Appearance
11.7 — Rockets
Reaching space is easy. Staying there is not.
The Kármán line, the conventional boundary of space, is at 100 km. Getting a payload to 100 km altitude and back down needs about 1.4 km/s. Getting it into orbit needs about 9.4 km/s.
\boxed{\text{Orbit is not about height. It is about sideways speed.}}
A useful way to see it: to be in orbit at 400 km you must be moving at 7.7 km/s horizontally. The energy in that speed is:
\frac{1}{2}v^2 = \frac{1}{2}(7670)^2 = 2.94\times10^{7}\ \text{J/kg}
Against the potential energy of the climb:
gh = (9.81)(400{,}000) = 3.92\times10^{6}\ \text{J/kg}
The horizontal speed carries 7.5 times as much energy as the altitude. Which is why suborbital flights like SpaceShipTwo reach 100 km with a fraction of the effort an orbital launch needs, and why atmospheric reentry is so violent — all 29 MJ/kg has to come back out.
Momentum, and why rockets work in vacuum
A rocket does not push against anything. It throws mass backwards, and conservation of momentum (Chapter 1.7) does the rest.
The persistent misconception that a rocket needs air to push against was printed in a New York Times editorial in 1920, mocking Robert Goddard for apparently not knowing that a rocket needs "something better than a vacuum against which to react". The paper published a correction in 1969, the day after Apollo 11 launched, noting that "it is now definitely established that a rocket can function in a vacuum" and regretting the error.
Rockets in fact work better in vacuum, because there is no back-pressure on the nozzle exit and no drag.
Deriving the rocket equation
Setup. At time t the rocket has mass m and velocity v. In time dt it expels mass |dm| at exhaust speed v_e relative to itself, and its own velocity rises to v+dv.
Momentum before:
p_i = mv
Momentum after — the rocket plus the expelled mass:
p_f = (m+dm)(v+dv)+(-dm)(v-v_e)
Note dm is negative since mass is being lost, so -dm is the positive mass expelled.
Expand:
p_f = mv+m\,dv+v\,dm+dm\,dv-v\,dm+v_e\,dm
Drop the second-order term dm\,dv, and the v\,dm terms cancel:
p_f = mv+m\,dv+v_e\,dm
With no external forces, p_f = p_i:
m\,dv+v_e\,dm = 0
dv = -v_e\frac{dm}{m}
Integrate from the initial mass m_0 to the final mass m_f:
\int_0^{\Delta v}dv = -v_e\int_{m_0}^{m_f}\frac{dm}{m}
\boxed{\Delta v = v_e\ln\frac{m_0}{m_f}}
The Tsiolkovsky rocket equation. Konstantin Tsiolkovsky derived it in 1897, a deaf schoolteacher in Kaluga working alone, and published it in 1903 — the same year as the Wright brothers' first flight.
He also worked out staging, orbital speeds, airlocks, closed-loop life support and space stations, decades before any of it was built. His epitaph reads: "Mankind will not remain on Earth forever, but in the pursuit of light and space will at first timidly penetrate beyond the limits of the atmosphere, and then conquer all the space around the Sun."
Reading the equation
Two variables, and the difference between them is the whole problem.
\Delta v = v_e\ln\frac{m_0}{m_f}
v_e appears linearly. The mass ratio appears inside a logarithm.
Doubling the exhaust speed doubles the delta-v. Doubling the mass ratio adds only v_e\ln 2 = 0.693v_e.
Invert it to see the cost:
\frac{m_0}{m_f} = e^{\Delta v/v_e}
| \Delta v/v_e | Mass ratio | Propellant fraction |
|---|---|---|
| 0.5 | 1.65 | 39 % |
| 1 | 2.72 | 63 % |
| 2 | 7.39 | 86 % |
| 3 | 20.1 | 95 % |
| 4 | 54.6 | 98 % |
| 5 | 148 | 99.3 % |
The exponential is brutal. At \Delta v = 4v_e, 98 % of the vehicle is propellant, leaving 2 % for the tanks, engines, structure, guidance and payload. That is not achievable, which is why staging exists.
Specific impulse
The standard measure of engine efficiency:
I_{sp} = \frac{v_e}{g_0}
with g_0 = 9.80665 m/s² by definition. The units are seconds, which is an accident of history — it comes from dividing thrust by propellant weight flow rather than mass flow — and it has the advantage of giving the same number in metric and imperial units.
Physical meaning: the time for which one kilogram of propellant can produce one kilogram-force of thrust.
| Engine | I_{sp} (s) | v_e (m/s) |
|---|---|---|
| Solid booster | 250 | 2450 |
| Kerosene/LOX (Merlin, sea level) | 282 | 2765 |
| Kerosene/LOX (vacuum) | 311 | 3050 |
| Methane/LOX (Raptor, vacuum) | 380 | 3727 |
| Hydrogen/LOX (RS-25, vacuum) | 452 | 4432 |
| Nuclear thermal (NERVA) | 850 | 8336 |
| Ion (Hall thruster) | 1600 | 15,700 |
| Ion (gridded, Dawn) | 3100 | 30,400 |
Hydrogen is the best chemical propellant for a reason that follows directly from thermodynamics. The exhaust velocity of a chemical rocket goes as:
v_e \propto \sqrt{\frac{T_c}{M}}
where T_c is the combustion temperature and M the mean molecular mass of the exhaust. Hydrogen–oxygen produces water vapour at M = 18, the lightest common exhaust, so its molecules move fastest for a given temperature (Chapter 3.2).
And its disadvantages are all practical. Liquid hydrogen has a density of 71 kg/m³ against kerosene's 810, so the tanks are eleven times larger. It boils at 20 K, requiring extreme insulation. And it leaks through almost anything, embrittles metals, and burns with an invisible flame.
This is why first stages usually use kerosene or methane and upper stages use hydrogen — the first stage needs thrust and compact tanks, the upper stage needs efficiency.
Staging
Single stage to orbit is theoretically possible and has never been done.
The arithmetic. For 9400 m/s with a kerosene engine at v_e = 3050 m/s:
\frac{m_0}{m_f} = e^{9400/3050} = e^{3.082} = 21.8
So 95.4 % of the vehicle must be propellant, leaving 4.6 % for everything else. The best real tank-and-structure fractions are about 4–6 %, which leaves nothing at all for payload.
With hydrogen at v_e = 4432:
\frac{m_0}{m_f} = e^{9400/4432} = e^{2.121} = 8.34
88 % propellant, 12 % for everything else. Feasible on paper, and hydrogen's low density makes the tanks so large and heavy that the structural fraction rises to eat the margin. The X-33 and VentureStar programmes failed on exactly this.
Staging solves it by throwing away the empty tanks.
\Delta v_{\text{total}} = \sum_i v_{e,i}\ln\frac{m_{0,i}}{m_{f,i}}
Delta-v adds across stages, and each stage's mass ratio is computed only for what it carries.
Worked example: a two-stage vehicle
Total mass 100 tonnes. Stage 1: 70 t propellant, 7 t structure, v_e = 3000 m/s. Stage 2: 20 t propellant, 2 t structure, v_e = 4400 m/s. Payload 1 t.
Stage 1:
m_0 = 100\ \text{t}, \qquad m_f = 100-70 = 30\ \text{t}
\Delta v_1 = 3000\ln\frac{100}{30} = 3000\times1.204 = 3612\ \text{m/s}
Stage 2, after discarding the 7 t of stage 1 structure:
m_0 = 23\ \text{t}, \qquad m_f = 23-20 = 3\ \text{t}
\Delta v_2 = 4400\ln\frac{23}{3} = 4400\times2.037 = 8963\ \text{m/s}
\Delta v_{\text{total}} = 3612+8963 = 12{,}575\ \text{m/s}
Comfortably above the 9400 needed.
Now the same 100 tonnes as a single stage — 90 t propellant, 9 t structure, 1 t payload, at v_e = 3000:
\Delta v = 3000\ln\frac{100}{10} = 3000\times2.303 = 6908\ \text{m/s}
Not enough. Staging turned a failure into a comfortable success, with the same total mass and the same payload.
Why more stages help less and less. Each staging event adds separation hardware and a second set of engines, and the delta-v gained per stage falls off. Two or three stages is the practical optimum, and almost every launcher ever flown uses two or three.
Thrust
F = \dot{m}v_e+(P_e-P_a)A_e
The first term is momentum flux — the main contribution. The second is the pressure difference across the nozzle exit.
The pressure term is why an engine performs better in vacuum. With P_a = 0 the term is entirely positive, adding roughly 10 % to the thrust.
And it is why nozzles have different shapes. A vacuum-optimised nozzle expands the exhaust much further, giving a large exit area — which is why the Merlin Vacuum engine's nozzle is enormous compared with the sea-level version, and why a sea-level engine would suffer flow separation and damage if fitted with one.
Thrust-to-weight ratio must exceed 1 at liftoff, or the rocket sits on the pad burning fuel. Typical launch values are 1.2 to 1.5, and higher is not obviously better because it increases aerodynamic loads and structural mass.
Worked example: Falcon 9 at liftoff. Nine Merlin engines at 845 kN each, vehicle mass 549 t:
F = 9\times845{,}000 = 7.605\times10^{6}\ \text{N}
W = 549{,}000\times9.81 = 5.386\times10^{6}\ \text{N}
\frac{F}{W} = 1.41
a = \frac{F-W}{m} = \frac{7.605-5.386}{0.549}\times10^{6}/10^{3} = 4.04\ \text{m/s}^2
Initial acceleration of 0.41 g — a rocket leaves the pad slowly, which surprises people watching for the first time. By the end of the first stage burn, with most of the propellant gone, the acceleration reaches about 4 g.
Losses
The rocket equation gives the ideal delta-v. Real ascent loses a substantial fraction.
Gravity losses. While thrusting, gravity removes g\cos\theta of acceleration, where \theta is the angle from vertical.
\Delta v_{\text{grav}} = \int g\cos\theta\,dt
Vertical flight is the worst case, costing 9.81 m/s for every second. Typical total: 1200–1500 m/s.
The fix is the gravity turn. Shortly after liftoff the rocket pitches slightly, and thereafter gravity itself curves the trajectory over — the vehicle simply follows its velocity vector with no side loads on the structure.
Two constraints fight each other. Pitch over too early and the aerodynamic loads at max-Q become dangerous; too late and gravity losses mount. The optimum is computed numerically for each vehicle.
Drag losses. Typically 100–200 m/s, and smaller than most people expect because the rocket is through the dense atmosphere within about 90 seconds.
Max-Q — maximum dynamic pressure, q = \frac{1}{2}\rho v^2 — occurs around 11–13 km, where the speed is rising fast and the density is falling fast. Vehicles throttle down through it, which is the origin of the "Go at throttle up" call.
Steering losses. Thrust not aligned with velocity, typically 50–100 m/s.
Total launch delta-v: about 9400 m/s to reach a 7700 m/s orbit (Chapter 11.5).
Reusability
The rocket equation is unforgiving about reuse, and the numbers are worth doing.
A recovered first stage must retain propellant for the boostback burn (if returning to the launch site), the entry burn, and the landing burn. Typically 20–30 % of its propellant.
And it carries landing hardware: legs, grid fins, and reinforced structure.
The cost is 30–40 % of payload to a given orbit. Falcon 9 lifts about 22.8 t expendable and 17.5 t with droneship recovery.
The economics. A Falcon 9 first stage is roughly 60 % of the vehicle cost. If refurbishment costs 10 % of a new stage and it flies twenty times, the amortised cost per flight falls dramatically. Individual boosters have now flown over twenty times.
Which is why the trade is worth it despite the payload penalty.
Beyond chemical
The rocket equation says the way forward is higher v_e, and chemical propulsion is near its ceiling — bounded by the energy available in chemical bonds (Chapter 10.1).
Nuclear thermal heats hydrogen with a reactor rather than combustion, reaching I_{sp} \approx 900 s. NERVA engines were built and ground-tested in the 1960s and worked. The programme was cancelled in 1973.
Ion propulsion accelerates ions electrostatically, reaching I_{sp} of 3000 s and more, with thrust measured in millinewtons.
Worked comparison. For a 5 km/s mission:
Chemical at v_e = 4400: mass ratio e^{1.136} = 3.11, so 68 % propellant.
Ion at v_e = 30{,}000: mass ratio e^{0.167} = 1.18, so 15 % propellant.
A factor of four less propellant — at the cost of months or years of thrusting, and of carrying the power supply.
Solar sails carry no propellant at all, so the rocket equation does not apply. Chapter 4.7 covered them.
And nuclear pulse propulsion — Project Orion — proposed detonating nuclear bombs behind a pusher plate, giving I_{sp} of 10,000 s or more with high thrust. The physics works and it was studied seriously from 1958 to 1965. The 1963 Partial Test Ban Treaty ended it.
Where this shows up in your life
Every satellite launch is this arithmetic.
GPS, weather forecasting, satellite television, Earth observation and satellite internet all exist because the rocket equation can be beaten with staging.
Falling launch costs have driven an increase from about 100 launches a year in 2010 to over 250 now, and the number of active satellites has grown roughly tenfold.
And the reason space remains expensive is the exponential. No engineering improvement changes \Delta v = v_e\ln(m_0/m_f), and the only lever with real leverage is v_e, which chemistry bounds.
What the next chapter fixes
Getting up is one problem. Coming back down under control, on a specified spot, with a vehicle that is nearly empty and has a thrust-to-weight ratio well above 1 even at minimum throttle, is a genuinely harder one. Chapter 11.8 states the powered descent problem properly, computes a suicide burn from scratch, explains why the guidance is solved as a convex optimisation, and describes what the control loop is doing every 100 milliseconds.