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

3.3.3 · D4 · Physics › Rocket Propulsion › Mass ratio m₀ - m_f — why it's so critical

Har jagah hum $g_{0} = 9.81 m/s^{2}$ use karte hain specific impulse ko $u = I_{s p} g_{0}$ se convert karne ke liye. Saare $ln$ natural logs hain (base $e$ ).

Level 1 — Recognition

Goal: definitions padho aur ek number ek formula mein plug karo.

Recall Solution 1.1

Hum kya karein: seedha do definitions apply karo. $R = \frac{m _{0}}{m _{f}} = \frac{5000}{1000} = 5.$ Kyun: $R$ sirf wet divided by dry hai — abhi tak koi logs nahi. Note karo $R = 5 > 1$ , jaisa hona chahiye. $ζ = 1 - \frac{1}{R} = 1 - \frac{1}{5} = \frac{4}{5} = 0.8.$ Isliye $R = 5$ hai aur liftoff mass ka $80%$ propellant hai.

Recall Solution 1.2

Hum kya karein: $Δ v = u ln R$ mein plug karo. $Δ v = u ln R = 3.0 \times ln (e) = 3.0 \times 1 = 3.0 km/s .$ $ln e = 1$ kyun: $ln$ poochhta hai " $e$ kis power par yeh number deta hai?" Kyunki $e = e^{1}$ , answer $1$ hai. Jab $R = e$ , tum exactly ek exhaust-speed ka $Δ v$ paate ho.

Recall Solution 1.3

$u = I_{s p} g_{0} = 320 \times 9.81 = 3139.2 m/s \approx 3.14 km/s .$ Kyun: seconds mein $I_{s p}$ bas $u$ hai jo $g_{0}$ se divide ho ke chhupa hai; wapas multiply karne se real speed milti hai. Dekho Specific Impulse Isp.

Level 2 — Application

Goal: equation ko invert karo, do steps chain karo.

Recall Solution 2.1

Hum kya karein: Tsiolkovsky ko invert karo, $R = e^{Δ v / u}$ . $R = e^{6.0/3.0} = e^{2} \approx 7.389.$ Exponential kyun: $Δ v = u ln R$ mein $R$ ke liye solve karne ka matlab log ko undo karna hai, aur $ln$ ka inverse $e^{(\cdot)}$ hai. Speed ka har extra " $u$ " $R$ ko $e$ se multiply karta hai. ( $R = 7.389 > 1$ ✓.)

Recall Solution 2.2

Step 1 — $m_{0}$ nikalo: $m_{0} = R m_{f} = 7.389 \times 800 = 5911.2 kg .$ Step 2 — propellant ke liye subtract karo: $m_{p} = m_{0} - m_{f} = 5911.2 - 800 = 5111.2 kg .$ Kyun: dry mass fixed hai; wet mass $R$ se upar force hota hai, aur propellant $m_{p}$ bas wahi hai jo dry mass hatane ke baad bachta hai. Tum dry rocket ke har kg ke liye lagbhag $6.4 kg$ fuel lete ho.

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: $R = e^{9.0/4.4145} = e^{2.039} \approx 7.68.$ Chaining kyun matter karta hai: tum step 1 skip nahi kar sakte — equation ko speed chahiye, seconds nahi.

Level 3 — Analysis

Goal: cases compare karo, ratios aur differences ke baare mein reason karo.

Recall Solution 3.1

Hum kya karein: do logs ka difference compute karo. $Δ (Δ v) = u ln 8 - u ln 4 = u ln \frac{8}{4} = u ln 2 \approx 0.693 u .$ Kyun yeh poora lesson hai: fuel double karne se sirf $ln 2 \approx 0.69$ exhaust-speeds juda — double speed nahi. Logs $R$ ki multiplication ko $Δ v$ ki addition mein convert karte hain. Delta-v Budget consequence dekho: speed kharidna exponentially expensive hota jaata hai.

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: $u$ mein $47%$ increase (3.0 se 4.4) ne useful (dry+payload) mass fraction ko lagbhag triple kar diya. Kyunki $R$ $1/ u$ mein exponential hai, engine ke chhote gains bahut zyaada faayda dete hain. Isliye hydrogen upper stages exist karte hain.

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: $7.389/2.718 = 2.718 = e$ , aur $20.09/7.389 = 2.718 = e$ . Kyun: har added $u$ exponent ko $+ 1$ se multiply karta hai, yaani $R$ ko $e$ se multiply karta hai. Woh constant multiplicative jump hi exponential wall hai — figure mein red curve (labelled axes: $Δ v / u$ across, $R$ up) jaldi vertical ho jaati hai.

Level 4 — Synthesis

Goal: staging, structural limits, aur derivation combine karo.

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: $7.58 < 9.4$ , isliye ek single kerosene stage LEO reach nahi kar sakta chahe tum usse kitna bhi bhar lo — structural floor $R$ ko cap karta hai. Yeh gap Multistage Rockets ki wajah hai: khaali tanks phenk do taaki agla stage ka $m_{f}$ (aur isliye $R$ ) reset ho jaye.

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 (do stages, $Δ v$ 's add hote hain): $Δ v_{t o t a l} = 2 \times 6.238 = 12.477 km/s .$ $R = 8$ ka single stage: sirf $6.238 km/s$ . Staging kyun jeetti hai: total mass ratio effectively $R^{2} = 64$ hai, jo $Δ v = u ln 64 = 2 u ln 8$ deta hai — single-stage speed se double, bina ek tank mein impossible $R = 64$ ke. Dead structure drop karna har stage ko ek fresh, achievable $R$ carry karne deta hai.

Recall Solution 4.3

Pehle sign convention (crucial): jab rocket fuel jalata hai, uska mass ghatta hai, isliye change $d m$ negative hai ( $d m < 0$ ). Actually ejected mass $- d m > 0$ hai. Momentum core $0 = m d v + u d m$ rearrange karne par milta hai: $d v = - u \frac{d m}{m} .$ Leading minus sign wahi hai jo $d v$ ko positive rakhta hai: kyunki $d m < 0$ , quantity $- d m / m > 0$ hai, isliye rocket speed up karta hai chahe mass gire. Integrate: $Δ v = - u \int_{m_{0}}^{m_{f}} \frac{d m}{m} = u ln \frac{m _{0}}{m _{f}} .$ Numbers: $R = 3000/1200 = 2.5$ , isliye $Δ v = 2.5 \times ln 2.5 = 2.5 \times 0.9163 = 2.291 km/s .$ Sign positive kyun flip hota hai: $\frac{d m}{m}$ ko bade $m_{0}$ se chhote $m_{f}$ tak integrate karna negative number deta hai ( $ln \frac{m _{f}}{m _{0}} < 0$ ); leading minus $Δ v$ ko positive banata hai. Rocket halka hone ke saath speed up karta hai.

Level 5 — Mastery

Goal: conversions, structure, aur payload trade-offs ke saath poora mission design.

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) Masses set up karo. Dry mass $m_{f} = m_{s t r u c t} + m_{p a y l o a d} = 0.06 m_{0} + 2000.$ Aur $m_{f} = m_{0} / R .$ Equate karo: $\frac{m _{0}}{R} = 0.06 m_{0} + 2000.$ Ab coefficient spell out karo: $\frac{1}{R} = \frac{1}{5.360} = 0.18657$ , isliye $\frac{m _{0}}{R} = 0.18657 m_{0} .$ Substitute karo: $0.18657 m_{0} = 0.06 m_{0} + 2000 \Rightarrow (0.18657 - 0.06) m_{0} = 2000 \Rightarrow 0.12657 m_{0} = 2000,$ $m_{0} = \frac{2000}{0.12657} = 15801 kg .$ (d) Propellant: $m_{f} = m_{0} / R = 15801/5.360 = 2948 kg$ , isliye $m_{p} = m_{0} - m_{f} = 15801 - 2948 = 12853 kg .$ Check: structure $= 0.06 \times 15801 = 948 kg$ , payload $2000 kg$ , sum $2948 = m_{f}$ . ✓ Consistent. Algebra kyun: dry mass dono sides par appear karta hai (yeh $m_{0}$ par structure fraction ke through depend karta hai), isliye hum seedha plug karne ki jagah ek linear equation solve karte hain.

Recall Solution 5.2

Naya $R$ : $R = e^{7.0/3.3354} = e^{2.099} \approx 8.158.$ Coefficient spell out karo: $\frac{1}{R} = \frac{1}{8.158} = 0.12259$ , isliye $\frac{m _{0}}{R} = 0.12259 m_{0} .$ 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: $25%$ zyaada $Δ v$ maangne par ( $5.6 \to 7.0$ ) poora rocket double ho gaya. Aur bhi bura, structure term $0.06 m_{0}$ shrinking $1/ R$ dry budget ka ek zyaada bada hissa khaata hai — denominator $0.06259$ chhota hai, isliye mass explode karta hai. Yeh nonlinearity hi poore parent note ka argument hai.

Recall Solution 5.3

Condition: $\frac{m _{0}}{R} = s m_{0} + m_{L}$ rearrange karo taaki $m_{0} (\frac{1}{R} - s) = m_{L} .$ Left factor positive hona chahiye ⇒ $1/ R > s$ ⇒ $R < 1/ s .$ 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 ko $7.0 < 9.383 km/s$ chahiye tha ✓ (feasible, isliye ek finite $m_{0}$ exist karta tha). Koi bhi $Δ v \geq 9.383 km/s$ ko $m_{0} \to \infty$ chahiye (ya halka structure, ya Multistage Rockets). Divergence kyun: jaise $R \to 1/ s$ , bracket $(1/ R - s) \to 0$ , isliye $m_{0} = m_{L} / (1/ R - s) \to \infty$ . Exponential wall literally ek asymptote ban jaati hai — koi payload ek stage mein us se aage nahi lift ho sakta.

Active recall

Recall Poori ladder ki one-line summary

Speed mass ratio ka log hai; mass ratio required speed ka exp hai; $Δ v$ 's add karo lekin mass ratios multiply karo; ek fixed structure fraction $s$ ek single stage ko $Δ v_{ma x} = u ln (1/ s)$ par cap karta hai. Aur hamesha $R > 1$ positive $Δ v$ ke liye ( $R = 1$ ⇒ koi fuel nahi ⇒ $Δ v = 0$ ).

Connections

Tsiolkovsky Rocket Equation — har problem $Δ v = u ln R$ ka application hai.
Specific Impulse Isp — L1/L2/L5 conversions $u = I_{s p} g_{0}$ .
Multistage Rockets — L4 dikhata hai kyun staging wall ko beat karta hai.
Delta-v Budget — L3/L5 mission- $Δ v$ requirements.
Conservation of Momentum — L4 derivation back-solve karta hai.
Exhaust Velocity and Thrust — $u$ ka physical source.