Staging events — separation dynamics, thrust tail-off: Worked examples

3.4.17 · D3 · Physics › Rocket Flight Mechanics › Staging events — separation dynamics, thrust tail-off

Kuch bhi compute karne se pehle, hum do formulas aur unke peeche ki picture restate karte hain, taaki kuch bhi bina samjhe use na ho.

Formula Tail-off force profile (yahan restate kiya gaya hai taaki tum yeh page kabhi chhodna na pado)

Jab tum "engine off" command dete ho $t = 0$ par, thrust usi waqt gayab nahi hoti — chamber mein bacha hua propellant jalता rehta hai aur dheere dheere fade hota hai. Hum us fade ko is tarah model karte hain: $F (t) = F_{0} e^{- t / τ} .$

$F_{0}$ = cutoff command ke instant par thrust (newtons $N$ mein ek dhakka) — fade ka starting height.
$τ$ ("tau") = tail-off time seconds mein — dribble kitni dheere khatam hoti hai. Ek $τ$ ke baad, thrust $F_{0} / e \approx 0.37 F_{0}$ tak gir jaati hai.
$e^{- t / τ}$ shrinking factor hai: $t = 0$ par yeh $1$ hai (full thrust), aur $0$ ki taraf decay karta hai. Impulse (deliver ki gayi total push) is curve ke neeche ka area hai — neeche ke har tail-off example mein yahi compute hota hai.

Scenario matrix

Har staging problem inhi cells mein se ek hai. Neeche ke worked examples mein se har ek us cell ke saath tagged hai jise woh cover karta hai.

Cell	Kya special hai	Covered by
A Ordinary tail-off	plain $F_{0}, τ$ finite, positive	Ex 1
B Tail-off, degenerate $τ \to 0$	instant cutoff limit	Ex 2
C Tail-off, large- $τ$ / partial window	sirf finite $t$ tak integrate karo, $\infty$ tak nahi	Ex 3
D Separation, unequal masses	general $m_{1} \neq = m_{2}$	Ex 4
E Separation, equal masses $m_{1} = m_{2}$	symmetric limit, $μ = m /2$	Ex 5
F Separation, extreme mass ratio	$m_{1} ≫ m_{2}$ (tiny upper stage) → $μ \to m_{2}$	Ex 6
G Clearance / collision-risk word problem	required $J_{se p}$ ko gap deadline se solve karo	Ex 7
H Combined real-world / exam twist	tail-off aur separation saath mein; sign bookkeeping	Ex 8
I Zero-input sanity	$J_{se p} = 0$ → kya hoga?	Ex 5 verify ke andar

Signs matter karte hain: hum forward (flight ki direction) ko positive lete hain. Upper stage ko $+$ push milta hai, discard ki gayi lower stage ko $-$ recoil.

Ex 1 — Cell A: ordinary tail-off

Worked example Plain case

Ek first-stage engine cutoff command par $F_{0} = 120 kN$ read karta hai. Tail-off time $τ = 0.6 s$ . Us instant par stage mass $m = 20000 kg$ hai. Tail-off impulse $J_{t ai l}$ aur un-commanded velocity kick $Δ v_{t ai l}$ nikalo.

Forecast: guess karo — extra $Δ v$ kuch cm/s hoga, kuch m/s, ya tens of m/s?

Units convert karo. $F_{0} = 120 kN = 120000 N$ . Yeh step kyun? Formula newtons use karta hai; 1000 ka stray factor sabse common numeric error hai.
Impulse = fade curve ke neeche ka area. Restate kiye gaye profile $F (t) = F_{0} e^{- t / τ}$ ka use karke, $J_{t ai l} = \int_{0}^{\infty} F_{0} e^{- t / τ} d t = F_{0} τ = 120000 \times 0.6 = 72000 N\cdots .$ Yeh step kyun? Impulse $F (t)$ ke neeche ka area hai; poori decaying curve ka area ek rectangle ke barabar hai jiska height $F_{0}$ aur width $τ$ ho.
Velocity kick. $Δ v_{t ai l} = \frac{J _{t ai l}}{m} = \frac{72000}{20000} = 3.6 m/s$ . Yeh step kyun? Impulse hi momentum mein change hai, $J = m Δ v$ , toh $Δ v = J / m$ .

Verify: Units — $\frac{N\cdots}{kg} = \frac{kg\cdotm/s ^{2} \cdot s}{kg} = m/s$ ✓. Magnitude $3.6$ m/s "kuch m/s" hai — beech wala forecast. Yahi real velocity hai jiske liye guidance budget karna padta hai.

Ex 2 — Cell B: instant-cutoff limit ( $τ \to 0$ )

Worked example Degenerate case jo log

hamesha sach maante hain Same engine, $F_{0} = 120 kN$ , lekin ab ek idealized valve imagine karo jo bilkul perfectly instantly band ho jaata hai: $τ \to 0$ . Model kya tail-off impulse predict karta hai?

Forecast: agar fade zero time leta hai, toh kya phir bhi koi "hidden" impulse hogi?

Limit lo. $J_{t ai l} = F_{0} τ \to 120000 \times 0 = 0 N\cdots$ . Yeh step kyun? Hum degenerate input test kar rahe hain yeh ensure karne ke liye ki formula apni boundary par sane behave kare.
Interpret karo. $τ = 0$ ke saath thrust sach mein ek switch hai — koi leftover propellant nahi jalta, koi extra impulse nahi. Yeh step kyun? Yeh impulse-momentum picture confirm karta hai: $F (t)$ curve ke neeche koi area nahi ⇒ koi momentum change nahi.

Verify: Jab $τ \to 0$ , $\int_{0}^{\infty} F_{0} e^{- t / τ} d t = F_{0} τ \to 0$ ✓. Toh "thrust is a switch" wala belief exactly $τ = 0$ special case hai — real engines mein $τ > 0$ hota hai, isliye parent note tumhe yeh assume karne se mana karta hai.

Ex 3 — Cell C: sirf tail ka kuch hissa count hota hai

Worked example Finite window, infinity tak nahi

Kabhi kabhi separation hoti hai pehle ki tail poori tarah die kare. Same $F_{0} = 120 kN$ , $τ = 0.6 s$ , lekin tum sirf pehle $t_{1} = 0.6 s$ (ek $τ$ ) mein deliver hua impulse chahte ho cutoff ke baad. Woh partial impulse $J (t_{1})$ nikalo.

Forecast: ek full $τ$ guzar gaya hai — tumhara kya khayal hai, tumne roughly aadha, do-tihaai, ya almost sab $72000$ N·s collect kar liya?

Integral set up karo finite top limit ke saath. $J (t_{1}) = \int_{0}^{t_{1}} F_{0} e^{- t / τ} d t .$ Yeh step kyun? Full-tail integral $\infty$ tak jaata hai; yahan physical window $t_{1}$ par cut short hai, toh top limit finite hai aur answer mein ek leftover term aayega.
Integral karo, step by step. $e^{- t / τ}$ ka antiderivative $- τ e^{- t / τ}$ hai (ise differentiate karke check karo: $\frac{d}{d t} [- τ e^{- t / τ}] = e^{- t / τ}$ ✓). Toh $J (t_{1}) = F_{0} [- τ e^{- t / τ}]_{0}^{t_{1}} = F_{0} (- τ e^{- t_{1} / τ} - (- τ e^{0})) = F_{0} τ (1 - e^{- t_{1} / τ}) .$ Yeh step kyun? Do limits plug in karke subtract karna exactly yahi hai jaise ek definite integral $0$ aur $t_{1}$ ke beech ka area collect karta hai; constant $τ$ bahar aa jaata hai aur do exponentials $(1 - e^{- t_{1} / τ})$ shape dete hain.
$t_{1} = τ$ plug karo taaki $t_{1} / τ = 1$ ho. $J (τ) = 72000 (1 - e^{- 1}) = 72000 \times 0.63212 \approx 45513 N\cdots .$ Yeh step kyun? Ek time-constant hamesha same fraction $1 - e^{- 1} \approx 63.2%$ deliver karta hai — yeh number yaad karne layak hai.

Verify: Fraction $= 1 - e^{- 1} = 0.632$ ✓, forecast "roughly two-thirds" se match karta hai. Bacha hua $36.8%$ $t_{1}$ ke baad dribble out hota hai, confirming karta hai ki Ex 1 ka total bada hai.

Figure 1 neeche exactly yeh split draw karta hai. Horizontal axis cutoff ke baad time $t$ (s) hai, vertical axis thrust $F$ (kN) hai. $0$ se $τ$ tak blue shaded area woh $63.2%$ hai jo humne abhi compute ki; yellow $t = τ$ line ke daayein faint pink tail baaki $36.8%$ hai.

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

Figure 1 — Thrust tail-off $F (t) = F_{0} e^{- t / τ}$ : impulse curve ke neeche ka area hai, ek time-constant par split kiya gaya.

Ex 4 — Cell D: separation, unequal masses

Worked example General two-body shove

Spent lower stage $m_{1} = 12000 kg$ , upper stage $m_{2} = 3000 kg$ . Separation springs $J_{se p} = 4500 N\cdots$ deliver karte hain. Reduced mass $μ$ , relative separation velocity $v_{r e l}$ , aur har stage ki apni $Δ v$ nikalo.

Forecast: push ke baad kaun sa stage faster move karega — bhaari wala ya halka wala?

Reduced mass. $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{12000 \times 3000}{15000} = \frac{3.6 \times 1 0 ^{7}}{15000} = 2400 kg$ . Yeh step kyun? Internal force ke under relative motion ek $μ$ mass ke particle ki tarah behave karta hai — Reduced Mass and Two-Body Problem dekho.
Relative velocity. $v_{r e l} = \frac{J _{se p}}{μ} = \frac{4500}{2400} = 1.875 m/s$ . Yeh step kyun? Yeh drift-apart speed hai, jo collision safety decide karta hai.
Individual kicks (forward $= +$ ): $Δ v_{2} = + \frac{J _{se p}}{m _{2}} = \frac{4500}{3000} = + 1.5 m/s$ (upper speeds up); $Δ v_{1} = - \frac{J _{se p}}{m _{1}} = - \frac{4500}{12000} = - 0.375 m/s$ (lower peeche reh jaata hai). Yeh step kyun? Newton's third law: har body par equal-and-opposite impulse.

Verify: $Δ v_{2} - Δ v_{1} = 1.5 - (- 0.375) = 1.875 = v_{r e l}$ ✓. Momentum check (Conservation of Momentum): $m_{2} Δ v_{2} + m_{1} Δ v_{1} = 3000 (1.5) + 12000 (- 0.375) = 4500 - 4500 = 0$ ✓. Halka upper stage faster move karta hai — forecast confirmed.

Ex 5 — Cell E: equal masses (aur Cell I: zero impulse)

Worked example Symmetric limit — plus ek zero-input sanity check

Do stages equal mass ke $m_{1} = m_{2} = 5000 kg$ . (a) Springs $J_{se p} = 4000 N\cdots$ deliver karte hain: $μ$ aur $v_{r e l}$ nikalo. (b) Phir degenerate case $J_{se p} = 0$ check karo.

Forecast: equal masses ke saath, $μ$ ek single stage mass se kaisa relate karta hai?

Equal masses ke liye reduced mass. $μ = \frac{m \cdot m}{m + m} = \frac{m}{2} = \frac{5000}{2} = 2500 kg$ . Yeh step kyun? Equal masses ek clean special case hai jaanna zaroori hai: $μ$ exactly har mass ka aadha hota hai.
Relative velocity (a). $v_{r e l} = \frac{4000}{2500} = 1.6 m/s$ . Yeh step kyun? Same formula, check karo ki yeh symmetric point par kaam karta hai.
Individual kicks. $Δ v_{2} = + \frac{4000}{5000} = + 0.8$ , $Δ v_{1} = - 0.8 m/s$ — symmetric, jaise expect kiya tha. Yeh step kyun? Equal masses ⇒ equal-and-opposite speeds, har ek $v_{r e l}$ ka exactly aadha.
Zero-input case (b). $J_{se p} = 0 \Rightarrow v_{r e l} = 0/2500 = 0 m/s$ . Yeh step kyun? Koi shove nahi ⇒ koi drift nahi ⇒ koi separation nahi. Stages saath ride karte hain aur ignition par collide karenge — woh danger jiske baare mein parent note warn karta hai.

Verify: $v_{r e l} = 0.8 - (- 0.8) = 1.6$ ✓. Momentum $5000 (0.8) + 5000 (- 0.8) = 0$ ✓. Aur $J_{se p} = 0 \Rightarrow v_{r e l} = 0$ ✓ (koi impulse nahi, koi motion nahi — formula gracefully degrade hota hai).

Ex 6 — Cell F: extreme mass ratio ( $m_{1} ≫ m_{2}$ )

Worked example Ek giant spent stage ek tiny upper stage ko kick off karta hai

Ek bada spent booster $m_{1} = 90000 kg$ ek chhote upper stage $m_{2} = 1000 kg$ se separate hota hai, $J_{se p} = 2000 N\cdots$ ke saath. $μ$ , $v_{r e l}$ nikalo, aur $μ$ ko $m_{2}$ se compare karo.

Forecast: jab ek body 90 guna bhaari ho, toh reduced mass halki mass ke paas hoti hai ya bhaari mass ke paas?

Reduced mass. $μ = \frac{90000 \times 1000}{91000} = \frac{9.0 \times 1 0 ^{7}}{91000} = 989.01 kg$ . Yeh step kyun? Limiting behaviour $m_{1} ≫ m_{2}$ test kar rahe hain.
$m_{2}$ se compare karo. $μ = 989 kg$ $m_{2} = 1000 kg$ ke $1%$ ke andar hai. Yeh step kyun? Yeh limit $μ \to m_{2}$ confirm karta hai jab $m_{1} \to \infty$ — bhaari body barely recoil karti hai, toh relative motion almost entirely light body ki apni motion hai.
Relative velocity. $v_{r e l} = \frac{2000}{989.01} = 2.0222 m/s$ . Yeh step kyun? Almost saari drift upper stage ki apni $Δ v_{2} = 2000/1000 = 2.0$ m/s hai; bhaari stage ka $Δ v_{1} = - 2000/90000 = - 0.0222$ m/s tiny hai.

Verify: $Δ v_{2} - Δ v_{1} = 2.0 - (- 0.02222) = 2.02222 = v_{r e l}$ ✓. Aur $μ / m_{2} = 989.01/1000 = 0.989$ ≈ 1, toh $μ \to m_{2}$ — forecast (light mass ke paas) confirmed.

Ex 7 — Cell G: collision-risk word problem (backwards solve karo)

Worked example "Springs kitni zyada push karein?"

Mission rule: upper stage ko engine ignite karne se pehle $d = 4 m$ ka gap open karna hoga $t_{c l e a r} = 2.0 s$ ke andar bolt-fire ke baad, warna plume spent stage se takraayega. Masses: $m_{1} = 8000 kg$ , $m_{2} = 2000 kg$ . Minimum separation impulse $J_{se p}$ kitna chahiye?

Forecast: kya needed impulse hundreds ya thousands of N·s mein hogi?

Required relative velocity. Short window ke dauran (gravity/drag neglected, jaise upar flag kiya), drift steady hai, toh $d = v_{r e l} t_{c l e a r}$ . Isliye $v_{r e l} \geq \frac{d}{t _{c l e a r}} = \frac{4}{2.0} = 2.0 m/s .$ Yeh step kyun? Clearance condition ek distance deadline ko ek minimum speed mein convert karta hai.
Reduced mass. $μ = \frac{8000 \times 2000}{10000} = 1600 kg$ . Yeh step kyun? Humein $μ$ chahiye required speed ko required impulse mein baadalne ke liye.
Impulse ke liye solve karo. $v_{r e l} = J_{se p} / μ$ se: $J_{se p} = μ v_{r e l} = 1600 \times 2.0 = 3200 N\cdots$ . Yeh step kyun? Yeh design number hai jo spring/retro-rocket deliver karna chahiye — at minimum.

Verify: Plug back: $v_{r e l} = 3200/1600 = 2.0$ m/s, gap $= 2.0 \times 2.0 = 4.0$ m ✓ = $d$ . Units: $kg \times m/s = N\cdots$ ✓. Koi bhi chhota impulse ⇒ gap $< 4$ m ⇒ collision — toh $3200$ N·s safety floor hai.

Ex 8 — Cell H: exam twist, tail-off aur separation saath mein

Worked example Combined scenario with sign bookkeeping

Staging se theek pehle poora vehicle (dono stages) $v_{0} = 2500 m/s$ par move kar raha hai. Upper-stage engine already cut ho chuka hai, lekin uska tail-off ek kick add karta hai. Data: $F_{0} = 30 kN$ , $τ = 0.5 s$ ; masses at separation $m_{1} = 6000 kg$ (spent), $m_{2} = 4000 kg$ (upper); separation impulse $J_{se p} = 5000 N\cdots$ . Tail-off aur separation push ke baad upper stage ki final velocity nikalo.

Forecast: tail-off aur separation dono upper stage ko forward push karte hain — kya uski final speed $2500$ m/s se upar ya neeche hogi, aur kitne se?

Pehle units convert karo. $F_{0} = 30 kN = 30000 N$ (same care jaise Ex 1 mein — kilonewtons ko seconds se multiply karne se pehle newtons banana zaroori hai). Yeh step kyun? Consistent SI units hi $J = F_{0} τ$ ko N·s mein nikaalte hain.
Upper stage par tail-off kick. Tail-off thrust upper stage (mass $m_{2}$ ) par act karta hai: $Δ v_{t ai l} = \frac{F _{0} τ}{m _{2}} = \frac{30000 \times 0.5}{4000} = \frac{15000}{4000} = 3.75 m/s .$ Yeh step kyun? Yeh Cell A ka tool hai upper stage ki apni mass par apply kiya — ek forward $(+)$ kick.
Upper stage par separation kick. $Δ v_{2} = + \frac{J _{se p}}{m _{2}} = \frac{5000}{4000} = + 1.25 m/s$ . Yeh step kyun? Upper stage equal-opposite spring impulse ka forward half receive karta hai.
Signed contributions add karo. Dono forward hain, toh $v_{f ina l} = v_{0} + Δ v_{t ai l} + Δ v_{2} = 2500 + 3.75 + 1.25 = 2505.0 m/s .$ Yeh step kyun? Same body par alag alag impulses se velocities simply add hoti hain (har ek $J / m$ hai); careful sign tracking dono ko $+$ rakhta hai.

Verify: $2500 + 3.75 + 1.25 = 2505.0$ m/s ✓ — start se upar, jaise forecast kiya. Sanity: spent stage ka separation recoil $Δ v_{1} = - 5000/6000 = - 0.8333$ m/s upper stage mein add nahi hota — yeh lower body ka hai. Sirf separation ka momentum: $4000 (1.25) + 6000 (- 0.8333) = 5000 - 5000 = 0$ ✓.

Recall Matrix ke paas quick self-test

Kis cell mein $μ = m /2$ hai? ::: Equal masses (Cell E). $m_{1} \to \infty$ jaane par, $μ \to$ ? ::: $m_{2}$ (Cell F, light stage). $τ \to 0$ jaane par, $J_{t ai l} \to$ ? ::: $0$ (Cell B, instant-cutoff limit). Ek time-constant total tail-off impulse ka kitna fraction deliver karta hai? ::: $1 - e^{- 1} \approx 63.2%$ (Cell C). Gap deadline ko minimum spring impulse mein baadalne ke liye use karo ::: $J_{se p} = μ d / t_{c l e a r}$ (Cell G). Short staging window ke dauran hum kaun se teen effects neglect karte hain? ::: gravity, drag, aur continued mass loss (constant-mass impulse-momentum model).

Yeh bhi dekho: Tsiolkovsky Rocket Equation, Multistage Rocket Optimization, Impulse-Momentum Theorem, Gravity Losses and Ascent Trajectory.