Exercises — Rendezvous and proximity operations — Clohessy-Wiltshire equations

KYA karna hai: seedha $n=\sqrt{\mu /R_0^3}$ mein plug karo. KYUN: $n$ hi har CW term ki clock set karta hai (saare trig arguments $nt$ hain). $R_0=6.778\times 10^6\text{m},R_0^3=3.113\times 10^{20}{\text{m}}^3.$ $n=\sqrt{\frac{3.986\times 10^{14}}{3.113\times 10^{20}}}=\sqrt{1.280\times 10^{-6}}=1.132\times 10^{-3}\text{rad/s}.$ $T=\frac{2\pi }{n}=\frac{6.2832}{1.132\times 10^{-3}}\approx 5551\text{s}\approx 92.5\text{min}.$ Woh ~92 min wala jaana-pehchana low-Earth-orbit period hai — ek accha sanity check.

Cross-track equation $\ddot{z}+n^{2}z=0$. Yeh SHM hai angular frequency $n$ ke saath (orbit rate ke barabar hi). Special feature: yeh $x$ aur $y$ se completely decoupled hai — ek cross-track nudge bas ek orbit mein ek baar oscillate karta hai aur in-plane motion ko kabhi affect nahi karta.

• $-2n\dot{y}$ Coriolis force se aata hai (yeh radial acceleration ko along-track velocity se couple karta hai) — Rotating Reference Frames — Coriolis and Centrifugal dekho.
• $-3n^{2}x$ tidal / gravity-gradient term hai: yeh constant gravity ($-n^2{R}_0$) aur centrifugal ($+n^2{R}_0$) ke cancel hone ke baad bachta hai, plus first-order gravity slope $+2n^{2}x$ aur centrifugal $+n^{2}x$ milke right-hand side par $+3n^{2}x$ banaate hain.

KYA: $x_0=y_0=0$, ${\dot{y}}_0=0$, ${\dot{x}}_0=1$ ko state solution mein substitute karo. $x(t)=\frac{\sin nt}{n}{\dot{x}}_0=\frac{\sin nt}{n},y(t)=-\frac{2}{n}(1-\cos nt){\dot{x}}_0=-\frac{2}{n}(1-\cos nt).$ $t=T/4$ par: $nt=n\cdot (2\pi /n)/4=\pi /2$, toh $\sin nt=1$, $\cos nt=0$. $x=\frac{1}{n}=\frac{1}{1.132\times 10^{-3}}\approx 883.5\text{m},$ $y=-\frac{2}{n}(1-0)=-2\times 883.5\approx -1767\text{m}.$ KAISA DIKHTA HAI: along-track excursion radial wale se do guna hai — woh 2:1 shape hi "football" ellipse hai. Neeche figure dekho.

State declare kiya: $x_0=-10$, $y_0=0$, ${\dot{x}}_0=0$, ${\dot{y}}_0=0$. Kyunki $y_0=0$ aur dono initial velocities zero hain, $y(t)$ mein har term drop ho jaata hai siwaaye us term ke jo $x_0$ se multiply hai. KYA: along-track secular term use karo. Ek period mein oscillatory $\sin nt$ zero par wapas aa jaata hai, toh sirf $-nt$ wala piece bachta hai: $\Delta y=y(T)-y_0=6(\sin nT-nT)x_0=-6nT{x}_0=-12\pi x_0.$ $12\pi$ KYUN: $nT=2\pi$, toh $6nT=12\pi$ — ek pure number, altitude se independent. $\Delta y=-12\pi (-10)=120\pi \approx +376.99\text{m per orbit}.$ Positive $\Delta y$: neeche hona (lower orbit, zyada fast) tumhe aage le jaata hai. Yahi phasing trick hai.

$z(t)=z_0\cos nt=50\cos nt.$ Pehla zero jab $nt=\pi /2$, yaani $t=\frac{\pi }{2n}=\frac{\pi }{2\times 1.132\times 10^{-3}}\approx 1388\text{s}$ (ek quarter period). Sahi lagta hai: SHM har quarter cycle mein zero cross karta hai.

KYA: $y(t)$ mein har wo term collect karo jo $t$ ke proportional hai (ye drift terms hain): $y_{\text{drift}}(t)=\underset{\text{from }6(\sin nt-nt)x_0}{\underbrace{-6nt{x}_0}}+\underset{\text{from }\frac{1}{n}(4\sin nt-3nt){\dot{y}}_0}{\underbrace{-3t{\dot{y}}_0}}.$ KYUN: ek closed (repeating) orbit $t$ ke saath grow nahi kar sakta, toh $t$ ka coefficient zero hona chahiye: $-6n{x}_0-3{\dot{y}}_0=0\Rightarrow \boxed{{\dot{y}}_0=-2n{x}_0}.$ Physical meaning: along-track speed ko radial offset ki wajah se hue orbit-energy difference ko exactly compensate karna hoga. Radial offset ka matlab hai thoda alag orbital energy → alag period → drift; $-2n{x}_0$ along-track velocity energy ko re-tune karta hai taaki relative period target ke period se match kare.

Substitute karte hain ($x_0=y_0=0,{\dot{y}}_0=0,{\dot{x}}_0=v_r$): $x(t)=\frac{v_r}{n}\sin nt,y(t)=-\frac{2v_r}{n}(1-\cos nt).$ Along-track centre shift karo: $y^{\prime }=y+\frac{2v_r}{n}$ lo, toh $y^{\prime }=\frac{2v_r}{n}\cos nt$. $a=v_r/n$ (radial amplitude) lo. Tab $x=a\sin nt,y^{\prime }=2a\cos nt.$ Sahi weights ke saath square karke add karo: ${\left(\frac{x}{a}\right)}^2+{\left(\frac{y^{\prime }}{2a}\right)}^2={\sin }^{2}nt+{\cos }^{2}nt=1.$ Yeh radial semi-axis $a$ aur along-track semi-axis $2a$ wala ellipse hai — exactly 2:1. Neeche figure ise trace karta hai.

Aage ka burn ${\dot{y}}_0>0$ set karta hai with $x_0=0$. Radial response dekho: $x(t)=\frac{2}{n}(1-\cos nt){\dot{y}}_0\ge 0.$ Toh aage push karna actually tumhe upar uthata hai (positive $x$, higher orbit). Ab secular drift term $-6ntx$ (jisme $x>0$ average par) $y$ ko time ke saath decrease karata hai — tum peechhe drift karte ho. Physically: forward burn → higher orbit → slower angular rate → tum race haar jaate ho aur peechhe reh jaate ho. Yahi Coriolis coupling $\ddot{x}-2n\dot{y}$ action mein hai.

Step 1 — hume ek orbit mein kitni drift chahiye? Ex 2.2 se, ek chaser jo fixed radial offset $x_0$ par parked hai (with bounded-motion along-track velocity ${\dot{y}}_0=-2n{x}_0$, taaki woh loop away na kare) ek orbit mein $\Delta y=-12\pi x_0$ drift karta hai. Hume $\Delta y=+2000\text{m}$ chahiye (2 km lag close karna = 2 km aage badhna): $2000=-12\pi x_0\Rightarrow x_0=-\frac{2000}{12\pi }\approx -53.05\text{m}.$ Target se ~53 m neeche jaane par tum ek lap mein 2 km aage aa jaate ho. Step 2 — KYUN ek matching along-track velocity offset ko constant rakhti hai. Ek raw radial velocity kick tumhe fixed altitude par nahi rakhta — woh tumhe loop karaata hai (Ex 2.1). Constant radial offset $x_0$ par rehne ke liye taaki clean $\Delta y=-12\pi x_0$ formula apply ho, initial state ko Ex 3.1 ki bounded-motion condition satisfy karni hogi: ${\dot{y}}_0=-2n{x}_0.$ Yeh tumhe level kyun rakhta hai: yeh along-track velocity relative orbital energy ko exactly re-tune karti hai taaki radial coordinate grow na kare — $x(t)$ $x_0$ ke around ek bounded oscillation rehta hai instead of drifting ke. Apni value substitute karte hain: ${\dot{y}}_0=-2n{x}_0=-2(1.132\times 10^{-3})(-53.05)\approx +0.1201\text{m/s}.$ Toh entry burn ek ~12 cm/s along-track impulse hai (plus jo chhota radial component $x_0$ set kare). Notice karo ki ${\dot{y}}_0=-2n{x}_0$ velocity condition hai; offset $x_0$ khud ek aisi position hai jahan tum pahunchte ho, koi aisi cheez nahi jo velocity akeli sustain kare — yahi naïve "${\dot{x}}_0=n{x}_0$" shortcut ka correction hai. Step 3 — end mein ruko. Ek orbit ke baad tum $+2$ km move kar chuke ho aur radial $x_0$ par wapas ho same ${\dot{y}}_0$ ke saath. Ek matching $-0.1201$ m/s along-track burn (aur chhota radial trim $x_0$ null karne ke liye) tumhe target ke saath level park kar deta hai. Compare karo Orbital Maneuvers — Hohmann Transfer se, jahan same phasing idea do altitude-changing burns ke roop mein execute hota hai.

Differentiate karte hain: $\dot{x}=an\cos nt$, $\dot{y}=\dot{y^{\prime }}=-2an\sin nt$. $nt=\pi /2$ par (loop ka top, $x=a$, $y^{\prime }=0$): $\dot{x}=0$, $\dot{y}=-2an$. Toh us instant par motion purely along-track hai speed $2an$ par. Ek single along-track burn jis magnitude ka $\Delta v=2an=2\times 100\times 1.132\times 10^{-3}=0.2264\text{m/s}\approx 22.6\text{cm/s}$ wahan velocity exactly cancel kar deta hai. Position abhi bhi $y^{\prime }$-shift se offset hai, toh tum ek small hop follow up karte ho — lekin velocity kill wahan sabse clean hai jahan velocity single-axis ho.

${\dot{x}}_0={\dot{y}}_0=y_0=0$ ke saath: $x(t)=(4-3\cos nt)x_0,y(t)=6(\sin nt-nt)x_0.$ $nt=2\pi$ par: $\cos 2\pi =1$, $\sin 2\pi =0$. $x(T)=(4-3\cdot 1)x_0=x_0✓(\text{radial returns}),$ $y(T)=6(0-2\pi )x_0=-12\pi x_0✓(\text{along-track drift}).$ Toh STM $x_0\mapsto (x_0,-12\pi x_0)$ map karta hai ek period mein. Yeh Ex 2.2 ka mathematically exact version hai: radial offset $x$ mein periodic hai lekin $y$ mein secular hai.

Step 1 — mean motion as a function of radius. Radius $R$ ki circular orbit ke liye, Kepler deta hai $n(R)=\sqrt{\frac{\mu }{R^3}}=\sqrt{\mu }{R}^{-3/2}.$ Step 2 — differentiate karke dekho ki rate radius ke saath kaise change hoti hai. Derivative kyun? Hum chahte hain ki jab $R$ neeche nudge kare toh angular rate kis direction mein jaati hai; $dn/dR$ ka sign exactly yahi batata hai. $\frac{dn}{dR}=-\frac{3}{2}\sqrt{\mu }{R}^{-5/2}<0.$ Derivative saare $R>0$ ke liye negative hai: $R$ decrease karna $n$ increase karta hai. Toh lower orbit ($R_0-\delta$, with $\delta >0$) target ki rate $n$ se zyada fast angle sweep karta hai. Step 3 — speedup ka first-order size. Linearise karte hain (ek Taylor step, Linearization and Taylor Expansion dekho): $n(R_0-\delta )\approx n-\frac{dn}{dR}{|}_{R_0}\delta =n+\frac{3}{2}\frac{n}{R_0}\delta >n.$ Step 4 — CW drift se connect karo. Radial offset $x_0<0$ wala chaser radius $R_0-|x_0|$ par baitha hai (yaani $\delta =|x_0|$). Step 3 se iske paas higher angular rate hai, toh woh steadily aage pull karta hai (forward, $+y$). CW formula agree karta hai: $\Delta y=-12\pi x_0$, aur $x_0<0$ ke liye yeh $\Delta y>0$ deta hai (forward). Toh linear CW secular drift exactly Kepler's speed-radius law ka first-order shadow hai — Kepler's Laws aur Two-Body Problem dekho.

CW circular reference ke baare mein two-body gravity ka first-order Taylor expansion hai. Jaise $x,y,z\to 0$ neglected terms $O(r^2/R_0^2)$ hain, jo linear terms se zyada fast vanish hote hain — toh CW us limit mein exact ho jaata hai. Eccentric orbits ke liye, $n$ ab constant nahi raha (angular rate true anomaly ke saath vary karta hai), toh constant-coefficient CW equations fail ho jaati hain; time-varying version Tschauner–Hempel Equations hai. Two-Body Problem bhi dekho.

$n=\sqrt{\mu /R_0^3}$; bounded condition ${\dot{y}}_0=-2n{x}_0$; drift per orbit $\Delta y=-12\pi x_0$; football is 2:1; cross-track is decoupled SHM; forward burn ⇒ fall behind. Agar tum yeh sab bina notes ke produce kar sako, toh tumne D4 master kar liya hai.