Aerobraking — gradual orbit lowering using atmospheric drag: Exercises

Before we start, here is the whole toolbox on one line, so no symbol appears unexplained:

Where each letter means:

$μ = GM$ — the planet's "gravity strength" (units $m^{3} / s^{2}$ ).
$r$ — how far you are from the planet's centre right now.
$a$ — the semi-major axis: half of (closest + farthest) $= \frac{1}{2} (r_{p} + r_{a})$ . Think "average size of the oval".
$v_{p}$ — speed at periapsis (the fast, low point).
$ρ$ — air density at your altitude; $H$ — the scale height, the height you must climb for density to drop by a factor $e \approx 2.718$ .

See Vis-Viva Equation, Orbital Energy and Semi-major Axis, Atmospheric Drag and Scale Height for the standalone build of each.

Figure — Aerobraking — gradual orbit lowering using atmospheric drag

The figure above is your compass for every problem: brake at the fast low point (red), the far point on the opposite side (blue) sinks.

Level 1 — Recognition

L1·Q1 — Which point moves?

A spacecraft on an ellipse fires a tiny retro-brake exactly at periapsis. Which orbital point drops the most: periapsis or apoapsis? State it and give the one-line reason.

Recall Solution

Apoapsis drops; periapsis stays essentially fixed. Why: the brake happens at $r_{p}$ , so that radius is where you already are — you can't fall below it in that instant. Removing speed there means you have less energy to climb the far side, so the opposite point ( $r_{a}$ ) comes down. The relation $Δ r_{a} \approx \frac{4 a ^{2} v _{p}}{μ} Δ v$ is written for $r_{a}$ precisely for this reason.

L1·Q2 — Read the sign

In $Δ r_{a} = \frac{4 a ^{2} v _{p}}{μ} Δ v$ , drag makes $Δ v$ negative. What is the sign of $Δ r_{a}$ , and what does that mean physically?

Recall Solution

All of $4$ , $a^{2}$ , $v_{p}$ , $μ$ are positive. So $Δ r_{a}$ has the same sign as $Δ v$ , i.e. negative → apoapsis goes down. Physically: every pass shrinks the far side of the orbit, exactly the shrinking-oval picture.

Level 2 — Application

L2·Q1 — One dip at Mars

Mars: $μ = 4.28 \times 1 0^{13} m^{3} / s^{2}$ . Orbit $a = 2.0 \times 1 0^{7}$ m, $v_{p} = 4000$ m/s. One pass removes $Δ v = - 1.0$ m/s. Find $Δ r_{a}$ .

Recall Solution

$Δ r_{a} = \frac{4 a ^{2} v _{p}}{μ} Δ v = \frac{4 ( 2.0 \times 1 0 ^{7} ) ^{2} ( 4000 )}{4.28 \times 1 0 ^{13}} (- 1.0)$ Numerator: $4 \times 4 \times 1 0^{14} \times 4000 = 6.4 \times 1 0^{18}$ . $Δ r_{a} = \frac{6.4 \times 1 0 ^{18}}{4.28 \times 1 0 ^{13}} (- 1.0) \approx - 1.50 \times 1 0^{5} m = - 150 km$ Meaning: a mere 1 m/s loss drops the far point ~150 km. Free, and huge.

L2·Q2 — Vis-viva speed check

For the same orbit, periapsis radius $r_{p} = 3.8 \times 1 0^{6}$ m. Use vis-viva to find the speed at periapsis and confirm it's near $4000$ m/s.

Recall Solution

Why vis-viva and not energy directly? We want a speed at a specific radius; vis-viva is the one formula that takes $(r, a)$ straight to $v$ . $v_{p}^{2} = μ (\frac{2}{r _{p}} - \frac{1}{a}) = 4.28 \times 1 0^{13} (\frac{2}{3.8 \times 1 0 ^{6}} - \frac{1}{2.0 \times 1 0 ^{7}})$ Inside: $\frac{2}{3.8 \times 1 0 ^{6}} = 5.263 \times 1 0^{- 7}$ , $\frac{1}{2.0 \times 1 0 ^{7}} = 5.0 \times 1 0^{- 8}$ ; difference $= 4.763 \times 1 0^{- 7}$ . $v_{p}^{2} = 4.28 \times 1 0^{13} \times 4.763 \times 1 0^{- 7} = 2.039 \times 1 0^{7} \Rightarrow v_{p} \approx 4515 m/s$ Same order as the quoted $4000$ m/s ✓ (the parent used a rounded value).

Level 3 — Analysis

L3·Q1 — Depth doubles the density

Density obeys $ρ = ρ_{0} e^{- h / H}$ with scale height $H = 11$ km at Mars. By how much (in km) must you lower periapsis altitude $h$ so that $ρ$ doubles?

Recall Solution

Why the exponential and why solve for the exponent? Doubling density is a ratio question; ratios of exponentials turn into differences of exponents — that's exactly what logarithms undo. We want $\frac{ρ _{new}}{ρ _{old}} = e^{- (h_{new} - h_{old}) / H} = 2$ . Lowering means $h_{new} < h_{old}$ , so let the drop be $d = h_{old} - h_{new} > 0$ : $e^{d / H} = 2 \Rightarrow d = H ln 2 = 11 \times 0.6931 \approx 7.6 km$ Meaning: a tiny 7.6 km deeper dip doubles the air you feel — that's why the corridor is knife-edge.

L3·Q2 — What that doubling does to heating

Heating rate scales as $\overset{q}{˙} \propto ρ^{1/2} v^{3}$ . At the same speed, if $ρ$ doubles, by what factor does $\overset{q}{˙}$ change?

Recall Solution

Speed $v$ (hence $v^{3}$ ) is unchanged, only $ρ^{1/2}$ moves: $\frac{q ˙ _{new}}{q ˙ _{old}} = (\frac{ρ _{new}}{ρ _{old}})^{1/2} = 2^{1/2} \approx 1.41$ A 41% heating jump for a 7.6 km slip — small in altitude, dangerous in watts.

Level 4 — Synthesis

L4·Q1 — Passes to circularize (rough count)

Start apoapsis $r_{a} = 3.62 \times 1 0^{7}$ m, target near-circular at $r_{a} = r_{p} = 3.8 \times 1 0^{6}$ m. Each pass drops apoapsis by $Δ r_{a} = - 150$ km (assume constant for estimate). How many passes $N$ ?

Recall Solution

Why treat $Δ r_{a}$ as constant? It's a first estimate; in reality $v_{p}$ and $a$ drift, but a fixed step gives the right order of magnitude. Total drop needed: $∣Δ r_{a}^{total} ∣ = 3.62 \times 1 0^{7} - 3.8 \times 1 0^{6} = 3.24 \times 1 0^{7} m = 32400 km$ $N = \frac{32400 km}{150 km/pass} = 216 passes$ Meaning: at ~one useful pass per orbit and multi-hour orbits, that's weeks to months — exactly the real MRO timescale. Slow by design.

L4·Q2 — Fuel you saved

Doing the same $Δ v$ chemically needs a burn. Suppose the equivalent single burn is $Δ v = 1200$ m/s, exhaust speed $v_{e} = 3000$ m/s, dry mass $m_{final} = 1000$ kg. Fuel required by the rocket equation?

Recall Solution

Why the rocket equation? It is the one law converting a required speed change into propellant mass, and it does so exponentially — the whole reason aerobraking pays. See Tsiolkovsky Rocket Equation. $m_{fuel} = m_{final} (e^{Δ v / v_{e}} - 1) = 1000 (e^{1200/3000} - 1) = 1000 (e^{0.4} - 1)$ $e^{0.4} = 1.4918$ , so $m_{fuel} = 1000 (0.4918) \approx 492 kg$ Meaning: aerobraking replaced ~490 kg of propellant with patience — the "hundreds of kg" the parent quoted.

Level 5 — Mastery

L5·Q1 — Design a corridor step

Your pass gives $a_{D} = 0.05 m/s^{2}$ at the current periapsis, and the thermal limit is $a_{D}^{m a x} = 0.04 m/s^{2}$ . You must raise periapsis so drag deceleration falls to the limit. Drag $a_{D} \propto ρ \propto e^{- h / H}$ ; $H = 11$ km. By how much $Δ h$ (km) must you raise $h$ ? (This is the "walk-out" burn logic.)

Recall Solution

Why only $ρ$ changes? In one small altitude nudge, $v$ at periapsis barely moves; the fierce lever is density. So $a_{D}$ tracks $ρ$ . $\frac{a _{D}^{m a x}}{a _{D}} = \frac{ρ _{new}}{ρ _{old}} = e^{- Δ h / H} = \frac{0.04}{0.05} = 0.8$ Raising periapsis (bigger $h$ ) shrinks $ρ$ , consistent with a ratio $< 1$ . Solve: $- \frac{Δ h}{H} = ln 0.8 \Rightarrow Δ h = - H ln 0.8 = - 11 \times (- 0.2231) \approx 2.45 km$ Meaning: lift periapsis just ~2.5 km (a whisper of a burn at apoapsis) to pull drag back inside the safe box. That's the real-time control loop of an aerobraking campaign.

L5·Q2 — Put it together: energy accounting

Show that one dip removing $Δ v = - 1.0$ m/s at $v_{p} = 4000$ m/s changes specific energy by $Δ ε = v_{p} Δ v$ , and confirm this predicts the same $Δ a$ that feeds $Δ r_{a} = 2Δ a$ .

Recall Solution

Step — energy change. From $ε = \frac{1}{2} v^{2} - μ / r$ , an impulse at fixed $r$ changes only kinetic term: $d (\frac{1}{2} v^{2}) = v d v$ . So $Δ ε = v_{p} Δ v = 4000 \times (- 1.0) = - 4000 J/kg .$ Step — energy to $a$ . Since $ε = - μ /2 a$ , differentiate: $Δ ε = \frac{μ}{2 a ^{2}} Δ a$ , so $Δ a = \frac{2 a ^{2}}{μ} Δ ε = \frac{2 ( 2.0 \times 1 0 ^{7} ) ^{2}}{4.28 \times 1 0 ^{13}} (- 4000) .$ Numerator $2 \times 4 \times 1 0^{14} = 8 \times 1 0^{14}$ ; divide: $\frac{8 \times 1 0 ^{14}}{4.28 \times 1 0 ^{13}} = 18.69$ ; times $- 4000$ : $Δ a \approx - 7.48 \times 1 0^{4} m = - 74.8 km .$ Step — to apoapsis. With $r_{p}$ fixed and $2 a = r_{p} + r_{a}$ , $Δ r_{a} = 2Δ a \approx - 150 km$ ✓ — matching L2·Q1. The two routes (direct $Δ r_{a}$ formula and the energy chain) agree, closing the loop.

Recall One-line self-test

Why lower apoapsis at periapsis and not at apoapsis? ::: At periapsis $v_{p}$ is largest, so $Δ ε = v_{p} Δ v$ removes the most energy per m/s, and it lowers the far point safely (periapsis stays out of trouble); braking at apoapsis would sink periapsis deeper into the air — dangerous.

Related builds: Hohmann Transfer · Orbital Energy and Semi-major Axis · Vis-Viva Equation · Atmospheric Drag and Scale Height.