Rendezvous and proximity operations — Clohessy-Wiltshire equations: Question bank

3.2.40 · D5 · Physics › Orbital Mechanics & Astrodynamics › Rendezvous and proximity operations — Clohessy-Wiltshire equ

Intuition Yeh solution

kahan se aati hai? (taaki secular terms magic na lagein) Tumhe ise memorise nahi karna — teen moves mein iska skeleton dekh sakte ho, aur wahi skeleton drift produce karta hai:

$y$ -equation ek perfect derivative hai. $\overset{y}{¨} + 2 n \overset{x}{˙} = 0$ ko ek baar integrate karo to milta hai $\overset{y}{˙} + 2 n x = \overset{y}{˙}_{0} + 2 n x_{0}$ (ek constant). Yahi linear coupling kaam kar rahi hai: along-track velocity radial position se bandhi hai.
$x$ -equation mein substitute karo. $\overset{y}{˙}$ ko replace karne par radial equation ban jaati hai $\overset{x}{¨} + n^{2} x = const$ — ek constant push wala simple harmonic oscillator. Iska solution hai (bounded) $sin n t, cos n t$ terms plus ek constant offset. Yahin se $x (t)$ ke $(4 - 3 cos n t)$ aur $(1 - cos n t)$ pieces paida hote hain.
$\overset{y}{˙} = (const) - 2 n x$ ko integrate karo $y$ paane ke liye. Constant wala part ek aisa term mein integrate hota hai jo $\propto t$ hai — time mein ek seedhi line. Woh $t$ -term hi secular drift hai. Yeh exist karta hai exactly isliye kyunki step 1 ka constant of integration ek non-zero average radial position ko $y$ mein wapas feed karta hai. Coupling nahi → constant nahi → drift nahi.

Intuition Secular drift ko isolate karna (traps se pehle yeh padho)

"Secular" ka matlab hai $y (t)$ ka woh part jo oscillate karne ki bajaye $t$ ke saath bound ke bina badhta hai. $sin n t, cos n t$ terms strip kar do (woh bounded rehte hain) aur sirf $t$ multiply karne wale pieces rakho: $y_{secular} (t) = - 6 n x_{0} t - 3 \overset{y}{˙}_{0} t .$ Dimensions check: $[n x_{0} t] = (1/s) (m) (s) = m$ ✓, aur $[\overset{y}{˙}_{0} t] = (m/s) (s) = m$ ✓ — dono lengths hain, isliye inhe add karna legal hai. Per-orbit net shift ( $T = 2 π / n$ ke upar, isliye $n T = 2 π$ ), $\overset{y}{˙}_{0} = 0$ rakhte hue: $Δ y = - 6 n x_{0} T = - 6 (2 π) x_{0} = - 12 π x_{0} .$ ⚠️ Do alag cheezein seedhi rakho. Drift rate (metres per second) $t$ ka coefficient hai, yaani $- 6 n x_{0}$ : iske units hain $(1/s) (m) = m/s$ , ek speed. Per-orbit shift $- 12 π x_{0}$ ek pure length hai, kyunki us speed ko ek period $T = 2 π / n$ se multiply karne par $n$ cancel ho jaata hai: $(- 6 n x_{0}) \times (2 π / n) = - 12 π x_{0}$ . $- 3 \overset{y}{˙}_{0} t$ term ka coefficient $- 3$ hai (koi $n$ nahi), kyunki $\overset{y}{˙}_{0} / n$ ek length hai aur $(4 sin n t - 3 n t)$ $- 3 n t$ contribute karta hai, jo overall $- 3 \overset{y}{˙}_{0} t$ deta hai — $t$ se multiply karne par phir ek clean length.

True or false — justify

A forward (+y) thrust makes the chaser move forward and catch up.

False. +y burn orbit ko raise karta hai; higher orbit slower hoti hai, isliye along-track Coriolis coupling (

\overset{y}{¨} + 2 n \overset{x}{˙} = 0

) ke through tum eventually peeche pad jaate ho.

The cross-track motion

z (t)

can drift away over many orbits.

False.

\overset{z}{¨} + n^{2} z = 0

pure simple harmonic motion hai; yeh hamesha ke liye ek bounded oscillation rehti hai jisme koi secular term nahi, kyunki

z

x, y

se puri tarah decouple hai.

Every in-plane orbit in the CW frame is a closed ellipse.

False. Sirf tabhi jab

\overset{y}{˙}_{0} = - 2 n x_{0}

ho. Warna

- 6 n x_{0} t

aur

- 3 \overset{y}{˙}_{0} t

terms

y

ko time mein linearly walk off kara dete hain.

CW equations hold for a target on a mildly eccentric orbit if you just use the instantaneous

n

False. Derivation ko

n

constant chahiye; ellipse par

n

(angular rate) position ke saath vary karta hai, isliye tumhe Tschauner–Hempel Equations use karne honge.

The constant gravity term and centrifugal term cancelling is a coincidence of algebra.

False. Yeh physics hai: target ek balanced circular orbit mein hai, isliye gravity exactly uski centripetal need supply karta hai (

μ / R_{0}^{2} = n^{2} R_{0}

). Wahi balance reason hai ki origin hover kar sakta hai.

Doubling the target's orbital radius

R_{0}

leaves the CW equation coefficients unchanged.

False. Saare coefficients

n = μ / R_{0}^{3}

par depend karte hain, jo

R_{0}

badhne par shrink hota hai, isliye poora relative-motion timescale slow ho jaata hai.

If you start exactly at the origin with zero relative velocity, you stay at the origin.

True. Saare initial conditions zero hain, isliye CW solution ka har term sab

t

ke liye zero hai — chaser target ke saath perfectly co-orbit kar raha hai.

The 2:1 "football" ellipse has its long axis along the radial direction.

False. Along-track (

y

) amplitude radial (

x

) amplitude se double hai, isliye long axis along-track direction mein hoti hai.

Spot the error

"Because

z

is SHM, the in-plane

x

–

y

motion is also just two independent SHMs."

Error yeh hai:

x

aur

y

Coriolis (

\pm 2 n \dot{\cdot}

) se coupled hain. Yeh independent oscillators nahi hain; solve karne ke liye

y

-equation ko integrate karke substitute karna padta hai, aur yahin se secular drift paida hoti hai.

"The secular drift appears because we kept nonlinear

1/ r^{2}

terms."

Error yeh hai: drift puri tarah ek linear effect hai — yeh linear CW system ke exact solution mein hi rehti hai. Yeh ek differential orbital period se aati hai, kisi nonlinearity se nahi.

"To hold position on V-bar you must continuously thrust radially."

Error yeh hai: V-bar par (

x_{0} = 0

, zero velocity) koi secular drift nahi hai aur first order par koi forcing nahi hai, isliye chaser passively loiter karta hai — koi continuous thrust ki zaroorat nahi (first order tak).

n = μ / R_{0}^{3}

is a separate empirical fit for CW."

Error yeh hai: yeh sirf Kepler's third law rearranged hai — gravity = centripetal,

μ / R_{0}^{2} = n^{2} R_{0}

. Koi nayi assumption nahi hai.

"Since the

sin n t

terms vanish after one period, the chaser returns to its start each orbit."

Error yeh hai: sirf oscillatory parts wapas aate hain; secular parts accumulate karte rehte hain. Per-orbit net along-track shift

- 12 π x_{0}

hai (with

\overset{y}{˙}_{0} = 0

), zero nahi, jab tak

x_{0} = 0

na ho aur

\overset{y}{˙}_{0}

bhi balanced na ho.

"CW gives the exact relative trajectory for the ISS and a nearby cargo ship."

Error yeh hai: CW ek first-order linearization hai jo sirf

r ≪ R_{0}

aur circular target orbit ke liye valid hai. Close range par yeh ek excellent approximation hai, kabhi exact nahi.

Why questions

Why do we bother with a rotating frame instead of a fixed inertial one?

Inertial frame mein target Earth ke around fly karta hai aur chaser ek badi curved path track karta hai; co-rotating target ko origin par pin kar deta hai taaki sirf chota relative motion — jo pilot ko matlab hai — bachta hai.

Why does a tiny radial offset

x_{0}

produce a large along-track drift?

Radial offset tumhara orbital radius change karta hai, aur isliye tumhara period; thoda alag period har orbit mein ek ever-growing along-track lead ya lag mein compound hota rahat hai (

- 6 n x_{0} t

Why is the bounded-motion condition specifically

\overset{y}{˙}_{0} = - 2 n x_{0}

Yeh exactly woh along-track velocity hai jo radial offset

x_{0}

se implied orbital period difference se match karti hai, isliye dono spacecraft ek period share karte hain aur secular

t

-terms cancel ho jaate hain.

Why is the

z

(cross-track) equation completely decoupled from

x, y

Rotation axis

\overset{z}{^}

hai, isliye Coriolis aur centrifugal sirf orbital plane mein act karte hain; plane ke perpendicular motion sirf linearized restoring gravity feel karta hai, jo clean SHM deta hai.

Why does the constant

- n^{2} R_{0}

appear and then vanish in the radial equation?

Yeh target ke radius par leading gravity term hai; yeh exactly centrifugal

+ n^{2} R_{0}

se cancel hota hai kyunki target force balance mein hai — sirf first-order tidal term

3 n^{2} x

bachta hai.

Why can CW be packaged as a State Transition Matrix?

Socho chhah-number state (position + velocity) ko 6D space mein ek point ki tarah. Kyunki CW forces us state mein linear hain, time mein aage evolve karna simply us space ko ek fixed tarike se stretch aur rotate karta hai — koi bhi term kabhi nonlinearly depend nahi karti jahan se shuru kiya. Isliye "time

t

par state" ek fixed matrix hai jo "time 0 par state" se multiply hoti hai, aur woh matrix (state transition matrix) har chaser ke liye same hai.

Edge cases

What happens to the CW solution if you set

n \to 0

(imagine no gravity/rotation)?

Saari coupling aur restoring terms vanish ho jaati hain; motion force-free straight lines ban jaati hai,

x = x_{0} + \overset{x}{˙}_{0} t

, etc. — physically expected limiting free-drift case.

What is the relative motion for purely cross-track initial conditions (

x_{0} = y_{0} = \overset{x}{˙}_{0} = \overset{y}{˙}_{0} = 0

Pure SHM sirf

z

mein: chaser frequency

n

par orbital plane ke upar aur neeche bob karta hai, hamesha ke liye in-plane ruka rehta hai.

What does the trajectory look like exactly on the bounded-motion boundary

\overset{y}{˙}_{0} = - 2 n x_{0}

Ek closed 2:1 ellipse (football) jo har orbit repeat karti hai — forward drift aur backward drift ke beech knife-edge.

What happens as the separation

r

grows toward

R_{0}

Linearization toot jaati hai: neglect kiye gaye

1/ r^{2}

nonlinear terms significant ho jaate hain, isliye CW accuracy kho deta hai aur full two-body relative dynamics integrate karna padta hai.

Zero radial offset but nonzero

\overset{y}{˙}_{0}

: is the motion bounded?

Nahi.

x_{0} = 0

ke saath bounded condition ko

\overset{y}{˙}_{0} = 0

chahiye; koi bhi nonzero

\overset{y}{˙}_{0}

- 3 \overset{y}{˙}_{0} t

secular term chhodta hai, isliye yeh drift karta hai.

If the target orbit were eccentric but you forced CW anyway, what error grows?

Neglected time-varying

n

aur radial acceleration errors cause karte hain jo eccentricity aur time ke saath badhte hain; Tschauner–Hempel form correctness restore karta hai.

Recall Fast self-check

Bounded in-plane motion ke liye kaun si condition chahiye? ::: $\overset{y}{˙}_{0} = - 2 n x_{0}$ . +y burn tumhari along-track position ka long-term kya karta hai? ::: Tumhe peeche kar deta hai (orbit raise karta hai, slow karta hai). Kaun sa coordinate decoupled SHM hai? ::: Cross-track $z$ . Radial offset se per-orbit net along-track shift ( $\overset{y}{˙}_{0} = 0$ ke saath)? ::: $Δ y = - 12 π x_{0}$ .