Δv = v_e · ln(m₀ - m_f) — understanding each term: Worked examples

This page hammers the parent equation $Δ v = v_{e} ln (m_{0} / m_{f})$ through every kind of situation it can throw at you. We map the space of cases first, then work one example per cell. If you meet a scenario in an exam, it lives somewhere in the table below.

Before anything, one reminder of the meaning of each symbol, because we use them on every line:

$Δ v$ = change in velocity the rocket can buy (m/s), a budget, not a final speed.
$v_{e}$ = effective exhaust velocity (m/s), how fast gas leaves relative to the rocket.
$m_{0}$ = wet mass (kg) = everything at the start, including all propellant.
$m_{f}$ = dry mass (kg) = everything left after the burn.
$R = m_{0} / m_{f}$ = the mass ratio, a pure number with no units.
$ln$ = the natural logarithm, the "shrinking machine" — see Natural Logarithm and Integration of 1/x.
$e \approx 2.718$ = the base of that logarithm; $e^{x}$ undoes $ln x$ .

The scenario matrix

The equation has only three knobs ( $v_{e}$ , $m_{0}$ , $m_{f}$ ), but the questions asked about it fall into these classes:

Cell	Case class	What is unknown	Degenerate / edge check
A	Forward: given all masses + $v_{e}$ , find $Δ v$	$Δ v$	—
B	Degenerate: no fuel burned, $m_{0} = m_{f}$	$Δ v$	$ln 1 = 0$
C	Inverse for fuel: given target $Δ v$ , find $R$	$R$ , propellant mass	needs $e^{x}$
D	Via specific impulse: $v_{e}$ not given, $I_{s p}$ is	$Δ v$	$v_{e} = g_{0} I_{s p}$
E	Limiting behaviour: $R \to \infty$ (all fuel, no dry mass)	ceiling of $Δ v$	$ln R \to \infty$ slowly
F	Two-stage: chain two burns	total $Δ v$	$Δ v$ add, ratios multiply
G	Word problem: real vehicle with losses	actual speed gained	subtract gravity loss
H	Exam twist: solve for $m_{f}$ given $Δ v$ , $m_{0}$ , $v_{e}$	$m_{f}$	invert the log twice

Each example below is tagged with its cell letter. Together they fill every row.

Cell A — the straight-ahead forecast

Worked example A · Given masses and

$v_{e}$ , find $Δ v$ A rocket has $m_{0} = 500, 000$ kg, $m_{f} = 100, 000$ kg, $v_{e} = 3000$ m/s. Find $Δ v$ .

Forecast: the ratio is 5, and $ln 5 \approx 1.6$ , so guess a bit under $3000 \times 1.6 \approx 4800$ m/s. Hold that number.

Compute the mass ratio $R = m_{0} / m_{f} = 500000/100000 = 5$ . Why this step? $Δ v$ depends on the ratio of masses, not their difference, so the ratio is the first thing to nail down.
Take the logarithm: $ln 5 = 1.609$ . Why this step? The equation feeds $R$ through $ln$ because velocity gain accumulates as $d v = - v_{e} d m / m$ ; the $1/ m$ integrates to a log.
Multiply by $v_{e}$ : $Δ v = 3000 \times 1.609 = 4828$ m/s. Why this step? Each fractional bit of mass lost buys $v_{e}$ times that fraction in speed.

Verify: units are $(m/s) \times (dimensionless) = m/s$ ✔. Our forecast (~4800) matches. Propellant burned $= m_{0} - m_{f} = 400, 000$ kg, less than the total — sane ✔.

Cell B — the degenerate "no burn"

Cell C — inverting for required fuel

Worked example C · Given target

$Δ v$ , find the mass ratio You must achieve $Δ v = 9400$ m/s (a typical low-Earth-orbit budget) with $v_{e} = 3400$ m/s. What mass ratio $R$ do you need, and what fraction of the rocket is propellant?

Forecast: $9400/3400 \approx 2.8$ , and $e^{2.8}$ is a big number — guess $R$ around 15, i.e. most of the rocket is fuel.

Isolate the log: divide both sides by $v_{e}$ . $ln R = \frac{Δ v}{v _{e}} = \frac{9400}{3400} = 2.765$ Why this step? We want $R$ , and it is trapped inside $ln$ ; peeling off $v_{e}$ first exposes the log.
Undo the log with $e^{x}$ : $R = e^{2.765} = 15.88$ Why this step? Because $e^{l n R} = R$ — exponentiation is the exact inverse of the natural log.
Propellant fraction: if $R = m_{0} / m_{f} = 15.88$ , then $m_{f} / m_{0} = 1/15.88 = 0.063$ , so propellant is $1 - 0.063 = 0.937$ , about $93.7%$ . Why this step? The dry fraction is the reciprocal of the ratio; everything else is fuel.

Verify: plug back — $3400 \times ln (15.88) = 3400 \times 2.765 = 9401$ m/s ≈ target ✔. Nearly 94% propellant is the famous "tyranny of the rocket equation."

Cell D — going through specific impulse

Worked example D ·

$v_{e}$ not given, only $I_{s p}$ An engine advertises $I_{s p} = 450$ s. The dry mass (structure + payload) is $m_{f} = 20, 000$ kg and it carries $300, 000$ kg of propellant. Find $Δ v$ . Use $g_{0} = 9.81$ m/s².

Forecast: $450$ s is a high-performance hydrogen engine, so $v_{e}$ is around 4400 m/s; with lots of fuel expect $Δ v$ over 10 km/s.

Convert $I_{s p}$ to exhaust velocity: $v_{e} = g_{0} I_{s p} = 9.81 \times 450 = 4414.5$ m/s. Why this step? The equation needs a speed, but $I_{s p}$ is measured in seconds; see Specific Impulse (Isp). The bridge is $v_{e} = g_{0} I_{s p}$ .
Build the wet mass: $m_{0} = 300000 + 20000 = 320000$ kg. Why this step? $m_{0}$ must include all propellant plus everything that stays.
Mass ratio $R = 320000/20000 = 16$ , and $ln 16 = 2.773$ . Why this step? Same core computation as always — feed the ratio through the log.
$Δ v = 4414.5 \times 2.773 = 12, 241$ m/s.

Verify: units — $(m/s^{2}) (s) = m/s$ for $v_{e}$ ✔, then $\times$ dimensionless ✔. Value comfortably exceeds LEO budget, as forecast ✔.

Cell E — the limiting ceiling

Worked example E · What is the most

$Δ v$ one stage can ever give? Suppose $v_{e} = 3400$ m/s is fixed. If you could magically shrink the dry mass toward zero (all fuel, no structure), what happens to $Δ v$ as $R \to \infty$ ?

Forecast: $ln$ grows without bound, so there is no hard ceiling — but it grows painfully slowly. Guess: doubling $R$ only adds a fixed lump of $Δ v$ .

Look at the growth of $ln R$ as $R$ increases. Each multiplication of $R$ by a fixed factor adds a fixed amount of $ln$ . Why this step? Because $ln (k R) = ln k + ln R$ — a constant $ln k$ is added no matter how big $R$ already is. See the figure.
Concretely, each time you double $R$ , $Δ v$ rises by $v_{e} ln 2 = 3400 \times 0.693 = 2356$ m/s — the same jump every doubling. Why this step? $ln (2 R) - ln R = ln 2$ , a constant, so the $Δ v$ step per doubling is constant.
Conclusion: $Δ v \to \infty$ as $R \to \infty$ , but only logarithmically. To gain another 2356 m/s you must double the whole mass ratio again — the cost explodes.

Verify: check the doubling law numerically. From $R = 8$ to $R = 16$ : $3400 (ln 16 - ln 8) = 3400 \times 0.693 = 2356$ m/s ✔. From $R = 16$ to $R = 32$ : same $2356$ m/s ✔. Equal steps confirm log growth.

Cell F — chaining two stages

Worked example F · Two-stage rocket, add the

$Δ v$ s Stage 1: $m_{0} = 100, 000$ kg, burns down to $m_{f} = 40, 000$ kg, $v_{e} = 3000$ m/s. Stage 2 (after dropping the empty stage-1 hardware): $m_{0} = 30, 000$ kg, $m_{f} = 8, 000$ kg, $v_{e} = 4000$ m/s. Find total $Δ v$ .

Forecast: each stage is a modest ratio (2.5 and 3.75), so each gives a couple of km/s; total maybe 6–8 km/s.

Stage 1: $R_{1} = 100000/40000 = 2.5$ , $Δ v_{1} = 3000 ln 2.5 = 3000 \times 0.9163 = 2749$ m/s. Why this step? Each stage obeys the same rocket equation independently.
Stage 2: $R_{2} = 30000/8000 = 3.75$ , $Δ v_{2} = 4000 ln 3.75 = 4000 \times 1.3218 = 5287$ m/s. Why this step? After separation the empty tank is gone, so stage 2 starts with a fresh, lighter wet mass — that is exactly why staging beats the log; see Multistage Rockets.
Add them: $Δ v_{total} = 2749 + 5287 = 8036$ m/s. Why this step? $Δ v$ is a change in velocity; sequential changes simply add.

Verify: ratios multiply into an effective $2.5 \times 3.75 = 9.375$ , but because the two stages have different $v_{e}$ you cannot use one log — you must add the two separately, which we did ✔. Total in the km/s range as forecast ✔.

Cell G — real-world word problem with losses

Worked example G · Real launch, subtract gravity loss

A booster has $m_{0} = 550, 000$ kg, $m_{f} = 150, 000$ kg, $v_{e} = 3100$ m/s. During its 150-second vertical climb it loses roughly $150 \times 9.81 = 1472$ m/s to gravity drag. What actual speed gain does the rocket end with?

Forecast: ideal $Δ v$ near 4000 m/s, minus ~1500 loss, so around 2500 m/s of real speed.

Ideal $Δ v$ : $R = 550000/150000 = 3.667$ , so $Δ v_{ideal} = 3100 ln 3.667 = 3100 \times 1.2993 = 4028$ m/s. Why this step? The Tsiolkovsky formula gives the budget assuming no external forces.
Subtract gravity loss: real speed gain $= 4028 - 1472 = 2556$ m/s. Why this step? Gravity pulls backward the whole climb; that momentum debt comes straight out of the ideal budget. $Δ v$ is a budget, not the final speed — this is the classic mistake the parent note warns about.

Verify: the real gain (2556) is less than ideal (4028) ✔ — losses always shrink the budget, never grow it. Order-of-magnitude matches forecast ✔.

Cell H — exam twist: solve for the dry mass

Worked example H · Given

$Δ v$ , $m_{0}$ , $v_{e}$ , find $m_{f}$ A design must hit $Δ v = 6000$ m/s with $v_{e} = 3200$ m/s and a fixed wet mass $m_{0} = 80, 000$ kg. What dry mass $m_{f}$ is allowed, and how much propellant does that imply?

Forecast: $6000/3200 \approx 1.9$ , so $R = e^{1.9} \approx 6.6$ ; dry mass is $m_{0}$ divided by that, roughly 12,000 kg.

Get the required log: $ln R = Δ v / v_{e} = 6000/3200 = 1.875$ . Why this step? Same isolation as Cell C — strip off $v_{e}$ to expose the log.
Exponentiate for the ratio: $R = e^{1.875} = 6.521$ . Why this step? $e^{x}$ undoes $ln$ , giving the pure mass ratio.
Solve for dry mass: $m_{f} = m_{0} / R = 80000/6.521 = 12, 268$ kg. Why this step? Since $R = m_{0} / m_{f}$ , rearranging gives $m_{f} = m_{0} / R$ .
Propellant needed: $m_{0} - m_{f} = 80000 - 12268 = 67, 732$ kg. Why this step? Propellant burned is always wet minus dry.

Verify: plug back — $3200 ln (80000/12268) = 3200 ln (6.521) = 3200 \times 1.875 = 6000$ m/s ✔. Dry mass matches forecast (~12,000 kg) ✔.

Wrap-up

Recall Which cell is each exam question?

"Find Δv from masses" ::: Cell A.
"Engine off, mass unchanged" ::: Cell B, answer 0.
"How much fuel for this Δv?" ::: Cell C (invert with $e^{x}$ ).
"Only $I_{s p}$ given" ::: Cell D (use $v_{e} = g_{0} I_{s p}$ ).
"Why does more fuel help so little?" ::: Cell E (log grows slowly).
"Two stages, total Δv?" ::: Cell F (add the Δv's).
"Real launch speed?" ::: Cell G (subtract gravity loss).
"Find the allowed dry mass" ::: Cell H (solve $m_{f} = m_{0} / e^{Δ v / v_{e}}$ ).

Connections

Parent topic — the equation itself.
Conservation of Momentum — the derivation these examples rest on.
Newton's Third Law — why throwing mass gives thrust.
Specific Impulse (Isp) — used in Cell D.
Multistage Rockets — the physics behind Cell F.
Thrust and Mass Flow Rate — differential cousin behind Cell G's climb.
Natural Logarithm and Integration of 1/x — why $ln$ and $e$ pair up.