Groundtrack analysis — swath, revisit: Question bank

3.2.38 · D5 · Physics › Orbital Mechanics & Astrodynamics › Groundtrack analysis — swath, revisit

Traps dhundne se pehle, yeh page apne har symbol ko define karta hai, isi page par, taaki kuch bhi andhe bharose se borrowed na ho. Neeche diya toolkit ek baar padh lo; figures har idea ko concrete banate hain.

Symbol toolkit (yahan se, zero se banaya)

Definition Woh symbols jo is page ko chahiye

$R_{E}$ — Earth ki radius, Earth ke center se surface tak ki doori (lagbhag $6371$ km).
$h$ — satellite altitude: satellite surface ke upar kitna ooncha udta hai. To Earth ke center se uski doori $R_{E} + h$ hai.
$i$ — inclination: orbit plane aur equator ke beech ka tilt angle (ek orbital element).
$ϕ$ — latitude: koi point equator se kitna north (+) ya south (−) hai, degrees mein.
$λ$ — longitude: koi point equator ke around kitna east (+) ya west (−) hai, degrees mein angle ke roop mein.
$ϕ_{m a x}$ — maximum latitude jo groundtrack reach karta hai (shadow ka sabse northernmost/southernmost point).
$Δ λ$ — longitude mein change: ek orbit se doosre mein crossing longitude kitna shift hota hai (negative = westward).
$T$ — orbital period (ek chakkar ka time), Kepler's third law se.
$ω_{\oplus}$ — Earth ki spin rate ( $\frac{2 π}{86164 s}$ , sidereal day use karke — neeche toolkit note dekho).
$S = ω_{\oplus} T$ — nodal spacing: har orbit mein equator crossing ka westward longitude shift $∣Δ λ ∣$ .
$η$ (eta) — sensor half-angle: camera seedha neeche se sideways mein kitna "dekh" sakta hai uska aadha angle, satellite par measure kiya gaya (swath formula ke liye radians mein rakho).
$λ_{s}$ — Earth ke center par swath half-angle: sensor jo ground arc cover karta hai uska aadha angle, Earth ke center se measure kiya gaya (neeche visually define kiya, swath formula ke liye radians mein).
$W = 2 R_{E} λ_{s}$ — swath width: sensor jo across-track ground distance image karta hai.
$δ = 360°/ N$ — track spacing: poore repeat cycle ke baad neighboring passes ke beech equator gap.
$N / D$ — orbits ( $N$ ) per sidereal days ( $D$ ), woh coprime ratio jo track ko exactly repeat karata hai.

Sidereal day, ek line mein: orbit plane stars ke against still rehta hai, isliye hum Earth ki spin bhi stars ke against measure karte hain — yahi sidereal day hai, $86164$ s, solar day ke $86400$ s se lagbhag $4$ minute chota (woh extra bit Earth ke Sun ke around orbit karne se aata hai). Isliye repeat ratio $N / D$ mein "days" $D$ sidereal hain, solar nahi. Dekho Earth Rotation & Sidereal Time.

Longitude sign convention (minus sign kyun hai)

Figure — Groundtrack analysis — swath, revisit

Figure dekho: east positive $λ$ hai, west negative $λ$ hai. Earth eastward spin karti hai, isliye do crossings ke beech ground fixed orbit ke neeche east mein slide kar chuka hota hai — matlab agla crossing west mein land karta hai, ek negative change $Δ λ$ . Yahi wajah hai ki $Δ λ = - ω_{\oplus} T$ mein minus hai, aur iski size nodal spacing $S = ∣Δ λ ∣ = ω_{\oplus} T$ hai.

Sine-wave groundtrack, concrete tarike se

Orbit ek great circle hai jo $i$ se tilted hai. Jab tum longitude $λ$ mein aage badhte ho, latitude $ϕ$ smoothly $+ i$ aur $- i$ ke beech rise aur fall karti hai. Spherical geometry se exact relation (tilted great circle ko latitude par project karna) yeh hai $ϕ = arcsin (sin i sin λ),$ aur equator ke paas (chota $λ$ , ya chota $i$ ) yeh approximate sine wave mein reduce ho jaata hai $ϕ (λ) \approx i sin λ,$ jo flat lat–lon map par literally ek sine wave hai. Red curve $i = 50°$ ke saath exact form hai; amplitude (iski height) hai $ϕ_{m a x} = i$ ( $λ = 90°$ daalo: $arcsin (sin i) = i$ ) — yahi wajah hai ki max latitude sirf inclination par depend karta hai, altitude par kabhi nahi.

Swath triangle (jahan se $W = 2 R_{E} λ_{s}$ aata hai)

Teen points ek triangle banate hain: Earth ka center $O$ , satellite $S a t$ (doori $R_{E} + h$ par), aur ground edge $G$ jahan tak camera just reach karta hai. Satellite par look-angle $η$ hai; $O$ par central angle $λ_{s}$ hai; $G$ par teesra angle $180° - (η + λ_{s})$ hai. Sine rule (side/opposite-angle-ka-sine sabke liye equal hota hai) deta hai $\frac{s i n η}{R _{E}} = \frac{s i n ( η + λ _{s} )}{R _{E} + h} \Rightarrow λ_{s} = arcsin (\frac{( R _{E} + h ) s i n η}{R _{E}}) - η,$ aur kyunki arc length $=$ radius $\times$ angle-radians-mein, poora swath $W = 2 R_{E} λ_{s}$ hai ( $λ_{s}$ radians mein). Sensor details ke liye dekho Remote Sensing Sensor Geometry. Sine rule kyun, cosine rule kyun nahi: sine rule har side ko uske opposite angle ke sine se link karta hai, aur yahan hum ek side jaante hain ( $R_{E}$ , known angle $η$ ke opposite) aur woh side chahiye ( $R_{E} + h$ , unknown $η + λ_{s}$ ke opposite) — ek known side/opposite-angle pair ko unknown side/opposite-angle pair se match karna exactly sine-rule pattern hai. Cosine rule ke liye instead do sides aur unke beech ka included angle chahiye, jo hamare paas nahi hai, isliye woh $λ_{s}$ directly isolate nahi kar sakta.

Repeat pattern aur J2 drift

Left: coprime cycle $N / D$ ke baad (jisme $D$ sidereal days mein count hota hai), bahut saare passes $N$ equally spaced strips mein interleave ho jaate hain jinka width $δ = 360°/ N$ hai; gap-free coverage ke liye swath $W$ ko satisfy karna hoga $W \geq δ$ . Right: J2 Earth ke equatorial bulge ko slowly puri orbit plane ko "regress" (rotate) karata hai — weeks mein yeh poore groundtrack pattern ko longitude mein shift kar deta hai, aur sirf tab repeat locked rehta hai jab regression tuned ho (jaise sun-synchronous design mein).

Sach ya jhooth — justify karo

Satellite ka orbit plane physically har orbit thoda west move karta hai, westward drift cause karta hai.

Galat. Orbit plane (first order tak) inertial space mein fixed hai; yeh Earth hai jo neeche eastward rotate karti hai, isliye agla crossing farther west dikhta hai. Plane khud ground ke peeche nahi bhagta.

Altitude

h

badhane se swath wide hoti hai lekin maximum latitude nahi badalti.

Sahi. Swath

h

ke saath badhti hai (sensor ka fixed view-angle

η

oonche se zyada ground subtend karta hai), lekin

ϕ_{m a x} = i

purely geometric hai — sirf inclination se set hota hai, altitude-independent.

Ek polar orbit (

i = 90°

) kaafi time dene par Earth ke har point ko image kar sakta hai.

Sahi.

ϕ_{m a x} = 90°

ke saath track dono poles tak pahunchta hai, aur westward drift (

S

) plus Earth ki spin eventually tracks ko har longitude par le jaata hai, isliye full coverage achievable hai.

Agar do orbits ka period same hai to unka groundtrack same hona chahiye.

Galat. Same

T

same nodal spacing

S

deta hai, lekin alag inclination alag latitude envelope (

\pm i

) deti hai, aur alag starting node poore pattern ko longitude mein shift karta hai. Same period ≠ same track.

86400 s solar day ki jagah 86164 s sidereal day use karne se thoda-sa off lekin harmless answer milta hai.

Galat. ~4 min/day ka error har orbit mein chota hai lekin repeat cycle mein accumulate hota hai aur integer

N / D

closure corrupt kar deta hai — track kabhi truly overlay nahi karta. Chota bias, barbad repeat math.

Ek repeat orbit mein

N / D

orbits per sidereal day ke liye,

N

aur

D

mein koi common factor nahi hona chahiye.

Sahi. Agar woh factor share karte, to pattern

N

se kam orbits mein close ho jaata; reduced (coprime) fraction hi true fundamental cycle hai.

Wider swath hamesha worst-case (full-cycle) revisit time kam karta hai.

Galat. Ek baar

W \geq δ

(track spacing) ho jaane par coverage already gap-free hai, isliye aur wide karna sirf overlap pile karta hai — yeh full repeat cycle nahi shortens karta, jo

N / D

se fixed hai. Us threshold ke neeche yeh help karta hai; isliye "hamesha" hi statement ko galat banata hai.

Retrograde orbit (

i > 90°

) ka groundtrack same numerical

i

wale prograde orbit se zyada latitude reach karta hai.

Galat.

i > 90°

ke liye,

ϕ_{m a x} = 180° - i

, jo 90° se kam hai.

i = 100°

par ek retrograde orbit

80°

latitude par top karta hai,

100°

par nahi (latitude physically

90°

se zyada ho hi nahi sakta).

Galti pakdo

"Revisit time orbital period ke equal hai, kyunki satellite har orbit mein wapas aa jaata hai."

Satellite usi orbital position par wapas aata hai, lekin Earth neeche ~

S

rotate kar chuki hoti hai, isliye woh ek alag jagah par hota hai. Revisit repeat cycle

N / D

aur swath se govern hoti hai, sirf

T

se nahi.

"Swath width aur track spacing same cheez hai — dono describe karte hain ki passes ground ko kaise cover karte hain."

Nahi. Swath

W

woh hai jo sensor across-track dekhta hai; spacing

δ = 360°/ N

neighboring passes ke beech geometric gap hai. Gap-free coverage ke liye

W \geq δ

chahiye — inhe compare kiya jaata hai, equal nahi samjha jaata.

"Kyunki orbit plane space mein fixed hai, equator crossing longitude har orbit mein same hota hai."

Plane fixed hai, lekin rotating Earth har crossing par ek naya longitude present karta hai; crossing longitude west mein

S = ω_{\oplus} T

shift hota hai har orbit mein. (Iske aalaawa, J2 nodal regression plane ko bhi slowly drift karta hai.)

"Max latitude

= i

hamesha hota hai, to

70°

-inclination orbit latitude

70°

reach karta hai aur

110°

wala

110°

reach karta hai."

Pehla sahi hai; doosra galat hai.

i > 90°

ke liye use karo

ϕ_{m a x} = 180° - i = 70°

. Latitude physically

90°

se zyada nahi ho sakta, isliye

\pm i

rule ko

\pm (180° - i)

mein switch karna padta hai.

"Globe jaldi fill karne ke liye, bas altitude badha do — zyada orbits per day milenge."

Ulta hai. Zyada altitude matlab lamba period (Kepler's third law), isliye kam orbits per day aur wider track spacing

S

— generally slower fill, faster nahi.

"Sun-synchronous orbit by definition har din apna groundtrack repeat karta hai."

Zarooori nahi. Sun-synchrony crossings ke local solar time ko fix karta hai (tuned nodal regression ke zariye), jo integer repeat

N / D

se ek alag condition hai. Ek sun-sync orbit ka 16-day repeat ho sakta hai, jaise Landsat ka.

"Swath formula sine rule use karta hai, to hum cosine rule use karke bhi same numbers le sakte the."

Sine rule isliye choose kiya jaata hai kyunki hamare paas ek known side apne opposite angle ke saath hai (

R_{E}

η

ke opposite hai) jo unknown side apne opposite angle ke saath match karta hai (

R_{E} + h

η + λ_{s}

ke opposite hai) — exactly sine-rule pattern. Cosine rule ke liye do sides aur unke beech ka included angle chahiye, jo hamare paas nahi hai, isliye woh

λ_{s}

directly isolate nahi kar sakta.

"Kyunki

λ_{s}

degrees mein ek chota angle hai, main sirf

2 R_{E}

ko degrees mein multiply karke swath km mein le sakta hoon."

Nahi —

W = 2 R_{E} λ_{s}

arc length hai aur radians demand karta hai. Degrees se multiply karne par answer

180/ π \approx 57

factor se inflate ho jaata hai. Pehle convert karo

λ_{s} \to λ_{s} \times π /180

Why questions (Kyun wale sawaal)

Groundtrack math ke liye solar day ki jagah sidereal day correct spin period kyun hai?

Orbit plane ek inertial (star-fixed) frame mein rehta hai; groundtrack Earth ki spin ko usi frame ke against compare karta hai. Stars ke relative Earth ki spin sidereal day hai (86164 s); solar day ke extra 236 s Earth ke Sun ke around orbital motion se aate hain, jo yahan irrelevant hai.

Flat (rectangular) map par groundtrack sine wave jaise kyun dikhta hai?

Tilted great circle latitude par project hota hai

ϕ = arcsin (sin i sin λ)

ke roop mein, jo smoothly

+ i

aur

- i

ke beech oscillate karta hai jaise longitude

λ

aage badhta hai; equator ke paas yeh

ϕ \approx i sin λ

mein flatten ho jaata hai, aur rectangular lat–lon grid par yeh amplitude

i

ka sinusoid trace karta hai.

Nodal spacing

S

Δ λ = - ω_{\oplus} T

mein minus sign kyun carry karta hai?

East ko positive longitude lene par, Earth eastward rotate karti hai, isliye ground ke relative agla crossing west mein land karta hai — ek negative longitude change

Δ λ

. Minus westward convention encode karta hai (sign-convention figure dekho);

S = ∣Δ λ ∣

iski magnitude hai.

ϕ_{m a x}

altitude se independent kyun hona chahiye?

Kyunki yeh is baat se set hota hai ki orbit plane equator se kitna tilt hai — ek fixed geometric angle

i

(

ϕ = arcsin (sin i sin λ)

mein

λ = 90°

daalo to

ϕ_{m a x} = i

milta hai). Us plane ke andar satellite ko ooncha ya neeche karna kabhi plane ka tilt nahi badalata.

Equator ke paas ka target generally high latitude wale target se zyada mushkil kyun hota hai revisit karne ke liye?

Poles ke paas, converging meridians tracks ko paas-paas pack karti hain (denser coverage), isliye high-latitude targets zyada baar dekhe jaate hain; equatorial targets wahan hote hain jahan tracks sabse zyada spread hoti hain — woh gaps avoid karne ke liye swath overlap par rely karte hain.

Gap-free global coverage ke liye

W \geq S

ki jagah

W \geq δ

kyun chahiye?

δ = 360°/ N

poore repeat cycle ke baad saare interleaved tracks ke beech ki spacing hai, jo single-orbit spacing

S

se finer hai. Coverage ko final packed pattern

δ

ke against judge kiya jaata hai, coarse per-orbit shift

S

ke against nahi.

Swath ka central half-angle

λ_{s}

, sensor half-angle

η

se horizon ke paas tezi se kyun badhta hai?

sin (η + λ_{s}) = \frac{R _{E} + h}{R _{E}} sin η

mein, jaise

η

badhta hai right side 1 ke paas pahunchti hai aur arcsine steep ho jaata hai — limb ke paas line of sight Earth ki curvature ko graze karti hai, isliye thoda extra look-angle ek bada ground arc sweep karta hai.

Edge cases

Perfectly equatorial orbit,

i = 0°

, ka groundtrack kya hoga?

ϕ_{m a x} = 0°

, isliye

ϕ = arcsin (sin 0° sin λ) = 0

sabhi

λ

ke liye: track equator ke along ek seedhi line hai, har orbit mein west mein

S

se drift karta hai lekin latitude

0

se kabhi nahi hata — koi sine wave nahi.

Jaise

i \to 90°

, "sine wave" ka kya hota hai?

Latitude swing

\pm 90°

tak grow karta hai, isliye track almost seedha north–south run karta hai, dono poles tak pahunchta hai; exact curve

ϕ = arcsin (sin i sin λ)

maximally "tall" ho jaata hai, rectangular map par poles ke paas almost vertical.

Agar swath bilkul track spacing ke equal hai,

W = δ

, to kaisi coverage milti hai?

Exactly seamless, zero-overlap coverage — har point ek baar image hota hai bina gaps aur bina redundancy ke. Yeh critical threshold hai; thodi bhi narrow aur ground ki stripes miss ho jaati hain.

Geostationary orbit (

T = 1

sidereal day,

i = 0

) ke liye, groundtrack kisme degenerate ho jaata hai?

S = 360°

per orbit matlab crossing same longitude par wapas aata hai — groundtrack equator par ek single stationary point mein collapse ho jaata hai. (Nonzero

i

ise ek chota figure-eight banata.)

Agar swath itni wide hai ki woh bilkul agle orbit mein har neighboring track ko overlap kar leti hai, to revisit kya hai?

Best-case revisit ek orbital period ke paas pahunchta hai — consecutive swaths already target ke neighborhood cover kar leti hain, isliye poore

N / D

cycle ka wait nahi karna padta.

Agar period

T

sidereal day ke relative irrational ho to repeat cycle ka kya hoga?

Koi integer

N, D

satisfy nahi karta

N T = D T_{Earth}

exactly, isliye track kabhi exactly repeat nahi karta — woh densely ek band fill karta hai aur sirf approximately revisit karta hai. True repeat orbits ke liye rational (ideally small-integer) ratio chahiye.

i = 180°

(fully retrograde equatorial) ke degenerate case mein

ϕ_{m a x}

kya hai?

ϕ_{m a x} = 180° - 180° = 0°

— phir se ek equatorial track, lekin satellite prograde orbit ke relative westward travel karta hai; groundtrack phir bhi equator ke saath chipka rehta hai.

J2 nodal regression long-term groundtrack ko kaise change karta hai, aur yeh asset kyun ho sakta hai?

Equatorial bulge orbit plane ko slowly rotate karta hai, poore pattern ko weeks mein longitude mein drift karta hai; uncorrected chhodne par yeh exact repeat tod deta hai, lekin tuned karne par yeh sun-synchrony power karta hai, crossing times ko Sun se lock karta hai.

Recall Jaane se pehle self-check

Which single quantity is set purely by inclination and nothing else? Maximum latitude $ϕ_{m a x}$ ::: $ϕ_{m a x} = i$ for $i \leq 90°$ , else $180° - i$ — altitude, period, aur swath sab iske liye irrelevant hain.