Visual walkthrough — Fairing separation — altitude, dynamic pressure requirements

Every symbol below is earned before it is used. If you have never seen the letter $\rho$ or the word "flux", start at Step 1 and do not skip.

Step 1 — What does "air pushing on a wall" even mean?

WHAT. Imagine you are the flat front of a rocket, a small patch of area $A$ (just a rectangle of surface, measured in square metres). You are flying forward into still air at speed $v$ (metres per second). In your own frame, the air rushes at you.

WHY. Before we can talk about heating or pressure, we must count how much air actually reaches the patch each second. Everything downstream — force, pressure, heat — is bookkeeping on that stream of air.

PICTURE. Look at the figure. In a slice of time $dt$ (a tiny sliver of a second), only the air inside a box of length $vdt$ can reach the patch — anything farther is too slow to arrive in time.

The box has length $\underset{\text{how far air travels}}{\underbrace{vdt}}$ and cross-section $\underset{\text{patch area}}{\underbrace{A}}$, so its volume is $Avdt$. Multiply by density to get the mass caught: $dm=\underset{{\text{kg per m}}^3}{\underbrace{\rho }}\cdot \underset{{\text{m}}^3\text{ of air}}{\underbrace{Avdt}}=\rho Avdt.$ Every term is doing one job: $\rho$ converts volume to mass, $Avdt$ is the swept volume.

Step 2 — From caught air to a pressure

WHAT. Each kilogram of that air was moving at speed $v$, so it carried momentum = mass × velocity. When it stops on the patch, it dumps that momentum.

WHY. A push (a force) is nothing but momentum handed over per second — this is Newton's second law written as $F=dp/dt$. So to find the force, we find the momentum arriving each second.

PICTURE. The arrows in the figure are momentum vectors; the wall catches them and they pile up as a push.

The momentum of the caught slug of air is $dp=dm\cdot v=(\rho Avdt)\cdot v=\rho A{v}^{2}dt.$ Here $dm$ (from Step 1) is the mass and the extra $v$ turns mass into momentum. Force is that per unit time, and pressure is force per unit area: $\frac{1}{A}\frac{dp}{dt}=\frac{1}{A}\cdot \frac{\rho A{v}^{2}dt}{dt}=\rho v^2.$ The $A$ and $dt$ both cancel — pressure does not care how big your patch is. What survives is $\rho v^2$: one $\rho$, two factors of $v$. Remember that count; the "two $v$'s" is the whole reason force scales as $v^2$.

Step 3 — Why we write $q=\frac{1}{2}\rho v^2$ (the ½ is not a fudge)

WHAT. The dynamic pressure $q$ is defined as $\boxed{q=\frac{1}{2}\rho v^2.}$

WHY the half? Two honest reasons, shown side by side in the figure:

1. The kinetic energy stored in one cubic metre of moving air is $\frac{1}{2}\rho v^2$ (the same $\frac{1}{2}m{v}^2$ you know, per unit volume). Dynamic pressure is literally that energy density.
2. It makes $q$ the clean coefficient in the aerodynamic force law $F=C_{D}qA$, where $C_D$ is a shape number.

PICTURE. Left panel: the raw momentum-flux $\rho v^2$. Right panel: the energy-density $\frac{1}{2}\rho v^2$. Same physics, factor-of-two apart.

Step 4 — Heat needs ONE MORE factor of $v$ (the crux)

WHAT. High up, air is so thin that molecules travel a long way between bumps — a molecule hits the rocket and flies off without touching another. That regime is free-molecular flow. Each molecule slams in and dumps its kinetic energy $\frac{1}{2}m{v}^2$ as heat.

WHY this is different from pressure. Pressure counted momentum per second. Heating counts energy per second. Energy is momentum-like flux multiplied by the extra speed at which that energy arrives. That single extra multiplication is where the third $v$ comes from.

PICTURE. The figure stacks the two bookkeepings: momentum flux (2 arrows of $v$) versus energy flux (3 arrows of $v$).

Mass arriving per second per area (from Step 1, drop the $A$): $\rho v$. Each kilogram carries kinetic energy $\frac{1}{2}{v}^2$. Multiply: $\dot{q}=\underset{\text{mass flux}}{\underbrace{\rho v}}\cdot \underset{\text{energy per kg}}{\underbrace{\frac{1}{2}{v}^2}}=\frac{1}{2}\rho v^3.$ Now count: one $\rho$, three $v$'s. Compare with $q$'s two $v$'s: $\dot{q}=\frac{1}{2}\rho v^3=\underset{=q}{\underbrace{\left(\frac{1}{2}\rho v^2\right)}}\cdot v=qv.$

Step 5 — Why altitude enters as an EXPONENTIAL

WHAT. The atmosphere does not thin out linearly; it thins exponentially: $\rho (h)={\rho }_0{e}^{-h/H}.$

WHY exponential? Each thin layer of air is squeezed by the weight of every layer above it. Add a fixed height $H$ and the same fraction of air is left — repeated fractional shrinking is exactly what $e^{-h/H}$ means. $H$ is the scale height ($\approx 7.5$ km): the climb needed to cut density to $1/e\approx 37\%$.

PICTURE. The curve free-falls: each 7.5 km step slices the height of the bar to a third.

Feed this into the heat flux: $\dot{q}(h)=\frac{1}{2}{\rho }_0{e}^{-h/H}{v}^3.$ Two forces fight: the huge $v^3$ (want to cook the payload) versus the collapsing $e^{-h/H}$ (wants to cool it). Altitude is the referee.

Step 6 — Solving for the jettison altitude

WHAT. Set the flux equal to the safe limit ${\dot{q}}_{\text{limit}}$ and ask: at what height does the exponential finally beat the $v^3$?

WHY the logarithm now. We have $h$ trapped inside $e^{-h/H}$. The natural log $\ln$ is the exact question "$e$ to what power gives this?" — the one tool that pulls $h$ back out of the exponent. That is why $\ln$ appears, not by taste but by necessity.

PICTURE. The flux curve crosses the horizontal safe-line at exactly one altitude — that crossing is $h_{\text{jett}}$.

Solve step by step. First isolate the density the limit demands: ${\dot{q}}_{\text{limit}}=\frac{1}{2}\rho v^3\Rightarrow \rho =\frac{2{\dot{q}}_{\text{limit}}}{v^3}.$ Then invert the exponential with a log: ${\rho }_0{e}^{-h/H}=\frac{2{\dot{q}}_{\text{limit}}}{v^3}\Rightarrow e^{-h/H}=\frac{2{\dot{q}}_{\text{limit}}}{{\rho }_0{v}^3}\Rightarrow \boxed{h_{\text{jett}}=H\ln \frac{\frac{1}{2}{\rho }_0{v}^3}{{\dot{q}}_{\text{limit}}}.}$

Plug in Example-2's numbers (${\dot{q}}_{\text{limit}}=1135$, $v=4000$, ${\rho }_0=1.2$, $H=7.5$): $h_{\text{jett}}=7.5\ln \frac{0.5\times 1.2\times (4000{)}^3}{1135}\approx 7.5\times 17.3\approx 130\text{km}.$

Step 7 — The edge cases (never let the reader hit an unshown scenario)

WHAT / WHY / PICTURE — all three degenerate limits on one figure.

The one-picture summary

One curve $\dot{q}(h)=\frac{1}{2}{\rho }_0{e}^{-h/H}{v}^3$ falling off a cliff, one horizontal safe-line ${\dot{q}}_{\text{limit}}$, one crossing point marked $h_{\text{jett}}\approx 130$ km. Everything on this page is that crossing. See also Dynamic Pressure (Max-Q) and Aerodynamic Heating & Free-Molecular Flow.

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