Groundtrack analysis — swath, revisit: Visual walkthrough

3.2.38 · D2 · Physics › Orbital Mechanics & Astrodynamics › Groundtrack analysis — swath, revisit

Step 1 — Do spinning cheezein, ek frame

KYA. Hamare paas do motions hain jinhe ek hi reference frame mein measure karna hai — woh frame jo door ke stars ke saath fixed hai (the "inertial" frame). Ek motion hai satellite ka apni orbit mein ghoomna. Doosri hai Earth ka apni axis par ghoomna.

KYUN star-fixed frame. Agar hum satellite ki orbit ko stars ke against measure karte lekin Earth ki spin ko Sun ke against, toh hum apples aur oranges compare kar rahe hote — dono numbers alag-alag clocks mein hote. Physics ek hi choice force karta hai: orbit plane stars ke relative fixed hai, isliye Earth ki spin bhi stars ke against clock karni hogi. (Yahi Earth Rotation & Sidereal Time wali sidereal-vs-solar-day subtlety hai.)

PICTURE. Figure dekho: stars ka faint dotted circle hila nahi. Uske against, blue satellite apni orbit sweep kar raha hai aur Earth ke equator par yellow arrow ghoom raha hai. Dono angles usi star background se padhte hain.

Figure — Groundtrack analysis — swath, revisit

Step 2 — Rates ko naam do: har cheez kitni tez ghoomi

KYA. Har rotation ko ek symbol do taaki hum arithmetic kar sakein.

$ω_{\oplus}$ (padho "omega-Earth") $=$ Earth kitne radians per second ghoomi. Ek poora chakkar $2 π$ radians ka hota hai, aur Earth ek sidereal day $T_{\oplus} = 86164 s$ mein yeh karta hai, toh $ω_{\oplus} = full turn \div time for it \frac{2 π}{T _{\oplus}} = \frac{2 π}{86164 s} .$ Yahan $2 π$ hai "radians mein measure kiya ek poora revolution" aur $86164 s$ hai "woh seconds jo yeh revolution stars ke against leta hai."
$T$ (padho "period") $=$ woh seconds jo satellite ko apni orbit ka ek chakkar lagane mein lagte hain. Yeh Orbital Period & Kepler's Third Law se aata hai — badi orbit, lamba $T$ .

KYUN yahan degrees nahi radians. Radians mein "arc length $=$ radius $\times$ angle" bina kisi conversion factor ke sach hota hai. Hum jaldi hi ek angle ko ground par distance mein convert karenge, aur radians woh currency hai jo yeh cleanly karne deti hai.

PICTURE. Figure ek hi clock ko do baar mark karta hai: blue wedge dikhata hai ki satellite time ke ek slice mein kitna aage bada; yellow wedge dikhata hai Earth usi slice mein kitni ghoomi. Hum abhi poochhne wale hain: satellite ke ek poore lap ke baad, yellow wedge kitna bada hua?

Step 3 — Ek lap baad, Earth neeche se ghoom gayi

KYA. Satellite ko us moment par freeze karo jab woh equator cross karta hai upar jaate hue (the "ascending node"). Use exactly ek lap udne do aur phir equator cross karne do upar jaate hue. Usme $T$ time laga. Usi $T$ time ke dauran, Earth itni ghoomi: $Δ = ω_{\oplus} T = rad per second ω_{\oplus} \times seconds per lap T = radians Earth turned in one lap.$ Yahan har symbol ab earned hai: $ω_{\oplus}$ Step 2 se, $T$ Step 2 se.

KYUN multiply karein. "Rate $\times$ time $=$ amount" — usi reason se jaise speed $\times$ time distance deta hai. Rate hai radians-per-second, time hai seconds, seconds cancel hote hain aur radians bachte hain.

PICTURE. Do snapshots side by side. Left mein, green subsatellite point equator par kisi longitude par baitha hai. Right mein (ek lap baad) orbit plane wahan hai jahan stars ke against tha, lekin Earth ka yellow reference mark eastward ghoom gaya hai $Δ$ angle se. Satellite equator cross karta hai usi inertial jagah par — lekin alag patch of ground ab neeche hai.

Step 4 — KYUN naya crossing WEST mein padta hai

KYA. Agla crossing pichle wale se west mein $Δ$ angle par appear hota hai. Hum direction record karne ke liye ek sign lagate hain: eastward-positive longitude, toh westward shift negative hai: $Δ λ_{node} = - ω_{\oplus} T .$ Yahan $λ$ (lambda) longitude hai, "kitna east" wala angle, aur subscript "node" yaad dilata hai ki hum equator crossing par measure kar rahe hain.

KYUN west aur east nahi. Earth eastward spin karti hai. Socho tum pehle crossing longitude par khade ho. Jab satellite ghumta raha, woh ground jis par tum khade the eastward ride kar gayi. Toh jab satellite usi star-fixed spot par wapas aata hai, woh ground jo ab wahan hai woh aur west se travel karke wahan aayi hogi — matlab crossing pehle se zyada westerly longitude par hai.

PICTURE. Ek single equator line. Red mark = pehla crossing. Green mark = doosra crossing, left (west) mein baitha. Yellow arrow ground ka eastward drift dikhata hai; red-se-green gap wohi amount hai, track ka westward jump banke dikhta hai.

Step 5 — Ek number lagao (LEO worked example)

KYA. Ek low orbit lo jisme $T \approx 5760 s$ (lagbhag $96$ minutes, altitude near $700$ km). $S = 360° \times \frac{5760}{86164} = 24.07° west per orbit.$

KYUN yeh useful form hai. $24.07°$ ek feel karne layak distance hai: lagbhag $24°$ longitude, equator par roughly $2670$ km. Paandrah aise jumps ( $15 \times 24° = 360°$ ) almost poori globe ko ek din mein tile kar dete hain — isliye ek LEO satellite tumhe $\sim 15$ passes daily deta hai.

PICTURE. Equator ek strip ke roop mein draw hai $0°$ se $360°$ tak. Successive crossings $0°, - 24°, - 48°, \dots$ par mark hain, leftward march karte, har ek orbit number ke saath label kiya. Tum dekh sakte ho ki woh wrap around karke interleave karte hain.

Step 6 — Edge case: kya hoga agar jump circle ko evenly divide kare?

KYA. Kabhi kabhi poore number of orbits $N$ ke baad, aur poore number of days $D$ ke baad, satellite exactly kisi purani track par wapas aa jaata hai. Yeh tab hota hai jab $time for N laps N T = time for D Earth turns D T_{\oplus} ⟹ \frac{N}{D} = \frac{T _{\oplus}}{T} = orbits per day,$ jahan $N$ aur $D$ ka koi common factor nahi (coprime).

KYUN yeh pattern close karta hai. $D$ sidereal days ke baad Earth exactly usi orientation mein wapas hai; $N$ orbits ke baad satellite exactly usi orbital position par wapas hai. Dono ek saath sach matlab green dot apna purana trail retrace karta hai — ek repeat groundtrack. Tracks tab equator ke around even spacing $δ = 360°/ N$ par baithte hain. Sun-synchronous survey satellites deliberately is tarah tune kiye jaate hain — dekho Sun-Synchronous Orbits aur Nodal Regression & J2 Perturbation ka drift bookkeeping.

PICTURE. Wahi equator strip. Jab jumps $360°$ ka ek rational fraction hain, marks ek finite comb of evenly spaced teeth par girte hain (yahan $N$ teeth), forever drift karne ki jagah. Figure dikhata hai teeth fill in karte hue jab tak pattern lock nahi ho jaata.

Step 7 — Degenerate cases: boundaries check karo

KYA & KYUN. Ek formula jis par tum trust karte ho use extremes par sensibly behave karna chahiye.

$T \to 0$ (impossibly fast orbit): $S = ω_{\oplus} T \to 0$ . Earth ek lap ke dauran muskil se ghoomti hai, toh track muskil se drift karta hai — sahi hai.
$T = T_{\oplus} = 86164 s$ (geosynchronous): $S = 360°$ , har orbit mein poora chakkar. Track har single orbit mein khud par wapas aata hai — groundtrack ek fixed figure-8 ban jaata hai (ya circular equatorial orbit ke liye ek point). Sahi hai.
$T > T_{\oplus}$ : $S > 360°$ ; jump ek full circle se past wrap karta hai, toh effective shift $S mod 360°$ hai — track east ki taraf drift karta bhi dikh sakta hai.
Inclination & Orbital Elements se $i$ inclination sirf latitude reach fix karta hai, kabhi $S$ nahi: maximum latitude $ϕ_{m a x} = i$ hai (ya retrograde ke liye $180° - i$ ), yeh ek purely geometric fact hai, $T$ se independent aur isliye $S$ se bhi independent.

PICTURE. Teen chhoti equator strips stack ki hui: $T$ tiny (dense clustered dots), $T = T_{\oplus}$ (sab dots ek longitude par stack), $T$ bada (dots overshooting aur wrapping). Ek nazar mein poori family dikh jaati hai.

Ek-picture summary

KYA yeh compress karta hai. Star-fixed frame $\Rightarrow$ dono spins wahan measure karo $\Rightarrow$ Earth $ω_{\oplus} T$ per lap ghoomi $\Rightarrow$ track utna west jump karta hai $\Rightarrow$ agar jump circle ka rational slice hai, toh track $N / D$ orbits-per-day ke baad repeat karta hai. Poora chain ek drawing mein.

Recall Feynman retelling — poora walkthrough plain words mein

Socho ek runner ek spinning merry-go-round ke around chakkar laga raha hai. Runner ke laps ko merry-go-round ki spin se fairly compare karne ke liye, tumhe bahar se dono dekhhne chahiye (room ki walls ke against — hamare "stars"). Har baar jab runner ek poora loop complete karta hai, merry-go-round thoda ghoom chuka hota hai, toh runner ek alag seat ke upar se guzarta hai — aur hamesha ek jo same taraf (westward, spin ke opposite) same angle se shift ho gayi, kyunki "kitna ghuma" sirf uski spin-rate hai times kitna ek loop liya. Agar woh shift ek poore circle ka neat fraction ho, toh poore number of loops ke baad runner wapas ek purani seat ke upar aata hai: trail repeat hoti hai. Woh repeat spacing, iske saath ki runner kitni door side mein dekh sakta hai (the swath), decide karta hai ki koi kitna wait kare dobara dekhe jaane ke liye.

Quick self-check

Recall

Earth ki spin ko Sun ke against nahi, stars ke against kyun clock karna chahiye? ::: Kyunki orbit plane stars ke relative fixed hai; dono motions ko same inertial frame mein compare karna hoga, isliye hum sidereal day $86164 s$ use karte hain. Period $T$ ke ek orbit mein, track kitna west jump karta hai? ::: $S = ω_{\oplus} T = 360° \times T /86164 s$ . $96$ -minute LEO ke liye $S$ kya hai? ::: Lagbhag $24.07°$ . Groundtrack exactly kab repeat karta hai? ::: Jab $N T = D T_{\oplus}$ coprime integers $N, D$ ke saath, yani orbits per day $= N / D$ . Geosynchronous period par $S$ ka kya hota hai? ::: $S = 360°$ — track har orbit mein khud par aata hai.