Δv = v_e · ln(m₀ - m_f) — understanding each term

The Tsiolkovsky rocket equation. It tells you how much change in speed (Δv) a rocket earns for burning a given amount of fuel.

The Big Picture

What each term means

Deriving it from scratch (WHY the log appears)

Setup. Work in an inertial frame, no gravity, no drag (ideal case). At some instant the rocket has mass $m$ and velocity $v$. In time $dt$ it ejects a small mass of exhaust and speeds up by $dv$.

Conserve momentum. Total momentum is constant because no external force acts.

Before: $p_{\text{before}} = m\,v$.

After: rocket mass drops by $|dm|$ (note $dm<0$ since mass decreases). The rocket now has mass $m+dm$ and velocity $v+dv$. The ejected chunk has mass $-dm$ and moves at velocity $v - v_e$ (its speed is $v_e$ backward relative to the rocket).

$p_{\text{after}} = (m+dm)(v+dv) + (-dm)(v - v_e)$

Why this step? We split the system into "rocket now" + "gas just thrown", each carrying its own momentum, so we can equate to the before-state.

Expand. $p_{\text{after}} = mv + m\,dv + v\,dm + dm\,dv - v\,dm + v_e\,dm$

Why this step? Multiply out and cancel the $v\,dm$ terms. The product $dm\,dv$ is a second-order infinitesimal → drop it.

$p_{\text{after}} = mv + m\,dv + v_e\,dm$

Equate before = after: $mv = mv + m\,dv + v_e\,dm$, so

$m\,dv + v_e\,dm = 0 \quad\Rightarrow\quad dv = -\,v_e\,\frac{dm}{m}$

Why this step? This is the heart: velocity gain is proportional to the fractional mass loss $dm/m$. The $1/m$ is exactly what breeds the logarithm.

Integrate from start ($v=v_i,\ m=m_0$) to end ($v=v_f,\ m=m_f$):

$\int_{v_i}^{v_f} dv = -v_e \int_{m_0}^{m_f} \frac{dm}{m}$

$v_f - v_i = -v_e\big[\ln m\big]_{m_0}^{m_f} = -v_e(\ln m_f - \ln m_0) = v_e \ln\frac{m_0}{m_f}$

Reading the equation (WHAT it says physically)

• If $m_0 = m_f$ (no fuel burned): $\ln 1 = 0 \Rightarrow \Delta v = 0$. ✔ makes sense.
• To double Δv at fixed $v_e$, you must square the mass ratio (since $\ln R^2 = 2\ln R$). Fuel demand explodes.
• Higher $v_e$ → more Δv per kg of fuel. This is why engineers chase high exhaust speed (ion engines, hot gases).

Worked examples

Common mistakes (steel-manned)

Connections

• Conservation of Momentum — the derivation's foundation.
• Newton's Third Law — the thrust mechanism.
• Specific Impulse (Isp) — links $v_e = g_0 I_{sp}$.
• Multistage Rockets — how staging beats the log's tyranny.
• Thrust and Mass Flow Rate — $F = v_e \,\dot m$, the differential cousin.
• Natural Logarithm and Integration of 1/x — why the log appears.

Concept Map

Hinglish (regional understanding)