Worked examples — Groundtrack analysis — swath, revisit

3.2.38 · D3 · Physics › Orbital Mechanics & Astrodynamics › Groundtrack analysis — swath, revisit

Yeh page Groundtrack analysis — swath, revisit ka drill-ground hai. Parent note ne tumhe char formulas diye; yahan hum unhe har possible corner mein push karte hain — small angles, big angles, straight-down looks, sideways looks, prograde aur retrograde orbits, exact-repeat trap, aur ek real-world word problem. Har example batata hai ki woh scenario matrix ka kaun sa cell fill karta hai.

Koi bhi symbol aane se pehle, ek reminder of the cast (sab parent note mein bane hain, lekin yahan restate kiye gaye hain taaki har worked example self-contained ho):

Recall Pura symbol list jo hum use karenge (tap karke re-read karo)

• $h$ ::: satellite altitude — satellite ki height Earth ki surface ke upar (centre ke upar nahi), km mein. Earth ke centre se orbit radius tab $a=R_E+h$ hoti hai.
• $R_E=6371$ km ::: Earth ki mean radius (surface se centre tak).
• $T$ ::: orbital period, seconds per one lap around Earth.
• $\eta$ ::: sensor half-look-angle — total field-of-view cone ka aadha, satellite par straight-down (nadir) se outermost sight-line tak measure kiya gaya.
• ${\lambda }_s$ ::: Earth-central half-angle — Earth ke centre par angle, subsatellite point se us ground edge tak jahan tak sensor dekh sakta hai.
• ${\varphi }_{\max }$ ::: highest latitude jo groundtrack kabhi reach karta hai (uska poleward turning point).
• $i$ ::: inclination — orbit plane ka equator ke relative tilt.
• $S=360°\times \frac{T}{86164}$ ::: agla equator crossing kitna west mein padta hai, degrees mein.
• $W=2R_E{\lambda }_s$ with ${\lambda }_s=\arcsin \left(\frac{(R_E+h)\sin \eta }{R_E}\right)-\eta$ ::: sensor geometry se swath width.
• Repeat: $NT=D{T}_{\text{Earth}}$, i.e. orbits/day $=N/D$; track gap $\delta =360°/N$ ::: jab pattern close hota hai.

Throughout: $R_E=6371$ km, $T_{\text{Earth}}=86164$ s (yeh sidereal day hai — kabhi bhi 86400 s solar day nahi).

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The scenario matrix

Is topic mein jo bhi problem aa sakti hai woh inhi cells mein se ek hai. Neeche ke examples [C#] se label kiye gaye hain.

#	Cell class	Kya special hai	Example
C1	Small look-angle swath	$\eta$ tiny → ${\lambda }_s$ almost linear in $\eta$	Ex 1
C2	Large look-angle swath	$(R_E+h)\sin \eta$ approaches $R_E$ → arcsin near its limit	Ex 2
C3	Degenerate: nadir-only ($\eta =0$)	swath zero ho jaata hai	Ex 3
C4	Limiting: grazing horizon	argument of arcsin hits $1$ → maximum possible swath	Ex 3
C5	Nodal spacing, low orbit	short $T$ → small westward $S$	Ex 4
C6	Nodal spacing, high orbit	long $T$ → large $S$ (altitude ↑ ⇒ $S$ ↑)	Ex 4
C7	Exact repeat $N/D$	integers, coprime, pattern close karte hain	Ex 5
C8	Gap test $W$ vs $\delta$	kya swath track gap cover karta hai?	Ex 6
C9	Prograde vs retrograde ${\varphi }_{\max }$	$i\le 90°$ vs $i>90°$, altitude-independent	Ex 7
C10	Real-world word problem	disaster-monitoring revisit requirement	Ex 8
C11	Exam twist	"konsa altitude 15-orbit-per-day repeat deta hai?"	Ex 9

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Example 1 — Small look-angle swath [C1]

Forecast: Ek tiny cone ke liye swath roughly $\eta$ ke proportional hona chahiye. Agar $15°$ ne $\approx 376$ km diya (parent note), toh guess karo ki $5°$ roughly ek third dega — around $120$ km. Dekho.

Figure dekho: satellite (top), Earth ka centre (bottom), aur ground edge ek triangle banate hain. Red ray sensor ki outermost sight-line hai, straight-down se $\eta$ tilt hoi hui. Woh surface par jo arc kaatti hai, woh half-swath hai.

1. Ratio factor. $\frac{R_E+h}{R_E}=\frac{7071}{6371}=1.1099$. Yeh step kyun? Sine rule $\frac{\sin \eta }{R_E}=\frac{\sin (\eta +{\lambda }_s)}{R_E+h}$ rearrange hoti hai $\sin (\eta +{\lambda }_s)=\frac{R_E+h}{R_E}\sin \eta$ mein; yeh factor multiplier hai.
2. Far angle sine. $\sin (\eta +{\lambda }_s)=1.1099\times \sin 5°=1.1099\times 0.08716=0.09674$. Kyun? Yeh satellite par straight-down aur tilted ray ke beech ka angle hai, altitude leverage se thoda bada ho gaya.
3. Sine undo karo. $\eta +{\lambda }_s=\arcsin (0.09674)=5.550°$, toh ${\lambda }_s=5.550°-5°=0.550°$. Arcsin kyun? $\arcsin$ jawab deta hai "kis angle ka sine yeh hai?" — yeh $\sin$ ko undo karta hai jo humne banaya tha. Yahan physical sum $\eta +{\lambda }_s$ sirf $5.55°$ hai, $90°$ se comfortably neeche, toh principal value arcsin return karta hai (jo hamesha $[-90°,90°]$ mein hoti hai) bilkul wahi angle hai jo chahiye — koi branch ambiguity nahi.
4. Arc length. $W=2R_E{\lambda }_s=2(6371)(0.550°\times \frac{\pi }{180})=2(6371)(0.009601)=122.3$ km. $\pi /180$ kyun? Arc length ke liye angle radians mein chahiye: $\text{arc}=R{\theta }_{\text{rad}}$.

Verify: $122$ km, $15°$ wale $376$ km ka roughly ek third hai — near-linear forecast se match karta hai (small angles mein $\sin \eta \approx \eta$, toh $W$ almost $\eta$ ke saath scale hota hai). Units: km × (dimensionless) = km. ✔

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Example 2 — Large look-angle swath [C2]

Forecast: Wide angles horizon ki taraf bahut door dekhte hain, toh swath linear scaling se bahut zyada bada hona chahiye — ek bada jump expect karo, shayad thousands of km, kyunki geometry bend ho rahi hai.

1. Far angle sine. $\sin (\eta +{\lambda }_s)=1.1099\times \sin 60°=1.1099\times 0.86603=0.96114$. Kyun? Same sine-rule rearrangement — lekin dhyan do result ab 1 ke kareeb hai, matlab ray almost Earth ke tangent hai.
2. Sine undo karo — aur branch valid hai yeh check karo. $\arcsin (0.96114)=73.98°$. Toh $\eta +{\lambda }_s=73.98°$, giving ${\lambda }_s=73.98°-60°=13.98°$. Principal value sahi kyun hai? Arcsin principle mein ya toh $73.98°$ ya uska supplement $180°-73.98°=106.02°$ return kar sakta hai (dono ka sine same hai). Hum $73.98°$ rakhte hain kyunki physical angle $\eta +{\lambda }_s$ satellite par triangle ke andar ka angle hai, aur woh $90°$ se neeche rehna chahiye: outermost ray hamesha nadir se bahar lean karti hai lekin kabhi horizon ke past double back nahi karti. Jab tak sensor zameen pe hit karta hai (Ex 3b dekho), $\eta +{\lambda }_s<90°$ hai, toh principal value arcsin deta hai — jo $[-90°,90°]$ mein hoti hai — automatically correct hai. Yahan $73.98°<90°$ confirm karta hai ki hum valid domain mein hain.
3. Arc length. $W=2(6371)(13.98°\times \frac{\pi }{180})=2(6371)(0.24404)=3109$ km. Kyun? Radians phir se; same $R\theta$ arc rule.

Verify: $5°\to 122$ km aur $15°\to 376$ km mein doubling-ish look-angles se near-proportional swaths mile; lekin $60°$ $\sim 3100$ km tak jump karta hai — woh nonlinear blow-up jo forecast mein expect kiya tha, kyunki $\sin (\eta +{\lambda }_s)$ $1$ ke kareeb aa raha hai. ✔

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Example 3 — Degenerate & limiting look-angles [C3][C4]

Forecast: (a) Bilkul straight down dekhna ek point hi dekhta hai, toh $W=0$. (b) Sabse bada swath tab hota hai jab ray Earth ke tangent ho — far angle $90°$ hit karta hai; guess karo kuch thousand km ka swath.

Case (a) — nadir, $\eta =0$:

1. General far-angle relation $\sin (\eta +{\lambda }_s)=\frac{R_E+h}{R_E}\sin \eta$ se shuru karo. $\eta =0$ substitute karne se left side ka argument collapse ho jaata hai: $\eta +{\lambda }_s=0+{\lambda }_s={\lambda }_s$, toh equation literally $\sin ({\lambda }_s)=1.1099\times \sin 0°=0$ ban jaati hai. $\sin$ ka argument $\eta +{\lambda }_s$ se sirf ${\lambda }_s$ kyun ho gaya? Form nahi badla — humne simply $\eta =0$ set kiya, aur $\eta +{\lambda }_s$ with $\eta =0$ hai hi ${\lambda }_s$. Toh $\sin {\lambda }_s=0\Rightarrow {\lambda }_s=0$, aur $W=2R_E(0)=0$ km. Physical kyun? Zero cone angle ka matlab hai sensor ka edge ray centre ray hi hai — koi width nahi. Formula gracefully zero par degrade ho jaata hai.

Case (b) — horizon graze:

1. Tangency condition. Ray Earth ke tangent hai jab ground edge par angle $90°$ ho, yani $\eta +{\lambda }_s=90°$. Tab $\sin (\eta +{\lambda }_s)=1$, uska maximum. $=1$ kyun? $\sin$ kabhi $1$ se exceed nahi kar sakta; sine-rule right side $\frac{(R_E+h)\sin \eta }{R_E}$ wahan cap hai. $\eta$ ko isse aage push karo toh arcsin ka koi answer nahi hoga — physically ray Earth miss kar leti hai (space mein dekhti hai). Yahi Ex 2 ke valid-branch discussion ki exact boundary bhi hai: $\eta +{\lambda }_s$ horizon par exactly $90°$ reach karta hai aur kabhi exceed nahi karta.
2. ${\eta }_{\max }$ solve karo. $\frac{(R_E+h)\sin {\eta }_{\max }}{R_E}=1\Rightarrow \sin {\eta }_{\max }=\frac{R_E}{R_E+h}=\frac{6371}{7071}=0.90101$, toh ${\eta }_{\max }=\arcsin (0.90101)=64.24°$. Kyun? Yeh sabse bada half-look-angle hai jo abhi bhi zameen ko touch karta hai — isse aage koi swath nahi.
3. Half-swath. ${\lambda }_s=90°-{\eta }_{\max }=90°-64.24°=25.76°$. Yeh step kyun? Tangency se $\eta +{\lambda }_s$ $90°$ par pinned hai, toh half-swath ground angle simply wahi bachta hai jo ${\eta }_{\max }$ subtract karne ke baad bachta hai — woh leftover subpoint se horizon tak ka arc hai.
4. Swath. $W=2(6371)(25.76°\times \frac{\pi }{180})=2(6371)(0.44961)=5729$ km. Yeh step kyun? Same arc-length rule $W=2R_E{\lambda }_s$ jaisi har doosri case; hum ${\lambda }_s$ double karte hain (subpoint ke dono sides) aur radians mein convert karte hain taaki $R\theta$ km mein length de.

Verify: (a) exactly $0$ deta hai — sensible degenerate limit. (b) Maximum swath $5729$ km ek real "edge of visible Earth" figure hai; $60°$ case ($3109$ km) se bada, jaise required, aur uska half-angle $25.76°$ 700 km se possible sabse bada ground reach hai. ✔

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Example 4 — Nodal spacing: low vs high orbit, aur sign convention [C5][C6]

Forecast: Ek lap mein zyada time = ek lap mein zyada Earth-spin = bada westward jump. Toh (b) $>$ (a).

Pehle sign convention. Parent likhta hai $\Delta {\lambda }_{\text{node}}=-{\omega }_{\oplus }T$: minus sign ka matlab hai agla crossing pichle se west mein land karta hai, kyunki Earth orbit ke neeche eastward spin karti hai. Hum magnitude $S={\omega }_{\oplus }T=360°\times T/86164$ report karte hain aur "west" direction yaad rakhte hain. Yeh drift purely Earth's rotation se aati hai, toh yeh har orbit ke liye hoti hai satellite ke travel direction se regardless — neeche retrograde note dekho.

1. Low orbit. $S_a=360°\times \frac{5400}{86164}=360°\times 0.062671=22.56°$ west. Yeh step kyun? $S=360°\times T/86164$: ek period mein Earth is fraction of a full turn ghoomti hai, aur itna west mein agla crossing land karta hai.
2. High orbit. $S_b=360°\times \frac{12000}{86164}=360°\times 0.139269=50.14°$ west. Yeh step kyun? Exact same formula reuse karte hain lekin longer period ke saath; kyunki $T$ bada hai, ek lap ke dauran Earth ek full turn ka bada fraction spin karti hai, toh agla crossing aur bhi west mein land karta hai.
3. Compare. $S_b/S_a=50.14/22.56=2.222$ — ratio exactly $12000/5400$ ke barabar hai, kyunki $S$ linear in $T$ hai. Kyun? $360°/86164$ ek constant multiplier hai, toh $S\propto T$; aur Kepler $T$ ko altitude ke saath badhata hai, toh altitude ↑ ⇒ $S$ ↑. Forecast confirmed.

Verify: $S_b-S_a=27.58°$, dono positive (westward), dono $360°$ se kam (ek lap in periods mein Earth ko twice wrap nahi kar sakta). ✔

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Example 5 — Exact repeat pattern [C7]

Forecast: $43/3\approx 14.3$ orbits/day → period thoda over 100 min. 43 strips globe wrap karenge toh spacing near $8°$.

1. Coprime check. $\gcd (43,3)=1$ ✔ (43 prime hai, 3 isse divide nahi karta). Kyun? Agar woh factor share karte toh pattern sooner close hota, toh "fundamental" cycle ke liye coprime $N,D$ chahiye.
2. Period. $NT=D{T}_{\text{Earth}}\Rightarrow T=\frac{D{T}_{\text{Earth}}}{N}=\frac{3\times 86164}{43}=\frac{258492}{43}=6011.4$ s $=100.2$ min. Kyun? $D$ sidereal days baad Earth re-oriented hai AUR $N$ laps baad satellite back hai — dono ek saath true honge toh track overlay karta hai.
3. Orbits per day. $N/D=43/3=14.333$. Yeh step kyun? Fraction $N/D$ daily lap-rate hai: yeh batata hai ki har din kitne crossings lay down hote hain, aur uska non-integer hona ($14.333$, koi whole number nahi) exactly wahi reason hai ki tracks $D$ days mein interleave karte hain instead of har single din usi line par retrace karne ke.
4. Track spacing. $\delta =360°/N=360°/43=8.372°$. Km mein equator par: $8.372°\times \frac{\pi }{180}\times 6371=0.14612\times 6371=930.9$ km. $360°/N$ kyun? $N$ crossings equator ($360°$) ko $N$ equal strips mein divide karte hain.

Verify: $43\times 6011.4=258490\approx 3\times 86164=258492$ s (rounding) ✔. Orbits/day $14.33$ period deta hai $86164/14.333=6011$ s — consistent. ✔

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Example 6 — Gap test: kya swath track gap cover karta hai? [C8]

Forecast: $376$ km swath vs $931$ km gap — clearly ek cycle mein enough nahi hai. Guess karo roughly $931/376\approx 3$ cycles chahiye.

1. Coverage condition. Gap-free ke liye $W\ge \delta$ chahiye. Yahan $376<930.9$, toh NO — ek cycle bands unseen chhodti hai. Kyun? Har pass ek $376$ km stripe paint karta hai; neighbours $931$ km apart baithte hain, unke beech $931-376=555$ km blind band chhodkar.
2. Kitne sub-cycles. Interleaved passes needed $=\lceil \delta /W\rceil =\lceil 930.9/376\rceil =\lceil 2.476\rceil =3$. $\lceil \rceil$ (round up) kyun? Stripe ka fraction nahi ho sakta; $2$ stripes ($752$ km) abhi bhi gap chhodti hain, toh $3$ chahiye.
3. Effective revisit. Agar ek repeat cycle $D=3$ days ka hai, toh full gap-free coverage mein roughly $3\times 3=9$ days lagte hain (teen offset cycles). Cycle length se multiply kyun karte hain? Teeno interleaved passes mein se har ek alag $3$-day repeat cycle se belong karta hai (orbit thoda nudge hoti hai taaki tracks pichli ones ke beech fall karti hain), toh gap fill karne mein $3$ full cycles lagte hain — aur har cycle $D=3$ days leti hai, giving $3\times 3=9$ days total.

Verify: $3$ stripes dete hain $3\times 376=1128$ km $>930.9$ km — gap overshoot (with overlap), toh $3$ sufficient hai jabki $2$ ($752$ km) fail hota hai. ✔

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Example 7 — Prograde vs retrograde max latitude [C9]

Recall from the symbol list: ${\varphi }_{\max }$ woh highest latitude hai jo groundtrack kabhi reach karta hai — map par sinusoid ka poleward turning point. Uska rule (parent note) hai ${\varphi }_{\max }=i$ for $i\le 90°$ aur ${\varphi }_{\max }=180°-i$ for $i>90°$.

Forecast: Max latitude purely inclination se set hoti hai, toh A $55°$ reach karta hai, aur B — $90°$ se past hone ke karan — $180°-i$ use karta hai, $82°$ reach karta hai. Altitude irrelevant hai.

1. Satellite A. $i=55°\le 90°$, toh ${\varphi }_{\max }=i=55°$ N aur S. Kyun? Ek great circle $55°$ equator ke tilted hoti hai toh latitude $55°$ par peak karti hai — tilted ring ka geometric top yahi hai.
2. Satellite B. $i=98°>90°$, toh ${\varphi }_{\max }=180°-i=180°-98°=82°$ N aur S. $180-i$ rule kyun? Ek retrograde orbit ($i>90°$) top ke upar tip karta hai; uske plane ka equator se tilt $180°-i$ measure hota hai. Toh $98°$ sirf $82°$ tak reach karta hai, $98°$ tak nahi (latitude kabhi $90°$ exceed nahi karta).
3. Altitude? Kisi bhi computation mein $h$ use nahi hua. ${\varphi }_{\max }$ purely geometric hai — 300 km ya 3000 km, same rehta hai.

Verify: A: $55°\le 90°$ ✔. B: $180-98=82°\le 90°$ ✔ (ek valid latitude). Dono altitude-independent — ek common exam trap se bacha. ✔

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Example 8 — Real-world word problem [C10]

Forecast: $29$ strips → $\delta \approx 12°\approx 1370$ km, $376$ km swath se bahut bada — toh ek single 2-day cycle probably gaps chhodega aur requirement fail karega.

1. Coprime & period. $\gcd (29,2)=1$ ✔. $T=\frac{2\times 86164}{29}=\frac{172328}{29}=5942.3$ s. Yeh step kyun? Same repeat condition $NT=D{T}_{\text{Earth}}$ jaisi Example 5 mein.
2. Track spacing. $\delta =360°/29=12.414°$; km mein: $12.414°\times \frac{\pi }{180}\times 6371=0.21667\times 6371=1380.4$ km. Is tarah convert kyun karte hain? Gap naturally ek angle hai ($360°$ ko $N$ strips mein split), lekin hum isse km mein measure kiye gaye swath se compare karte hain, toh angle ko ground distance mein turn karte hain arc-length rule $\text{arc}=R_E{\theta }_{\text{rad}}$ se — aur $\theta$ radians mein hona chahiye, isliye $\pi /180$.
3. Gap test. $W=376$ km $<1380.4$ km ⇒ gaps remain ek 2-day cycle ke andar. Gap fill karne ke liye passes needed: $\lceil 1380.4/376\rceil =\lceil 3.67\rceil =4$. Kyun? $3$ stripes $=1128$ km abhi bhi $1380$ km se short; $4$ chahiye.
4. Verdict. Full coverage mein $\sim 4$ interleaved 2-day cycles $\approx 8$ days $\gg 2$ days lagte hain. Requirement NOT met. Fix hint: swath wide karo (bada $\eta$) ya denser repeat choose karo (bada $N$) taaki $\delta$ shrink ho $W$ se neeche.

Verify: $4\times 376=1504$ km $>1380.4$ km ✔ (4 fills, 3 doesn't). $8\text{days}>2\text{days}$ ⇒ fails, forecast se match karta hai. ✔

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Example 9 — Exam twist: 15 orbits/day ke liye altitude nikalo [C11]

Forecast: $15$ orbits/day → period $\approx 86164/15\approx 5744$ s $\approx 96$ min → ek Low Earth Orbit, altitude shayad $500$–$600$ km.

1. Required period. $NT=D{T}_{\text{Earth}}$ with $N=15,D=1$: $T=86164/15=5744.3$ s. Yeh step kyun? $15$ equal laps exactly ek sidereal day mein fit hone chahiye track daily repeat ke liye, toh hum repeat condition solve karte hain har lap ke liye allowed period ke liye.
2. Kepler ko $a$ ke liye invert karo. $T=2\pi \sqrt{a^3/\mu }$ se: $a^3=\frac{\mu T^2}{4{\pi }^2}$. Yeh step kyun? Hum $T$ jaante hain (step 1 se) aur orbit radius $a$ chahiye, toh period law ko algebraically $a$ ke liye solve karte hain — dono sides square karo aur rearrange karo. Orbital Period & Kepler's Third Law dekho.
3. Plug in. $a^3=\frac{3.986\times 10^5\times (5744.3{)}^2}{4{\pi }^2}=\frac{3.986\times 10^5\times 3.29970\times 10^7}{39.478}=\frac{1.31527\times 10^{13}}{39.478}=3.3316\times 10^{11}{\text{km}}^3$. Yeh step kyun? Right side numerically evaluate karne se $a^3$ milta hai; saare inputs known constants hain, toh yeh sirf arithmetic hai.
4. Cube root. $a=(3.3316\times 10^{11}{)}^{1/3}=6931.6$ km. Cube root kyun? Yeh $a^3$ undo karta hai; $a$ ek length hai toh hume real cube root chahiye.
5. Altitude. $h=a-R_E=6931.6-6371=560.6$ km. Yeh step kyun? $a$ Earth ke centre se measure hota hai, lekin altitude $h$ surface ke upar height hai, toh Earth ka radius $R_E$ subtract karte hain.

Verify: $a=6931.6$ se $T$ recompute karo: $T=2\pi \sqrt{6931.6^3/3.986\times 10^5}=2\pi \sqrt{3.3312\times 10^{11}/3.986\times 10^5}=2\pi \sqrt{8.3573\times 10^5}=2\pi (914.2)=5744$ s ✔ — exactly $86164/15$. Altitude $\approx 561$ km ek sensible LEO hai, forecast se match karta hai. ✔

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Recall Pure matrix ki ek-line summary

Konsa formula kaun sa cell own karta hai? ::: Swath $W=2R_E{\lambda }_s$ C1–C4 own karta hai (look-angle extremes); nodal spacing $S=360°T/86164$ C5–C6 own karta hai; repeat $N/D$ + $\delta =360°/N$ C7, C11 own karta hai; gap test $W\ge \delta$ C8, C10 own karta hai; ${\varphi }_{\max }=i$ or $180°-i$ C9 own karta hai.

Yeh bhi dekho: Orbital Period & Kepler's Third Law, Sun-Synchronous Orbits, Inclination & Orbital Elements, Remote Sensing Sensor Geometry, Earth Rotation & Sidereal Time, Nodal Regression & J2 Perturbation.