Rocket Equation (Tsiolkovsky)

The situation

A rocket in free space (no gravity, no air resistance) ejects exhaust gas at speed $u$ relative to the rocket. The rocket starts with mass $m_0$ and velocity $v_0$. After burning fuel until mass reduces to $m$, what is the rocket's velocity $v$?

$v - v_0 = u\ln\frac{m_0}{m}$

Derivation

At time $t$: rocket has mass $M$ and velocity $v$.

In a small time $dt$: rocket ejects mass $|dM|$ (so rocket mass decreases: $dM < 0$).

The exhaust moves at velocity $v - u$ (rocket velocity minus exhaust speed relative to rocket).

Conservation of momentum:

Before: total momentum $= Mv$

After: rocket has mass $M + dM$ and velocity $v + dv$. Exhaust has mass $-dM$ and velocity $v - u$.

$Mv = (M + dM)(v + dv) + (-dM)(v - u)$

$Mv = Mv + M \, dv + v \, dM + dM \, dv - v \, dM + u \, dM$

Drop the second-order term $dM \, dv$:

$0 = M \, dv + u \, dM$

$M \, dv = -u \, dM$

$dv = -u \frac{dM}{M}$

Integrate from $m_0$ to $m$ (mass) and $v_0$ to $v$ (velocity):

$\int_{v_0}^v dv = -u\int_{m_0}^m \frac{dM}{M}$

$v - v_0 = -u[\ln M]_{m_0}^m = -u(\ln m - \ln m_0) = u\ln\frac{m_0}{m}$

$\boxed{v - v_0 = u\ln\frac{m_0}{m}}$

Understanding the result

Since $m_0 > m$ (mass decreases as fuel burns), $\ln\frac{m_0}{m} > 0$ — the rocket speeds up.

$\Delta v = u\ln\frac{m_0}{m}$

• Larger exhaust speed $u$ → more $\Delta v$ per kg of fuel
• Larger mass ratio $m_0/m$ → more $\Delta v$ (but diminishing returns due to the logarithm)

The mass ratio problem

To achieve $\Delta v = u$ (speed equal to exhaust speed):

$\frac{m_0}{m} = e \approx 2.718$

About 63% of the initial mass must be fuel.

To achieve $\Delta v = 2u$:

$\frac{m_0}{m} = e^2 \approx 7.39$

About 86% must be fuel.

To achieve $\Delta v = 3u$:

$\frac{m_0}{m} = e^3 \approx 20$

About 95% must be fuel. The diminishing returns of the logarithm make very high velocities extremely costly in fuel mass. This is the fundamental challenge of rocket propulsion.

With gravity

In a gravitational field $g$ (rocket launching vertically):

$v - v_0 = u\ln\frac{m_0}{m} - gt$

The gravity term $-gt$ reduces the velocity gain. This is why multi-stage rockets are used — discarding empty fuel tanks reduces the mass being accelerated.