Launch window — phasing with target orbit: Worked examples

3.2.39 · D3 · Physics › Orbital Mechanics & Astrodynamics › Launch window — phasing with target orbit

Yeh page parent note ki phase-angle machinery ko har possible case ke through drill karti hai. Numbers se pehle, hum scenarios ka pura grid lay out karte hain taaki aap dekh sako ki koi gap nahi hai — phir hum har cell ko hit karte hain.

Hum parent se sirf do equations reuse karte hain, toh aayiye pehle unhe plain words mein restate karte hain.

Do symbols ko compute karne se pehle ek picture chahiye. Figure dekho: chaser (teal) radius $r_{1}$ par hai, target (plum) $r_{2}$ par. Phase angle $ϕ$ launch instant par Earth ke centre par measure kiya gaya plum-to-teal angle hai. Positive $ϕ$ = target aage hai (motion ki direction mein aage); negative $ϕ$ = target peeche hai.

Figure — Launch window — phasing with target orbit

The scenario matrix

Har phasing problem in cells mein se ek (ya blend) hoti hai. Har row ek alag behaviour hai jo formula produce karta hai — alag sign, alag limit, ya alag real-world flavour.

Cell	Kya cheez ise distinct banati hai	Sign of $ϕ$	Covered by
A. Outward, big jump ( $r_{2} ≫ r_{1}$ )	target bahut slow, long transfer mein kam hilta hai	large +	Ex 1 (LEO→GEO)
B. Outward, small jump ( $r_{2} ≳ r_{1}$ )	target almost utni hi fast jitna arc cover karna hai	small +	Ex 2
C. Equal radii ( $r_{2} = r_{1}$ )	degenerate: koi transfer nahi chahiye, $T_{syn} \to \infty$	$ϕ = 0$	Ex 3
D. Inward transfer ( $r_{2} < r_{1}$ )	target faster hai, $π$ se zyada cover karta hai	− (trails)	Ex 4
E. Extreme inward ( $r_{2} ≪ r_{1}$ )	target arrival point ko lap kar leta hai, $ϕ$ bahut 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	$ϕ$ ko physical lead ke roop mein $[0, 2 π)$ mein kaise report karein	wrap	Ex 8

The worked examples

Poore mein, $μ_{\oplus} = 3.986 \times 1 0^{5} km^{3} / s^{2}$ jab tak koi alag body na bataya jaaye.

Worked example Example 1 — Cell A: LEO → GEO (big outward jump)

Chaser at $r_{1} = 6678$ km (300 km LEO), target at $r_{2} = 42164$ km (GEO). Forecast: GEO satellites bahut slow chalte hain. Guess karo: kya target ko bahut aage hona chahiye ya thoda aage?

$a_{t} = \frac{6678 + 42164}{2} = 24421$ km. Yeh step kyun? Semi-major axis perigee ( $r_{1}$ ) aur apogee ( $r_{2}$ ) ka mean hota hai.
Ratio $\frac{r _{1} + r _{2}}{2 r _{2}} = \frac{a _{t}}{r _{2}} = \frac{24421}{42164} = 0.5792$ . Kyun? Yeh ratio exactly woh fraction hai jo control karta hai ki target kitna sweep karta hai.
$(0.5792)^{3/2} = 0.4408$ . $3/2$ power kyun? Yeh Kepler se aata hai: sweep angle $\propto (a_{t} / r_{2})^{3/2}$ .
$ϕ = π (1 - 0.4408) = π (0.5592) = 1.757 rad = 100. 6^{\circ}$ . Kyun? $ϕ = π - Δ θ_{2}$ aur $Δ θ_{2} = π (0.4408) = 79. 4^{\circ}$ ... ruko — check karo: target sweeps $π \cdot 0.4408 = 1.385$ rad $= 79. 4^{\circ}$ , toh use $180 - 79.4 = 100. 6^{\circ}$ lead karna hoga. ✓

Verify: $Δ θ_{2} \approx 79. 4^{\circ} < 18 0^{\circ}$ → target lags → $ϕ > 0$ . Positive large lead, Cell A se match karta hai. GEO bird launch par lagbhag $10 0^{\circ}$ aage hota hai.

Worked example Example 3 — Cell C: equal radii (degenerate)

$r_{1} = r_{2} = 7000$ km. Forecast: aap already target ki orbit par ho. $ϕ$ kya hona chahiye? Aur window kitni baar aayegi?

$\frac{r _{1} + r _{2}}{2 r _{2}} = \frac{2 r _{2}}{2 r _{2}} = 1$ . Kyun? dono radii identical hain.
$1^{3/2} = 1$ , toh $ϕ = π (1 - 1) = 0$ . Kyun? koi transfer arc nahi matlab koi lead required nahi — aap milte ho sirf tab jab already co-located ho.
Synodic: $n_{1} = n_{2}$ , toh $∣ n_{1} - n_{2} ∣ = 0$ aur $T_{syn} = 2 π /0 \to \infty$ . Kyun? identical speeds kabhi ek doosre ke relative drift nahi karte.

Verify: $ϕ = 0$ sign matrix ka crossover hai; $T_{syn} = \infty$ kehta hai "agar aap already lined up nahi ho, toh ek pure co-orbital Hohmann kabhi re-phase nahi karega" — isliye co-orbital rendezvous ko phasing orbit ki zarurat hoti hai (drift build karne ke liye deliberately $r$ change karo). ✓

Worked example Example 4 — Cell D: inward transfer (target faster)

Chaser at $r_{1} = 42164$ km (GEO), target at $r_{2} = 6678$ km (LEO). Hum descend karte hain. Forecast: inner target faster hai. Kya use launch par humse aage hona chahiye ya peeche?

$a_{t} = (42164 + 6678) /2 = 24421$ km. Kyun? Ex 1 jaisa hi semi-major axis — ellipse same shape ka hai, bas doosri taraf travel kiya gaya.
$\frac{a _{t}}{r _{2}} = \frac{24421}{6678} = 3.657$ . >1 kyun? ab $a_{t} > r_{2}$ kyunki target radius small wali hai.
$(3.657)^{3/2} = 6.993$ . $3/2$ kyun? Kepler sweep exponent, same rule.
$ϕ = π (1 - 6.993) = π (- 5.993) = - 18.83 rad$ . Negative kyun? target $π$ se bahut zyada sweep karta hai (woh kai baar lap karta hai) toh use peeche start karna hoga.

Verify: $ϕ < 0$ → target trails, exactly Cell D. Raw radians ( $- 18.83$ ) $2 π$ se zyada hain kyunki fast LEO target hamare slow descent ke dauran arrival point ko kai baar loop karta hai; physical lead $ϕ mod 2 π$ hai (Ex 8 dekhein). ✓

Worked example Example 6 — Cell F: synodic period, ISS-like chase

Chaser at $r_{1} = 6778$ km (400 km), target at $r_{2} = 6795$ km. Forecast: almost identical orbits. Kya window har orbit aayegi, ya rarely?

$n_{1} = μ / r_{1}^{3} = 3.986 \times 1 0^{5} /677 8^{3} = 1.1315 \times 1 0^{- 3} rad/s$ . Kyun? Mean Motion and Orbital Period se mean motion.
$n_{2} = 3.986 \times 1 0^{5} /679 5^{3} = 1.1272 \times 1 0^{- 3} rad/s$ . Kyun? same rule, bada radius → slower.
$∣ n_{1} - n_{2} ∣ = 4.25 \times 1 0^{- 6} rad/s$ . Kyun subtract? relative drift rate angular speeds ka difference hai.
$T_{syn} = 2 π /4.25 \times 1 0^{- 6} = 1.48 \times 1 0^{6} s \approx 17.1$ days. Itna lamba kyun? nearly equal $n$ → tiny drift → $2 π$ ka relative drift bahut time leta hai.

Verify: $\approx 17$ days parent note ke ISS figure se match karta hai. Isliye real ISS visits multiple phasing orbits use karte hain alignment speed up karne ke liye, bajaye puri Synodic Period wait karne ke. ✓

Worked example Example 7 — Cell G: real-world word problem (Earth → Mars)

Ek probe Earth ki orbit ( $r_{1} = 1.496 \times 1 0^{8}$ km) se Mars ( $r_{2} = 2.279 \times 1 0^{8}$ km) ke liye Sun ke around jaata hai, $μ_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ . Departure par Mars ka lead angle kya hona chahiye? Forecast: Mars Earth se slower hai; GEO case ki tarah ek large positive lead guess karo.

$a_{t} = (1.496 + 2.279) \times 1 0^{8} /2 = 1.8875 \times 1 0^{8}$ km. Kyun? do heliocentric radii ka mean.
$\frac{a _{t}}{r _{2}} = \frac{1.8875}{2.279} = 0.8282$ . Kyun? controlling ratio (note karo $μ_{⊙}$ appear nahi karta — woh cancel ho jaata hai).
$(0.8282)^{3/2} = 0.7538$ . $3/2$ kyun? Kepler sweep exponent.
$ϕ = π (1 - 0.7538) = π (0.2462) = 0.7735 rad = 44. 3^{\circ}$ . Positive kyun? Mars ~259-day transfer ke dauran $π$ se kam sweep karta hai, toh use lead karna hoga.

Verify: famous real Mars-transfer phase angle lagbhag $4 4^{\circ}$ hai — hamara clean geometric result wahan pahunchta hai. $μ_{⊙}$ exactly cancel hua jaisa parent ke mistake-callout ne warn kiya tha. ✓

Worked example Example 8 — Cell H: exam twist, negative

$ϕ$ ko physical lead mein wrap karna Ex 4 se hamein GEO→LEO descent ke liye $ϕ = - 18.83$ rad mila. Ek exam puchta hai: "Launch par target ki physical angular position $[0, 36 0^{\circ})$ range mein batao." Forecast: raw radians ek lap ke baad meaningless hain. Jo bachta hai woh hai $ϕ$ modulo $2 π$ .

Mod $2 π$ reduce karo: $- 18.83 + 4 (2 π) = - 18.83 + 25.133 = 6.303$ rad. $2 π$ baar baar kyun add karo? angular position sirf full turns tak matter karta hai.
Abhi bhi $> 2 π$ hai? $6.303 - 2 π = 6.303 - 6.283 = 0.0199$ rad. Phir kyun? $6.303 > 2 π = 6.283$ , ek aur wrap.
Convert karo: $0.0199 \times 180/ π = 1.1 4^{\circ}$ . Kyun? radians → degrees.

Verify: physical lead $\approx 1.1 4^{\circ}$ hai — ek small positive angle, yeh dikhata hai ki ek wildly negative raw $ϕ$ bhi ek concrete launch geometry mein reduce hota hai. Exams mein hamesha $ϕ mod 36 0^{\circ}$ report karo. ✓

Recall Matrix ko self-test karo

Kaun sa cell $ϕ = 0$ deta hai? ::: Cell C — equal radii ( $r_{1} = r_{2}$ ), koi transfer arc nahi chahiye. Outward transfer mein $ϕ$ ka sign kya hota hai? ::: Positive — target ko lead karna hota hai. Inward transfer mein $ϕ$ ka sign kya hota hai? ::: Negative — target ko trail karna hota hai (woh faster hai aur $π$ se zyada sweep karta hai). Earth→Mars phase angle $μ_{⊙}$ se independent kyun hai? ::: Kyunki $μ$ ratio mein cancel ho jaata hai; $ϕ$ radii ki pure geometry hai. $- 2 π$ se neeche $ϕ$ ko kaise report karte hain? ::: $[0, 2 π)$ mein physical launch lead paane ke liye modulo $2 π$ reduce karo. ISS $T_{syn} \approx 17$ days kyun hai? ::: Chaser aur target radii almost equal hain → tiny $∣ n_{1} - n_{2} ∣$ → bahut slow relative drift.