Exercises — Reentry corridor — angle of attack constraints

• $V$ ::: the vehicle's speed through the air (m/s) — the length of its velocity vector.
• $V_E$ ::: the entry speed, i.e. the value of $V$ at the moment the vehicle first meets the sensible atmosphere (the "interface"). For LEO return $\approx 7.8\text{ km/s}$; for lunar return $\approx 11\text{ km/s}$.
• $m$ ::: the mass of the vehicle (kg) — appears in Newton's law $F=ma$ for the trajectory equations.
• $C_{L\alpha }$ ::: the lift-curve slope ($/\text{rad}$) — how many units of lift coefficient $C_L$ you gain per radian of angle of attack, i.e. $C_L\approx C_{L\alpha }\alpha$. Bigger $C_{L\alpha }$ = the shape makes lift more eagerly.
• $C_{D0}$ ::: the zero-lift drag coefficient (dimensionless) — the drag the body has even at $\alpha =0$, purely from its bluntness/shape.
• $k$ ::: the induced-drag factor (dimensionless) — the penalty coefficient in $C_D\approx C_{D0}+k{C}_L^2$; it says how fast drag grows as you demand more lift. (All from Lift and Drag Coefficients.)

Undershoot (the steep boundary). Diving fast means the vehicle reaches dense low air while still moving quickly, so dynamic pressure $q=\frac{1}{2}\rho V^2$ and drag spike together. Undershoot is the limit boundary (g-load + heating). Overshoot is the skip-out boundary — the shallow one, where you glance off the thin upper air and leave the atmosphere again.

• $\alpha$ sets the size of both lift $L$ and drag $D$ (through $C_L(\alpha )$ and $C_D(\alpha )$).
• $\sigma$ sets the direction of the (fixed-magnitude) lift — how much points up, $L\cos \sigma$, versus sideways, $L\sin \sigma$.

Think of $\alpha$ as the throttle on the aero force and $\sigma$ as the steering wheel that rotates it.

On descent $\gamma <0$: the velocity vector points below the local horizontal (nose-down relative to horizon). $\gamma =0$ means you are flying exactly horizontally — level flight, neither climbing nor sinking at that instant. $\gamma >0$ would mean the path is curving back up (the start of a skip).

Use the parent's boxed results: ${\alpha }_{\text{opt}}=\frac{1}{C_{L\alpha }}\sqrt{\frac{C_{D0}}{k}}=\frac{1}{1.8}\sqrt{\frac{0.08}{0.4}}=\frac{1}{1.8}(0.4472)=0.2485\text{ rad}.$ Convert: $0.2485\times \frac{180}{\pi }\approx 14.2^{\circ }$. ${\left(\frac{L}{D}\right)}_{\max }=\frac{1}{2\sqrt{k{C}_{D0}}}=\frac{1}{2\sqrt{0.4\cdot 0.08}}=\frac{1}{2(0.1789)}=\mathrm{2.795.}$ WHY the peak sits exactly there (the core idea). Look at Figure s01 below. The yellow dotted line is the numerator alone, $C_{L\alpha }\alpha$ — pure lift, rising in a straight line. The blue curve is the full $L/D$. At small $\alpha$ the blue curve hugs the yellow line: adding $\alpha$ buys almost pure lift, because the drag denominator is still dominated by the constant $C_{D0}$ and the ${\alpha }^2$ term is negligible. As $\alpha$ grows, the $k{C}_{L\alpha }^2{\alpha }^2$ drag term wakes up and starts eating the gains, so the blue curve peels away and bends over. The red dot is the exact balance point: one more unit of $\alpha$ adds as much drag (denominator) as lift (numerator), so the slope of $L/D$ is momentarily flat — that flat top is why the derivative vanishes at ${\alpha }_{\text{opt}}$. Left of it you are wasting available lift; right of it drag wins.

The tool we use (and where it comes from). The peak-deceleration formula ${\left|\frac{dV}{dt}\right|}_{\max }=\frac{V_E^2|\sin {\gamma }_E|}{2eH}$ is not dropped in — it is derived from scratch in problem L4.1 below (Allen–Eggers; see also Allen–Eggers Ballistic Reentry). For now use it as a ready tool: it says the worst deceleration a purely ballistic (no-lift) capsule feels depends only on entry speed squared, how steeply it dives ($\sin {\gamma }_E$), the atmosphere's scale height $H$, and the constant $e$. Numerator: $(7800{)}^2\sin 5^{\circ }=6.084\times 10^7\times 0.08716=5.303\times 10^6$. Denominator: $2(2.71828)(7000)=3.806\times 10^4$. ${\left|\frac{dV}{dt}\right|}_{\max }=\frac{5.303\times 10^6}{3.806\times 10^4}=139.3{\text{ m/s}}^2=\frac{139.3}{9.81}\approx 14.2g.$ Why $e$ appears: the density grows like $e^{h/H}$ while speed bleeds off; their product $\rho V^2$ peaks exactly once, and the peak value carries a factor $1/e$ from that crossover (proved in L4.1).

$\frac{\Delta {\gamma }_B}{\Delta {\gamma }_A}=\frac{(L/D{)}_B}{(L/D{)}_A}=\frac{1.5}{0.3}=5.$ B's corridor is 5× wider. More lift lets you both pull out of a steeper dive (dodge undershoot) and dig into a shallow one (dodge overshoot) — it widens both boundaries at once.

WHAT we do: differentiate the quotient and set it to zero. Let $f(\alpha )=\frac{N}{D}$ with $N=C_{L\alpha }\alpha$, $D=C_{D0}+k{C}_{L\alpha }^2{\alpha }^2$. Quotient rule: $f^{\prime }=\frac{N^{\prime }D-N{D}^{\prime }}{D^2}$. The sign of $f^{\prime }$ is the sign of the numerator $N^{\prime }D-N{D}^{\prime }=C_{L\alpha }(C_{D0}+k{C}_{L\alpha }^2{\alpha }^2)-C_{L\alpha }\alpha (2k{C}_{L\alpha }^2\alpha )=C_{L\alpha }(C_{D0}-k{C}_{L\alpha }^2{\alpha }^2).$ Setting to zero: $C_{D0}=k{C}_{L\alpha }^2{\alpha }^2\Rightarrow {\alpha }_{\text{opt}}=\frac{1}{C_{L\alpha }}\sqrt{C_{D0}/k}$. ✓ Why it's a maximum: the bracket $C_{D0}-k{C}_{L\alpha }^2{\alpha }^2$ is positive for small $\alpha$ (function rising) and negative for large $\alpha$ (function falling). Rising-then-falling ⇒ a peak. Plug back: $D$ at optimum $=C_{D0}+C_{D0}=2C_{D0}$, and $N=C_{L\alpha }\cdot \frac{1}{C_{L\alpha }}\sqrt{C_{D0}/k}=\sqrt{C_{D0}/k}$, so ${\left(\frac{L}{D}\right)}_{\max }=\frac{\sqrt{C_{D0}/k}}{2C_{D0}}=\frac{1}{2\sqrt{k{C}_{D0}}}.✓$

$\dot{\gamma }$ becomes most negative when $L\cos \sigma$ is most negative, i.e. $\cos \sigma =-1$, so $\boxed{\sigma =180^{\circ }}.$ At $\sigma =180^{\circ }$ the lift points straight down, adding to gravity to bend the path back into the atmosphere. Why: bank rotates the lift vector without changing its length; $\cos \sigma$ runs from $+1$ (lift fully up) through $0$ (lift fully sideways) to $-1$ (lift fully down). "Commit to entry" = roll to $180^{\circ }$. The $\frac{m{V}^2}{R+h}\cos \gamma$ term (centrifugal, from riding the curved Earth) and gravity are fixed here — only the lift term is under our control.

Heating carries $V^3$; deceleration carries $V^2$. So a fractional speed increase $\delta$ multiplies heating by $(1+\delta {)}^3$ but deceleration by only $(1+\delta {)}^2$. Heating is more speed-sensitive. Consequence: for lunar/interplanetary return ($V_E\approx 11\text{ km/s}$) the corridor is squeezed mainly by the heat shield, not by crew g-tolerance; for low-Earth-orbit return ($V_E\approx 7.8\text{ km/s}$) g-load is the more even competitor. Faster returns are "heating-limited." (See Aerodynamic Heating and Stanton Number.)

WHAT: we want the altitude where $\rho V^2$ is largest. Write $u=e^{-h/H}$ so $\rho ={\rho }_{0}u$ and let $B=\frac{{\rho }_{0}H}{2\beta |\sin {\gamma }_E|}$, giving $V=V_E{e}^{-Bu}$, hence $\rho V^2={\rho }_{0}u{V}_E^2{e}^{-2Bu}.$ WHY: deceleration magnitude $\propto \rho V^2$, so its peak is the peak deceleration. Differentiate w.r.t. $u$ (a monotone function of $h$): $\frac{d}{du}\left(u{e}^{-2Bu}\right)=e^{-2Bu}(1-2Bu)=0\Rightarrow u_{\text{pk}}=\frac{1}{2B}$. Then ${\rho }_{\text{pk}}={\rho }_0{u}_{\text{pk}}=\frac{{\rho }_0}{2B}=\frac{{\rho }_0}{2}\cdot \frac{2\beta |\sin {\gamma }_E|}{{\rho }_{0}H}=\frac{\beta |\sin {\gamma }_E|}{H}.$ ✓ At $u_{\text{pk}}$: the exponent $2B{u}_{\text{pk}}=1$, so $V_{\text{pk}}=V_E{e}^{-1/2}$, and ${\left|\dot{V}\right|}_{\max }=\frac{{\rho }_{\text{pk}}{V}_{\text{pk}}^2}{2\beta }=\frac{1}{2\beta }\cdot \frac{\beta |\sin {\gamma }_E|}{H}\cdot V_E^2{e}^{-1}=\frac{V_E^2|\sin {\gamma }_E|}{2eH}.✓$ WHAT it looks like (read Figure s02 below). The blue curve is density $\rho$ — it climbs steadily as you fall (right-to-left, since altitude decreases). The yellow curve is $V^2$ — high up it is still near the full entry speed, but low down it has bled away to almost nothing. Deceleration $\propto \rho V^2$ (the red curve) is the product of a rising thing and a falling thing, so it must have a single hump: it is small high up (no air) and small low down (no speed), peaking in between. The green dashed line marks that peak; algebra pins it exactly where the speed has dropped to $V_E/\sqrt{e}$ (i.e. $2Bu=1$) — which is precisely why the $1/e$ factor lands in the final formula.

Set peak deceleration $=8g=78.48{\text{ m/s}}^2$ and solve for $\sin {\gamma }_E$: $|\sin {\gamma }_E|=\frac{2eH\cdot 8g}{V_E^2}=\frac{2(2.71828)(7000)(78.48)}{(7600{)}^2}=\frac{2.987\times 10^6}{5.776\times 10^7}=\mathrm{0.05171.}$ $|{\gamma }_E|=\arcsin (0.05171)=2.964^{\circ }\approx 2.96^{\circ }.$ So anything steeper than $\approx -3.0^{\circ }$ exceeds the $8g$ crew limit — that is the undershoot edge for this vehicle.

Corridor width: $\frac{\Delta {\gamma }_B}{\Delta {\gamma }_A}=\frac{1.4}{0.35}=4.0$ — Option B's corridor is 4× wider, so guidance errors and atmospheric uncertainty are far more forgiving. Deceleration: more lift lets B fly a lofted, gentler trajectory, spreading energy loss over more time ⇒ lower peak $g$ and lower peak heating rate (though possibly larger total heat load because it stays hot longer — Aerodynamic Heating and Stanton Number). The price: a lifting body needs a more complex, heavier, asymmetric heat shield and active guidance (Bank Angle Modulation and Guidance); the blunt capsule is lighter, simpler, self-stabilizing, and cheaper. Verdict: for a razor-thin lunar-return corridor, B's 4× margin is worth the mass — but capsules (Apollo, Orion) won historically because active bank modulation gave enough corridor from a simple, robust shape.

(a) ${\left|\dot{V}\right|}_{\max }=\frac{V_E^2|\sin {\gamma }_E|}{2eH}=\frac{(11000{)}^2\sin 6.5^{\circ }}{2(2.71828)(7200)}=\frac{1.21\times 10^8\times 0.11320}{3.914\times 10^4}=\frac{1.370\times 10^7}{3.914\times 10^4}=349.9{\text{ m/s}}^2.$ In $g$: $349.9/9.81=35.7g$. (b) $35.7g>12g$ — yes, it exceeds a $12g$ limit. (Historically Apollo peaked near ~7g in practice because the vehicle lifts and flies a lofted profile — this ballistic estimate is the worst-case, purely-ballistic upper bound, which is exactly why they used $L/D\approx 0.3$ lift to shave it down.) (c) Steepest angle at $12g$: set the formula equal to $12g=117.72{\text{ m/s}}^2$ and solve for $\sin {\gamma }_E$: $|\sin {\gamma }_E|=\frac{2eH(12g)}{V_E^2}=\frac{2(2.71828)(7200)(117.72)}{(11000{)}^2}=\frac{4.609\times 10^6}{1.21\times 10^8}=0.03809,$ $|{\gamma }_E|=\arcsin (0.03809)=2.183^{\circ }\approx 2.18^{\circ }.$ Reading it: on a purely ballistic path even $6.5^{\circ }$ is far too steep — the whole point of lift + bank modulation is to make a survivable corridor exist where ballistics alone says "no." A crewed vehicle simply cannot enter this steep without active lift to loft the trajectory and spread the deceleration.