Aerobraking — gradual orbit lowering using atmospheric drag: Visual walkthrough

This is the visual companion to Aerobraking — gradual orbit lowering using atmospheric drag. We rebuild its master result $Δ r_{a} \approx \frac{4 a ^{2} v _{p}}{μ} Δ v$ from absolute zero. Every letter in that formula is earned below.

Step 1 — What an orbit even is (the oval and its two special points)

WHAT. A spacecraft in orbit traces an ellipse — a squashed circle. Two points matter:

Periapsis, the closest point to the planet. Call its distance from the planet's centre $r_{p}$ (" $r$ for radius, $p$ for periapsis").
Apoapsis, the farthest point. Its distance is $r_{a}$ .

WHY. Everything about aerobraking is a tug-of-war between these two points: drag lives at the bottom (thick air), and we want to change the top. So we must name them before we can move them.

PICTURE. Look at the figure: the planet sits at one focus (not the centre!) of the oval. The short green line is $r_{p}$ ; the long magenta line is $r_{a}$ .

Figure — Aerobraking — gradual orbit lowering using atmospheric drag

Step 2 — Giving the orbit ONE number: its energy

WHAT. A moving object near a planet has two energy stores:

Kinetic (energy of motion), which per unit mass is $\frac{1}{2} v^{2}$ , where $v$ is speed.
Gravitational potential (energy of position), which per unit mass is $- \frac{μ}{r}$ .

Here $μ = GM$ ("mu") is the planet's gravitational strength: $G$ is the universal gravity constant and $M$ the planet's mass. Bigger planet → bigger $μ$ → stronger pull.

Add the two stores to get the specific mechanical energy: $ε = motion \frac{v ^{2}}{2} position - \frac{μ}{r}$

WHY this tool. Speed and distance both change constantly as the craft swings around — but their combination $ε$ does not change under pure gravity. A quantity that stays constant is a gift: it labels the entire orbit with a single frozen number. That is why we reach for energy instead of tracking $v$ and $r$ separately.

PICTURE. The figure shows the two energy bars trading height as the craft moves: near periapsis the motion bar (fast) is tall and the position bar is deep; near apoapsis they swap — but their sum stays flat (the dashed line).

Step 3 — The magic link: energy tells you the orbit's SIZE

WHAT. For any closed (bound) orbit, the frozen energy number connects straight to the size $a$ : $ε = - \frac{μ}{2 a}$

WHY. We want a lever that turns "energy" into "how big is the orbit." Here it is. Read the signs carefully:

$ε$ is negative (Step 2).
Make $ε$ more negative (take energy away) → the right side must grow more negative too → $2 a$ must shrink → the orbit gets smaller.

That is the seed of aerobraking in one line: remove energy, shrink the orbit.

PICTURE. The figure stacks three orbits of decreasing $a$ . As the energy bar drops lower (more negative), the drawn oval visibly shrinks. Same planet, same focus — just a tighter loop.

Recall Where does

$ε = - μ /2 a$ come from? Evaluate $ε = v^{2} /2 - μ / r$ at periapsis and apoapsis. Plug $r_{p} = a (1 - e)$ , $r_{a} = a (1 + e)$ (with $e$ the eccentricity, how squashed the oval is) and the speeds at those points. All the $r$ -dependence and $e$ -dependence cancels, leaving only $- μ /2 a$ . We take the result on trust here; the Orbital Energy and Semi-major Axis note does the full algebra.

Step 4 — How fast are you anywhere? (Vis-viva)

WHAT. Combine Step 2's definition with Step 3's link. Set them equal: $\frac{v ^{2}}{2} - \frac{μ}{r} = - \frac{μ}{2 a}$ Solve for $v^{2}$ (multiply by 2, move the $μ / r$ across): $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ Term by term: $v$ is your speed at the moment you are at distance $r$ ; $a$ is the fixed orbit size; $μ$ scales everything by planet strength. This is the vis-viva equation ("living force").

WHY this tool. We need the periapsis speed $v_{p}$ later, because drag pushes against motion and the amount it steals depends on how fast you were going. Vis-viva is the one formula that hands us $v$ at any point without solving the full motion.

PICTURE. The figure plots $v$ against position around one orbit: a tall spike at periapsis ( $r$ small → $2/ r$ large → $v$ large) and a shallow dip at apoapsis. The red dot marks $v_{p}$ , the fastest point — exactly where the atmosphere waits.

Step 5 — A tiny brake at the bottom: how much does the orbit shrink?

WHAT. Drag removes a small slice of speed at periapsis. Write it $Δ v$ (a negative number, since it's a loss). How much does energy change?

Start from the kinetic part $\frac{1}{2} v^{2}$ . A calculus fact (the "how a square changes" rule): if $v$ changes by a tiny $Δ v$ , then $\frac{1}{2} v^{2}$ changes by $v Δ v$ . At periapsis $v = v_{p}$ , so: $Δ ε = v_{p} Δ v$

$v_{p}$ = periapsis speed (big, from Step 4).
$Δ v$ = speed lost (small, negative).
$Δ ε$ = energy lost (negative → orbit shrinks, as promised).

WHY this tool (why $v Δ v$ , why not just $Δ v$ ?). We ask: "when speed drops a hair, how much energy drops?" Energy depends on speed squared, and the change in a square is not the change in the thing — it's the thing times its change. That factor $v_{p}$ is precisely why braking at high speed (periapsis) is efficient: the same 1 m/s loss removes far more energy than it would at slow apoapsis.

Now turn energy change into size change. From Step 3, $ε = - μ /2 a$ ; the same "how it changes" rule gives $Δ ε = \frac{μ}{2 a ^{2}} Δ a$ . Set the two expressions for $Δ ε$ equal and solve: $Δ a = \frac{2 a ^{2}}{μ} Δ ε = \frac{2 a ^{2} v _{p}}{μ} Δ v$

PICTURE. The figure shows the ellipse before (dashed) and after (solid) one brake. The periapsis point barely moves; the whole oval deflates toward it.

Step 6 — Why the FAR point falls by 4× (the payoff)

WHAT. Recall Step 1: $2 a = r_{p} + r_{a}$ . The brake happens at periapsis, so $r_{p}$ is pinned — it does not change. Take the change of both sides, with $Δ r_{p} = 0$ : $2 Δ a = = 0 Δ r_{p} + Δ r_{a} \Rightarrow Δ r_{a} = 2 Δ a$ Substitute Step 5's $Δ a$ : $Δ r_{a} = \frac{4 a ^{2} v _{p}}{μ} Δ v$ The whole change in $a$ gets doubled into apoapsis because periapsis absorbs none of it.

WHY this is the safe direction. The near side (where thick air lives) stays put; the far side quietly drops each pass. You never accidentally plunge deeper.

PICTURE. The figure fixes periapsis with a green pin and shows the apoapsis arrow retreating inward. A little arrow at the bottom ( $Δ v$ ) causes a long arrow at the top ( $Δ r_{a} = 4 \times$ scale of $Δ a$ ).

Step 7 — The degenerate & edge cases (never let the reader fall off the map)

WHAT / WHY / PICTURE, one panel each in the figure:

Circular orbit ( $r_{p} = r_{a}$ ). Here the ellipse is a perfect circle, $e = 0$ . There is no "far side" distinct from the near side — braking anywhere lowers the whole radius uniformly. Aerobraking is finished here; this is the target.
Brake at apoapsis instead ( $Δ v$ at the top). Flip the logic: now $r_{a}$ is pinned, so $Δ r_{p} = 2Δ a$ — you'd lower periapsis into denser air. Dangerous mid-campaign, but exactly how you end aerobraking: a burn here raises $r_{p}$ out of the atmosphere.
$Δ v = 0$ (a pass above the air). If periapsis skims too high, $ρ \approx 0$ , so no speed is lost and $Δ r_{a} = 0$ . The orbit is unchanged — a wasted, harmless pass.
Too deep ( $Δ v$ too large). Heating $\overset{q}{˙} \propto ρ^{1/2} v^{3}$ and pressure $\propto ρ v^{2}$ explode. The formula still gives a big $Δ r_{a}$ , but the craft may not survive — the reason campaigns stay shallow.

The one-picture summary

Everything above collapses into one chain: remove speed at the bottom → remove energy → shrink $a$ → drop the top by $4 \times$ . The final figure walks the whole causal arrow in a single frame, ellipse and energy bar side by side.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a stone on a string swinging in a big lopsided loop around your head. The closest swoop is periapsis — it's moving fastest there. The far swing-out is apoapsis. Now the stone's total oomph — motion plus height — is one fixed number for the whole loop; call it energy. That single number secretly decides how big the loop is: less energy, tighter loop. So if you brush the stone against something at the fast bottom point and steal a tiny bit of speed, you steal a lot of energy (because energy grows with speed squared, and at the bottom the speed is huge). Stealing energy tightens the loop — but here's the pretty part: since you touched it right at the bottom, the bottom can't move. All the tightening has to show up on the far side, which drops by four times as much as the loop's half-size changed. Do that gently, pass after pass, and your big oval quietly deflates into a neat circle — and you never spent a drop of fuel.

Recall check

Which quantity stays constant over a whole orbit and lets us label it with one number?

The specific mechanical energy

ε = v^{2} /2 - μ / r

Why does a 1 m/s loss at periapsis remove more energy than at apoapsis?

Because

Δ ε = v Δ v

and

v_{p}

(periapsis speed) is large.

A brake at periapsis lowers which point, and by what factor of

Δ a

Apoapsis, by

Δ r_{a} = 2Δ a

(since

r_{p}

is pinned).

How do you safely end an aerobraking campaign?

Brake at apoapsis to raise periapsis out of the atmosphere.