Exercises — Aerocapture — using atmosphere to decelerate into orbit

The sign of $\epsilon$ is the whole story:

• $\epsilon >0$ → hyperbola (escapes),
• $\epsilon =0$ → parabola (marginal),
• $\epsilon <0$ → ellipse (captured).

Here $\epsilon =+2.0>0$, so it is on a hyperbola and will not be captured on its own — it flies past. That positive energy is exactly what drag must remove.

Plug straight in: $\epsilon =-\frac{\mu }{2a}=-\frac{4.28\times 10^4}{2\times 12000}=-\frac{4.28\times 10^4}{2.4\times 10^4}=-1.783{\text{km}}^2/{\text{s}}^2.$ Negative → captured. Good, this is a bound orbit. Why the minus: a bound orbit sits in a "well"; you'd need to add energy to climb out to zero (escape).

$a_D=\frac{\rho v^2}{2\beta }.$

• $\rho \uparrow$ (thicker air, deeper dive) → more braking, linearly.
• $v\uparrow$ → more braking, but as $v^2$ (double the speed → four times the drag).
• $\beta \uparrow$ (denser/smaller craft) → less braking, because $\beta$ is in the denominator; a heavy compact craft resists the same air more.

Far away ($r\to \infty$) the potential term $-\mu /r\to 0$, so all the energy is kinetic: ${\epsilon }_{in}=\frac{v_{\infty }^2}{2}=\frac{4^2}{2}=8.0{\text{km}}^2/{\text{s}}^2.$ Positive, as every arriving craft is — that $+8.0$ is the "excess" drag must delete.

Since periapsis radius $r_p$ barely changes during the quick pass, only the speed drops: $\Delta v_{aero}=v_{in}-v_{out}=10.2-9.4=0.8\text{km/s}.$ This $0.8$ km/s of braking came from air, not fuel — that is the propellant you never launched.

$\frac{\rho (45)}{\rho (0)}=e^{-45/15}=e^{-3}\approx \mathrm{0.0498.}$ So about 20× thinner ($1/0.0498\approx 20$). This steep exponential is why a few km of altitude change the drag dramatically — the razor's-edge corridor (see Scale Height & Exponential Atmosphere).

(a) Arrival energy ${\epsilon }_{in}=v_{\infty }^2/2=3^2/2=4.5{\text{km}}^2/{\text{s}}^2$. $|\Delta \epsilon |={\epsilon }_{in}-{\epsilon }_{out}=4.5-(-1)=5.5{\text{km}}^2/{\text{s}}^2.$ You cross from $+4.5$ down to $-1$. (b) Invert $\epsilon =-\mu /(2a)$: $a=-\frac{\mu }{2{\epsilon }_{out}}=-\frac{4.28\times 10^4}{2\times (-1)}=2.14\times 10^4\text{km}.$ A loose, big ellipse — exactly what you want so you do not over-brake.

$a_D=\rho v^2/(2\beta )$, and $\rho ,v$ are identical, so $\frac{a_{D,A}}{a_{D,B}}=\frac{{\beta }_B}{{\beta }_A}=\frac{200}{50}=4.$ A brakes 4× harder at the same altitude — it is "fluffier" (more area per mass). For B to brake as hard, it must find $4\times$ the density, i.e. dive deeper — which means hotter. This is why low-$\beta$ (big blunt) shields are safer. See Ballistic Coefficient.

Keep SI throughout (metres, seconds): $a_D=\frac{\rho v^2}{2\beta }=\frac{(2\times 10^{-5})(6000{)}^2}{2\times 80}=\frac{(2\times 10^{-5})(3.6\times 10^7)}{160}=\frac{720}{160}=4.5{\text{m/s}}^2.$ In g: $4.5/9.81=0.459g\approx 0.46g$. Gentle here — but this is only the instantaneous peak; drag acts over the whole pass to remove the km/s of $\Delta v$.

(a) Energy is conserved from infinity to periapsis (drag hasn't acted yet at entry into the atmosphere; use it at periapsis with pre-drag energy ${\epsilon }_{in}=v_{\infty }^2/2$). From $\epsilon =v^2/2-\mu /r$: $v_{in}=\sqrt{2\left({\epsilon }_{in}+\frac{\mu }{r_p}\right)}=\sqrt{2\left(\frac{3^2}{2}+\frac{3.986\times 10^5}{6500}\right)}.$ Inside: $\frac{9}{2}=4.5$ and $\frac{3.986\times 10^5}{6500}=61.32$. Sum $=65.82$, times 2 $=131.65$. $v_{in}=\sqrt{131.65}=11.47\text{km/s}.$ (b) Same $r_p$, but now with ${\epsilon }_{out}=-8$: $v_{out}=\sqrt{2\left({\epsilon }_{out}+\frac{\mu }{r_p}\right)}=\sqrt{2(-8+61.32)}=\sqrt{2\times 53.32}=\sqrt{106.65}=10.33\text{km/s}.$ (c) $\Delta v_{aero}=v_{in}-v_{out}=11.47-10.33=1.15\text{km/s}$ of free braking. (See Vis-viva Equation for the $v(r)$ relation used.)

From the Tsiolkovsky Rocket Equation, a burn of $\Delta v$ needs mass fraction $\frac{\Delta m}{m}=1-e^{-\Delta v/v_e}=1-e^{-1.15/3}=1-e^{-0.3833}.$ $e^{-0.3833}=0.6816$, so $\frac{\Delta m}{m}=1-0.6816=0.318\approx 32\%.$ Nearly a third of the arrival mass not hauled as propellant — the whole economic case for aerocapture in one number.

The higher you fly (larger $h$), the thinner the air, the smaller $a_D$. So:

• $h_{\max }$ (highest allowed) corresponds to the smallest allowed drag $a_{\min }$.
• $h_{\min }$ (lowest allowed) corresponds to the largest allowed drag $a_{\max }$.

Because $a_D\propto e^{-h/H}$, the ratio of the two drag limits maps to an altitude difference: $\frac{a_{\max }}{a_{\min }}=\frac{e^{-h_{\min }/H}}{e^{-h_{\max }/H}}=e^{(h_{\max }-h_{\min })/H}=e^{-\Delta h/H}.$ Solve for the corridor width $|\Delta h|=h_{\min }-h_{\max }$ magnitude: $|\Delta h|=H\ln \frac{a_{\max }}{a_{\min }}=8\ln \frac{6}{3}=8\ln 2=8\times 0.6931=5.55\text{km}.$ The whole safe corridor is only ≈ 5.5 km tall. Miss it by a couple of km and you either skip back to escape or overshoot into a crash — this razor's-edge is why aerocapture demands precise navigation and lift steering (bank-angle control), not just "aim deep." Look at the figure: the drag band is narrow because the density curve is so steep.

(a) Same drag: ${\rho }_X{v}_X^2={\rho }_Y{v}_Y^2$, so ${\rho }_Y={\rho }_X\frac{v_X^2}{v_Y^2}={\rho }_0\frac{v_0^2}{(v_0/2{)}^2}={\rho }_0\times 4=4{\rho }_0.$ Slower Y must dive into 4× denser air to brake as hard. (b) Heating ratio: $\frac{{\dot{q}}_Y}{{\dot{q}}_X}=\frac{\sqrt{{\rho }_Y}{v}_Y^3}{\sqrt{{\rho }_X}{v}_X^3}=\sqrt{\frac{4{\rho }_0}{{\rho }_0}}{\left(\frac{v_0/2}{v_0}\right)}^3=2\times \frac{1}{8}=\mathrm{0.25.}$ Y heats at only a quarter the rate. Lesson: the cube on speed dominates the half-power on density — braking at lower speed (even in thicker air) is far gentler thermally. This is exactly why entering fast is the hard case (see Atmospheric Entry & Heating): heating, not braking, is the binding constraint.

• Sign of $\epsilon$ decides captured (−) vs escaping (+). ::: True — magnitude only sets orbit tightness.
• $v_{\infty }$ vs $v_{in}$? ::: $v_{in}=\sqrt{2(v_{\infty }^2/2+\mu /r_p)}$ — gravity speeds you up before periapsis.
• Which $v$ units in $a_D$? ::: m/s (SI), because $\rho$ and $\beta$ are metric.
• Which radius for air density? ::: altitude $h=r-R_{\text{planet}}$, not $r$.
• Why is heating the limit, not braking? ::: $\dot{q}\propto v^3$ beats $a_D\propto v^2$ — speed hurts heat most.