3.4.18
Fairing (payload shroud) woh nose cone hai jo satellite ko atmospheric ascent ke dauran aerodynamic aur thermal loads se bachata hai. Jab rocket sensible atmosphere ke upar pahunch jaata hai, tab yeh dead weight ban jaata hai. Toh guidance question yeh hai: jitna jaldi ho sake, lekin itna jaldi bhi nahi. Zyada jaldi drop karo → aero heating aur pressure payload ko damage kar sakti hai. Zyada der se drop karo → poore burn mein bekar mass carry karte raho, Δv waste hogi. Yahan deciding quantity hai free-molecular heat flux , jo almost poori tarah dynamic pressure q
Definition Fairing jettison conditionFairings tab separate ki jaati hain jab exposed payload par free-molecular aerodynamic heating flux mission-defined threshold se neeche aa jaaye, typically
q ˙ free-mol ≤ 1135 W/m 2 ( ≈ 0.1 BTU/ft 2 s ) dynamic-pressure limit q ≤ q jett 110–140 km altitude par hoti hai.
Ek launch program requirement likhne ke do "official" tarike:
Altitude-based: h ≳ 110 km Flux/pressure-based: jab q q ˙
Dono ek hi underlying physics ke proxies hain: atmosphere itni patli aur slow-heating honi chahiye.
v 3 bahut fast hoti hai (often > 3 ρ v 3
Toh optimum exactly boundary par hai: jis instant q ˙
Worked example Example 1 — Jettison par dynamic pressure nikaloEk rocket h = 115 v = 4000 ρ 0 = 1.2 kg/m 3 H = 7.5 Step 1 — density: ρ = 1.2 e − 115/7.5 = 1.2 e − 15.33 ≈ 1.2 × 2.2 × 1 0 − 7 = 2.6 × 1 0 − 7 kg/m 3 . Yeh step kyun? Altitude sirf exponential density ke through aati hai; velocity given hai.
Step 2 — dynamic pressure: q = 2 1 ρ v 2 = 0.5 × 2.6 × 1 0 − 7 × ( 4000 ) 2 = 2.1 Pa . Kyun: q = 2 1 ρ v 2 Step 3 — heat flux: q ˙ = q v = 2.1 × 4000 ≈ 8.4 × 1 0 3 W/m 2 . jettison ke liye bahut jaldi hai ; thoda aur upar wait karo.
Worked example Example 2 — Jettison altitude solve karoThreshold q ˙ limit = 1135 W/m 2 v = 4000 ρ 0 = 1.2 H = 7.5 Step 1 — required density: q ˙ = 2 1 ρ v 3 ⇒ ρ = v 3 2 q ˙ = ( 4000 ) 3 2 × 1135 = 6.4 × 1 0 10 2270 = 3.5 × 1 0 − 8 kg/m 3 . Step 2 — density law ko invert karo: h = H ln ρ ρ 0 = 7.5 ln 3.5 × 1 0 − 8 1.2 = 7.5 ln ( 3.4 × 1 0 7 ) = 7.5 × 17.3 ≈ 130 km . Kyun: h Answer: 130 km ke paas jettison karo — real missions ke saath consistent hai.
Worked example Example 3 — Δv savedUpper stage: I s p = 450 m 0 = 20 , 000 m f = 6000 m f r = 1000 Δ v = 450 ( 9.81 ) ln 6000 20000 = 4415 × 1.204 = 5316 Δ v ′ = 4415 ln 5000 19000 = 4415 × 1.335 = 5893 Gain ≈ 577 m/s sirf fairing nahi dhone se — orbit insertion ke liye kaafi matter karta hai.
q v 2 Kyun sahi lagta hai: aerodynamic force sach mein C D q A ∝ v 2 Fix: Heat energy per time hai = force-like flux × velocity → q ˙ ∝ ρ v 3 v
Common mistake "Mass bachane ke liye fairing jitni neeche ho sake utni neeche drop karo."Kyun sahi lagta hai: Δv math mein dead mass jaldi drop karna faydemand hai.
Fix: heating limit se neeche payload cook ya crush ho jaayega. Optimum flux limit par hai, usse neeche nahi — yeh constrained optimization hai, "jitna jaldi ho sake" nahi.
Common mistake "Altitude akele decide karti hai."Kyun sahi lagta hai: launch manuals "~110 km" quote karte hain.
Fix: 110 km sirf ρ v 3 zyada jettison altitude chahiye kyunki v 3
Fairing jettison timing fundamentally kaun si quantity govern karti hai? Free-molecular convective heat flux q ˙ ≈ 2 1 ρ v 3
Dynamic pressure ko momentum flux se derive karo. Area A d m = ρ A v d t ρ A v 2 d t ρ v 2 q = 2 1 ρ v 2
Heat flux v 3 v 2 Force flux ∼ ρ v 2 ρ v 3
Typical fairing jettison altitude? ~110–140 km, jahan atmospheric density kaafi drop ho jaati hai.
Fairing orbit par jettison kyun nahi karte? Dead mass carry karna Δv waste karta hai (rocket equation); jaise hi heating safe ho drop karo.
Flux limit given ho toh jettison altitude ka formula? h = H ln q ˙ l imi t ρ 0 v 3 /2 H ≈ 7.5
Requirements likhne ke do equivalent tarike? Altitude-based (h > 110 q q ˙
Recall Feynman: 12-saal ke bacche ko explain karoRocket ek pointed "topi" pehanti hai satellite ko hawa se bachane ke liye, jo rub hoke car ki khidki se bahar haath nikalne jaisi garam ho jaati hai — lekin bahut zyada. Jab rocket itna upar ud jaaye ki hawa almost khatam ho jaaye, tab topi sirf bhari ho jaati hai bekar. Toh computer wait karta hai jab tak hawa itni patli na ho jaaye ki satellite jale nahi, phir topi utar deta hai halke se udne ke liye. Jaldi nahi (satellite hurt hogi), der bhi nahi (rocket thak jaayegi isse uthate uthate).
Mnemonic Physics yaad karo"Force loves v 2 v 3 ρ
(q ∝ v 2 q ˙ ∝ v 3 ρ
Dynamic Pressure (Max-Q) — wahi q = 2 1 ρ v 2 Atmospheric Density Model — Scale Height — ρ ( h ) = ρ 0 e − h / H Rocket Equation & Mass Ratios — mass drop karne ka Δv payoff quantify karta hai.Aerodynamic Heating & Free-Molecular Flow — ρ v 3 Ascent Trajectory Optimization — jahan jettison constraint guidance mein enter karti hai.
Dynamic pressure q = half rho v squared
Density rho = rho0 exp minus h over H
Altitude criterion 110-140 km