Systems with variable mass — rocket equation derivation preview: Visual walkthrough

Step 1 — What is "momentum" and why we watch it

WHAT. We label everything in one fixed ground frame (a camera bolted to the launchpad) and read off the total momentum of everything in the picture.

WHY momentum and not force? Force needs a fixed mass to be simple. Momentum $p = m v$ works even while mass changes, because we can add up the $m v$ of every piece separately. That is the whole reason we track $p$ .

PICTURE. One arrow, the rocket, drifting right at speed $v$ . Its momentum is the blue bar underneath — length $= m v$ .

Figure — Systems with variable mass — rocket equation derivation preview

Step 2 — Draw the box around a fixed lump of stuff

WHAT. We refuse to track "the rocket" (its mass shrinks — cheating). Instead we draw a dashed box around the rocket and the tiny puff of fuel it is about to spit out. This box holds the same total matter for the whole instant $d t$ (a tiny slice of time).

WHY this box? Newton's honest law $F_{ext} = \frac{d p}{d t}$ is only allowed on a system whose membership doesn't change. Our dashed box qualifies: no matter enters or leaves the box; the fuel just moves from one part of the box to another.

PICTURE. The dashed box holds two things about to separate: the main rocket (mass $m + d m$ , where $d m < 0$ because it's losing mass) and a red puff (mass $- d m > 0$ ).

Step 3 — How fast does the puff actually move? (relative velocity)

WHAT. We pin down the puff's ground-frame velocity: $v - u$ .

WHY relative and not ground speed? Because $u$ is a fixed property of the engine (how hot and fast the nozzle throws gas, ~2000–4500 m/s), not something that changes as the rocket accelerates. Building the derivation on $u$ keeps it honest for every speed the rocket reaches.

PICTURE. Two arrows from the split point: the rocket now at $v + d v$ (a hair faster), the red puff at $v - u$ (backward relative to the ship, so slower or even negative in the ground frame).

Step 4 — Write the momentum a heartbeat later

WHAT. At time $t + d t$ the box holds two moving pieces. Add their momenta.

WHY add them? Momentum is additive: total $p$ of the box is just the sum of each piece's $m v$ . Since the box is closed, this sum is a legitimate thing to feed into $F_{ext} d t = d p$ .

PICTURE. Two stacked momentum bars — rocket bar $(m + d m) (v + d v)$ and puff bar $(- d m) (v - u)$ — whose total we will compare to the single bar from Step 1.

Step 5 — Expand, and watch two terms kill each other

WHAT. Multiply everything out:

$p (t + d t) = m v + m d v + v d m + d m d v - v d m + u d m$

WHY expand? To spot cancellation. Look at the two circled terms: $+ v d m$ (from the rocket) and $- v d m$ (from the puff). They are equal and opposite — the ground-frame speed $v$ of the departing mass cancels itself out.

WHY that matters. After cancellation the only speed left describing the exhaust is $u$ , the relative one. The picture proves the relative velocity is what does the pushing — not the ground speed.

PICTURE. The six terms laid in a row; the two $v d m$ terms drawn in green with a strike-through, and the tiny $d m d v$ term shaded out as "second-order, negligible".

After cancelling $v d m$ and dropping $d m d v$ :

$d p = p (t + d t) - p (t) = m d v + u d m$

Step 6 — Bring in the outside force → the master equation

WHAT. Newton's honest law on the closed box: $F_{ext} d t = d p$ . Substitute $d p = m d v + u d m$ and divide by $d t$ .

WHY $F = \overset{p}{˙}$ and not $F = ma$ ? $F = ma$ assumes fixed mass. Our box is fixed-mass by construction, so $F_{ext} = \frac{d p}{d t}$ is exactly legal here — this is the impulse–momentum theorem applied over the instant $d t$ .

PICTURE. A free-body sketch: one external arrow $F_{ext}$ on the box, the internal thrust arrow emerging as $- u \overset{m}{˙}$ , and the resulting acceleration $a$ on the rocket.

Step 7 — Turn off gravity and integrate → Tsiolkovsky

WHAT. In deep space $F_{ext} = 0$ , so $m d v = - u d m$ , i.e. $d v = - u \frac{d m}{m}$ . Sum every tiny gain from launch $(v_{0}, m_{0})$ to burnout $(v_{f}, m_{f})$ .

WHY a logarithm appears. Each kilogram burned pushes the remaining, ever-smaller mass. The gain per kg is $- u \frac{d m}{m}$ — a change divided by the current amount. Adding up "change over current amount" is exactly what a logarithm measures. That is why $\int \frac{d m}{m} = ln m$ .

PICTURE. The curve $\frac{1}{m}$ vs $m$ ; the shaded area from $m_{f}$ to $m_{0}$ equals $ln \frac{m _{0}}{m _{f}}$ , and each thin strip is one kilogram's ever-growing contribution as $m$ shrinks leftward.

Step 8 — The edge cases (never leave the reader stranded)

WHAT & WHY. Every degenerate input, drawn once so no scenario surprises you.

No fuel burned ( $m_{f} = m_{0}$ ): $ln \frac{m _{0}}{m _{0}} = ln 1 = 0 \Rightarrow Δ v = 0$ . Nothing thrown, no speed gained. ✔
Burn everything to a point ( $m_{f} \to 0$ ): $ln \frac{m _{0}}{m _{f}} \to + \infty$ . Infinite $Δ v$ — but impossible, since real rockets carry engines and structure that can't be burned. The math warns you the payoff diverges only in a fantasy.
Launch against gravity ( $F_{ext} = - m g$ ): master equation becomes $m \overset{v}{˙} = u ∣ \overset{m}{˙} ∣ - m g$ . If $u ∣ \overset{m}{˙} ∣ = m g$ , then $a = 0$ — the rocket hovers. Lift-off needs thrust > weight: $u ∣ \overset{m}{˙} ∣ > m_{0} g$ .
Slow burn / big mass ratio contrast: doubling $u$ doubles $Δ v$ linearly, but doubling $Δ v$ by fuel alone needs the mass ratio squared — the tyranny of the rocket equation.

PICTURE. Four mini-panels: $m_{f} = m_{0}$ (flat), $m_{f} \to 0$ (blow-up), hover balance, and lift-off with thrust arrow beating weight arrow.

The one-picture summary

The whole derivation on one canvas: box the closed lump → split into rocket + puff → the $v d m$ terms cancel → only relative $u$ survives → integrate $d m / m$ → logarithm.

Recall Feynman retelling of the whole walkthrough

Picture yourself on a frictionless skateboard holding bricks. First we agree to watch momentum — mass times speed — because it still adds up nicely even while you're getting lighter (Step 1). We draw a dashed box around you and the next brick together, so nothing sneaks in or out during one throw (Step 2). You throw the brick backward at some fixed speed relative to you — that's $u$ , a property of your arm, not of how fast you're already rolling (Step 3). We tally momentum a heartbeat later: you a bit faster, brick flying back (Step 4). When we multiply it all out, the part that depends on your ground speed cancels itself — proof that only the relative throw speed matters (Step 5). Feed that into Newton's law for the closed box and out pops the master equation: your acceleration equals outside forces plus a forward push $u ∣ \overset{m}{˙} ∣$ (Step 6). Add up every throw from full to empty; because each throw pushes a lighter you, the gains compound like interest — that's the logarithm, $Δ v = u ln (m_{0} / m_{f})$ (Step 7). Finally we check the corners: no bricks thrown = no speed; throwing all your mass = fantasy infinite speed; and to beat gravity your throwing must out-push your weight (Step 8).

Recall Quick self-test

In Step 5, which two terms cancel and what remains? ::: $+ v d m$ and $- v d m$ cancel; only the relative-speed term $u d m$ survives. Why integrate $d m / m$ instead of $d m$ ? ::: Each kg pushes the ever-smaller remaining mass, so gain per kg is (change ÷ current amount) — summing that gives a logarithm. What does the master equation reduce to in deep space? ::: $m d v = - u d m$ , which integrates to $Δ v = u ln (m_{0} / m_{f})$ . Hover condition when launching? ::: $u ∣ \overset{m}{˙} ∣ = m g$ gives $a = 0$ ; lift-off needs $u ∣ \overset{m}{˙} ∣ > m_{0} g$ .