Launch window — phasing with target orbit: Question bank

3.2.39 · D5 · Physics › Orbital Mechanics & Astrodynamics › Launch window — phasing with target orbit

Shuru karne se pehle, neeche use kiye gaye har symbol ko yahan earn kiya gaya hai — koi bhi term bina explanation ke nahi aayega.

Definition Is page ke symbols (ek baar padho, phir refer karo)

Earth ke centre ke around do circular tracks imagine karo, ek andar, ek bahar.

$r_{1}$ ::: chaser's orbit radius (andar wala runner, Earth ke centre se measure kiya hua).
$r_{2}$ ::: target's orbit radius (bahar wala runner, Earth ke centre se).
$a_{t}$ ::: transfer ellipse ka semi-major axis, $a_{t} = \frac{r _{1} + r _{2}}{2}$ — iska perigee $r_{1}$ ko touch karta hai, apogee $r_{2}$ ko, isliye yeh dono radii ka average hai.
$t_{t}$ ::: transfer time — chaser kitne time tak half-ellipse ke along coast karta hai, $t_{t} = π a_{t}^{3} / μ$ (ek Hohmann-ellipse period ka aadha). Dekho Hohmann Transfer Orbit.
$n_{1}, n_{2}$ ::: chaser aur target ki mean motions (angular speeds), $n_{i} = μ / r_{i}^{3}$ — har runner Earth ke centre ke around kitni tezi se angle sweep karta hai. Dekho Mean Motion and Orbital Period.
$μ$ ::: Earth ka gravitational parameter (mass $\times$ $G$ ); yeh set karta hai ki cheezein kitni tezi se orbit karti hain.

Recall Teen phasing symbols

$ϕ$ ::: phase angle — launch instant par target kitna aage (degrees ya radians mein, Earth ke centre se) hona chahiye; closed form $ϕ = π [1 - (\frac{r _{1} + r _{2}}{2 r _{2}})^{3/2}]$ . $Δ θ_{2}$ ::: woh arc jo target transfer coast ke dauran sweep karta hai, $Δ θ_{2} = n_{2} t_{t}$ ; aur $ϕ = π - Δ θ_{2}$ . $T_{syn}$ ::: synodic period, $T_{syn} = \frac{2 π}{∣ n _{1} - n _{2} ∣}$ — yahi relative geometry (aur isliye wahi window) kitni baar wapas aati hai.

Intuition

$ϕ = π - Δ θ_{2}$ kyun — picture Figure dekho. Chaser inner orbit se neeche se nikalta hai aur half ellipse coast karke upar jaata hai — perigee se apogee tak woh arc hamesha exactly $π$ (180°) hota hai, kyunki half ellipse hamesha aadha sky hota hai chahe kitna bhi stretched ho. Meeting point woh upar wala apogee hai. Toh target, lead angle $ϕ$ se start karke, bacha hua arc $Δ θ_{2}$ sweep karta hai taaki same moment par upar pahunche. Kyunki chaser-arc + kuch nahi = target-start + target-arc same point par land karta hai: $ϕ + Δ θ_{2} = π$ , yaani $ϕ = π - Δ θ_{2}$ .

Figure mein: do black circles chaser ( $r_{1}$ ) aur target ( $r_{2}$ ) orbits hain; red half-ellipse transfer path hai; neeche wala black dot chaser ka start (perigee) hai aur upar wala red dot meeting point (apogee) hai. Outer circle par black dot woh target hai launch par, lead angle $ϕ$ aage baitha hua; woh chhota red arc jo use abhi bhi travel karna hai woh $Δ θ_{2}$ hai. Picture se seedha padho: $ϕ + Δ θ_{2} = π$ .

True or false — justify karo

True or false: Launch window poori tarah is baat se decide hoti hai ki rocket kab fuel ho kar ready hai.

False — readiness zaroori hai par kaafi nahi; window geometry (site plane ke neeche) aur timing (target lead angle

ϕ

par) se set hoti hai. Galat instant par ek ready rocket bahut zyada fuel waste karta hai.

True or false: Transfer orbit ko target ki current position par aim karna chahiye.

False — coast ke dauran target kai km/s move karta hai, isliye tum uski future position par aim karte ho; yahi lead angle

ϕ

encode karta hai.

True or false: Ek heavier central body (bada

μ

) required phase angle

ϕ

ko change karta hai.

False —

μ

ϕ

mein cancel ho jaata hai kyunki

ϕ

geometric radii ka ratio hai. Bada

μ

sirf transfer time aur synodic period ko chhota karta hai, angle ko nahi.

True or false: Ek chaser ke liye jo higher orbit (

r_{2} > r_{1}

) mein ja raha hai, launch par target hamesha chaser se aage hota hai.

False — "hamesha" blanket galat hai kyunki

ϕ

sign change kar sakta hai; yeh positive (target leads) tabhi rehta hai jab

Δ θ_{2} < π

ho.

r_{1}

ke thoda upar

r_{2}

ke liye yeh hold karta hai aur lead chhota hota hai, lekin claim as stated guaranteed nahi hai.

True or false: Ek inner target (

r_{2} < r_{1}

) ke liye required

ϕ

negative ho sakta hai, matlab target launch par chaser se peeche hota hai.

True — inner orbit faster hai aur transfer ke dauran $π$ se zyada arc sweep karta hai, isliye

Δ θ_{2} > π

aur

ϕ = π - Δ θ_{2} < 0

, yaani target peeche se start karta hai.

True or false: Do almost-equal circular orbits ek bahut chhota synodic period dete hain, isliye windows aksar aati hain.

False — almost-equal radii ka matlab almost-equal mean motions hai, isliye relative drift

∣ n_{1} - n_{2} ∣

tiny hai; geometry dheere repeat hoti hai, giving a long

T_{syn}

(isliye ISS chases mein minutes nahi balki days lagte hain).

True or false: Target ka orbital plane match karna launch instant define karne ke liye kaafi hai.

False — plane match sirf direction fix karta hai (site ko plane ke neeche rotate karna hoga). Phir bhi tumhe phasing chahiye taaki tum tab pahuncho jab target meeting point par ho. Dono coincide hone chahiye.

True or false: Chaser Hohmann transfer ke dauran hamesha exactly

18 0^{\circ}

sweep karta hai chahe radii kuch bhi ho.

True — ek Hohmann transfer perigee se apogee tak half ellipse hota hai, jo hamesha aadha sky hota hai,

r_{1}

aur

r_{2}

se independent.

Error dhundho

Error dhundho: "Target

Δ θ_{2}

sweep karta hai, chaser bhi

Δ θ_{2}

sweep karta hai, isliye woh automatically milt jaate hain."

Chaser hamesha

π

(half ellipse) sweep karta hai, $Δ θ_{2}$ nahi. Rendezvous ke liye zaroori hai ki target apogee point par pahunche, isliye

ϕ = π - Δ θ_{2}

— dono arcs generally alag hote hain.

Error dhundho: "

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

Denominator

2 r_{1}

nahi,

2 r_{2}

hona chahiye (upar boxed closed form se compare karo) — arc

Δ θ_{2}

target ki motion hai, isliye yeh target radius

r_{2}

se scale hota hai (

n_{2} = μ / r_{2}^{3}

ke zariye).

Error dhundho: "Kyunki

μ

t_{t}

mein aur

n_{2}

mein appear karta hai, toh yeh

ϕ

mein bhi hona chahiye."

Jab tum

n_{2} t_{t}

multiply karte ho, toh

n_{2} = μ / r_{2}^{3}

mein

μ

cancel ho jaata hai transfer time

t_{t} = π a_{t}^{3} / μ

(with

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

) ke

μ

se, sirf ek pure geometric ratio bachti hai. Pieces mein appear karna matlab product mein survive karna nahi hota.

Error dhundho: "Synodic period

T_{syn} = 2 π / (n_{1} + n_{2})

hai."

Yeh

2 π /∣ n_{1} - n_{2} ∣

hai — dono bodies same direction mein orbit karti hain, isliye jo matter karta hai woh relative rate (difference) hai, sum nahi. Sum counter-rotating bodies ko describe karta.

Error dhundho: "GEO door hai, isliye transfer lamba hai, isliye phase angle chhota hai."

Ulta — lamba transfer matlab slow GEO target barely move karta hai (

Δ θ_{2} \approx 7 9^{\circ}

), isliye use zyada aage start karna padta hai (

ϕ \approx 10 1^{\circ}

). Lamba transfer ⇒ bada lead, chhota nahi.

Error dhundho: "Humein ek Hohmann burn perfectly timed chahiye, isliye ISS ke liye phasing orbits unnecessary hain."

ISS synodic period ~days ka hai; ek single Hohmann almost kabhi exactly line up nahi karta, isliye real missions

ϕ

ko jagah par lane ke liye kai phasing orbits use karte hain. Dekho Rendezvous and Phasing Orbits.

Why questions

Target ko seedha aim karne ki bajay launch ko lead kyun karna padta hai?

Kyunki coast time

t_{t}

non-zero hai aur target orbit karta rehta hai; tum future point par milte ho, aur required head-start exactly

ϕ = π - Δ θ_{2}

hai.

Phase-angle formula mein exponent

3/2

kyun hai?

Yeh Kepler's third law se aata hai — orbital period

a^{3/2}

ke saath scale karta hai (dekho Kepler's Third Law). Transfer time ka target period se ratio wohi

3/2

power inherit karta hai.

Window repeat kyun hoti hai, aur specifically

T_{syn}

period ke saath?

Do bodies relative rate

∣ n_{1} - n_{2} ∣

par drift karti hain; jab woh full

2 π

drift kar leti hain toh same relative geometry — aur isliye same

ϕ

— wapas aati hai, jisme time lagta hai

2 π /∣ n_{1} - n_{2} ∣

Thoda pehle ya baad launch karke simply plane change kyun avoid nahi kiya ja sakta?

Pehle/baad launch karna sirf phasing shift karta hai, plane ko nahi; plane condition Earth ke rotation se set hoti hai jo site ko target ke orbital plane ke neeche le jaati hai, yeh ek independent geometric fact hai (dekho Orbital Plane and Inclination).

ϕ

ko central body (Earth ke centre) par kyun measure kiya jaata hai, launch site par nahi?

Ek orbit par angular position focus (Earth's centre) ke baare mein define ki jaati hai; surface se measure karne par Earth ka radius aur rotation mix ho jaata, clean geometric relation ko muddy kar deta.

Ek inner target ko launch par chaser se peeche kyun hona chahiye?

Inner orbit faster hai, isliye transfer ke dauran target

π

se zyada arc overtake karta hai; peeche se start karna use transfer ke doosre end par chaser se milne ke liye swing karne deta hai.

Edge cases

Edge case:

r_{1} = r_{2}

(same orbit).

ϕ

kya hai?

Ratio

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

hai, isliye

1^{3/2} = 1

aur

ϕ = π (1 - 1) = 0

. Koi lead nahi chahiye — lekin transfer degenerate hai (koi altitude change nahi), isliye yeh phasing-only problem hai, Hohmann nahi.

Edge case:

r_{1} = r_{2}

— synodic period ka kya hota hai?

n_{1} = n_{2}

isliye

∣ n_{1} - n_{2} ∣ = 0

aur

T_{syn} \to \infty

. Identical orbits kabhi drift nahi karte, isliye "window" ya toh hamesha open hai ya kabhi repeat nahi hoti — phase karne ke liye tumhe actively apna orbit change karna hoga (ek phasing orbit).

Edge case: target orbit chaser se bahut badi hai (

r_{2} ≫ r_{1}

ϕ

kya approach karta hai?

Ratio

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

, isliye

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

— slow distant target ko well over

9 0^{\circ}

lead karna padta hai.

Edge case: agar plane condition aur phasing condition kisi given day mein kabhi coincide na karein?

Toh us din koi window nahi hogi; tum wait karte ho jab tak phasing (

T_{syn}

period) aur ground-track alignment (period ~1 sidereal day) wapas agreement mein drift karein, jo kai cycles le sakta hai.

Edge case: transfer single Hohmann nahi balki bi-elliptic ya Lambert arc hai — kya

ϕ = π - Δ θ_{2}

tab bhi hold karta hai?

Nahi — woh formula assume karta hai ki chaser exactly

π

sweep karta hai. Doosri geometries ke liye

π

ki jagah chaser ka actual swept angle aur

Δ θ_{2}

ki jagah true transfer time par target ki motion rakhte hain; general timing problem Lambert's Problem hai.

Edge case: kya

ϕ

magnitude mein

18 0^{\circ}

se zyada ho sakta hai?

Agar

Δ θ_{2}

negative ho ya target ek full extra revolution past sweep kare, toh magnitudes

18 0^{\circ}

se zyada ho sakti hain; practice mein

ϕ

36 0^{\circ}

modulo reduce kiya jaata hai physically nearest lead tak, kyunki sirf circle par position matter karti hai.