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

Every problem uses the master equation from the parent note: $Δ v = v_{e} ln (\frac{m _{0}}{m _{f}}), R = \frac{m _{0}}{m _{f}}$ where $v_{e}$ = exhaust speed relative to the rocket, $m_{0}$ = initial mass, $m_{f}$ = final mass, $R$ = mass ratio. If any of those words feel shaky, re-read the parent before continuing. We also use $ln$ = the natural logarithm (the inverse of $e^{x}$ : "what power of $e$ gives this number?") and its inverse $e^{x}$ (see Conservation of Momentum and Newton's Second Law (momentum form) for the physics underneath).

A quick numerical anchor you will reuse everywhere:

$ln 2 = 0.6931$
$ln 4 = 2 ln 2 = 1.3863$
$ln 10 = 2.3026$
$e^{1} = 2.7183$

Level 1 — Recognition

Can you read the equation and identify its pieces?

Recall Solution L1·1

Wrong. $m_{f}$ is the final total mass (dry rocket plus any leftover fuel), not the fuel burned.

$Δ v$ = the change in the rocket's speed (in $m/s$ ).
$v_{e}$ = exhaust speed relative to the rocket (constant).
$m_{0}$ = total mass at the start (structure + payload + all propellant).
$m_{f}$ = total mass at the end (structure + payload + unburned propellant). The fuel burned is $m_{0} - m_{f}$ , a different quantity.

Recall Solution L1·2

$R = \frac{m _{0}}{m _{f}} = \frac{12 , 000}{3000} = 4.$ Why it's dimensionless: both masses are in kg, so the units cancel — $R$ is a pure number.

Recall Solution L1·3

$Δ v = v_{e} ln R = 2500 \times ln 4 = 2500 \times 1.3863 = 3466 m/s \approx 3.47 km/s .$

Level 2 — Application

Can you run the equation forwards and backwards?

Recall Solution L2·1

$R = \frac{50 , 000}{10 , 000} = 5, ln 5 = 1.6094.$ $Δ v = 3200 \times 1.6094 = 5150 m/s \approx 5.15 km/s .$

Recall Solution L2·2

Rearrange by undoing the log with $e^{x}$ (that's why we use the exponential here — it is the exact inverse of $ln$ ): $\frac{Δ v}{v _{e}} = ln R \Rightarrow R = e^{Δ v / v_{e}} = e^{6000/4000} = e^{1.5} = 4.482.$ So the rocket must be about $4.48 \times$ heavier at launch than at burnout.

Recall Solution L2·3

Propellant fraction is $f = \frac{m _{0} - m _{f}}{m _{0}} = 1 - \frac{m _{f}}{m _{0}} = 1 - \frac{1}{R} = 1 - \frac{1}{4.482} = 0.7769.$ About 77.7% of the launch mass must be fuel.

Recall Solution L2·4

From the parent note, $F_{t h r u s t} = v_{e} \overset{m}{˙}$ : $F = 2800 \times 250 = 700, 000 N = 700 kN .$

Level 3 — Analysis

Can you reason about how the pieces respond to change?

Recall Solution L3·1

Compare mass ratio $R$ with $2 R$ : $Δ v (2 R) - Δ v (R) = v_{e} ln (2 R) - v_{e} ln R = v_{e} [ln 2 + ln R - ln R] = v_{e} ln 2.$ The $ln R$ terms cancel, leaving a constant $v_{e} ln 2$ independent of $R$ . This is the log's fingerprint: multiplying the input adds a fixed amount to the output. $v_{e} ln 2 = 3000 \times 0.6931 = 2079 m/s \approx 2.08 km/s .$ Look at Figure s01: the orange curve flattens — the same horizontal doubling gives ever-smaller-looking (but numerically equal) vertical steps.

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

Recall Solution L3·2

We need $Δ v > v_{e}$ , i.e. $v_{e} ln R > v_{e} \Rightarrow ln R > 1 \Rightarrow R > e \approx 2.718.$ So any mass ratio above $\approx 2.72$ makes the rocket faster than its own exhaust speed. It works because the rocket receives many repeated pushes, each on an ever-lighter body — the pushes accumulate past $v_{e}$ .

Recall Solution L3·3

Baseline: $Δ v = 3000 ln 4 = 3000 (1.3863) = 4159 m/s$ . (a) $v_{e} \to 3300$ : $Δ v = 3300 \times 1.3863 = 4575 m/s$ . Gain $= 416 m/s$ (a flat 10% because $v_{e}$ multiplies). (b) $R \to 4.4$ : $ln 4.4 = 1.4816$ , so $Δ v = 3000 \times 1.4816 = 4445 m/s$ . Gain $= 286 m/s$ . Raising $v_{e}$ wins. Why: $v_{e}$ scales $Δ v$ linearly (it sits outside the log), but $R$ only enters through the slowly-growing log. This is exactly why engineers chase higher Specific Impulse (better $v_{e}$ ) so aggressively.

Level 4 — Synthesis

Can you combine the equation with staging, thrust, and losses?

Recall Solution L4·1

(a) $Δ v = 3000 ln 16 = 3000 (2.7726) = 8318 m/s$ . (b) Stage burns add up: $Δ v = 3000 ln 4 + 3000 ln 4 = 3000 (ln 4 + ln 4) = 3000 ln (4 \times 4) = 3000 ln 16 = 8318 m/s$ . They tie — if the mass ratios truly multiply to the same total. The real advantage of staging is that dropping empty tanks lets each stage achieve a high ratio it could never reach while dragging dead structural mass, so real multistage rockets get a larger effective total $R$ than a single stage ever could. Here we held $R_{tot}$ fixed by construction, so the log's additivity shows the tie.

Recall Solution L4·2

The real velocity gain is the ideal minus the gravity loss term: $Δ v_{actual} = Δ v_{ideal} - g t = 6000 - (9.8) (120) = 6000 - 1176 = 4824 m/s .$ About 4.82 km/s. Gravity alone ate nearly 1.18 km/s — this is why launches pitch over early to spend less time fighting straight-up.

Recall Solution L4·3

(i) Burn time = propellant divided by flow rate: $t_{b} = \frac{M _{p}}{m ˙} = \frac{18 , 000}{300} = 60 s .$ (ii) Thrust $= v_{e} \overset{m}{˙} = 3000 \times 300 = 900, 000 N = 900 kN .$ (iii) Initial mass $m_{0} = m_{f} + M_{p} = 6000 + 18, 000 = 24, 000 kg$ , so $R = 24, 000/6000 = 4$ : $Δ v = 3000 ln 4 = 3000 (1.3863) = 4159 m/s \approx 4.16 km/s .$

Level 5 — Mastery

Can you derive, invert, and reason at the edge cases?

Recall Solution L5·1

Rearrange to isolate the fractional mass change: $d v = - v_{e} \frac{d m}{m} .$ Integrate from $(v_{i}, m_{0})$ to $(v_{f}, m_{f})$ : $\int_{v_{i}}^{v_{f}} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m} .$ The right integrand $1/ m$ is the only function whose antiderivative is $ln m$ — that is why the log is forced upon us; nothing else was chosen. Thus $v_{f} - v_{i} = - v_{e} (ln m_{f} - ln m_{0}) = v_{e} ln \frac{m _{0}}{m _{f}},$ where the minus sign flipped the ratio right-side up (since $m_{0} > m_{f}$ , the log stays positive). See Variable Mass Systems for the general framework.

Recall Solution L5·2

Required ratio: $R = e^{9400/3000} = e^{3.1333} = 22.95$ . Let launch mass $m_{0} = 1$ (normalise). Then $m_{f} = 1/ R = 0.04357$ . Split propellant $M_{p}$ and structure $M_{s} = ε M_{p}$ . Burned mass $= M_{p} = m_{0} - m_{f} - (payload)$ ... work it cleanly: propellant burned $= m_{0} - m_{f} = 1 - 0.04357 = 0.95643$ only if payload $\to 0$ and structure counts inside $m_{f}$ . Structure required $= ε M_{p} = 0.08 \times 0.95643 = 0.07651$ . But $m_{f} = 0.04357$ must hold all the leftover mass (structure + payload). Since required structure $0.07651 > m_{f} = 0.04357$ , the structure alone won't fit inside the allowed final mass. Conclusion: no payload is possible — the design is impossible as a single stage. This is the classic "tyranny of the rocket equation," and the reason Multistage Rockets exist.

Recall Solution L5·3

(a) $m_{f} = m_{0} \Rightarrow R = 1 \Rightarrow ln 1 = 0 \Rightarrow Δ v = 0.$ Sensible: burn no fuel, gain no speed. (b) $m_{f} \to 0 \Rightarrow R \to \infty \Rightarrow ln R \to \infty \Rightarrow Δ v \to \infty.$ Mathematically unbounded — but physically impossible because real rockets have irreducible structural mass, so $m_{f}$ never reaches 0. The log's slow growth also means even huge $R$ gives only modestly larger $Δ v$ (Figure s01). (c) $v_{e} = 0 \Rightarrow Δ v = 0$ for any $R$ . If gas leaves at zero relative speed it carries no momentum away, so by Conservation of Momentum the rocket gains nothing. Both levers ( $v_{e}$ and $R$ ) must be non-trivial.

Recall Solution L5·4

$Δ v_{tot} = v_{e} ln R_{1} + v_{e} ln R_{2} = v_{e} ln (R_{1} R_{2}) = v_{e} ln 9.$ $= 3500 \times 2.1972 = 7690 m/s \approx 7.69 km/s,$ the same for any split with $R_{1} R_{2} = 9$ . What breaks it in reality: each stage carries its own structural mass and (for lower stages) the dead weight of the stages above it, so the achievable $R_{i}$ per stage is not free — optimal staging chooses splits that balance these structural penalties. The clean additivity here is the idealised skeleton.

Recall One-line self-check

Why do stage $Δ v$ 's add while stage mass ratios multiply? ::: Because $ln (R_{1} R_{2}) = ln R_{1} + ln R_{2}$ — the log turns the product of ratios into a sum of velocity contributions.