Mass ratio m₀ - m_f — why it's so critical: Exercises

Throughout we use $g_{0} = 9.81 m/s^{2}$ for converting specific impulse via $u = I_{s p} g_{0}$ . All $ln$ are natural logs (base $e$ ).

Level 1 — Recognition

Goal: read the definitions and plug one number into one formula.

Recall Solution 1.1

What we do: apply the two definitions directly. $R = \frac{m _{0}}{m _{f}} = \frac{5000}{1000} = 5.$ Why: $R$ is just wet over dry — no logs yet. Note $R = 5 > 1$ , as it must be. $ζ = 1 - \frac{1}{R} = 1 - \frac{1}{5} = \frac{4}{5} = 0.8.$ So $R = 5$ and $80%$ of the liftoff mass is propellant.

Recall Solution 1.2

What we do: plug into $Δ v = u ln R$ . $Δ v = u ln R = 3.0 \times ln (e) = 3.0 \times 1 = 3.0 km/s .$ Why $ln e = 1$ : $ln$ asks " $e$ to what power gives this number?" Since $e = e^{1}$ , the answer is $1$ . When $R = e$ , you gain exactly one exhaust-speed of $Δ v$ .

Recall Solution 1.3

$u = I_{s p} g_{0} = 320 \times 9.81 = 3139.2 m/s \approx 3.14 km/s .$ Why: $I_{s p}$ in seconds is just $u$ hidden behind a division by $g_{0}$ ; multiplying back recovers a real speed. See Specific Impulse Isp.

Level 2 — Application

Goal: invert the equation, chain two steps.

Recall Solution 2.1

What we do: invert Tsiolkovsky, $R = e^{Δ v / u}$ . $R = e^{6.0/3.0} = e^{2} \approx 7.389.$ Why exponential: solving $Δ v = u ln R$ for $R$ means undoing the log, and the inverse of $ln$ is $e^{(\cdot)}$ . Every extra " $u$ " of speed multiplies $R$ by $e$ . ( $R = 7.389 > 1$ ✓.)

Recall Solution 2.2

Step 1 — find $m_{0}$ : $m_{0} = R m_{f} = 7.389 \times 800 = 5911.2 kg .$ Step 2 — subtract to get propellant: $m_{p} = m_{0} - m_{f} = 5911.2 - 800 = 5111.2 kg .$ Why: the dry mass is fixed; the wet mass is forced up by $R$ , and propellant $m_{p}$ is simply what's left after removing the dry mass. You carry about $6.4 kg$ of fuel per kg of dry rocket.

Recall Solution 2.3

Step 1 — exhaust speed: $u = 450 \times 9.81 = 4414.5 m/s = 4.4145 km/s .$ Step 2 — mass ratio: $R = e^{9.0/4.4145} = e^{2.039} \approx 7.68.$ Why the chaining matters: you cannot skip step 1 — the equation needs a speed, not seconds.

Level 3 — Analysis

Goal: compare cases, reason about ratios and differences.

Recall Solution 3.1

What we do: compute the difference of two logs. $Δ (Δ v) = u ln 8 - u ln 4 = u ln \frac{8}{4} = u ln 2 \approx 0.693 u .$ Why this is the whole lesson: doubling the fuel added only $ln 2 \approx 0.69$ exhaust-speeds — not double the speed. Logs turn multiplication of $R$ into addition of $Δ v$ . See the Delta-v Budget consequence: buying speed gets exponentially expensive.

Recall Solution 3.2

Engine A: $R_{A} = e^{9.4/3.0} = e^{3.133} \approx 22.95.$ Dry fraction $1/ R_{A} \approx 0.0436$ ( $4.4%$ ). Engine B: $R_{B} = e^{9.4/4.4} = e^{2.136} \approx 8.47.$ Dry fraction $1/ R_{B} \approx 0.1181$ ( $11.8%$ ). Analysis: a $47%$ increase in $u$ (from 3.0 to 4.4) nearly tripled the useful (dry+payload) mass fraction. Because $R$ is exponential in $1/ u$ , small engine gains pay off hugely. This is why hydrogen upper stages exist.

Figure — Mass ratio m₀ - m_f — why it's so critical

Recall Solution 3.3

$Δ v = u : R = e^{1} = 2.718, Δ v = 2 u : R = e^{2} = 7.389, Δ v = 3 u : R = e^{3} = 20.09.$ Ratio check: $7.389/2.718 = 2.718 = e$ , and $20.09/7.389 = 2.718 = e$ . Why: each added $u$ multiplies the exponent by $+ 1$ , i.e. multiplies $R$ by $e$ . That constant multiplicative jump is the exponential wall — the red curve in the figure (labelled axes: $Δ v / u$ across, $R$ up) gets vertical fast.

Level 4 — Synthesis

Goal: combine staging, structural limits, and the derivation.

Recall Solution 4.1

Max $R$ : $R_{ma x} = \frac{m _{0}}{m _{f}} = \frac{1}{0.08} = 12.5.$ Max $Δ v$ : $Δ v_{ma x} = u ln R_{ma x} = 3.0 \times ln 12.5 = 3.0 \times 2.526 = 7.58 km/s .$ Synthesis: $7.58 < 9.4$ , so a single kerosene stage cannot reach LEO no matter how you fill it — the structural floor caps $R$ . This gap is the reason for Multistage Rockets: throw away empty tanks so the next stage's $m_{f}$ (and thus $R$ ) resets.

Recall Solution 4.2

Per stage: $Δ v_{s t a g e} = 3.0 \times ln 8 = 3.0 \times 2.079 = 6.238 km/s .$ Total (two stages, $Δ v$ 's add): $Δ v_{t o t a l} = 2 \times 6.238 = 12.477 km/s .$ Single stage of $R = 8$ : only $6.238 km/s$ . Why staging wins: the total mass ratio is effectively $R^{2} = 64$ , giving $Δ v = u ln 64 = 2 u ln 8$ — double the single-stage speed, without needing an impossible $R = 64$ in one tank. Dropping dead structure lets each stage carry a fresh, achievable $R$ .

Recall Solution 4.3

Sign convention first (crucial): as the rocket burns fuel, its mass decreases, so the change $d m$ is negative ( $d m < 0$ ). The mass actually ejected is $- d m > 0$ . Rearranging the momentum core $0 = m d v + u d m$ gives: $d v = - u \frac{d m}{m} .$ The leading minus sign is what keeps $d v$ positive: since $d m < 0$ , the quantity $- d m / m > 0$ , so the rocket speeds up even though mass falls. Integrate: $Δ v = - u \int_{m_{0}}^{m_{f}} \frac{d m}{m} = u ln \frac{m _{0}}{m _{f}} .$ Numbers: $R = 3000/1200 = 2.5$ , so $Δ v = 2.5 \times ln 2.5 = 2.5 \times 0.9163 = 2.291 km/s .$ Why the sign flips positive: integrating $\frac{d m}{m}$ from a big $m_{0}$ down to a small $m_{f}$ gives a negative number ( $ln \frac{m _{f}}{m _{0}} < 0$ ); the leading minus makes $Δ v$ positive. The rocket speeds up as it lightens.

Level 5 — Mastery

Goal: full mission design with conversions, structure, and payload trade-offs.

Recall Solution 5.1

(a) Exhaust speed: $u = 340 \times 9.81 = 3335.4 m/s = 3.3354 km/s .$ (b) Mass ratio: $R = e^{5.6/3.3354} = e^{1.679} \approx 5.360.$ (c) Set up masses. Dry mass $m_{f} = m_{s t r u c t} + m_{p a y l o a d} = 0.06 m_{0} + 2000.$ Also $m_{f} = m_{0} / R .$ Equate: $\frac{m _{0}}{R} = 0.06 m_{0} + 2000.$ Now spell out the coefficient: $\frac{1}{R} = \frac{1}{5.360} = 0.18657$ , so $\frac{m _{0}}{R} = 0.18657 m_{0} .$ Substituting: $0.18657 m_{0} = 0.06 m_{0} + 2000 \Rightarrow (0.18657 - 0.06) m_{0} = 2000 \Rightarrow 0.12657 m_{0} = 2000,$ $m_{0} = \frac{2000}{0.12657} = 15801 kg .$ (d) Propellant: $m_{f} = m_{0} / R = 15801/5.360 = 2948 kg$ , so $m_{p} = m_{0} - m_{f} = 15801 - 2948 = 12853 kg .$ Check: structure $= 0.06 \times 15801 = 948 kg$ , payload $2000 kg$ , sum $2948 = m_{f}$ . ✓ Consistent. Why the algebra: the dry mass appears on both sides (it depends on $m_{0}$ through the structure fraction), so we solve a linear equation rather than plug blindly.

Recall Solution 5.2

New $R$ : $R = e^{7.0/3.3354} = e^{2.099} \approx 8.158.$ Spell out the coefficient: $\frac{1}{R} = \frac{1}{8.158} = 0.12259$ , so $\frac{m _{0}}{R} = 0.12259 m_{0} .$ Solve: $0.12259 m_{0} = 0.06 m_{0} + 2000 \Rightarrow (0.12259 - 0.06) m_{0} = 2000.$ $0.06259 m_{0} = 2000 \Rightarrow m_{0} = \frac{2000}{0.06259} = 31954 kg .$ Growth factor: $31954/15801 = 2.02.$ Mastery insight: asking for $25%$ more $Δ v$ ( $5.6 \to 7.0$ ) doubled the whole rocket. Worse, the structure term $0.06 m_{0}$ eats an ever-larger slice of the shrinking $1/ R$ dry budget — the denominator $0.06259$ is small, so mass explodes. This nonlinearity is the argument of the whole parent note.

Recall Solution 5.3

Condition: rearrange $\frac{m _{0}}{R} = s m_{0} + m_{L}$ to $m_{0} (\frac{1}{R} - s) = m_{L} .$ Left factor must be positive ⇒ $1/ R > s$ ⇒ $R < 1/ s .$ Max $R$ : $R_{ma x} = 1/ s = 1/0.06 = 16.67.$ Max $Δ v$ : $Δ v_{ma x} = u ln (1/ s) = 3.3354 \times ln (16.67) = 3.3354 \times 2.813 = 9.383 km/s .$ Check: Problem 5.2 wanted $7.0 < 9.383 km/s$ ✓ (feasible, hence a finite $m_{0}$ existed). Any $Δ v \geq 9.383 km/s$ needs $m_{0} \to \infty$ (or a lighter structure, or Multistage Rockets). Why the divergence: as $R \to 1/ s$ , the bracket $(1/ R - s) \to 0$ , so $m_{0} = m_{L} / (1/ R - s) \to \infty$ . The exponential wall becomes a literal asymptote — no payload can be lifted past it in one stage.

Active recall

Recall One-line summary of the whole ladder

Speed is the log of the mass ratio; mass ratio is the exp of the required speed; add $Δ v$ 's but multiply mass ratios; a fixed structure fraction $s$ caps a single stage at $Δ v_{ma x} = u ln (1/ s)$ . And always $R > 1$ for a positive $Δ v$ ( $R = 1$ ⇒ no fuel ⇒ $Δ v = 0$ ).

Connections

Tsiolkovsky Rocket Equation — every problem is an application of $Δ v = u ln R$ .
Specific Impulse Isp — L1/L2/L5 conversions $u = I_{s p} g_{0}$ .
Multistage Rockets — L4 shows why staging beats the wall.
Delta-v Budget — L3/L5 mission- $Δ v$ requirements.
Conservation of Momentum — L4 back-solves the derivation.
Exhaust Velocity and Thrust — the physical source of $u$ .