Rocket Calculator

Apply the Tsiolkovsky rocket equation to find delta-v, mass ratio and propellant mass from specific impulse and vehicle masses.

100% FreeNo sign-upPrivate — runs in your browser

Ideal rocket equation: it assumes constant specific impulse, no gravity or drag losses and instantaneous burns, so real mission delta-v budgets are larger.

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Delta-v (m/s)

Delta-v (m/s)
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Delta-v (km/s)
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Mass ratio (wet / dry)
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Propellant mass (kg)
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How to use this tool

1
Enter specific impulse

Use seconds: around 300 s for a storable chemical rocket, around 450 s for hydrogen-oxygen, thousands for electric propulsion.

2
Enter wet and dry mass

Wet mass includes propellant; dry mass is structure plus payload after burnout.

3
Read delta-v

This is the velocity change the propellant can deliver in free space, before gravity and drag losses.

4
Check the mass ratio

Delta-v grows only logarithmically with mass ratio, so each extra km/s costs disproportionately more propellant.

Formula & worked example

Tsiolkovsky: delta-v = Isp x g0 x ln(m0 / mf), with g0 = 9.80665 m/s², m0 the initial mass and mf the final mass.

Isp = 300 s, wet mass 1,000 kg, dry mass 200 kg: delta-v = 300 x 9.80665 x ln(5) = 4,734.96 m/s (4.735 km/s) from 800 kg of propellant.

Limitations — read this before you rely on it

Idealised model: constant Isp, no gravity loss, no drag, no staging or finite-burn effects. Real launch delta-v budgets add roughly 1.5-2 km/s of losses for reaching low Earth orbit.

Frequently asked questions

Each increment of velocity needs propellant that itself must be accelerated, so returns diminish: doubling the mass ratio adds a fixed delta-v, not double.

Thrust per unit weight flow of propellant, in seconds. Higher Isp means more delta-v from the same propellant mass.

Low Earth orbit needs about 9.4 km/s including losses, even though orbital speed itself is about 7.8 km/s.

Yes: dropping empty tanks resets the mass ratio, which is why multi-stage rockets beat single-stage ones for the same payload.

Yes, as long as both masses use the same unit, because only their ratio enters the equation.