Optimal staging — equal mass ratios (for same Isp): Exercises

3.3.46 · D4 · Physics › Rocket Propulsion › Optimal staging — equal mass ratios (for same Isp)

Yeh page Optimal staging — equal mass ratios (for same Isp) ke liye ek graded workout hai. Har problem apna difficulty level batati hai aur ek full solution hide karti hai jise aapko sirf genuine attempt ke baad hi kholna chahiye. Agar koi symbol unfamiliar lage, toh yeh signal hai ki parent note ya uske prerequisites ko revisit karo: Tsiolkovsky rocket equation, Staging basics, Structural fraction, Payload fraction.

Poore time, hum in tools par rely karte hain:

Recall Teen formulas jo aap baar baar use karoge

Tsiolkovsky (ek burn): $Δ v = v_{e} ln R$ , jahaan $R = \frac{m _{before}}{m _{after}}$ mass ratio hai aur $v_{e}$ exhaust speed hai. Asli staging constraint (har stage ka leftover mass payload tak chain karo): $\frac{m _{p}}{M _{0}} = \prod_{i = 1}^{n} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} . (⋆)$ $n$ identical stages ke liye optimal single ratio ( $ϵ$ = structural fraction): $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}, k \equiv (\frac{m _{p}}{M _{0}})^{1/ n} .$

Level 1 — Recognition

Yeh test karte hain ki kya aap formulas padh sakte ho aur numbers plug in kar sakte ho. Koi cleverness required nahi.

Exercise 1.1 — Mass ratio padho

Ek stage $M_{before} = 8000$ kg par ignite hoti hai aur $M_{after} = 2000$ kg par burn out hoti hai. Iska mass ratio $R$ nikalo aur jo $Δ v$ yeh deliver karta hai woh bhi, agar $v_{e} = 2500$ m/s ho.

Recall Solution

Hum kya karte hain: mass ratio bas "before over after" hai. $R = \frac{8000}{2000} = 4.$ $ln$ kyun? Rocket equation kehta hai ki mass ka har equal fractional loss utni hi additive velocity deta hai — mass fractions ka yeh repeated multiplication logarithm ke neeche addition ban jaata hai. Toh: $Δ v = v_{e} ln R = 2500 \cdot ln 4 = 2500 \cdot 1.3863 = 3466 m/s .$

Exercise 1.2 — Masses se structural fraction nikalo

Ek stage mein $m_{f} = 4500$ kg fuel aur $m_{s} = 500$ kg structure hai. Structural fraction $ϵ$ compute karo.

Recall Solution

Definition: $ϵ$ structure divided by (structure + fuel) hai — stage ka "dead weight fraction". $ϵ = \frac{m _{s}}{m _{s} + m _{f}} = \frac{500}{500 + 4500} = \frac{500}{5000} = 0.1.$

Exercise 1.3 — Optimum ko words mein batao

Blanks fill karo (check karne ke liye reveal karo):

n

stages jo same

ϵ

aur same

v_{e}

share karte hain, unke liye maximum

Δ v

tab hota hai jab saare mass ratios

equal hon,

R_{1} = R_{2} = \dots = R_{n} = R^{*}

Woh tool jo yeh prove karta hai (kyunki hum ek quantity ko ek constraint ke under maximize karte hain) woh hai

Lagrange multipliers.

Level 2 — Application

Ab aap do ya teen formulas ko sequence mein assemble karte ho.

Exercise 2.1 — Do-stage rocket ke liye Optimal $R^{*}$

$m_{p} = 800$ kg, $M_{0} = 12800$ kg, $ϵ = 0.08$ , do stages. $R^{*}$ nikalo.

Recall Solution

Step 1 — $k$ nikalo. $k$ payload fraction ka $n$ -th root hai: $k = (\frac{m _{p}}{M _{0}})^{1/2} = (\frac{800}{12800})^{1/2} = (0.0625)^{1/2} = 0.25.$ Step 2 — optimal-ratio formula mein plug karo. $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k} = \frac{1}{0.08 + 0.92 \cdot 0.25} = \frac{1}{0.08 + 0.23} = \frac{1}{0.31} = 3.226.$ Yeh formula kyun use karein aur $(M_{0} / m_{p})^{1/ n}$ kyun nahi? Woh simpler form sirf tab kaam karta hai jab $ϵ = 0$ ho. Yahaan structure real hai, toh hume woh version use karna hai jo $ϵ$ include karta hai.

Exercise 2.2 — Optimal staging se total $Δ v$

Exercise 2.1 ko $v_{e} = 3200$ m/s ke saath continue karo. $Δ v_{total}$ nikalo.

Recall Solution

Dono stages $R^{*} = 3.226$ share karti hain, toh $Δ v_{total} = n v_{e} ln R^{*} = 2 \cdot 3200 \cdot ln 3.226 = 6400 \cdot 1.1714 = 7497 m/s .$

Exercise 2.3 — Stage-1 ka fuel reconstruct karo

Exercise 2.1 ke rocket ke liye, stage 1 kitna fuel burn karta hai?

Recall Solution

Hum kya use karte hain: ek stage mein burn hua fuel initial mass times lost fraction hai, $m_{f} = M_{before} (1 - 1/ R)$ . Stage 1, $M_{0} = 12800$ kg par ignite hoti hai: $m_{f 1} = 12800 (1 - \frac{1}{3.226}) = 12800 (1 - 0.31) = 12800 \cdot 0.69 = 8832 kg .$

Level 3 — Analysis

Yahaan aap scenarios compare karte ho, ya formula ko "backwards" run karte ho.

Exercise 3.1 — Real structure ke saath equal vs unequal

Do stages, $ϵ = 0.1$ , $v_{e} = 3000$ m/s, $M_{0} = 10000$ kg, $m_{p} = 1000$ kg. Parent note ne equal solution $R^{*} = 2.60$ find ki thi jo $Δ v = 5734$ m/s deti thi. Ab unequal split $R_{1} = 3.5$ try karo. Constraint $(⋆)$ use karke forced $R_{2}$ solve karo, aur $Δ v$ compute karo. Kaun jeetega?

Recall Solution

Step 1 — do stages ke liye $(⋆)$ likho. $\frac{m _{p}}{M _{0}} = \frac{1 - ϵ R _{1}}{( 1 - ϵ ) R _{1}} \cdot \frac{1 - ϵ R _{2}}{( 1 - ϵ ) R _{2}} = 0.1.$ Step 2 — $R_{1} = 3.5$ plug karo. $\frac{1 - 0.1 \cdot 3.5}{0.9 \cdot 3.5} = \frac{0.65}{3.15} = 0.20635.$ Step 3 — doosra factor solve karo. $\frac{1 - 0.1 R _{2}}{0.9 R _{2}} = \frac{0.1}{0.20635} = 0.48462.$ Cross-multiply: $1 - 0.1 R_{2} = 0.43616 R_{2} \Rightarrow 1 = 0.53616 R_{2} \Rightarrow R_{2} = 1.8651$ . Step 4 — $Δ v$ compute karo. $Δ v = 3000 (ln 3.5 + ln 1.8651) = 3000 (1.2528 + 0.6234) = 3000 \cdot 1.8762 = 5629 m/s .$ Conclusion: $5629 < 5734$ . Equal ratios jeette hain ~105 m/s se. Yeh woh strict advantage hai jo sirf tab appear hota hai jab $ϵ > 0$ ho — exactly wahi jo Lagrange argument predict karta hai.

Exercise 3.2 — Single-stage ceiling

$ϵ = 0.1$ ke saath, ek single stage maximum kitna mass ratio reach kar sakta hai (payload $\to 0$ )? Kya ek single stage $M_{0} / m_{p} = 30$ reach kar sakta hai?

Recall Solution

$n = 1$ lo, toh $k = m_{p} / M_{0}$ . Jaise $m_{p} \to 0$ , $k \to 0$ : $R_{m a x} = \frac{1}{ϵ + ( 1 - ϵ ) \cdot 0} = \frac{1}{ϵ} = \frac{1}{0.1} = 10.$ Ek single stage is $ϵ$ ke saath kabhi bhi $R = 10$ se aage nahi ja sakta. $M_{0} / m_{p} = 30$ reach karne ke liye $R = 30 > 10$ chahiye: ek stage mein impossible. Yahi pura reason hai ki staging exist karti hai — aap structure shed karte ho taaki $1/ ϵ$ ceiling se aage jaa sako.

Exercise 3.3 — Compounding ceiling ko beat karta hai

Do identical stages, har ek $R = 5.5$ ke saath. Ideal ( $ϵ \to 0$ ) limit mein pair kaunsa effective mass ratio $M_{0} / m_{p}$ achieve karta hai? 3.2 ke single-stage ceiling se compare karo.

Recall Solution

$ϵ \to 0$ limit mein constraint $\prod R_{i} = M_{0} / m_{p}$ hai, toh ratios multiply hote hain: $\frac{M _{0}}{m _{p}} = R_{1} R_{2} = 5.5 \times 5.5 = 30.25.$ Har individual stage ka $R = 5.5$ comfortably $R_{m a x} = 10$ ceiling se neeche hai, phir bhi saath milke $30$ se zyada achieve karte hain — ek stage ke liye impossible. Staging mass ratios ko compound karta hai.

Level 4 — Synthesis

Multi-part problems jo poore chapter ko stitch karte hain.

Exercise 4.1 — Full three-stage design

Ek three-stage rocket design karo: $M_{0} = 27000$ kg, $m_{p} = 1000$ kg, $ϵ = 0.1$ per stage, $v_{e} = 3000$ m/s. (a) $R^{*}$ nikalo. (b) $Δ v_{total}$ nikalo. (c) Stage 2 mein enter karne wala mass $M_{1}$ compute karo.

Recall Solution

(a) $k = (m_{p} / M_{0})^{1/3} = (1000/27000)^{1/3} = (0.037037)^{1/3} = 0.33333$ . $R^{*} = \frac{1}{0.1 + 0.9 \cdot 0.33333} = \frac{1}{0.1 + 0.3} = \frac{1}{0.4} = 2.5.$ (b) $Δ v = n v_{e} ln R^{*} = 3 \cdot 3000 \cdot ln 2.5 = 9000 \cdot 0.91629 = 8247 m/s .$ (c) Har stage mass ko factor $\frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}} = k = 0.33333$ se multiply karta hai: $M_{1} = M_{0} \cdot k = 27000 \cdot 0.33333 = 9000 kg .$ (Check: $M_{2} = 9000 \cdot 0.3333 = 3000$ , $M_{3} = 3000 \cdot 0.3333 = 1000 = m_{p}$ . ✓)

Exercise 4.2 — Kya 4th stage worth it hai?

Wahi rocket as 4.1 lekin ab $n = 4$ stages mein split karo ( $ϵ = 0.1$ each). $Δ v$ compute karo aur 3-stage answer se compare karo. Interpret karo.

Recall Solution

$k = (1000/27000)^{1/4} = (0.037037)^{0.25} = 0.43869$ . $R^{*} = \frac{1}{0.1 + 0.9 \cdot 0.43869} = \frac{1}{0.1 + 0.39482} = \frac{1}{0.49482} = 2.0209.$ $Δ v = 4 \cdot 3000 \cdot ln 2.0209 = 12000 \cdot 0.70359 = 8443 m/s .$ Compare: 3 stages ne 8247 m/s diya, 4 stages 8443 m/s deta hai — sirf 196 m/s (~2.4%) ka gain. Har extra stage kam help karta hai (diminishing returns) jabki real hardware, interstages aur complexity add karta hai jo model ignore karta hai. Real rockets 2–3 stages par rok lete hain.

Exercise 4.3 — Required $M_{0}$ back-solve karo

Aapko $Δ v_{total} = 9000$ m/s chahiye, $n = 2$ , $ϵ = 0.05$ , $v_{e} = 3400$ m/s, $m_{p} = 500$ kg ke saath. Kitna launch mass $M_{0}$ required hai?

Recall Solution

Step 1 — needed $R^{*}$ per stage. $Δ v = 2 v_{e} ln R^{*}$ : $ln R^{*} = \frac{9000}{2 \cdot 3400} = 1.32353 \Rightarrow R^{*} = e^{1.32353} = 3.7566.$ Step 2 — per-stage mass factor $k$ . $k = \frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}} = \frac{1 - 0.05 \cdot 3.7566}{0.95 \cdot 3.7566} = \frac{0.81217}{3.5688} = 0.22758.$ Step 3 — constraint invert karo. $m_{p} / M_{0} = k^{2}$ , toh $M_{0} = \frac{m _{p}}{k ^{2}} = \frac{500}{0.2275 8 ^{2}} = \frac{500}{0.051795} = 9653 kg .$

Level 5 — Mastery

Yeh demand karte hain ki aap derive karo, sirf plug na karo.

Exercise 5.1 — Scratch se equal ratios prove karo (2 stages, $ϵ \to 0$ )

$ϵ \to 0$ ke saath constraint $R_{1} R_{2} = C$ (fixed) hai. $f = ln R_{1} + ln R_{2}$ maximize karo aur calculus use karke prove karo ki optimum $R_{1} = R_{2} = C$ hai.

Recall Solution

Constraint eliminate karo. $R_{2} = C / R_{1}$ , toh $f (R_{1}) = ln R_{1} + ln \frac{C}{R _{1}} = ln R_{1} + ln C - ln R_{1} = ln C .$ Surprise: $f$ , $R_{1}$ mein constant hai! $ϵ \to 0$ ideal case mein har split same $Δ v$ deta hai (yeh parent note ki "equal beats unequal is not strict here" se match karta hai). Optimum degenerate hai — koi bhi split, including $R_{1} = R_{2} = C$ , optimal hai. Equal ratios ka theorem phir bhi kyun naam leta hai: $R_{i}$ ke equal hone ki strict preference sirf tab appear hoti hai jab $ϵ > 0$ constraint ko nonlinear banata hai (Ex 3.1 dekho). Equal ratios woh split hai jo dono regimes mein survive karta hai, isliye yeh safe design rule hai.

Exercise 5.2 — Lagrange derivation dikhata hai ki saare $R_{i}$ ek equation satisfy karte hain

General $ϵ$ ke liye, hum $f = \sum_{i} ln R_{i}$ maximize karte hain subject to $g = \sum_{i} ln \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} = const$ . Dikhao ki $\partial g / \partial R_{i}$ sirf $R_{i}$ par depend karta hai, aur conclude karo ki saare $R_{i}$ equal hain.

Recall Solution

$g$ ka ek term differentiate karo. $g_{i} = ln (1 - ϵ R_{i}) - ln (1 - ϵ) - ln R_{i}$ likho. Tab $\frac{\partial g}{\partial R _{i}} = \frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}} .$ Aur $\partial f / \partial R_{i} = 1/ R_{i}$ . Lagrange condition $\nabla f = λ \nabla g$ har $i$ ke liye kehta hai: $\frac{1}{R _{i}} = λ (\frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}}) .$ Key observation: is equation mein har quantity sirf $R_{i}$ ki function hai only (kyunki $ϵ$ aur $λ$ shared constants hain). Toh saare $n$ stages ek unknown mein identical equation satisfy karte hain. Aise well-behaved equation ka ek physical root hota hai ⇒ $R_{1} = R_{2} = \dots = R^{*}$ . Equal mass ratios optimal hain. $■$

Exercise 5.3 — Sensitivity: galat $ϵ$ aapko kitna cost karta hai

Aapne Ex 4.1 rocket design kiya tha ( $n = 3$ , planned $ϵ = 0.1$ , $R^{*} = 2.5$ , $Δ v = 8247$ m/s). Manufacturing heavier aayi: actual $ϵ = 0.12$ . $Δ v$ recompute karo (same $M_{0}, m_{p}, n$ rakho lekin $ϵ = 0.12$ ke saath), nayi optimal $Δ v$ aur loss nikalo.

Recall Solution

Same $k = (1000/27000)^{1/3} = 0.33333$ (unchanged — yeh sirf $m_{p} / M_{0}$ aur $n$ par depend karta hai). $R_{new}^{*} = \frac{1}{0.12 + 0.88 \cdot 0.33333} = \frac{1}{0.12 + 0.29333} = \frac{1}{0.41333} = 2.4194.$ $Δ v_{new} = 3 \cdot 3000 \cdot ln 2.4194 = 9000 \cdot 0.88354 = 7952 m/s .$ Loss: $8247 - 7952 = 295$ m/s. Structural fraction mein 2-percentage-point ki rise lagbhag 300 m/s cost karti hai — yeh dikhata hai ki real design mein lightweight structures (Structural fraction) par itna obsess kyun kiya jaata hai.

Recall Master check — ek ek karke reveal karo

$n$ equal- $I s p$ stages ke liye optimal rule ::: saare mass ratios equal, $R_{1} = \dots = R_{n} = R^{*}$ . $R^{*}$ ka formula ::: $R^{*} = 1/ (ϵ + (1 - ϵ) k)$ jahaan $k = (m_{p} / M_{0})^{1/ n}$ . Single-stage ceiling ::: $R_{m a x} = 1/ ϵ$ . Hume Lagrange kyun chahiye, sirf AM–GM kyun nahi ::: $ϵ > 0$ ke liye product $\prod R_{i}$ fixed nahi hota; asli constraint $(⋆)$ nonlinear hai.