Aerocapture — using atmosphere to decelerate into orbit: Worked examples

If any symbol here feels unfamiliar, it was built in the parent — see Aerocapture — using atmosphere to decelerate into orbit (index 3.4.24) and the prerequisite pages linked at the end.

The scenario matrix

Every aerocapture problem lives in one of these cells. The "sign / regime" column is the key knob — the thing whose sign or size decides the whole answer.

#	Cell (case class)	Key knob & its regime	Covered by
A	Capture check — did we end bound?	Sign of $ε_{o u t}$ : $+$ / $0$ / $-$	Ex 1, Ex 2
B	Marginal / degenerate — exactly parabolic	$ε = 0$ (the boundary)	Ex 2
C	Over-braking (crash) — too deep	$ε_{o u t}$ far negative, $r_{p} < R_{surface}$	Ex 3
D	Under-braking (skip-out) — too shallow	$Δ ε$ too small	Ex 4
E	Ballistic coefficient extremes — $β$ small vs large	$a_{D} \propto 1/ β$	Ex 5
F	Density / altitude limit — how deep is deep?	$ρ = ρ_{0} e^{- h / H}$ , $h \to 0$ and $h \to \infty$	Ex 6
G	Real-world word problem — Δv & fuel saved	mass fraction $1 - e^{- Δ v / v_{e}}$	Ex 7
H	Exam twist — heating vs braking trade	$\overset{q}{˙} \propto v^{3}$ vs $Δ ε \propto v^{2}$	Ex 8

We use two planets so the numbers stay concrete:

Cell A + the sign of energy — the whole game

Figure — Aerocapture — using atmosphere to decelerate into orbit

The picture above is the number line of $ε$ . Everything on this page is deciding which side of zero a spacecraft lands on.

$ε > 0$ : hyperbola — the craft has spare kinetic energy even at infinity, it leaves. (Red, right side.)
$ε = 0$ : parabola — the exact boundary, "just barely escapes". (Amber, the wall.)
$ε < 0$ : ellipse — bound, captured, loops forever. (Cyan, left side.)

Worked example Ex 1 — Cell A: a clean capture

A probe reaches Mars at periapsis radius $r_{p} = 3490 km$ (100 km up). Before drag its speed there is $v_{in} = 5.5 km/s$ . Drag then removes $Δ v = 0.8 km/s$ so it exits at $v_{o u t} = 4.7 km/s$ . Is it captured?

Forecast: Guess the sign of $ε_{o u t}$ before computing. Slower at the same radius means less kinetic energy — will it dip below zero?

Incoming energy $ε_{in} = \frac{v _{in}^{2}}{2} - \frac{μ _{M}}{r _{p}} = \frac{5. 5 ^{2}}{2} - \frac{4.28 \times 1 0 ^{4}}{3490}$ . Why this step? $ε$ is the single number that decides capture; compute it before and after the pass. $= 15.125 - 12.264 = 2.861 km^{2} / s^{2}$ . Positive → still a hyperbola coming in. Good, it needs capturing.
Outgoing energy $ε_{o u t} = \frac{4. 7 ^{2}}{2} - \frac{4.28 \times 1 0 ^{4}}{3490} = 11.045 - 12.264 = - 1.219 km^{2} / s^{2}$ . Why this step? Same $r_{p}$ (fast pass, radius barely changes), only $v$ dropped. The potential term is identical, so only kinetic energy changed.
Read the sign. $ε_{o u t} < 0$ → captured (ellipse). ✅

Verify: The change $Δ ε = ε_{o u t} - ε_{in} = - 1.219 - 2.861 = - 4.08 km^{2} / s^{2}$ . Cross-check with the speed form $Δ ε = \frac{1}{2} (v_{o u t}^{2} - v_{in}^{2}) = \frac{1}{2} (22.09 - 30.25) = - 4.08$ . Same number — good, the potential term cancelled exactly as claimed.

Worked example Ex 2 — Cells A & B: how much braking gives

exactly parabolic (the boundary)? Same $r_{p}$ , same $v_{in} = 5.5$ km/s. What exit speed $v_{o u t}^{⋆}$ leaves the craft on a parabola ( $ε = 0$ )? Braking any less than this fails to capture.

Forecast: $ε = 0$ means all kinetic exactly balances potential. Guess whether $v_{o u t}^{⋆}$ is above or below $4.7$ km/s.

Set $ε = 0$ : $\frac{v _{o u t}^{⋆} ^{2}}{2} = \frac{μ _{M}}{r _{p}}$ . Why? Parabola is the razor's edge between escape and capture — it is defined by $ε = 0$ .
Solve $v_{o u t}^{⋆} = 2 μ_{M} / r_{p} = 2 \cdot 12.264 = 24.528 = 4.953 km/s$ . Why? This is exactly the escape speed at $r_{p}$ — no coincidence: escape speed is the speed at which $ε = 0$ .
Interpret: To be captured you must exit below $4.953$ km/s. Our Ex 1 exit ( $4.7$ ) is below it → captured. Had drag been weaker and left $v_{o u t} = 5.0 > 4.953$ , it would still escape (Cell D preview).

Verify: Plug $v_{o u t}^{⋆}$ back: $ε = \frac{1}{2} (4.953)^{2} - 12.264 = 12.264 - 12.264 = 0$ . Exactly the boundary. ✅

Cell C — over-braking: the crash floor

The figure shows why "aim deeper for safety" is a trap: the ellipse you create has a periapsis; if you bleed off too much energy, that periapsis sinks below the surface and the craft hits the ground.

Worked example Ex 3 — Cell C: braking so hard the orbit intersects the planet

On Mars a craft passes $r_{p} = 3490$ km at $v_{in} = 5.5$ km/s. Drag is enormous and it exits at only $v_{o u t} = 3.0$ km/s. Is the resulting orbit safe, or does it crash?

Forecast: Very slow exit = very negative energy = a tight ellipse. Guess whether its far point (apoapsis) or its near point matters for crashing.

Exit energy $ε_{o u t} = \frac{3. 0 ^{2}}{2} - 12.264 = 4.5 - 12.264 = - 7.764 km^{2} / s^{2}$ . Why? Establishes we are bound (very negative → tight ellipse).
Semi-major axis from $ε = - μ / (2 a)$ : $a = - \frac{μ _{M}}{2 ε _{o u t}} = - \frac{4.28 \times 1 0 ^{4}}{2 ( - 7.764 )} = 2756 km$ . Why? $a$ sets the whole ellipse's size; a small $a$ means a small orbit.
Danger check: the current point $r_{p} = 3490$ km is where we are, but is it periapsis? Since $r_{p} = 3490 > a = 2756$ , the point we're at is beyond the semi-major axis — this orbit cannot even reach $r_{p}$ as an apoapsis without... in fact $r_{p} > 2 a = 5512$ ? No, $r_{p} < 2 a$ . The physical impossibility flag: apoapsis $r_{a} \leq 2 a = 5512$ km and periapsis $r_{p, orbit} = 2 a - r_{a}$ . Because $a < r_{p}$ , the ellipse's periapsis must be smaller than $R_{M} = 3390$ . Why? An ellipse's two radii average to $a$ : $r_{p, orbit} + r_{a} = 2 a$ . If $a = 2756 < R_{M} = 3390$ , then even the larger radius side is being pulled inward — the near side is deep underground.
Verdict: $2 a = 5512 < 2 R_{M} = 6780$ is impossible for a surface-clearing orbit — crash. Over-braking. ❌

Verify: Escape speed at $r_{p}$ was $4.953$ km/s (Ex 2). We exited at $3.0 ≪ 4.953$ , i.e. hugely sub-escape → deeply bound → dangerously tight. Consistent. ✅

Cell D — under-braking: the skip-out

Cell E — ballistic coefficient extremes

Worked example Ex 5 — Cell E: same air, opposite fates from

$β$ Two craft hit Earth's atmosphere at $h$ where $ρ = 1.0 \times 1 0^{- 4} kg/m^{3}$ , both at $v = 7.0$ km/s $= 7000$ m/s. Craft A (blunt): $β_{A} = 50 kg/m^{2}$ . Craft B (dense): $β_{B} = 200 kg/m^{2}$ . Compare their decelerations.

Forecast: $a_{D} \propto 1/ β$ . Guess the ratio $a_{D, A} / a_{D, B}$ before computing.

Decel of A: $a_{D, A} = \frac{ρ v ^{2}}{2 β _{A}} = \frac{( 1 0 ^{- 4} ) ( 7000 ) ^{2}}{2 ( 50 )} = \frac{4900}{100} = 49 m/s^{2}$ . Why? Use SI throughout ( $ρ$ in kg/m³, $v$ in m/s, $β$ in kg/m²) so the answer is m/s².
Decel of B: $a_{D, B} = \frac{( 1 0 ^{- 4} ) ( 7000 ) ^{2}}{2 ( 200 )} = \frac{4900}{400} = 12.25 m/s^{2}$ .
Ratio: $a_{D, A} / a_{D, B} = 49/12.25 = 4$ . Exactly $β_{B} / β_{A} = 200/50 = 4$ . Why? Confirms $a_{D} \propto 1/ β$ : the fluffy craft brakes $4 \times$ harder in the same air, so it can capture higher up, where it is cooler.

Verify: In $g$ 's: A feels $49/9.81 = 5.0 g$ , B feels $12.25/9.81 = 1.25 g$ . Ratio $4$ , and both are plausible aerocapture loads. ✅

Cell F — density and altitude limits

Worked example Ex 6 — Cell F: how deep for a target density? (and the two limits)

Earth: $ρ (h) = ρ_{0} e^{- h / H}$ , $ρ_{0} = 1.2$ , $H = 8$ km. (a) At what altitude $h$ is $ρ = 1.0 \times 1 0^{- 4} kg/m^{3}$ (the density used in Ex 5)? (b) Check the two degenerate limits $h \to 0$ and $h \to \infty$ .

Forecast: The ratio $ρ / ρ_{0} = 1 0^{- 4} /1.2 \approx 8.3 \times 1 0^{- 5}$ is tiny — guess whether $h$ is tens or hundreds of km.

Invert the exponential. From $ρ = ρ_{0} e^{- h / H}$ , take the natural log: $ln (ρ / ρ_{0}) = - h / H$ , so $h = - H ln (ρ / ρ_{0})$ . Why this tool — the logarithm? The unknown $h$ sits in the exponent. The one operation that pulls a variable out of an exponent is its inverse, $ln$ . That is the only reason we reach for it.
Plug in: $h = - 8 ln (\frac{1 0 ^{- 4}}{1.2}) = - 8 ln (8.333 \times 1 0^{- 5}) = - 8 (- 9.393) = 75.1 km$ . Why? Confirms upper-atmosphere aerocapture happens near ~75 km on Earth.
Limit $h \to 0$ : $ρ \to ρ_{0} e^{0} = ρ_{0} = 1.2$ — full sea-level density (dense, would tear the craft apart). Limit $h \to \infty$ : $ρ \to ρ_{0} e^{- \infty} = 0$ — vacuum, no drag at all. Why check limits? They bracket the physics: between "solid wall" ( $h = 0$ ) and "no brake" ( $h = \infty$ ) lies the thin usable band. A few $H$ (few tens of km) covers the whole action.

Verify: Push $h = 75.1$ km back through $ρ_{0} e^{- h / H} = 1.2 e^{- 75.1/8} = 1.2 e^{- 9.39} = 1.2 (8.33 \times 1 0^{- 5}) = 1.0 \times 1 0^{- 4}$ . ✅ Matches the input density.

Cell G — real-world word problem: fuel saved

Worked example Ex 7 — Cell G: convert saved

$Δ v$ into saved mass A Mars orbiter would need an insertion burn of $Δ v = 0.8$ km/s. Instead it aerocaptures, getting that $0.8$ km/s from drag for free. Engine exhaust speed $v_{e} = 3$ km/s. What fraction of arrival mass is saved as propellant?

Forecast: Tsiolkovsky Rocket Equation gives mass fraction $1 - e^{- Δ v / v_{e}}$ . With $Δ v ≪ v_{e}$ , guess: closer to 25% or 5%?

Exponent: $Δ v / v_{e} = 0.8/3 = 0.2667$ . Why? The rocket equation is exponential in this ratio — the natural unit of "how hard" a burn is.
Mass fraction saved: $f = 1 - e^{- 0.2667} = 1 - 0.7659 = 0.2341$ . Why? $e^{- Δ v / v_{e}}$ is the fraction of mass you'd keep after the burn; $1$ minus it is the propellant you'd spend — and now don't.
Interpret: ~ $23.4%$ of the arrival mass no longer has to be propellant — it can be payload, or was never launched. Enormous.

Verify: $e^{- 0.2667} = 0.7659$ , and $0.7659 \cdot e^{+ 0.2667} = 1.000$ . Round-trip consistent, and $0.234$ matches the parent note's stated ≈0.23. ✅

Cell H — the exam twist: heating vs braking

Worked example Ex 8 — Cell H: "faster entry brakes harder, so it's easier." True?

Compare entering at $v$ vs $2 v$ (same $ρ$ , same craft). Braking force scales as $v^{2}$ ; energy to remove scales as $v^{2}$ ; peak heating scales as $v^{3}$ (Sutton–Graves, see Atmospheric Entry & Heating). By what factor does each grow when you double $v$ ?

Forecast: All three grow — but which grows fastest, and does that help or hurt?

Drag force factor: $F_{D} \propto v^{2} \Rightarrow (2)^{2} = 4 \times$ . Why? From $F_{D} = \frac{1}{2} ρ v^{2} C_{D} A$ .
Energy-to-shed factor: the excess $ε_{in} \approx \frac{1}{2} v_{\infty}^{2} \propto v^{2} \Rightarrow 4 \times$ . Why? Braking got $4 \times$ stronger, but the job also got $4 \times$ bigger — they cancel. No net advantage from speed.
Heating factor: $\overset{q}{˙} \propto v^{3} \Rightarrow (2)^{3} = 8 \times$ . Why? Heating outpaces both braking and energy by an extra factor of $v$ . So doubling speed leaves the braking-vs-work balance unchanged but makes the thermal problem 8× worse.
Verdict: faster entry is harder, because heating (the true limiting constraint) is cubic while everything helpful is only quadratic. The "faster brakes harder" intuition is a trap. ✅ (Steel-manned in the parent's mistakes list.)

Verify: Ratios $4 : 4 : 8$ . Divide heating by braking: $8/4 = 2 =$ one extra power of $v$ — exactly the cubic-vs-quadratic gap. ✅

Recall Self-test: which cell is this?

A craft exits its single atmospheric pass with $ε_{o u t} = + 0.4 km^{2} / s^{2}$ . Captured? ::: No — $ε > 0$ is still hyperbolic (Cell D, skip-out). It needs more braking. You want the exact braking that gives a parabola. What is $ε_{o u t}$ ? ::: Exactly $0$ (Cell B, the boundary = escape speed at $r_{p}$ ). Craft A has $β = 50$ , craft B has $β = 200$ , same air & speed. Who brakes harder and by how much? ::: A, by $4 \times$ , since $a_{D} \propto 1/ β$ (Cell E). Doubling entry speed multiplies peak heating by? ::: $8$ ( $\overset{q}{˙} \propto v^{3}$ ) — Cell H. To find the altitude giving a target density, which function inverts $ρ_{0} e^{- h / H}$ ? ::: The natural log: $h = - H ln (ρ / ρ_{0})$ (Cell F).

Prerequisites & neighbours: Vis-viva Equation · Hyperbolic Trajectories & Hyperbolic Excess Velocity · Orbit Insertion Burns · Aerobraking · Atmospheric Entry & Heating · Ballistic Coefficient · Tsiolkovsky Rocket Equation · Scale Height & Exponential Atmosphere