This page drills the phase-angle machinery from the parent note across every case it can throw at you. Before any numbers, we lay out the full grid of scenarios so you can see there are no gaps — then we hit each cell.
We reuse only two equations from the parent, so let us restate them in plain words first.
Two symbols need a picture before we compute. Look at the figure: the chaser (teal) sits at radius r 1 , the target (plum) at r 2 . The phase angle ϕ is the plum-to-teal angle measured at Earth's centre at the launch instant. Positive ϕ = target is ahead (further along the direction of motion); negative ϕ = target is behind .
Every phasing problem is one (or a blend) of these cells. Each row is a distinct behaviour the formula produces — different sign, different limit, or a different real-world flavour.
Cell
What makes it distinct
Sign of ϕ
Covered by
A. Outward, big jump (r 2 ≫ r 1 )
target very slow, moves little during long transfer
large +
Ex 1 (LEO→GEO)
B. Outward, small jump (r 2 ≳ r 1 )
target nearly as fast as arc it must cover
small +
Ex 2
C. Equal radii (r 2 = r 1 )
degenerate: no transfer needed, T syn → ∞
ϕ = 0
Ex 3
D. Inward transfer (r 2 < r 1 )
target is faster , covers more than π
− (trails)
Ex 4
E. Extreme inward (r 2 ≪ r 1 )
target laps the arrival point, ϕ very negative
large −
Ex 5
F. Synodic — near-equal orbits
tiny $
n_1-n_2
$ → huge window gap
G. Real-world word problem
Mars departure phasing, planetary μ
+
Ex 7
H. Exam twist — negative-to-positive wrap
how to report ϕ as a physical lead in [ 0 , 2 π )
wrap
Ex 8
Intuition Why sign flips at
r 2 = r 1
The target sweeps Δ θ 2 = π ( a t / r 2 ) 3/2 during the transfer, and the chaser always sweeps exactly π . If the target sweeps less than π it lags the meeting point, so it must start ahead (ϕ > 0 ). If it sweeps more than π it would overshoot, so it must start behind (ϕ < 0 ). The crossover is exactly Δ θ 2 = π , which happens when a t = r 2 , i.e. r 1 = r 2 . That is why outward = lead, inward = trail .
Throughout, μ ⊕ = 3.986 × 1 0 5 km 3 / s 2 unless a different body is named.
Worked example Example 1 — Cell A: LEO → GEO (big outward jump)
Chaser at r 1 = 6 678 km (300 km LEO), target at r 2 = 42 164 km (GEO).
Forecast: GEO satellites crawl. Guess: does the target need to be far ahead or just ahead?
a t = 2 6678 + 42164 = 24 421 km. Why this step? Semi-major axis is the mean of perigee (r 1 ) and apogee (r 2 ).
Ratio 2 r 2 r 1 + r 2 = r 2 a t = 42164 24421 = 0.5792 . Why? This ratio is exactly the fraction that controls how much the target sweeps.
( 0.5792 ) 3/2 = 0.4408 . Why the 3/2 power? It comes from Kepler: sweep angle ∝ ( a t / r 2 ) 3/2 .
ϕ = π ( 1 − 0.4408 ) = π ( 0.5592 ) = 1.757 rad = 100. 6 ∘ . Why? ϕ = π − Δ θ 2 and Δ θ 2 = π ( 0.4408 ) = 79. 4 ∘ ... wait — check: target sweeps π ⋅ 0.4408 = 1.385 rad = 79. 4 ∘ , so it must lead by 180 − 79.4 = 100. 6 ∘ . ✓
Verify: Δ θ 2 ≈ 79. 4 ∘ < 18 0 ∘ → target lags → ϕ > 0 . Positive large lead, matches Cell A. The GEO bird sits about 10 0 ∘ ahead at launch.
Worked example Example 2 — Cell B: small outward jump
r 1 = 6 700 km, r 2 = 7 000 km.
Forecast: radii almost equal. Will ϕ be tiny or big?
a t = ( 6700 + 7000 ) /2 = 6850 km. Why? mean of the two radii.
r 2 a t = 7000 6850 = 0.97857 . Why? the controlling ratio again.
( 0.97857 ) 3/2 = 0.96820 . Why 3/2 ? Kepler sweep exponent.
ϕ = π ( 1 − 0.96820 ) = π ( 0.03180 ) = 0.0999 rad = 5.7 2 ∘ . Why positive? target sweeps just under π , so a small lead is enough.
Verify: small positive lead — sits between Cell C (0 ∘ ) and Cell A (10 0 ∘ ), exactly as a small-jump case should. Units: radians → degrees via × 180/ π . ✓
Worked example Example 3 — Cell C: equal radii (degenerate)
r 1 = r 2 = 7 000 km.
Forecast: you are already on the target's orbit. What should ϕ be? And how often does the window come?
2 r 2 r 1 + r 2 = 2 r 2 2 r 2 = 1 . Why? both radii identical.
1 3/2 = 1 , so ϕ = π ( 1 − 1 ) = 0 . Why? no transfer arc means no lead is required — you meet only if already co-located.
Synodic: n 1 = n 2 , so ∣ n 1 − n 2 ∣ = 0 and T syn = 2 π /0 → ∞ . Why? identical speeds never drift relative to each other.
Verify: ϕ = 0 is the crossover of the sign matrix; T syn = ∞ says "if you are not already lined up, a pure co-orbital Hohmann never re-phases you" — which is why co-orbital rendezvous needs a phasing orbit (deliberately change r to build drift). ✓
Worked example Example 4 — Cell D: inward transfer (target faster)
Chaser at r 1 = 42 164 km (GEO), target at r 2 = 6 678 km (LEO). We descend.
Forecast: the inner target is faster . Should it lead or trail us at launch?
a t = ( 42164 + 6678 ) /2 = 24 421 km. Why? same semi-major axis as Ex 1 — the ellipse is the same shape, just travelled the other way.
r 2 a t = 6678 24421 = 3.657 . Why >1? now a t > r 2 because the target radius is the small one.
( 3.657 ) 3/2 = 6.993 . Why 3/2 ? Kepler sweep exponent, same rule.
ϕ = π ( 1 − 6.993 ) = π ( − 5.993 ) = − 18.83 rad . Why negative? target sweeps far more than π (it laps many times) so it must start behind .
Verify: ϕ < 0 → target trails, exactly Cell D. The raw radians (− 18.83 ) exceed 2 π because the fast LEO target loops the arrival point several times during our slow descent; the physical lead is ϕ mod 2 π (see Ex 8). ✓
Worked example Example 5 — Cell E: extreme inward
r 1 = 42 164 km (GEO), r 2 = 6 678 km, but imagine a hypothetically even smaller r 2 = 1 700 km (skimming a tiny moon).
Forecast: tiniest inner orbit = fastest target. Expect the most negative ϕ ?
a t = ( 42164 + 1700 ) /2 = 21 932 km. Why? mean radius.
r 2 a t = 1700 21932 = 12.90 . Why so large? target radius is tiny.
( 12.90 ) 3/2 = 46.34 . Why 3/2 ? Kepler.
ϕ = π ( 1 − 46.34 ) = − 142.4 rad . Why hugely negative? the tiny fast orbit laps the meeting point ~22 times during the descent.
Verify: magnitude far exceeds Ex 4's 18.83 — smaller inner radius → more laps → more negative, monotone as the matrix predicts for Cell E. ✓
Worked example Example 6 — Cell F: synodic period, ISS-like chase
Chaser at r 1 = 6 778 km (400 km), target at r 2 = 6 795 km.
Forecast: almost identical orbits. Will the window come every orbit, or rarely?
n 1 = μ / r 1 3 = 3.986 × 1 0 5 /677 8 3 = 1.1315 × 1 0 − 3 rad/s . Why? mean motion from Mean Motion and Orbital Period .
n 2 = 3.986 × 1 0 5 /679 5 3 = 1.1272 × 1 0 − 3 rad/s . Why? same rule, larger radius → slower.
∣ n 1 − n 2 ∣ = 4.25 × 1 0 − 6 rad/s . Why subtract? relative drift rate is the difference of angular speeds.
T syn = 2 π /4.25 × 1 0 − 6 = 1.48 × 1 0 6 s ≈ 17.1 days. Why so long? nearly equal n → tiny drift → 2 π of relative drift takes ages.
Verify: ≈ 17 days matches the parent note's ISS figure. This is why real ISS visits use multiple phasing orbits to speed up the alignment rather than waiting a full Synodic Period . ✓
Worked example Example 7 — Cell G: real-world word problem (Earth → Mars)
A probe leaves Earth's orbit (r 1 = 1.496 × 1 0 8 km) for Mars (r 2 = 2.279 × 1 0 8 km) around the Sun, μ ⊙ = 1.327 × 1 0 11 km 3 / s 2 . What lead angle must Mars have at departure?
Forecast: Mars is slower than Earth; guess a large positive lead like the GEO case.
a t = ( 1.496 + 2.279 ) × 1 0 8 /2 = 1.8875 × 1 0 8 km. Why? mean of the two heliocentric radii.
r 2 a t = 2.279 1.8875 = 0.8282 . Why? controlling ratio (note μ ⊙ does not appear — it cancels).
( 0.8282 ) 3/2 = 0.7538 . Why 3/2 ? Kepler sweep exponent.
ϕ = π ( 1 − 0.7538 ) = π ( 0.2462 ) = 0.7735 rad = 44. 3 ∘ . Why positive? Mars sweeps less than π during the ~259-day transfer, so it must lead.
Verify: the famous real Mars-transfer phase angle is about 4 4 ∘ — our clean geometric result lands there. μ ⊙ cancelled exactly as the parent's mistake-callout warned. ✓
Worked example Example 8 — Cell H: exam twist, wrapping a negative
ϕ into a physical lead
From Ex 4 we got ϕ = − 18.83 rad for the GEO→LEO descent. An exam asks: "State the physical angular position of the target at launch, in the range [ 0 , 36 0 ∘ ) ."
Forecast: raw radians are meaningless past one lap. What survives is ϕ modulo 2 π .
Reduce mod 2 π : − 18.83 + 4 ( 2 π ) = − 18.83 + 25.133 = 6.303 rad. Why add 2 π repeatedly? angular position only matters up to full turns.
Still > 2 π ? 6.303 − 2 π = 6.303 − 6.283 = 0.0199 rad. Why again? 6.303 > 2 π = 6.283 , one more wrap.
Convert: 0.0199 × 180/ π = 1.1 4 ∘ . Why? radians → degrees.
Verify: the physical lead is ≈ 1.1 4 ∘ — a small positive angle, showing that even a wildly negative raw ϕ reduces to a concrete launch geometry. Always report ϕ mod 36 0 ∘ on exams. ✓
Recall Self-test the matrix
Which cell gives ϕ = 0 ? ::: Cell C — equal radii (r 1 = r 2 ), no transfer arc needed.
Outward transfer sign of ϕ ? ::: Positive — target must lead.
Inward transfer sign of ϕ ? ::: Negative — target must trail (it is faster and sweeps more than π ).
Why is the Earth→Mars phase angle independent of μ ⊙ ? ::: Because μ cancels in the ratio; ϕ is pure geometry of radii.
How do you report a ϕ below − 2 π ? ::: Reduce modulo 2 π into [ 0 , 2 π ) to get the physical launch lead.
Why is ISS T syn ≈ 17 days? ::: Chaser and target radii nearly equal → tiny ∣ n 1 − n 2 ∣ → very slow relative drift.
Mnemonic The one-line rule
Out = lead (+), In = trail (−), Equal = zero, Near-equal = wait forever.