Library / Propulsion / PRP-01

Propulsion

Tsiolkovsky rocket equation

Velocity change a rocket can achieve from its exhaust velocity and the ratio of full to empty mass.

Ref: PRP-01
Rev: v1
Field: Propulsion
Result: Δv [m/s]

Δ v = I_{sp} g_{0} \ln (\frac{m_{0}}{m_{f}})

with g₀ = 9.80665 m/s² and ln the natural logarithm. The product I_spg₀ is the effective exhaust velocity.

A rocket speeds up by throwing mass backwards. The faster the exhaust and the larger the share of the vehicle that is propellant, the more speed it gains. But extra propellant pays back less and less, because you must also accelerate the propellant you have not yet burned.

Calculator

Solve for

ΔvVelocity changeComputed

m/s

I_spSpecific impulse

200450

m₀Initial (wet) mass

6,00030,000

m_fFinal (burnout) mass

1,5005,000

Exhaust velocity v_e2,942 m/s
Mass ratio m₀/m_f3.333 ×
Propellant share70 %

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Velocity change (m/s) current value 3,542 m/s

m₀Initial (wet) massKey driver

2,0406,770

m_fFinal (burnout) mass

2,0405,580

I_spSpecific impulse

2,3605,310

Δv is linear in I_sp but only logarithmic in the mass ratio. A 10% better engine gives 10% more Δv; 10% more propellant gives noticeably less.
Every extra v_e of Δv multiplies the required mass ratio by e (about 2.72). This exponential is why staging exists.
A mass ratio of 2 gives Δv of about 0.69 v_e. A mass ratio of 10 gives about 2.3 v_e.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (m/s)	Local step	Change in result
m₀ Initial (wet) mass	6,000 to 30,000 kg	2,039	6,774	4,735	+10%	+7.92 %
m_f Final (burnout) mass	1,500 to 5,000 kg	2,039	5,581	3,542	+10%	−7.92 %
I_sp Specific impulse	200 to 450 s	2,361	5,313	2,952	+10%	+10 %

Sweep of Initial (wet) mass

Initial (wet) mass (kg)	Velocity change (m/s)
6,000	2,039
7,337	2,631
8,972	3,223
10,970	3,815
13,420	4,407
16,410	4,999
20,060	5,590
24,530	6,182
30,000	6,774

Sweep one variable

Vary

Velocity change (m/s)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Δv

Velocity change

Result · m/s

What the equation gives you. Everything else feeds it.

I_sp

Specific impulse

s · typical 200 to 450

Engine efficiency: thrust per unit weight flow of propellant. Δv is directly proportional to I_sp: 10% better engine, 10% more Δv.

m₀

Initial (wet) mass

kg · typical 6,000 to 30,000

Everything at ignition: structure, payload and propellant. Enters through ln(m₀/m_f), so returns diminish: each extra kilogram of propellant buys less Δv than the one before.

m_f

Final (burnout) mass

kg · typical 1,500 to 5,000

Everything left after the burn: structure and payload. Only the ratio m₀/m_f matters, so a lighter structure lifts Δv as surely as more propellant does.

Worked example with the values above

Effective exhaust velocity: v_e = I_spg₀ = 300 × 9.80665 = 2,942 m/s.
Mass ratio: m₀/m_f = 10,000 / 3,000 = 3.333.
Natural logarithm: ln(3.333) = 1.204.
Δv = v_e × ln(m₀/m_f) = 2,942 × 1.204 = 3,542 m/s.

Assumptions and limits

Ideal rocket: constant exhaust velocity, thrust along the direction of motion.
No gravity, drag or steering losses. For a launch to low Earth orbit these cost roughly 1.5 to 2 km/s.
Single stage. For several stages, apply the equation to each stage and add the Δv values.
I_sp is defined with the standard g₀ = 9.80665 m/s², wherever the rocket is.

Common mistakes

Using log₁₀ instead of the natural logarithm.
Treating I_sp in seconds as a velocity. The exhaust velocity is I_sp × g₀, in m/s.
Using dry mass as m₀, or propellant mass as m_f. m₀ is everything at ignition; m_f is everything at burnout.
Reusing the vehicle mass for later stages. Each stage has its own m₀ and m_f.

Statistics and uncertainty

The equation is deterministic, but its inputs carry measurement uncertainty. To first order, independent relative errors in I_sp and in ln(m₀/m_f) combine in quadrature. Mass errors are amplified when the mass ratio is close to 1, because the logarithm is then small. Test-stand I_sp values are themselves estimates with a spread.

Where it is used

Δv budgeting for missiles, boosters and orbital manoeuvres.
Comparing propulsion options at the same mass ratio, for example solid motors against liquid engines.
Finding the propellant fraction needed for a target velocity change.

Maths you need

Natural logarithm: ln and e^x are inverses; ln(a/b) = ln a − ln b
Units and g₀: I_sp in seconds × 9.80665 gives a velocity in m/s
Ratios: only the ratio of two masses matters, so any mass unit works

Related formulas

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