Rendezvous and proximity operations — Clohessy-Wiltshire equations: Worked examples

3.2.40 · D3 · Physics › Orbital Mechanics & Astrodynamics › Rendezvous and proximity operations — Clohessy-Wiltshire equ

Scenario matrix

Har CW problem asal mein yeh choose karna hai ki kaunse initial numbers non-zero hain aur unka sign kya hai. Yeh table har case class list karti hai. Neeche ke examples us cell ke saath tagged hain jo woh cover karte hain, aur "Figure" column batata hai ki kaun se draw kiye gaye hain (ek case tab draw hota hai jab path ki geometry insight deti hai; algebra-only cases numerically check kiye gaye hain).

#	Case class	Kya special hai	Covered by	Figure
C1	Pure cross-track, $z_{0} \neq = 0$ , sab in-plane $= 0$	Decoupled SHM, koi drift nahi	Ex 1	s01
C2	Radial offset $x_{0} > 0$ (higher orbit)	Secular drift, walk-off ka sign	Ex 2	s02
C3	Radial offset $x_{0} < 0$ (lower orbit)	Opposite-sign drift, phasing	Ex 3	s02 ka mirror
C4	Bounded-motion condition $\overset{y}{˙}_{0} = - 2 n x_{0}$	Football ellipse, zero drift	Ex 4	s03
C5	Sirf along-track offset, $y_{0} \neq = 0$ , zero velocity	Degenerate: chaser hovers	Ex 5	(fixed point)
C6	Pure radial velocity kick $\overset{x}{˙}_{0} \neq = 0$	Burn se ellipse, net drift nahi	Ex 6	2:1 like s03
C7	Pure along-track velocity kick $\overset{y}{˙}_{0} \neq = 0$	Tangential burn se strong drift	Ex 7	s04
C8	Limiting case $t \to 0$ aur $t = T$ (ek period)	Formulas ke edges par sanity check	Ex 8	(algebraic)
C9	Real-world word problem (ISS resupply catch-up)	Full numeric design	Ex 9	s02 use karta hai
C10	Exam twist: radial + tangential combine karke drift cancel karo	Burn ke liye solve karo	Ex 10	2:1 like s03

Poore note mein, ek low-Earth target orbit of radius $R_{0} = 6.778 \times 1 0^{6} m$ lo (lagbhag $400$ km altitude). Tab $n = \frac{3.986 \times 1 0 ^{14}}{( 6.778 \times 1 0 ^{6} ) ^{3}} \approx 1.131 \times 1 0^{- 3} s^{- 1},$ aur ek orbital period hai $T = 2 π / n \approx 5556 s \approx 92.6 min$ . Hum yeh numbers numeric examples mein reuse karte hain.

Worked examples

Worked example Example 1 — Pure cross-track oscillation (cell C1)

Chaser $z_{0} = 50$ m orbital plane se bahar start karta hai, baaki sab zero. Ek quarter period ke baad, $t = T /4$ , yeh kahan hai, aur wahan iska cross-track speed kya hai?

Forecast: Compute karne se pehle andaaza lagao — kya yeh drift ho jaayega, ya wapas aayega? $t = T /4$ par yeh kis direction mein move kar raha hoga?

$z$ solution use karo $\overset{z}{˙}_{0} = 0$ ke saath: $z (t) = z_{0} cos n t$ . Yeh step kyun? Cross-track $x, y$ se fully decoupled hai (parent ka $\overset{z}{¨} + n^{2} z = 0$ spring par mass ki equation jaisa hai), isliye sirf yeh ek line chahiye.
$t = T /4$ par $n t = \frac{n}{4} \cdot \frac{2 π}{n} = \frac{π}{2}$ hai, toh $cos n t = 0$ . Isse $z (T /4) = 50 cos (π /2) = 0$ m. Yeh step kyun? Quarter period exactly wahan hai jahan ek jhoolta pendulum middle se guzarta hai — position zero, speed maximum.
Speed: $\overset{z}{˙} (t) = - z_{0} n sin n t$ , toh $\overset{z}{˙} (T /4) = - 50 \cdot n \cdot 1 = - 0.0566$ m/s. Yeh step kyun? Cosine differentiate karne se sine milta hai; $n t = π /2$ par sine $1$ hai, apni peak par — crossing par maximum speed confirm karta hai.

Verify: $z$ ka amplitude kabhi $50$ m se zyada nahi hoga aur motion periodic hai — koi $n t$ term nahi aata, isliye secular drift nahi, bilkul expected SHM ki tarah. Units: $z_{0} [m] \times n [s^{- 1}] = m/s$ . ✓

Neeche wala figure ek full period mein $z (t)$ plot karta hai. Step 2 isme dekho: green curve exactly $t = T /4$ par zero cross karti hai (red dot), aur wahi crossing hai jahan slope sabse steep hai — step 3 ki maximum speed. Notice karo ki curve $t = T$ par $+ 50$ m par wapas aati hai: bounded, no drift, cell C1 ki pehchaan.

Figure — Rendezvous and proximity operations — Clohessy-Wiltshire equations

Worked example Example 2 — Radial offset, higher orbit (cell C2)

Chaser $x_{0} = + 100$ m radially outward park kiya hua hai, sab velocities zero. Ek full period $T$ ke baad yeh along-track kitna drift hua hai, aur kis direction mein?

Forecast: Higher orbit = slower. Kya tumhara andaaza hai ki chaser peeche reh jaayega ( $y$ zyada negative) ya aage nikal jaayega?

Parent ke $y$ solution se along-track drift, $\overset{x}{˙}_{0} = \overset{y}{˙}_{0} = y_{0} = 0$ ke saath: $y (t) = 6 (sin n t - n t) x_{0}$ . Yeh step kyun? Sirf $x_{0}$ term bachta hai, aur yahi secular $- 6 n t x_{0}$ piece carry karta hai.
$t = T$ par, $n t = 2 π$ , toh $sin n t = 0$ aur $y (T) = 6 (0 - 2 π) x_{0} = - 12 π x_{0}$ . Yeh step kyun? Oscillation ek full period ke baad zero par wapas aati hai; sirf linear-in- $t$ walk-off bachta hai, jo clean number $12 π$ deta hai.
Numerically $y (T) = - 12 π (100) = - 3769.9$ m $\approx - 3.77$ km. Yeh step kyun? $x_{0} = + 100$ plug karo; negative sign ka matlab hai chaser peeche khisak gaya (along-track negative), jo "higher orbit is slower" ke saath consistent hai.

Verify: Sign check — $x_{0} > 0$ se $y < 0$ (peeche reh jaata hai). $100$ m radial nudge se ek orbit mein $3.77$ km along-track walk hua: chhoti si wajah, bada asar, parent ki secular drift warning se match karta hai. ✓

Figure LVLH plane mein actual $(y, x)$ path trace karta hai. Blue dot se shuru karo ( $x = + 100$ m, $y = 0$ ); ek orbit ke baad chaser red dot par 3.77 km peeche land karta hai — bhaale radial coordinate $+ 100$ m par wapas aa gayi. Non-closing, open curve exactly "secular drift" jaisi dikhti hai: isse Example 4 ke closed loop se compare karo.

Worked example Example 4 — Bounded football orbit (cell C4)

$x_{0} = + 50$ m radially bahar se shuru karo. Woh along-track velocity $\overset{y}{˙}_{0}$ choose karo jo sab drift khatam kar de, aur resulting path describe karo.

Forecast: Exactly ek magic $\overset{y}{˙}_{0}$ hai. Kya yeh positive hai ya negative? Kaisi shape banti hai?

Bounded-motion condition (parent): $\overset{y}{˙}_{0} = - 2 n x_{0}$ . Yeh step kyun? Yeh value $x_{0}$ aur $\overset{y}{˙}_{0}$ contributions ke beech secular $n t$ terms cancel karti hai, sirf pure trig bachti hai — ek closed loop.
$\overset{y}{˙}_{0} = - 2 (1.131 \times 1 0^{- 3}) (50) = - 0.1131$ m/s (retrograde along-track). Yeh step kyun? $x_{0} > 0$ ke saath walk-off balance karne ke liye negative along-track kick chahiye.
$\overset{x}{˙}_{0} = 0, y_{0} = 0$ ke saath, solution ban jaata hai $x (t) = (4 - 3 cos n t) x_{0} + \frac{2}{n} (1 - cos n t) \overset{y}{˙}_{0}$ . $\overset{y}{˙}_{0} = - 2 n x_{0}$ substitute karne par milta hai $x (t) = x_{0} cos n t$ , aur correspondingly $y (t) = - 2 x_{0} sin n t$ . Yeh step kyun? Mushkil constants collapse ho jaate hain; path ek ellipse hai jiska radial amplitude $x_{0}$ aur along-track amplitude $2 x_{0}$ hai.
Yahi hai 2:1 football ellipse — along-track swing radial swing ki do guni hai.

Verify: $t = T$ par, $cos n t = 1, sin n t = 0$ , toh $x (T) = x_{0} = 50$ m, $y (T) = 0$ — chaser apne start par wapas. Closed loop, zero drift. Along-track amplitude $2 \times 50 = 100$ m 2:1 ratio confirm karta hai. ✓

Figure woh closed ellipse draw karta hai. s02 ki open curve ke ulta, yeh apne start par wapas aati hai (green dot) — drift step 2 ke retrograde kick ne cancel kar di. Axes measure karo: radial half-width $50$ m, along-track half-width $100$ m — step 4 ka 2:1 ratio, seedha picture se padho.

Worked example Example 5 — Degenerate along-track offset (cell C5)

Chaser along-track axis par target ke directly peeche $y_{0} = - 200$ m baitha hai, sab velocities zero, $x_{0} = 0$ . Kisi bhi time $t$ ke baad yeh kahan hai?

Forecast: Kya yeh wahan rehta hai, ya "peeche" hona drift cause karta hai?

$x_{0} = 0, \overset{x}{˙}_{0} = 0, \overset{y}{˙}_{0} = 0$ ke saath solutions reduce ho jaate hain $x (t) = 0$ aur $y (t) = y_{0}$ mein. Yeh step kyun? Dono formulas ke har term ko $x_{0}$ ya kisi velocity se multiply kiya gaya hai; sab zero hain, toh sirf standalone $y_{0}$ bachta hai.
Toh $x (t) = 0, y (t) = - 200$ m har $t$ ke liye — ek fixed point. Yeh step kyun? Along-track axis mein koi restoring force nahi hai (ek baar $\overset{x}{˙} = 0$ ho jaaye toh $\overset{y}{¨} + 2 n \overset{x}{˙} = 0$ mein koi $y$ term nahi hai), isliye pure along-track offset (first order tak) ek equilibrium hai.

Verify: $\overset{y}{¨} + 2 n \overset{x}{˙} = 0$ mein plug karo: $\overset{x}{˙} = 0$ aur $\overset{y}{¨} = 0$ ✓; plug into $\overset{x}{¨} - 2 n \overset{y}{˙} - 3 n^{2} x = 0$ : sab terms zero ✓. Isliye V-bar ( $y$ -axis) docking ke liye stable loiter line hai. Koi figure zaroori nahi — trajectory $y$ -axis par ek stationary point hai, jo plot us sentence se zyada clear nahi kar sakta ki "yeh hilta nahi". ✓

Worked example Example 6 — Pure radial velocity kick (cell C6)

Origin se ( $x_{0} = y_{0} = 0$ ), ek radial burn $\overset{x}{˙}_{0} = + 0.05$ m/s fire karo. Motion trace karo; kya yeh drift karta hai?

Forecast: Radial "up" burn — kya chaser hamesha ke liye nikal jaata hai, ya loop karke wapas aata hai?

Sirf $\overset{x}{˙}_{0}$ non-zero ke saath in-plane solution: $x (t) = \frac{x ˙ _{0}}{n} sin n t$ , $y (t) = - \frac{2 x ˙ _{0}}{n} (1 - cos n t)$ . Yeh step kyun? Dono surviving terms purely oscillatory hain — koi $n t$ factor nahi — isliye secular drift nahi hai.
Radial amplitude $= \overset{x}{˙}_{0} / n = 0.05/ (1.131 \times 1 0^{- 3}) = 44.2$ m. Yeh step kyun? $sin n t$ ki peak $1$ hai, toh sabse bada radial excursion $\overset{x}{˙}_{0} / n$ hai.
Along-track term $- \frac{2 x ˙ _{0}}{n} (1 - cos n t)$ $0$ aur $- 2 (44.2) = - 88.4$ m ke beech swing karta hai, phir wapas aata hai. Yeh step kyun? $(1 - cos n t)$ $0$ se $2$ range karta hai; motion phir se ek closed 2:1 ellipse hai, bas origin se shift hokar — geometrically s03 football jaisi shape, shifted.

Verify: $t = T$ par: $sin n t = 0, 1 - cos n t = 0$ , toh $x = 0, y = 0$ — ghar wapas, drift nahi. Sirf radial burn se kabhi secular walk-off nahi hota (yeh sirf $x_{0}$ aur $\overset{y}{˙}_{0}$ se hota hai). ✓

Worked example Example 7 — Pure along-track velocity kick (cell C7)

Origin se, ek tangential burn $\overset{y}{˙}_{0} = + 0.05$ m/s (forward) fire karo. Ek period ke baad along-track drift?

Forecast: "Aage push karo toh aage jao" — parent ne ise ek trap kaha tha. Asal mein kya hota hai?

Sirf $\overset{y}{˙}_{0}$ non-zero ke saath along-track solution: $y (t) = \frac{1}{n} (4 sin n t - 3 n t) \overset{y}{˙}_{0}$ . Yeh step kyun? Andar ka $- 3 n t$ ek secular term hai, toh ek forward burn se drift hoti hai.
$t = T$ par: $sin n t = 0$ , toh $y (T) = \frac{1}{n} (- 3 \cdot 2 π) \overset{y}{˙}_{0} = - \frac{6 π}{n} \overset{y}{˙}_{0}$ . Yeh step kyun? Ek full loop ke baad sirf linear piece bachti hai.
Numerically $y (T) = - \frac{6 π}{1.131 \times 1 0 ^{- 3}} (0.05) = - 833.7$ m $\approx - 0.83$ km. Yeh step kyun? Sign dekho: ek forward burn ne backward net shift produce ki — Coriolis coupling ne orbit raise ki aur chaser slow kar diya.

Verify: Sign negative hai bawajood positive (forward) burn ke — yahi parent ki "target ki taraf fire karo aur peeche reh jaao" wali galti quantitative roop mein hai. Magnitude: $6 π / n = 6 π / (1.131 \times 1 0^{- 3}) \approx 1.667 \times 1 0^{4}$ , times $0.05$ se $\approx 834$ m milta hai. ✓

Figure yeh path trace karta hai: green start se chaser loop karta hai lekin close nahi karta — ek orbit ke baad yeh $834$ m peeche baitha hai (orange dot), us forward direction ke ulta jisme push kiya gaya tha. Yeh parent ki "fire forward, fall behind" warning ka visual proof hai.

Worked example Example 8 — Limiting checks

$t \to 0$ aur $t = T$ (cell C8) Confirm karo ki formulas do edges par theek behave karti hain: release ka instant, aur ek full period.

Forecast: $t = 0$ par answer initial state ke bilkul barabar hona chahiye. Andaaza lagao ki kya drift terms wahan problem create karte hain.

$t = 0$ par: $cos n t = 1, sin n t = 0, n t = 0$ . Tab $x (0) = (4 - 3) x_{0} = x_{0}$ , $y (0) = y_{0}$ , $z (0) = z_{0}$ . Yeh step kyun? Ek solution ko $t = 0$ par apni initial condition reproduce karni hi chahiye; $4 - 3 = 1$ combination built-in check hai ki yeh aise karta hai.
$t = 0$ par velocities: differentiate karo aur $t = 0$ set karo taaki $\overset{x}{˙} (0) = \overset{x}{˙}_{0}$ , $\overset{y}{˙} (0) = \overset{y}{˙}_{0}$ mile. Yeh step kyun? Sab chhe state components recover hone chahiye, confirm karte hue ki closed form ODEs ka genuine solution hai, sirf baad mein valid approximation nahi.
$t = T$ par ( $n t = 2 π$ ): har $cos \to 1, sin \to 0$ , sirf $x_{0}$ -driven $- 12 π x_{0}$ aur $\overset{y}{˙}_{0}$ -driven $- \frac{6 π}{n} \overset{y}{˙}_{0}$ $y$ mein bachte hain; baaki apne start par wapas aate hain. Yeh step kyun? Yeh exactly identify karta hai ki kaun se inputs drift cause karte hain (radial offset aur tangential velocity) aur kaun nahi karte (radial velocity, cross-track) — system ka full-period fingerprint.

Verify: $x (0) = x_{0}$ aur $z (0) = z_{0}$ exactly recover hue; $t = T$ ke sirf wahi surviving terms hain jo Examples 2 aur 7 mein identify kiye gaye do secular ones hain. Poore matrix mein consistent. ✓

Worked example Example 9 — ISS resupply catch-up (real-world, cell C9)

Ek cargo vehicle ISS se $y_{0} = - 2$ km peeche hai ( $R_{0} = 6.778 \times 1 0^{6}$ m). Radial-offset phasing burn use karke, exactly ek orbit mein gap close karne ke liye kaun sa radial offset $x_{0}$ chahiye, aur kya yeh upar hai ya neeche?

Forecast: Gap peeche hai, toh aage badhna hoga. Kya higher ya lower orbit mein jaana chahiye?

Ek radial offset se ek orbit mein net along-track shift: $Δ y = - 12 π x_{0}$ (Example 2). Yeh step kyun? Yahi drift-per-period fingerprint hai; oscillatory parts ek full loop mein cancel ho jaate hain.
Hamein $Δ y = + 2000$ m chahiye ( $- 2000$ m gap close karne ke liye). Solve karo $+ 2000 = - 12 π x_{0}$ : $x_{0} = - \frac{2000}{12 π} = - 53.05$ m. Yeh step kyun? Positive required drift $x_{0} < 0$ force karta hai — ek lower orbit, jo faster hai, chaser ko aage kheenchti hai.
Interpretation: ISS se $\approx 53$ m neeche jao, ek orbit coast karo ( $\approx 92.6$ min), aur 2 km gap close ho jaata hai. Phir match karne ke liye wapas raise karo. Yeh step kyun? Offset altitude par depend nahi karta (kyunki $n$ aur $T$ cancel hokar pure number $12 π$ de dete hain), isliye yeh ek robust rule of thumb hai.

Verify: $Δ y = - 12 π (- 53.05) = + 2000.0$ m ✓. Sign: lower orbit ( $x_{0} < 0$ ) → forward drift ( $Δ y > 0$ ) → gap close. Units: $[m]$ throughout. Yahi phasing s02 figure ke mirror mein miniature roop mein dikhti hai (open curve forward ki taraf opening). ✓

Worked example Example 10 — Exam twist: combined burn se drift cancel karo (cell C10)

Ek chaser mein unwanted radial offset $x_{0} = + 80$ m hai (baaki sab zero). Drift hone ke bajay, $t = 0$ par ek single along-track burn $\overset{y}{˙}_{0}$ apply karo taaki motion drift-free ban jaaye. $\overset{y}{˙}_{0}$ aur resulting ellipse amplitudes nikalo.

Forecast: Example 4 se tumhe magic condition pata hai — lekin $x_{0} > 0$ ke liye sign predict karo.

Drift-free ke liye chahiye $\overset{y}{˙}_{0} = - 2 n x_{0}$ (bounded-motion condition). Yeh step kyun? Yahi woh unique tangential velocity hai jo donon secular terms simultaneously cancel karti hai; yeh endless drift ko closed loop mein convert kar deti hai.
$\overset{y}{˙}_{0} = - 2 (1.131 \times 1 0^{- 3}) (80) = - 0.1810$ m/s. Yeh step kyun? $x_{0} > 0$ ke saath burn retrograde (negative $y$ ) hai, walk-off roka jaata hai.
Resulting path: $x (t) = x_{0} cos n t$ (radial amplitude $80$ m), $y (t) = - 2 x_{0} sin n t$ (along-track amplitude $160$ m) — ek 2:1 football. Yeh step kyun? Magic $\overset{y}{˙}_{0}$ substitute karne par constants exactly Example 4 ki tarah collapse ho jaate hain, naye $x_{0}$ ke saath scale hokar; shape wohi closed ellipse hai jaise s03, scale up hui.

Verify: $t = T$ par: $x (T) = x_{0} = 80$ m, $y (T) = 0$ — koi net drift nahi ✓. Amplitude ratio $160 : 80 = 2 : 1$ ✓. Ek single scalar burn ne drift ko tame kar diya, classic exam result. Yeh matrix ki last cell close karta hai: har scenario ab worked hai. ✓

Recall

Recall Kaun si initial conditions secular drift cause karti hain, aur kaun si nahi?

Radial offset $x_{0}$ (via $- 6 n t x_{0}$ ) aur along-track velocity $\overset{y}{˙}_{0}$ (via $- 3 n t \overset{y}{˙}_{0}$ ) drift cause karte hain ::: jabki radial velocity $\overset{x}{˙}_{0}$ , cross-track $z_{0}, \overset{z}{˙}_{0}$ , aur pure along-track offset $y_{0}$ NAHI karte. Ek radial offset $x_{0}$ se ek orbit mein drift kitni? ::: $Δ y = - 12 π x_{0}$ (lower orbit, $x_{0} < 0$ , tumhe aage le jaata hai). Closed (drift-free) in-plane loop ke liye ek condition? ::: $\overset{y}{˙}_{0} = - 2 n x_{0}$ , jo 2:1 football ellipse deta hai. Forward tangential burn se tum peeche kyun reh jaate ho? ::: Yeh orbit raise karta hai; higher orbit slower hoti hai, aur Coriolis velocity ko rotate karta hai — net $- \frac{6 π}{n} \overset{y}{˙}_{0}$ drift $\overset{y}{˙}_{0} > 0$ ke liye backward hoti hai.

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