Δv and the Rocket Equation

Tsiolkovsky's equation, specific impulse, mass ratio and staging

Δv as the mission budget

In space you cannot buy time. You cannot buy distance. You cannot buy altitude, or a different inclination, or an earlier arrival. There is exactly one purchasable commodity, and it is change in velocity. Δv. Everything a spacecraft will ever do — launch, transfer, rendezvous, land, station-keep, deorbit — is denominated in it. A mission's entire capability is one number in metres per second, fixed at the moment the tanks are sealed, and it only ever goes down. So far this course has treated velocity as free. Every orbit in the last five units came with whatever speed it needed at every point, and we read those speeds off the page as though they were facts about geometry. It is not. It is a purchase, and here is what makes the purchase brutal: The exchange rate between Δv and propellant is exponential. Not linear. Not quadratic. Exponential — and exponentials are not the kind of thing you engineer your way around. Want twice the Δv? You do not need twice the propellant. Depending on your engine, you may need ten times. Or a hundred. Or a mass of propellant exceeding the mass of the solar system, which is not a joke: for interstellar Δv with chemical engines, the required mass ratio genuinely exceeds the number of atoms available.

Watch first · presented by Shoaib Aziz

The numbers in this film are the labs' own: at an exhaust speed of 3.05 km/s, 3, 6, 9 and 15 km/s of Δv need mass ratios of 2.7, 7.2, 19 and 137; putting 1,000 kg into GEO (3.91 km/s) takes 2,604 kg of propellant at 311 s but 283 kg with a Hall thruster at 1,600 s, which then fires for 574 days; every factor of 10 in mass ratio costs the same 7.02 km/s; and 100 t of propellant with a 1 t payload reaches 8.3 km/s as one stage but 10.0 km/s as two.

Start with the one thing a rocket can actually do. It has no ground to push against and no air to bite. It has mass, and it can throw that mass away in a chosen direction. Conservation of momentum does the rest.

Throw mass

dm

backwards at exhaust speed

v_e

, and the remaining vehicle gains forward momentum. Integrate that over the burn — and note that the vehicle is getting lighter as it goes, which is the whole subtlety:

m\,dv = -v_{e}\,dm \quad\Longrightarrow\quad \int_{v_0}^{v_f} dv = -v_{e}\int_{m_0}^{m_f}\frac{dm}{m}

Momentum bookkeeping over an infinitesimal instant of a burn. The minus sign says the vehicle loses mass while gaining speed.

\boxed{\;\Delta v = v_{e} \ln\frac{m_0}{m_f} = I_{sp}\,g_{0} \ln\frac{m_0}{m_f}\;}

Tsiolkovsky's rocket equation, 1903. m₀ is wet mass, m_f is dry mass. Written by a deaf schoolteacher in a log cabin outside Kaluga, half a century before anything flew.

A planet is the cradle of mind, but one cannot live in a cradle forever.

– Konstantin Tsiolkovsky, who published the rocket equation in 1903

💡Mass ratio as a function of Δv

As written, the equation looks harmless: Δv grows with the logarithm of the mass ratio. Logarithms are gentle. But you never get to choose the mass ratio. You are handed a required Δv by orbital mechanics, and you must find the mass ratio that delivers it. So invert:

\frac{m_0}{m_f} = e^{\Delta v / v_e}

Now it is an exponential, and the required mass explodes. With a good chemical engine (

v_e \approx 3.05

km/s): Δv = 3 km/s → mass ratio 2.7. Propellant is 63% of liftoff mass. Routine. Δv = 6 km/s → mass ratio 7.2. Propellant is 86%. Hard. Δv = 9 km/s → mass ratio 19.3. Propellant is 95%. This is a launch vehicle, and it is why they look the way they do. Δv = 15 km/s → mass ratio 135. For every kilogram delivered, 134 kg of propellant. Not buildable in one stage. Δv = 30 km/s → mass ratio 18,000. Absurd. And notice that each of those steps was a fixed 3 km/s. The cost of each additional kilometre per second depends on everything you have already bought.

Figure 6.1

Mass ratio against Δv for three engine classes. Solid boosters flatten out early. Cryogenic hydrogen buys another few km/s. Ion propulsion at 30 km/s exhaust speed keeps the curve nearly flat all the way across — but delivers so little thrust that the burn takes months.

The specific impulse

I_{sp}

is how engines are compared, and it is just the exhaust speed divided by

g_0 = 9.80665

m/s². The units come out as seconds, which is historical rather than meaningful — it survives because it is the same number whether you work in metric or imperial.

Engine	I_sp (s)	v_e (km/s)	Thrust	Where it is used
Solid booster	250	2.45	Enormous	First stages, escape systems
Hypergolic (N₂O₄/MMH)	320	3.14	High	Upper stages, in-space, storable for years
Kerosene/LOX	340	3.33	Enormous	Falcon 9 first stage, Saturn V first stage
Hydrogen/LOX	450	4.41	High	Upper stages, Centaur, Saturn V upper stages
Hall-effect thruster	1,600	15.7	Tiny	Station-keeping, orbit raising over months
Gridded ion	3,000	29.4	Minuscule	Deep-space probes, Dawn, BepiColombo
Nuclear thermal	900	8.83	High	Proposed, repeatedly, since 1955

⚠Exhaust speed versus thrust

Looking at that table, the obvious question is why anyone uses chemical engines at all when ion thrusters have five times the exhaust speed. The answer is power. Kinetic power in the exhaust goes as

\tfrac{1}{2}\dot{m}v_e^2

, so for a fixed available power, doubling

v_e

quarters the mass flow — and thrust is

\dot{m}v_e

, so thrust halves. High exhaust speed and high thrust are in direct competition, and the referee is the power supply. A chemical rocket carries its power in the propellant, which is why it can produce meganewtons. An ion engine runs on solar panels and produces about as much thrust as a sheet of paper resting on your hand — around 90 mN for a large Hall thruster. So the choice is made by the mission, not by the numbers: Leaving a planet needs thrust exceeding weight, or you never lift off. Chemical, no alternative. Already in orbit and not in a hurry lets you fire for months. Ion wins overwhelmingly — Dawn accumulated 11 km/s of Δv with 425 kg of xenon, which chemically would have needed several tonnes. Needing a precise burn at a precise instant — a planetary capture — needs it done in minutes, not months. Chemical again. This is why most spacecraft carry both.

The exponential has exactly one escape hatch, and it is not a better engine. It is throwing away the tanks.

A single stage carries its empty tanks, engines and structure all the way to the end of the burn, and every kilogram of that is dead weight riding on the mass ratio. Staging discards the structure the instant its propellant is gone, so the next stage starts with a much smaller

m_f

. The Δv adds:

\Delta v_{\text{total}} = \sum_{k} v_{e,k}\ln\frac{m_{0,k}}{m_{f,k}}

Total Δv is the sum over stages. Each stage sees only the mass above it, which is why the multiplication of small mass ratios beats one large one.

Figure 6.2

The same payload and the same total propellant, flown as one stage and as three. The single stage drags empty tankage to burnout and falls short. The three-stage vehicle sheds structure twice and reaches orbit with margin. Staging is not an optimisation — for chemical launch from Earth, it is the difference between possible and impossible.

Figure 6.3

A launch vehicle drawn as a mass budget rather than a rocket. Propellant dominates so completely that payload is a sliver at the top — typically 2–4% of liftoff mass to LEO. Everything about the shape of a rocket is this diagram made physical.

Finally, the map every mission designer keeps in their head: the Δv budget. Each leg of a journey has a price, the prices add, and the total tells you whether the mission exists.

Leg	Δv	Note
Earth surface → LEO	9.3–10.0	Only 7.8 is orbital speed; the rest is gravity and drag losses
LEO → GTO	2.44	The first Hohmann burn
GTO → GEO	1.47	Circularisation at apogee
LEO → GEO (total)	3.86	Missions budget ~3.9 with margin; add a plane change from most launch sites
LEO → escape (C₃ = 0)	3.22	√2 − 1 times circular speed
LEO → lunar transfer	3.1	Cheaper than escape, because the Moon is not at infinity
LEO → Mars transfer	3.6	Plus 2.1 to capture at Mars, or 0 if you aerobrake
GEO station-keeping	0.05/year	For fifteen years — this is often the tank-sizing driver
Deorbit from LEO	0.1–0.2	Almost free; the atmosphere does the rest

Figure 6.4

The solar system as a subway map, with Δv for fares. Note what dominates: getting off Earth costs more than everything after it combined. Once you are in LEO you are, in the often-quoted phrase, halfway to anywhere.

📐Worked Example: one spacecraft, two engines, two very different rockets

A 1000 kg satellite must go from LEO to GEO. The ideal Hohmann cost is 3.86 km/s; budget 3.91 km/s with margin. Option A — bipropellant, $I_{sp} = 311$ s.

v_e = 311 \times 9.80665 = 3050

m/s. Mass ratio

= e^{3910/3050} = e^{1.282} = 3.604

. Wet mass

= 1000 \times 3.604 = 3604

kg, so propellant

= \mathbf{2604}

kg. You must launch 3.6 tonnes to deliver 1 tonne. Burn time: minutes. Option B — Hall thruster, $I_{sp} = 1600$ s.

v_e = 15{,}690

m/s. Mass ratio

= e^{3910/15690} = e^{0.2492} = 1.283

. Wet mass

= 1283

kg, so propellant

= \mathbf{283}

kg. A saving of 2.3 tonnes — most of a launch slot, worth tens of millions of dollars. The catch. At 90 mN of thrust and an average mass of about 1140 kg (it falls from 1283 to 1000 as the xenon is used), the acceleration is about

7.9 \times 10^{-5}

m/s². Accumulating 3910 m/s takes about

5.0 \times 10^7

s — 574 days, over a year and a half of continuous firing, spiralling slowly outward through the Van Allen belts the whole time, degrading solar arrays and earning no revenue. This is a real commercial decision, made repeatedly since about 2015, and operators now routinely choose the ion option and eat the delay — because the launch cost saved exceeds the revenue lost.

🛰The [Apollo lunar module](https://en.wikipedia.org/wiki/Apollo_Lunar_Module) ascent stage

The Apollo lunar module ascent stage — the part that lifted two astronauts off the Moon — had a dry mass of about 2,150 kg and carried roughly 2,350 kg of propellant. A mass ratio just over two, for a Δv of about 1.85 km/s. Work the exponential backwards through the whole mission and the arithmetic is staggering. Every kilogram landed on the Moon required roughly 125 kilograms on the launch pad: propellant to lift the propellant to lift the propellant, compounding at every stage. Which is why Apollo's engineers fought over grams. Removing the ascent stage's seats — the crew flew standing up, tethered — saved about 36 kg. Multiply by 125 and that decision saved four and a half tonnes on the pad. The rocket equation is why spacecraft engineering has the culture it has.

↓But there is a problem

So Δv is the currency, the exchange rate is exponential, and every metre per second is fought for. Which makes the next question unavoidable and enormously valuable: given two orbits, what is the cheapest possible way between them? Not a good way. Not a way that works. The provably cheapest one, with a proof that nothing else can beat it. Remarkably, there is an answer, and it is beautiful. It was worked out in 1925 by a German architect who designed office buildings for a living and did orbital mechanics in the evenings, four decades before anyone could fly it, and it is still the first calculation any mission designer runs.