Reentry corridor — angle of attack constraints: Worked examples

3.4.20 · D3 · Physics › Rocket Flight Mechanics › Reentry corridor — angle of attack constraints

Yeh page ek case-by-case tour hai. Parent note ne formulas banaye the; yahan hum unpe har tarah ka input fire karte hain — steep, shallow, zero lift, maximum lift, degenerate bank angles, aur ek exam twist — taaki koi bhi scenario tumhe surprise na kare.

Scenario matrix

Is topic ka har problem in cells mein se kisi ek mein aata hai. Neeche ke examples mein cell ka label diya gaya hai.

Cell	Kya vary karta hai	Physical question	Example
A. Steep entry ( $γ_{E}$ large negative)	undershoot side	Kya peak $g$ limit se zyada hai?	Ex 1, Ex 6
B. Shallow entry ( $γ_{E}$ near $0$ )	overshoot side	Kya yeh skip out karega?	Ex 2
C. Optimal $α$	steering gain peak	Sabse wide corridor config	Ex 3
D. Zero / degenerate $α$	$α = 0$	Koi lift nahi — pure ballistic	Ex 4
E. Bank knob limits ( $σ = 0, 9 0^{\circ}, 18 0^{\circ}$ )	lift direction	Full-up / knife-edge / full-down	Ex 5
F. Limiting speed ( $V_{E} \to$ orbital vs. slow)	$V_{E}^{2}$ ke saath scaling	Entry speed $g$ ko kaise magnify karta hai	Ex 6
G. Real-world word problem	full mission numbers	Ek survivable $γ_{E}$ chuno	Ex 7
H. Exam twist	corridor width vs. $L / D$	Capsule vs. lifting body compare karo	Ex 8

Hum poore mein yeh constant symbols reuse karte hain (sab parent mein defined hain):

$V_{E}$ = entry speed, $γ_{E}$ = entry flight-path angle (negative = nose-down),
$H$ = atmospheric scale height, $e \approx 2.718$ (Euler's number, $e^{- h / H}$ atmosphere se),
$C_{Lα}, C_{D 0}, k$ = lift-slope, zero-lift drag, induced-drag factor.

Cell A + C + F — steep/optimal/speed cases

Worked example Example 1 — Steep entry g-check (Cell A)

Statement. Ek capsule $V_{E} = 7.8 km/s$ (low-Earth-orbit speed) par enter karta hai, $H = 7.2 km$ , $γ_{E} = - 8^{\circ}$ . Peak deceleration $g$ 's mein estimate karo ( $g = 9.81 m/s^{2}$ lo).

Forecast: Apollo ke $- 6. 5^{\circ}$ se steeper hai lekin entry slower hai — kya peak $g$ Apollo ke $\sim 36 g$ se zyada ya kam hoga? (Do effects ladte hain: shallower speed kam karta hai, steeper angle badhata hai.)

Step 1. Exponential atmosphere ke liye Allen–Eggers peak-deceleration formula use karo: $\frac{d V}{d t}_{m a x} = \frac{V _{E}^{2} ∣ s i n γ _{E} ∣}{2 eH} .$ Yeh step kyun? Yeh $V^{2} ρ$ product ka closed-form peak hai — dekho Allen–Eggers Ballistic Reentry. Isme sirf entry speed, entry angle, aur scale height chahiye.

Step 2. Numbers plug karo. $V_{E} = 7800 m/s$ , $sin 8^{\circ} = 0.1392$ , $H = 7200 m$ : $\frac{( 7800 ) ^{2} ( 0.1392 )}{2 ( 2.718 ) ( 7200 )} = \frac{6.084 \times 1 0 ^{7} \times 0.1392}{3.914 \times 1 0 ^{4}} = \frac{8.47 \times 1 0 ^{6}}{3.914 \times 1 0 ^{4}} \approx 216 m/s^{2} .$ Yeh step kyun? Pehle km ko m mein convert karo, warna answer $1 0^{6}$ se off hoga — ek classic unit trap.

Step 3. $g$ mein convert karo: $216/9.81 \approx 22 g$ .

Verify: Apollo mein $V_{E} = 11$ km/s, $γ_{E} = - 6. 5^{\circ}$ pe $\sim 36 g$ mila tha. Humara vehicle slower hai ( $7.8$ vs $11$ km/s, aur $g \propto V_{E}^{2}$ toh speed ratio $(7.8/11)^{2} = 0.50$ ise half karta hai) lekin steeper hai ( $sin 8^{\circ} / sin 6. 5^{\circ} = 1.23$ ise badhata hai). Net $\approx 36 \times 0.50 \times 1.23 = 22 g$ . ✓ Consistent.

Worked example Example 3 — Optimal angle of attack, widest corridor (Cell C)

Statement. Ek lifting body mein $C_{Lα} = 2.0 / rad$ , $C_{D 0} = 0.05$ , $k = 0.4$ hai. $α_{opt}$ (degrees mein) aur $(L / D)_{m a x}$ nikalo.

Forecast: Kam zero-lift drag $C_{D 0} = 0.05$ ke saath, kya tum expect karte ho ki $L / D$ parent ke capsule value $\approx 2.24$ se upar ya neeche hoga?

Step 1. $α_{opt} = \frac{1}{C _{Lα}} \frac{C _{D 0}}{k} = \frac{1}{2.0} \frac{0.05}{0.4} = \frac{1}{2.0} (0.3536) = 0.1768 rad .$ Yeh step kyun? Yahan marginal drag, marginal lift benefit ke barabar hota hai — linear-over-quadratic $L / D$ ka peak (dekho Lift and Drag Coefficients).

Step 2. Convert karo: $0.1768 \times \frac{180}{π} = 10. 1^{\circ}$ .

Step 3. $(\frac{L}{D})_{m a x} = \frac{1}{2 k C _{D 0}} = \frac{1}{2 0.4 \times 0.05} = \frac{1}{2 ( 0.1414 )} = 3.54.$

Verify: Parent ke $0.10$ se kam $C_{D 0}$ ⇒ zyada $L / D$ . $3.54 > 2.24$ . ✓ Ek sleeker body ka corridor wider hota hai, $Δ γ_{corridor} \propto L / D$ .

Worked example Example 6 — Entry speed double karna (Cell F, limiting behaviour)

Statement. Same vehicle aur angle, lekin ek hyperbolic return $V_{E} = 15 km/s$ ko $V_{E} = 7.5 km/s$ se compare karo, dono $γ_{E} = - 6^{\circ}$ , $H = 7.2$ km par. Peak $g$ kis factor se change hogi?

Forecast: Speed double hoti hai ( $\times 2$ ). Peak $g \propto V_{E}^{2}$ — padhne se pehle factor guess karo.

Step 1. $∣ d V / d t ∣_{m a x} \propto V_{E}^{2}$ se, ratio hai $(15/7.5)^{2} = 4$ . Yeh step kyun? $γ_{E}$ aur $H$ identical hain, toh cancel ho jaate hain; sirf $V_{E}^{2}$ bachta hai.

Step 2. Fast case ke liye sanity magnitude: $\frac{( 15000 ) ^{2} s i n 6 ^{\circ}}{2 ( 2.718 ) ( 7200 )} = \frac{2.25 \times 1 0 ^{8} \times 0.1045}{3.914 \times 1 0 ^{4}} = \frac{2.352 \times 1 0 ^{7}}{3.914 \times 1 0 ^{4}} \approx 601 m/s^{2} \approx 61 g .$

Verify: Slow case hai $601/4 = 150 m/s^{2} \approx 15 g$ . Factor 4, jaise forecast kiya tha. ✓ Yahi wajah hai ki lunar-return vehicles (higher $V_{E}$ ) LEO returns ki tulna mein undershoot g-limit ke bahut kareeb hote hain.

Cell B — shallow / skip-out case

Worked example Example 2 — Kya yeh entry capture ke liye bahut shallow hai? (Cell B)

Statement. $L / D = 0.3$ wala capsule $γ_{E} = - 1. 2^{\circ}$ par lift-up ( $σ = 0$ ) fly karta hai. Atmosphere ke top par cross-track equation hai: $mV \overset{γ}{˙} = L cos σ - m g cos γ + \frac{m V ^{2}}{R + h} cos γ .$ Kya trajectory upar curve karti hai (skip) ya neeche (capture)?

Forecast: Shallow angle + lift upar pointing + high speed centrifugal term — sab path ko upar push karte hain. Guess: skip.

Step 1. $m$ se divide karo aur $V \approx V_{E} = 7.8$ km/s, $h \approx 90$ km, $R = 6371$ km, $g \approx 9.5 m/s^{2}$ par bracket ka sign check karo: $\overset{γ}{˙} \propto upar \frac{L}{m} cos σ - neeche g cos γ + upar \frac{V ^{2}}{R + h} cos γ .$ Yeh step kyun? Signs, magnitudes nahi, decide karte hain skip vs. capture — hume sirf yeh chahiye ki bracket kis taraf jhukta hai.

Step 2. Centrifugal term akele: $\frac{V ^{2}}{R + h} = \frac{( 7800 ) ^{2}}{6.461 \times 1 0 ^{6}} = 9.41 m/s^{2}$ , essentially $g$ ke barabar. $cos γ \approx 1$ ke saath, gravity aur centrifugal nearly cancel ho jaate hain. Yeh step kyun? Near-orbital speed par vehicle almost "orbit mein" hota hai — gravity muskil se ise neeche bend karti hai.

Step 3. Un dono ke cancel hone se, bachi hui $L cos σ > 0$ (lift-up) $\overset{γ}{˙} > 0$ banati hai: nose utha, path upar curve karta hai → skip-out.

Verify: $γ_{E} = - 1. 2^{\circ}$ (bahut shallow) aur lift-up ke saath, physics Bank Angle Modulation and Guidance rule se match karta hai: skip rokne ke liye $σ = 18 0^{\circ}$ ki taraf roll karna padega. Forecast confirm — yeh entry skip karti hai jab tak banked na ho. ✓

Cell D — zero lift, degenerate input

Worked example Example 4 — Pure ballistic entry,

$α = 0$ (Cell D) Statement. Ek spherical probe $α = 0$ par fly karta hai, toh $C_{L} = C_{Lα} \cdot 0 = 0$ . $L / D$ kya hai, cross-track equation kis chiz mein reduce hoti hai, aur kya yeh vehicle steer kar sakta hai?

Forecast: $α = 0$ lift kill kar deta hai. $L / D$ guess karo aur corridor wide hai ya single line.

Step 1. $L = \frac{1}{2} ρ V^{2} A C_{L} (α)$ jahan $C_{L} = 0 \Rightarrow L = 0$ , toh $L / D = 0$ . Yeh step kyun? Lift, $α$ mein zero ke paas linear hai; zero angle par ek sphere symmetry se zero net lift produce karta hai.

Step 2. Cross-track equation mein $L = 0$ set karo: $mV \overset{γ}{˙} = - m g cos γ + \frac{m V ^{2}}{R + h} cos γ .$ Bank $σ$ gayab ho gaya — $cos σ$ ek zero ko multiply karta hai. Koi steering knob nahi bachi. Yeh step kyun? Agar lift zero hai, toh ise rotate karna (banking) kuch nahi karta: $0 cos σ = 0$ har $σ$ ke liye.

Step 3. Corridor width $Δ γ_{corridor} \propto L / D = 0$ . "Corridor" essentially ek single entry angle mein collapse ho jaata hai jo tumhe passively hit karna padta hai.

Verify: Yeh Terminal Velocity and Ballistic Coefficient se match karta hai — ek purely ballistic body ka corridor sabse narrow hota hai aur yeh mid-flight errors correct nahi kar sakti (early ICBM RVs, kuch sample-return capsules). Forecast confirm: $L / D = 0$ , no steering. ✓

Cell E — teen degenerate bank angles

Worked example Example 5 — Full-up, knife-edge, full-down bank (Cell E)

Statement. Ek vehicle mein constant lift magnitude $L$ hai. Teen special bank angles $σ = 0^{\circ}, 9 0^{\circ}, 18 0^{\circ}$ par vertical lift component $L cos σ$ compute karo, aur bolo ki har ek $\overset{γ}{˙}$ ke saath kya karta hai.

Forecast: $cos$ run karta hai $+ 1 \to 0 \to - 1$ . Guess karo kaun sa angle "entry ke liye commit" karta hai.

Step 1. $σ = 0^{\circ}$ : $cos 0 = + 1$ , vertical lift $= + L$ (full up). $\overset{γ}{˙}$ ko upar push karta hai — diving resist karta hai, steep entry se pull out karne ke liye best.

Step 2. $σ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0$ , vertical lift $= 0$ . Saari lift sideways hai (ground track turn karta hai, range/azimuth change karta hai) lekin $\overset{γ}{˙}$ ke liye kuch nahi karta — "knife-edge" jo sirf left/right steer karta hai. Yeh step kyun? Yeh Bank Angle Modulation and Guidance ka geometric heart hai: bank fixed lift vector ko vertical (corridor control) aur horizontal (range control) ke beech split karta hai.

Step 3. $σ = 18 0^{\circ}$ : $cos 18 0^{\circ} = - 1$ , vertical lift $= - L$ (full down). $\overset{γ}{˙}$ ko neeche force karta hai — "entry ke liye commit" maneuver jo skip rokta hai (yeh exactly Example 2 ka fix hai).

Verify: $cos 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ , $cos 18 0^{\circ} = - 1$ . Teen cases poore vertical-lift range $[- L, + L]$ ko span karte hain, har steering possibility cover karte hain. ✓ Full-down entry ke liye commit karta hai, jaise forecast kiya tha.

Figure — Reentry corridor — angle of attack constraints

Cell G — real-world word problem

Worked example Example 7 — Ek survivable entry angle chuno (Cell G)

Statement. Mission rule: crew peak deceleration $30 g$ se neeche rehni chahiye. Return speed $V_{E} = 11 km/s$ , $H = 7.2$ km. Steepest allowed $∣ γ_{E} ∣$ kya hai (undershoot boundary)?

Forecast: Apollo ne $- 6. 5^{\circ}$ par $\sim 36 g$ diya tha. $30 g$ limit lower hai, toh allowed angle $6. 5^{\circ}$ se shallow hona chahiye.

Step 1. Peak- $g$ formula ko $∣ sin γ_{E} ∣$ ke liye invert karo: $∣ sin γ_{E} ∣ = \frac{2 eH}{V _{E}^{2}} \frac{d V}{d t}_{m a x} .$ Yeh step kyun? G-limit set karta hai deceleration; hum woh angle solve karte hain jo ise just reach karta hai.

Step 2. $∣ d V / d t ∣_{m a x} = 30 \times 9.81 = 294.3 m/s^{2}$ rakho: $∣ sin γ_{E} ∣ = \frac{2 ( 2.718 ) ( 7200 )}{( 11000 ) ^{2}} (294.3) = \frac{3.914 \times 1 0 ^{4}}{1.21 \times 1 0 ^{8}} (294.3) = 3.235 \times 1 0^{- 4} \times 294.3 = 0.09521.$

Step 3. $∣ γ_{E} ∣ = arcsin (0.09521) = 5.4 6^{\circ}$ . Steepest survivable entry $\approx - 5. 5^{\circ}$ .

Verify: Apollo ke $- 6. 5^{\circ}$ (jo $36 g > 30 g$ deta hai) se shallow, exactly jaise forecast kiya tha. Consistency check: $γ_{E} = - 6. 5^{\circ}$ par yahi formula $∣ sin ∣ = 0.1132 \Rightarrow 36 g$ deta hai, parent ke Example 3 se match karta hai. ✓

Cell H — exam twist

Worked example Example 8 — Capsule vs. lifting body corridor width (Cell H)

Statement. "Corridor half-width $Δ γ \propto L / D$ se scale karta hai. Ek capsule ka $L / D = 0.3$ hai; ek lifting body ka $L / D = 1.5$ hai. Agar capsule ka corridor $\pm 0. 5^{\circ}$ wide hai, toh lifting body ka corridor (roughly) kitna wide hai?"

Forecast: $L / D$ $1.5/0.3 = 5 \times$ jump karta hai. Corridor width guess karo.

Step 1. Proportionality: $Δ γ_{LB} = Δ γ_{cap} \times \frac{( L / D ) _{LB}}{( L / D ) _{cap}} = 0. 5^{\circ} \times \frac{1.5}{0.3} .$ Yeh step kyun? Humein bataya gaya hai ki relationship linear hai; sirf ratio matter karta hai, baaki sab constants cancel ho jaate hain.

Step 2. $0. 5^{\circ} \times 5 = 2. 5^{\circ}$ , yani $\pm 2. 5^{\circ}$ .

Verify: Paanch guna wider — ek lifting body paanch guna entry-angle error maaf karta hai. Yahi wajah hai ki Shuttle ( $L / D \approx 1$ – $2$ ) ek runway target kar sakta tha jabki capsules ko wide ocean splashdown zone chahiye. Ratio $1.5/0.3 = 5$ exactly. ✓ Forecast confirm.

Recall Self-check

$V_{E} = 11$ km/s par $30 g$ limit ke liye steepest survivable $γ_{E}$ kya hai? ::: lagbhag $- 5. 5^{\circ}$ $α = 0$ hone par $L / D$ aur corridor width kya hoti hai? ::: dono zero — koi lift nahi, koi steering nahi $σ = 9 0^{\circ}$ par lift $\overset{γ}{˙}$ ke saath kya karta hai? ::: kuch nahi; $cos 9 0^{\circ} = 0$ , lift purely sideways hai Agar entry speed fixed angle par double ho jaaye, toh peak $g$ kis factor se change hoti hai? ::: $\times 4$ (kyunki $g \propto V_{E}^{2}$ ) Kaun sa bank angle "entry ke liye commit" karta hai aur skip rokta hai? ::: $σ = 18 0^{\circ}$ (lift neeche point karta hai)