Groundtrack analysis — swath, revisit: Exercises

3.2.38 · D4 · Physics › Orbital Mechanics & Astrodynamics › Groundtrack analysis — swath, revisit

Yeh page ek self-testing ladder hai. Har problem ko pehle khud solve karo, phir solution dekho. Levels "kya tumhe formula yaad hai" se shuru hokar "kya tum scratch se mission design kar sakte ho" tak jaate hain. Agar koi symbol unfamiliar lage, toh pehle parent note pe jaao.

Constants jo poore exercises mein use honge (ek baar yahan state kar dete hain, taaki koi symbol bina introduce kiye na aaye):

$R_{E} = 6371 km$ — Earth ki mean radius (centre se surface tak ki distance).
$T_{\oplus} = 86164 s$ — sidereal day, stars ke relative Earth ka ek spin (dekho Earth Rotation & Sidereal Time).
$μ = 398600 km^{3} / s^{2}$ — Earth ka gravitational parameter, woh number jo Orbital Period & Kepler's Third Law ke zariye orbital period set karta hai.
$ω_{\oplus} = \frac{2 π}{86164 s} = 7.292 \times 1 0^{- 5} rad/s$ — Earth ki spin rate.

Jab bhi altitude $h$ ki circular orbit ka period chahiye, hum Kepler's third law use karte hain: $T = 2 π \frac{a ^{3}}{μ}, a = R_{E} + h,$ jahaan $a$ orbital radius hai (Earth ke centre se satellite tak). Hum yahi tool use karte hain, koi guess nahi, kyunki period sirf geometry ( $a$ ) aur gravity ( $μ$ ) se fix hoti hai — kuch aur nahi.

Definition Do spacings jo tum constantly use karoge — Level 1 se pehle inhe samjho

$S$ = nodal spacing: pichle orbit ke equator crossing se current orbit ka equator crossing kitna west shift hota hai. Parent note isko radians mein $S = ω_{\oplus} T$ deta hai. Degrees mein padhne ke liye hum sirf angle convert karte hain: ek full turn $2 π$ radians bhi hai aur $360°$ bhi, toh $\frac{360°}{2 π}$ se multiply karo: $S = ω_{\oplus} T = \frac{2 π}{T _{\oplus}} T [rad] \times \frac{360°}{2 π} S = 360° \times \frac{T}{T _{\oplus}} [deg] .$ $2 π$ cancel ho jaata hai — isliye degree form mein $2 π$ nahi hota aur sirf $360° \times T / T_{\oplus}$ bachta hai.
$δ$ = closed-cycle track spacing: poore $N$ orbits ke repeat cycle ke baad, $N$ distinct tracks $360°$ equator ko $N$ equal strips mein tile kar deti hain, toh har strip $δ = 360°/ N$ wide hoti hai. $S$ ek orbit ka jump hai; $δ$ final tightened gap hai. Inhe alag rakho.

Level 1 — Recognition

L1.1

Words aur symbol mein batao ki subsatellite point kya hota hai, aur inclination $i = 82°$ wali groundtrack ki maximum latitude ka formula likho.

Recall Solution — L1.1

Subsatellite point woh jagah hai Earth ki surface pe jo satellite ke bilkul neeche hoti hai — woh point jahaan Earth ke centre se satellite ke through jaane wali straight line surface ko pierce karti hai. Kyunki $i = 82° \leq 90°$ , hum top branch use karte hain: $ϕ_{m a x} = min (82°, 98°) = i = 82°$ . Track $+ 82°$ aur $- 82°$ latitude ke beech oscillate karti hai aur kabhi poles tak nahi pahunchi.

L1.2

Ek orbit ka period $T = 6000 s$ hai. Westward nodal spacing $S$ degrees mein compute karo.

Recall Solution — L1.2

Parent ke radian form se shuru karo aur degrees mein convert karo (jaisa upar setup kiya hai): kyunki $S = ω_{\oplus} T = \frac{2 π}{T _{\oplus}} T$ radians hai, aur ek turn $2 π$ rad bhi hai aur $360°$ bhi, $2 π$ cancel ho jaata hai aur bachta hai: $S = 360° \times \frac{T}{T _{\oplus}} .$ Convert kyun: hum ek human-readable angle degrees mein chahte hain, radians mein nahi. Substitute karo: $S = 360° \times \frac{6000}{86164} = 25.07°.$ Har equator crossing pichle wale se $\approx 25°$ west mein land karti hai.

L1.3

Ek satellite exactly $N / D = 15/1$ orbits per day karta hai. Woh repeat condition jo yeh satisfy karta hai batao aur kitne din baad track repeat hogi.

Recall Solution — L1.3

Repeat condition hai $N T = D T_{\oplus}$ jahaan $N = 15, D = 1$ coprime hain. Track $D = 1$ sidereal day baad (15 orbits mein) repeat hogi. Upar define ki gayi closed-cycle spacing $δ = 360°/ N$ use karte hue, equatorial track spacing $δ = 360°/15 = 24°$ hai.

Level 2 — Application

L2.1

Ek Low-Earth orbit altitude $h = 567 km$ pe hai. Iska period $T$ , phir nodal spacing $S$ nikalo, aur estimate karo ki ek din mein kitne equator crossings hote hain.

Recall Solution — L2.1

Step 1 (period). Orbital radius $a = R_{E} + h = 6371 + 567 = 6938 km$ . $T = 2 π \frac{a ^{3}}{μ} = 2 π \frac{693 8 ^{3}}{398600} = 5760 s (\approx 96 min) .$ Yahan Kepler kyun: period free nahi hoti — altitude $a$ fix karta hai, aur $a$ $T$ fix karta hai. Step 2 (spacing). $S = 360° \times \frac{5760}{86164} = 24.07°.$ Step 3 (crossings/day). Orbits per day $= T_{\oplus} / T = 86164/5760 = 14.96 \approx 15$ crossings — $360°/24.07° = 14.96$ se match karta hai. Consistent hai.

L2.2

$h = 700 km$ pe ek nadir-pointing imager ka half field-of-view $η = 20°$ hai. Earth-central half-angle $λ_{s}$ aur swath width $W$ nikalo.

Figure — Groundtrack analysis — swath, revisit

Recall Solution — L2.2

Hum sine rule use karte hain triangle Earth-centre → satellite → ground-edge pe (figure dekho: lavender edge $R_{E} + h$ hai, dashed slate edge $R_{E}$ hai, coral ray sensor line of sight hai, angle $η$ satellite pe hai aur $λ_{s}$ Earth ke centre pe). Sine rule isliye use karte hain kyunki hume do lengths ( $R_{E}$ , $R_{E} + h$ ) aur ek angle ( $η$ ) pata hai aur far angle chahiye — exactly wahi jo sine rule deta hai. Step 1. $sin (η + λ_{s}) = \frac{R _{E} + h}{R _{E}} sin η = \frac{7071}{6371} sin 20° = 1.1099 \times 0.3420 = 0.3796.$ Step 2. $η + λ_{s} = arcsin (0.3796) = 22.31° \Rightarrow λ_{s} = 2.31°.$ Step 3. Arc length ke liye radians mein convert karo ( $arc = R θ_{rad}$ ): $W = 2 R_{E} λ_{s} = 2 \times 6371 \times (2.31 \times \frac{π}{180}) = 2 \times 6371 \times 0.04032 = 514 km .$

L2.3

L2.1 ki orbit ke liye ( $S = 24.07°$ ), ek sensor ka swath $W = 376 km$ hai. Equatorial track gap ko kilometres mein convert karo aur decide karo ki kya ek din ke passes equator ko gap-free cover karte hain.

Recall Solution — L2.3

Neighbouring same-day passes ke beech equator pe gap spacing $S$ ko arc length mein convert karke milta hai. Equatorial arc ka ek degree hota hai: $\frac{2 π R _{E}}{360°} = \frac{2 π \times 6371}{360} = 111.2 km/deg .$ Gap $= 24.07° \times 111.2 = 2677 km$ . Kyunki swath $W = 376 km ≪ 2677 km$ hai, ek din ke passes bahut bade gaps chhodte hain — coverage ke liye multi-day repeat pattern zaruri hai jo tracks ko interleave kare.

Level 3 — Analysis

L3.1

Ek satellite $D = 3$ days mein $N = 43$ orbits ke baad apni groundtrack repeat karta hai. Nikalo (a) orbits per day, (b) exact repeat period seconds mein, (c) required orbital period $T$ , (d) equatorial track spacing $δ$ degrees aur km mein.

Recall Solution — L3.1

$g cd (43, 3) = 1$ , toh fraction already lowest terms mein hai — achha hai, warna true cycle chhoti hoti. (a) Orbits/day $= N / D = 43/3 = 14.333$ . (b) Repeat period $= D T_{\oplus} = 3 \times 86164 = 258492 s$ . (c) $N T = D T_{\oplus}$ se, dono sides ko $N$ se divide karo $T$ isolate karne ke liye ( $T$ chahiye, toh $N$ se multiply undo karo): $T = \frac{D T _{\oplus}}{N} = \frac{258492}{43} = 6011.4 s$ . (d) $δ = 360°/ N = 360°/43 = 8.372°$ . Km mein: $8.372° \times 111.2 = 931 km$ . $δ$ mein $N$ kyun hai, $D$ nahi: poore cycle mein $N$ distinct tracks $360°$ equator ko $N$ equal strips mein tile karte hain.

L3.2

Do candidate imagers L3.1 ki orbit pe fly karte hain. Sensor A ka $W = 800 km$ hai, Sensor B ka $W = 600 km$ hai. Equatorial track spacing $δ = 931 km$ hai (L3.1 se). Kaun sa sensor 3-day cycle mein gap-free equatorial coverage achieve karta hai, aur dono ke liye coverage overlap ya shortfall kya hai?

Recall Solution — L3.2

Gap-free coverage ke liye $W \geq δ = 931 km$ chahiye.

Sensor A: $W = 800 < 931$ → shortfall $= 931 - 800 = 131 km$ uncovered strip adjacent tracks ke beech. Gap-free nahi hai.
Sensor B: $W = 600 < 931$ → shortfall $= 331 km$ . Aur bura. Koi bhi 3 days mein equator gap-free cover nahi karta; ya toh swath 931 km se zyada badhaao ya mission ko longer repeat cycle accept karni hogi (bada $N$ ⇒ chhota $δ$ ).

L3.3

Qualitatively aur phir numerically, Kepler's law use karke dikhao ki altitude $h_{1} = 567 km$ se $h_{2} = 1200 km$ tak badhane se nodal spacing $S$ increase hoti hai. Dono $S$ values do.

Recall Solution — L3.3

Qualitative: bada $a$ ⇒ bada $T$ (Kepler) ⇒ Earth har orbit mein zyada rotate karti hai ⇒ bada $S = ω_{\oplus} T$ . Altitude aur spacing saath saath badhte hain. Numerical:

$a_{1} = 6938 km$ : $T_{1} = 5760 s$ , $S_{1} = 24.07°$ (L2.1 se).
$a_{2} = 6371 + 1200 = 7571 km$ : $T_{2} = 2 π 757 1^{3} /398600 = 6558 s$ , toh $S_{2} = 360° \times 6558/86164 = 27.40°$ . Sach mein $S_{2} > S_{1}$ hai. ✔

Level 4 — Synthesis

L4.1

Design task. Tum ek 1-day repeat orbit ( $D = 1$ ) chahte ho exactly $N = 14$ orbits per day ke saath, circular. Required period $T$ , semi-major axis $a$ , aur altitude $h$ nikalo.

Recall Solution — L4.1

Step 1 (repeat condition se period). Kyun: repeat condition $N T = D T_{\oplus}$ fix karta hai ki ek orbit kitni der ki honi chahiye taaki $N$ orbits ke baad exactly $D$ days guzar jayein. $N$ se divide karke $T$ solve karo: $T = \frac{86164}{14} = 6154.57 s$ . Step 2 ( $a$ ke liye Kepler invert karo). Invert kyun: hume $T$ pata hai aur $a$ chahiye, toh hum $T = 2 π a^{3} / μ$ algebraically undo karte hain: dono sides square karo ( $T^{2} = 4 π^{2} a^{3} / μ$ ) square root hatane ke liye, $μ$ se multiply aur $4 π^{2}$ se divide karo $a^{3}$ isolate karne ke liye, phir cube root lao $a$ isolate karne ke liye: $a = (\frac{μ T ^{2}}{4 π ^{2}})^{1/3} = (\frac{398600 \times 6154.5 7 ^{2}}{4 π ^{2}})^{1/3} = 7270 km .$ Step 3 (altitude). $R_{E}$ kyun subtract karo: $a$ Earth ke centre se measure hota hai, lekin altitude surface ke upar ki height hai, toh ek Earth radius hatao: $h = a - R_{E} = 7270 - 6371 = 899 km$ . $\approx 899 km$ pe ek circular orbit 14 revolutions ke baad har ek sidereal day mein apna track close karti hai.

L4.2

L4.1 extend karo: us 1-day, $N = 14$ cycle mein gap-free equatorial coverage ke liye kaun sa swath $W$ (aur matching sensor half-angle $η$ $h = 899 km$ pe) chahiye?

Recall Solution — L4.2

Step 1 (required swath). $δ$ kyun, $S$ kyun nahi: coverage closed-cycle spacing se decide hoti hai. $δ = 360°/14 = 25.714°$ . Km mein: $25.714° \times 111.2 = 2860 km$ . Toh $W \geq 2860 km$ chahiye — ek wide-swath imager. Step 2 ( $λ_{s}$ back out karo). $2 R_{E}$ se kyun divide karo: swath relation $W = 2 R_{E} λ_{s}$ $λ_{s}$ mein linear hai, toh divide karke isolate karo: $λ_{s} = \frac{W}{2 R _{E}} = \frac{2860}{2 \times 6371} = 0.2245 rad = 12.86°.$ Step 3 ( $η$ back out karo). Swath relation $sin (η + λ_{s}) = \frac{R _{E} + h}{R _{E}} sin η$ invert karo $a / R_{E} = 7270/6371 = 1.1411$ ke saath: $sin (η + 12.86°) = 1.1411 sin η .$ Angle-sum identity se expand karo: $sin η cos 12.86° + cos η sin 12.86° = 1.1411 sin η$ . Har term ko $cos η$ se divide karo (isliye $tan$ aata hai — yeh do unknowns $sin η, cos η$ ko ek quantity $tan η$ mein collapse karta hai): $tan η = \frac{s i n 12.86°}{1.1411 - c o s 12.86°} = \frac{0.2225}{1.1411 - 0.9749} = \frac{0.2225}{0.1662} = 1.339 \Rightarrow η = 53.2°.$ $53°$ half-angle bahut wide hai (horizon ke paas grazing), jo dikhata hai ki LEO se 1-day gap-free coverage physically demanding hai.

Level 5 — Mastery

L5.1

Ek Sun-synchronous, Landsat-jaisi design $N = 233$ orbits $D = 16$ days mein use karti hai. (a) $T$ nikalo. (b) $a$ aur $h$ nikalo. (c) $δ$ km mein nikalo. (d) Sensor swath $W = 185 km$ hai — kya equator pe coverage gap-free hai, aur margin kya hai?

Recall Solution — L5.1

(a) $N$ se kyun divide karo: repeat condition $N T = D T_{\oplus}$ kehta hai $N$ orbits exactly $D$ sidereal days fill karein, toh ek orbit $T = \frac{D T _{\oplus}}{N} = \frac{16 \times 86164}{233} = 5916.6 s$ tak chalti hai. (b) Kepler kyun invert karo (square, isolate, cube-root): period $a$ se $T = 2 π a^{3} / μ$ ke through aati hai; isko ulta chalane se $a = (\frac{μ T ^{2}}{4 π ^{2}})^{1/3} = (\frac{398600 \times 5916. 6 ^{2}}{4 π ^{2}})^{1/3} = 7077 km$ milta hai. $R_{E}$ kyun subtract karo: altitude surface ke upar ki height hai, $h = 7077 - 6371 = 706 km$ . (c) $\times 111.2$ kyun: equatorial arc ke ek degree ko km mein turn karne ke liye km-per-degree factor $2 π R_{E} /360 = 111.2 km/deg$ se multiply karte hain. $δ = 360°/233 = 1.5451° \Rightarrow δ = 1.5451 \times 111.2 = 171.8 km$ . (d) $W = 185 km \geq δ = 171.8 km$ → gap-free, overlap margin $185 - 171.8 = 13.2 km$ ke saath (har strip apne neighbour se $\approx 13$ km overlap karti hai). Revisit $\leq 16$ days equator pe har jagah.

L5.2

Capstone. L5.1 orbit ke liye inclination $i = 98.2°$ hai (retrograde, Sun-synchronous). (a) Groundtrack maximum kaun si latitude reach karta hai? (b) Latitude $ϕ = 60°$ pe, adjacent equatorial tracks converge hoti hain — ground spacing $cos ϕ$ factor se shrink hoti hai. Pehle derive karo kyun factor $cos ϕ$ hai, phir $60°$ N pe effective track spacing compute karo aur confirm karo ki swath ab generously overlap karti hai.

Recall Solution — L5.2

(a) Kaun sa branch: kyunki $i = 98.2° > 90°$ orbit retrograde hai, toh $ϕ_{m a x} = 180° - i$ use karo (equivalently $min (98.2°, 81.8°)$ ): $ϕ_{m a x} = 180° - 98.2° = 81.8°$ . Track $\pm 81.8°$ latitude tak sweep karti hai, polar caps unimaged chhodte hue — Sun-synchronous near-polar orbits ki yeh signature hai. Note karo ki branch essential hai: naive " $ϕ_{m a x} = i = 98.2°$ " claim karega ek latitude North Pole se aage, jo impossible hai.

(b) — Pehle $cos ϕ$ factor derive karo. Fig 2 dekho. Do neighbouring tracks ek fixed longitude difference $Δ λ$ se separated hain (Earth ki spin axis ke around ek angle). Hum inke beech ground distance chahte hain, aur yahan key geometric fact hai: constant latitude $ϕ$ ka circle equator se chhota hota hai. Iska radius $R_{E}$ nahi balki $r (ϕ) = R_{E} cos ϕ$ hota hai — kyunki latitude circle sphere ka ek slice hai jo axis ke $R_{E} sin ϕ$ upar liya gaya hai, aur right triangle (axis, radius-to-surface, horizontal radius) se horizontal radius $R_{E} cos ϕ$ hota hai. $cos$ kyun, $sin$ kyun nahi: horizontal radius woh side hai jo Earth ke centre pe latitude angle $ϕ$ se adjacent hai, aur adjacent/hypotenuse $= cos ϕ$ hota hai. Ek fixed longitude gap $Δ λ$ is chhote circle pe chhota arc subtend karta hai: $ground gap (ϕ) = r (ϕ) Δ λ = R_{E} cos ϕ Δ λ = = δ at equator R_{E} Δ λ \times cos ϕ .$ Toh effective spacing $δ_{ϕ} = δ cos ϕ$ hai — wahi longitude spacing sirf ek chhote circle pe land karti hai higher latitude pe. Fig 2 mein dono tracks ke beech double-headed arrow visibly shrink hota dikhta hai jaise tum upar jaate ho, aur dotted guide lines meridians ko pole ki taraf crowding dikhate hain.

$ϕ = 60°$ pe compute karo: $δ_{ϕ} = δ cos ϕ = 171.8 \times cos 60° = 171.8 \times 0.5 = 85.9 km .$ $W = 185 km ≫ 85.9 km$ ke saath, tracks $60°$ pe double se zyada overlap karti hain — high-latitude coverage equator se kaafi denser (aur revisit faster) hoti hai. Isliye Sun-synchronous missions mein polar-region imaging ek strength hai: wahi convergence jo $ϕ_{m a x}$ limit karti hai woh coverage ko iske bilkul tak tight bhi kar deti hai.

Recall Self-check: kaun sa formula kaun sa sawaal answer karta hai?

Spacing per orbit ::: $S = 360° \times T / T_{\oplus}$ Track spacing after full repeat ::: $δ = 360°/ N$ Ground spacing at latitude $ϕ$ ::: $δ_{ϕ} = δ cos ϕ$ Period from altitude ::: $T = 2 π a^{3} / μ$ Repeat condition ::: $N T = D T_{\oplus}$ , $N, D$ coprime Swath width ::: $W = 2 R_{E} λ_{s}$ , $λ_{s} = arcsin (\frac{( R _{E} + h ) s i n η}{R _{E}}) - η$ Max latitude ::: $ϕ_{m a x} = min (i, 180° - i)$

Prerequisite links: Inclination & Orbital Elements · Nodal Regression & J2 Perturbation · Remote Sensing Sensor Geometry · Orbital Period & Kepler's Third Law