Launch window — phasing with target orbit: Visual walkthrough

Step 1 — Two circles and a "lead" between two runners

WHAT: we place both runners on their tracks and mark the angle between them. WHY: rendezvous is entirely about angles at the centre — who is ahead, by how much. Distances only matter through how they set the speeds (Step 3). PICTURE: the two runners, the centre, and the wedge-shaped angle $ϕ$ from chaser to target.

Related vault ideas: Hohmann Transfer Orbit, Rendezvous and Phasing Orbits.

Step 2 — The bridge between the circles: the transfer half-ellipse

WHAT: we connect the two circles with the cheapest possible arc — a half-ellipse (a Hohmann Transfer Orbit). WHY: we cannot teleport from the inner to the outer circle; we must coast along a curve that meets both. The Hohmann half-ellipse is the fuel-cheapest such curve. PICTURE: perigee at $r_{1}$ , apogee at $r_{2}$ , the dashed half-ellipse joining them.

Step 3 — How fast does each runner sweep? (mean motion)

We need a "sweeping speed": how many radians per second a runner covers around the centre.

WHAT: we compute $n_{1}$ and $n_{2}$ , one for each circle, from its radius. WHY: to know where the target drifts to during our flight, we must know how fast it sweeps. This is the only place the planet's strength enters — and, spoiler, it will cancel. PICTURE: two runners with speed arrows; the inner one's arrow is longer (inner orbits are faster).

Step 4 — How long is the flight? (transfer time)

Apply this to our transfer ellipse (semi-major axis $a_{t}$ ). Call the time for one full loop of that ellipse $T_{ellipse}$ : $T_{ellipse} = 2 π \frac{a _{t}^{3}}{μ}$

$T_{ellipse}$ = the period of the transfer ellipse — how long it would take to fly the whole ellipse, both halves.

WHAT: we take half of $T_{ellipse}$ . WHY: the Hohmann path is only near-touch → far-touch — exactly half a loop. By the ellipse's mirror symmetry about its major axis, that half takes half the time. PICTURE: the full faint ellipse split by its major axis; only the top half (our coast) is solid, labelled "half the period."

Step 5 — How far does the target drift while we fly?

WHAT: multiply the target's sweep-rate $n_{2}$ (from Step 3) by the flight time $t_{t}$ (from Step 4). WHY: angle covered = rate × time. This is the target's head-start consumption during our coast. PICTURE: the target at launch, an arc $Δ θ_{2}$ , and the target at arrival.

$Δ θ_{2} = n_{2} t_{t} = target rate n_{2} \frac{μ}{r _{2}^{3}} \cdot flight time t_{t} π \frac{a _{t}^{3}}{μ} = π (\frac{a _{t}}{r _{2}})^{3/2} = π (\frac{r _{1} + r _{2}}{2 r _{2}})^{3/2}$

$Δ θ_{2}$ = radians the target sweeps during our coast.
$\frac{r _{1} + r _{2}}{2 r _{2}} = \frac{a _{t}}{r _{2}}$ = transfer-size ÷ target-size.
Outward chase ( $r_{2} > r_{1}$ , our running picture): then $a_{t} = \frac{r _{1} + r _{2}}{2} < r_{2}$ , so the ratio is $< 1$ and $Δ θ_{2} < π$ . The slow outer target covers less than a half-lap.
Inward chase ( $r_{2} < r_{1}$ ): then $a_{t} = \frac{r _{1} + r _{2}}{2} > r_{2}$ , so the ratio is $> 1$ and $Δ θ_{2} > π$ . We return to this in Step 7 — do not assume $Δ θ_{2} \leq π$ in general.

Step 6 — The rendezvous condition (the payoff)

WHAT: we demand that when we reach the far touch-point (apogee of the transfer), the target is there too. WHY: the chaser flies exactly $π$ radians (half the sky) from one touch-point to the other. To meet, the target — which starts $ϕ$ ahead — must, after drifting $Δ θ_{2}$ , land on that same arrival point. PICTURE: launch snapshot (target leads by $ϕ$ ) beside arrival snapshot (both at apogee).

At arrival both sit at the arrival point. Bookkeeping of angles (all in radians): $target’s head start ϕ + target drifts Δ θ_{2} = chaser’s sweep π$

Solve for the head start:

$ϕ = π - Δ θ_{2} = π [1 - (\frac{r _{1} + r _{2}}{2 r _{2}})^{3/2}]$

Step 7 — Every case: outer target, inner target, degenerate

The sign of $ϕ$ flips depending on where the target sits. We must cover all of them.

WHAT: test the ratio $\frac{r _{1} + r _{2}}{2 r _{2}}$ against $1$ . WHY: the ratio decides whether $Δ θ_{2}$ is less than, equal to, or greater than $π$ — which sets the sign of $ϕ$ . PICTURE: three panels — outward chase ( $ϕ > 0$ ), same-orbit degenerate ( $ϕ = 0$ ), inward chase ( $ϕ < 0$ ).

Case	Radii	Ratio $\frac{r _{1} + r _{2}}{2 r _{2}}$	$Δ θ_{2}$	$ϕ$	Meaning
Outward chase	$r_{2} > r_{1}$	$< 1$	$< π$	$> 0$	target leads
Same orbit (degenerate)	$r_{2} = r_{1}$	$= 1$	$= π$	$= 0$	already together; no lead
Inward chase	$r_{2} < r_{1}$	$> 1$	$> π$	$< 0$	target trails

Step 8 — How often does the window come back? (synodic period)

WHAT: find how long until this exact phasing repeats. WHY: if you miss the instant, you wait for the geometry to return — that wait is the Synodic Period. PICTURE: the two runners drifting apart; a full $2 π$ of relative drift returns the same picture.

The one-picture summary

One frame carries the whole story: two circles, the half-ellipse bridge, the chaser sweeping $π$ , the target eating $Δ θ_{2} = π (\frac{r _{1} + r _{2}}{2 r _{2}})^{3/2}$ , and the launch-instant lead $ϕ = π - Δ θ_{2}$ that makes them collide at the arrival point.

Recall Feynman retelling — the whole walkthrough in plain words

You and a slower friend run on two circular tracks around a lake, your friend on the outer one. You want to throw a ball so it lands on your friend — but you must throw along a curved path (the half-ellipse) that starts at your track and ends at theirs. That curved throw takes a fixed number of seconds ( $t_{t}$ ). During those seconds your friend keeps running and covers some arc ( $Δ θ_{2}$ ). Meanwhile the ball itself swings exactly halfway around the lake ( $π$ radians). For the ball and your friend to arrive together, your friend must start far enough ahead so that "their start-lead plus how far they run equals the half-lap the ball travels": $ϕ + Δ θ_{2} = π$ . That leftover lead $ϕ = π - Δ θ_{2}$ is the launch window. If your friend is on a bigger, slower track, they barely move, so they must start almost a quarter-lap ahead (LEO→GEO ≈ 100°). If they're on your own track, no lead is needed ( $ϕ = 0$ ). If they're on an inner, faster track, they run more than half a lap and you launch when they're behind you ( $ϕ < 0$ ). And how strong the lake's "gravity" ( $μ$ ) is never matters for the angle — it speeds up ball and friend equally, so it cancels. Miss the moment? Just wait one synodic period and the picture comes back.

Recall Quick self-check

Why does the target sweep less than $π$ for an outer orbit? ::: Because $r_{2} > r_{1}$ makes $a_{t} = (r_{1} + r_{2}) /2 < r_{2}$ , so the ratio is $< 1$ ; the slower outer orbit covers only part of a half-lap in the flight time. What equation states the rendezvous condition? ::: $ϕ + Δ θ_{2} = π$ — the target's lead plus its drift equals the chaser's half-lap (radians). Sign of $ϕ$ for an inner target ( $r_{2} < r_{1}$ )? ::: Negative — the target trails, because it sweeps more than $π$ during transfer.

See also: Lambert's Problem (the general "connect two points in a chosen time" problem this special-cases), Orbital Plane and Inclination (the other half of a real launch window).