Fairing separation — altitude, dynamic pressure requirements: Worked examples

See the parent for theory: 3.4.18 parent. Building blocks: Dynamic Pressure (Max-Q), Atmospheric Density Model — Scale Height, Rocket Equation & Mass Ratios, Aerodynamic Heating & Free-Molecular Flow, Ascent Trajectory Optimization.

Before we start — symbols, units, and one definition

The scenario matrix

Every problem this topic can pose falls into one of these cells. The examples below are labelled by cell so you can see the whole map is covered.

Cell	What varies / the trap	Example
A. Forward	Given $h, v$ → find $q$ and $\overset{q}{˙}$ , decide jettison	Ex 1
B. Inverse (altitude)	Given flux limit → solve for $h_{jett}$	Ex 2
C. Inverse (velocity)	Given $h$ and limit → what $v$ trips it	Ex 3
D. Zero / degenerate	$v = 0$ , or $h \to \infty$ ( $ρ \to 0$ ) — no heating	Ex 4
E. Limiting / scaling	How does $h_{jett}$ shift if $v$ doubles? (log behaviour)	Ex 5
F. Trajectory twist	Steeper/faster ascent → higher jettison altitude	Ex 6
G. $Δ v$ trade	Mass penalty of carrying vs dropping the fairing	Ex 7
H. Word / real-world	Mission engineer's "is it safe yet?" call	Ex 8

Setting the stage — one picture for all cases

Figure s01 plots, on a single vertical logarithmic scale, how two quantities fall as altitude rises (for a fixed $v = 4000 m/s$ ):

the cyan curve is dynamic pressure $q = \frac{1}{2} ρ v^{2}$ (in Pa),
the amber curve is heat flux $\overset{q}{˙} = \frac{1}{2} ρ v^{3}$ (in $W/m^{2}$ ).

The horizontal axis is altitude $h$ in km (60→160). Notice both curves are straight lines on the log scale — that is the exponential density law $ρ = ρ_{0} e^{- h / H}$ showing itself (a log of an exponential is a straight line). The amber curve sits a factor $v = 4000$ above the cyan one, exactly the relation $\overset{q}{˙} = q v$ . The dashed white horizontal line marks the safe limit $1135 W/m^{2}$ ; the dotted white vertical line where the amber curve crosses it is $h_{jett} \approx 130 km$ — read that intersection and you have read the answer to Example 2.

Figure — Fairing separation — altitude, dynamic pressure requirements

Cell A — Forward evaluation

Worked example Example 1 — Is it safe to jettison at 115 km?

A rocket at $h = 115 km$ flies at $v = 4000 m/s$ . Find $q$ and $\overset{q}{˙}$ , then decide.

Forecast: guess — will $\overset{q}{˙}$ be above or below $1135 W/m^{2}$ ? (Hint: $v^{3}$ is enormous.)

Step 1 — density from altitude. $ρ = ρ_{0} e^{- h / H} = 1.2 e^{- 115/7.5} = 1.2 e^{- 15.33} \approx 2.6 \times 1 0^{- 7} kg/m^{3} .$ Why this step? Altitude enters the physics only through density; nothing else on this page depends on $h$ directly. (Here $h / H = 115/7.5$ — km cancels km, as promised.)

Step 2 — dynamic pressure. $q = \frac{1}{2} ρ v^{2} = 0.5 \times 2.6 \times 1 0^{- 7} \times 400 0^{2} \approx 2.1 Pa .$ Why this step? $q$ is the aerodynamic "punch"; it is the coefficient in the drag force $F = C_{D} q A$ (with $C_{D}$ and $A$ defined above), so it is the natural bridge to heat flux.

Step 3 — heat flux. $\overset{q}{˙} = q v = 2.1 \times 4000 \approx 8.4 \times 1 0^{3} W/m^{2} .$ Why this step? Heat is energy/time = force-like flux × velocity, so we multiply $q$ by one more $v$ (the free-molecular $v^{3}$ signature).

Decision: $8.4 \times 1 0^{3} > 1135$ → too early. Wait higher.

Verify: Units of $\overset{q}{˙}$ : $[Pa] \cdot [m/s] = \frac{N}{m ^{2}} \cdot \frac{m}{s} = \frac{N\cdotm}{m ^{2} s} = \frac{J}{m ^{2} s} = W/m^{2}$ . ✓ Correct units for a flux.

Cell B — Inverse for altitude

Worked example Example 2 — At what altitude does it become safe?

Same $v = 4000 m/s$ . Find $h_{jett}$ where $\overset{q}{˙} = 1135 W/m^{2}$ .

Forecast: Ex 1 said 115 km is too hot. Guess: higher or lower than 130 km?

Step 1 — required density. From $\overset{q}{˙} = \frac{1}{2} ρ v^{3}$ : $ρ = \frac{2 q ˙}{v ^{3}} = \frac{2 \times 1135}{400 0 ^{3}} = \frac{2270}{6.4 \times 1 0 ^{10}} \approx 3.5 \times 1 0^{- 8} kg/m^{3} .$ Why this step? We want the density at which heating exactly hits the limit; invert the flux formula for $ρ$ .

Step 2 — invert the density law. $h = H ln \frac{ρ _{0}}{ρ} = 7.5 ln \frac{1.2}{3.5 \times 1 0 ^{- 8}} = 7.5 ln (3.4 \times 1 0^{7}) \approx 7.5 \times 17.35 \approx 130 km .$ Why this step? The log turns a gigantic density ratio ( $3.4 \times 1 0^{7}$ ) into a friendly number ( $\approx 17$ ); that's the whole reason we use logarithms here.

Answer: jettison just above 130 km (strict rule: at 130 km we are still at the limit, so climb slightly past it). This lies inside the parent note's quoted band of 110–140 km for real launch programs, so the number is physically reasonable.

Verify: Plug 130 km back: $ρ = 1.2 e^{- 130/7.5} = 1.2 e^{- 17.33} \approx 3.6 \times 1 0^{- 8}$ , then $\overset{q}{˙} = \frac{1}{2} ρ v^{3} \approx 1150 W/m^{2} \approx 1135$ . ✓ (Small rounding.)

Cell C — Inverse for velocity

Worked example Example 3 — Fixed altitude, what speed trips the limit?

At $h = 120 km$ , what velocity makes $\overset{q}{˙}$ exactly $1135 W/m^{2}$ ?

Forecast: density is thin at 120 km, so must the rocket be slow or fast to still reach the limit?

Step 1 — density at 120 km. $ρ = 1.2 e^{- 120/7.5} = 1.2 e^{- 16} \approx 1.35 \times 1 0^{- 7} kg/m^{3} .$ Why this step? Fix the altitude → fix the density; only $v$ is unknown now.

Step 2 — solve flux for $v$ . $\overset{q}{˙} = \frac{1}{2} ρ v^{3} \Rightarrow v = (\frac{2 q ˙}{ρ})^{1/3} = (\frac{2 \times 1135}{1.35 \times 1 0 ^{- 7}})^{1/3} .$ Why this step? Isolate $v$ ; because $\overset{q}{˙} \propto v^{3}$ (the free-molecular signature defined above), we take a cube root, not a square root.

Step 3 — evaluate. $\frac{2270}{1.35 \times 1 0 ^{- 7}} \approx 1.68 \times 1 0^{10}$ , and $(1.68 \times 1 0^{10})^{1/3} \approx 2560 m/s .$

Answer: with the strict rule, $v \geq 2560 m/s$ at 120 km is still unsafe; only $v < 2560 m/s$ is safe.

Verify: back-substitute $v = 2560$ : $\frac{1}{2} (1.35 \times 1 0^{- 7}) (2560)^{3} \approx 1132 W/m^{2} \approx 1135$ . ✓

Cell D — Zero and degenerate inputs

Worked example Example 4 — The edge cases:

$v = 0$ and $h \to \infty$ (a) Rocket momentarily at rest relative to air. (b) Rocket in deep space, $ρ \to 0$ .

Forecast: which of these can still cook the payload? (Neither — see why.)

Step 1 — $v = 0$ . $\overset{q}{˙} = \frac{1}{2} ρ \cdot 0^{3} = 0.$ Why this step? No motion → no molecules slam the surface → zero convective heating, regardless of how thick the air is.

Step 2 — $h \to \infty$ . $ρ = ρ_{0} e^{- h / H} \to 0$ , so $\overset{q}{˙} = \frac{1}{2} ρ v^{3} \to 0$ for any finite $v$ . Why this step? This is the whole point of jettison: get high enough that density kills the heating even at huge speed.

Step 3 — the dangerous corner. Heating is only large when both $ρ$ and $v$ are appreciable. The product $ρ v^{3}$ is the true master variable; either factor going to zero makes $\overset{q}{˙} \to 0$ . Why this step? Prevents the classic error of watching only altitude or only speed.

Verify: $lim_{v \to 0} \frac{1}{2} ρ v^{3} = 0$ and $lim_{h \to \infty} \frac{1}{2} ρ_{0} e^{- h / H} v^{3} = 0$ . ✓ Both limits vanish.

Cell E — Limiting / scaling behaviour

Worked example Example 5 — If the rocket is twice as fast, how much higher must it climb?

Threshold and $ρ_{0}, H$ fixed. Compare $h_{jett}$ for $v$ and for $2 v$ .

Forecast: twice the speed → double the altitude? Guess, then check the log.

Step 1 — write the jettison altitude. $h_{jett} = H ln \frac{\frac{1}{2} ρ _{0} v ^{3}}{q ˙ _{limit}} .$ Why this step? This is the closed form from the parent; only the $v^{3}$ inside the log changes.

Step 2 — take the difference for $2 v$ vs $v$ . $Δ h = H [ln (2 v)^{3} - ln v^{3}] = H ln \frac{( 2 v ) ^{3}}{v ^{3}} = H ln 8 = 3 H ln 2.$ Why this step? Subtracting logs isolates the change; the density and threshold constants cancel.

Step 3 — number. $Δ h = 3 \times 7.5 \times ln 2 = 22.5 \times 0.693 \approx 15.6 km .$ Why this step? Turns the scaling law into a concrete "climb ~16 km more."

Answer: doubling speed raises jettison altitude by only ~15.6 km, NOT by doubling — because altitude enters through the logarithm. A huge speed change costs a modest height change.

Verify: $3 H ln 2 = 3 (7.5) (0.6931) \approx 15.6 km$ . ✓

Cell F — Trajectory twist

Figure s02 plots the jettison altitude $h_{jett}$ on the vertical axis in kilometres against the velocity $v$ at jettison on the horizontal axis in metres per second (range 2500→6000), using the closed form $h_{jett} = H ln \frac{\frac{1}{2} ρ _{0} v ^{3}}{q ˙ _{limit}}$ . The cyan curve rises slowly (it is a logarithm), and the two marked dots are the two trajectories of this example: the slower amber dot (A, at $v = 3500 m/s$ ) sits lower, the faster white dot (B, at $v = 5000 m/s$ ) sits higher. Look at how little the height (km) climbs for a big jump in speed (m/s) — that gentle slope is the whole lesson of the example. Follow the calculation, then check your numbers against those two dots.

Worked example Example 6 — Steep-and-fast vs shallow-and-slow

Trajectory A reaches jettison conditions at $v_{A} = 3500 m/s$ , trajectory B at $v_{B} = 5000 m/s$ (steeper, faster). Which jettisons higher, and by how much?

Forecast: the faster trajectory — higher or lower jettison altitude? (Recall $v^{3}$ .)

Step 1 — altitude for each. $h = H ln \frac{\frac{1}{2} ρ _{0} v ^{3}}{q ˙ _{limit}}$ with $\frac{1}{2} ρ_{0} = 0.6$ . Why this step? Same formula, two velocities; the difference is what we want. This is the curve drawn in s02.

Step 2 — trajectory A. $\frac{0.6 \times 350 0 ^{3}}{1135} = \frac{0.6 \times 4.29 \times 1 0 ^{10}}{1135} \approx 2.27 \times 1 0^{7}$ , so $h_{A} = 7.5 ln (2.27 \times 1 0^{7}) \approx 7.5 \times 16.94 \approx 127 km .$ (This is the amber dot in s02.)

Step 3 — trajectory B. $\frac{0.6 \times 500 0 ^{3}}{1135} = \frac{0.6 \times 1.25 \times 1 0 ^{11}}{1135} \approx 6.61 \times 1 0^{7}$ , so $h_{B} = 7.5 ln (6.61 \times 1 0^{7}) \approx 7.5 \times 18.01 \approx 135 km .$ (The white dot in s02.)

Answer: the faster trajectory B jettisons higher (135 vs 127 km) — a difference of $Δ h = 3 H ln (5000/3500) = 3 (7.5) ln (1.4286) \approx 8 km$ . This is exactly the parent's warning: "steeper, faster → higher jettison altitude."

Verify: $h_{B} - h_{A} = 3 H ln (v_{B} / v_{A}) = 22.5 ln (1.4286) \approx 8.0 km$ , matching $135 - 127$ . ✓

Cell G — The $Δ v$ trade

Worked example Example 7 — How much

$Δ v$ does dropping the fairing save? Upper stage: $I_{s p} = 450 s$ , $m_{0} = 20000 kg$ , $m_{f} = 6000 kg$ , fairing $m_{f r} = 1000 kg$ . See Rocket Equation & Mass Ratios for the underlying equation.

Forecast: guess the saving — tens, hundreds, or thousands of m/s?

Step 1 — carry-through $Δ v$ . $Δ v = I_{s p} g_{0} ln \frac{m _{0}}{m _{f}} = 450 \times 9.81 \times ln \frac{20000}{6000} = 4415 \times 1.204 \approx 5316 m/s .$ Why this step? The rocket equation converts a mass ratio into velocity budget. Here we keep the fairing through the whole burn, so it inflates both $m_{0}$ and $m_{f}$ .

Step 2 — drop-early $Δ v$ . Both masses drop by $m_{f r}$ : $Δ v^{'} = 4415 ln \frac{19000}{5000} = 4415 \times 1.335 \approx 5893 m/s .$ Why this step? Removing $1000 kg$ from both ends raises the ratio $19000/5000 = 3.8$ above $20000/6000 = 3.33$ ; the higher ratio buys more $Δ v$ .

Step 3 — the gain. $Δ v_{saved} = 5893 - 5316 \approx 577 m/s .$ Why this step? This is the payoff that pushes engineers to jettison as early as the heat limit allows.

Answer: ~577 m/s freed — enough to change an orbit insertion.

Verify: $4415 (ln 3.8 - ln 3.333) = 4415 \times 0.1307 \approx 577 m/s$ . ✓

Cell H — Real-world call

Worked example Example 8 — Flight director's "go/no-go"

Telemetry: $h = 112 km$ , $v = 4200 m/s$ . Payload limit $\overset{q}{˙}_{limit} = 1135 W/m^{2}$ . Is it "GO" for fairing sep right now?

Forecast: 112 km sounds "high enough" per the manual — but check the flux, not the altitude.

Step 1 — density. $ρ = 1.2 e^{- 112/7.5} = 1.2 e^{- 14.93} \approx 3.9 \times 1 0^{- 7} kg/m^{3} .$ Why this step? Convert the reported altitude into the density that actually drives heating.

Step 2 — heat flux. $\overset{q}{˙} = \frac{1}{2} ρ v^{3} = 0.5 \times 3.9 \times 1 0^{- 7} \times 420 0^{3} \approx 0.5 \times 3.9 \times 1 0^{- 7} \times 7.41 \times 1 0^{10} \approx 1.45 \times 1 0^{4} W/m^{2} .$ Why this step? The single number that decides safety.

Step 3 — compare. $1.45 \times 1 0^{4} ≫ 1135$ → NO-GO (strict rule: even equality would be no-go, and we are far above). The "112 km rule" is only a proxy; the real trajectory is fast enough that $v^{3}$ keeps heating high. Why this step? Demonstrates the parent's mistake #3: altitude alone does not decide it. See also Ascent Trajectory Optimization.

Answer: Do not jettison. Wait until the flux drops — for this speed that's near 133 km (from Ex 2's method with $v = 4200$ ).

Verify: $\frac{1}{2} (3.9 \times 1 0^{- 7}) (4200)^{3} \approx 1.45 \times 1 0^{4} W/m^{2} > 1135$ . ✓ NO-GO confirmed.

Recall Quick self-test across the matrix

Which cell? "Given $h$ and the limit, find the speed that trips heating." ::: Cell C (inverse for velocity, cube root). Doubling $v$ raises $h_{jett}$ by how much? ::: $3 H ln 2 \approx 15.6 km$ — only a log-sized change. Why can a 112 km "GO by altitude" still be NO-GO? ::: Because $\overset{q}{˙} \propto ρ v^{3}$ ; a fast trajectory keeps flux above limit even there. Dropping a 1000 kg fairing saved how much $Δ v$ in Ex 7? ::: About 577 m/s. Is $\overset{q}{˙} = \overset{q}{˙}_{limit}$ safe? ::: No — strict rule: equality is the last unsafe instant; jettison only when strictly below.