How this instrument works
Delta-v is not a speed anything travels at. It is a budget — how much a vehicle can alter its own motion before propellant runs dry, banked as a number and spent burn by burn. Momentum supplies it: hurl a kilogram of exhaust backwards at v_e and whatever remains gains forward speed, so m·dv = −v_e·dm. Integrate from wet mass down to dry and a logarithm falls out. Logarithms grow grudgingly, which is why velocity gets exponentially dearer — each further v_e worth of Δv multiplies mass ratio by e, near 2.72 times more vehicle for every increment.
Konstantin Tsiolkovsky, a provincial schoolteacher in Kaluga, deaf since a childhood bout of scarlet fever, worked this out on 10 May 1897 and published it in 1903 in a paper on exploring space by reaction devices. He was not first. William Moore, an instructor at Woolwich, derived essentially this relation for military rockets in a treatise of 1813; Robert Goddard reached it again in 1919, Hermann Oberth in 1923, both unaware. Tsiolkovsky's name stuck because he alone drew out what it implies — that chemical propellants cap what one tank can ever do, and that staging is less an engineering preference than an arithmetic obligation.
Everything absent from those two symbols matters. This expression describes a vehicle alone in empty space, thrust aligned with motion, exhaust velocity steady from ignition to burnout, no gravity or atmosphere arguing back. A launch settles all three bills separately: 1.5 to 2 km/s lost fighting gravity during ascent, a few hundred m/s more to drag and steering, so a rocket needing 7.8 km/s of orbital speed must carry nearer 9.4 km/s of budget. Duration is missing too. Five km/s delivered by a Hall thruster across eight patient months and five km/s delivered by a chemical stage in four minutes read identically here, and could hardly differ more in practice.
- Enter Effective exhaust velocity in m/s. Holding specific impulse in seconds instead? Multiply by 9.80665 — 300 s becomes 2942 m/s.
- Set Wet mass (fuelled) to everything aboard at ignition: structure, engines, payload and propellant together.
- Set Dry mass (burnt out) to whatever survives once tanks are empty. It must stay below wet mass, or a check will stop you.
- Read Delta-v in m/s, or flip that field's unit menu to km/h for a figure that sits closer to intuition.
- Only mass ratio enters, so scaling both masses by any common factor leaves your answer untouched — 1000 and 500 kg behave exactly as 2 and 1 t.
Worked example — a kick stage that burns half its mass
A compact upper stage masses 1000 kg fuelled and 500 kg once burnt out, and its engine throws exhaust at 3000 m/s. Mass ratio first: 1000 ⁄ 500 = 2. Natural logarithm of 2 is 0.6931472. Multiply through and Δv = 3000 × 0.6931472 = 2079.44 m/s. Half of what ignited now trails behind as exhaust; 2.08 km/s is what that half purchased.
Useful capability, that. Circularising out of a geostationary transfer orbit costs roughly 1.5 km/s, so this stage does that job with margin left over. Then try doubling it. Reaching 4158.88 m/s on identical exhaust demands a mass ratio of 4 — dry mass cut to 250 kg, propellant up to 750. Another 2079 m/s on top wants ratio 8, dry mass 125 kg. Every fixed increment costs half of all that still remains, and there sits the whole argument for throwing away spent tanks partway up rather than hauling them onward.
Questions
Why quote delta-v in m/s when it is not a speed?
Because it is a change of speed, totalled over a mission without regard to direction. A tug parked in low Earth orbit holding 2 km/s of delta-v is not moving at 2 km/s relative to anything in particular; it can rearrange its motion by that much altogether, whether spent in one long burn or twenty short ones. Mission planners add these figures like money in a ledger — so much to raise apoapsis, so much to change plane, so much held back for rendezvous and margin.
What is effective exhaust velocity, and how does specific impulse relate?
They are one quantity in two costumes: v_e = I_sp × 9.80665, so an engine rated 300 s exhales at 2942 m/s. It is called effective because it folds the nozzle's pressure-thrust term into a single equivalent speed rather than reporting raw gas velocity. Typical values spread widely — a solid booster near 2,600 m/s, kerosene and oxygen around 3,000 to 3,400, hydrogen and oxygen in vacuum about 4,430, and a Hall-effect thruster anywhere from 15,000 to 20,000 m/s.
How much delta-v does reaching orbit take?
Roughly 9.4 km/s from Earth's surface to low orbit, although orbital speed up there is only 7.8 km/s. That gap is gravity losses, drag and steering — none of which this equation knows about, so add them to your requirement by hand before sizing tanks. Onward legs are cheaper: about 3.1 km/s from low orbit onto a lunar trajectory, near 3.6 km/s for a Mars transfer, and roughly 1.5 km/s to circularise at geostationary altitude.
If only mass ratio matters, why does staging help?
Because dry mass refuses to shrink freely. Tanks, plumbing and engines are sized for propellant that has already gone, and a vehicle carrying twenty times its burnout mass in propellant is close to unbuildable in one piece. Staging evades that ceiling by discarding hardware whose work is done: apply this equation once per stage and add the results, each later stage beginning from a smaller, cleaner m₀. Saturn V's three stages summed to something near 12 km/s that no single tank could have reached.
When does this formula stop being true?
Whenever one of its three silent assumptions breaks: constant exhaust velocity, thrust collinear with motion, and no external force. Ascent through atmosphere violates all three at once, which is why launch budgets are padded rather than computed straight from here. Push exhaust toward light speed and even the algebra changes — the relativistic version reads Δv = c·tanh[(v_e ⁄ c)·ln(m₀ ⁄ m_f)], which reduces to this one at ordinary speeds.
Can I enter masses in tonnes instead of kilograms?
Yes, so long as both mass fields share whichever unit you pick, because only their ratio ever enters the logarithm. Wet 2 t against dry 1 t returns precisely what 1000 kg against 500 kg returns. Exhaust velocity is different: it sets the scale of the answer outright, so that field needs a genuine speed in m/s or km/h. Note also that dry mass above wet mass is refused, since a negative logarithm would imply a vehicle gaining propellant mid-burn.