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

This page is a drill floor. The parent derivation proved the formula once. Here we throw every kind of question at it — forward, backward, degenerate, real-world, and exam-trap — so you never meet a scenario you have not already seen.

Everything rests on one equation you must be able to read three ways:

Before any symbol is used, recall the plain-word meaning: $v_{e}$ = speed of exhaust gas relative to the rocket (metres per second); $m_{0}$ = mass at the start of the burn; $m_{f}$ = mass at the end; $Δ v$ = the change in the rocket's own speed. The natural log $ln$ is the question "e to the what gives this number?" — it is the exact inverse of $e^{x}$ , which is why Faces 1 and 2 undo each other.

The scenario matrix

Every question this topic can ask lives in one of these cells. Below the table, Examples 1–8 each stamp the cell(s) they cover.

Cell	What makes it different	Covered by
A. Forward, basic	given $m_{0}, m_{f}, v_{e}$ → find $Δ v$	Ex 1
B. Backward (mass ratio)	given $Δ v, v_{e}$ → find $R = m_{0} / m_{f}$ or fuel fraction	Ex 2
C. Backward (find $v_{e}$ )	given $Δ v$ and masses → find $v_{e}$	Ex 3
D. Doubling / log-scaling	show a fixed boost per multiplicative fuel step	Ex 4
E. Degenerate: no fuel burned	$m_{f} = m_{0} \Rightarrow R = 1$	Ex 5
F. Limiting: burn everything	$m_{f} \to$ payload only, and $m_{f} \to 0$	Ex 5
G. $Δ v > v_{e}$ "impossible" case	rocket outruns its own exhaust	Ex 6
H. Real-world word problem	staging, add-the- $Δ v$ logic	Ex 7
I. Exam twist: gravity loss	non-ideal $Δ v$ , subtract $g t$	Ex 8

Related tools appear as they are needed: Specific Impulse (Ex 3), Thrust and Mass Flow Rate (Ex 8), Multistage Rockets (Ex 7), Gravity Loss and Drag Loss (Ex 8).

Example 1 — Cell A: forward, basic

Worked example

A rocket ejects gas at $v_{e} = 2500 m/s$ . Its mass falls from $m_{0} = 12, 000 kg$ to $m_{f} = 3000 kg$ . Find $Δ v$ .

Forecast: The ratio is $12000/3000 = 4$ . Will $Δ v$ be around $v_{e}$ , much less, or much more? Guess before reading.

Step 1 — Form the mass ratio. $R = \frac{m _{0}}{m _{f}} = \frac{12000}{3000} = 4$ . Why this step? The formula uses only the ratio, never the raw kilograms — so we collapse two numbers into one.

Step 2 — Apply Face 1. $Δ v = v_{e} ln R = 2500 \times ln 4$ . Why this step? We know both masses and $v_{e}$ , so Face 1 (find the speed) is the direct route.

Step 3 — Evaluate. $ln 4 = 1.3863$ , so $Δ v = 2500 \times 1.3863 = 3466 m/s$ . Why this step? Turn the symbolic answer into a number a mission planner can use.

Verify: Units: $m/s \times (dimensionless log) = m/s$ ✓. Sanity: $Δ v = 3466 > v_{e} = 2500$ , which is fine because $R = 4 > e \approx 2.718$ (we'll unpack this in Ex 6). $Δ v \approx 3.47 km/s$ .

Example 2 — Cell B: work backward to a fuel fraction

Worked example

You must reach $Δ v = 6.0 km/s$ with an engine of $v_{e} = 3.0 km/s$ . What fraction of the launch mass must be propellant?

Forecast: $Δ v$ is twice $v_{e}$ . Will the fuel fraction be near 50%, 75%, or above 85%? Guess.

Step 1 — Solve for the ratio using Face 2. $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{6.0/3.0} = e^{2} = 7.389$ . Why this step? We know the speed we want, not the masses — so we run the equation backwards through $e^{x}$ , the inverse of $ln$ .

Step 2 — Convert to leftover fraction. $\frac{m _{f}}{m _{0}} = \frac{1}{7.389} = 0.1353$ . Why this step? "Fraction of mass remaining" is the reciprocal of the mass ratio — this is the dry+payload share.

Step 3 — Propellant fraction is everything else. $f_{p r o p} = 1 - \frac{m _{f}}{m _{0}} = 1 - 0.1353 = 0.8647$ . Why this step? Total mass = propellant + what's left; whatever isn't left was burned.

Verify: Plug back: $ln (1/0.1353) = ln (7.39) = 2.0$ , times $v_{e} = 3.0 = 6.0 km/s$ ✓. ≈ 86.5% must be propellant. A $Δ v$ of twice $v_{e}$ already demands the rocket be almost all fuel — the log's warning.

Example 3 — Cell C: solve for the exhaust velocity (with specific impulse)

Worked example

A stage carries $m_{0} = 500 kg$ , ends at $m_{f} = 200 kg$ , and this burn measured a speed gain of $Δ v = 3200 m/s$ . What was $v_{e}$ ? Then express it as a Specific Impulse $I_{s p} = v_{e} / g_{0}$ with $g_{0} = 9.81 m/s^{2}$ .

Forecast: $R = 2.5$ , so $ln R \approx 0.916$ . Is $v_{e}$ bigger or smaller than $Δ v$ ? Guess.

Step 1 — Rearrange Face 1 for $v_{e}$ . $v_{e} = \frac{Δ v}{ln ( m _{0} / m _{f} )}$ . Why this step? The one unknown is $v_{e}$ ; divide it free.

Step 2 — Numbers. $R = 500/200 = 2.5$ , $ln 2.5 = 0.9163$ , so $v_{e} = \frac{3200}{0.9163} = 3492 m/s$ . Why this step? Now every quantity but $v_{e}$ is known.

Step 3 — Convert to specific impulse. $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{3492}{9.81} = 356 s$ . Why this step? Engineers quote engine quality in seconds; $I_{s p}$ is just $v_{e}$ scaled by gravity so the units become time.

Verify: Reverse-check: $v_{e} ln R = 3492 \times 0.9163 = 3200 m/s$ ✓. Units of $I_{s p}$ : $(m/s) / (m/s^{2}) = s$ ✓. $v_{e} \approx 3490 m/s$ , $I_{s p} \approx 356 s$ — a good chemical rocket.

Example 4 — Cell D: the doubling law made concrete

Worked example

With $v_{e} = 3000 m/s$ , tabulate $Δ v$ for mass ratios $R = 2, 4, 8, 16$ . Show each doubling of $R$ adds the same $Δ v$ .

Forecast: Will the extra $Δ v$ from $R : 2 \to 4$ be more, less, or equal to that from $R : 8 \to 16$ ? Guess.

Step 1 — Compute each. $Δ v (2) = 3000 ln 2 = 2079 m/s$ $Δ v (4) = 3000 ln 4 = 4159 m/s$ $Δ v (8) = 3000 ln 8 = 6238 m/s$ $Δ v (16) = 3000 ln 16 = 8318 m/s$ Why this step? Direct Face 1 at four ratios.

Step 2 — Take the gaps. Each step up: $4159 - 2079 = 2080$ , $6238 - 4159 = 2079$ , $8318 - 6238 = 2080$ . Why this step? Differences expose the pattern the raw numbers hide.

Step 3 — Explain it. $ln (2 R) - ln (R) = ln 2$ always, so every doubling adds $v_{e} ln 2 = 3000 \times 0.6931 = 2079 m/s$ , no matter where you start. Why this step? The constant gap is the diminishing-returns message: fuel must multiply to add speed.

Verify: $v_{e} ln 2 = 2079 m/s$ matches all three gaps ✓. Curve in the figure is concave — the hallmark of a logarithm.

Example 5 — Cells E & F: degenerate and limiting inputs

Worked example

Same engine $v_{e} = 3000 m/s$ . Three edge cases: (a) No fuel burned: $m_{f} = m_{0}$ . (b) Payload-only survivor: $m_{0} = 10, 000 kg$ , $m_{f} = 1000 kg$ (90% fuel). (c) Theoretical limit $m_{f} \to 0$ .

Forecast: Which of these gives $Δ v = 0$ , which gives a big finite number, which "blows up"?

Step 1 — Case (a). $R = m_{0} / m_{0} = 1$ , and $ln 1 = 0$ , so $Δ v = 0$ . Why this step? If you throw away nothing, you gain nothing — the formula must return zero, and it does. This is the sanity anchor for the whole equation.

Step 2 — Case (b). $R = 10000/1000 = 10$ , $Δ v = 3000 ln 10 = 3000 \times 2.3026 = 6908 m/s$ . Why this step? A realistic "spend 90% on fuel" case — note it still only buys ≈2.3 exhaust-speeds.

Step 3 — Case (c). As $m_{f} \to 0$ , $R \to \infty$ and $ln R \to \infty$ , so $Δ v \to \infty$ . Why this step? Mathematically unbounded — but physically impossible: you can never have zero mass (there's always the engine, tank, payload). The limit warns that the equation's promise is only cashed if you could ship pure fuel, which you can't.

Verify: (a) $ln 1 = 0$ exactly ✓. (b) $ln 10 = 2.3026$ , product $= 6908 m/s$ ✓. (c) is a limit, not a number — the divergence is the point.

Example 6 — Cell G: beating your own exhaust ( $Δ v > v_{e}$ )

Worked example

"The rocket can't go faster than the gas it throws." Disprove it. With $v_{e} = 2000 m/s$ , what mass ratio makes $Δ v = v_{e}$ exactly? What ratio makes $Δ v = 2 v_{e}$ ?

Forecast: Is the crossover ratio around 2, around $e \approx 2.7$ , or around 10? Guess.

Step 1 — Set $Δ v = v_{e}$ . Then $v_{e} = v_{e} ln R \Rightarrow ln R = 1 \Rightarrow R = e = 2.718$ . Why this step? We hunt the exact ratio where the rocket matches its exhaust speed — dividing by $v_{e}$ isolates $ln R = 1$ .

Step 2 — Interpret. For any $R > e \approx 2.718$ , we get $Δ v > v_{e}$ . Example 1 already had $R = 4 > e$ , so its $3466 m/s$ genuinely exceeded $v_{e} = 2500$ . Why this step? This nails when the "impossible" happens — a modest 2.72:1 mass ratio.

Step 3 — Set $Δ v = 2 v_{e}$ . $ln R = 2 \Rightarrow R = e^{2} = 7.389$ . Why this step? Shows you can even double exhaust speed — for a price of $R \approx 7.4$ (matching Ex 2's ratio!).

Why it isn't magic: the gas leaves at $v_{e}$ relative to the rocket at each instant. But the rocket receives millions of tiny pushes; they accumulate. A person on a frictionless cart throwing balls at 5 m/s each will, after enough balls, be moving faster than 5 m/s.

Verify: $e^{1} = 2.718$ , $e^{2} = 7.389$ ✓; at $R = 4$ , $ln 4 = 1.386 > 1$ so $Δ v > v_{e}$ ✓.

Example 7 — Cell H: real-world staging word problem

Worked example

A two-stage rocket (Multistage Rockets). Stage 1: $v_{e} = 2800 m/s$ , burns from $m_{0} = 100, 000 kg$ to $m_{1} = 40, 000 kg$ (then the empty first stage is dropped). Stage 2: $v_{e} = 3400 m/s$ , burns from $25, 000 kg$ down to $8, 000 kg$ . Find the total $Δ v$ .

Forecast: Total $Δ v$ = sum of two logs. Will it beat the ≈6.9 km/s single-stage-90% case from Ex 5? Guess.

Step 1 — Stage 1 $Δ v$ . $R_{1} = 100000/40000 = 2.5$ , $Δ v_{1} = 2800 ln 2.5 = 2800 \times 0.9163 = 2566 m/s$ . Why this step? Each stage is its own rocket-equation problem, with its own $m_{0}, m_{f}, v_{e}$ .

Step 2 — Stage 2 $Δ v$ . $R_{2} = 25000/8000 = 3.125$ , $Δ v_{2} = 3400 ln 3.125 = 3400 \times 1.1394 = 3874 m/s$ . Why this step? Fresh masses because the dead first stage was jettisoned — that's the whole benefit of staging.

Step 3 — Add them. $Δ v_{t o t} = 2566 + 3874 = 6440 m/s$ . Why this step? Speed gains from sequential burns simply add — $Δ v$ is additive even though $R$ is multiplicative.

Verify: $ln 2.5 = 0.9163$ , $ln 3.125 = 1.1394$ ; products $2566$ and $3874 m/s$ ; sum $6440 m/s$ ✓. Staging let us discard dead weight mid-flight, which is why two modest stages rival one extreme single stage.

Example 8 — Cell I: exam twist with gravity loss

Worked example

A rocket burns straight up for $t = 90 s$ . Ideal $Δ v_{i d e a l}$ comes from $v_{e} = 3000 m/s$ , $R = m_{0} / m_{f} = 4$ . But during that time gravity steals $Δ v_{g r a v} = g_{0} t$ with $g_{0} = 9.81 m/s^{2}$ (Gravity Loss and Drag Loss). Find the actual speed gained.

Forecast: Gravity loss over 90 s — is it a small nibble or a huge bite out of the ideal $Δ v$ ?

Step 1 — Ideal $Δ v$ . $Δ v_{i d e a l} = 3000 ln 4 = 3000 \times 1.3863 = 4159 m/s$ . Why this step? The Tsiolkovsky equation gives the frictionless, gravity-free speed — our baseline.

Step 2 — Gravity loss. $Δ v_{g r a v} = g_{0} t = 9.81 \times 90 = 883 m/s$ . Why this step? While thrusting upward, gravity pulls down the whole time; integrating a constant $g_{0}$ over the burn gives $g_{0} t$ . (This is the $- \int g d t$ term the parent's fine print warned about.)

Step 3 — Actual gain. $Δ v_{a c t u a l} = 4159 - 883 = 3276 m/s$ . Why this step? Real speed = ideal minus what gravity confiscated.

Verify: $9.81 \times 90 = 882.9 m/s$ ; $4159 - 883 = 3276 m/s$ ✓. Units all m/s ✓. Gravity swallowed ≈21% of the ideal $Δ v$ — which is why rockets pitch over early to spend less time fighting straight-up gravity.

Recall Which cell does each example hit?

Ex 1 → A (forward basic) ::: given masses, find $Δ v$ Ex 2 → B (backward to fuel fraction) ::: given $Δ v$ , find $R$ Ex 3 → C (find $v_{e}$ ) + $I_{s p}$ ::: rearrange for exhaust speed Ex 4 → D (doubling law) ::: each fuel doubling adds $v_{e} ln 2$ Ex 5 → E and F (degenerate + limits) ::: $R = 1$ gives 0, $m_{f} \to 0$ diverges Ex 6 → G ( $Δ v > v_{e}$ ) ::: crossover at $R = e$ Ex 7 → H (staging word problem) ::: two rocket equations, added Ex 8 → I (gravity loss twist) ::: subtract $g_{0} t$

Active Recall Flashcards

To solve backward (know

Δ v

, want the mass ratio), which function do you apply?

The exponential:

m_{0} / m_{f} = e^{Δ v / v_{e}}

— the inverse of

ln

Above what mass ratio does

Δ v

exceed

v_{e}

R > e \approx 2.718

, since

ln R > 1

there.

When no fuel is burned (

m_{f} = m_{0}

), what is

Δ v

Exactly

0

, because

ln 1 = 0

How do

Δ v

values from successive rocket stages combine?

They add:

Δ v_{t o t} = Δ v_{1} + Δ v_{2} + \dots

How is gravity loss over a burn of duration

t

estimated for a vertical climb?

Δ v_{g r a v} = g_{0} t

, subtracted from the ideal

Δ v

Each doubling of the mass ratio adds how much

Δ v

A fixed

v_{e} ln 2

, independent of the starting ratio.