Aerodynamic heating during reentry — stagnation point heat flux Chapman equation: Worked examples

Every symbol used below is already defined in the parent note. As a one-line reminder so you never have to scroll:

The scenario matrix

Every question this formula can throw is one of these case classes. The final column names the worked example that covers that cell.

#	Case class	What is being tested	Covered by
A	Baseline full calculation	Plug all three knobs, get a number	Example 1
B	Scale one knob up/down (ratios)	$V^{3}$ sensitivity — "double the speed"	Example 2
C	Scale nose radius (ratios)	$R_{n}^{- 1/2}$ sensitivity — bluntness	Example 3
D	Zero / degenerate input	$V = 0$ , $ρ = 0$ , $R_{n} \to \infty$ , $R_{n} \to 0$ limits	Example 4
E	Limiting behaviour along a trajectory	Falling $V$ × rising $ρ$ → a peak (derived)	Example 5
F	Inverse problem	Given $\overset{q}{˙}$ , solve for a knob	Example 6
G	Real-world word problem	Mars entry, unit juggling	Example 7
H	Exam-style twist	Convective vs radiative crossover	Example 8

We work them in order. Cells D and E get figures because their behaviour is geometric (curves, limits).

Example 1 — Cell A: the baseline plug-in

Step 1 — Cube the speed. $V^{3} = (7.8 \times 1 0^{3})^{3} = 4.746 \times 1 0^{11} .$ Why this step? $V$ enters as $V^{3}$ , so it must be cubed as a whole — cube first, keep the powers of ten tidy.

Step 2 — Take the two square roots. $R_{n} = 0.3 = 0.5477, ρ_{\infty} = 1 0^{- 3} = 0.03162.$ Why this step? Both appear under a root; doing them separately keeps the arithmetic honest (parent's advice).

Step 3 — Assemble. $\overset{q}{˙}_{s} = \frac{1.83 \times 1 0 ^{- 4}}{0.5477} \times 0.03162 \times 4.746 \times 1 0^{11} .$ $= 3.341 \times 1 0^{- 4} \times 0.03162 \times 4.746 \times 1 0^{11} \approx 5.01 \times 1 0^{6} W/m^{2} .$ Why this step? Multiply left-to-right; the $1 0^{11}$ carries the magnitude.

Step 4 — Convert. $5.01 \times 1 0^{6} W/m^{2} \div 1 0^{4} = 501 W/cm^{2} .$

Example 2 — Cell B: double the speed

Step 1 — Use the ratio, not a fresh calculation. $\frac{q ˙ _{2}}{q ˙ _{1}} = (\frac{V _{2}}{V _{1}})^{3} = 2^{3} = 8.$ Why this step? Everything else is unchanged, so it cancels — you never re-plug $R_{n}$ or $ρ$ . Ratios are the fast, error-proof way.

Step 2 — Apply. $501 W/cm^{2} \times 8 = 4008 W/cm^{2} .$

Example 3 — Cell C: quadruple the nose radius

Step 1 — Ratio again. $\frac{q ˙ _{2}}{q ˙ _{1}} = (\frac{R _{n, 1}}{R _{n, 2}})^{1/2} = (\frac{0.3}{1.2})^{1/2} = (\frac{1}{4})^{1/2} = \frac{1}{2} .$ Why this step? $R_{n}$ sits under a root in the denominator, so a factor-4 growth becomes a factor-2 shrink — the sign of the exponent tells the direction.

Step 2 — Apply. $501 \times \frac{1}{2} = 250.5 W/cm^{2} .$

Example 4 — Cell D: zero and degenerate inputs

How to read the figure below. The horizontal axis is speed $V$ (arbitrary units); the vertical axis is the stagnation flux $\overset{q}{˙}_{s}$ in the same arbitrary units, with $R_{n}$ and $ρ_{\infty}$ held fixed. The red curve is the cube $\overset{q}{˙}_{s} \propto V^{3}$ . The one black dot marks the origin, the $V = 0$ limit we test in Step 1 — notice the curve leaves it perfectly flat (its slope $3 V^{2}$ is zero there) before climbing steeply. The figure is the picture of Steps 1 and the "climbs as a cube" fact; the $ρ$ , $R_{n}$ limits are algebra we do beneath it.

Figure — Aerodynamic heating during reentry — stagnation point heat flux Chapman equation

Step 1 — $V = 0$ (vehicle at rest). $\overset{q}{˙}_{s} = \frac{C}{R _{n}} ρ_{\infty} \cdot 0^{3} = 0.$ Why this step? No motion → no ordered kinetic energy $\frac{1}{2} V^{2}$ to dump → no convective heating. On the figure this is the red curve touching down at the black dot at the origin. ✔

Step 2 — $ρ_{\infty} = 0$ (vacuum / above the atmosphere). $\overset{q}{˙}_{s} = \frac{C}{R _{n}} \cdot 0 \cdot V^{3} = 0.$ Why this step? No gas means no molecules to shock and no medium to carry heat to the wall — convective heating literally needs a fluid. This is why a satellite in orbit doesn't burn despite $V \approx 7.8 km/s$ : up there $ρ_{\infty}$ is essentially $0$ . ✔

Step 3 — $R_{n} \to \infty$ (perfectly flat nose). Treat $\overset{q}{˙}_{s}$ as a function of $R_{n}$ only (fix $V, ρ$ ) and take the formal limit: $lim_{R_{n} \to \infty} \frac{C ρ _{\infty} V ^{3}}{R _{n}} = C ρ_{\infty} V^{3} \cdot lim_{R_{n} \to \infty} \frac{1}{R _{n}} = C ρ_{\infty} V^{3} \cdot 0 = 0.$ Why this step? $1/ R_{n} \to 0$ . Physically: an infinite radius means the velocity gradient $β \propto 1/ R_{n} \to 0$ , the boundary layer grows without bound, and heat can't diffuse across it. Blunter is always cooler — taken to the extreme, flat is coolest. ✔

Step 4 — $R_{n} \to 0$ (perfectly sharp nose). Same one-variable function, opposite end: $lim_{R_{n} \to 0^{+}} \frac{C ρ _{\infty} V ^{3}}{R _{n}} = C ρ_{\infty} V^{3} \cdot lim_{R_{n} \to 0^{+}} \frac{1}{R _{n}} = C ρ_{\infty} V^{3} \cdot (+ \infty) = + \infty.$ Why this step? As the denominator $R_{n}$ shrinks toward zero the fraction grows without bound. The formula predicts an idealised sharp nose melts instantly — the maths behind "never fly a sharp reentry nose." ✔

Example 5 — Cell E: the peak along a real descent (derived)

Before we touch calculus, we must fix every symbol and its units, because this example juggles four separate quantities that all look alike. Here is the complete bookkeeping.

Definition The five symbols in this example (with units)

$h$ — altitude above the ground, in metres ( $m$ ). This is the "where along the fall" the question asks about; large $h$ = high up, $h = 0$ = ground.
$ρ$ — local air density at that altitude, in $kg/m^{3}$ . Small high up, large low down.
$H$ — the atmospheric scale height, in metres ( $m$ ): the vertical distance over which density falls by a factor $e \approx 2.718$ . It is a fixed property of the atmosphere (for Earth $H \approx 7000 m$ ). It links $h$ and $ρ$ through the model $ρ (h) = ρ_{0} e^{- h / H}$ .
$γ$ — the flight-path angle: how steeply below horizontal the trajectory dives, an angle (dimensionless, measured in radians). Assumed constant for a straight ballistic plunge.
$β$ — the ballistic coefficient, in $kg/m^{2}$ (mass per frontal area per drag coefficient). Big $β$ = heavy, streamlined, hard to brake. See Ballistic coefficient and deceleration.

Note there is no separate " $H_{eff}$ " — that phantom is gone; there is only the one scale height $H$ .

Step 0 — Derive the speed–density relation (Allen–Eggers), and its domain. For a steep ballistic entry we assume: (i) gravity is negligible next to drag, (ii) the flight-path angle $γ$ stays constant (a straight-line plunge), (iii) lift is zero. Newton's second law along the path is then just drag, $m \frac{d V}{d t} = - \frac{1}{2} ρ V^{2} C_{D} A, β \equiv \frac{m}{C _{D} A} (units kg/m^{2}) .$ Why introduce $β$ ? It bundles the vehicle's mass $m$ , drag coefficient $C_{D}$ and frontal area $A$ into one number, so the equation depends on the vehicle only through $β$ .

Change the independent variable from time $t$ to altitude $h$ using the geometry of a straight dive, $d h = - V sin γ d t$ (falling, so $h$ decreases as $t$ grows). Substituting removes $t$ : $\frac{d V}{V} = \frac{ρ d h}{2 β s i n γ} .$ Now use the atmosphere model $ρ (h) = ρ_{0} e^{- h / H}$ , whose differential is $d ρ = - \frac{ρ}{H} d h$ , i.e. $ρ d h = - H d ρ$ . This is exactly where the scale height $H$ enters — it is the conversion factor between an altitude step $d h$ and a density step $d ρ$ . Substituting: $\frac{d V}{V} = - \frac{H d ρ}{2 β s i n γ} .$ Integrate from entry (where $ρ \approx 0$ high up, $V = V_{e}$ ) down to a general $ρ$ : $ln \frac{V}{V _{e}} = - \frac{H ρ}{2 β s i n γ} \Rightarrow V (ρ) = V_{e} exp (- \frac{H ρ}{2 β sin γ}) .$ Domain of validity: steep, high-speed, drag-dominated, constant- $γ$ descent — the classic Allen–Eggers regime. It breaks down for shallow lifting entries (Shuttle-type) where $γ$ varies and gravity matters. Note $H$ now sits inside the exponent, so it will carry through to the answer.

Step 1 — Write the flux as a function of the single variable $ρ$ . Insert $V (ρ)$ into $\overset{q}{˙} = A_{0} ρ^{1/2} V^{3}$ (writing $A_{0} = C / R_{n}$ , a constant here): $\overset{q}{˙} (ρ) = A_{0} V_{e}^{3} ρ^{1/2} exp (- \frac{3 H ρ}{2 β s i n γ}) .$ Why this step? To maximise we need $\overset{q}{˙}$ as a function of one variable; $ρ$ is that variable and $V$ rides on it. Define the shorthand $k \equiv \frac{3 H}{2 β sin γ}$ (units $m^{3} / kg$ , so $k ρ$ is dimensionless — a good units check).

Step 2 — Differentiate and set to zero. With $\overset{q}{˙} = A_{0} V_{e}^{3} ρ^{1/2} e^{- k ρ}$ , $\frac{d q ˙}{d ρ} = A_{0} V_{e}^{3} ρ^{- 1/2} e^{- k ρ} (\frac{1}{2} - k ρ) .$ Why this step? Product rule on $ρ^{1/2}$ and $e^{- k ρ}$ . The exponential and $ρ^{- 1/2}$ never vanish for $ρ > 0$ , so the peak sits where the bracket is zero.

Step 3 — Solve for $ρ_{peak}$ , keeping $H$ explicit. $\frac{1}{2} - k ρ_{peak} = 0 \Rightarrow ρ_{peak} = \frac{1}{2 k} = \frac{1}{2} \cdot \frac{2 β s i n γ}{3 H} = \frac{β sin γ}{3 H} .$ Why this step? The bracket is linear in $ρ$ , so it has a single root — one interior peak, exactly as forecast. The scale height $H$ appears in the denominator because it entered through $k$ ; there is no separate " $H_{eff}$ " — it was always this one $H$ . The 3 is the ghost of the exponent $3$ on $V$ . Units check: $[β] / [H] = (kg/m^{2}) / m = kg/m^{3}$ ✔ — a density, as it must be.

Step 4 — Convert the peak density back to an altitude $h_{peak}$ (the actual question). The question asked where along the fall, i.e. an altitude, not a density. Invert the atmosphere model $ρ = ρ_{0} e^{- h / H}$ : $e^{- h / H} = \frac{ρ}{ρ _{0}} \Rightarrow h = H ln \frac{ρ _{0}}{ρ} .$ Plug $ρ = ρ_{peak} = β sin γ / (3 H)$ : $h_{peak} = H ln (\frac{3 H ρ _{0}}{β sin γ}) .$ Why this step? This is the piece the earlier draft skipped. It turns the abstract $ρ_{peak}$ into a real altitude the engineer can point to on the trajectory. Because $β sin γ / (3 H)$ is a small density, the logarithm is large and positive, so $h_{peak}$ is high up — confirming the forecast.

Step 5 — Sanity number. Take Earth-ish values $H = 7000 m$ , $ρ_{0} = 1.2 kg/m^{3}$ , $β = 100 kg/m^{2}$ , $γ = 3 0^{\circ}$ so $sin γ = 0.5$ . Then $ρ_{peak} = \frac{100 \times 0.5}{3 \times 7000} = 2.38 \times 1 0^{- 3} kg/m^{3}$ and $h_{peak} = 7000 ln (\frac{1.2}{2.38 \times 1 0 ^{- 3}}) = 7000 \times ln (504) = 7000 \times 6.22 \approx 4.35 \times 1 0^{4} m \approx 44 km .$ Why this step? A concrete altitude (~44 km, high in the stratosphere) makes "intermediate, high up" tangible — well above the ~10 km where a ballistic capsule feels peak g-load.

How to read the figure below. Horizontal axis: descent depth (0 = top of atmosphere, 1 = deep). Vertical axis: each quantity normalised to its own maximum so all three fit. Black dashed = the rising factor $ρ^{1/2}$ ; black dotted = the falling factor $V^{3}$ ; the red curve is their product $\overset{q}{˙}_{s}$ , and the red dot marks the interior maximum we just solved for algebraically.

Example 6 — Cell F: inverse problem (solve for a knob)

Step 1 — Rearrange the formula for $R_{n}$ . Starting from $\overset{q}{˙}_{s} = \frac{C ρ _{\infty} V ^{3}}{R _{n}}$ , isolate the root: $R_{n} = \frac{C ρ _{\infty} V ^{3}}{q ˙ _{s}} \Rightarrow R_{n} = (\frac{C ρ _{\infty} V ^{3}}{q ˙ _{s}})^{2} .$ Why this step? The unknown lives under a root in the denominator; solve algebraically so we can plug once.

Step 2 — Compute the numerator group. $C ρ_{\infty} V^{3} = 1.83 \times 1 0^{- 4} \times 0.03162 \times 4.746 \times 1 0^{11} = 2.746 \times 1 0^{6} .$ Why this step? This is the "flux the nose would feel if $R_{n} = 1$ " — a handy intermediate.

Step 3 — Divide and square. $R_{n} = \frac{2.746 \times 1 0 ^{6}}{3.0 \times 1 0 ^{6}} = 0.9153, R_{n} = 0.915 3^{2} \approx 0.838 m .$

Example 7 — Cell G: real-world word problem (Mars)

Step 1 — Convert speed to SI. $V = 6.0 \times 1 0^{3} m/s$ , so $V^{3} = 2.16 \times 1 0^{11}$ . Why this step? The formula is pure SI; a leftover "km" would inflate the answer by $1 0^{9}$ . This is the parent's top warning.

Step 2 — Roots. $R_{n} = 1.5 = 1.2247$ , $ρ_{\infty} = 2 \times 1 0^{- 4} = 0.01414$ .

Step 3 — Assemble. $\overset{q}{˙}_{s} = \frac{1.83 \times 1 0 ^{- 4}}{1.2247} \times 0.01414 \times 2.16 \times 1 0^{11} = 4.56 \times 1 0^{5} W/m^{2} = 45.6 W/cm^{2} .$

Example 8 — Cell H: exam twist (convective vs radiative crossover)

Worked example When does radiative heating overtake convective?

Convective heating scales as $\overset{q}{˙}_{c} = C R_{n}^{- 1/2} ρ^{1/2} V^{3}$ (grows as $R_{n}^{- 1/2}$ ). Radiative heating (the glow of the shocked gas — see the parent's footnote and Radiative heating at hypersonic speeds) follows an empirical power law $\overset{q}{˙}_{r} = K R_{n} ρ^{a} V^{b}$ , where:

$K$ is an empirical constant (fitted to gas-radiation data, units chosen so $\overset{q}{˙}_{r}$ comes out in $W/m^{2}$ ),
$a \approx 1.2$ is the density exponent, and
$b \approx 8$ (can be $8$ – $15$ ) is the very steep velocity exponent that makes radiation matter only at high speed.

The single feature we need is the sign of the $R_{n}$ exponent: convective carries $R_{n}^{- 1/2}$ (falls with radius), radiative carries $R_{n}^{+ 1}$ (rises with radius). For fixed $ρ, V$ , at what nose radius $R_{*}$ are they equal, and which dominates for very blunt noses?

Forecast: Convective falls with $R_{n}$ , radiative rises with $R_{n}$ — opposite slopes, so they must cross at one $R_{*}$ . For very blunt noses (large $R_{n}$ ) radiative wins. Guess: a crossover radius exists; beyond it, radiation dominates.

Step 1 — Set them equal (symbolically). Group all the fixed $ρ, V$ pieces into constants $a_{c} = C ρ^{1/2} V^{3}$ and $a_{r} = K ρ^{a} V^{b}$ (so $a, b, K$ are now buried inside the single number $a_{r}$ ). Then $a_{c} R_{*}^{- 1/2} = a_{r} R_{*}^{+ 1} \Rightarrow \frac{a _{c}}{a _{r}} = R_{*}^{3/2} \Rightarrow R_{*} = (\frac{a _{c}}{a _{r}})^{2/3} .$ Why this step? Collecting $R_{n}$ powers ( $- 1/2$ minus $+ 1$ gives exponent $3/2$ ) turns two curves into one clean crossover equation. The messy exponents $a, b$ never resurface — only $a_{c} / a_{r}$ matters.

Step 2 — Numeric example. Suppose at some altitude the grouped constants come out to $a_{c} = 2.0 \times 1 0^{6}$ and $a_{r} = 1.0 \times 1 0^{5}$ (SI, so that $\overset{q}{˙}$ is in $W/m^{2}$ with $R_{n}$ in metres). Then $R_{*} = (\frac{2.0 \times 1 0 ^{6}}{1.0 \times 1 0 ^{5}})^{2/3} = 2 0^{2/3} \approx 7.37 m .$ Why this step? A concrete crossover radius makes the trade-off tangible.

Step 3 — Interpret. For $R_{n} < 7.37 m$ convective dominates (making it blunter helps). For $R_{n} > 7.37 m$ radiative dominates (making it blunter now hurts). There is an optimum, not "blunter is always better."

Recall

Recall Test yourself on every cell

Doubling reentry speed multiplies $\overset{q}{˙}_{s}$ by what factor? ::: $2^{3} = 8$ . Quadrupling the nose radius does what to $\overset{q}{˙}_{s}$ ? ::: Multiplies it by $4^{- 1/2} = \frac{1}{2}$ (halves it). Why doesn't an orbiting satellite ( $V \approx 7.8$ km/s) burn up? ::: $ρ_{\infty} \approx 0$ up there, so $\overset{q}{˙}_{s} \propto ρ_{\infty} \to 0$ — no gas to shock or carry heat. What does the formula predict for a perfectly sharp nose ( $R_{n} \to 0$ )? ::: $\overset{q}{˙}_{s} \to + \infty$ — it diverges, so sharp reentry noses are forbidden. Where is the peak of $ρ^{1/2} V^{3}$ located as a density, and how do you find it? ::: Differentiate $ρ^{1/2} e^{- k ρ}$ w.r.t. $ρ$ , set to zero, get $ρ_{peak} = 1/ (2 k) = β sin γ /3 H$ . How do you turn that peak density into an altitude? ::: Invert $ρ = ρ_{0} e^{- h / H}$ to get $h_{peak} = H ln (3 H ρ_{0} / β sin γ)$ — a high altitude. Why does peak heating occur higher than peak deceleration? ::: $V^{3}$ (falling) collapses faster than $ρ^{1/2}$ (rising), so the product peaks early, high up. For very blunt noses, which heating eventually dominates? ::: Radiative — it grows as $R_{n}^{+ 1}$ while convective falls as $R_{n}^{- 1/2}$ ; they cross at $R_{*} = (a_{c} / a_{r})^{2/3}$ .