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

This page builds $Δ v = v_{e} ln (m_{0} / m_{f})$ from nothing but a picture of a rocket throwing stuff backward. Every symbol is earned before it is used. Follow the figures — they carry the argument; the words just point at them.

We are going deeper than the parent note: there the algebra flew by. Here we watch it happen.

Step 0 — What is a rocket even doing?

PICTURE. Look at the rocket below. The whole thing (rocket + fuel) has some mass. A little puff of gas is about to leave the tail.

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

The big blue block is the rocket right now: mass $m$ , moving right at speed $v$ .
The small red puff at the tail is the fuel it's about to eject.
The word "speed" here means "how many metres it covers each second" — nothing fancier.

Everything below is just accounting for the momentum of these two pieces.

Step 1 — Name the quantities (define every symbol on the picture)

PICTURE. The same rocket, now labelled with an instant "before" and an instant "after" a single tiny puff.

Before (top): one object, mass $m$ , speed $v$ .
After (bottom): rocket is now mass $m + d m$ (lighter, since $d m < 0$ ) at speed $v + d v$ ; the puff is mass $- d m$ at speed $v - v_{e}$ .

Step 2 — Why momentum, and why it stays constant

Why is this the right tool and not, say, energy? Because the exhaust carries away messy heat and kinetic energy we can't easily track — but momentum is exactly conserved with no leaks. It is the cleanest ledger available.

PICTURE. The "before" system and the "after" system as two sides of a balance scale — total momentum must weigh the same on both pans.

total motion before p_{before} = total motion after p_{after}

Step 3 — Write the "before" momentum

WHAT. Before the puff leaves, there is one object: mass $m$ , speed $v$ .

WHY. With everything still together, the whole system's momentum is just one product.

p_{before} = current mass m \cdot current speed v

PICTURE. One arrow, length $m v$ , pointing right.

Step 4 — Write the "after" momentum (split into two pieces)

WHAT. After the puff, we have two objects. Add their momenta.

WHY. Momentum is additive: total motion = rocket's motion + puff's motion. We must count both, or the ledger won't balance.

p_{after} = lighter rocket (m + d m) faster (v + d v) + mass of puff (- d m) puff’s speed to a bystander (v - v_{e})

$(m + d m)$ — the rocket after losing the sliver; smaller than $m$ because $d m < 0$ .
$(v + d v)$ — the rocket after its little kick.
$(- d m)$ — a positive mass: the sliver that left.
$(v - v_{e})$ — the sliver flies backward relative to the rocket, so a stationary bystander clocks it at $v$ minus the exhaust speed.

PICTURE. Two arrows now: a long forward one for the rocket, a short (possibly backward) one for the puff.

Step 5 — Multiply out and cancel (the algebra, watched closely)

WHAT. Expand $p_{after}$ and simplify.

p_{after} = m v + m d v + cancels v d m + drop: tiny \times tiny d m d v cancels - v d m + v_{e} d m

$v d m$ and $- v d m$ cancel — one from expanding the rocket term, one from the puff term.
$d m d v$ is a tiny number times another tiny number = negligibly small (second order). We discard it. This is legal precisely because both are infinitesimals; a tiny $^{2}$ vanishes faster than a tiny.

What survives:

p_{after} = m v + m d v + v_{e} d m

WHY. Now set $p_{before} = p_{after}$ (the scale from Step 2):

m v = m v + m d v + v_{e} d m

The $m v$ on each side cancels, leaving the heart of the whole subject:

m d v + v_{e} d m = 0 ⟹ d v = - v_{e} \frac{d m}{m}

$d v$ — the speed you gain.
$- v_{e}$ — bigger exhaust speed $\Rightarrow$ bigger kick; minus sign because losing mass ( $d m < 0$ ) gives a positive $d v$ .
$\frac{d m}{m}$ — the star: not the raw mass lost, but the mass lost as a fraction of what's there.

PICTURE. A term-by-term callout of $d v = - v_{e} d m / m$ , with the fraction $d m / m$ circled.

Step 6 — Add up all the tiny kicks (integrate)

WHAT. One sliver gave $d v = - v_{e} d m / m$ . The real burn is billions of slivers, from the full rocket ( $m = m_{0}$ , $v = v_{i}$ ) to the empty one ( $m = m_{f}$ , $v = v_{f}$ ). Adding infinitely many infinitesimals = integrate.

\int_{v_{i}}^{v_{f}} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m}

Left side: all the tiny speed-ups summed $= v_{f} - v_{i} = Δ v$ .
Right side: $v_{e}$ is constant (assumption), pulled out front; the sum of $\frac{1}{m} d m$ is a $ln$ .

v_{f} - v_{i} = - v_{e} [ln m]_{m_{0}}^{m_{f}} = - v_{e} (ln m_{f} - ln m_{0}) = v_{e} (ln m_{0} - ln m_{f})

Using the log rule $ln a - ln b = ln (a / b)$ :

Δ v = v_{e} ln (\frac{m _{0}}{m _{f}})

PICTURE. The area under the curve $- v_{e} / m$ from $m_{0}$ down to $m_{f}$ — that shaded area is $Δ v$ . Watch it grow slowly as $m_{f}$ shrinks.

$m_{0}$ — wet mass (full rocket).
$m_{f}$ — dry mass (empty rocket: structure + payload).
$ln (m_{0} / m_{f})$ — the log of the ratio $R = m_{0} / m_{f}$ , dimensionless.

Step 7 — Every edge case, drawn

PICTURE. The curve $Δ v = v_{e} ln R$ against mass ratio $R$ , with each special case marked.

No fuel burned, $m_{0} = m_{f}$ : then $R = 1$ , and $ln 1 = 0$ , so $Δ v = 0$ . ✔ You threw nothing, you gained nothing.
Just a whisker of fuel, $R$ slightly above 1: $ln R \approx R - 1$ (the curve starts as a straight line of slope 1). For tiny burns, $Δ v \approx v_{e} (R - 1)$ — it does look linear at first. The log's cruelty only shows up for big ratios.
Huge mass ratio, $R \to \infty$ (almost all fuel): $ln R$ grows, but agonisingly slowly. To double $Δ v$ you must square $R$ (since $ln R^{2} = 2 ln R$ ). This is the "tyranny of the rocket equation" — the reason for Multistage Rockets.
Impossible input, $m_{f} = 0$ : dry mass zero means no structure, no engine, nothing physical. $ln (m_{0} /0) = ln \infty = \infty$ : infinite $Δ v$ , which is nonsense — you can never eject literally all your mass. The equation correctly warns you by blowing up.
Bigger $v_{e}$ : the whole curve stretches vertically — more $Δ v$ for the same ratio. This is why engineers chase exhaust speed (see Specific Impulse (Isp), where $v_{e} = g_{0} I_{s p}$ ).

The one-picture summary

PICTURE. The entire derivation on one canvas: puff ejected → momentum ledger balances → $d v = - v_{e} d m / m$ → sum the slivers (area) → $Δ v = v_{e} ln (m_{0} / m_{f})$ .

Recall Feynman: retell the whole walkthrough

You're floating in space on a heavy sled full of bricks. (Step 0–1) Throw one brick backward and you drift forward a hair. (Step 2–4) The universe keeps a "motion account" (momentum) that must balance: your forward drift exactly pays for the brick's backward flight. (Step 5) Do the bookkeeping and a beautiful thing falls out — the speed you gain from one throw depends not on the brick's raw weight, but on that weight compared to how heavy you still are: $d v = - v_{e} d m / m$ . Throw a brick when you're heavy, small kick; throw it when you're nearly empty, big kick. (Step 6) Add up every throw from full sled to empty sled, and summing all those "fraction of yourself" cuts spells a logarithm. (Step 7) The verdict: $Δ v = v_{e} ln (m_{0} / m_{f})$ . Throwing bricks faster ( $v_{e}$ ) helps enormously; carrying more bricks helps only slowly, because those bricks are themselves heavy to haul — the log makes sure of it.

Connections

Conservation of Momentum — the ledger that balances in Steps 2–5.
Newton's Third Law — why throwing gas back pushes you forward (Step 0).
Natural Logarithm and Integration of 1/x — why summing $d m / m$ becomes $ln$ (Step 6).
Specific Impulse (Isp) — how $v_{e}$ is measured in practice.
Multistage Rockets — the engineering escape from Step 7's tyranny.
Thrust and Mass Flow Rate — the same physics written per-second: $F = v_{e} \overset{m}{˙}$ .

In the derivation, what does

d m / m

represent and why does it matter?

The fractional mass lost; summing it gives the logarithm.

Why is momentum (not energy) the right conserved quantity here?

Momentum is exactly conserved with no external force; exhaust energy is messy to track.

What happens to

Δ v

when

m_{0} = m_{f}

R = 1

ln 1 = 0

, so

Δ v = 0

— no fuel burned.

Why does

m_{f} = 0

give infinite

Δ v

, and is it physical?

ln (m_{0} /0) = \infty

; not physical — you can't eject all your mass.

To double

Δ v

at fixed

v_{e}

, what must happen to the mass ratio?

It must be squared, since

ln R^{2} = 2 ln R