Systems with variable mass — rocket equation derivation preview: Worked examples

The scenario matrix

Every variable-mass problem is one cell of this grid. The columns ask what is happening to the mass and the outside world; the rows ask what quantity you want.

Cell	Sign of $\overset{m}{˙}$	External force $F_{ext}$	What you compute	Example
A Deep-space $Δ v$	$\overset{m}{˙} < 0$ (loses)	$0$	Burnout speed via $ln$	Ex 1
B Thrust + accel now	$\overset{m}{˙} < 0$	$0$	$F_{thrust}$ , instantaneous $a$	Ex 2
C Launch vs gravity	$\overset{m}{˙} < 0$	$- m g$	Lift-off / hover condition	Ex 3
D Mass gained	$\overset{m}{˙} > 0$ (accretes)	given	Force to keep it moving	Ex 4
E Degenerate $u = 0$	$\overset{m}{˙} < 0$	$0$	Shows no thrust	Ex 5
F Degenerate $m_{f} \to m_{0}$	$\overset{m}{˙} < 0$	$0$	$Δ v \to 0$ limit	Ex 6
G Mass-ratio design (inverse)	$\overset{m}{˙} < 0$	$0$	Solve for required $m_{0} / m_{f}$	Ex 7
H Word problem / exam twist	mixed	mixed	Multi-step reasoning	Ex 8

The two "sign of $\overset{m}{˙}$ " cases (loses vs gains) are the deep split. Everything else is which force you add or which unknown you solve for.

Figure — Systems with variable mass — rocket equation derivation preview

Cell A — Deep-space burnout speed

Worked example Ex 1 (Cell A) — the pure Tsiolkovsky case

A probe at rest in deep space: $m_{0} = 5000$ kg, burns to $m_{f} = 2000$ kg, exhaust speed $u = 3000$ m/s. Find its final speed.

Forecast: guess before computing — will it be more or less than $u = 3000$ m/s? (The log usually punishes you; jot a number.)

No external force, so use $Δ v = u ln \frac{m _{0}}{m _{f}}$ . Why this step? $F_{ext} = 0$ collapses the master equation straight to Tsiolkovsky — no gravity term to carry.
Mass ratio: $\frac{m _{0}}{m _{f}} = \frac{5000}{2000} = 2.5$ . Why this step? Everything the equation needs is squeezed into this single dimensionless number — the fuel story reduced to one ratio.
$Δ v = 3000 ln 2.5 = 3000 \times 0.9163 = 2748.9$ m/s. Why this step? Started at rest, so $v_{f} = Δ v$ . The $ln$ turns "burned 60% of the mass" into "reached only $0.92 u$ ."

Verify: $ln 2.5 < 1$ , so $Δ v < u$ — matches "the log is unkind." Units: $(m/s) \times (dimensionless) = m/s$ . ✓

Cell B — Thrust and instantaneous acceleration

Worked example Ex 2 (Cell B) — force and

$a$ at one instant The Ex 1 probe burns at $∣ \overset{m}{˙} ∣ = 40$ kg/s. Find the thrust, and its acceleration at the moment its mass is $3000$ kg.

Forecast: thrust is constant during the burn, but is $a$ constant? Guess yes/no.

$F_{thrust} = u ∣ \overset{m}{˙} ∣ = 3000 \times 40 = 120000$ N $= 1.2 \times 1 0^{5}$ N. Why this step? Thrust depends only on how fast and how much you throw — $u$ and $∣ \overset{m}{˙} ∣$ — not on how much rocket is left.
Deep space $\Rightarrow F_{ext} = 0$ , so $a = \frac{F _{thrust}}{m}$ with the current mass. Why this step? Newton's law uses the mass you have right now, and that is shrinking.
At $m = 3000$ kg: $a = \frac{1.2 \times 1 0 ^{5}}{3000} = 40$ m/s $^{2}$ . Why this step? Same thrust, smaller mass ⇒ bigger $a$ . So $a$ is not constant — it climbs as fuel burns.

Verify: at start ( $m = 5000$ ) $a = 1.2 \times 1 0^{5} /5000 = 24$ m/s $^{2}$ $< 40$ m/s $^{2}$ . Acceleration rose as mass fell — consistent. Units: $N/kg = m/s^{2}$ . ✓

Cell C — Launch against gravity

Worked example Ex 3 (Cell C) — will it even lift off?

A rocket on the pad: $m_{0} = 4000$ kg, $u = 2000$ m/s, burn rate $∣ \overset{m}{˙} ∣ = 25$ kg/s. Take $g = 10$ m/s $^{2}$ . Does it lift off? If not, what burn rate would?

Forecast: compute weight in your head first — is thrust bigger?

Master equation with gravity: $m \frac{d v}{d t} = u ∣ \overset{m}{˙} ∣ - m g$ . Why this step? Now $F_{ext} = - m g$ (weight points down), so we keep that term instead of dropping it.
Thrust: $u ∣ \overset{m}{˙} ∣ = 2000 \times 25 = 50000$ N. Weight: $m_{0} g = 4000 \times 10 = 40000$ N. Why this step? Lift-off needs the upward push to beat the downward weight at the heaviest moment — the start.
$50000 > 40000$ , so it lifts. Initial acceleration $a = \frac{50000 - 40000}{4000} = 2.5$ m/s $^{2}$ . Why this step? Net upward force divided by current mass gives the rocket's launch acceleration.
Hover burn rate (thrust = weight): $u ∣ \overset{m}{˙} ∣_{hover} = m_{0} g \Rightarrow ∣ \overset{m}{˙} ∣_{hover} = \frac{40000}{2000} = 20$ kg/s. Why this step? Setting $a = 0$ isolates the break-even burn rate below which it never leaves the pad.

Verify: thrust-to-weight $= 50000/40000 = 1.25 > 1$ ⇒ lift-off, matches parent's condition $u ∣ \overset{m}{˙} ∣ > m_{0} g$ . At $∣ \overset{m}{˙} ∣ = 20$ kg/s thrust $= 40000 =$ weight ⇒ $a = 0$ hover. ✓

Cell D — Mass being GAINED (accretion / loading)

This is the mirror image: now $\overset{m}{˙} > 0$ .

Worked example Ex 4 (Cell D) — conveyor belt loading

Sand drops vertically onto a horizontal belt at rate $\frac{d m}{d t} = + 10$ kg/s. The belt moves at constant $v = 3$ m/s. What horizontal force keeps it going at that speed?

Forecast: the belt runs at constant speed — is the required force zero (since $d v = 0$ )? Guess.

Master equation: $m \frac{d v}{d t} = F_{ext} - u \frac{d m}{d t}$ . Here the incoming sand has zero horizontal velocity, and it joins matter moving at $v$ , so its speed relative to the belt is $u = v = 3$ m/s (the belt must drag it up from $0$ to $3$ ). Why this step? $u$ is always the relative speed between the transferred mass and the body. Falling sand starts with no forward speed, so relative to the belt it lags by $v$ .
Constant speed ⇒ $\frac{d v}{d t} = 0$ , so $0 = F_{ext} - u \frac{d m}{d t}$ , giving $F_{ext} = u \frac{d m}{d t}$ . Why this step? With no acceleration, the applied force does one job only: accelerate the newly added mass up to belt speed.
$F_{ext} = v \frac{d m}{d t} = 3 \times 10 = 30$ N. Why this step? Every second you must give $10$ kg an extra $3$ m/s; the force delivers that momentum flow.

Verify: momentum flux check: $\frac{d p}{d t} = v \frac{d m}{d t} = 3 \times 10 = 30$ N. Because $\overset{m}{˙} > 0$ , the mass term is a drag — you push forward even at steady speed. Units: $(m/s) (kg/s) = kg\cdotm/s^{2} = N$ . ✓

Cell E — Degenerate: $u = 0$

Worked example Ex 5 (Cell E) — dropping mass with zero relative speed

A cart of mass $m_{0} = 200$ kg glides in deep space at $v = 6$ m/s. It gently releases mass (lets it fall away with zero speed relative to the cart, i.e. $u = 0$ ), losing $80$ kg. Final cart speed?

Forecast: does dumping a third of its mass speed it up? Guess before reading.

Thrust $= - u \frac{d m}{d t}$ with $u = 0$ ⇒ thrust $= 0$ . Why this step? This is exactly the parent's first mistake made deliberately: dropping mass at the cart's own velocity produces no push.
With no external force and no thrust, $\frac{d v}{d t} = 0$ : speed unchanged. Why this step? Momentum of the closed system was already carried at $v$ ; splitting it into two pieces both moving at $v$ changes nobody's velocity.
$v_{f} = 6$ m/s. (Or by Tsiolkovsky: $Δ v = 0 \cdot ln (200/120) = 0$ .) Why this step? Setting $u = 0$ zeroes the whole Tsiolkovsky expression — the equation itself confirms "no relative speed, no gain."

Verify: $Δ v = 0 \times ln (200/120) = 0$ , so $v_{f} = 6$ m/s. This is the degenerate case that exposes why $u$ (not $v$ ) drives everything. ✓

Cell F — Degenerate: $m_{f} \to m_{0}$ (almost no fuel)

Worked example Ex 6 (Cell F) — the tiny-burn limit

A satellite $m_{0} = 1000$ kg fires a thruster with $u = 3000$ m/s but ejects only $m_{ej} = 5$ kg, so $m_{f} = 995$ kg. Find $Δ v$ , and compare to the crude estimate " $u \times (fraction of mass ejected)$ ."

Forecast: for a small burn, will $ln (m_{0} / m_{f})$ be close to the simple fraction $m_{ej} / m_{0}$ ? Guess.

Exact: $Δ v = u ln \frac{m _{0}}{m _{f}} = 3000 ln \frac{1000}{995} = 3000 ln (1.005025)$ . Why this step? Even a tiny burn uses the same log — no separate "small" formula needed.
$ln (1.005025) = 0.0050126$ , so $Δ v = 15.04$ m/s. Why this step? Plugging the near-1 ratio through the log gives the honest answer.
Crude estimate: $u \cdot \frac{m _{ej}}{m _{0}} = 3000 \times \frac{5}{1000} = 15.0$ m/s. Why this step? For small $x$ , $ln (1 + x) \approx x$ (see Logarithms and Exponential Growth); here $x = m_{ej} / m_{f}$ is tiny, so log ≈ fraction.

Verify: exact $15.04$ vs crude $15.0$ — agree to $< 0.3%$ . As $m_{f} \to m_{0}$ , $ln \to 0$ and $Δ v \to 0$ : the limiting behaviour is smooth, no blow-up. ✓

Cell G — Inverse design: solve for the mass ratio

Worked example Ex 7 (Cell G) — how much fuel do I need?

A mission demands $Δ v = 6000$ m/s with an engine of $u = 2500$ m/s in deep space. What mass ratio $m_{0} / m_{f}$ is required? If the dry (empty) mass is $m_{f} = 1500$ kg, what is the fuelled mass?

Forecast: $Δ v$ is more than twice $u$ — will the mass ratio be around 2, or much bigger? Guess.

Invert Tsiolkovsky: $Δ v = u ln \frac{m _{0}}{m _{f}} \Rightarrow \frac{m _{0}}{m _{f}} = e^{Δ v / u}$ . Why this step? To undo a $ln$ you apply its inverse, the exponential $e^{(\cdot)}$ — see Logarithms and Exponential Growth. This is where the "tyranny" becomes visible.
$\frac{Δ v}{u} = \frac{6000}{2500} = 2.4$ , so $\frac{m _{0}}{m _{f}} = e^{2.4} = 11.02$ . Why this step? The exponential blows a modest ratio $Δ v / u = 2.4$ up into an 11-fold mass ratio — exponentially expensive fuel.
Fuelled mass: $m_{0} = 11.02 \times 1500 = 16530$ kg. Why this step? Multiply the dry mass by the required ratio to get the launch mass; over $15000$ kg of that is fuel.

Verify: back-substitute: $2500 ln (16530/1500) = 2500 ln 11.02 = 2500 \times 2.3997 = 5999$ m/s $\approx 6000$ . ✓ Ratio $> 1$ , mass positive. Units consistent. ✓

Cell H — Exam-style word twist

Worked example Ex 8 (Cell H) — the two-stage trap

A student claims: "A single-stage rocket with $u = 2500$ m/s can never reach $Δ v = 9000$ m/s because that needs mass ratio $e^{3.6} \approx 36.6$ , which is unbuildable." Then they split it into two stages, each giving $Δ v_{1} = Δ v_{2} = 4500$ m/s, and claim the total mass ratio needed is smaller. Check both claims. Do the stage $Δ v$ 's add?

Forecast: do the two $4500$ 's add to $9000$ ? And is the split really cheaper? Guess both.

Single stage: $m_{0} / m_{f} = e^{9000/2500} = e^{3.6} = 36.6$ . Why this step? Same inverse-Tsiolkovsky as Ex 7 — confirms the "unbuildable" ratio.
$Δ v$ 's add because velocity is additive across sequential burns: $Δ v_{total} = Δ v_{1} + Δ v_{2} = 4500 + 4500 = 9000$ m/s. ✓ Why this step? Each stage starts where the last left off; speeds pile on linearly (unlike the mass, which multiplies).

Each stage ratio: $e^{4500/2500} = e^{1.8} = 6.05$ . Two identical stages ⇒ product of ratios $= 6.05 \times 6.05 = 36.6$ .

Why this step? Mass ratios multiply when stages are chained, and $6.05 \times 6.05 = e^{1.8} \cdot e^{1.8} = e^{3.6} = 36.6$ — the same total. 4. So staging does not reduce the ideal mass ratio; its real benefit is throwing away dead structural mass between burns (not captured in this idealised sum). Why this step? The exam trap is thinking the log "helps" by splitting; algebraically $e^{a} e^{b} = e^{a + b}$ , so the ideal cost is identical. The genuine saving is dropping empty tanks.

Verify: $6.05 \times 6.05 = 36.6 = e^{3.6}$ , and $ln (36.6) \times 2500 = 3.6 \times 2500 = 9000$ m/s. Claims: " $Δ v$ add" ✓ true; "split is ideally cheaper" ✗ false. ✓

Recall Quick self-test on the matrix

Which cell has zero thrust despite mass leaving? ::: Cell E ( $u = 0$ ) — no relative speed, no push. When does the mass term act as a drag instead of thrust? ::: Cell D — when $\overset{m}{˙} > 0$ (mass arriving). How do you go from a required $Δ v$ to a mass ratio? ::: Take $e^{Δ v / u}$ — the inverse of the $ln$ . Do chained stages reduce the ideal total mass ratio? ::: No — ratios multiply, $e^{a} e^{b} = e^{a + b}$ ; only shedding dead mass helps.

Back to the parent topic · related: Conservation of Linear Momentum, Impulse–Momentum Theorem, Relative Velocity, Logarithms and Exponential Growth.