Exercises — Reentry mechanics — ballistic coefficient β = m - (C_D A)

WHAT: substitute into $\beta =m/(C_{D}A)$. WHY: $\beta$ is mass divided by the "aerodynamic footprint" $C_{D}A$ — nothing else to compute. $\beta =\frac{40}{0.2\times 0.01}=\frac{40}{0.002}=20000{\text{ kg/m}}^2.$ A tiny, heavy dart → huge $\beta$ → a "cannonball" that will punch deep.

WHAT: compare the two numbers. WHY: small $\beta$ = light-for-its-size = "beach ball" that the air stops early and high; big $\beta$ = "bowling ball" that plunges deep. Body P (small $\beta =45$) is the feather — stopped higher, gentler heating. Body Q is the cannonball.

WHAT: track units. $C_D$ is dimensionless, so it contributes nothing. $A$ is in ${\text{m}}^2$, $m$ in $\text{kg}$. $[\beta ]=\frac{\text{kg}}{(\text{dimensionless})\cdot {\text{m}}^2}=\frac{\text{kg}}{{\text{m}}^2}.✓$

WHAT: use $v(h)=v_e\exp \left(-\frac{\rho H}{2\beta \sin \gamma }\right)$. WHY the exponential? Because density grows exponentially as you descend, and the drag integral of an exponential is itself an exponential — that is exactly what Allen–Eggers integrated in the parent note. $\sin 90^{\circ }=1$. Exponent: $-\frac{\rho H}{2\beta \sin \gamma }=-\frac{1.0\times 7000}{2\times 6000\times 1}=-\frac{7000}{12000}=-\mathrm{0.5833.}$ $v=7000e^{-0.5833}=7000\times 0.558\approx 3906\text{ m/s}.$ It has lost about 44% of its speed — a middling $\beta$, half feather half cannonball.

WHAT: use $a_{\max }=\frac{v_e^2\sin \gamma }{2eH}$. WHY no $\beta$? The parent note showed the $\beta$ cancels when you maximise over density — the magnitude of peak-g depends only on speed and angle. $a_{\max }=\frac{(6000{)}^2\times \sin 40^{\circ }}{2\times 2.71828\times 7000}=\frac{3.6\times 10^7\times 0.6428}{38056}\approx 608{\text{ m/s}}^2.$ In g's: $608/9.81\approx 62g$.

WHAT: use the peak condition ${\rho }^{\ast }=\frac{\beta \sin \gamma }{H}$ from the parent derivation. WHY: setting $da/d\rho =0$ gives this single special density where the trade-off between "more air" and "less speed left" balances. ${\rho }^{\ast }=\frac{6000\times 1}{7000}=0.857{\text{ kg/m}}^3.$ That is roughly $70\%$ of sea-level density — this heavy body peaks deep and low. (Units: $\beta$ is ${\text{kg/m}}^2$ and $H$ is $\text{m}$, so $\beta /H$ has units ${\text{kg/m}}^3$ — a density, as it must.)

WHAT: compute both speeds' ratio $v_B/v_A$. WHY the setup: the exponent of $v(h)$ is $-\rho H/(2\beta \sin \gamma )=-N/\beta$, where $N=\rho H/(2\sin \gamma )$ collects everything except $\beta$. Since both bodies see the same $\rho$, $H$ and $\gamma$ at a given altitude, they share the same $N$; only $\beta$ differs. We are told to evaluate at the altitude where $N=3500{\text{ kg/m}}^2$. $\frac{v_A}{v_e}=e^{-N/{\beta }_A}=e^{-3500/500}=e^{-7}=9.12\times 10^{-4},$ $\frac{v_B}{v_e}=e^{-N/{\beta }_B}=e^{-3500/5000}=e^{-0.7}=\mathrm{0.497.}$ Ratio $v_B/v_A=e^{-0.7}/e^{-7}=e^{6.3}\approx 545$. The heavier body B is about 545× faster at that altitude — dramatic, because $\beta$ sits in an exponent.

WHAT: subtract the two peak altitudes. WHY: $h^{\ast }$ depends on $\beta$ only through $\ln (1/\beta )$, so a factor-of-10 change is a fixed additive shift. $h_A^{\ast }-h_B^{\ast }=H\ln \frac{{\rho }_0}{{\rho }_A^{\ast }}-H\ln \frac{{\rho }_0}{{\rho }_B^{\ast }}=H\ln \frac{{\rho }_B^{\ast }}{{\rho }_A^{\ast }}=H\ln \frac{{\beta }_B}{{\beta }_A}.$ $=7000\times \ln 10=7000\times 2.3026\approx 16118\text{ m}.$ B peaks about 16.1 km lower than A.

The chalkboard sketch below plots altitude $h$ (km, vertical axis) against air density $\rho$ (kg/m³, horizontal axis). The pale-yellow curve is the exponential atmosphere $\rho ={\rho }_0{e}^{-h/H}$: dense (right) near the ground, thin (left) up high. The two dashed lines mark each body's peak-deceleration density ${\rho }^{\ast }=\beta \sin \gamma /H$ — body A (blue, small $\beta$) peaks at low density high up; body B (pink, 10× $\beta$) peaks at 10× the density, far lower. The vertical double-arrow between them is exactly the $H\ln 10=16.1\text{ km}$ shift we just computed, reinforcing that a decade of $\beta$ costs one "log-decade" of altitude.

WHAT: separate peak-g from heating. WHY: $a_{\max }=v_e^2\sin \gamma /(2eH)$ has no $\beta$, so peak-g is identical. But heating scales with air density at the deceleration point, and the altitude of the peak, $h^{\ast }=H\ln ({\rho }_0/{\rho }^{\ast })$, DOES depend on $\beta$. The high-$\beta$ warhead peaks deep in dense air ⇒ far more heating despite equal g. False — same g, very different heat load. (See Aerodynamic heating and thermal protection systems.)

WHAT: rearrange the force balance and identify $\beta$. WHY: the same combination $m/(C_{D}A)=\beta$ that appeared in reentry reappears here — the parent note flagged this link to Terminal velocity. Start from $\frac{1}{2}\rho v_t^2{C}_{D}A=mg$. Divide by $C_{D}A$: $\frac{1}{2}\rho v_t^2=\frac{mg}{C_{D}A}=\beta g\Rightarrow v_t^2=\frac{2\beta g}{\rho }\Rightarrow v_t=\sqrt{\frac{2\beta g}{\rho }}.$ Numbers: $v_t=\sqrt{\frac{2\times 45\times 9.81}{1.0}}=\sqrt{882.9}\approx 29.7\text{ m/s}.$ A light body settles to ~30 m/s — parachute-like. This is why low-$\beta$ = gentle.

WHAT: invert ${\rho }^{\ast }=\beta \sin \gamma /H$ to get the needed $\beta$, then invert $\beta =m/(C_{D}A)$ for $A$. WHY: we are running the physics backwards — from a desired outcome (peak altitude) to a design variable (area). Step 1 — required $\beta$: $\beta =\frac{{\rho }^{\ast }H}{\sin \gamma }=\frac{0.05\times 7000}{\sin 30^{\circ }}=\frac{350}{0.5}=700{\text{ kg/m}}^2.$ Step 2 — required $A$: $A=\frac{m}{C_D\beta }=\frac{4000}{1.4\times 700}=\frac{4000}{980}\approx 4.08{\text{ m}}^2.$ A ~4 m² heat shield puts the peak at a nicely thin ${\rho }^{\ast }=0.05$, high up.

WHAT: compute $\beta$ before and after; take the ratio. WHY: $\beta \propto 1/(C_{D}A)$ at fixed mass, so the ratio is just the inverse ratio of footprints. ${\beta }_{\text{before}}=\frac{5200}{16.9}=307.7,{\beta }_{\text{after}}=\frac{5200}{2.6}=2000.$ $\frac{{\beta }_{\text{after}}}{{\beta }_{\text{before}}}=\frac{16.9}{2.6}=\mathrm{6.5.}$ $\beta$ jumps 6.5× — the body suddenly becomes far more "cannonball" and plunges. (This is the ballistic side of Skip vs ballistic vs lifting reentry.)

WHAT: differentiate $a$ with respect to $\rho$, set to zero, back-substitute. WHY differentiate? The peak of a smooth curve is where its slope is zero — calculus is the tool that finds "the worst instant" without guessing (see Newton's second law for where $a$ came from). Write $a(\rho )=C\rho e^{-k\rho }$ with $C=v_e^2/(2\beta )$ and $k=H/(\beta \sin \gamma )$. Product rule: $\frac{da}{d\rho }=C(e^{-k\rho }-k\rho e^{-k\rho })=C{e}^{-k\rho }(1-k\rho ).$ The exponential is never zero, so $\frac{da}{d\rho }=0\Rightarrow 1-k{\rho }^{\ast }=0\Rightarrow {\rho }^{\ast }=1/k=\frac{\beta \sin \gamma }{H}.$ Back-substitute. Note $k{\rho }^{\ast }=1$, so $e^{-k{\rho }^{\ast }}=e^{-1}$: $a_{\max }=C{\rho }^{\ast }{e}^{-1}=\frac{v_e^2}{2\beta }\cdot \frac{\beta \sin \gamma }{H}\cdot \frac{1}{e}=\frac{v_e^2\sin \gamma }{2eH}.$ The $\beta$ in $C$ cancels the $\beta$ in ${\rho }^{\ast }$. Peak-g is blind to $\beta$. $■$

WHAT: evaluate the $\sin \gamma$ factor across its whole domain $0^{\circ }\le \gamma \le 90^{\circ }$. WHY: the contract says cover every case — $\sin \gamma$ runs from 0 to 1, so peak-g runs from 0 to its max.

• $\gamma =90^{\circ }$ (vertical dive): $\sin \gamma =1$, largest peak-g $=v_e^2/(2eH)$. Brutal, brief, deep. Warhead territory.
• $0^{\circ }<\gamma <90^{\circ }$ (typical entry): $\sin \gamma$ between 0 and 1 scales peak-g proportionally. Capsules pick a small $\gamma$ (~5–7°) to keep g's survivable.
• $\gamma \to 0^{\circ }$ (grazing, approached from above): $\sin \gamma \to 0$, so $a_{\max }\to 0$. Physically the body barely dips into the atmosphere and slows very gently — but this is the skip regime: too shallow and the vehicle bounces back out of the atmosphere instead of committing to descent (see Skip vs ballistic vs lifting reentry). The Allen–Eggers straight-line assumption also breaks here, because the path is no longer approximately straight — the flight path curves and the vehicle may re-ascend. So the formula's "$a_{\max }\to 0$" is correct as a limit but the underlying model stops being trustworthy.
• $\gamma =0$ exactly (fully degenerate): the descent rate is $dh/dt=-v\sin \gamma =0$, so altitude never decreases — the vehicle skims horizontally and never actually reenters. The formula correctly returns zero deceleration ($a_{\max }=0$), which is the honest answer: with no downward motion there is no plunge into denser air to decelerate you. The model is simply not applicable to a body that never descends.

WHAT: send $\beta$ to its two extremes, holding $\sin \gamma >0$ fixed and positive. WHY: limits reveal the "ideal" endpoints the design lives between.

• $\beta \to \infty$ (infinitely dense cannonball): exponent $=-\rho H/(2\beta \sin \gamma )\to 0^{-}$, so $v\to v_e{e}^0=v_e$. The body ignores the air entirely and keeps its full entry speed all the way down.
• $\beta \to 0^{+}$ (infinitely light feather): with $\sin \gamma >0$ the exponent $\to -\infty$, so $v\to v_e{e}^{-\infty }=0$. The body is stopped essentially instantly by the first whiff of atmosphere. Real vehicles sit between these poles; choosing $\beta$ is choosing where on this spectrum you land.

Peak-g formula has no $\beta$; $\beta$ only sets the altitude/heating; $\beta$ lives in an exponent so its effect on retained speed is exponential, not linear. Which of those three did you get wrong on this page? ::: Revisit that level's mistake callout.