Exercises — Rendezvous and proximity operations — Clohessy-Wiltshire equations

WHAT we do: plug straight into $n=\sqrt{\mu /R_0^3}$. WHY: $n$ sets the clock for every CW term (all the trig arguments are $nt$). $R_0=6.778\times 10^6\text{m},R_0^3=3.113\times 10^{20}{\text{m}}^3.$ $n=\sqrt{\frac{3.986\times 10^{14}}{3.113\times 10^{20}}}=\sqrt{1.280\times 10^{-6}}=1.132\times 10^{-3}\text{rad/s}.$ $T=\frac{2\pi }{n}=\frac{6.2832}{1.132\times 10^{-3}}\approx 5551\text{s}\approx 92.5\text{min}.$ That ~92 min is the familiar low-Earth-orbit period — a good sanity check.

The cross-track equation $\ddot{z}+n^{2}z=0$. It is SHM with angular frequency $n$ (same as the orbit rate). Special feature: it is completely decoupled from $x$ and $y$ — a cross-track nudge just oscillates back and forth once per orbit and never affects in-plane motion.

• $-2n\dot{y}$ comes from the Coriolis force (it couples radial acceleration to along-track velocity) — see Rotating Reference Frames — Coriolis and Centrifugal.
• $-3n^{2}x$ is the tidal / gravity-gradient term: it is the leftover after the constant gravity ($-n^2{R}_0$) and centrifugal ($+n^2{R}_0$) cancel, plus the first-order gravity slope $+2n^{2}x$ and centrifugal $+n^{2}x$ combining to $+3n^{2}x$ on the right-hand side.

WHAT: substitute $x_0=y_0=0$, ${\dot{y}}_0=0$, ${\dot{x}}_0=1$ into the state solution. $x(t)=\frac{\sin nt}{n}{\dot{x}}_0=\frac{\sin nt}{n},y(t)=-\frac{2}{n}(1-\cos nt){\dot{x}}_0=-\frac{2}{n}(1-\cos nt).$ At $t=T/4$: $nt=n\cdot (2\pi /n)/4=\pi /2$, so $\sin nt=1$, $\cos nt=0$. $x=\frac{1}{n}=\frac{1}{1.132\times 10^{-3}}\approx 883.5\text{m},$ $y=-\frac{2}{n}(1-0)=-2\times 883.5\approx -1767\text{m}.$ WHAT IT LOOKS LIKE: the along-track excursion is twice the radial one — that 2:1 shape is the "football" ellipse. See the figure below.

State declared: $x_0=-10$, $y_0=0$, ${\dot{x}}_0=0$, ${\dot{y}}_0=0$. Because $y_0=0$ and both initial velocities are zero, every term in $y(t)$ drops except the one multiplying $x_0$. WHAT: use the along-track secular term. Over one period the oscillatory $\sin nt$ returns to zero, so only the $-nt$ piece survives: $\Delta y=y(T)-y_0=6(\sin nT-nT)x_0=-6nT{x}_0=-12\pi x_0.$ WHY $12\pi$: $nT=2\pi$, so $6nT=12\pi$ — a pure number, independent of altitude. $\Delta y=-12\pi (-10)=120\pi \approx +376.99\text{m per orbit}.$ Positive $\Delta y$: being below (lower orbit, faster) walks you forward. That is the phasing trick.

$z(t)=z_0\cos nt=50\cos nt.$ First zero when $nt=\pi /2$, i.e. $t=\frac{\pi }{2n}=\frac{\pi }{2\times 1.132\times 10^{-3}}\approx 1388\text{s}$ (a quarter period). Makes sense: SHM crosses zero every quarter cycle.

WHAT: collect every term in $y(t)$ that is proportional to $t$ (these are the drift terms): $y_{\text{drift}}(t)=\underset{\text{from }6(\sin nt-nt)x_0}{\underbrace{-6nt{x}_0}}+\underset{\text{from }\frac{1}{n}(4\sin nt-3nt){\dot{y}}_0}{\underbrace{-3t{\dot{y}}_0}}.$ WHY: a closed (repeating) orbit cannot grow with $t$, so the coefficient of $t$ must vanish: $-6n{x}_0-3{\dot{y}}_0=0\Rightarrow \boxed{{\dot{y}}_0=-2n{x}_0}.$ Physical meaning: the along-track speed must exactly compensate the orbit-energy difference caused by the radial offset. A radial offset means a slightly different orbital energy → different period → drift; the $-2n{x}_0$ along-track velocity re-tunes the energy so the relative period matches the target's.

Substituting ($x_0=y_0=0,{\dot{y}}_0=0,{\dot{x}}_0=v_r$): $x(t)=\frac{v_r}{n}\sin nt,y(t)=-\frac{2v_r}{n}(1-\cos nt).$ Shift the along-track centre: let $y^{\prime }=y+\frac{2v_r}{n}$, so $y^{\prime }=\frac{2v_r}{n}\cos nt$. Let $a=v_r/n$ (radial amplitude). Then $x=a\sin nt,y^{\prime }=2a\cos nt.$ Square and add with the right weights: ${\left(\frac{x}{a}\right)}^2+{\left(\frac{y^{\prime }}{2a}\right)}^2={\sin }^{2}nt+{\cos }^{2}nt=1.$ That is an ellipse with radial semi-axis $a$ and along-track semi-axis $2a$ — exactly 2:1. The figure below traces it.

A forward burn sets ${\dot{y}}_0>0$ with $x_0=0$. Look at the radial response: $x(t)=\frac{2}{n}(1-\cos nt){\dot{y}}_0\ge 0.$ So a forward push actually raises you (positive $x$, higher orbit). Now the secular drift term $-6ntx$ (with $x>0$ on average) makes $y$ decrease over time — you drift backward. Physically: forward burn → higher orbit → slower angular rate → you lose the race and fall behind. This is the Coriolis coupling $\ddot{x}-2n\dot{y}$ in action.

Step 1 — how much drift do we need per orbit? From Ex 2.2, a chaser parked at a fixed radial offset $x_0$ (with the bounded-motion along-track velocity ${\dot{y}}_0=-2n{x}_0$, so it does not loop away) drifts $\Delta y=-12\pi x_0$ per orbit. We need $\Delta y=+2000\text{m}$ (close a 2 km lag = move forward by 2 km): $2000=-12\pi x_0\Rightarrow x_0=-\frac{2000}{12\pi }\approx -53.05\text{m}.$ A ~53 m drop below the target moves you 2 km forward in one lap. Step 2 — WHY a matching along-track velocity keeps the offset constant. A raw radial velocity kick does not hold you at a fixed altitude — it makes you loop (Ex 2.1). To sit at the constant radial offset $x_0$ so that the clean $\Delta y=-12\pi x_0$ formula applies, the initial state must satisfy the bounded-motion condition from Ex 3.1: ${\dot{y}}_0=-2n{x}_0.$ Why this holds you level: this along-track velocity exactly re-tunes the relative orbital energy so the radial coordinate does not grow — $x(t)$ stays a bounded oscillation about $x_0$ rather than drifting. Substituting our value: ${\dot{y}}_0=-2n{x}_0=-2(1.132\times 10^{-3})(-53.05)\approx +0.1201\text{m/s}.$ So the entry burn is a ~12 cm/s along-track impulse (plus whatever tiny radial component sets $x_0$). Notice ${\dot{y}}_0=-2n{x}_0$ is the velocity condition; the offset $x_0$ itself is a position you arrive at, not something a velocity alone sustains — that is the correction to the naïve "${\dot{x}}_0=n{x}_0$" shortcut. Step 3 — stop at the end. After one orbit you have moved $+2$ km and are back at radial $x_0$ with the same ${\dot{y}}_0$. A matching $-0.1201$ m/s along-track burn (and small radial trim to null $x_0$) parks you level with the target. Compare with Orbital Maneuvers — Hohmann Transfer, where the same phasing idea is executed as two altitude-changing burns.

Differentiate: $\dot{x}=an\cos nt$, $\dot{y}=\dot{y^{\prime }}=-2an\sin nt$. At $nt=\pi /2$ (top of loop, $x=a$, $y^{\prime }=0$): $\dot{x}=0$, $\dot{y}=-2an$. So at that instant the motion is purely along-track at speed $2an$. A single along-track burn of magnitude $\Delta v=2an=2\times 100\times 1.132\times 10^{-3}=0.2264\text{m/s}\approx 22.6\text{cm/s}$ exactly cancels the velocity there. Position is still offset by $y^{\prime }$-shift, so you follow up with a small hop — but the velocity kill is cleanest where the velocity is single-axis.

With ${\dot{x}}_0={\dot{y}}_0=y_0=0$: $x(t)=(4-3\cos nt)x_0,y(t)=6(\sin nt-nt)x_0.$ At $nt=2\pi$: $\cos 2\pi =1$, $\sin 2\pi =0$. $x(T)=(4-3\cdot 1)x_0=x_0✓(\text{radial returns}),$ $y(T)=6(0-2\pi )x_0=-12\pi x_0✓(\text{along-track drift}).$ So the STM maps $x_0\mapsto (x_0,-12\pi x_0)$ over one period. This is the mathematically exact version of Ex 2.2: radial offset is periodic in $x$ but secular in $y$.

Step 1 — mean motion as a function of radius. For a circular orbit of radius $R$, Kepler gives $n(R)=\sqrt{\frac{\mu }{R^3}}=\sqrt{\mu }{R}^{-3/2}.$ Step 2 — differentiate to see how rate changes with radius. Why a derivative? We want the direction the angular rate moves when we nudge $R$ down; the sign of $dn/dR$ answers exactly that. $\frac{dn}{dR}=-\frac{3}{2}\sqrt{\mu }{R}^{-5/2}<0.$ The derivative is negative for all $R>0$: decreasing $R$ increases $n$. So a lower orbit ($R_0-\delta$, with $\delta >0$) sweeps angle faster than the target at rate $n$. Step 3 — first-order size of the speedup. Linearising (a Taylor step, see Linearization and Taylor Expansion): $n(R_0-\delta )\approx n-\frac{dn}{dR}{|}_{R_0}\delta =n+\frac{3}{2}\frac{n}{R_0}\delta >n.$ Step 4 — connect to the CW drift. A chaser at radial offset $x_0<0$ sits at radius $R_0-|x_0|$ (i.e. $\delta =|x_0|$). By Step 3 it has a higher angular rate, so it steadily pulls ahead (forward, $+y$). The CW formula agrees: $\Delta y=-12\pi x_0$, and for $x_0<0$ this gives $\Delta y>0$ (forward). So the linear CW secular drift is exactly the first-order shadow of Kepler's speed-radius law — see Kepler's Laws and Two-Body Problem.

CW is the first-order Taylor expansion of two-body gravity about a circular reference. As $x,y,z\to 0$ the neglected terms are $O(r^2/R_0^2)$, which vanish faster than the linear terms — so CW becomes exact in that limit. For eccentric orbits, $n$ is no longer constant (angular rate varies with true anomaly), so the constant-coefficient CW equations fail; the time-varying version is the Tschauner–Hempel Equations. See also Two-Body Problem.

$n=\sqrt{\mu /R_0^3}$; 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. If you can produce each without notes, you have mastered D4.