Lesson 3 of 14

Flight Performance

What exhaust velocity buys: the rocket equation and its exponential wall, why every orbital rocket stages, how gravity and air steal 1.5 km/s on the way up, and the Δv map that mission planners use instead of distance.

7 min4 live simulations
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The question

A Falcon 9 weighs 550 tonnes on the pad and puts about 20 tonnes into low orbit. The rest — more than 95% — is propellant and hardware thrown away on the way up. Why must a rocket to orbit be almost entirely propellant?

By the end of this lesson you will see the answer as one curve: a logarithm that refuses to be hurried.

The rocket equation

In Lesson 1 you threw bricks off a cart. Make the throws continuous, and the momentum balance for a rocket of mass mm ejecting mass at relative speed vev_e reads m dv=−ve dmm\,dv = -v_e\,dm. Integrate from the starting mass m0m_0 to the final mass mfm_f and you get the most famous equation in astronautics, first written down by Konstantin Tsiolkovsky in 1903:

Δv=ve ln⁡m0mf\cdv{\Delta v} = \cvel{v_e}\,\ln\frac{m_0}{m_f}

It says two things. First, the velocity a rocket can gain is proportional to its exhaust velocity. Second, it grows only with the logarithm of the Mass ratioInitial mass over final (burnout) mass, m₀/m_f.. To add another vev_e worth of speed you must multiply the mass ratio by e≈2.7e \approx 2.7. Again and again.

ExploreThe rocket equation is an exponential wall
0%20%40%60%80%95%0481216propellant fraction of liftoff massΔv (km/s)Low Earth orbit (with losses) · 9.4 km/s · needs 94.5%LEO → escape · 12.6 km/s · needs 98.0%Surface → Moon landing · 15.4 km/s · needs 99.1%6.14 km/spropellantdry + payload
330 s · vₑ 3.24 km/s
85.0% · m₀/m_f 6.7
Δv6.14 km/s
Mass ratio m₀/m_f6.7
Structure + payload15.0%of liftoff mass
Not enough for orbit. Either carry more propellant — which means a lighter structure — or find a faster exhaust.
Challenges0/3
  • Reach LEO in one stage with kerosene (Isp ≤ 340 s).
  • Reach LEO in one stage with no more than 88% propellant.
  • Reach the Moon’s surface in one stage — and notice what it takes.
Δv = vₑ ln(m₀/m_f). The horizontal axis is how much of the rocket’s starting mass is propellant. Every extra km/s of Δv costs exponentially more — which is why exhaust velocity, sitting outside the logarithm, is so precious.

Reaching low Earth orbit takes about 9.4 km/s of Δv once losses are counted. With kerosene at an average Isp of about 330 s (ve≈3.2v_e \approx 3.2 km/s), that is a mass ratio of e9.4/3.2≈19e^{9.4/3.2} \approx 19: the rocket must be 95% propellant before it carries anything at all. Hydrogen, at 450 s, brings this down to 88% — but hydrogen is so light that its tanks are huge, and tanks weigh something. That tension runs through the whole of rocket design.

Staging: stop carrying empty tanks

The problem with a single stage is that it has to carry its entire structure — every empty tank and every engine — all the way to orbit. Once a tank is empty it is dead mass, and each kilogram of dead mass must still be accelerated to 7.8 km/s.

StagingDiscarding empty tanks and engines during flight so the remaining vehicle does not have to keep accelerating dead mass. throws the dead mass away. Each stage fires, empties, and falls off, and the next stage starts over with a much smaller rocket to push. The rocket equation applies stage by stage, and the Δvs simply add:

Δvtotal=∑ive,i ln⁡m0,imf,i\cdv{\Delta v_{total}} = \sum_i \cvel{v_{e,i}}\,\ln\frac{m_{0,i}}{m_{f,i}}

DesignStaging: stop carrying empty tanks
12345012345number of stagespayload fraction (%)no3.19%3.85%4.11%4.24%stage 1: 26 tstage 2: 4.6 tpayload: 1 t
2
340 s
8.0%
dry mass / (dry + propellant) of each stage
9.4 km/s
Payload fraction3.19%
Liftoff mass per tonne of payload31 t
Gain from one more stage×1.21diminishing returns
Challenges0/3
  • Make a single-stage-to-orbit vehicle possible at 9.4 km/s. What did it take?
  • Put 3.5% of liftoff mass into orbit (9.4 km/s) with Isp ≤ 340 s and ε ≥ 8%.
  • Find a case where adding a stage gains less than 5%.
Identical stages, each with the same Isp and structural coefficient ε (dry mass as a fraction of the stage), sharing the Δv equally. The left chart is payload fraction — tonnes of payload per 100 tonnes at liftoff. The drawing on the right is the vehicle, sized by the cube root of mass so you can see all of it.

The first extra stage makes an enormous difference: it can turn "impossible" into "3% payload". The third helps less, and the fourth less still, while every stage adds engines, separation systems, ignition events and failure modes. That is why nearly all orbital launchers have two or three stages. Single-stage-to-orbit remains possible on paper — the lab shows exactly what it demands of structure and Isp.

Going deeper
Optimum staging: why identical stages share Δv equally

For nn stages with the same exhaust velocity and the same structural coefficient ε=ms/(ms+mp)\varepsilon = m_s/(m_s + m_p), each stage with mass ratio RR carries a payload ratio

π=1/R−ε1−ε,\pi = \frac{1/R - \varepsilon}{1 - \varepsilon},

and the vehicle's overall payload fraction is πn\pi^n. Maximizing πn\pi^n subject to n veln⁡R=Δvn\,v_e \ln R = \Delta v gives equal mass ratios in every stage — each stage delivers an equal share of the Δv. Real vehicles differ: upper stages usually have higher Isp (vacuum nozzles, often hydrogen) and worse structural coefficients, so the optimum shifts more of the Δv to the upper stage, and the Lagrange-multiplier solution in Sutton §4.7 handles the general case.

Estimate first
What percentage of a Falcon 9's liftoff mass (about 550 t) ends up in low Earth orbit as payload, when it is flown without recovering the booster?
0.10110100
Drag along the line to place your guess — each tick is ten times the last.

Flying through air and gravity

The rocket equation assumes a rocket floating in empty space. A real launch fights two extra enemies:

  • Gravity. While the rocket climbs, part of its thrust only holds it up. The Gravity loss∫ g sin γ dt: velocity spent holding the rocket up against gravity while it climbs. Short, powerful burns lose less. is ∫gsin⁡γ dt\int g \sin\gamma\,dt, where γ\gamma is the flight-path angle above horizontal. It shrinks if you burn faster (more thrust) and if you turn horizontal sooner.
  • Air. The Drag loss∫ D/m dt: velocity spent pushing through the atmosphere. Typically 100–200 m/s for a large launcher. ∫D/m dt\int D/m\,dt grows if you fly fast while the air is still thick. For big rockets it is surprisingly small, 50–150 m/s, but the aerodynamic load peaks sharply at Max-QThe moment of maximum dynamic pressure q = ½ρv², when aerodynamic loads peak — usually around 10–14 km altitude., where dynamic pressure q=12ρv2q = \tfrac12\rho v^2 is largest.

The two pull in opposite directions. A rocket that climbs slowly wastes Δv holding itself up; a rocket that accelerates hard low down meets thick air at high speed. Real ascents are a compromise, and the classic shape is the Gravity turnAn ascent in which, after a small initial pitch-over, gravity itself bends the trajectory while the thrust stays aligned with the velocity.: rise vertically, tip over by a few degrees, and then let gravity itself bend the trajectory while the thrust stays aligned with the velocity, so the vehicle never flies sideways through the air.

FlyFly a two-stage rocket to orbit
04008001200160020002400050100150200250downrange distance (km)altitude (km)Kármán line · 100 kmtarget orbitmax-Q 31 kPastagingT+531 s · 200 km · 7.80 km/s
losses: 2281 m/s of 10912 m/s ideal Δvgravity 1111steering 1151scale: full width = 5 km/s
1.30
3.0°
12.0 t
Gravity loss1111 m/s∫ g sin γ dt
Drag loss19 m/s
Steering loss1151 m/s
Max-Q31 kPaat 11 km
Peak acceleration5.4 g
Orbit! 6.6 t of stage-2 propellant left over: that is payload margin.
Challenges0/3
  • Reach orbit with 12 t of payload.
  • Reach orbit with total losses under 1.6 km/s and max-Q under 45 kPa.
  • Put 22 t into orbit.
A Falcon-9-class vehicle launched east. You choose its liftoff thrust-to-weight and how far it tips over after clearing the tower; from then on stage 1 flies a gravity turn, and stage 2 steers itself into a 200 km circular orbit if it can. Every metre per second spent fighting gravity, air and its own steering is Δv not spent on orbital speed.

Play with the two controls and you will find a narrow valley. Too little thrust-to-weight and gravity eats you alive. Too small a kick and the rocket goes nearly straight up, then has to spend Δv turning horizontal later. Too big a kick and a slow rocket falls over. Too much thrust and you slam into max-Q and pull uncomfortable g. The best trajectories lose about 1.5 km/s in total. That is why "9.4 km/s to orbit" is quoted, when orbital speed is only 7.8 km/s.

Orbits and the Δv map

Once in space, distances stop mattering and Δv is everything. A circular orbit at radius rr needs speed v=μ/rv = \sqrt{\mu/r} — 7.8 km/s just above the atmosphere. Escape needs 2\sqrt{2} times that. Changing orbit costs Δv at each engine burn, and burns are most efficient where the vehicle is already moving fastest (the Oberth effect).

Mission planners therefore draw the solar system as a subway map, with Δv on each line:

PlanThe Δv subway map
9.402.441.473.150.901.873.220.392.103.80Earth surfaceLow Earth orbitGTOGEOTrans-lunarLow lunar orbitMoon surfaceEarth escapeMars transferLow Mars orbitMars surface5.92 km/s from low Earth orbitnumbers in km/s · click a white-ringed station
Mission Δv5.92 km/s
Mass ratio needed3.8
Mass that arrives26.1%of what started, before subtracting tanks and engines
Challenges0/2
  • Price a Moon landing from the ground with storable propellants.
  • Compare GEO with the Moon’s surface, starting from LEO.
Mission planners budget in Δv, not distance: the Moon is 380 000 km away but only ~6 km/s from low Earth orbit. Pick a destination; the route lights up and the rocket equation tells you how much of your starting mass arrives.

The map makes some counter-intuitive facts obvious. Geostationary orbit, 36 000 km up, costs almost as much as landing on the Moon. Landing on Mars is cheap only because its atmosphere does most of the braking. And every kilometre per second compounds: the fraction of your mass that arrives is e−Δv/vee^{-\Delta v/v_e}.

Check your intuition
A spacecraft in low Earth orbit has a hydrogen stage with Isp 450 s. Roughly what fraction of its mass (tanks, engine and payload together) can land on the Moon, if the stage could do it all?
Recap
What to carry forward
  • Rocket equation: Δv=veln⁡(m0/mf)\Delta v = v_e \ln(m_0/m_f). Δv grows linearly with exhaust velocity but only logarithmically with mass ratio.
  • LEO needs ~9.4 km/s, so a single kerosene stage would be ~95% propellant. Staging discards dead mass; the gains diminish after two or three stages.
  • Launch costs gravity loss (climbing), drag loss (air) and steering loss: together about 1.5 km/s for a good trajectory.
  • The gravity turn keeps thrust aligned with velocity; a narrow window of thrust-to-weight and pitch-over makes orbit.
  • In space, missions are budgeted in Δv, and the fraction of mass that arrives is e−Δv/vee^{-\Delta v/v_e}.

Next: where exhaust velocity comes from. Why hydrogen, with a cooler flame than kerosene, still leaves the nozzle 30% faster.

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