Exercises — Fairing separation — altitude, dynamic pressure requirements

What: apply the definition $q=\frac{1}{2}\rho v^2$ straight from Step 1 of the parent note. Why this formula: $q$ is the kinetic-energy density of the oncoming air; it needs a density and a speed, both given. $q=\frac{1}{2}(5\times 10^{-4})(2000{)}^2=0.5\times 5\times 10^{-4}\times 4\times 10^6=1000\text{Pa}.$ Answer: $q=1000\text{Pa}=1\text{kPa}$.

What: use $\dot{q}=qv$ (Step 2 of the parent note). Why: heat flux is force-flux times velocity; since $q$ already carries $\rho v^2$, one more factor of $v$ gives $\rho v^3/2$. $\dot{q}=qv=1000\times 2000=2\times 10^6{\text{W/m}}^2.$ Answer: $\dot{q}=2.0\times 10^6{\text{W/m}}^2$ — far above the 1135 limit, so far too low/dense to jettison.

What: exponential density law $\rho ={\rho }_0{e}^{-h/H}$. Why: the atmosphere thins by a constant factor every scale height, so the altitude enters only through the exponent $-h/H$. Units check: $h=30$ km and $H=7.5$ km share the unit km, so $h/H=4$ is a pure number. $\rho =1.2e^{-30/7.5}=1.2e^{-4}=1.2\times 0.0183=0.0220{\text{kg/m}}^3.$ Answer: $\rho \approx 2.2\times 10^{-2}{\text{kg/m}}^3$.

Step 1 — density: the altitude enters only through the exponential thinning of the air, so we start there. With $h$ and $H$ both in km, $\rho =1.2e^{-100/7.5}=1.2e^{-13.33}$. Since $e^{-13.33}=1.62\times 10^{-6}$, $\rho =1.2\times 1.62\times 10^{-6}=1.94\times 10^{-6}{\text{kg/m}}^3.$ Step 2 — heat flux: use $\dot{q}=\frac{1}{2}\rho v^3$. Why this formula and not $\frac{1}{2}\rho v^2$: we want the heat the payload absorbs, which is energy per second per area — not the aerodynamic force. In free-molecular flow each molecule dumps its kinetic energy $\frac{1}{2}{v}^2$ on impact, and the mass arriving per second per area is $\rho v$; multiply them and you get $\frac{1}{2}\rho v^3$. That extra factor of $v$ over the pressure $\frac{1}{2}\rho v^2$ is exactly what makes fast vehicles dangerous. $\dot{q}=0.5\times 1.94\times 10^{-6}\times (3500{)}^3=0.5\times 1.94\times 10^{-6}\times 4.29\times 10^{10}=4.16\times 10^4{\text{W/m}}^2.$ Decision: $4.16\times 10^4\gg 1135$ → do not jettison yet; climb higher where $\rho$ keeps dropping. Answer: $\dot{q}\approx 4.2\times 10^4{\text{W/m}}^2$, too high.

Step 1 — required density: invert $\dot{q}=\frac{1}{2}\rho v^3$ to find the density at which the flux just equals the limit. Why invert this way: the speed $v$ is set by the trajectory, and the flux ceiling is fixed by the mission; the only free variable that brings $\dot{q}$ down is $\rho$. So we solve the heating law for $\rho$ instead of for $\dot{q}$. $\rho =\frac{2{\dot{q}}_{\text{limit}}}{v^3}=\frac{2\times 1135}{(3500{)}^3}=\frac{2270}{4.29\times 10^{10}}=5.29\times 10^{-8}{\text{kg/m}}^3.$ Step 2 — invert density law: $h=H\ln \frac{{\rho }_0}{\rho }$ (result comes out in km because $H$ is in km). Why the log: density is tied to altitude by $\rho ={\rho }_0{e}^{-h/H}$, an exponential; the only way to pull $h$ out of an exponent is the natural log, which undoes the exponential. As a bonus it converts the terrifying density ratio ($2\times 10^7$) into the tame number $16.9$. $h=7.5\ln \frac{1.2}{5.29\times 10^{-8}}=7.5\ln (2.27\times 10^7)=7.5\times 16.94=127\text{km}.$ Answer: $h_{\text{jett}}\approx 127\text{km}$.

Carry-through Δv: $\Delta v=I_{sp}{g}_0\ln \frac{m_0}{m_f}$. $\Delta v=460\times 9.81\times \ln \frac{25000}{7000}=4512.6\times \ln (3.571)=4512.6\times 1.2730=5745\text{m/s}.$ Drop-early Δv: subtract $m_{fr}$ from both masses: $\Delta v^{\prime }=4512.6\times \ln \frac{23800}{5800}=4512.6\times \ln (4.103)=4512.6\times 1.4117=6371\text{m/s}.$ Why both masses drop: the fairing is gone before the burn, so it is neither initial nor final propellant mass — removing it from both raises the mass ratio $m_0/m_f$, and $\ln$ of a bigger ratio is bigger. Answer: gain $=6371-5745\approx 626\text{m/s}$.

Shared density: $\rho =1.2e^{-115/7.5}=1.2e^{-15.33}=1.2\times 2.19\times 10^{-7}=2.63\times 10^{-7}{\text{kg/m}}^3.$ Flux A: ${\dot{q}}_A=0.5\times 2.63\times 10^{-7}\times (3000{)}^3=0.5\times 2.63\times 10^{-7}\times 2.7\times 10^{10}=3.55\times 10^3{\text{W/m}}^2.$ Flux B: ${\dot{q}}_B=0.5\times 2.63\times 10^{-7}\times (5000{)}^3=0.5\times 2.63\times 10^{-7}\times 1.25\times 10^{11}=1.64\times 10^4{\text{W/m}}^2.$ Ratio: ${\dot{q}}_B/{\dot{q}}_A=(5000/3000{)}^3=(5/3{)}^3=\mathrm{4.63.}$ Interpretation: at the same 115 km, B heats the payload 4.6× harder purely because $\dot{q}\propto v^3$. Both are above 1135, but B needs to climb noticeably higher before it is safe. Altitude is only a proxy for $\rho$; the real criterion is $\rho v^3$. Answer: ${\dot{q}}_A\approx 3.6\times 10^3$, ${\dot{q}}_B\approx 1.6\times 10^4{\text{W/m}}^2$, ratio $=4.63$.

Key trick: the difference kills every constant term. Since $\ln \frac{{\rho }_0{v}^3/2}{\dot{q}}=\ln ({\rho }_0/2\dot{q})+3\ln v$, the only $v$-dependence is $3\ln v$. $\Delta h=H\cdot 3(\ln v_2-\ln v_1)=3H\ln \frac{v_2}{v_1}=3\times 7.5\times \ln \frac{5000}{3000}.$ $\Delta h=22.5\times \ln (1.667)=22.5\times 0.5108=11.5\text{km}.$ Why $3\ln$: the $v^3$ inside the log becomes a coefficient $3$ in front of $\ln v$ — the same reason the flux ratio was cubed in L3.1. Answer: the faster vehicle must jettison about 11.5 km higher.

What happens: inside the log, ${\rho }_0{v}^3/2\to 0$, so $\ln (\text{something}\to 0^{+})\to -\infty$, giving $h_{\text{jett}}\to -\infty$. Physical reading: a vehicle that is barely moving deposits almost no kinetic energy per molecule, so its heat flux is below the limit at any altitude — even the ground. A negative "required" altitude just means "the constraint is already satisfied; no altitude margin needed." The formula is correct but its output must be clamped: $h_{\text{jett}}=\max (0,\dots )$. Answer: $h_{\text{jett}}\to -\infty$; interpret as "safe everywhere," clamp to $h\ge 0$.

(a) Altitude. Required density: Why density first: the flux limit and the speed are both fixed, so — exactly as in L2.2 — the only knob that brings $\dot{q}$ to the limit is the density. Solve the heating law for $\rho$, then convert that density to an altitude. $\rho =\frac{2\times 1135}{(4200{)}^3}=\frac{2270}{7.409\times 10^{10}}=3.064\times 10^{-8}{\text{kg/m}}^3.$ $h=7.5\ln \frac{1.2}{3.064\times 10^{-8}}=7.5\ln (3.916\times 10^7)=7.5\times 17.483=131.1\text{km}.$ The log again undoes the exponential atmosphere so we can read off the altitude directly (in km, since $H$ is in km). (b) Δv gain. $I_{sp}{g}_0=455\times 9.81=4463.6\text{m/s}$. Why we compare two mass ratios: the rocket equation says Δv depends only on the ratio $m_0/m_f$. Carrying the fairing keeps both masses high; dropping it before the burn lowers both by $m_{fr}$, which raises the ratio and therefore the achievable Δv. The gain is the difference between the two. Carry-through: $\Delta v=4463.6\ln \frac{22000}{6500}=4463.6\times 1.2192=5442\text{m/s}.$ Drop early: $\Delta v^{\prime }=4463.6\ln \frac{20900}{5400}=4463.6\times 1.3527=6039\text{m/s}.$ Gain: $6039-5442\approx 597\text{m/s}$. Answer: jettison at $\approx 131\text{km}$; Δv gain $\approx 597\text{m/s}$.

Why flux scales $\propto \rho$ here: the heating law is $\dot{q}=\frac{1}{2}\rho v^3$. At a fixed speed, $v^3$ is a constant multiplier, so $\dot{q}$ is directly proportional to $\rho$ — halve the density, halve the flux. This lets us compare two altitudes by comparing their densities alone. Density at 121 km: the ratio to the 131.1 km case uses only the altitude drop (both altitudes in km, so the exponent is unitless): $\frac{\rho (121)}{\rho (131.1)}=e^{(131.1-121)/7.5}=e^{10.1/7.5}=e^{1.347}=\mathrm{3.85.}$ Since $\dot{q}\propto \rho$ at fixed $v$, the flux is 3.85× the limit, i.e. $\approx 4370{\text{W/m}}^2$. Reading: dropping just 10 km early nearly quadruples the heating — the exponential atmosphere makes the constraint sharp. The extra Δv is not worth cooking the payload. Answer: flux $\approx 3.85\times$ limit $\approx 4.4\times 10^3{\text{W/m}}^2$ — forbidden.

Set up the two competing effects as functions of jettison altitude $h$:

1. Δv benefit is monotonically decreasing in $h$. Δv saved comes from shedding $m_{fr}$: the earlier (lower $h$) you drop, the longer you fly light, so $\Delta v_{\text{saved}}(h)$ decreases as $h$ increases. There is no interior maximum — the benefit only wants $h$ smaller.
2. Heating constraint is a hard floor. $\dot{q}(h)=\frac{1}{2}{\rho }_0{e}^{-h/H}{v}^3$ decreases with $h$. The payload survives only where $\dot{q}(h)\le {\dot{q}}_{\text{limit}}$, i.e. $h\ge h_{\text{jett}}$.

Combine: we minimise "wasted Δv" (equivalently maximise Δv saved) subject to $h\ge h_{\text{jett}}$. The objective pushes $h$ down; the constraint blocks it at $h=h_{\text{jett}}$. A minimiser of a strictly monotone objective on a half-line $[h_{\text{jett}},\infty )$ sits at the endpoint. Therefore the optimum is exactly $h_{\text{opt}}=h_{\text{jett}}$, where $\dot{q}={\dot{q}}_{\text{limit}}$. Why the boundary: whenever a benefit strictly favours one direction and safety blocks it, the best you can do is press right up against the wall. Dropping earlier violates safety; dropping later needlessly throws away Δv. The equality $\dot{q}={\dot{q}}_{\text{limit}}$ is the optimum. This connects to Ascent Trajectory Optimization: the whole trajectory is shaped so that this boundary is reached as early in wall-clock time as the climb allows.

Required density is fixed (it depends only on $v$ and the flux limit): $\rho =3.064\times 10^{-8}{\text{kg/m}}^3$ from L4.1. Only the inversion $h=H\ln ({\rho }_0/\rho )$ shifts, and the pure-number log $\ln ({\rho }_0/\rho )=17.483$ is constant. $h_{\min }=7.0\times 17.483=122.4\text{km},h_{\max }=8.0\times 17.483=139.9\text{km}.$ Spread: about $122\text{km}$ to $140\text{km}$ — a $17.5\text{km}$ band, matching the "110–140 km" range quoted in the parent note. Conservative choice: a larger $H$ means the atmosphere stays denser to higher altitude, so you must wait higher to reach the same low density. Designing to $H=8.0$ km (jettison at $\approx 140$ km) guarantees safety even on a "puffy" atmosphere day; the payload is never over-heated, at the cost of a little extra Δv. Safety dominates → pick the high end. Answer: $h_{\text{jett}}\in [122,140]\text{km}$; design conservatively to $\approx 140$ km.

(a) Altitude. Required density: $\rho =\frac{2\times 1135}{(4600{)}^3}=\frac{2270}{9.7336\times 10^{10}}=2.332\times 10^{-8}{\text{kg/m}}^3.$ Invert the density law (result in km): $h=7.5\ln \frac{1.2}{2.332\times 10^{-8}}=7.5\ln (5.146\times 10^7)=7.5\times 17.756=133.2\text{km}.$ (b) Δv gain. $I_{sp}{g}_0=462\times 9.81=4532.2\text{m/s}$. Carry-through: $\Delta v=4532.2\ln \frac{30000}{8000}=4532.2\times 1.3218=5991\text{m/s}.$ Drop early: $\Delta v^{\prime }=4532.2\ln \frac{28500}{6500}=4532.2\times 1.4784=6701\text{m/s}.$ Gain: $6701-5991\approx 710\text{m/s}$. (c) Decision: jettison the fairing exactly at $\approx 133$ km, the instant $\dot{q}$ falls to 1135 W/m² — the earliest altitude that is thermally safe, which by L5.1 is also the Δv-optimal point. Answer: $h\approx 133\text{km}$; Δv gain $\approx 710\text{m/s}$; drop at the flux-limit boundary.

L1: plug into $q=\frac{1}{2}\rho v^2$ and $\dot{q}=qv$. L2: rearrange for altitude, apply rocket equation. L3: $\dot{q}\propto v^3$ makes speed dominate; handle the $v\to 0$ limit. L4: exponential density makes the constraint sharp. L5: the optimum lives exactly on the heating boundary and you design to the conservative (large-$H$) extreme.