Lesson 1 of 14

What Makes a Rocket

Momentum, not air, is what a rocket pushes on. Thrust, specific impulse and exhaust velocity from first principles — and a map of every way humans have found to throw mass backwards fast.

9 min3 live simulations
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The question

A jet engine pushes on air. A car pushes on the road. A rocket in space has nothing at all around it. What is it pushing against?

By the end of this lesson you should be able to answer that in one sentence — and see why a single number, the exhaust velocity, rules almost everything that follows.

Throw something backwards

Forget engines for a moment. You are standing on a cart floating in empty space, with a pile of bricks. There is no ground to push on and no air. The only thing you can do is throw bricks.

Every brick you throw backwards carries momentum away with it, and since the total momentum of you-plus-bricks must stay zero, you move forwards. That is the entire principle of the rocket. The "something" a rocket pushes against is its own propellant.

PlayPush against nothing: throw the bricks
010203040506002468bricks thrown so far (kg)speed (m/s)as a stream (a rocket): u·ln(m₀/m) = 4.58 m/s
5.0 m/s
Your arm is the engine; u is its exhaust velocity
Final speed, this way3.00 m/s60% of your throw speed
Stream limit4.58 m/sthe rocket equation
Mass ratio2.5100 kg at start / 40 kg at end
Challenges0/3
  • Throw all the bricks at once — then try again as a stream of sand. Which wins?
  • Reach 5 m/s. (Which slider matters most?)
  • Get within 1% of the rocket-equation limit.
You and your cart (40 kg) float in space with 60 kg of bricks. Every brick you throw backwards pushes you forwards — no ground, no air, just momentum. Choose how to throw them, and watch how the answer depends on throw speed far more than on anything else.

Play with it for a minute, because three surprises are hiding in the plot.

Throw speed matters most. Double your throw speed u\cvel{u} and your final speed doubles exactly. Nothing else you can change is that powerful.

Many small throws beat one big one. Throwing all 60 kg at once gets you 60% of your throw speed. Throwing it as a continuous stream of sand gets you 92%. The reason: each later throw pushes a lighter cart, so the same brick buys more velocity. A rocket engine is the limit of infinitely many infinitely small throws.

More bricks help less and less. The dashed pink curve bends over. The final speed of the stream is

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

— the rocket equation, which we will spend all of Lesson 3 on. For now, notice the logarithm: carrying twice as much propellant does not give twice the speed.

Thrust is pressure on metal

A real engine doesn't throw bricks; it throws hot gas, about a tonne of it per second for a large booster, at 3 km/s. But how does throwing gas actually push the rocket? The honest answer is surprisingly concrete: the gas pushes on the inside of the engine, and the engine is bolted to the rocket.

ExploreWhere does thrust actually push?
control volumeair pₐexit pₑ = 1.56 barchamber p_c = 97 barthrust
97 bar
8.0
0.0 km · pₐ 1.013 bar
momentum ṁvₑ917 kNpressure (pₑ−pₐ)Aₑ26 kNtotal thrust943 kN
Mass flow ṁ325 kg/sset by p_c and the throat
Exhaust velocity vₑ2819 m/s
Thrust F943 kNmomentum 917 + pressure 26
Specific impulse296 sF / (ṁ g₀)
Under-expanded: the gas leaves still above ambient pressure and keeps expanding outside, where it can no longer push on the nozzle.
Challenges0/3
  • Make the pressure thrust negative.
  • At sea level with p_c = 97 bar, find the area ratio that gives maximum thrust.
  • Get more than 1.5 MN out of this throat in vacuum.
The engine is a pressure vessel with a hole in it. Purple arrows show the gas pressure on the inside walls (length ∝ local pressure). Add up their forward components and you get the thrust — which is exactly the momentum flux ṁvₑ plus the pressure mismatch at the exit, (pₑ − pₐ)Aₑ.

The combustion chamber is a pressure vessel with a hole in one end. Gas presses on every inside surface. On the side walls the forces cancel. But the injector face at the front has no hole opposite it, so there is a large net push forwards. The converging walls near the throat lean the wrong way and push back a little; the diverging bell leans the right way and pushes forward again. Add them all up and you get the ThrustF = ṁvₑ + (pₑ − pₐ)Aₑ: the momentum of the exhaust plus the pressure imbalance across the nozzle exit..

Summing wall pressures is hard. Fortunately a control volume drawn around the whole engine gives the same answer in one line:

F=m˙ ve+(pe−pa) AeF = \cfuel{\dot m}\,\cvel{v_e} + (\cpress{p_e} - \cair{p_a})\,A_e

The first term is the momentum thrust: mass per second times the velocity it leaves with. The second is the pressure thrust: if the gas leaves at a pressure pe\cpress{p_e} different from the air pressure pa\cair{p_a} outside, the exit plane is pushed on by the mismatch. At sea level with a large nozzle it is negative — the atmosphere pushes the engine backwards. Climb, and the air falls away, so the same engine makes more thrust.

Going deeper
The control-volume derivation of the thrust equation

Draw a box around the engine that cuts through the exit plane. The only momentum crossing the boundary is the exhaust leaving through AeA_e at velocity vev_e, carrying momentum flux m˙ve\dot m v_e. The forces on the box are the thrust FF the rocket structure exerts on the engine (equal and opposite to what the engine exerts on the rocket), the exhaust pressure pep_e acting over AeA_e, and the ambient pressure pap_a acting over everything else. Since pap_a acting over a closed surface has no net effect, what remains is pap_a's missing contribution over the exit, and the momentum balance in the flight direction gives

F−(pe−pa)Ae=m˙ve.F - (p_e - p_a)A_e = \dot m v_e.

That is the whole thing. Every wall-pressure arrow in the widget is hidden inside m˙ve\dot m v_e.

Two numbers that describe any rocket

Because thrust has two parts, engineers bundle it into one velocity: the effective exhaust velocity c=F/m˙c = F/\dot m. It equals the real exhaust velocity when the nozzle is perfectly matched to the air around it.

Divide it by standard gravity g0=9.807g_0 = 9.807 m/s² and you get the propellant's miles-per-gallon, the Specific impulse IspThrust per unit weight flow of propellant, in seconds: Isp = F/(ṁg₀) = vₑ/g₀. The propellant’s “miles per gallon”.:

Isp=Fm˙ g0=cg0I_{sp} = \frac{F}{\cfuel{\dot m}\,g_0} = \frac{\cvel{c}}{g_0}

It is measured in seconds, and it is the most-quoted number in rocketry. A kerosene engine at sea level manages about 300 s; hydrogen in vacuum about 450 s; a Hall thruster 1600 s. Multiply by ten (strictly, 9.81) to get the exhaust velocity in m/s.

Check your intuition
Why is specific impulse measured in seconds?

The other key number is the Total impulseThrust integrated over burn time, I = ∫F dt (N·s). The same impulse can be a short kick or a long gentle push., I=∫F dtI = \int F\,dt: the whole push the engine delivers. The same impulse can be a violent two-minute shove from a solid booster or a gentle year-long whisper from an ion engine; how it is delivered is the question of what the mission needs.

Estimate first
The Saturn V's F-1 engine made 6.77 MN of thrust at sea level with an Isp of 263 s. How many kilograms of propellant did one engine burn every second?
101001,00010,000100,000
Drag along the line to place your guess — each tick is ten times the last.

Every way to throw mass

Now step back. Anything that throws mass backwards fast is a rocket, and people have found many ways to do it. They sort themselves by where the energy comes from:

  • Chemical rockets store the energy in the propellant itself: a liquid bipropellant engine, a solid motor, a hybrid. The propellant is both the energy source and the reaction mass, which makes them compact and enormously powerful — but caps the exhaust velocity at around 4.5 km/s, because that is all the energy the chemistry holds.
  • Electric thrusters get their energy from outside — solar panels or a reactor — and use it to accelerate a propellant electrically: by heating it (resistojets, arcjets), with electric fields (ion engines), or with crossed electric and magnetic fields (Hall thrusters). The exhaust can be ten times faster, but the thrust is limited by the power supply.
  • Nuclear thermal rockets heat hydrogen with a fission reactor. No combustion means the exhaust can be pure, light hydrogen: roughly double the Isp of chemical engines.
ExploreThe atlas of propulsion
3010030010003000100 µN10 mN1 N100 N10 kN1 MNspecific impulse Isp (s) — exhaust velocity ≈ 10 × Isp m/sthrust (N)jet power 1 Wjet power 1 kWjet power 1 MWjet power 1 GWCold gasMonopropellantSolid motorsLiquid bipropellantNuclear thermalResisto- & arcjetsHall thrustersGridded ionPulsed plasmaF-1 (Saturn V)
F-1 (Saturn V)7.77 MNLOX/RP-1
Specific impulse304 svₑ ≈ 3.0 km/s
Jet power ½Fvₑ12 GWsupplied by the propellant itself
Five of these lifted the Saturn V. Each burned 2.6 tonnes of propellant per second.
Challenges0/3
  • Visit eight engines (1/8).
  • Find the engine with the greatest jet power.
  • Find the engine whose exhaust is more than 40 km/s.
Every rocket ever flown sits somewhere on this map. Across the bottom-right, electric thrusters: enormous exhaust velocity, tiny thrust. Up the left, chemical rockets: huge thrust, modest Isp. The dashed diagonals are lines of constant jet power ½Fvₑ — and they explain the whole shape. Click any engine.

The atlas has a striking shape: it is roughly a diagonal band from top-left to bottom-right. The dashed lines explain it. The kinetic power carried off in the jet is

Pjet=12 m˙ ve2=12 F ve,P_{jet} = \tfrac12\,\dot m\,v_e^2 = \tfrac12\,F\,v_e,

so at fixed power, thrust and exhaust velocity trade one for the other. Chemical engines sit on the gigawatt lines because their energy source is the propellant, and a tonne per second of propellant can release gigawatts. An electric thruster sits on the kilowatt lines because that is what its solar panels can deliver.

Going deeper
Energy and efficiency: how much of the fire ends up as motion?

Sutton separates two efficiencies. The internal efficiency is the fraction of the chemical energy that becomes kinetic energy of the jet: for a good chemical engine about 50–70%, the rest leaving as heat in the exhaust. The propulsive efficiency is the fraction of the jet's kinetic energy that ends up as kinetic energy of the vehicle:

ηp=2 (v/c)1+(v/c)2.\eta_p = \frac{2\,(v/c)}{1 + (v/c)^2}.

It reaches 100% when the vehicle moves at exactly the exhaust velocity — then the exhaust is left motionless in space, carrying no energy away. At liftoff (v=0v = 0) the propulsive efficiency is zero: all the energy goes into the exhaust. This is why a rocket is so wasteful on the pad and so efficient near burnout.

The map of this course

The rest of the course follows one launch vehicle and its engines from end to end. Each lesson lives in one part of the machine:

The mapOne system, thirteen questions
Every lesson lives somewhere in this picture. Hover a chapter to see where — or tap a part of the machine to find its chapter.
Recap
What to carry forward
  • A rocket pushes against its own propellant: momentum thrown backwards moves the vehicle forwards. No air or ground is needed.
  • Thrust F=m˙ve+(pe−pa)AeF = \dot m v_e + (p_e - p_a)A_e: momentum thrust plus pressure thrust. Physically it is the gas pressing on the injector face and nozzle walls.
  • Specific impulse Isp=F/(m˙g0)I_{sp} = F/(\dot m g_0), in seconds, measures how much push each kilogram buys. Exhaust velocity ≈ 10 × Isp in m/s.
  • The rocket equation Δv=veln⁡(m0/mf)\Delta v = v_e \ln(m_0/m_f) rewards exhaust velocity linearly but mass only logarithmically.
  • Propulsion families are sorted by energy source: chemical (energy in the propellant, huge thrust), electric (external power, huge exhaust velocity), nuclear.

Next: the part of the engine that turns 3500 K gas into a 3 km/s jet. The nozzle, and the strange fact that to make supersonic gas go faster, you have to give it more room.

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