Worked examples — Aerobraking — gradual orbit lowering using atmospheric drag

This page is the "throw everything at it" companion to the parent note. We march through every kind of case the master relations can produce: a normal brake, the zero case, the degenerate circular case, the sign flip (braking at apoapsis), the limiting deep-dive, a real mission-style word problem, and an exam twist. Before we start, let us pin down every symbol and tool this page uses.

See also: Vis-Viva Equation, Orbital Energy and Semi-major Axis, Atmospheric Drag and Scale Height, Tsiolkovsky Rocket Equation, Aerocapture, Hohmann Transfer.

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The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The "Applied at" column names where the $\Delta v$ acts — the single most important fact for deciding which orbital point moves.

Cell	What makes it special	Sign of $\Delta v$	Applied at	Covered by
A. Normal brake at periapsis	Standard shallow dip, small speed loss	$\Delta v<0$	periapsis	Ex 1
B. Zero input	Drag removes no speed (grazes above atmosphere)	$\Delta v=0$	periapsis	Ex 2
C. Degenerate orbit	Already circular, $r_p=r_a$, $e=0$	$\Delta v<0$	the (only) point	Ex 3
D. Sign flip	Brake at apoapsis instead	$\Delta v<0$	apoapsis	Ex 4
E. Limiting / deep-dive	Periapsis dropped one scale height — density & heating blow up	$\Delta v\ll 0$	periapsis	Ex 5
F. Real word problem	Full campaign: many passes, count them	$\Delta v<0$ per pass	periapsis	Ex 6
G. Exam twist	Given $\Delta v$ from drag force, find both $\Delta a$ and $\Delta r_a$	$\Delta v<0$	periapsis	Ex 7
H. End-of-campaign burn	Raise periapsis out of atmosphere	$\Delta v>0$	apoapsis	Ex 8

Throughout, unless stated, we use Mars: $\mu =4.28\times 10^{13}{\text{m}}^3/{\text{s}}^2$.

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Example 1 — Cell A: the normal periapsis brake

Forecast: Guess first — will apoapsis drop by more or less than $1$ km? (The factor of $4$ and the big $a^2$ should tip you off.)

1. Pick the right relation. Why this step? We want the change of the far point from a small speed loss at the near point — that is exactly what $\Delta r_a\approx \frac{4a^2{v}_p}{\mu }\Delta v$ answers, and nothing else does it in one line.
2. Substitute numbers. Why this step? Plug in to turn symbols into metres. $\Delta r_a=\frac{4(2.0\times 10^7{)}^2(4000)}{4.28\times 10^{13}}(-1.0)$
3. Crunch the powers. Why this step? $(2.0\times 10^7{)}^2=4.0\times 10^{14}$, so the numerator is $4\cdot 4.0\times 10^{14}\cdot 4000=6.4\times 10^{18}$. $\Delta r_a=\frac{6.4\times 10^{18}}{4.28\times 10^{13}}(-1.0)\approx -1.50\times 10^5\text{m}=-150\text{km}$

Verify: Units: $\frac{({\text{m}}^2)(\text{m/s})(\text{m/s})}{{\text{m}}^3/{\text{s}}^2}=\frac{{\text{m}}^4/{\text{s}}^2}{{\text{m}}^3/{\text{s}}^2}=\text{m}$ ✓. A tiny $1$ m/s loss dropped apoapsis $150$ km — that is the "leverage" that makes aerobraking worth the wait. Matches the parent note's estimate.

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Example 2 — Cell B: the zero case

Forecast: If the atmosphere touches nothing, does the orbit change at all?

1. Substitute $\Delta v=0$. Why this step? This is the boundary case — it tells us the formula behaves sensibly at zero input, which any good formula must. $\Delta a=\frac{2a^2{v}_p}{\mu }\cdot 0=0,\Delta r_a=\frac{4a^2{v}_p}{\mu }\cdot 0=0$
2. Interpret. Why this step? No energy removed ($\Delta \epsilon =v_p\Delta v=0$) means $\epsilon$ unchanged, so $a=-\mu /2\epsilon$ unchanged, so the whole ellipse is identical.

Verify: Pure gravity conserves energy — a pass with no drag must return to the same orbit. The formula gives exactly $0$. ✓ This is why controllers "walk out" by raising periapsis: once dips graze empty space, the campaign stops progressing on its own.

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Example 3 — Cell C: the degenerate circular orbit

Forecast: In a circle there is no distinct apoapsis. Does the $\Delta r_a$ formula still make sense?

1. Find the circular speed. Why this step? For a circle $a=r$, so vis-viva gives $v^2=\mu (2/r-1/r)=\mu /r$. This is the speed we brake from; we need it as our $v$ in the $\Delta a$ relation. $v_c=\sqrt{\frac{\mu }{r}}=\sqrt{\frac{4.28\times 10^{13}}{2.0\times 10^7}}=\sqrt{2.14\times 10^6}\approx 1463\text{m/s}$
2. Apply the $\Delta a$ relation. Why this step? Helper 1, $\Delta a=\frac{2a^2{v}_c}{\mu }\Delta v$, still holds — it came from energy, which doesn't care about eccentricity. $\Delta a=\frac{2(2.0\times 10^7{)}^2(1463)}{4.28\times 10^{13}}(-1.0)\approx -2.735\times 10^4\text{m}\approx -27.4\text{km}$
3. What happens to "apoapsis"? Why this step? This is the degenerate subtlety. The instant you brake, the orbit is no longer circular ($e$ jumps up off $0$): the brake point becomes the new apoapsis (you slowed at the top of what is now an ellipse), and a new periapsis appears $\frac{1}{2}$ orbit later, lower down. So $r_a$ stays $\approx r$ (you're at it), and it's the periapsis that drops. The naive $\Delta r_a$ formula, which assumed the brake is at periapsis, does not apply — its assumption is violated.

Verify: $v_c^2=\mu /r=2.14\times 10^6$, and $1463^2=2.14\times 10^6$ ✓. The lesson: the $\Delta r_a$ formula is only for a brake at periapsis; the degenerate circle is exactly where that assumption breaks, so we fall back to the energy relation for $\Delta a$.

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Example 4 — Cell D: the sign flip (brake at apoapsis)

Forecast: The parent note's "forecast-then-verify" said braking at apoapsis lowers periapsis. By how much compared with Example 1?

1. Use the $\Delta a$ relation with $v=v_a$. Why this step? Helper 1 works at any point; here the brake is at apoapsis, so we use the apoapsis speed $v_a$. $\Delta a=\frac{2a^2{v}_a}{\mu }\Delta v=\frac{2(2.0\times 10^7{)}^2(800)}{4.28\times 10^{13}}(-1.0)\approx -1.495\times 10^4\text{m}\approx -15.0\text{km}$
2. Figure out which point moves. Why this step? We are at apoapsis, so $r_a$ is fixed. From $2a=r_p+r_a$ (the semi-major-axis definition), a change in $a$ with $r_a$ held fixed means all of it goes into $r_p$: $\Delta r_p=2\Delta a$. $\Delta r_p=2\Delta a\approx -29.9\text{km}$
3. Compare with Example 1. Why this step? At periapsis $v_p=4000$; at apoapsis $v_a=800$ — five times slower. Since $\Delta a\propto v$, braking at apoapsis is 5× less effective and it lowers the dangerous end.

The figure below (Figure 1) makes step 2 visual. The blue solid curve is the original ellipse; the planet (orange) sits at the focus. The red arrow is the brake $\Delta v<0$ applied at apoapsis (left, red dot). Because we hold the apoapsis point fixed and remove energy, the ellipse shrinks inward on the near side only: the gray dashed curve shares the same apoapsis but has a lower periapsis (green). Notice the near end (right) is pulled toward the planet while the far end barely moves — the mirror image of a periapsis brake.

Figure 1 — Braking at apoapsis lowers the periapsis (Cell D).

Verify: Ratio of effectiveness $=v_p/v_a=4000/800=5$, and comparing the size changes: Example 1's $\Delta a=\frac{1}{2}\Delta r_a=-75$ km versus this $-15$ km, giving the same $5\times$ ✓. This is exactly why real drag lives at periapsis: it's both safer and more efficient.

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Example 5 — Cell E: the limiting deep-dive

Forecast: Dropping just $11$ km — surely a modest change? Watch the exponential.

1. Density factor. Why this step? Lowering altitude by $\Delta h=-H$ multiplies density by $e^{-(h-H)/H}/e^{-h/H}=e^{+1}$. The exponential is the whole story. $\frac{{\rho }_{\text{new}}}{{\rho }_{\text{old}}}=e^1\approx 2.718$
2. Speed loss per pass. Why this step? From the parent's drag deceleration $a_D=\frac{\rho v^2{C}_{D}A}{2m}$ (symbols in Helper 3), we have $a_D\propto \rho$, so $\Delta v$ per pass scales with $\rho$ (speed $v_p$ roughly unchanged for a small $r_p$ move). $\Delta v_{\text{new}}\approx 2.718\times (-1.0)=-2.72\text{m/s}$
3. Dynamic pressure and heating. Why this step? Dynamic pressure is $q=\frac{1}{2}\rho v^2$, so at fixed $v$ it scales as $\rho$ ($\times e=2.72$). The heating-rate law $\dot{q}\propto {\rho }^{1/2}{v}^3$ is the standard convective stagnation-point (Sutton–Graves) model used in Aerocapture: heat delivered per second grows like the mass flux of air hitting the nose ($\propto \rho v$) times the energy per unit mass it carries ($\propto v^2$), with a ${\rho }^{-1/2}$ correction from the boundary layer — net ${\rho }^{1/2}{v}^3$. At fixed $v$ this scales as ${\rho }^{1/2}$ ($\times e^{1/2}=1.65$). $q:\times 2.72,\dot{q}:\times 1.65$

Verify: $e^1=2.718$, $e^{0.5}=1.649$ ✓. A mere $11$ km deeper nearly triples the braking but also nearly triples dynamic pressure — this is why "just dip deeper to finish faster" is dangerous, and why the safe corridor is narrow. Limiting behaviour: as $r_p\to$ dense atmosphere, $\rho$ grows without bound and the craft overheats.

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Example 6 — Cell F: the real word problem

Forecast: Months-long campaign — hundreds of passes? Tens? Guess an order of magnitude.

1. Total apoapsis to remove. Why this step? Distance to cover sets the workload. $\Delta r_a^{\text{total}}=3.35\times 10^7-6.9\times 10^6=2.66\times 10^7\text{m}$
2. Divide by per-pass drop. Why this step? Constant-drop approximation gives a first estimate (real drops shrink as $a$ shrinks, so this is a lower bound). $N\approx \frac{2.66\times 10^7}{1.5\times 10^5}\approx 177\text{passes}$
3. Turn passes into calendar time. Why this step? Use Helper 2, the period $T=2\pi \sqrt{a^3/\mu }$, at the early (largest) orbit: $T=2\pi \sqrt{\frac{(2.0\times 10^7{)}^3}{4.28\times 10^{13}}}=2\pi \sqrt{\frac{8.0\times 10^{21}}{4.28\times 10^{13}}}=2\pi \sqrt{1.869\times 10^8}\approx 8.59\times 10^4\text{s}\approx 23.9\text{h}$ So $177$ passes at roughly a day each $\Rightarrow$ several months (early passes are longest; the orbit period shrinks as $a$ falls, so the real campaign is somewhat shorter).

Verify: $2.66\times 10^7/1.5\times 10^5=177.3$ ✓. $T=2\pi \sqrt{8.0\times 10^{21}/4.28\times 10^{13}}=2\pi \sqrt{1.869\times 10^8}\approx 8.59\times 10^4$ s ✓. Order of magnitude "hundreds of passes over months" matches real MGS/MRO campaigns. ✓

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Example 7 — Cell G: the exam twist (from drag force)

Forecast: Small deceleration, short time — will $\Delta v$ be around $1$ m/s (like Example 1)?

1. Turn deceleration into $\Delta v$. Why this step? $\Delta v$ is deceleration times the time it acts (impulse per unit mass), and it's negative because it slows the craft. $\Delta v=-a_{D}t=-(0.02)(50)=-1.0\text{m/s}$
2. Compute $\Delta a$. Why this step? Helper 1 converts the speed loss (at periapsis, so $v=v_p$) into an orbit-size change. $\Delta a=\frac{2a^2{v}_p}{\mu }\Delta v=\frac{2(4.0\times 10^{14})(4000)}{4.28\times 10^{13}}(-1.0)\approx -7.48\times 10^4\text{m}\approx -74.8\text{km}$
3. Compute $\Delta r_a$. Why this step? Since the brake is at periapsis, $r_p$ is fixed and $2a=r_p+r_a$ gives $\Delta r_a=2\Delta a$. $\Delta r_a=2\Delta a\approx -1.50\times 10^5\text{m}=-150\text{km}$

Verify: $\Delta v=-1.0$ m/s reproduces Example 1's $-150$ km ✓ — good, the two routes (given $\Delta v$ vs given drag force) agree. Units of $a_{D}t$: $({\text{m/s}}^2)(\text{s})=\text{m/s}$ ✓.

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Example 8 — Cell H: ending the campaign

Forecast: A positive (prograde) burn — does it raise the near end or the far end?

1. Find the apoapsis radius, then $v_a$ from vis-viva. Why this step? We need the true speed at the point where we burn, and vis-viva gives speed anywhere once we know $r$ and $a$. Apoapsis: $r_a=2a-r_p=2(6.95\times 10^6)-3.5\times 10^6=1.04\times 10^7$ m. $v_a=\sqrt{\mu \left(\frac{2}{r_a}-\frac{1}{a}\right)}=\sqrt{4.28\times 10^{13}\left(\frac{2}{1.04\times 10^7}-\frac{1}{6.95\times 10^6}\right)}\approx 1.63\times 10^3\text{m/s}$
2. Apply Helper 1 at apoapsis. Why this step? We burn at $r_a$, so use $v=v_a$; positive $\Delta v$ raises energy, so $a$ grows. $\Delta a=\frac{2a^2{v}_a}{\mu }\Delta v=\frac{2(6.95\times 10^6{)}^2(1633)}{4.28\times 10^{13}}(+2.0)\approx +7.37\times 10^3\text{m}\approx +7.4\text{km}$
3. Which point rises? Why this step? We're at apoapsis, so $r_a$ is fixed; $2a=r_p+r_a\Rightarrow \Delta r_p=2\Delta a$. $\Delta r_p=2\Delta a\approx +14.7\text{km}$

Verify: $v_a\approx 1633$ m/s from vis-viva (checked below). Positive $\Delta v$ → positive $\Delta r_p$: periapsis rises out of the atmosphere, exactly the "walk-out" the parent note describes. Sign is correct: raising periapsis ends aerobraking.

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Recall

Recall Which cell does "brake at apoapsis" hit, and why is it the wrong tool for lowering apoapsis?

Cell D. Braking at apoapsis lowers periapsis (the point opposite the burn), and since $v_a\ll v_p$ it is also far less efficient. To lower apoapsis you must brake at periapsis.

Recall Why does the

$\Delta r_a$ formula fail for a circular orbit? Because it assumes the brake happens at periapsis of an ellipse. A circle has no distinct periapsis; braking makes the brake point the new apoapsis, so periapsis (not apoapsis) is what drops. Use the energy relation $\Delta a=\frac{2a^{2}v}{\mu }\Delta v$ instead.