic beats Hohmann when r₂/r₁ > 11.94
Solution
  1. Ratio: r₂/r₁ = 1.05×10⁸ / 7×10⁶ = 15 > 11.94 ✓
  2. Bi-elliptic uses an intermediate very high apoapsis (like 200×r₁) to leverage Oberth effect.
  3. At the extreme apoapsis, only a tiny Δv is needed (very slow speed there).
  4. Total Δv can be 10–15% less than Hohmann for very large orbit changes!
  5. Trade-off: much longer transfer time (months instead of hours).
Answerr₂/r₁ = 15 > 11.94 → bi-elliptic wins, saving ~12% Δv vs Hohmann.
8 · Unit 05 Complete Mission Architecture & System Design
5.1 Full Mission Δv Budget — End-to-End Design

A mission Δv budget accounts for every maneuver from launch to mission end. For a crewed Mars mission, the budget spans launch, TLI/TMI burns, orbital insertion, descent, surface operations, ascent, and return. Mastering budget construction is the final skill that ties together all preceding topics.

Mission Total Δv_mission = Σ(all maneuver Δv) + contingency (typically +5–10%)
Worked Examples
EXAMPLE 1LEO communications satellite — complete Δv budget
Given / Segments
  • Ascent to LEO: 9,400 m/s
  • Hohmann to GEO: 3,900 m/s
  • Stationkeeping (10yr): 500 m/s
  • Contingency: +5%
Solution
  1. Mission total before contingency: 9400 + 3900 + 500 = 13,800 m/s
  2. With 5% contingency: 13,800 × 1.05 = 14,490 m/s
  3. This dictates the mass ratio of the entire launch vehicle system.
AnswerTotal Δv budget = 14,490 m/s for a 10-year GEO communications satellite.
EXAMPLE 2Mars mission — one-way budget (crewed)
Given / Mission Phases
  • Earth ascent: 9,400 m/s
  • Trans-Mars injection: 3,600 m/s
  • Mars orbit insertion: 2,100 m/s
  • Mars descent: 500 m/s (aerobraking assists)
Solution
  1. Total one-way: 9400 + 3600 + 2100 + 500 = 15,600 m/s
  2. With 10% contingency: 15,600 × 1.10 = 17,160 m/s
  3. This doesn't include the return trip — round trip needs ~30,000+ m/s total!
  4. This is why Mars missions are proposed as 1.5-year stays to wait for planetary alignment.
AnswerOne-way Mars budget = 17,160 m/s. Full round-trip ~30,000–35,000 m/s.
EXAMPLE 3Design a 3-stage rocket to meet a 14,490 m/s Δv budget
Given
  • Δv_total needed: 14,490 m/s
  • v_e stages 1&2: 3,100 m/s (kerosene)
  • v_e stage 3: 4,400 m/s (hydrogen)
  • Payload: 1,000 kg
Solution
  1. Split Δv: Stage 1: 5,500 m/s | Stage 2: 4,990 m/s | Stage 3: 4,000 m/s
  2. Stage 3 mass ratio: e^(4000/4400) = e^0.909 = 2.48 → fuel = 1000×1.48 = 1,480 kg, total = 2,480 kg
  3. Stage 2 mass ratio: e^(4990/3100) = e^1.61 = 5.00 → total = 2,480×5.0 = 12,400 kg
  4. Stage 1 mass ratio: e^(5500/3100) = e^1.774 = 5.89 → total = 12,400×5.89 = 73,036 kg
Answer73 tonne rocket delivers 1 tonne to GEO — 1.37% payload fraction, typical!
EXAMPLE 4How reusability changes the economics (SpaceX model)
Given
  • Stage 1 cost (new): $40M
  • Stage 1 refurbishment: $2M per flight
  • Propellant cost: $300,000
  • Reuse Δv penalty: landing burn ~400 m/s (less payload)
Solution
  1. Non-reusable cost per flight: $40M + $300K = $40.3M
  2. Reusable (10 flights): $40M/10 + $2M + $300K = $4M + $2M + $300K = $6.3M
  3. Savings: $40.3M − $6.3M = $34M saved per flight!
  4. Landing burn penalty: 400 m/s less Δv → ~15% less payload to orbit (acceptable trade-off).
AnswerReusability saves $34M per flight — 15% payload loss is well worth it.
EXAMPLE 5Capstone design — design a rocket to your specifications
Design Challenge
  • Mission: Place 200 kg to ISS (400 km circular orbit)
  • v_e available: 3,000 m/s
  • Stages allowed: 2
Solution Framework
  1. Δv needed: 9,400 m/s | Split: 5,000 m/s (S1) + 4,400 m/s (S2)
  2. Stage 2 mass ratio: e^(4400/3000) = e^1.467 = 4.335
  3. With 200 kg payload + 20% structural fraction: m_S2_total = 200×4.335/(1−0.2×(4.335−1)) = ~1,050 kg
  4. Stage 1 mass ratio: e^(5000/3000) = e^1.667 = 5.296
  5. S1 total: 1050 × 5.296 / (1−structural) ≈ 7,000 kg launch mass
  6. Payload fraction: 200/7000 = 2.9% — realistic for 2-stage kerosene!
Design Result~7,000 kg rocket delivers 200 kg to ISS — 2.9% payload fraction. Mission achievable!
Rocket Science Physics II & III · 7th & 8th Grade Courses · 10 Units · 40 Topics · 200+ Worked Examples
Builds on 6th Grade Foundations · Prerequisite Chain: Newton → Vectors → I_sp → Orbital Mechanics → Mission Architecture