Rocket Equation Calculator
Calculate delta-v from the Tsiolkovsky rocket equation using mass ratio and exhaust velocity.
Rocket Equation Calculator
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Formula: Δv = vₑ × ln(m₀/m₁)
Worked example — 6,908 m/s
Formula
Δv = vₑ × ln(m₀/m₁)
The Tsiolkovsky rocket equation relates delta-v to exhaust velocity and mass ratio. vₑ = Isp × g₀. Higher mass ratio = more delta-v but diminishing returns.
Worked Examples
Example 1: Simple rocket
Problem:vₑ=3000 m/s, wet=100,000 kg, dry=10,000 kg
Solution:Δv = 3000 × ln(10) = 3000 × 2.303 = 6908 m/s
Result:6,908 m/s
Frequently Asked Questions
What is delta-v?
Delta-v is the total change in velocity a rocket can achieve. Getting to Low Earth Orbit requires ~9.4 km/s of delta-v.
What is the relationship between Isp and exhaust velocity?
Specific impulse (Isp) relates to exhaust velocity by: vₑ = Isp × g₀, where g₀ = 9.80665 m/s². A chemical rocket with Isp of 450 s (like the Space Shuttle main engine) has an exhaust velocity of about 4,413 m/s.
Why does the Tsiolkovsky equation use a natural logarithm?
As a rocket burns fuel, its mass decreases, so each unit of fuel produces slightly more acceleration than the last. The natural logarithm of the mass ratio captures this compounding effect, describing how total delta-v accumulates as propellant is expelled continuously.
What mass ratio is needed to reach orbit?
To achieve LEO delta-v of ~9,400 m/s with a kerosene-oxygen engine (vₑ ≈ 3,000 m/s), you need a mass ratio of e^(9400/3000) ≈ 23. This means 22 kg of propellant for every 1 kg of dry mass — illustrating why staged rockets are essential for reaching orbit.
How does staging get around the exponential propellant penalty?
Each stage is solved with its own mass ratio and the resulting delta-v values are simply added. Two stages that each burn kerosene-oxygen at vₑ = 3,000 m/s with a mass ratio of 6 deliver 3,000 × ln(6) ≈ 5,375 m/s apiece, about 10,750 m/s combined. One stage delivering that same 10,750 m/s would need a mass ratio of e^(10750/3000) ≈ 36, because dropping spent tanks removes dry mass that the later burn would otherwise keep accelerating.
What is the difference between sea-level and vacuum specific impulse?
Ambient air pressure acts against the nozzle exit plane, so the same engine delivers a lower effective exhaust velocity in the atmosphere than in vacuum. A kerosene-oxygen engine quoted near 282 s at sea level typically rises to roughly 311 s in vacuum, and upper-stage nozzles use much larger expansion ratios precisely because they never face back-pressure. Size first stages with the sea-level or trajectory-averaged figure and upper stages with the vacuum value.
Why must gravity and drag losses be added separately to the calculated delta-v?
The Tsiolkovsky derivation assumes momentum exchange between vehicle and exhaust with no external forces acting. A real launch thrusts while gravity pulls the vehicle back and while the lower atmosphere resists it, so the ideal delta-v here describes what the propellant can supply, not what the trajectory demands. Orbital speed at low Earth orbit is about 7.8 km/s, yet budgets run near 9.3–9.5 km/s; the gap is gravity loss, drag and steering.
How much propellant does a small station-keeping burn actually consume?
Invert the equation: the required mass ratio is e^(Δv/vₑ). For a 100 m/s burn with vₑ = 3,000 m/s, m₀/m₁ = e^0.0333 ≈ 1.034, so the propellant used is 1 − 1/1.034 ≈ 3.3% of the starting mass. Because the exponent is small the cost stays close to linear at this scale; the punishing exponential behaviour only appears once Δv approaches or exceeds the exhaust velocity.
Why can ion thrusters reach large delta-v with very little propellant?
Electric propulsion operates at specific impulses of roughly 1,500–4,000 s, which corresponds to exhaust velocities of about 14.7–39.2 km/s once multiplied by g₀ = 9.80665 m/s². Since delta-v scales directly with vₑ, even a modest mass ratio near 1.3 buys several km/s. The trade-off is thrust: these engines produce millinewtons, so burns run for months and they cannot launch from the ground or perform time-critical manoeuvres.
What does propellant mass fraction tell you that the mass ratio does not?
Propellant mass fraction is (m₀ − m₁)/m₀, which equals 1 − 1/MR, the share of departure mass that is fuel and oxidiser. It converts the ratio into an intuitive percentage: a mass ratio of 10 means 90% propellant, and 23 means about 95.7%. Because dry mass has to cover tanks, engines, structure and payload together, that percentage shows at a glance how little margin remains for hardware.
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Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer · Editorial policy
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