Exercises — Rocket staging — series staging, parallel staging

Hum kya karein: Tsiolkovsky mein numbers daalo. Mass ratio $=\frac{m_0}{m_f}=\frac{20000}{5000}=4$. $\Delta v=3000\times \ln (4)=3000\times 1.3863\approx 4159\text{ m/s}.$ Matlab kya hai: rocket ignition par $4\times$ bhari thi; woh ratio (raw masses nahi) hi speed gain decide karta hai.

Series (sequential). Series staging ki defining feature yeh hai ki stages ek ek karke jalti hain aur har ek ko drop kiya jaata hai pehle ki agla jale. Parallel staging mein core aur boosters saath mein liftoff se jalte.

Hum kya karein: hardware split karo, phir $m_0$ aur $m_f$ banao.

• Propellant $m_p=0.9\times 10000=9000$ kg.
• Structure $m_s=0.1\times 10000=1000$ kg.
• $m_0=S+\text{payload}=10000+2000=12000$ kg (full).
• $m_f=m_s+\text{payload}=1000+2000=3000$ kg (fuel jal gaya, tank abhi bhi attached hai). $\Delta v=3200\times \ln \left(\frac{12000}{3000}\right)=3200\times \ln (4)\approx 4436\text{ m/s}.$

Top-down kyun kaam karein? Har stage ka "payload" uske upar ki sab cheez hai, toh pehle upar ke masses jaanne chahiye. Figure dekho: stage 2, stage 1 par baithi hai, aur dono capsule ke neeche hain.

Stage 2 (sirf 1000 kg capsule carry karta hai):

• $m_{p,2}=0.88\times 6000=5280$ kg, $m_{s,2}=0.12\times 6000=720$ kg.
• $m_0=6000+1000=7000$ kg; $m_f=720+1000=1720$ kg. $\Delta v_2=3600\ln \left(\frac{7000}{1720}\right)=3600\times 1.4036\approx 5053\text{ m/s}.$

Stage 1 (stage 2 ka poora hardware + capsule $=6000+1000=7000$ kg carry karta hai):

• $m_{p,1}=0.86\times 60000=51600$ kg, $m_{s,1}=0.14\times 60000=8400$ kg.
• $m_0=60000+7000=67000$ kg; $m_f=8400+7000=15400$ kg. $\Delta v_1=2900\ln \left(\frac{67000}{15400}\right)=2900\times 1.4708\approx 4265\text{ m/s}.$

Total: $\Delta v=5053+4265\approx 9318$ m/s — low-Earth orbit ke liye zaruri ~9.4 km/s ke kareeb agar gravity losses jodon.

Phase 1 — sab saath jalte hain.

• $m_0=m_c+2m_b+m_{upper}=30000+2(15000)+4000=64000$ kg.
• Mass remove hua $=2m_{p,b}+m_{p,c}^{(1)}=2(13000)+9000=35000$ kg.
• $m_1=64000-35000=29000$ kg. $\Delta v_1=3000\ln \left(\frac{64000}{29000}\right)=3000\times 0.7916\approx 2375\text{ m/s}.$

Booster jettison: $2m_{s,b}=2(2000)=4000$ kg ke khale booster shells drop karo.

• $m_2=m_1-4000=29000-4000=25000$ kg.

Phase 2 — core apna bacha hua fuel jalaata hai.

• Remaining core propellant $m_{p,c}^{(2)}=27000-9000=18000$ kg.
• $m_{2,f}=m_{s,c}+m_{upper}=3000+4000=7000$ kg.
• Check: $m_2=m_{s,c}+m_{p,c}^{(2)}+m_{upper}=3000+18000+4000=25000$ kg. ✓ $\Delta v_2=4200\ln \left(\frac{25000}{7000}\right)=4200\times 1.2730\approx 5347\text{ m/s}.$

Total: $\Delta v=2375+5347\approx 7722$ m/s.

Zaruri mass ratio: Tsiolkovsky ulta karo. Kyunki $\Delta v=v_e\ln R$ hai, $R=m_0/m_f$ ke liye solve karo: $R=e^{\Delta v/v_e}=e^{9400/3500}=e^{2.6857}\approx \mathrm{14.67.}$ Yeh kya maangta hai: $m_f$ zyada se zyada $m_0/14.67\approx 0.0682m_0$ honi chahiye. Toh khali mass (structure + payload) liftoff mass ka zyada se zyada $6.82\%$ ho sakti hai. Contradiction: structure akela hi $ϵ=8\%$ stage hardware kha jaata hai, aur payload uske upar aur add hota hai. Kyunki $6.82\%<8\%$, khali mass akele hi budget se zyada hai koi bhi payload add karne se pehle. Impossible. Conclusion: tum payload bhi carry nahi kar sakte aur $m_f$ itna chhota bhi nahi rakh sakte — beech mein dead structure drop karna (staging) hi ek raasta hai. Yeh Payload fraction optimization se connect hota hai.

Do stages: har ek $\Delta v_i=3000\ln 6$ deta hai, aur gains add hote hain: $\Delta v=2\times 3000\ln 6=6000\times 1.7918\approx 10751\text{ m/s}.$ Single stage same $R=6$ par: sirf $3000\ln 6\approx 5375$ m/s. Doosra stage kya buy karta hai: yeh $\Delta v$ ko double kar deta hai, kyunki woh apne favourable mass ratio ke saath fresh start karta hai ek spent tank kheenchne ki jagah. Yeh compounding effect hai: equal ratios ke logs add karna ek giant ratio force karne ki koshish se behtar hai (jise structure allow nahi karta, jaise L4.1 mein dekha).

Step 1 — mass ratio per stage (Tsiolkovsky ulta karo): $R=e^{\Delta v_i/v_e}=e^{4500/3400}=e^{1.3235}\approx \mathrm{3.757.}$

Step 2 — $R$ ko hardware se relate karo. Hardware mass $S$ aur payload $P$ wali stage ke liye: $R=\frac{m_0}{m_f}=\frac{S+P}{ϵS+P},ϵ=\mathrm{0.10.}$ $S$ ke liye solve karo: cross-multiply $R(ϵS+P)=S+P$, jo deta hai $S(1-Rϵ)=P(R-1)\Rightarrow S=\frac{P(R-1)}{1-Rϵ}.$

Step 3 — stage 2 $P=1000$ kg carry karta hai:

Blow-up: jaise $R$ $1/ϵ=10$ ki taraf badhta hai, denominator $1-Rϵ\to 0^{+}$, toh zaruri hardware mass $S\to \infty$. Physically, $1/ϵ$ ka mass ratio matlab hoga ki poori stage propellant hai sirf khali structure hi final mass ke roop mein — koi bhi payload ke liye jagah nahi. Tum infinite fuel ki zarurat lete infinite tanks uthane ke liye: parent note wala vicious cycle, precisely bana diya. $R$ ki ceiling: kyunki $R$ $1/ϵ=10$ se neeche rehna chahiye, sabse bada single-stage $\Delta v$ hai

Series staging stages ko jalati hai ::: ek ek karke, har ek ko drop karti hai pehle ki agla jale Parallel staging core aur boosters ko jalati hai ::: liftoff se ek saath Parallel burn mein effective exhaust velocity weight ki jaati hai ::: thrust se Woh mass ratio jo ek single stage kabhi exceed nahi kar sakta ::: $1/ϵ$ (structural fraction ka ulta) $\Delta v$ target hit karne ke liye Tsiolkovsky ulta karo as ::: $R=e^{\Delta v/v_e}$