Reentry corridor — angle of attack constraints: Worked examples

This page is a case-by-case tour. The parent note built the formulas; here we fire every kind of input at them — steep, shallow, zero lift, maximum lift, degenerate bank angles, and an exam twist — so that no scenario surprises you.

The scenario matrix

Every problem in this topic lives in one of these cells. The examples below are labelled with the cell they cover.

Cell	What varies	The physical question	Example
A. Steep entry ( $γ_{E}$ large negative)	undershoot side	Does peak $g$ exceed the limit?	Ex 1, Ex 6
B. Shallow entry ( $γ_{E}$ near $0$ )	overshoot side	Does it skip out?	Ex 2
C. Optimal $α$	steering gain peak	Widest corridor config	Ex 3
D. Zero / degenerate $α$	$α = 0$	No lift — 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)	scaling with $V_{E}^{2}$	How entry speed magnifies $g$	Ex 6
G. Real-world word problem	full mission numbers	Pick a survivable $γ_{E}$	Ex 7
H. Exam twist	corridor width vs. $L / D$	Compare capsule vs. lifting body	Ex 8

We reuse these constant symbols throughout (all defined in the parent):

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

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

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

Statement. A capsule enters at $V_{E} = 7.8 km/s$ (low-Earth-orbit speed), $H = 7.2 km$ , $γ_{E} = - 8^{\circ}$ . Estimate the peak deceleration in $g$ 's (take $g = 9.81 m/s^{2}$ ).

Forecast: steeper than Apollo's $- 6. 5^{\circ}$ but slower entry — will peak $g$ be above or below Apollo's $\sim 36 g$ ? (Two effects fight: shallower speed lowers it, steeper angle raises it.)

Step 1. Use the Allen–Eggers peak-deceleration formula for an exponential atmosphere: $\frac{d V}{d t}_{m a x} = \frac{V _{E}^{2} ∣ s i n γ _{E} ∣}{2 eH} .$ Why this step? This is the closed-form peak of the $V^{2} ρ$ product — see Allen–Eggers Ballistic Reentry. It needs only entry speed, entry angle, and scale height.

Step 2. Plug numbers. $V_{E} = 7800 m/s$ , $sin 8^{\circ} = 0.1392$ , $H = 7200 m$ : $\frac{( 7800 ) ^{2} ( 0.1392 )}{2 ( 2.718 ) ( 7200 )} = \frac{6.084 \times 1 0 ^{7} \times 0.1392}{3.914 \times 1 0 ^{4}} = \frac{8.47 \times 1 0 ^{6}}{3.914 \times 1 0 ^{4}} \approx 216 m/s^{2} .$ Why this step? Convert km to m first, else the answer is off by $1 0^{6}$ — a classic unit trap.

Step 3. Convert to $g$ : $216/9.81 \approx 22 g$ .

Verify: Apollo at $V_{E} = 11$ km/s, $γ_{E} = - 6. 5^{\circ}$ gave $\sim 36 g$ . Our vehicle is slower ( $7.8$ vs $11$ km/s, and $g \propto V_{E}^{2}$ so the speed ratio $(7.8/11)^{2} = 0.50$ halves it) but steeper ( $sin 8^{\circ} / sin 6. 5^{\circ} = 1.23$ raises it). 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. A lifting body has $C_{Lα} = 2.0 / rad$ , $C_{D 0} = 0.05$ , $k = 0.4$ . Find $α_{opt}$ (in degrees) and $(L / D)_{m a x}$ .

Forecast: With low zero-lift drag $C_{D 0} = 0.05$ , do you expect $L / D$ above or below the capsule value $\approx 2.24$ from the parent?

Step 1. $α_{opt} = \frac{1}{C _{Lα}} \frac{C _{D 0}}{k} = \frac{1}{2.0} \frac{0.05}{0.4} = \frac{1}{2.0} (0.3536) = 0.1768 rad .$ Why this step? This is where marginal drag equals marginal lift benefit — the peak of the linear-over-quadratic $L / D$ (see Lift and Drag Coefficients).

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

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: Lower $C_{D 0}$ than the parent's $0.10$ ⇒ higher $L / D$ . $3.54 > 2.24$ . ✓ A sleeker body has a wider corridor, $Δ γ_{corridor} \propto L / D$ .

Worked example Example 6 — Doubling entry speed (Cell F, limiting behaviour)

Statement. Same vehicle and angle, but compare a hyperbolic return $V_{E} = 15 km/s$ against $V_{E} = 7.5 km/s$ , both at $γ_{E} = - 6^{\circ}$ , $H = 7.2$ km. By what factor does peak $g$ change?

Forecast: Speed doubles ( $\times 2$ ). Peak $g \propto V_{E}^{2}$ — guess the factor before reading on.

Step 1. From $∣ d V / d t ∣_{m a x} \propto V_{E}^{2}$ , the ratio is $(15/7.5)^{2} = 4$ . Why this step? $γ_{E}$ and $H$ are identical, so they cancel; only the $V_{E}^{2}$ survives.

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

Verify: Slow case is $601/4 = 150 m/s^{2} \approx 15 g$ . A factor 4, as forecast. ✓ This is why lunar-return vehicles (higher $V_{E}$ ) sit much closer to the undershoot g-limit than LEO returns.

Cell B — the shallow / skip-out case

Worked example Example 2 — Is this entry too shallow to capture? (Cell B)

Statement. A capsule with $L / D = 0.3$ enters at $γ_{E} = - 1. 2^{\circ}$ flying lift-up ( $σ = 0$ ). At the top of the atmosphere the cross-track equation is $mV \overset{γ}{˙} = L cos σ - m g cos γ + \frac{m V ^{2}}{R + h} cos γ .$ Does the trajectory curve up (skip) or down (capture)?

Forecast: shallow angle + lift pointing up + high speed centrifugal term — all push the path upward. Guess: skip.

Step 1. Divide by $m$ and check the sign of the bracket at $V \approx V_{E} = 7.8$ km/s, $h \approx 90$ km, $R = 6371$ km, $g \approx 9.5 m/s^{2}$ : $\overset{γ}{˙} \propto up \frac{L}{m} cos σ - down g cos γ + up \frac{V ^{2}}{R + h} cos γ .$ Why this step? Signs, not magnitudes, decide skip vs. capture — we only need which way the bracket leans.

Step 2. The centrifugal term alone: $\frac{V ^{2}}{R + h} = \frac{( 7800 ) ^{2}}{6.461 \times 1 0 ^{6}} = 9.41 m/s^{2}$ , essentially equal to $g$ . With $cos γ \approx 1$ , gravity and centrifugal nearly cancel. Why this step? At near-orbital speed the vehicle is almost "in orbit" — gravity barely bends it down.

Step 3. With those two cancelling, the leftover $L cos σ > 0$ (lift-up) makes $\overset{γ}{˙} > 0$ : the nose lifts, the path curves up → skip-out.

Verify: At $γ_{E} = - 1. 2^{\circ}$ (very shallow) and lift-up, the physics matches the Bank Angle Modulation and Guidance rule: to stop skipping you must roll toward $σ = 18 0^{\circ}$ . Forecast confirmed — this entry skips unless banked. ✓

Cell D — zero lift, degenerate input

Worked example Example 4 — Pure ballistic entry,

$α = 0$ (Cell D) Statement. A spherical probe flies at $α = 0$ , so $C_{L} = C_{Lα} \cdot 0 = 0$ . What is $L / D$ , what does the cross-track equation reduce to, and can this vehicle steer?

Forecast: $α = 0$ kills lift. Guess $L / D$ and whether the corridor is wide or a single line.

Step 1. $L = \frac{1}{2} ρ V^{2} A C_{L} (α)$ with $C_{L} = 0 \Rightarrow L = 0$ , so $L / D = 0$ . Why this step? Lift is linear in $α$ near zero; a sphere at zero angle produces zero net lift by symmetry.

Step 2. Set $L = 0$ in the cross-track equation: $mV \overset{γ}{˙} = - m g cos γ + \frac{m V ^{2}}{R + h} cos γ .$ Bank $σ$ has vanished — $cos σ$ multiplies a zero. No steering knob remains. Why this step? If lift is zero, rotating it (banking) does nothing: $0 cos σ = 0$ for every $σ$ .

Step 3. Corridor width $Δ γ_{corridor} \propto L / D = 0$ . The "corridor" collapses to essentially a single entry angle you must hit passively.

Verify: This matches Terminal Velocity and Ballistic Coefficient — a purely ballistic body has the narrowest corridor and cannot correct errors mid-flight (early ICBM RVs, some sample-return capsules). Forecast confirmed: $L / D = 0$ , no steering. ✓

Cell E — the three degenerate bank angles

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

Statement. A vehicle has constant lift magnitude $L$ . Compute the vertical lift component $L cos σ$ at the three special bank angles $σ = 0^{\circ}, 9 0^{\circ}, 18 0^{\circ}$ , and say what each does to $\overset{γ}{˙}$ .

Forecast: $cos$ runs $+ 1 \to 0 \to - 1$ . Guess which angle "commits to entry."

Step 1. $σ = 0^{\circ}$ : $cos 0 = + 1$ , vertical lift $= + L$ (full up). Pushes $\overset{γ}{˙}$ up — resists diving, best for pulling out of a steep entry.

Step 2. $σ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0$ , vertical lift $= 0$ . All lift is sideways (turns the ground track, changes range/azimuth) but does nothing to $\overset{γ}{˙}$ — the "knife-edge" that only steers left/right. Why this step? This is the geometric heart of Bank Angle Modulation and Guidance: bank splits the fixed lift vector between vertical (corridor control) and horizontal (range control).

Step 3. $σ = 18 0^{\circ}$ : $cos 18 0^{\circ} = - 1$ , vertical lift $= - L$ (full down). Forces $\overset{γ}{˙}$ down — the "commit to entry" maneuver that stops a skip (this is exactly Example 2's fix).

Verify: $cos 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ , $cos 18 0^{\circ} = - 1$ . The three cases span the full vertical-lift range $[- L, + L]$ , covering every steering possibility. ✓ Full-down commits to entry, as forecast.

Figure — Reentry corridor — angle of attack constraints

Cell G — real-world word problem

Worked example Example 7 — Pick a survivable entry angle (Cell G)

Statement. Mission rule: crew peak deceleration must stay below $30 g$ . Return speed $V_{E} = 11 km/s$ , $H = 7.2$ km. What is the steepest allowed $∣ γ_{E} ∣$ (the undershoot boundary)?

Forecast: Apollo used $- 6. 5^{\circ}$ for $\sim 36 g$ . A $30 g$ limit is lower, so the allowed angle must be shallower than $6. 5^{\circ}$ .

Step 1. Invert the peak- $g$ formula for $∣ sin γ_{E} ∣$ : $∣ sin γ_{E} ∣ = \frac{2 eH}{V _{E}^{2}} \frac{d V}{d t}_{m a x} .$ Why this step? The g-limit sets the deceleration; we solve for the angle that just reaches it.

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

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

Verify: Shallower than Apollo's $- 6. 5^{\circ}$ (which gave $36 g > 30 g$ ), exactly as forecast. Check consistency: at $γ_{E} = - 6. 5^{\circ}$ this same formula gives $∣ sin ∣ = 0.1132 \Rightarrow 36 g$ , matching the parent's Example 3. ✓

Cell H — the exam twist

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

Statement. "The corridor half-width scales as $Δ γ \propto L / D$ . A capsule has $L / D = 0.3$ ; a lifting body has $L / D = 1.5$ . If the capsule's corridor is $\pm 0. 5^{\circ}$ wide, how wide (roughly) is the lifting body's?"

Forecast: $L / D$ jumps by $1.5/0.3 = 5 \times$ . Guess the corridor width.

Step 1. Proportionality: $Δ γ_{LB} = Δ γ_{cap} \times \frac{( L / D ) _{LB}}{( L / D ) _{cap}} = 0. 5^{\circ} \times \frac{1.5}{0.3} .$ Why this step? We're told the relationship is linear; only the ratio matters, all other constants cancel.

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

Verify: Five times wider — a lifting body forgives five times the entry-angle error. This is why the Shuttle ( $L / D \approx 1$ – $2$ ) could target a runway while capsules need a wide ocean splashdown zone. Ratio $1.5/0.3 = 5$ exactly. ✓ Forecast confirmed.

Recall Self-check

Steepest survivable $γ_{E}$ for $V_{E} = 11$ km/s at the $30 g$ limit? ::: about $- 5. 5^{\circ}$ What is $L / D$ and the corridor width when $α = 0$ ? ::: both zero — no lift, no steering At $σ = 9 0^{\circ}$ , what does lift do to $\overset{γ}{˙}$ ? ::: nothing; $cos 9 0^{\circ} = 0$ , lift is purely sideways If entry speed doubles at fixed angle, peak $g$ changes by what factor? ::: $\times 4$ (since $g \propto V_{E}^{2}$ ) Which bank angle "commits to entry" and stops a skip? ::: $σ = 18 0^{\circ}$ (lift points down)