Rendezvous and proximity operations — Clohessy-Wiltshire equations

1. Setting up the frame (WHY a rotating frame?)

WHY rotating? Because the target itself is accelerating (curving around Earth). If we used a non-rotating frame the target would fly away. By co-rotating, the target stays put and only the relative motion survives — exactly what a docking pilot sees.

The circular target orbit has radius $R_0$ and mean motion $n=\sqrt{\frac{\mu}{R_0^3}},$ which is just Kepler's law: gravity provides the centripetal force, $\dfrac{\mu}{R_0^2}=n^2R_0$ .

2. Derivation from scratch (HOW)

In a frame rotating with angular velocity $\vec\omega = n\,\hat z$ , the acceleration of a point at position $\vec r$ (measured from Earth's center) obeys Newton, but written in the rotating frame it gains fictitious terms:

\ddot{\vec r}_{\text{rot}} = \vec g(\vec r) - 2\,\vec\omega\times\dot{\vec r}_{\text{rot}} - \vec\omega\times(\vec\omega\times\vec r).

The three added pieces (from left) are gravity, Coriolis, and centrifugal.

Let the chaser be at $\vec r = (R_0+x)\hat x + y\,\hat y + z\,\hat z$ (target at $R_0\hat x$ ).

Step A — Gravity, linearized. Why? Gravity is $-\mu\vec r/r^3$ ; we only want first order in $(x,y,z)$ .

\vec g = -\frac{\mu}{r^3}\vec r,\qquad r=\sqrt{(R_0+x)^2+y^2+z^2}.

Expand $r^{-3}\approx R_0^{-3}\left(1-\tfrac{3x}{R_0}\right)$ to first order. Then

g_x \approx -\frac{\mu}{R_0^2}+\frac{2\mu}{R_0^3}x,\quad g_y\approx -\frac{\mu}{R_0^3}y,\quad g_z\approx -\frac{\mu}{R_0^3}z.

Why this step? The $-\mu/R_0^2$ constant is exactly cancelled by centrifugal on the target, and $\mu/R_0^3=n^2$ .

Step B — Coriolis $-2\vec\omega\times\dot{\vec r}$ with $\vec\omega=n\hat z$ :

-2n\hat z\times(\dot x,\dot y,\dot z)=(+2n\dot y,\,-2n\dot x,\,0).

Step C — Centrifugal $-\vec\omega\times(\vec\omega\times\vec r)=n^2(x,y,0)$ (points outward in the orbital plane).

Step D — Add them. Component by component, using $\mu/R_0^3=n^2$ :

$x$ : $\ddot x = \big(-n^2R_0 + 2n^2x\big) + 2n\dot y + n^2x + n^2R_0$ $\Rightarrow \boxed{\ddot x - 2n\dot y - 3n^2 x = 0}$ (the $-n^2R_0$ from gravity and $+n^2R_0$ from centrifugal cancel — this is why the target hovers!)
$y$ : $\ddot y = -n^2 y - 2n\dot x + n^2 y \Rightarrow \boxed{\ddot y + 2n\dot x = 0}$
$z$ : $\ddot z = -n^2 z \Rightarrow \boxed{\ddot z + n^2 z = 0}$

3. Solving (HOW — closed form)

The $z$ equation is SHM: $z(t)=z_0\cos nt + \dfrac{\dot z_0}{n}\sin nt$ .

For the in-plane pair, integrate the $y$ -equation once: $\dot y + 2n x = \dot y_0 + 2n x_0 \;(\text{const}).$ Substitute into the $x$ -equation to get $\ddot x + n^2 x = \text{const}$ , another driven SHM. Solving:

Figure — Rendezvous and proximity operations — Clohessy-Wiltshire equations

4. Worked examples

Worked example Example 1 — Football orbit (bounded relative motion)

A chaser starts at $x_0=0$ , $y_0=0$ , $\dot x_0 = v_r$ , and we want a closed loop (no drift). Condition: kill the secular term. From the $x$ solution's constant piece, bounded motion needs $\dot y_0 = -2n\,x_0.$ Why? That makes the $nt$ terms vanish, leaving pure ellipse. With $x_0=0$ : choose $\dot y_0=0$ . Then $x(t)=\frac{v_r}{n}\sin nt$ , $y(t)=\frac{2v_r}{n}(\cos nt -1)$ . Why this step? Plug $x_0=y_0=\dot y_0=0$ into the formulas. Result is a 2:1 ellipse (along-track amplitude twice radial) — the classic "football/racetrack" a chaser flies.

Worked example Example 3 — Time to close an along-track gap using radial offset

You are $y_0=-1$ km behind, want to catch up. A phasing trick: drop to a slightly lower orbit by giving a radial offset $x_0<0$ so the secular term $6(\sin nt - nt)x_0$ carries you forward. Net drift per full orbit. Over one period $T=2\pi/n$ , the oscillatory $\sin nt$ returns to zero, so the net along-track shift comes only from the $-nt$ piece: $\Delta y = 6(\underbrace{\sin nT}_{=0}-nT)\,x_0 = -6nT\,x_0 = -12\pi\,x_0.$ Why this step? $nT=2\pi$ , so $-6nT\,x_0 = -12\pi x_0$ per orbit — a big along-track walk from a tiny radial offset. (Note $\Delta y>0$ for $x_0<0$ : a lower orbit moves you forward.) To close $1000$ m in one orbit we need $|\,{-12\pi x_0}\,|=1000\Rightarrow |x_0|=\dfrac{1000}{12\pi}\approx 26.5$ m. Why this step? Just invert $\Delta y=-12\pi x_0$ ; the $n$ and $T$ cancelled into the pure number $12\pi$ , so the required offset is independent of orbit altitude.

5. Steel-manned mistakes

6. Active recall

What frame are CW equations written in?

The LVLH / Hill frame — centered on the target, rotating at mean motion

n

(x radial, y along-track, z cross-track).

State the three CW equations.

\ddot x -2n\dot y -3n^2x=0

;

\ddot y +2n\dot x=0

;

\ddot z + n^2z=0

Why does the target hover at the origin?

The constant gravity term

-n^2R_0

exactly cancels the centrifugal term

+n^2R_0

in the radial equation.

What is

n

in terms of

\mu,R_0

n=\sqrt{\mu/R_0^3}

(from

\mu/R_0^2 = n^2R_0

Which motion is a decoupled SHM?

The cross-track

z

\ddot z + n^2 z=0

, oscillating at frequency

n

What causes the coupling between x and y?

The Coriolis force (

\pm 2n\dot{\cdot}

) from working in a rotating frame.

Condition for bounded (drift-free) in-plane motion?

\dot y_0 = -2n\,x_0

What is the shape of bounded relative motion?

A 2:1 ellipse ("football/racetrack") — along-track amplitude twice the radial amplitude.

Where does secular along-track drift come from?

The

-6nt\,x_0

and

-3nt\,\dot y_0

terms; a radial offset changes orbital period → endless along-track walk.

Net along-track drift per orbit from a radial offset

x_0

\Delta y = -6nT\,x_0 = -12\pi\,x_0

(since

nT=2\pi

Two key assumptions of CW?

Circular target orbit (constant

n

) and small relative separation (linearization of gravity).

Recall Feynman: explain it to a 12-year-old

Two race cars go around a circular track. One (the target) stays in its lane; you (the chaser) are just beside it. Instead of watching the whole track, put a camera on the target car and only watch how you slide next to it. In space, if you nudge "up" (away from Earth), your lane becomes slower, so you slowly slip backward. If you nudge "forward", you climb higher and again fall behind! The CW equations are the simple rulebook for this weird sliding when you're near a friend in orbit. The up-down bobbing is a simple bounce; the up-forward motions are tangled together.

7. Connections

Two-Body Problem — the exact nonlinear system we linearized.
Kepler's Laws — source of $n=\sqrt{\mu/R_0^3}$ and why higher orbits are slower.
Rotating Reference Frames — Coriolis and Centrifugal — origin of the coupling terms.
Tschauner–Hempel Equations — the eccentric-orbit generalization.
Orbital Maneuvers — Hohmann Transfer — global cousin of local phasing.
Linearization and Taylor Expansion — the mathematical engine behind the derivation.
State Transition Matrix — packaging the CW solution for guidance.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho do spacecraft lagbhag ek hi circular orbit mein hain — ek "target" aur ek "chaser". Poore planet ke around dono ko track karna faltu hai; smart trick ye hai ki hum camera target par laga dete hain aur ek rotating frame (LVLH/Hill frame) mein sirf relative motion dekhte hain. Is frame mein target origin par chup-chaap baitha rehta hai, aur chaser aas-paas ghoomta hai. Gravity ka messy $1/r^2$ small distances par linear ban jaata hai — bas yahi Clohessy–Wiltshire ka jaadu hai.

Derivation ka dil: rotating frame mein Coriolis aur centrifugal fictitious forces aati hain. Gravity ko target ke paas Taylor expand karke first order terms rakho, $\mu/R_0^3=n^2$ use karo, aur teen simple equations milti hain. Cross-track $z$ toh seedha SHM hai ( $\ddot z+n^2z=0$ ). In-plane $x$ (radial) aur $y$ (along-track) Coriolis ki wajah se coupled ho jaate hain. Sabse important counter-intuitive baat: agar tum "up" (radial) nudge doge, tumhari orbit slow ho jaati hai aur tum peeche gir jaate ho — isko secular drift ( $-6nt\,x_0$ ) kehte hain. Ek poore orbit mein net drift $\Delta y=-12\pi x_0$ hoti hai — chhoti radial offset se bada along-track shift.

Practical importance: ISS docking, satellite servicing, debris capture — sab yahi maths use karte hain. Agar bounded (drift-free) loop chahiye toh condition yaad rakho: $\dot y_0=-2nx_0$ , isse famous 2:1 "football" ellipse banti hai. Aur agar aage nikalna hai toh radial offset use karke secular drift ko apne favour mein use karo. Yaad rakho assumptions: orbit circular ho aur distance chhoti ho, warna Tschauner–Hempel ya full nonlinear equations chahiye.