Launch window — phasing with target orbit: Visual walkthrough

3.2.39 · D2 · Physics › Orbital Mechanics & Astrodynamics › Launch window — phasing with target orbit

Step 1 — Do circles aur do runners ke beech ek "lead"

KYA: hum dono runners ko unke tracks par rakhte hain aur unके beech angle mark karte hain. KYU: rendezvous poori tarah centre par angles ke baare mein hai — kaun aage hai, kitna. Distances sirf is liye matter karti hain ki woh speeds set karti hain (Step 3). PICTURE: do runners, centre, aur chaser se target tak wedge-shaped angle $ϕ$ .

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

Step 2 — Circles ke beech ka pul: transfer half-ellipse

KYA: hum do circles ko sabse saste arc se jodte hain — ek half-ellipse (ek Hohmann Transfer Orbit). KYU: hum inner se outer circle par teleport nahi kar sakte; hume ek aisi curve par coast karna hoga jo dono se mile. Hohmann half-ellipse woh sabse fuel-cheap curve hai. PICTURE: perigee $r_{1}$ par, apogee $r_{2}$ par, dashed half-ellipse dono ko jodte hue.

Step 3 — Har runner kitni tezi se sweep karta hai? (mean motion)

Hume ek "sweeping speed" chahiye: ek runner centre ke around ek second mein kitne radians cover karta hai.

KYA: hum $n_{1}$ aur $n_{2}$ compute karte hain, har circle ke liye ek, uske radius se. KYU: yeh jaanne ke liye ki target hamare flight ke dauran kahan drift karta hai, hume pata hona chahiye ki woh kitni tezi se sweep karta hai. Yahi ek jagah hai jahan planet ki strength enter hoti hai — aur, spoiler, yeh cancel ho jayegi. PICTURE: do runners speed arrows ke saath; inner wale ka arrow lamba hai (inner orbits tez hoti hain).

Step 4 — Flight kitni lambi hai? (transfer time)

Ise apne transfer ellipse (semi-major axis $a_{t}$ ) par apply karo. Us ellipse ke ek complete loop ke time ko $T_{ellipse}$ kaho: $T_{ellipse} = 2 π \frac{a _{t}^{3}}{μ}$

$T_{ellipse}$ = transfer ellipse ki period — agar poora ellipse, dono halves, fly kiya jata to kitna time lagta.

KYA: hum $T_{ellipse}$ ka aadha lete hain. KYU: Hohmann path sirf near-touch → far-touch hai — bilkul aadha loop. Ellipse ki major axis ke baare mein mirror symmetry ki wajah se, woh aadha aadha time leta hai. PICTURE: apne major axis se vibhajit poora faint ellipse; sirf upar wala aadha (hamari coast) solid hai, "half the period" labelled.

Step 5 — Hamare fly karne ke dauran target kitna drift karta hai?

KYA: target ki sweep-rate $n_{2}$ (Step 3 se) ko flight time $t_{t}$ (Step 4 se) se multiply karo. KYU: angle covered = rate × time. Yeh hamare coast ke dauran target ka head-start consumption hai. PICTURE: launch par target, ek arc $Δ θ_{2}$ , aur arrival par target.

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

$Δ θ_{2}$ = hamare coast ke dauran target ke radians.
$\frac{r _{1} + r _{2}}{2 r _{2}} = \frac{a _{t}}{r _{2}}$ = transfer-size ÷ target-size.
Outward chase ( $r_{2} > r_{1}$ , hamari running picture): to $a_{t} = \frac{r _{1} + r _{2}}{2} < r_{2}$ , isliye ratio $< 1$ hai aur $Δ θ_{2} < π$ . Slow outer target aadhe lap se kam cover karta hai.
Inward chase ( $r_{2} < r_{1}$ ): to $a_{t} = \frac{r _{1} + r _{2}}{2} > r_{2}$ , isliye ratio $> 1$ hai aur $Δ θ_{2} > π$ . Hum Step 7 mein is par wapas aate hain — generally $Δ θ_{2} \leq π$ mat maano.

Step 6 — Rendezvous condition (payoff)

KYA: hum demand karte hain ki jab hum far touch-point (transfer ka apogee) par pahunchein, target bhi wahin ho. KYU: chaser exactly $π$ radians (aadha aasman) ek touch-point se doosre tak fly karta hai. Milne ke liye, target — jo $ϕ$ aage se shuru karta hai — $Δ θ_{2}$ drift karne ke baad, usi arrival point par land karna chahiye. PICTURE: launch snapshot (target $ϕ$ se lead karta hai) arrival snapshot ke paas (dono apogee par).

Arrival par dono arrival point par hain. Angles ki bookkeeping (sab radians mein): $target ka head start ϕ + target drift karta hai Δ θ_{2} = chaser ka sweep π$

Head start ke liye solve karo:

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

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

$ϕ$ ka sign is baat par depend karta hai ki target kahan hai. Hume sab cover karne chahiye.

KYA: ratio $\frac{r _{1} + r _{2}}{2 r _{2}}$ ko $1$ ke against test karo. KYU: ratio decide karta hai ki $Δ θ_{2}$ $π$ se kam, barabar, ya zyada hai — jo $ϕ$ ka sign set karta hai. PICTURE: teen panels — outward chase ( $ϕ > 0$ ), same-orbit degenerate ( $ϕ = 0$ ), inward chase ( $ϕ < 0$ ).

Case	Radii	Ratio $\frac{r _{1} + r _{2}}{2 r _{2}}$	$Δ θ_{2}$	$ϕ$	Matlab
Outward chase	$r_{2} > r_{1}$	$< 1$	$< π$	$> 0$	target leads
Same orbit (degenerate)	$r_{2} = r_{1}$	$= 1$	$= π$	$= 0$	already together; koi lead nahi
Inward chase	$r_{2} < r_{1}$	$> 1$	$> π$	$< 0$	target trails

Step 8 — Window kitni baar wapas aata hai? (synodic period)

KYA: dhundho ki yahi exact phasing dobara kab hogi. KYU: agar woh instant miss ho jaye, tum geometry ke wapas aane ka intezaar karte ho — woh intezaar Synodic Period hai. PICTURE: do runners alag hote hue; relative drift ka ek poora $2 π$ wahi picture wapas laata hai.

Ek-picture summary

Ek frame poori kahani carry karta hai: do circles, half-ellipse bridge, chaser $π$ sweep karta hua, target $Δ θ_{2} = π (\frac{r _{1} + r _{2}}{2 r _{2}})^{3/2}$ khaata hua, aur launch-instant lead $ϕ = π - Δ θ_{2}$ jo unhe arrival point par collide karwaata hai.

Recall Feynman retelling — poora walkthrough seedhe alfazon mein

Tum aur ek dheema dost ek lake ke around do circular tracks par daud rahe ho, tumhara dost bahar wale par. Tum ek ball throw karna chahte ho taaki woh tumhare dost par gire — lekin tumhe ek curved path (half-ellipse) ke along throw karna hai jo tumhare track se shuru hoti hai aur unke track par khatam hoti hai. Woh curved throw ek fixed number of seconds leta hai ( $t_{t}$ ). Un seconds ke dauran tumhara dost daudta rehta hai aur kuch arc cover karta hai ( $Δ θ_{2}$ ). Isi dauran ball khud lake ke bilkul aadhe around swing karti hai ( $π$ radians). Ball aur tumhare dost ke saath aane ke liye, tumhare dost ko itna aage shuru hona chahiye ki "unka start-lead plus woh kitna daudte hain, ball jo half-lap travel karti hai uske barabar ho": $ϕ + Δ θ_{2} = π$ . Woh bacha hua lead $ϕ = π - Δ θ_{2}$ hi launch window hai. Agar tumhara dost bade, slow track par hai, woh muskil se hilte hain, isliye unhe almost quarter-lap aage shuru hona chahiye (LEO→GEO ≈ 100°). Agar woh tumhare apne track par hain, koi lead nahi chahiye ( $ϕ = 0$ ). Agar woh inner, tez track par hain, woh aadhe lap se zyada daudte hain aur tum tab launch karte ho jab woh tumhare peeche hoon ( $ϕ < 0$ ). Aur lake ki "gravity" ( $μ$ ) kitni strong hai yeh angle ke liye kabhi matter nahi karta — yeh ball aur dost dono ko equally speed up karta hai, isliye cancel ho jaata hai. Moment miss kar diya? Bas ek synodic period intezaar karo aur picture wapas aa jaayegi.

Recall Quick self-check

Outer orbit ke liye target $π$ se kam kyun sweep karta hai? ::: Kyunki $r_{2} > r_{1}$ se $a_{t} = (r_{1} + r_{2}) /2 < r_{2}$ hota hai, isliye ratio $< 1$ hai; slow outer orbit flight time mein sirf half-lap ka ek hissa cover karti hai. Rendezvous condition kaun sa equation state karta hai? ::: $ϕ + Δ θ_{2} = π$ — target ki lead plus uska drift, chaser ke half-lap ke barabar hai (radians). Inner target ( $r_{2} < r_{1}$ ) ke liye $ϕ$ ka sign? ::: Negative — target trails karta hai, kyunki transfer ke dauran woh $π$ se zyada sweep karta hai.

Yeh bhi dekho: Lambert's Problem (woh general "do points ko chosen time mein connect karo" problem jiska yeh special case hai), Orbital Plane and Inclination (ek real launch window ka doosra aadha).