Tsiolkovsky Rocket Equation Calculator

Q: Can I get more Δv by making the rocket bigger?

Not by itself. Only the ratio of the masses appears, so scaling everything up by a factor of a thousand changes nothing at all. You get more Δv by making the dry part lighter relative to the propellant, or by raising the exhaust speed, and the second is far more effective because it is outside the logarithm.

Δv = v_e ln(m₀/m_f). The speed you can gain grows only as the logarithm of the mass ratio, which is the cruellest arithmetic in engineering and the reason a rocket is a thin shell wrapped round a tank. Put in a mass ratio and get the Δv; put in a Δv you want and get the propellant fraction it demands.

Last updated 11 Nov 2026

Delta-v from exhaust speed and mass ratio, and the mass ratio a target needs

The ideal rocket equation, with no gravity and no drag in it

How the engine is specifiedThe two are the same number in different clothes: v_e = I_sp × g₀, with g₀ the standard gravity of 9.806 65 m/s² fixed by the 3rd CGPM in 1901. Switching between them here converts rather than resets, so you can type one and read the other off.

Effective exhaust velocity (m/s)The speed of the exhaust relative to the rocket, which is what actually appears in the equation. The default 2982.2 m/s is the Saturn V F-1 engine’s vacuum figure: NASA publishes its vacuum I_sp as 304.1 s, and 304.1 × 9.806 65 = 2982.2.

Specific impulse (s)In seconds, because it is defined as thrust divided by weight flow rather than mass flow. See the thrust and specific impulse page for what that second really measures.

Mass at the start of the burn m₀ (kg)Everything: structure, engines, payload and all the propellant you are about to burn. Sometimes called the wet mass.

Mass at the end of the burn m_f (kg)Everything that is left when the engine shuts down: structure, engines, payload, and any propellant you did not burn. Must not be larger than the starting mass — a rocket cannot gain mass by firing its engine, and the page refuses rather than returning a negative Δv.

A Δv you are aiming for (m/s)Runs the equation backwards: what mass ratio and what propellant fraction would this Δv demand? The default 11 180 m/s is the escape speed at the Earth’s surface, √(2GM/R) with the IAU nominal values — not a real launch budget, which is larger still because of gravity and drag losses, but a clean sourced number to aim at. The escape velocity page computes it.

The rocket with its propellant shaded to your own mass fraction. Drop the mass ratio and the green block shrinks towards nothing while the blue Δv arrow on the left shortens with it — but not in step, because Δv follows the logarithm. The dashed red bar on the right is the propellant fraction your target Δv would demand; when it towers over the green one beside it, that gap is the honest reason rockets are almost entirely tank.

4799.7m/sExample

A stage starts at 10 000 kg and burns down to 2000 kg. Its engine has an effective exhaust velocity of 2982.2 m/s — the Saturn V F-1’s vacuum figure, from an I_sp of 304.1 s. How much Δv does it get?

Momentum conservation, integrated over a burn

Δv = v_e ln(m₀/m_f) · m₀/m_f = e^Δv/v_e · v_e = I_spg₀

Δv: the change of speed the burn can produce, in m/s, in empty space with nothing pulling on the rocket. It is a budget, not a speed: a vehicle’s Δv says what manoeuvres it can afford.
v_e: effective exhaust velocity in m/s. This is the only property of the engine that appears, and it sets the scale of everything — a better engine helps linearly, more propellant only logarithmically.
m₀: mass at the start of the burn, propellant included.
m_f: mass at the end of the burn. m₀/m_f is the mass ratio, and it is the only way the masses enter — a 10 tonne and a 10 000 tonne rocket with the same ratio get the same Δv.
ln: the natural logarithm, not log base 10. Using log₁₀ by mistake divides every answer by 2.3026, which is a wrong number that still looks like a plausible rocket.

Worked example

A stage starts at 10 000 kg and burns down to 2000 kg. Its engine has an effective exhaust velocity of 2982.2 m/s — the Saturn V F-1’s vacuum figure, from an I_sp of 304.1 s. How much Δv does it get?
Mass ratio first. m₀/m_f = 10 000 ÷ 2000 = 5. This is the only way the masses enter the equation, so a 10-tonne stage and a 10 000-tonne stage with the same ratio get exactly the same answer.
Take the natural logarithm. ln 5 = 1.6094. Natural, not base 10: log₁₀5 = 0.6990, and using it would give 2084 m/s instead, which is wrong by a factor of 2.3026 and does not look wrong.
Multiply by the exhaust speed. Δv = 2982.2 × 1.6094 = 4799.7 m/s, or 4.80 km/s.
Look at what that cost. 8000 kg of propellant — 80% of the whole stage — bought 1.61 exhaust speeds. The first 63.2% of the mass bought the first exhaust speed on its own; the remaining 16.8% bought only 0.61 more.
Now run it backwards. Suppose you wanted 11 180 m/s, the escape speed at the Earth’s surface. The mass ratio needed is e^11180/2982.2 = e^3.749 = 42.47, so the propellant fraction would have to be 1 − 1/42.47 = 97.65%, and the 10-tonne stage could arrive with only 235 kg of everything — tank, engine, structure and payload together.
That is the argument for a better engine, and for staging. Swap in a hydrogen engine at I_sp 430 s (v_e = 4216.9 m/s) and the same 11 180 m/s needs a mass ratio of only 14.17, so 92.9% propellant instead of 97.65%. The exhaust speed sits outside the logarithm, which is why improving it is worth so much more than adding tankage.

What each mass ratio buys, at an exhaust speed of 2982.2 m/s

Read the first and last columns together and the cruelty is obvious. Going from half propellant to 90% propellant — nearly doubling the fraction of the rocket that is fuel — multiplies the Δv by only 3.3. Going from 90% to 99% takes a tenfold increase in mass ratio and buys one more factor of 2. **The row worth memorising is the third**: burning 63.2% of the rocket, 1 − 1/e of it, buys exactly one exhaust speed.
Mass ratio m₀/m_f	Propellant fraction	Δv (m/s)	Δv in exhaust speeds
1.5	33.3%	1209	0.405
2	50.0%	2067	0.693
e = 2.718	63.2%	2982	1.000
5	80.0%	4800	1.609
10	90.0%	6867	2.303
20	95.0%	8934	2.996
100	99.0%	13 734	4.605

Real engines, and the mass ratio each would need for 11 180 m/s

Specific impulses from a NASA Marshall review of liquid propulsion systems; the last two columns are this page. Two things to notice. First, every engine has a sea-level figure and a higher vacuum one, because the atmosphere pushes back on the exhaust — which is the thrust page’s subject. Second, hydrogen is in a class of its own: 430 s against 304 s cuts the mass ratio needed from 42 to 14. That is the whole reason upper stages burn hydrogen despite the tanks it needs.
Engine and propellant	Isp (s)	vₑ (m/s)	Mass ratio needed	Propellant fraction
F-1, LO2/RP-1, at sea level	265.4	2602.7	73.4	98.6%
F-1, LO2/RP-1, in vacuum	304.1	2982.2	42.5	97.6%
RD-180, LO2/RP-1, at sea level	311.9	3058.7	38.7	97.4%
RD-180, LO2/RP-1, in vacuum	338.4	3318.6	29.1	96.6%
Vulcain Mk-1, LO2/LH2, in vacuum	430.0	4216.9	14.2	92.9%

Why the logarithm is the enemy, and what can be done about it

A rocket works by throwing mass backwards. There is nothing to push against in space, so the only way to change speed is conservation of momentum: throw some of yourself away, fast, in the opposite direction. Each kilogram thrown back at v_e gives the rest of the rocket a shove — but the rocket is lighter afterwards, so the next kilogram shoves a smaller rocket harder. Integrating that over a whole burn is where the logarithm comes from, and Konstantin Tsiolkovsky wrote it down in 1903.

The equation has two things in it and one of them is far more valuable than the other. The exhaust speed sits outside the logarithm, so doubling it doubles the Δv. The mass ratio sits inside, so doubling it adds only v_e ln 2, about 0.69 of an exhaust speed, no matter where you started. A 10% better engine is worth a lot. A 10% bigger tank is worth almost nothing.

Here is the shape of the problem in one line. To gain one exhaust speed you must burn 1 − 1/e = 63.2% of the rocket. To gain two, 86.5%. To gain three, 95.0%. To gain four, 98.2%. Each extra exhaust speed costs you the same fraction of what is left, so the dry mass — engines, tanks, structure and payload together — is squeezed towards zero. Real launch vehicles need three or four exhaust speeds to reach orbit, which is why NASA’s own introduction says 90% of a launcher is propellant and the payload is about 1%.

Staging is the way out, and it is a trick about dry mass rather than about the equation. The equation is not cheated by staging — each stage still obeys it exactly. What staging does is throw away empty tanks the moment they are empty, so the next stage is not dragging them. Two stages of mass ratio 5 give 1.609 + 1.609 = 3.219 exhaust speeds, while one stage would need a mass ratio of 25 to do the same, and a single tank that is 96% propellant is far harder to build than two that are 80%.

What this equation does not include matters as much as what it does. It assumes no gravity, no atmosphere, and a burn that takes no time. A real launch loses speed to gravity for as long as the engine is running — roughly g times the burn duration, if you are going straight up — and loses more to drag low down. Those losses are why the real Δv budget to reach low orbit is well above the orbital speed itself. The number this page returns is an ideal ceiling; everything real is below it.

Where this goes wrong, every time: using log base 10. On most calculators the key marked log is base 10 and ln is the natural one, and swapping them divides every answer by 2.3026. The result is not absurd — 2084 m/s instead of 4800 m/s still looks like a rocket — so nothing warns you. The second trap is putting the propellant mass in the denominator instead of the final mass. The ratio is start over finish, and it is always greater than 1.

Frequently asked questions

Why is the rocket equation logarithmic?

Because the rocket gets lighter as it burns. The momentum given to each small lump of exhaust is the same, but the mass it has to shove is shrinking, so the speed gained per kilogram of propellant rises as the burn goes on. Adding up dv = v_e dm/m over the burn gives a logarithm — that integral is literally the definition of ln. It is the same reason compound interest has logarithms in it.

Does the rocket equation work on Earth, with gravity?

It gives you an upper bound, not the answer. The equation has no gravity and no air in it, so a launch always achieves less than it predicts: some of the thrust is spent holding the rocket up rather than accelerating it, and some is spent pushing air out of the way. Engineers call these gravity losses and drag losses and subtract them from the ideal Δv. In deep space, where there is nothing to lose to, the equation is essentially exact.

What is Δv actually, if it is not a speed?

It is a budget. A spacecraft with 2 km/s of Δv left can change its velocity by 2 km/s in total, in any combination of manoeuvres — one big burn, or twenty small ones. It is the currency of spaceflight, which is why mission planners talk about the Δv “cost” of going somewhere. The Hohmann transfer page prices one such trip.

Can I get more Δv by making the rocket bigger?

Not by itself. Only the ratio of the masses appears, so scaling everything up by a factor of a thousand changes nothing at all. You get more Δv by making the dry part lighter relative to the propellant, or by raising the exhaust speed, and the second is far more effective because it is outside the logarithm.

Why do rockets have stages?

To stop carrying empty tanks. Once a tank is dry it is pure dead weight, and the equation charges you for it in m_f for the rest of the flight. Dropping it resets the mass ratio for what is left. Two stages at a mass ratio of 5 each deliver the same Δv as a single stage at a mass ratio of 25 — and a 96%-propellant single tank is a much harder engineering problem than two 80% ones.

My answer is negative. What did I do wrong?

You put the final mass above the initial one, so the ratio is below 1 and its logarithm is negative. This page refuses that case rather than printing a negative Δv, because there is no physical situation it describes — a rocket does not gain mass by firing its engine. Check which mass is which: m₀ is the heavier one, at the start.

Related calculators

References

NASA Glenn Research Center, Beginner’s Guide to Aeronautics: Ideal Rocket Equation. Writes the equation as Δu = V_eq ln(MR) with MR the propellant mass ratio, relates the equivalent exhaust velocity to specific impulse by V_eq = I_sp g₀, and states that “90% of the weight of a rocket going to orbit is propellant weight”, with the payload about 1%.
NASA Glenn Research Center, Beginner’s Guide to Aeronautics: Specific Impulse. Thrust F = ṁV_e + A_e(p_e − p₀), equivalent velocity V_eq = F/ṁ, and I_sp = V_eq/g₀ = F/(ṁg₀). The page checks the units itself: (m/s) ÷ (m/s²) = s.
R. Ballard (NASA Marshall, SLS Liquid Engines Office), Liquid Propulsion Systems — Evolution & Advancements, NASA NTRS 20140002716. Source of every engine figure quoted here: F-1 LO2/RP-1 I_sp 265.4 s at sea level and 304.1 s in vacuum with an engine thrust-to-weight of 82 and 94; RD-180 311.9 and 338.4 s at 70.3 and 76.4; Vulcain Mk-1 LO2/LH2 430 s in vacuum.
BIPM, Declaration on the unit of mass and on the definition of weight; conventional value of g_n, Declaration 2 of the 3rd CGPM (1901). Fixes standard gravity at 980.665 cm/s², that is 9.806 65 m/s² exactly — a convention, not a measurement of the gravity anywhere. doi:10.59161/CGPM1901DECL2E
E. Mamajek et al., IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties, arXiv:1510.07674. Source of every GM and radius used here: nominal solar mass parameter 1.3271244 × 10²⁰ m³ s⁻², terrestrial 3.986004 × 10¹⁴, jovian 1.2668653 × 10¹⁷; nominal solar radius 6.957 × 10⁸ m, terrestrial equatorial radius 6.3781 × 10⁶ m, jovian equatorial 7.1492 × 10⁷ m. The resolution is explicit that these are conversion factors fixed by convention, not measurements with uncertainties.