Reentry corridor — angle of attack constraints: Question bank

The symbols you need (built here, from zero)

Everything below reuses a small cast of symbols. Before you touch the questions, meet them all — each is a plain-English idea first, a symbol second.

Recall Where does

$α_{opt}$ come from? (the differentiation, step by step) The ratio to maximize is $\frac{L}{D} = \frac{C _{Lα} α}{C _{D 0} + k C _{Lα}^{2} α ^{2}}$ . Why differentiate? A maximum is a flat spot: the slope $d (L / D) / d α$ passes through zero there. Call the numerator $u = C_{Lα} α$ and the denominator $w = C_{D 0} + k C_{Lα}^{2} α^{2}$ . The quotient rule says $\frac{d}{d α} \frac{u}{w} = \frac{u ^{'} w - u w ^{'}}{w ^{2}}$ , with $u^{'} = C_{Lα}$ and $w^{'} = 2 k C_{Lα}^{2} α$ . So the numerator of the derivative is $u^{'} w - u w^{'} = C_{Lα} (C_{D 0} + k C_{Lα}^{2} α^{2}) - (C_{Lα} α) (2 k C_{Lα}^{2} α) .$ Expand the two products: $= C_{Lα} C_{D 0} + C_{Lα} k C_{Lα}^{2} α^{2} - 2 k C_{Lα}^{3} α^{2} = C_{Lα} C_{D 0} + k C_{Lα}^{3} α^{2} - 2 k C_{Lα}^{3} α^{2} .$ The two $α^{2}$ terms combine ( $+ 1 - 2 = - 1$ ), leaving $C_{Lα} C_{D 0} - k C_{Lα}^{3} α^{2}$ . A maximum needs this $= 0$ ; divide out the common $C_{Lα}$ : $C_{D 0} - k C_{Lα}^{2} α^{2} = 0 \Rightarrow C_{D 0} = k C_{Lα}^{2} α^{2} .$ Solving for $α$ (take the positive root, since a physical angle of attack is positive): $α_{opt} = \frac{C _{D 0}}{k C _{Lα}^{2}} = \frac{1}{C _{Lα}} \frac{C _{D 0}}{k}$ Physically: the peak is where the fixed drag $C_{D 0}$ equals the lift-induced drag $k C_{Lα}^{2} α^{2}$ — added lift and added drag exactly trade off.

Recall Why peak deceleration

$\propto ∣ sin γ_{E} ∣$ and heating $\propto ρ V^{3}$ Deceleration: drag deceleration is $D / m = \frac{1}{2} ρ V^{2} A C_{D} / m$ (dividing the drag force by the mass $m$ gives an acceleration). As the vehicle plunges, altitude drops at rate $\dot{h} = V sin γ$ , so the steepness $sin γ_{E}$ controls how fast $ρ = ρ_{0} e^{- h / H}$ thickens per second. Carrying this through the exponential atmosphere (the Allen–Eggers Ballistic Reentry result) the single peak works out to $\propto V_{E}^{2} ∣ sin γ_{E} ∣$ : the $V_{E}^{2}$ from dynamic pressure, the $∣ sin γ_{E} ∣$ from how sharply you drive into thicker air. Heating: convective heating rate at the nose scales as $\overset{q}{˙} \propto ρ V^{3}$ (see Aerodynamic Heating and Stanton Number) — the $ρ$ from boundary-layer physics and the $V^{3}$ because heat flux is roughly (energy flux $\frac{1}{2} ρ V^{2} \cdot V$ ) modified by the Stanton number; the extra power of $V$ over drag makes heating the more speed-sensitive limit.

True or false — justify

The corridor is a range of angle-of-attack values

False. The corridor is fundamentally a range of entry flight-path angles

γ_{E}

(and speeds);

α

and

σ

are the actuators you use to stay inside it, not the corridor itself.

A steeper (more negative)

γ_{E}

always means a higher peak deceleration

True. Peak deceleration scales with

∣ sin γ_{E} ∣

, which grows as the dive steepens, so plunging harder into denser air produces a bigger g-spike — that is exactly the undershoot boundary.

Increasing angle of attack always increases lift-to-drag ratio

False. Lift rises linearly (

C_{L} \approx C_{Lα} α

) but the induced drag rises quadratically (

k C_{L}^{2}

), so past

α_{opt}

the ratio

L / D

actually falls.

A capsule with

L / D \approx 0.3

has a narrower corridor than a lifting body with

L / D \approx 1.5

True. Corridor width scales roughly with

L / D

, because more available lift lets the vehicle both pull out of a steeper dive and fight a shallow skip.

Once the engines are off, the crew has no way to change the trajectory

False. Lift (set by

α

) and its direction (set by bank

σ

) are aerodynamic steering that work with engines cold — that is the whole point of the corridor problem.

Shallow entry is safest because it gives the least heating

False. Too shallow and you skip back out of the atmosphere (or badly overshoot the landing site); the shallow side has its own hard boundary, so "least heating" is not automatically "safe."

At the optimal angle of attack, the marginal lift gained exactly balances the marginal drag added

True. Setting

d (L / D) / d α = 0

gives

C_{D 0} = k C_{Lα}^{2} α^{2}

, the point where an extra bit of

α

stops improving the ratio because added drag cancels added lift.

Dynamic pressure

q = \frac{1}{2} ρ V^{2}

keeps rising all the way down through reentry

False.

ρ

grows but

V

bleeds off from drag, so the product

ρ V^{2}

rises, peaks once, then falls — that single peak is what fixes the maximum g and heating.

Banking to

σ = 18 0^{\circ}

changes the magnitude of the lift force

False. Bank only rotates the fixed-magnitude lift vector about the velocity; at

σ = 18 0^{\circ}

the same-size lift simply points downward, driving

\overset{γ}{˙} < 0

Spot the error

"To survive a steep entry, just increase

α

to its maximum — more lift means you get pulled out of the dive fastest."

Beyond

α_{opt}

L / D

drops and drag (hence heating and g-load) rises; you also risk stall. Max

α

neither maximizes control nor minimizes heating.

"The undershoot boundary is a skip limit and the overshoot boundary is a heating limit."

Reversed. Undershoot (too steep) is the heating/g limit; overshoot (too shallow) is the skip-out limit.

"Since

γ

is measured below the horizon,

γ > 0

during reentry."

On descent the velocity points below the horizon, so

γ < 0

throughout entry; a positive

γ

would mean climbing.

"Angle of attack

α

is measured from the local horizontal, just like flight path angle."

α

is measured between the body axis and the velocity vector, not the horizon;

γ

is the one measured from the horizontal.

"Peak heating rate scales the same way as peak deceleration,

\propto V^{2}

Heating rate scales as

\overset{q}{˙} \propto ρ V^{3}

— a stronger dependence on speed — so it tightens the steep boundary even more sharply than the g-limit does.

"To stop a skip-out, roll to

σ = 9 0^{\circ}

so all the lift points sideways."

σ = 9 0^{\circ}

the vertical lift

L cos σ \to 0

, which merely removes the upward push; to actively force the nose back down you need

σ \to 18 0^{\circ}

cos σ \to - 1

and lift points downward.

"The

\frac{m V ^{2}}{R + h}

term is drag written in a different form."

It is the centrifugal effect of a vehicle of mass

m

moving at speed

V

on a planet of radius

R

at altitude

h

; it lives in the cross-track (perpendicular) equation, whereas drag lives in the along-track equation and slows

V