Visual walkthrough — Rendezvous and proximity operations — Clohessy-Wiltshire equations

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Step 1 — Two ships, one origin (WHAT is the frame?)

WHAT. Picture Earth as a big dot. Two spacecraft circle it in almost the same path: the target (the thing we dock to) and the chaser (us). Instead of watching both loop the planet, we glue a camera to the target and let it spin with the target. In this "chase-cam" the target never moves — it sits at the origin — and we only watch the chaser wobble nearby.

WHY. The target is constantly curving (accelerating) around Earth. In a still frame it would zoom off-screen. A frame that turns at the same rate keeps it pinned, so only the small relative separation survives — which is all a docking pilot ever cares about.

PICTURE. The three axes are:

• $x$ = radial, straight up away from Earth,
• $y$ = along-track, the direction the ships are flying,
• $z$ = cross-track, out of the page (the orbital plane's normal).

This is the LVLH / Hill frame. It is a rotating reference frame, so fake forces will appear — that is the price of a moving camera, and we pay it in Step 4.

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Step 2 — Where $n$ comes from (WHY that spin rate?)

WHAT. The frame spins at exactly the rate the target orbits. For a circular orbit that rate is fixed. We call it $n$ and pin it down now, before any equation uses it.

WHY. On a circle, gravity is the only inward pull, and it must supply exactly the centripetal force needed to keep bending the path into a circle. Balancing those two gives $n$. This is just Kepler's third law in disguise.

PICTURE. Gravity pulls the target in with strength $\mu /R_0^2$. Circular motion needs an inward pull of $n^2{R}_0$. Set them equal:

$$\underset{\text{gravity pull}}{\underbrace{\frac{\mu }{R_0^2}}}=\underset{\text{needed to circle}}{\underbrace{n^2{R}_0}}⟹n=\sqrt{\frac{\mu }{R_0^3}}.$$

Here $\mu$ is the planet's gravitational parameter (mass $\times$ Newton's $G$) — a single number packaging "how strong is this planet's gravity."

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Step 3 — Gravity, but only the change (WHY Taylor-expand?)

WHAT. Real gravity on the chaser is $\vec{g}=-\mu \vec{r}/r^3$, where $\vec{r}$ points from Earth to the chaser. That $1/r^3$ is nasty and nonlinear. We replace it by a straight-line approximation good for small $(x,y,z)$.

WHY this tool — the Taylor expansion? We only ever need how gravity differs between target and chaser when they are metres apart on a kilometres-wide orbit. When a change is tiny, the honest thing is linearization: keep the value plus its first slope, throw away the curved rest. It answers exactly the question "how much does gravity tilt as I step off the target?" — no more.

PICTURE. The chaser sits at $\vec{r}=(R_0+x)\hat{x}+y\hat{y}+z\hat{z}$. Its distance from Earth is

$$r=\sqrt{(R_0+x{)}^2+y^2+z^2}.$$

For small offsets, $r^{-3}\approx R_0^{-3}\left(1-\frac{3x}{R_0}\right)$ — the $x$ (radial) term dominates because moving up/down changes your altitude, while sideways moves barely change your distance to Earth. Multiplying out to first order:

$$g_x\approx \underset{\text{baseline pull}}{\underbrace{-\frac{\mu }{R_0^2}}}+\underset{\text{tidal stretch}}{\underbrace{\frac{2\mu }{R_0^3}x}},g_y\approx \underset{\text{sideways squeeze}}{\underbrace{-\frac{\mu }{R_0^3}y}},g_z\approx \underset{\text{sideways squeeze}}{\underbrace{-\frac{\mu }{R_0^3}z}}.$$

Read the arrows: along $x$ gravity stretches you apart (positive coefficient — go higher, gravity weakens, you drift up more); along $y$ and $z$ it pinches you back toward the orbital line.

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Step 4 — The two fake forces (WHY a moving camera invents them)

WHAT. Because our camera spins, Newton's law picks up two extra apparent forces: the Coriolis force and the centrifugal force. We compute each on the chaser.

WHY this tool — the cross product $\times$? Rotation acts sideways: spin the camera and a moving object seems to curve at right angles to its velocity. The cross product $\vec{a}\times \vec{b}$ is the machine that produces a vector perpendicular to both — exactly the sideways kick rotation delivers. The frame spins about $z$, so $\vec{\omega }=n\hat{z}$.

PICTURE — Coriolis. It acts only on things that are moving in the frame:

$$-2\vec{\omega }\times \dot{\vec{r}}=-2n\hat{z}\times (\dot{x},\dot{y},\dot{z})=(\underset{\text{along-track speed kicks radial}}{\underbrace{+2n\dot{y}}},\underset{\text{radial speed kicks along-track}}{\underbrace{-2n\dot{x}}},0).$$

Notice: moving forward ($\dot{y}$) pushes you up ($x$); moving up ($\dot{x}$) pushes you backward ($-y$). This right-angle swap is the whole reason "burn toward the target" backfires.

PICTURE — Centrifugal. It acts on position, flinging you outward in the spin plane:

$$-\vec{\omega }\times (\vec{\omega }\times \vec{r})=n^2(x,y,0).$$

Both these fictitious forces are explained in full in Rotating Reference Frames — Coriolis and Centrifugal.

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Step 5 — The great cancellation (WHY the target hovers)

WHAT. Add the three contributions — real gravity (Step 3), Coriolis (Step 4), centrifugal (Step 4) — component by component. Swap every $\mu /R_0^3$ for $n^2$ using Step 2.

WHY. The target must sit still at the origin. That can only happen if the leftover constant force on it is zero. Watch the constants collide.

PICTURE — the $x$ (radial) equation. Line up every piece acting along $x$:

$$\ddot{x}=\underset{\text{gravity baseline}}{\underbrace{-n^2{R}_0}}\underset{\text{tidal}}{\underbrace{+2n^{2}x}}\underset{\text{Coriolis}}{\underbrace{+2n\dot{y}}}\underset{\text{centrifugal}}{\underbrace{+n^{2}x}}\underset{\text{centrifugal on }{R}_0}{\underbrace{+n^2{R}_0}}.$$

The two big constants $-n^2{R}_0$ and $+n^2{R}_0$ annihilate — the target, sitting at $x=0$, feels nothing. This is the hover. Collecting the rest:

$$\boxed{\ddot{x}-2n\dot{y}-3n^{2}x=0}(2n^{2}x+n^{2}x=3n^{2}x).$$

PICTURE — the $y$ (along-track) equation. Gravity gives $-n^{2}y$, centrifugal gives $+n^{2}y$ — those cancel too — and Coriolis leaves $-2n\dot{x}$:

$$\boxed{\ddot{y}+2n\dot{x}=0}.$$

PICTURE — the $z$ (cross-track) equation. No centrifugal (it lives in the $x,y$ plane), no Coriolis coupling, just the pinch $-n^{2}z$:

$$\boxed{\ddot{z}+n^{2}z=0}.$$

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Step 6 — Reading the solution's drift (WHY coasting fails)

WHAT. Solving the boxed pair (integrate the $y$-equation once, feed it into $x$, solve the driven oscillator) gives the closed forms below. We only read them here — the algebra lives in the parent note.

$$x(t)=(4-3\cos nt)x_0+\frac{\sin nt}{n}{\dot{x}}_0+\frac{2}{n}(1-\cos nt){\dot{y}}_0,$$ $$y(t)=\underset{\text{grows forever}}{\underbrace{6(\sin nt-nt)x_0}}+y_0-\frac{2}{n}(1-\cos nt){\dot{x}}_0+\underset{\text{grows forever}}{\underbrace{\frac{1}{n}(4\sin nt-3nt){\dot{y}}_0}}.$$

WHY it matters. Two terms contain a bare $nt$ that never stops growing — the secular drift. A small radial offset $x_0$ or forward speed ${\dot{y}}_0$ makes you march along-track without end.

PICTURE. Plotting the chaser's path in the $x$–$y$ plane: with the wrong initial velocity it spirals away (drift); with the magic choice ${\dot{y}}_0=-2n{x}_0$ the $nt$ terms vanish and the path closes into a tidy 2:1 ellipse — the famous "football."

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Step 7 — The degenerate & edge cases (so nothing surprises you)

WHAT. Every special starting condition, drawn, so you never hit an unshown scenario.

• $x_0=0,{\dot{y}}_0=0$ (pure radial kick ${\dot{x}}_0=v_r$): both drift terms vanish → bounded ellipse $x=\frac{v_r}{n}\sin nt,y=\frac{2v_r}{n}(\cos nt-1)$.
• $x_0=0$, all velocities zero, $y_0\ne 0$ (V-bar loiter): every trig term dies, no $x_0$ to drive drift → chaser just sits at $y=y_0$. This is why you park on the along-track axis before docking.
• $x_0<0$ (drop to a lower orbit): the $-6nt{x}_0$ term is now positive → you creep forward, catching up. Over one full period ($nT=2\pi$) the oscillatory part returns to zero, leaving a net $\Delta y=-12\pi x_0$.
• $z$ alone: pure SHM, independent of everything, $z(t)=z_0\cos nt+\frac{{\dot{z}}_0}{n}\sin nt$ — it can never drift, only oscillate.

WHY. Each case is just "which terms survive when these initials are zero." Reading survivors is faster than resolving the ODE every time.

PICTURE. Four little trajectories side by side: the football, the parked dot, the forward-creeping spiral, and the flat cross-track sine.

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The one-picture summary

This single frame stacks the whole story: Earth and the spinning frame (Step 1–2), the three forces meeting on the chaser with the two constants cancelling (Step 3–5), and the resulting football-or-drift path (Step 6–7). Trace it left to right and you have re-derived the CW equations from a blank page.

The state-marching machinery that packages $x(t),y(t),z(t)$ into a single matrix lives in State Transition Matrix; the eccentric-orbit generalisation is the Tschauner–Hempel Equations; and the underlying orbital dynamics come from the Two-Body Problem and Orbital Maneuvers — Hohmann Transfer.

Recall Feynman retelling — say it back in plain words

We sat a camera on the target and spun it so the target froze at centre. Because the camera spins, two make-believe forces appear: Coriolis (kicks you sideways when you move) and centrifugal (flings you outward). We wrote real gravity but kept only its tiny difference between the two ships. Adding gravity plus the two fake forces, the two giant constant pushes cancelled — that is why the target just hovers. What's left is three neat equations: the up/down and forward/back ones are locked together by Coriolis, and the sideways one is a plain spring. Solving them, a hidden term grows with time: a small altitude offset walks you endlessly along-track. That is both the trap ("don't just coast") and the tool ("drop lower to catch up"). Kill that term with ${\dot{y}}_0=-2n{x}_0$ and your path closes into a 2:1 football.

Recall Quick self-test

Why does the target hover at the origin? ::: The constant gravity term $-n^2{R}_0$ and the centrifugal term $+n^2{R}_0$ cancel exactly. Which tool converts the $1/r^3$ gravity into linear terms, and why? ::: Taylor expansion / linearization — because we only need gravity's small change over a tiny separation. Why does a forward ($+y$) burn make you fall behind? ::: Coriliois turns radial/along-track velocities into each other; a forward push raises your orbit, slowing you. What choice of initial velocity kills the drift? ::: ${\dot{y}}_0=-2n{x}_0$. Net along-track shift over one orbit from a radial offset $x_0$? ::: $\Delta y=-12\pi x_0$.