Tsiolkovsky rocket equation — full first-principles derivation from momentum: Visual walkthrough

We work in one dimension the whole time: everything moves along a horizontal line, "right" is positive, "left" is negative. That's it — no angles, no triangles, just a number line with arrows on it.

Step 0 — Meeting the characters (before any physics)

WHAT. Three quantities run this whole story. Let me name them by pointing at a picture, not by formula.

$m$ — the mass of the rocket right now (the rocket plus every drop of fuel still inside it). Measured in kilograms.
$v$ — the velocity of the rocket right now, measured from the ground. A number on our line: positive = moving right.
$v_{e}$ — the effective exhaust velocity: how fast the gas shoots out as seen by an astronaut riding the rocket. Always a fixed positive number, set by the engine nozzle. Crucially — measured relative to the rocket, not the ground.

WHY these three and nothing else. Momentum is "mass times velocity". To do momentum bookkeeping we only ever need how much stuff and how fast it goes. Those are $m$ and $v$ . The engine adds one fact of its own — the throw-speed $v_{e}$ . Three numbers, whole problem.

PICTURE. The rocket sits on the number line moving right at $v$ ; a little nozzle at the back is where gas will come out at $v_{e}$ relative to the ship.

Figure — Tsiolkovsky rocket equation — full first-principles derivation from momentum

Step 1 — Freeze the system NOW and count its momentum

WHAT. Take a snapshot at time $t$ . Everything — the rocket body and the little puff of fuel that is about to be fired — is still one lump, mass $m$ , moving together at speed $v$ .

$p (t) = whole mass, kg m \cdot its speed, m/s v$

WHY. We must count the whole system before anything is thrown, so we have something to compare against afterward. The fuel that's about to leave is still inside right now, so it rides at $v$ and counts as part of $m v$ . Forgetting this is the classic slip — you cannot let mass "appear from nowhere" a moment later.

PICTURE. One black arrow of momentum $p (t) = m v$ pointing right. That's our "before" total; hold onto it.

Step 2 — Let a tiny time $d t$ pass: the split

WHAT. A tiny sliver of time $d t$ later, the rocket has fired one small puff of gas out the back. The single lump has become two objects:

The rocket, now slightly lighter and slightly faster.
The gas puff, now flying backward.

Let $d m$ be the change in the rocket's mass. The rocket got lighter, so $d m$ is negative. The mass that left is therefore $- d m$ , a positive amount.

$rocket mass = m + d m (smaller, since d m < 0)$ $gas mass = - d m (positive)$

WHY the negative sign. $m$ means the rocket's mass, and the rocket is losing weight. When a quantity shrinks, its change is negative — that's just what "change" means. So $d m < 0$ , and the ejected chunk is its mirror, $- d m > 0$ . Getting this sign right is what makes the final answer come out positive instead of upside-down.

PICTURE. The "before" lump on top; the "after" split into rocket (still going right) and a small gas blob heading the other way.

Step 3 — How fast does the gas move in the GROUND frame?

WHAT. The engine throws the gas at $v_{e}$ relative to the rocket, pointing backward. But our momentum ledger is written from the ground. So we need the gas's ground speed.

$v_{gas, ground} = rocket’s ground speed v - backward throw speed v_{e} = v - v_{e}$

WHY subtract. Imagine standing on the rocket: you see the gas leave at $v_{e}$ to the left. Now step onto the ground, where the rocket itself is drifting right at $v$ . To convert "speed seen from the rocket" into "speed seen from the ground," you add the rocket's own velocity. The gas's rocket-frame velocity is $- v_{e}$ (leftward), so its ground velocity is $v + (- v_{e}) = v - v_{e}$ . This is why we insisted $v_{e}$ is a relative speed back in Step 0 — it's what keeps the two frames straight.

PICTURE. A little velocity-addition diagram: on the rocket the gas reads $- v_{e}$ ; add the rocket's own $+ v$ ; the ground meter reads $v - v_{e}$ .

Step 4 — Count the momentum AFTER the split

WHAT. Total "after" momentum = rocket's new momentum + gas's momentum.

$p (t + d t) = new rocket mass (m + d m) new rocket speed (v + d v) + gas mass (- d m) gas ground speed (v - v_{e})$

Here $d v$ is the tiny speed the rocket gained in time $d t$ — the very thing we're hunting for.

WHY. Momentum is additive: when a system splits, the total is just the sum over the pieces. We built each piece in Steps 2 and 3 — now we simply write "rocket part plus gas part."

PICTURE. Two momentum arrows side by side: a big black one for the rocket (rightward), a small one for the gas (leftward), their tail-to-tip sum equal to the single "before" arrow from Step 2.

Step 5 — Expand, and watch terms cancel

WHAT. Multiply everything out:

$p (t + d t) = m v + m d v + v d m + tiny \times tiny d m d v - v d m + v_{e} d m$

Two things happen:

$+ v d m$ and $- v d m$ cancel exactly.
$d m d v$ is a tiny number times a tiny number — negligibly small, so we drop it.

$\Rightarrow p (t + d t) = m v + m d v + v_{e} d m$

WHY drop $d m d v$ . Both $d m$ and $d v$ are "as small as we like." Their product is doubly small — like $0.001 \times 0.001 = 0.000001$ . As we shrink $d t$ toward zero, this term dies far faster than the others, so it contributes nothing in the limit. This is the standard move of calculus: keep first-order, discard second-order.

PICTURE. The six expanded terms laid out; the two $v d m$ terms struck through (they annihilate) and the $d m d v$ term greyed out and crossed (too small to matter). What survives is highlighted.

Step 6 — Apply the one law: momentum is conserved

WHAT. No gravity, no drag → no outside push → total momentum can't change:

$p (t + d t) = p (t)$ $m v + m d v + v_{e} d m = m v$

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

$0 = m d v + v_{e} d m$

WHY. This is Newton's law in momentum form: $F_{e x t} = d p / d t$ . With $F_{e x t} = 0$ the momentum simply does not move — before equals after. Everything from here on is pure calculus; all the physics lives in this one line, which is really Conservation of Momentum in disguise.

PICTURE. The "before" arrow and "after" arrow drawn identical length, an equals sign between — momentum unchanged — and the leftover balance $m d v = - v_{e} d m$ shown as a see-saw.

Step 7 — Add up all the tiny puffs (integrate)

WHAT. Rearrange the balance to isolate one speed-gain:

$d v = - v_{e} \frac{d m}{m}$

Each puff gives a sliver of speed $d v$ . To get the total $Δ v$ , add up every sliver from launch ( $m = m_{0}$ , $v = v_{i}$ ) to burnout ( $m = m_{f}$ , $v = v_{f}$ ). "Adding up infinitely many slivers" is exactly what the integral $\int$ does:

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

WHY a logarithm shows up. We need the running-total of $\frac{d m}{m}$ — a fractional change. The function whose small changes look like "change divided by current value" is the natural logarithm: $\int \frac{d m}{m} = ln m$ . That is not a coincidence we impose; it's forced by the balance in Step 6. The log is the answer to the question "what adds up fractional changes?"

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

Using $ln A - ln B = ln (A / B)$ and flipping the fraction to kill the minus sign (legal because $m_{0} > m_{f}$ ):

PICTURE. The area under the curve $1/ m$ from $m_{f}$ to $m_{0}$ shaded — that shaded area is $ln (m_{0} / m_{f})$ , and multiplying it by $v_{e}$ gives $Δ v$ .

This same balance, read per second instead of totalled, becomes thrust $F = v_{e} \overset{m}{˙}$ .

Step 8 — The edge cases (so you never hit a wall)

WHAT & WHY — four corners of the formula:

Burn no fuel: $m_{f} = m_{0} \Rightarrow R = 1 \Rightarrow ln 1 = 0 \Rightarrow Δ v = 0$ . No throw, no push. ✔ Sanity holds.
Burn everything (impossible ideal): $m_{f} \to 0 \Rightarrow R \to \infty \Rightarrow ln R \to \infty$ . The formula promises unlimited speed — but only if the rocket could weigh nothing at the end. It can't; there's always structure left. This is why $m_{f}$ never truly reaches zero.
Can $Δ v$ beat $v_{e}$ ? Yes! As soon as $R > e \approx 2.718$ , $ln R > 1$ , so $Δ v > v_{e}$ . The rocket outruns its own exhaust because it gets pushed again and again.
Doubling the fuel share: going $R \to 2 R$ adds a fixed $v_{e} ln 2$ each time, never doubling $Δ v$ . Brutal diminishing returns — the reason for Multistage Rockets.

PICTURE. The curve $Δ v = v_{e} ln R$ : flat-zero at $R = 1$ , crossing the $v_{e}$ line at $R = e$ , and rising ever more slowly — with the equal-height "staircase" showing each doubling of $R$ adds the same step.

The one-picture summary

PICTURE. The whole derivation on a single canvas: (1) the intact rocket at $v$ , (2) the split into lighter rocket $(v + d v)$ and backward gas $(v - v_{e})$ , (3) the balance $m d v = - v_{e} d m$ , (4) the integral sign gathering slivers, (5) the boxed result $Δ v = v_{e} ln (m_{0} / m_{f})$ — an arrow flowing left-to-right through all five.

Recall Feynman retelling of the whole walkthrough

You're floating in space holding a bag of baseballs — total weight is your mass $m$ , and you're gliding at speed $v$ . Snapshot one: you and every ball move together; that's your momentum, mass times speed. Now throw one ball backward. You split into two: a slightly lighter you nudged forward, and one ball flying back. From the ground the ball's speed is your speed minus how hard you threw it, $v - v_{e}$ . Count the momentum after the throw and set it equal to before — because space has nothing to push on, the total can't change. Multiply it all out and the messy middle bits either cancel or are too small to keep; what's left is a clean deal: the speed you gain equals the throw-speed times the fraction of your mass you just tossed. Do that over and over, adding up every tiny fraction — and "adding up fractions of your shrinking mass" is precisely what a logarithm counts. So your total speed-up is throw-speed times the log of (start weight over end weight). Throw nothing, gain nothing. Throw almost everything, and you can even scream past the speed of your own baseballs — because each throw shoves a lighter-and-lighter you a little harder.

Recall

In one sentence, where does the logarithm come from? ::: From integrating $d m / m$ — the fractional mass change — and $\int d m / m = ln m$ . Why is the gas's ground velocity $v - v_{e}$ and not $- v_{e}$ ? ::: Because $v_{e}$ is measured relative to the moving rocket; adding the rocket's own ground velocity $v$ gives $v - v_{e}$ . At what mass ratio does $Δ v$ first exceed $v_{e}$ ? ::: At $R = e \approx 2.718$ , where $ln R = 1$ . What single physical law powers the entire derivation? ::: Conservation of momentum (zero external force in free space).