Staging events — separation dynamics, thrust tail-off: Exercises

3.4.17 · D4 · Physics › Rocket Flight Mechanics › Staging events — separation dynamics, thrust tail-off

Shuru karne se pehle, yahan ek card par poora toolkit hai taaki neeche koi bhi symbol "unexplained" na lage.

Yahan woh geometry hai jis par aap separation problems ke liye rely karenge.

Figure — Staging events — separation dynamics, thrust tail-off

Level 1 — Recognition

Goal: tool ka naam lo aur plug in karo. Abhi koi traps nahi.

Exercise 1.1

Ek engine thrust $F_{0} = 40 kN$ rakhta hai aur cutoff command ke baad uski thrust tail-off time $τ = 0.25 s$ ke saath decay karti hai. Tail-off impulse $J_{t ai l}$ kya hai?

Recall Solution

KAUN SA tool: tail-off impulse decaying-thrust curve ke neeche ka area hai, jo height $F_{0}$ aur width $τ$ ke rectangle ke barabar hota hai. $J_{t ai l} = F_{0} τ = 40000 \times 0.25 = 10000 N\cdots = 10 kN\cdots .$ $F_{0} τ$ kyun aur koi fancy cheez kyun nahi: exponential $e^{- t / τ}$ ko $0$ se $\infty$ tak integrate karne par exactly $τ$ milta hai, isliye sirf $F_{0} \times τ$ bachta hai.

Exercise 1.2

Do stages: spent lower $m_{1} = 6000 kg$ , upper $m_{2} = 3000 kg$ . Reduced mass $μ$ compute karo.

Recall Solution

Reduced mass kyun: do stages ek doosre par push karti hain (internal force), isliye unki relative motion ek aisi particle ki tarah behave karti hai jiska mass $μ$ ho. $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{6000 \times 3000}{6000 + 3000} = \frac{1.8 \times 1 0 ^{7}}{9000} = 2000 kg .$ Gaur karo ki $μ = 2000$ dono masses se chota hai — yeh hamesha sach hota hai. Yeh lighter body ke zyada kareeb hota hai.

Level 2 — Application

Goal: do formulas ko chain karo.

Exercise 2.1

$m = 4000 kg$ mass wale ek upper stage ko tail-off suffer hoti hai jisme $F_{0} = 50 kN$ , $τ = 0.3 s$ hai. Uncommanded velocity kick $Δ v_{t ai l}$ nikalo.

Recall Solution

Step 1 — impulse (KYA/KYUN): tail ke neeche ka area $J_{t ai l} = F_{0} τ = 50000 \times 0.3 = 15000 N\cdots$ hai. Step 2 — impulse ko speed mein badlo: Impulse-Momentum Theorem se, $J = m Δ v$ , isliye $Δ v_{t ai l} = \frac{J _{t ai l}}{m} = \frac{15000}{4000} = 3.75 m/s .$ Yeh 3.75 m/s free velocity hai jise guidance ko predict karna hoga, warna burnout speed galat ho jaayegi.

Exercise 2.2

Springs spent stage $m_{1} = 6000 kg$ aur upper $m_{2} = 3000 kg$ ke beech $J_{se p} = 2400 N\cdots$ deliver karti hain. $v_{r e l}$ aur $t = 2.0 s$ ke baad gap nikalo.

Recall Solution

Step 1: Exercise 1.2 se $μ = 2000 kg$ reuse karo. Step 2 — relative speed: $v_{r e l} = \frac{J _{se p}}{μ} = \frac{2400}{2000} = 1.2 m/s .$ Step 3 — gap (constant drift, YEH KAISA DIKHTA HAI: figure mein orange arrow bas lambi hoti jaati hai): $d = v_{r e l} t = 1.2 \times 2.0 = 2.4 m .$

Level 3 — Analysis

Goal: effects alag karo, conservation check karo, "enough hai ya nahi" ke baare mein reason karo.

Exercise 3.1

Same masses ( $m_{1} = 6000$ , $m_{2} = 3000$ kg) aur $J_{se p} = 2400 N\cdots$ . Har stage ki apni velocity change nikalo, phir verify karo ki total momentum conserved hai.

Recall Solution

Newton's third law: har body ko equal-and-opposite impulse milta hai.

Upper (forward push milta hai): $Δ v_{2} = + \frac{J _{se p}}{m _{2}} = \frac{2400}{3000} = + 0.8 m/s .$
Lower (backward recoil): $Δ v_{1} = - \frac{J _{se p}}{m _{1}} = - \frac{2400}{6000} = - 0.4 m/s .$
Relative check: $Δ v_{2} - Δ v_{1} = 0.8 - (- 0.4) = 1.2 m/s = v_{r e l}$ ✓.
Momentum check (Conservation of Momentum): $m_{2} Δ v_{2} + m_{1} Δ v_{1} = 3000 (0.8) + 6000 (- 0.4) = 2400 - 2400 = 0$ ✓. Internal forces kisi ke centre of mass ko nahi hilaate.

Exercise 3.2

Mission rule: stages ko upper engine light karne se pehle $t_{c l e a r} = 1.5 s$ ke andar $d = 3.0 m$ ka gap kholna chahiye. $μ = 2000 kg$ ke saath, springs ko minimum kitna $J_{se p}$ deliver karna chahiye?

Recall Solution

Step 1 — required relative speed: $v_{r e l} \geq \frac{d}{t _{c l e a r}} = \frac{3.0}{1.5} = 2.0 m/s .$ Step 2 — separation formula ko invert karo: kyunki $v_{r e l} = J_{se p} / μ$ , $J_{se p} \geq μ v_{r e l} = 2000 \times 2.0 = 4000 N\cdots .$ Isse kum hua toh plume abhi-tak-clear-na-hui stage se takra sakta hai.

Level 4 — Synthesis

Goal: tail-off + separation, ya separation + rocket equation combine karo.

Exercise 4.1

$m = 4000 kg$ mass wale ek upper stage ko pehle tail-off ( $F_{0} = 50 kN$ , $τ = 0.3 s$ ) hoti hai jo forward speed add karti hai, phir springs ( $J_{se p} = 3600 N\cdots$ ) use spent stage $m_{1} = 6000 kg$ se aage dhakka deti hain. (a) In dono events se upper stage ka total velocity gain, aur (b) $v_{r e l}$ compute karo.

Recall Solution

(a) Same body par do forward pushes add hote hain:

Tail-off kick: $Δ v_{t ai l} = F_{0} τ / m = 15000/4000 = 3.75 m/s$ (Ex 2.1 se).
Separation push upper stage par: $Δ v_{2} = J_{se p} / m_{2} = 3600/4000 = 0.9 m/s$ .
Upper-stage ka total gain $= 3.75 + 0.9 = 4.65 m/s .$ (b) Relative speed reduced mass use karta hai $m_{1} = 6000$ , $m_{2} = 4000$ ke saath: $μ = \frac{6000 \times 4000}{6000 + 4000} = \frac{2.4 \times 1 0 ^{7}}{1 0 ^{4}} = 2400 kg, v_{r e l} = \frac{3600}{2400} = 1.5 m/s .$ Kyun yeh alag hain: $Δ v_{2} = 0.9$ m/s woh hai jitna sirf upper stage speed up hota hai; $v_{r e l} = 1.5$ m/s woh hai jitni tezi se dono alag hote hain (isme lower stage ka backward recoil bhi count hota hai).

Exercise 4.2

Ek upper stage wet mass $m_{0} = 9000 kg$ se dry mass $m_{f} = 3000 kg$ tak $I_{s p} = 340 s$ ke saath burn karta hai (lo $g_{0} = 9.81 m/s^{2}$ ). Exhaust speed $v_{e}$ aur Tsiolkovsky Rocket Equation se stage ka $Δ v$ nikalo.

Recall Solution

Step 1 — exhaust speed: $v_{e} = I_{s p} g_{0} = 340 \times 9.81 = 3335.4 m/s .$ Step 2 — rocket equation (LOG KYUN: yeh count karta hai ki fuel nikalne par mass ratio continuously kaise shrink hota hai): $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 3335.4 \times ln \frac{9000}{3000} = 3335.4 \times ln 3.$ $ln 3 = 1.0986$ ke saath: $Δ v = 3335.4 \times 1.0986 \approx 3664.4 m/s .$

Level 5 — Mastery

Goal: poori reasoning, edge cases, design decisions.

Exercise 5.1 (design margin)

Springs $J_{se p} = 4000 N\cdots$ deliver karne ke liye certified hain. Stages: $m_{1} = 8000 kg$ , $m_{2} = 2000 kg$ . Ignition tab tak wait karni hai jab tak $d = 3.5 m$ ka gap na ho, aur flight plan sirf $t_{c l e a r} = 1.8 s$ allow karta hai. Kya design clear karta hai? Margin batao.

Recall Solution

Step 1 — $μ$ : $μ = \frac{8000 \times 2000}{10000} = 1600 kg .$ Step 2 — achieved relative speed: $v_{r e l} = \frac{4000}{1600} = 2.5 m/s .$ Step 3 — allowed time mein achieved gap: $d_{a c h} = v_{r e l} t_{c l e a r} = 2.5 \times 1.8 = 4.5 m .$ Step 4 — verdict: required $3.5$ m tha, achieved $4.5$ m hai → clear karta hai, margin $4.5 - 3.5 = 1.0 m$ ke saath (requirement se lagbhag $29%$ zyada). ✓

Exercise 5.2 (limiting / degenerate case)

Separation formula $v_{r e l} = J_{se p} / μ$ lo jahan $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ hai. Maano spent lower stage upper stage ke comparison mein bahut badi hai: $m_{2}$ aur $J_{se p}$ fixed rakhte hue $m_{1} \to \infty$ . $v_{r e l}$ kis cheez ke paas jaata hai, aur iska physical matlab kya hai?

Recall Solution

Limit lo: $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{m _{2}}{1 + m _{2} / m _{1}} m_{1} \to \infty m_{2} .$ Isliye $v_{r e l} \to \frac{J _{se p}}{m _{2}} = Δ v_{2}$ . Physical meaning: infinitely heavy lower stage bilkul recoil nahi karti ( $Δ v_{1} = - J_{se p} / m_{1} \to 0$ ). Tab saari relative motion upper stage ki apni motion hai; separation speed upper stage ki apni speed change ke barabar ho jaati hai. Numbers se check karo: $m_{2} = 2000$ , $J_{se p} = 4000$ → $v_{r e l} \to 2.0 m/s$ . Yeh wohi "battleship se bullet chalao toh recoil nahi hota" wala idea hai: ship almost nahi hilti.

Exercise 5.3 (cutoff timing correction)

Ek stage ko ek precise burnout speed hit karni hai. Guidance ek tail-off kick $Δ v_{t ai l} = F_{0} τ / m$ predict karta hai. $F_{0} = 60 kN$ , $τ = 0.4 s$ , $m = 5000 kg$ ke saath, velocity terms mein commanded cutoff ko kitna pehle move karna padega, aur agar us waqt stage $a = 25 m/s^{2}$ par accelerate kar rahi hai, toh tail-off kick cancel karne ke liye time mein roughly kitna pehle cutoff command karna chahiye?

Recall Solution

Step 1 — unwanted kick: $Δ v_{t ai l} = \frac{F _{0} τ}{m} = \frac{60000 \times 0.4}{5000} = 4.8 m/s .$ Guidance ko target se yeh subtract karna hoga. Step 2 — speed error ko time-advance mein convert karo: cutoff ke paas stage $a$ rate se speed gain karti hai, isliye $δ t$ pehle cutoff karne par $a δ t$ speed remove hoti hai. Set karo $a δ t = Δ v_{t ai l}$ : $δ t = \frac{Δ v _{t ai l}}{a} = \frac{4.8}{25} = 0.192 s .$ Interpretation: shutdown ko lagbhag $0.19$ s pehle command karo taaki tail-off dribble aakhri $4.8$ m/s exactly fill kar sake. Gaur karo ki $δ t = 0.192 s$ , $τ = 0.4$ s se chota hai — aap poore tail-off duration se advance nahi karte, sirf utna jitna time normal thrust ko same $4.8$ m/s build karne mein lagta hai.

Recall Quick self-test (cloze)

Tail-off impulse == $F_{0} τ$ == ke barabar hota hai kyunki $F_{0} e^{- t / τ}$ ka area $0$ se $\infty$ tak == $τ$ hai. Separation relative speed reduced mass== $μ = m_{1} m_{2} / (m_{1} + m_{2})$ use karta hai, jo hamesha dono masses se smaller hota hai. Ek tail-off kick $Δ v_{t ai l}$ cancel karne ke liye aap cutoff == $Δ v_{t ai l} / a$ == se advance karte ho, na ki $τ$ se.