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

This page is a self-test ladder for the parent note Δv = v_e · ln(m₀/m_f). Each level climbs in difficulty. Try each problem with the solution collapsed, then open it.

Before you start, keep these anchored (every symbol from the parent, in plain words):

Definition The symbols you'll reuse

$Δ v$ = change in speed the rocket can buy by burning fuel (a budget, m/s).
$v_{e}$ = exhaust velocity, how fast gas leaves the nozzle relative to the rocket (m/s).
$m_{0}$ = wet mass (everything at the start, including all fuel).
$m_{f}$ = dry mass (everything left after the burn — structure + payload).
$R = m_{0} / m_{f}$ = the mass ratio, a pure number with no units.
$I_{s p}$ = specific impulse — a way of rating an engine's efficiency, measured in seconds. Bigger $I_{s p}$ means more push per kilogram of fuel. It connects to exhaust velocity by $v_{e} = g_{0} I_{s p}$ — see Specific Impulse (Isp).
$g_{0}$ = standard gravity, the fixed constant $9.81$ m/s² used only to convert specific impulse into a speed. It is not the local gravity of any planet — just a bookkeeping number.
Core relation: $Δ v = v_{e} ln R$ , and its inverse $R = e^{Δ v / v_{e}}$ .

Two tools we lean on constantly, so let's re-earn them here:

Intuition The two moves you'll make over and over

$ln$ ("log") is a shrinking machine: it turns the ratio $R$ into a much smaller number. $ln 1 = 0$ , $ln e = 1$ , $ln R$ grows very slowly as $R$ grows. Why use it? Because the velocity gain built up as $d v = - v_{e} \frac{d m}{m}$ . Here $m$ is the rocket's instantaneous mass at that moment of the burn, and $d m$ is the tiny mass change in one instant. Since the rocket is losing mass, $d m < 0$ (it's negative), and the minus sign out front flips that back so $d v$ comes out positive — the rocket speeds up. The $1/ m$ factor means each kilogram thrown gives a bigger kick to a lighter rocket, and adding up (integrating) $1/ m$ is exactly what produces $ln$ — see Natural Logarithm and Integration of 1/x.
$e^{x}$ ("exponential") undoes $ln$ . If $ln R = x$ , then $R = e^{x}$ . We use it whenever we're given a needed $Δ v$ and must find the mass ratio — we run the equation backward.

The two little plots below are worth a long look — they are the whole personality of this equation.

Figure — Δv = v_e · ln(m₀ - m_f) — understanding each term

Level 1 — Recognition

L1.1

Which symbol in $Δ v = v_{e} ln (m_{0} / m_{f})$ is dimensionless (has no units), and why?

Recall Solution

The ratio $R = m_{0} / m_{f}$ is dimensionless. Why: it is mass divided by mass (kg ÷ kg), so the units cancel. That is essential — you cannot take $ln$ of something with units. So $ln R$ is also a pure number, and $v_{e} ln R$ correctly carries the units of $v_{e}$ : metres per second. ✔

L1.2

A rocket has $m_{0} = 400, 000$ kg wet and burns $250, 000$ kg of propellant. What is $m_{f}$ ?

Recall Solution

$m_{f}$ is what remains after the burn, not what was burned. $m_{f} = m_{0} - (propellant burned) = 400, 000 - 250, 000 = 150, 000 kg .$

L1.3

If no fuel is burned ( $m_{0} = m_{f}$ ), what is $Δ v$ ? Show the one-line reason.

Recall Solution

$R = m_{0} / m_{f} = 1$ , and $ln 1 = 0$ , so $Δ v = v_{e} \cdot 0 = 0 m/s .$ Why it makes sense: no mass thrown backward ⇒ no forward kick (from Conservation of Momentum).

Level 2 — Application

L2.1

$m_{0} = 600, 000$ kg, $m_{f} = 120, 000$ kg, $v_{e} = 3200$ m/s. Find $Δ v$ .

Recall Solution

Ratio first (that's what the log eats): $R = \frac{600 , 000}{120 , 000} = 5.$ Why compute $R$ first? The equation says $Δ v$ depends only on the ratio of wet to dry mass, never on the raw kilograms separately. Two rockets with masses $(600, 000, 120, 000)$ and $(5, 1)$ get the same $Δ v$ . So we collapse the two masses into the single number $R = 5$ before feeding it to the log — that's the only quantity the log cares about. $Δ v = 3200 \cdot ln 5 = 3200 \times 1.60944 \approx 5150 m/s .$

L2.2

You need $Δ v = 8000$ m/s with $v_{e} = 3000$ m/s. What mass ratio $R$ is required?

Recall Solution

Run the equation backward with the exponential (the tool that undoes $ln$ ): $ln R = \frac{Δ v}{v _{e}} = \frac{8000}{3000} = 2.6667.$ Why exponentiate after dividing? Dividing by $v_{e}$ peels off the multiplier and leaves $ln R$ alone — but that's the log of the ratio, not the ratio itself. The ratio $R$ is still trapped inside the log. To free it we apply $e^{(\cdot)}$ to both sides, because $e^{l n R} = R$ exactly — the exponential is the one operation that cancels a natural log. Skipping it leaves you with $ln R \approx 2.67$ , which is a logarithm, not a mass ratio. $R = e^{2.6667} \approx 14.39.$ Meaning: about $14.4$ kg of wet mass for every $1$ kg reaching the target — the fuel fraction is huge.

L2.3

An engine has specific impulse $I_{s p} = 380$ s. Find its exhaust velocity $v_{e}$ (use $g_{0} = 9.81$ m/s²).

Recall Solution

From Specific Impulse (Isp), $v_{e} = g_{0} I_{s p}$ , where $I_{s p}$ is the specific impulse and $g_{0} = 9.81$ m/s² is the standard-gravity conversion constant, both defined in the symbol list: $v_{e} = 9.81 \times 380 \approx 3728 m/s .$ Why this link exists: $I_{s p}$ is just $v_{e}$ measured in "seconds of thrust per weight of fuel"; multiplying by $g_{0}$ converts it back to a speed.

Level 3 — Analysis

L3.1

A rocket achieves $Δ v_{1}$ with mass ratio $R$ . If you square the mass ratio to $R^{2}$ (keeping $v_{e}$ fixed), how does the new $Δ v_{2}$ compare to $Δ v_{1}$ ?

Recall Solution

$Δ v_{2} = v_{e} ln (R^{2}) = v_{e} \cdot 2 ln R = 2 (v_{e} ln R) = 2 Δ v_{1} .$ Why does $ln (R^{2}) = 2 ln R$ ? This is the power rule of logarithms: $ln (a^{b}) = b ln a$ . It follows because a log turns exponents into multipliers (squaring a number adds its log to itself, i.e. doubles it). See Natural Logarithm and Integration of 1/x. Interpretation: squaring the ratio only doubles $Δ v$ . To double your speed budget you must square how much heavier the wet rocket is than the dry one — the "tyranny of the rocket equation." This is exactly why the parent note calls the log the villain.

L3.2

Two designs must both deliver $Δ v = 6000$ m/s. Design A: $v_{e} = 3000$ m/s. Design B: $v_{e} = 4500$ m/s. Find each required mass ratio and comment on which needs less fuel per kg of payload.

Recall Solution

Design A: $ln R_{A} = 6000/3000 = 2.000$ , so $R_{A} = e^{2} \approx 7.389$ . Design B: $ln R_{B} = 6000/4500 = 1.3333$ , so $R_{B} = e^{1.3333} \approx 3.793$ . Comment: the higher- $v_{e}$ engine (B) needs a mass ratio of only $\sim 3.8$ versus $\sim 7.4$ — roughly half the wet-to-dry ratio for the same job. Higher exhaust speed buys $Δ v$ far more cheaply. This is the whole appeal of ion engines.

L3.3

A single stage has $m_{f} = 30, 000$ kg and carries $270, 000$ kg of propellant, $v_{e} = 3400$ m/s. (a) Find $Δ v$ . (b) A mission needs $10, 000$ m/s. By how much does this single stage fall short?

Recall Solution

(a) $m_{0} = 270, 000 + 30, 000 = 300, 000$ kg, so $R = 300, 000/30, 000 = 10$ . $Δ v = 3400 \cdot ln 10 = 3400 \times 2.302585 \approx 7829 m/s .$ (b) Shortfall $= 10, 000 - 7829 = 2171$ m/s. The single stage cannot reach the budget — a hint toward Multistage Rockets.

Level 4 — Synthesis

L4.1 — Two-stage rocket

A rocket flies in two stages, each firing to depletion. Total $Δ v$ is the sum of each stage's $Δ v$ (velocities add).

Stage 1: $v_{e, 1} = 3000$ m/s, wet $= 500, 000$ kg, dry-at-separation $= 200, 000$ kg.
Stage 2: $v_{e, 2} = 4000$ m/s, wet $= 90, 000$ kg, dry $= 30, 000$ kg.

Find the total $Δ v$ .

The figure below tracks the velocity building up across both burns. Read it left to right: the lavender line is Stage 1's climb, the mint arrow marks where the spent Stage-1 shell is dropped, and the coral line is Stage 2's steeper climb (lighter rocket, faster exhaust). The dots are the cumulative speed at each milestone — the final dot is the total we compute below.

Recall Solution

Why add? Each stage produces its own change in velocity; changes in velocity in the same direction simply add (from the integration of $d v$ over each burn). Stage 1: $R_{1} = 500, 000/200, 000 = 2.5$ , $Δ v_{1} = 3000 \cdot ln 2.5 = 3000 \times 0.916291 \approx 2749 m/s .$ Stage 2: $R_{2} = 90, 000/30, 000 = 3$ , $Δ v_{2} = 4000 \cdot ln 3 = 4000 \times 1.098612 \approx 4394 m/s .$ Total: $Δ v = Δ v_{1} + Δ v_{2} \approx 2749 + 4394 = 7144 m/s .$

L4.2 — Why staging beats one big tank

Take the staged rocket of L4.1 and ask: what if we tried to do the whole job as a single stage — no dropping of spent hardware mid-flight? A true single stage must carry all its structure the entire way, so its dry mass is the combined dead weight of both stages' hardware plus the payload. Suppose that honest single-stage dry mass is $m_{f} = 90, 000$ kg (both stage shells + payload), total wet $= 500, 000$ kg, and use $v_{e} = 3500$ m/s (the plain average $\frac{3000 + 4000}{2}$ of the two engines, since a single rocket has one engine type). Find $Δ v$ and compare to the staged total $7144$ m/s from L4.1.

Recall Solution

$R = 500, 000/90, 000 = 5.5 \overline{5}$ . $Δ v = 3500 \cdot ln (5.5556) = 3500 \times 1.714798 \approx 6002 m/s .$ Compare: the honest single stage gives only $\approx 6002$ m/s versus the staged $7144$ m/s — staging wins by over $1100$ m/s. Why: the single stage must haul all its dead structure ( $90, 000$ kg) to the very end, keeping $R$ low. Staging drops the heavy Stage-1 shell partway, so the final push works on a much lighter rocket — a bigger effective ratio for the last leg. Dry mass is the enemy, and the log punishes carrying it — see Multistage Rockets.

Level 5 — Mastery

L5.1 — Design backward from a budget

A Mars-transfer stage must deliver $Δ v = 5700$ m/s. The engine gives $I_{s p} = 340$ s ( $g_{0} = 9.81$ m/s²). The payload + structure that must survive is $m_{f} = 25, 000$ kg. (a) Find $v_{e}$ . (b) Find the required mass ratio $R$ . (c) Find the required wet mass $m_{0}$ and the propellant mass needed.

Recall Solution

(a) $v_{e} = g_{0} I_{s p} = 9.81 \times 340 = 3335.4$ m/s. (b) Run backward: $ln R = Δ v / v_{e} = 5700/3335.4 = 1.70880$ , so $R = e^{1.70880} \approx 5.522.$ (c) $m_{0} = R \cdot m_{f} = 5.522 \times 25, 000 \approx 138, 050$ kg. Propellant $= m_{0} - m_{f} \approx 138, 050 - 25, 000 = 113, 050$ kg. Sense check: propellant is $\sim 82%$ of the wet mass — heavy, but far from the $94%$ of the harder LEO example in the parent. Lower $Δ v$ demand ⇒ gentler ratio. ✔

L5.2 — Sensitivity of the design

Using L5.1's engine ( $v_{e} = 3335.4$ m/s), the budget creeps up from $5700$ to $6300$ m/s (a $600$ m/s increase). Keeping $m_{f} = 25, 000$ kg, how much extra propellant is needed? Comment on why a modest $Δ v$ bump costs so much.

Recall Solution

New $ln R^{'} = 6300/3335.4 = 1.88872$ , so $R^{'} = e^{1.88872} \approx 6.611$ . New $m_{0}^{'} = 6.611 \times 25, 000 \approx 165, 280$ kg. New propellant $= 165, 280 - 25, 000 = 140, 280$ kg. Extra propellant $= 140, 280 - 113, 050 \approx 27, 230$ kg. Why so costly: because $R = e^{Δ v / v_{e}}$ is exponential in $Δ v$ . A linear $+ 600$ m/s in the budget multiplies the ratio by $e^{600/3335.4} \approx 1.197$ — a $\sim 20%$ jump in wet mass. Each extra m/s taxes you exponentially. This is the mastery-level statement of the tyranny of the rocket equation.

Wrap-up recall

Recall Cover and answer

How do you find $R$ when you're given $Δ v$ and $v_{e}$ ? ::: $R = e^{Δ v / v_{e}}$ (exponential undoes the log).
Why do multi-stage $Δ v$ values add? ::: Each stage integrates $d v$ separately; same-direction velocity changes sum.
Why does a $+ 600$ m/s budget bump cost so much fuel? ::: Mass ratio is exponential in $Δ v$ , so fuel compounds.
What converts $I_{s p}$ to $v_{e}$ ? ::: Multiply by $g_{0}$ : $v_{e} = g_{0} I_{s p}$ .

Connections

Parent topic (Hinglish)
Conservation of Momentum — foundation of the derivation.
Specific Impulse (Isp) — the $v_{e} = g_{0} I_{s p}$ conversion used in L2.3, L5.
Multistage Rockets — the L4 staging results.
Natural Logarithm and Integration of 1/x — why $ln$ and $e^{x}$ appear.
Thrust and Mass Flow Rate — the differential cousin $F = v_{e} \overset{m}{˙}$ .