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

The scenario matrix

Every CW problem is really a choice of which initial numbers are non-zero and which sign they carry. This table lists every case class. The examples below are tagged with the cell they cover, and the "Figure" column says which ones are drawn (a case is drawn whenever the geometry of the path carries the insight; the algebra-only cases are checked numerically instead).

#	Case class	What is special	Covered by	Figure
C1	Pure cross-track, $z_{0} \neq = 0$ , all in-plane $= 0$	Decoupled SHM, no drift	Ex 1	s01
C2	Radial offset $x_{0} > 0$ (higher orbit)	Secular drift, sign of walk-off	Ex 2	s02
C3	Radial offset $x_{0} < 0$ (lower orbit)	Opposite-sign drift, phasing	Ex 3	mirror of s02
C4	Bounded-motion condition $\overset{y}{˙}_{0} = - 2 n x_{0}$	Football ellipse, zero drift	Ex 4	s03
C5	Along-track offset only, $y_{0} \neq = 0$ , zero velocity	Degenerate: chaser hovers	Ex 5	(fixed point)
C6	Pure radial velocity kick $\overset{x}{˙}_{0} \neq = 0$	Ellipse from a burn, no net drift	Ex 6	2:1 like s03
C7	Pure along-track velocity kick $\overset{y}{˙}_{0} \neq = 0$	Strong drift from a tangential burn	Ex 7	s04
C8	Limiting case $t \to 0$ and $t = T$ (one period)	Sanity of the formulas at the edges	Ex 8	(algebraic)
C9	Real-world word problem (ISS resupply catch-up)	Full numeric design	Ex 9	uses s02
C10	Exam twist: combine radial + tangential to cancel drift	Solve for the burn	Ex 10	2:1 like s03

Throughout, take a low-Earth target orbit of radius $R_{0} = 6.778 \times 1 0^{6} m$ (about $400$ km altitude). Then $n = \frac{3.986 \times 1 0 ^{14}}{( 6.778 \times 1 0 ^{6} ) ^{3}} \approx 1.131 \times 1 0^{- 3} s^{- 1},$ and one orbital period is $T = 2 π / n \approx 5556 s \approx 92.6 min$ . We reuse these numbers in the numeric examples.

The worked examples

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

The chaser starts $z_{0} = 50$ m out of the orbital plane, everything else zero. Where is it after a quarter period, $t = T /4$ , and what is its cross-track speed there?

Forecast: Guess before computing — does it drift away, or come back? Which way is it moving at $t = T /4$ ?

Use the $z$ solution with $\overset{z}{˙}_{0} = 0$ : $z (t) = z_{0} cos n t$ . Why this step? Cross-track is fully decoupled from $x, y$ (the parent's $\overset{z}{¨} + n^{2} z = 0$ is the equation of a mass on a spring), so we only need the one line.
At $t = T /4$ we have $n t = \frac{n}{4} \cdot \frac{2 π}{n} = \frac{π}{2}$ , so $cos n t = 0$ . Thus $z (T /4) = 50 cos (π /2) = 0$ m. Why this step? A quarter period is exactly where a swinging pendulum passes through the middle — position zero, speed maximum.
Speed: $\overset{z}{˙} (t) = - z_{0} n sin n t$ , so $\overset{z}{˙} (T /4) = - 50 \cdot n \cdot 1 = - 0.0566$ m/s. Why this step? Differentiating the cosine gives a sine; at $n t = π /2$ the sine is $1$ , its peak — confirming maximum speed at the crossing.

Verify: Amplitude of $z$ never exceeds $50$ m and the motion is periodic — no $n t$ term appears, so no secular drift, exactly as expected for pure SHM. Units: $z_{0} [m] \times n [s^{- 1}] = m/s$ . ✓

The figure below plots this $z (t)$ over one full period. Read step 2 off it: the green curve crosses zero exactly at $t = T /4$ (the red dot), and that crossing is where the slope is steepest — the maximum speed of step 3. Notice the curve returns to $+ 50$ m at $t = T$ : bounded, no drift, the signature of cell C1.

Figure — Rendezvous and proximity operations — Clohessy-Wiltshire equations

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

The chaser is parked $x_{0} = + 100$ m radially outward, all velocities zero. How far along-track has it drifted after one full period $T$ , and in which direction?

Forecast: Higher orbit = slower. Do you think the chaser falls behind ( $y$ more negative) or pulls ahead?

Along-track drift from the parent's $y$ solution with $\overset{x}{˙}_{0} = \overset{y}{˙}_{0} = y_{0} = 0$ : $y (t) = 6 (sin n t - n t) x_{0}$ . Why this step? Only the $x_{0}$ term survives, and it carries the secular $- 6 n t x_{0}$ piece.
At $t = T$ , $n t = 2 π$ , so $sin n t = 0$ and $y (T) = 6 (0 - 2 π) x_{0} = - 12 π x_{0}$ . Why this step? The oscillation returns to zero after a full period; only the linear-in- $t$ walk-off remains, giving the clean number $12 π$ .
Numerically $y (T) = - 12 π (100) = - 3769.9$ m $\approx - 3.77$ km. Why this step? Plug $x_{0} = + 100$ ; the negative sign means the chaser slid backward (along-track negative), consistent with "higher orbit is slower".

Verify: Sign check — $x_{0} > 0$ gives $y < 0$ (falls behind). A $100$ m radial nudge produced a $3.77$ km along-track walk in one orbit: tiny cause, huge effect, matching the parent's warning about secular drift. ✓

The figure traces the actual $(y, x)$ path in the LVLH plane. Start at the blue dot ( $x = + 100$ m, $y = 0$ ); after one orbit the chaser lands at the red dot 3.77 km behind — even though the radial coordinate came back to $+ 100$ m. The open, non-closing curve is exactly what "secular drift" looks like: contrast it with the closed loop of Example 4.

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

Start $x_{0} = + 50$ m radially out. Choose the along-track velocity $\overset{y}{˙}_{0}$ that kills all drift, and describe the resulting path.

Forecast: There is exactly one magic $\overset{y}{˙}_{0}$ . Is it positive or negative? What shape results?

The bounded-motion condition (parent) is $\overset{y}{˙}_{0} = - 2 n x_{0}$ . Why this step? This value makes the secular $n t$ terms cancel between the $x_{0}$ and $\overset{y}{˙}_{0}$ contributions, leaving pure trig — a closed loop.
$\overset{y}{˙}_{0} = - 2 (1.131 \times 1 0^{- 3}) (50) = - 0.1131$ m/s (retrograde along-track). Why this step? With $x_{0} > 0$ we need a negative along-track kick to balance the walk-off.
With $\overset{x}{˙}_{0} = 0, y_{0} = 0$ , the solution becomes $x (t) = (4 - 3 cos n t) x_{0} + \frac{2}{n} (1 - cos n t) \overset{y}{˙}_{0}$ . Substituting $\overset{y}{˙}_{0} = - 2 n x_{0}$ gives $x (t) = x_{0} cos n t$ , and correspondingly $y (t) = - 2 x_{0} sin n t$ . Why this step? The messy constants collapse; the path is an ellipse with radial amplitude $x_{0}$ and along-track amplitude $2 x_{0}$ .
That is the 2:1 football ellipse — along-track swing is twice the radial swing.

Verify: At $t = T$ , $cos n t = 1, sin n t = 0$ , so $x (T) = x_{0} = 50$ m, $y (T) = 0$ — the chaser returns to its start. Closed loop, zero drift. Along-track amplitude $2 \times 50 = 100$ m confirms the 2:1 ratio. ✓

The figure draws that closed ellipse. Unlike the open curve of s02, this one returns to its start (green dot) — the drift has been cancelled by the retrograde kick of step 2. Measure the axes: radial half-width $50$ m, along-track half-width $100$ m — the 2:1 ratio of step 4, read straight off the picture.

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

Chaser sits $y_{0} = - 200$ m directly behind the target on the along-track axis, all velocities zero, $x_{0} = 0$ . Where is it after any time $t$ ?

Forecast: Does it stay put, or does being "behind" cause it to drift?

With $x_{0} = 0, \overset{x}{˙}_{0} = 0, \overset{y}{˙}_{0} = 0$ the solutions reduce to $x (t) = 0$ and $y (t) = y_{0}$ . Why this step? Every term in both formulas is multiplied by $x_{0}$ or by a velocity; all are zero, so only the standalone $y_{0}$ survives.
So $x (t) = 0, y (t) = - 200$ m for all $t$ — a fixed point. Why this step? The along-track axis carries no restoring force (there is no $y$ term in $\overset{y}{¨} + 2 n \overset{x}{˙} = 0$ once $\overset{x}{˙} = 0$ ), so a pure along-track offset is (to first order) an equilibrium.

Verify: Plug into $\overset{y}{¨} + 2 n \overset{x}{˙} = 0$ : $\overset{x}{˙} = 0$ and $\overset{y}{¨} = 0$ ✓; plug into $\overset{x}{¨} - 2 n \overset{y}{˙} - 3 n^{2} x = 0$ : all terms zero ✓. This is why the V-bar ( $y$ -axis) is a stable loiter line for docking. No figure is needed — the trajectory is a single stationary point on the $y$ -axis, which a plot cannot make clearer than the sentence "it does not move". ✓

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

From the origin ( $x_{0} = y_{0} = 0$ ), fire a radial burn $\overset{x}{˙}_{0} = + 0.05$ m/s. Trace the motion; does it drift?

Forecast: A radial "up" burn — does the chaser leave forever, or loop back?

In-plane solution with only $\overset{x}{˙}_{0}$ non-zero: $x (t) = \frac{x ˙ _{0}}{n} sin n t$ , $y (t) = - \frac{2 x ˙ _{0}}{n} (1 - cos n t)$ . Why this step? Both surviving terms are purely oscillatory — no $n t$ factor — so there is no secular drift.
Radial amplitude $= \overset{x}{˙}_{0} / n = 0.05/ (1.131 \times 1 0^{- 3}) = 44.2$ m. Why this step? The peak of $sin n t$ is $1$ , so the largest radial excursion is $\overset{x}{˙}_{0} / n$ .
Along-track term $- \frac{2 x ˙ _{0}}{n} (1 - cos n t)$ swings between $0$ and $- 2 (44.2) = - 88.4$ m, then returns. Why this step? $(1 - cos n t)$ ranges $0$ to $2$ ; the motion is again a closed 2:1 ellipse, just centred off the origin — geometrically the same shape as the s03 football, shifted.

Verify: At $t = T$ : $sin n t = 0, 1 - cos n t = 0$ , so $x = 0, y = 0$ — back home, no drift. A radial burn alone never causes secular walk-off (only $x_{0}$ and $\overset{y}{˙}_{0}$ do). ✓

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

From the origin, fire a tangential burn $\overset{y}{˙}_{0} = + 0.05$ m/s (forward). Along-track drift after one period?

Forecast: "Push forward to go forward" — the parent flagged this as a trap. What actually happens?

Along-track solution with only $\overset{y}{˙}_{0}$ non-zero: $y (t) = \frac{1}{n} (4 sin n t - 3 n t) \overset{y}{˙}_{0}$ . Why this step? The $- 3 n t$ inside is a secular term, so a forward burn does cause drift.
At $t = T$ : $sin n t = 0$ , so $y (T) = \frac{1}{n} (- 3 \cdot 2 π) \overset{y}{˙}_{0} = - \frac{6 π}{n} \overset{y}{˙}_{0}$ . Why this step? Only the linear piece survives after a full loop.
Numerically $y (T) = - \frac{6 π}{1.131 \times 1 0 ^{- 3}} (0.05) = - 833.7$ m $\approx - 0.83$ km. Why this step? Note the sign: a forward burn produced a backward net shift — the Coriolis coupling raised the orbit and slowed the chaser.

Verify: Sign is negative despite a positive (forward) burn — this is precisely the parent's "fire toward the target and fall behind" mistake made quantitative. Magnitude: $6 π / n = 6 π / (1.131 \times 1 0^{- 3}) \approx 1.667 \times 1 0^{4}$ , times $0.05$ gives $\approx 834$ m. ✓

The figure traces this path: from the green start the chaser loops but does not close — after one orbit it sits $834$ m behind (orange dot), the opposite of the forward direction it was pushed. This is the visual proof of the parent's "fire forward, fall behind" warning.

Worked example Example 8 — Limiting checks

$t \to 0$ and $t = T$ (cell C8) Confirm the formulas behave correctly at the two edges: the instant of release, and one full period.

Forecast: At $t = 0$ the answer must equal the initial state exactly. Guess whether the drift terms cause trouble there.

At $t = 0$ : $cos n t = 1, sin n t = 0, n t = 0$ . Then $x (0) = (4 - 3) x_{0} = x_{0}$ , $y (0) = y_{0}$ , $z (0) = z_{0}$ . Why this step? A solution must reproduce its own initial condition at $t = 0$ ; the $4 - 3 = 1$ combination is the built-in check that it does.
Velocities at $t = 0$ : differentiate and set $t = 0$ to get $\overset{x}{˙} (0) = \overset{x}{˙}_{0}$ , $\overset{y}{˙} (0) = \overset{y}{˙}_{0}$ . Why this step? All six state components must be recovered, confirming the closed form is a genuine solution of the ODEs, not an approximation valid only later.
At $t = T$ ( $n t = 2 π$ ): every $cos \to 1, sin \to 0$ , leaving only $x_{0}$ -driven $- 12 π x_{0}$ and $\overset{y}{˙}_{0}$ -driven $- \frac{6 π}{n} \overset{y}{˙}_{0}$ in $y$ ; the rest return to their start. Why this step? This isolates exactly which inputs cause drift (radial offset and tangential velocity) and which do not (radial velocity, cross-track) — a full-period fingerprint of the system.

Verify: $x (0) = x_{0}$ and $z (0) = z_{0}$ recovered exactly; the only surviving $t = T$ terms are the two secular ones identified in Examples 2 and 7. Consistent across the whole matrix. ✓

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

A cargo vehicle is $y_{0} = - 2$ km behind the ISS ( $R_{0} = 6.778 \times 1 0^{6}$ m). Using a radial-offset phasing burn, what radial offset $x_{0}$ closes the gap in exactly one orbit, and is it up or down?

Forecast: The gap is behind, so you must move forward. Should you go to a higher or lower orbit?

Net along-track shift per orbit from a radial offset: $Δ y = - 12 π x_{0}$ (Example 2). Why this step? This is the drift-per-period fingerprint; the oscillatory parts cancel over one full loop.
We need $Δ y = + 2000$ m (to close a $- 2000$ m gap). Solve $+ 2000 = - 12 π x_{0}$ : $x_{0} = - \frac{2000}{12 π} = - 53.05$ m. Why this step? Positive required drift forces $x_{0} < 0$ — a lower orbit, which is faster, pulling the chaser forward.
Interpretation: drop $\approx 53$ m below the ISS, coast one orbit ( $\approx 92.6$ min), and the 2 km gap closes. Then re-raise to match. Why this step? The offset is independent of altitude (the $n$ and $T$ cancelled into the pure number $12 π$ ), so this is a robust rule of thumb.

Verify: $Δ y = - 12 π (- 53.05) = + 2000.0$ m ✓. Sign: lower orbit ( $x_{0} < 0$ ) → forward drift ( $Δ y > 0$ ) → gap closes. Units: $[m]$ throughout. This is the phasing shown, in miniature, by the mirror of the s02 figure (open curve opening forward). ✓

Worked example Example 10 — Exam twist: cancel the drift with a combined burn (cell C10)

A chaser has an unwanted radial offset $x_{0} = + 80$ m (all else zero). Instead of drifting away, apply a single along-track burn $\overset{y}{˙}_{0}$ at $t = 0$ so the motion becomes drift-free. Find $\overset{y}{˙}_{0}$ and the resulting ellipse amplitudes.

Forecast: From Example 4 you already know the magic condition — but predict the sign for $x_{0} > 0$ .

Drift-free requires $\overset{y}{˙}_{0} = - 2 n x_{0}$ (bounded-motion condition). Why this step? This is the unique tangential velocity that cancels both secular terms simultaneously; it converts endless drift into a closed loop.
$\overset{y}{˙}_{0} = - 2 (1.131 \times 1 0^{- 3}) (80) = - 0.1810$ m/s. Why this step? With $x_{0} > 0$ the burn is retrograde (negative $y$ ), braking the walk-off.
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) — a 2:1 football. Why this step? Substituting the magic $\overset{y}{˙}_{0}$ collapses the constants exactly as in Example 4, scaled by the new $x_{0}$ ; the shape is the same closed ellipse as s03, scaled up.

Verify: At $t = T$ : $x (T) = x_{0} = 80$ m, $y (T) = 0$ — no net drift ✓. Amplitude ratio $160 : 80 = 2 : 1$ ✓. A single scalar burn tamed the drift, the classic exam result. This closes the last cell of the matrix: every scenario is now worked. ✓

Recall

Recall Which initial conditions cause secular drift, and which do not?

Radial offset $x_{0}$ (via $- 6 n t x_{0}$ ) and along-track velocity $\overset{y}{˙}_{0}$ (via $- 3 n t \overset{y}{˙}_{0}$ ) cause drift ::: while radial velocity $\overset{x}{˙}_{0}$ , cross-track $z_{0}, \overset{z}{˙}_{0}$ , and pure along-track offset $y_{0}$ do NOT. Drift per orbit from a radial offset $x_{0}$ ? ::: $Δ y = - 12 π x_{0}$ (lower orbit, $x_{0} < 0$ , moves you forward). The one condition for a closed (drift-free) in-plane loop? ::: $\overset{y}{˙}_{0} = - 2 n x_{0}$ , giving a 2:1 football ellipse. Why does a forward tangential burn make you fall behind? ::: It raises your orbit; a higher orbit is slower, and Coriolis rotates the velocity — the net $- \frac{6 π}{n} \overset{y}{˙}_{0}$ drift is backward for $\overset{y}{˙}_{0} > 0$ .

Related: Rotating Reference Frames — Coriolis and Centrifugal · Tschauner–Hempel Equations · Orbital Maneuvers — Hohmann Transfer · State Transition Matrix · Two-Body Problem · Kepler's Laws · Linearization and Taylor Expansion