Launch window — phasing with target orbit: Question bank

Before we start, every symbol used below gets earned right here — no term appears unexplained.

Definition The symbols this page uses (read once, refer back)

Picture two circular tracks around Earth's centre, one inside the other.

$r_{1}$ ::: the chaser's orbit radius (the inner runner, measured from Earth's centre).
$r_{2}$ ::: the target's orbit radius (the outer runner, from Earth's centre).
$a_{t}$ ::: the transfer ellipse's semi-major axis, $a_{t} = \frac{r _{1} + r _{2}}{2}$ — its perigee touches $r_{1}$ , its apogee touches $r_{2}$ , so it is the average of the two radii.
$t_{t}$ ::: the transfer time — how long the chaser coasts along the half-ellipse, $t_{t} = π a_{t}^{3} / μ$ (half a Hohmann-ellipse period). See Hohmann Transfer Orbit.
$n_{1}, n_{2}$ ::: the mean motions (angular speeds) of chaser and target, $n_{i} = μ / r_{i}^{3}$ — how fast each runner sweeps angle around Earth's centre. See Mean Motion and Orbital Period.
$μ$ ::: Earth's gravitational parameter (mass $\times$ $G$ ); it sets how fast things orbit.

Recall The three phasing symbols

$ϕ$ ::: the phase angle — how far ahead (in degrees or radians, from Earth's centre) the target must sit at the launch instant; closed form $ϕ = π [1 - (\frac{r _{1} + r _{2}}{2 r _{2}})^{3/2}]$ . $Δ θ_{2}$ ::: the arc the target sweeps during the transfer coast, $Δ θ_{2} = n_{2} t_{t}$ ; and $ϕ = π - Δ θ_{2}$ . $T_{syn}$ ::: the synodic period, $T_{syn} = \frac{2 π}{∣ n _{1} - n _{2} ∣}$ — how often the same relative geometry (and thus the same window) comes back.

Intuition Why

$ϕ = π - Δ θ_{2}$ — the picture Look at the figure. The chaser leaves the inner orbit at the bottom and coasts half an ellipse to the top — that arc from perigee to apogee is always exactly $π$ (180°), because half an ellipse is half the sky no matter how stretched it is. The meeting point is that top apogee. So the target, starting at lead angle $ϕ$ , must sweep the remaining arc $Δ θ_{2}$ to arrive at the top at the same moment. Since chaser-arc + nothing = target-start + target-arc land on the same point: $ϕ + Δ θ_{2} = π$ , i.e. $ϕ = π - Δ θ_{2}$ .

In the figure: the two black circles are the chaser ( $r_{1}$ ) and target ( $r_{2}$ ) orbits; the red half-ellipse is the transfer path; the black dot at the bottom is the chaser's start (perigee) and the red dot at the top is the meeting point (apogee). The black dot on the outer circle is the target at launch, sitting a lead angle $ϕ$ ahead; the short red arc it must still travel is $Δ θ_{2}$ . Read straight off the picture: $ϕ + Δ θ_{2} = π$ .

True or false — justify

True or false: The launch window is decided entirely by when the rocket is fuelled and ready.

False — readiness is necessary but not sufficient; the window is set by geometry (site under the plane) and timing (target at lead angle

ϕ

). A ready rocket at the wrong instant wastes huge fuel.

True or false: You should aim the transfer orbit at the target's current position.

False — the target moves several km/s during the coast, so you aim at its future position; that is exactly what the lead angle

ϕ

encodes.

True or false: A heavier central body (larger

μ

) changes the required phase angle

ϕ

False —

μ

cancels in

ϕ

because

ϕ

is a ratio of geometric radii. Larger

μ

only shortens the transfer time and synodic period, not the angle.

True or false: For a chaser going to a higher orbit (

r_{2} > r_{1}

), the target is always ahead of the chaser at launch.

False — the blanket "always" is wrong because

ϕ

can change sign; it stays positive (target leads) only while

Δ θ_{2} < π

. For

r_{2}

just above

r_{1}

that holds and the lead is small, but the claim as stated is not guaranteed.

True or false: For an inner target (

r_{2} < r_{1}

) the required

ϕ

can be negative, meaning the target trails the chaser at launch.

True — an inner target moves faster and sweeps more than

π

during the transfer, so

Δ θ_{2} > π

and

ϕ = π - Δ θ_{2} < 0

, i.e. the target starts behind.

True or false: Two nearly-equal circular orbits give a very short synodic period, so windows come often.

False — nearly-equal radii mean nearly-equal mean motions, so relative drift

∣ n_{1} - n_{2} ∣

is tiny; the geometry repeats slowly, giving a long

T_{syn}

(why ISS chases take days, not minutes).

True or false: Matching the target's orbital plane is enough to define the launch instant.

False — plane match fixes direction only (the site must rotate under the plane). You still need phasing so you arrive when the target is at the meeting point. Both must coincide.

True or false: The chaser always sweeps exactly

18 0^{\circ}

during a Hohmann transfer regardless of the radii.

True — a Hohmann transfer is half an ellipse from perigee to apogee, which is always half the sky, independent of

r_{1}

and

r_{2}

Spot the error

Spot the error: "The target sweeps

Δ θ_{2}

, the chaser sweeps

Δ θ_{2}

too, so they meet automatically."

The chaser always sweeps

π

(half ellipse), not

Δ θ_{2}

. Rendezvous requires the target to arrive at the apogee point, so

ϕ = π - Δ θ_{2}

— the two arcs are generally different.

Spot the error: "

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

The denominator should be

2 r_{2}

, not

2 r_{1}

(compare the boxed closed form above) — the arc

Δ θ_{2}

is the target's motion, so it is scaled by the target radius

r_{2}

(via

n_{2} = μ / r_{2}^{3}

Spot the error: "Since

μ

appears in

t_{t}

and in

n_{2}

, it must appear in

ϕ

When you multiply

n_{2} t_{t}

, the

μ

n_{2} = μ / r_{2}^{3}

cancels the

μ

in the transfer time

t_{t} = π a_{t}^{3} / μ

(with

a_{t} = (r_{1} + r_{2}) /2

), leaving a pure geometric ratio. Appearing in the pieces does not mean surviving in the product.

Spot the error: "The synodic period is

T_{syn} = 2 π / (n_{1} + n_{2})

It is

2 π /∣ n_{1} - n_{2} ∣

— both bodies orbit the same way, so what matters is the relative rate (a difference), not the sum. The sum would describe counter-rotating bodies.

Spot the error: "GEO is far away, so the transfer is long, so the phase angle is small."

The opposite — a long transfer means the slow GEO target barely moves (

Δ θ_{2} \approx 7 9^{\circ}

), so it must start far ahead (

ϕ \approx 10 1^{\circ}

). Long transfer ⇒ large lead, not small.

Spot the error: "We need one Hohmann burn timed perfectly, so phasing orbits are unnecessary for the ISS."

The ISS synodic period is ~days; a single Hohmann almost never lines up exactly, so real missions use several phasing orbits to nudge

ϕ

into place. See Rendezvous and Phasing Orbits.

Why questions

Why must the launch be led rather than aimed straight at the target?

Because the coast time

t_{t}

is non-zero and the target keeps orbiting; you meet at the future point, and the required head-start is exactly

ϕ = π - Δ θ_{2}

Why does the phase-angle formula contain the exponent

3/2

It comes from Kepler's third law — orbital period scales as

a^{3/2}

(see Kepler's Third Law). The ratio of transfer time to target period inherits that same

3/2

power.

Why does the window repeat, and repeat with period

T_{syn}

specifically?

The two bodies drift apart at relative rate

∣ n_{1} - n_{2} ∣

; when they have drifted a full

2 π

the same relative geometry — and hence the same

ϕ

— recurs, taking time

2 π /∣ n_{1} - n_{2} ∣

Why can't a plane change simply be avoided by launching a bit early or late?

Launching early/late only shifts phasing, not the plane; the plane condition is set by Earth's rotation carrying the site under the target's orbital plane, an independent geometric fact (see Orbital Plane and Inclination).

Why is

ϕ

measured at the central body (Earth's centre) and not at the launch site?

Angular position on an orbit is defined about the focus (Earth's centre); measuring from the surface would mix in Earth's radius and rotation, muddying the clean geometric relation.

Why does an inner target need to be behind the chaser at launch?

An inner orbit is faster, so during the transfer the target overtakes more than

π

of arc; starting behind lets it swing around to meet the chaser at the transfer's other end.

Edge cases

Edge case:

r_{1} = r_{2}

(same orbit). What is

ϕ

The ratio

\frac{r _{1} + r _{2}}{2 r _{2}} = 1

, so

1^{3/2} = 1

and

ϕ = π (1 - 1) = 0

. No lead needed — but the transfer is degenerate (no altitude change), so this is a phasing-only problem, not a Hohmann.

Edge case:

r_{1} = r_{2}

— what happens to the synodic period?

n_{1} = n_{2}

∣ n_{1} - n_{2} ∣ = 0

and

T_{syn} \to \infty

. Identical orbits never drift, so the "window" is either always open or never repeats — you must actively change your orbit to phase (a phasing orbit).

Edge case: the target orbit is very much larger than the chaser's (

r_{2} ≫ r_{1}

). What does

ϕ

approach?

The ratio

\frac{r _{1} + r _{2}}{2 r _{2}} \to \frac{1}{2}

, so

ϕ \to π (1 - 0.354) \approx 0.646 π \approx 11 6^{\circ}

— the slow distant target must lead by well over

9 0^{\circ}

Edge case: what if the plane condition and the phasing condition never coincide in a given day?

Then there is no window that day; you wait until the phasing (period

T_{syn}

) and the ground-track alignment (period ~1 sidereal day) drift back into agreement, which may take many cycles.

Edge case: the transfer is not a single Hohmann but a bi-elliptic or Lambert arc — does

ϕ = π - Δ θ_{2}

still hold?

No — that formula assumes the chaser sweeps exactly

π

. For other geometries you replace

π

with the chaser's actual swept angle and

Δ θ_{2}

with the target's motion over the true transfer time; the general timing problem is Lambert's Problem.

Edge case: could

ϕ

exceed

18 0^{\circ}

in magnitude?

Δ θ_{2}

is negative or the target sweeps past a full extra revolution, magnitudes can exceed

18 0^{\circ}

; in practice

ϕ

is reduced modulo

36 0^{\circ}

to the physically nearest lead, since only position on the circle matters.