Fairing separation — altitude, dynamic pressure requirements: Worked examples

3.4.18 · D3 · Physics › Rocket Flight Mechanics › Fairing separation — altitude, dynamic pressure requirements

Theory ke liye parent dekho: 3.4.18 parent. Building blocks: Dynamic Pressure (Max-Q), Atmospheric Density Model — Scale Height, Rocket Equation & Mass Ratios, Aerodynamic Heating & Free-Molecular Flow, Ascent Trajectory Optimization.

Shuru karne se pehle — symbols, units, aur ek definition

Scenario matrix

Is topic ka har problem in cells mein se kisi ek mein aata hai. Neeche ke examples cell ke hisaab se label hain taaki dekh sako ki poora map cover hua hai.

Cell	Kya vary karta hai / trap	Example
A. Forward	Given $h, v$ → find $q$ aur $\overset{q}{˙}$ , decide jettison	Ex 1
B. Inverse (altitude)	Given flux limit → solve for $h_{jett}$	Ex 2
C. Inverse (velocity)	Given $h$ aur limit → kaunsa $v$ limit trip karta hai	Ex 3
D. Zero / degenerate	$v = 0$ , ya $h \to \infty$ ( $ρ \to 0$ ) — no heating	Ex 4
E. Limiting / scaling	Agar $v$ double ho toh $h_{jett}$ kitna shift hoga? (log behaviour)	Ex 5
F. Trajectory twist	Steeper/faster ascent → higher jettison altitude	Ex 6
G. $Δ v$ trade	Fairing carry karne vs drop karne ka mass penalty	Ex 7
H. Word / real-world	Mission engineer ka "is it safe yet?" call	Ex 8

Stage set karna — ek picture sabke liye

Figure s01 ek single vertical logarithmic scale par dikhata hai ki jaise altitude badhti hai (fixed $v = 4000 m/s$ ke liye) do quantities kaise girte hain:

cyan curve hai dynamic pressure $q = \frac{1}{2} ρ v^{2}$ (Pa mein),
amber curve hai heat flux $\overset{q}{˙} = \frac{1}{2} ρ v^{3}$ ( $W/m^{2}$ mein).

Horizontal axis altitude $h$ km mein hai (60→160). Notice karo dono curves log scale par straight lines hain — yeh exponential density law $ρ = ρ_{0} e^{- h / H}$ khud ko dikha rahi hai (ek exponential ka log straight line hota hai). Amber curve cyan wali se $v = 4000$ factor upar hai, exactly relation $\overset{q}{˙} = q v$ ki tarah. Dashed white horizontal line safe limit $1135 W/m^{2}$ mark karti hai; dotted white vertical line jahan amber curve us se milti hai woh $h_{jett} \approx 130 km$ hai — us intersection ko padho aur tumhare paas Example 2 ka jawab hai.

Figure — Fairing separation — altitude, dynamic pressure requirements

Cell A — Forward evaluation

Worked example Example 1 — Kya 115 km par jettison karna safe hai?

Ek rocket $h = 115 km$ par $v = 4000 m/s$ se fly kar raha hai. $q$ aur $\overset{q}{˙}$ find karo, phir decide karo.

Forecast: guess karo — kya $\overset{q}{˙}$ $1135 W/m^{2}$ se upar hoga ya neeche? (Hint: $v^{3}$ enormous hai.)

Step 1 — altitude se density. $ρ = ρ_{0} e^{- h / H} = 1.2 e^{- 115/7.5} = 1.2 e^{- 15.33} \approx 2.6 \times 1 0^{- 7} kg/m^{3} .$ Yeh step kyun? Altitude physics mein sirf density ke through enter karti hai; is page par koi aur cheez directly $h$ par depend nahi karti. (Yahan $h / H = 115/7.5$ — km cancel km, jaise promise kiya tha.)

Step 2 — dynamic pressure. $q = \frac{1}{2} ρ v^{2} = 0.5 \times 2.6 \times 1 0^{- 7} \times 400 0^{2} \approx 2.1 Pa .$ Yeh step kyun? $q$ aerodynamic "punch" hai; yeh drag force $F = C_{D} q A$ mein coefficient hai (upar $C_{D}$ aur $A$ define hain), isliye yeh heat flux ka natural bridge hai.

Step 3 — heat flux. $\overset{q}{˙} = q v = 2.1 \times 4000 \approx 8.4 \times 1 0^{3} W/m^{2} .$ Yeh step kyun? Heat = energy/time = force-like flux × velocity, isliye hum $q$ ko ek aur $v$ se multiply karte hain (free-molecular $v^{3}$ signature).

Decision: $8.4 \times 1 0^{3} > 1135$ → bahut jaldi hai. Aur upar ruko.

Verify: $\overset{q}{˙}$ ke units: $[Pa] \cdot [m/s] = \frac{N}{m ^{2}} \cdot \frac{m}{s} = \frac{N\cdotm}{m ^{2} s} = \frac{J}{m ^{2} s} = W/m^{2}$ . ✓ Flux ke liye correct units.

Cell B — Altitude ke liye Inverse

Worked example Example 2 — Kis altitude par safe ho jaata hai?

Same $v = 4000 m/s$ . $h_{jett}$ find karo jahan $\overset{q}{˙} = 1135 W/m^{2}$ .

Forecast: Ex 1 ne kaha 115 km bahut hot hai. Guess karo: 130 km se upar hai ya neeche?

Step 1 — required density. $\overset{q}{˙} = \frac{1}{2} ρ v^{3}$ se: $ρ = \frac{2 q ˙}{v ^{3}} = \frac{2 \times 1135}{400 0 ^{3}} = \frac{2270}{6.4 \times 1 0 ^{10}} \approx 3.5 \times 1 0^{- 8} kg/m^{3} .$ Yeh step kyun? Hum woh density chahte hain jis par heating exactly limit hit kare; flux formula ko $ρ$ ke liye invert karo.

Step 2 — density law invert karo. $h = H ln \frac{ρ _{0}}{ρ} = 7.5 ln \frac{1.2}{3.5 \times 1 0 ^{- 8}} = 7.5 ln (3.4 \times 1 0^{7}) \approx 7.5 \times 17.35 \approx 130 km .$ Yeh step kyun? Log ek gigantic density ratio ( $3.4 \times 1 0^{7}$ ) ko ek friendly number ( $\approx 17$ ) mein badal deta hai; isi liye hum yahan logarithms use karte hain.

Answer: 130 km se thoda upar jettison karo (strict rule: 130 km par hum abhi limit par hain, isliye thoda aur chadho). Yeh parent note ke 110–140 km ke quoted band ke andar aata hai jo real launch programs ke liye hai, toh number physically reasonable hai.

Verify: 130 km plug back karo: $ρ = 1.2 e^{- 130/7.5} = 1.2 e^{- 17.33} \approx 3.6 \times 1 0^{- 8}$ , phir $\overset{q}{˙} = \frac{1}{2} ρ v^{3} \approx 1150 W/m^{2} \approx 1135$ . ✓ (Thodi rounding.)

Cell C — Velocity ke liye Inverse

Worked example Example 3 — Fixed altitude, kaunsi speed limit trip karti hai?

$h = 120 km$ par, kaunsi velocity $\overset{q}{˙}$ ko exactly $1135 W/m^{2}$ banati hai?

Forecast: 120 km par density patli hai, toh kya rocket ko limit tak pahunchne ke liye slow ya fast hona chahiye?

Step 1 — 120 km par density. $ρ = 1.2 e^{- 120/7.5} = 1.2 e^{- 16} \approx 1.35 \times 1 0^{- 7} kg/m^{3} .$ Yeh step kyun? Altitude fix karo → density fix karo; ab sirf $v$ unknown hai.

Step 2 — flux ko $v$ ke liye solve karo. $\overset{q}{˙} = \frac{1}{2} ρ v^{3} \Rightarrow v = (\frac{2 q ˙}{ρ})^{1/3} = (\frac{2 \times 1135}{1.35 \times 1 0 ^{- 7}})^{1/3} .$ Yeh step kyun? $v$ ko isolate karo; kyunki $\overset{q}{˙} \propto v^{3}$ (upar define ki gayi free-molecular signature), hum cube root lete hain, square root nahi.

Step 3 — evaluate karo. $\frac{2270}{1.35 \times 1 0 ^{- 7}} \approx 1.68 \times 1 0^{10}$ , aur $(1.68 \times 1 0^{10})^{1/3} \approx 2560 m/s .$

Answer: strict rule se, 120 km par $v \geq 2560 m/s$ abhi bhi unsafe hai; sirf $v < 2560 m/s$ safe hai.

Verify: $v = 2560$ back-substitute karo: $\frac{1}{2} (1.35 \times 1 0^{- 7}) (2560)^{3} \approx 1132 W/m^{2} \approx 1135$ . ✓

Cell D — Zero aur degenerate inputs

Worked example Example 4 — Edge cases:

$v = 0$ aur $h \to \infty$ (a) Rocket momentarily hawa ke relative rest mein. (b) Rocket deep space mein, $ρ \to 0$ .

Forecast: in mein se kaunsa payload ko abhi bhi cook kar sakta hai? (Koi nahi — kyun dekho.)

Step 1 — $v = 0$ . $\overset{q}{˙} = \frac{1}{2} ρ \cdot 0^{3} = 0.$ Yeh step kyun? Koi motion nahi → koi molecule surface se nahi takraate → zero convective heating, chahe hawa kitni bhi thick ho.

Step 2 — $h \to \infty$ . $ρ = ρ_{0} e^{- h / H} \to 0$ , toh $\overset{q}{˙} = \frac{1}{2} ρ v^{3} \to 0$ kisi bhi finite $v$ ke liye. Yeh step kyun? Yahi jettison ka poora point hai: itna upar jao ki density heating ko khatam kar de chahe speed kitni bhi ho.

Step 3 — khatarnak corner. Heating sirf tab badi hoti hai jab dono $ρ$ aur $v$ appreciable hoon. Product $ρ v^{3}$ asli master variable hai; koi bhi factor zero jaane par $\overset{q}{˙} \to 0$ . Yeh step kyun? Yeh classic error rokta hai ki sirf altitude ya sirf speed dekho.

Verify: $lim_{v \to 0} \frac{1}{2} ρ v^{3} = 0$ aur $lim_{h \to \infty} \frac{1}{2} ρ_{0} e^{- h / H} v^{3} = 0$ . ✓ Dono limits vanish ho jaate hain.

Cell E — Limiting / scaling behaviour

Worked example Example 5 — Agar rocket twice as fast ho, toh kitna aur upar chadna padega?

Threshold aur $ρ_{0}, H$ fixed. $v$ aur $2 v$ ke liye $h_{jett}$ compare karo.

Forecast: double speed → double altitude? Guess karo, phir log check karo.

Step 1 — jettison altitude likho. $h_{jett} = H ln \frac{\frac{1}{2} ρ _{0} v ^{3}}{q ˙ _{limit}} .$ Yeh step kyun? Yeh parent se closed form hai; sirf log ke andar $v^{3}$ badalta hai.

Step 2 — $2 v$ vs $v$ ka difference lo. $Δ h = H [ln (2 v)^{3} - ln v^{3}] = H ln \frac{( 2 v ) ^{3}}{v ^{3}} = H ln 8 = 3 H ln 2.$ Yeh step kyun? Logs subtract karna change ko isolate karta hai; density aur threshold constants cancel ho jaate hain.

Step 3 — number. $Δ h = 3 \times 7.5 \times ln 2 = 22.5 \times 0.693 \approx 15.6 km .$ Yeh step kyun? Scaling law ko ek concrete "~16 km aur chadho" mein badle.

Answer: speed double karne par jettison altitude sirf ~15.6 km badhti hai, double NAHI hoti — kyunki altitude logarithm ke through enter karti hai. Ek badi speed change sirf ek modest height change maangti hai.

Verify: $3 H ln 2 = 3 (7.5) (0.6931) \approx 15.6 km$ . ✓

Cell F — Trajectory twist

Figure s02 jettison altitude $h_{jett}$ ko vertical axis par kilometres mein plot karta hai versus jettison par velocity $v$ ko horizontal axis par metres per second mein (range 2500→6000), closed form $h_{jett} = H ln \frac{\frac{1}{2} ρ _{0} v ^{3}}{q ˙ _{limit}}$ use karke. Cyan curve dheere badhti hai (yeh logarithm hai), aur do marked dots is example ki do trajectories hain: slower amber dot (A, $v = 3500 m/s$ par) neeche hai, faster white dot (B, $v = 5000 m/s$ par) upar hai. Dekho speed (m/s) mein bade jump ke liye height (km) kitni kam badhti hai — woh gentle slope hi is example ka poora lesson hai. Calculation follow karo, phir apne numbers un dono dots se check karo.

Worked example Example 6 — Steep-and-fast vs shallow-and-slow

Trajectory A jettison conditions $v_{A} = 3500 m/s$ par reach karti hai, trajectory B $v_{B} = 5000 m/s$ par (steeper, faster). Kaun upar jettison karta hai, aur kitne se?

Forecast: faster trajectory — zyada upar jettison karega ya kam? ( $v^{3}$ yaad karo.)

Step 1 — har ek ke liye altitude. $h = H ln \frac{\frac{1}{2} ρ _{0} v ^{3}}{q ˙ _{limit}}$ with $\frac{1}{2} ρ_{0} = 0.6$ . Yeh step kyun? Same formula, do velocities; difference wahi hai jo chahiye. Yahi woh curve hai jo s02 mein draw ki gayi hai.

Step 2 — trajectory A. $\frac{0.6 \times 350 0 ^{3}}{1135} = \frac{0.6 \times 4.29 \times 1 0 ^{10}}{1135} \approx 2.27 \times 1 0^{7}$ , toh $h_{A} = 7.5 ln (2.27 \times 1 0^{7}) \approx 7.5 \times 16.94 \approx 127 km .$ (Yeh s02 mein amber dot hai.)

Step 3 — trajectory B. $\frac{0.6 \times 500 0 ^{3}}{1135} = \frac{0.6 \times 1.25 \times 1 0 ^{11}}{1135} \approx 6.61 \times 1 0^{7}$ , toh $h_{B} = 7.5 ln (6.61 \times 1 0^{7}) \approx 7.5 \times 18.01 \approx 135 km .$ (s02 mein white dot.)

Answer: faster trajectory B upar jettison karta hai (135 vs 127 km) — difference $Δ h = 3 H ln (5000/3500) = 3 (7.5) ln (1.4286) \approx 8 km$ hai. Yeh exactly parent ki warning hai: "steeper, faster → higher jettison altitude."

Verify: $h_{B} - h_{A} = 3 H ln (v_{B} / v_{A}) = 22.5 ln (1.4286) \approx 8.0 km$ , $135 - 127$ se match karta hai. ✓

Cell G — $Δ v$ trade

Worked example Example 7 — Fairing drop karne se kitna

$Δ v$ bachta hai? Upper stage: $I_{s p} = 450 s$ , $m_{0} = 20000 kg$ , $m_{f} = 6000 kg$ , fairing $m_{f r} = 1000 kg$ . Underlying equation ke liye dekho Rocket Equation & Mass Ratios.

Forecast: saving guess karo — tens, hundreds, ya thousands of m/s?

Step 1 — carry-through $Δ v$ . $Δ v = I_{s p} g_{0} ln \frac{m _{0}}{m _{f}} = 450 \times 9.81 \times ln \frac{20000}{6000} = 4415 \times 1.204 \approx 5316 m/s .$ Yeh step kyun? Rocket equation mass ratio ko velocity budget mein convert karta hai. Yahan hum fairing ko poore burn mein saath rakhte hain, toh yeh $m_{0}$ aur $m_{f}$ dono ko inflate karta hai.

Step 2 — drop-early $Δ v$ . Dono masses $m_{f r}$ se girte hain: $Δ v^{'} = 4415 ln \frac{19000}{5000} = 4415 \times 1.335 \approx 5893 m/s .$ Yeh step kyun? Dono ends se $1000 kg$ remove karna ratio $19000/5000 = 3.8$ ko $20000/6000 = 3.33$ se upar le jaata hai; zyada ratio zyada $Δ v$ deta hai.

Step 3 — gain. $Δ v_{saved} = 5893 - 5316 \approx 577 m/s .$ Yeh step kyun? Yahi woh payoff hai jo engineers ko jitni jaldi heat limit allow kare utni jaldi jettison karne par push karta hai.

Answer: ~577 m/s free — orbit insertion change karne ke liye kaafi.

Verify: $4415 (ln 3.8 - ln 3.333) = 4415 \times 0.1307 \approx 577 m/s$ . ✓

Cell H — Real-world call

Worked example Example 8 — Flight director ka "go/no-go"

Telemetry: $h = 112 km$ , $v = 4200 m/s$ . Payload limit $\overset{q}{˙}_{limit} = 1135 W/m^{2}$ . Kya abhi fairing sep ke liye "GO" hai?

Forecast: 112 km manual ke hisaab se "kaafi upar" lagta hai — lekin altitude nahi, flux check karo.

Step 1 — density. $ρ = 1.2 e^{- 112/7.5} = 1.2 e^{- 14.93} \approx 3.9 \times 1 0^{- 7} kg/m^{3} .$ Yeh step kyun? Reported altitude ko us density mein convert karo jo actually heating drive karti hai.

Step 2 — heat flux. $\overset{q}{˙} = \frac{1}{2} ρ v^{3} = 0.5 \times 3.9 \times 1 0^{- 7} \times 420 0^{3} \approx 0.5 \times 3.9 \times 1 0^{- 7} \times 7.41 \times 1 0^{10} \approx 1.45 \times 1 0^{4} W/m^{2} .$ Yeh step kyun? Woh single number jo safety decide karta hai.

Step 3 — compare karo. $1.45 \times 1 0^{4} ≫ 1135$ → NO-GO (strict rule: equality bhi no-go hoti, aur hum bahut upar hain). "112 km rule" sirf ek proxy hai; asli trajectory itni fast hai ki $v^{3}$ heating high rakhta hai. Yeh step kyun? Parent ka mistake #3 demonstrate karta hai: akeli altitude decide nahi karti. Dekho bhi Ascent Trajectory Optimization.

Answer: Jettison mat karo. Flux girne tak ruko — is speed ke liye woh 133 km ke paas hai (Ex 2 ke method se $v = 4200$ ke saath).

Verify: $\frac{1}{2} (3.9 \times 1 0^{- 7}) (4200)^{3} \approx 1.45 \times 1 0^{4} W/m^{2} > 1135$ . ✓ NO-GO confirmed.

Recall Matrix ke paas quick self-test

Kaunsa cell? "Given $h$ aur limit, woh speed find karo jo heating trip kare." ::: Cell C (velocity ke liye inverse, cube root). $v$ double karne par $h_{jett}$ kitna badhta hai? ::: $3 H ln 2 \approx 15.6 km$ — sirf log-sized change. Kyun ek 112 km "GO by altitude" abhi bhi NO-GO ho sakta hai? ::: Kyunki $\overset{q}{˙} \propto ρ v^{3}$ ; fast trajectory flux ko wahan bhi limit se upar rakhti hai. Ex 7 mein 1000 kg fairing drop karne se kitna $Δ v$ bacha? ::: Lagbhag 577 m/s. Kya $\overset{q}{˙} = \overset{q}{˙}_{limit}$ safe hai? ::: Nahi — strict rule: equality last unsafe instant hai; sirf tab jettison karo jab strictly below ho.