Exercises — Launch window — phasing with target orbit

$\varphi$ is the angular lead of the target over the chaser at the instant of the departure burn that guarantees rendezvous at the transfer's arrival point.

• $\varphi >0$: target must be ahead of the chaser (in the direction of motion).
• $\varphi <0$: target must be behind the chaser.
• $\varphi =0$: they must be at the same angular position at launch. It is measured at the central body (Earth's centre), in radians.

The gravitational parameter ==$\mu$== cancels from $\varphi$ (the angle is pure geometry — a ratio of radii). It does not cancel from the transfer time $t_t=\pi \sqrt{a_t^3/\mu }$ nor from the synodic period $T_{\text{syn}}=2\pi /|n_1-n_2|$, because those measure how long things take, and time is set by the strength of gravity.

Exactly $180^{\circ }=\pi$ radians. The Hohmann transfer is half an ellipse — perigee to apogee — and perigee and apogee sit on opposite ends of the major axis, i.e. $180^{\circ }$ apart across the focus (Earth's centre).

WHAT: compute the ratio, raise to $3/2$, subtract from 1, multiply by $\pi$. $\frac{r_1+r_2}{2r_2}=\frac{7000+8000}{2\cdot 8000}=\frac{15000}{16000}=0.9375$ $0.9375^{3/2}=0.9077$ $\varphi =\pi (1-0.9077)=\pi (0.0923)=0.2900\text{rad}=16.6^{\circ }$ WHY positive: the target orbit is only slightly higher, so during transfer it covers just under $180^{\circ }$ ($\Delta {\theta }_2=163.4^{\circ }$), leaving a small positive lead. ✅

$a_t=\frac{7000+8000}{2}=7500\text{km}$ $t_t=\pi \sqrt{\frac{a_t^3}{\mu }}=\pi \sqrt{\frac{7500^3}{3.986\times 10^5}}$ $7500^3=4.2188\times 10^{11}$, divide by $\mu$: $1.0584\times 10^6$, square root $=1028.8$ s, times $\pi$: $t_t\approx 3232\text{s}\approx 53.9\text{min}$

Target mean motion: $n_2=\sqrt{\mu /r_2^3}=\sqrt{3.986\times 10^5/8000^3}=\sqrt{7.785\times 10^{-7}}=8.823\times 10^{-4}$ rad/s. $\Delta {\theta }_2=n_2{t}_t=8.823\times 10^{-4}\times 3232=2.851\text{rad}=163.4^{\circ }$ Then $\varphi =180^{\circ }-163.4^{\circ }=16.6^{\circ }$. ✅ Matches L2.1 — the closed formula and the direct time computation agree, because $\mu$ cancelled between them.

$\frac{r_1+r_2}{2r_2}=\frac{42164+6678}{2\cdot 6678}=\frac{48842}{13356}=3.6570$ $3.6570^{3/2}=6.9926$ $\varphi =\pi (1-6.9926)=\pi (-5.9926)=-18.83\text{rad}$ Reduce mod $2\pi$: $-18.83+6\pi =-18.83+18.850=+0.020$ rad $\approx +1.1^{\circ }$... but the physically meaningful statement is the raw value. WHY negative and huge: the inner LEO target is fast — during the long transfer it laps around many times ($\Delta {\theta }_2\approx 6.99\pi \approx 1259^{\circ }$). So $\varphi$ is a large negative number: the target must be far "behind" (equivalently, wraps around multiple times).

The honest interpretation: for an inner target you must add integer multiples of $2\pi$ to place it in $[0,2\pi )$. Take $\varphi \mathrm{mod}2\pi$: $-18.83+6\pi =0.0192$ rad $=1.1^{\circ }$. So at launch the LEO target should sit essentially at the chaser's angular position (a hair ahead). The sign of the un-reduced $\varphi$ still correctly encodes "target covers more than $\pi$." ✅

As $r_2\to r_1$: $\frac{r_1+r_2}{2r_2}\to \frac{2r_1}{2r_1}=1$, and $1^{3/2}=1$, so $\varphi \to \pi (1-1)=0.$ Meaning: if the two orbits are (almost) the same, the transfer takes (almost) exactly half a period, during which the target covers (almost) exactly $180^{\circ }$ — the same $180^{\circ }$ the chaser covers. So there is nothing to lead: they must start at the same angular position. This is exactly the near-co-orbital ISS case, where phasing is razor-thin — see Rendezvous and Phasing Orbits.

As $r_2\to \infty$, $\frac{r_1+r_2}{2r_2}\to \frac{r_2}{2r_2}=\frac{1}{2}$, and $(1/2{)}^{3/2}=0.3536$. $\varphi \to \pi (1-0.3536)=\pi (0.6464)=2.031\text{rad}=116.4^{\circ }.$ So it does not reach $\pi$ — it saturates near $116^{\circ }$. WHY: even for an infinitely far, infinitely slow target, the ratio floors at $1/2$ (perigee $r_1$ is negligible so $a_t\approx r_2/2$). The target still creeps forward a little during the (very long) transfer, so the lead maxes out around $116^{\circ }$, not the full $180^{\circ }$. The GEO example ($100.6^{\circ }$) is on the way to this ceiling.

(a) $n_1=\sqrt{\mu /r_1^3}=\sqrt{3.986\times 10^5/6778^3}$. $6778^3=3.1137\times 10^{11}$; ratio $=1.2802\times 10^{-6}$; $n_1=1.1315\times 10^{-3}$ rad/s. $6798^3=3.1414\times 10^{11}$; ratio $=1.2689\times 10^{-6}$; $n_2=1.1265\times 10^{-3}$ rad/s. (b) $|n_1-n_2|=5.01\times 10^{-6}$ rad/s (chaser is faster, being lower). $T_{\text{syn}}=\frac{2\pi }{5.01\times 10^{-6}}=1.254\times 10^6\text{s}=14.5\text{days}.$ (c) The two orbits differ by only 20 km, so their $n$ differ by $0.4\%$. The relative drift is tiny, so the one correct phasing repeats only every ~2 weeks. Waiting that long is unacceptable, so real missions use several phasing orbits to nudge the phase quickly instead. This is the Synodic Period speaking. ✅

$\frac{r_1+r_2}{2r_2}=\frac{6778+6798}{2\cdot 6798}=\frac{13576}{13596}=0.99853$ $0.99853^{3/2}=0.99779$ $\varphi =\pi (1-0.99779)=\pi (0.00221)=0.00693\text{rad}=0.397^{\circ }.$ Yes — consistent with L3.2: nearly-equal radii give an almost-zero lead. The target need only sit about $0.4^{\circ }$ ahead. Combined with L4.1's 14.5-day synodic period, this shows why the whole window is so restrictive here. ✅

(a) $a_t=(6700+26560)/2=16630$ km. (b) $t_t=\pi \sqrt{16630^3/\mu }$. $16630^3=4.5999\times 10^{12}$; $/\mu =1.1540\times 10^7$; $\sqrt{}=3396.9$ s; $\times \pi =10672$ s $=2.965$ h. (c) $\frac{r_1+r_2}{2r_2}=\frac{33260}{53120}=0.6262$; $0.6262^{3/2}=0.4954$; $\varphi =\pi (1-0.4954)=\pi (0.5046)=1.585$ rad $=90.8^{\circ }$. Interpretation: during the ~3-hour transfer, the GPS target only moves $\Delta {\theta }_2=180^{\circ }-90.8^{\circ }=89.2^{\circ }$, so it must start $\approx 91^{\circ }$ ahead. Sits neatly between the LEO ($5.7^{\circ }$) and GEO ($100.6^{\circ }$) parent examples. ✅

WHAT: define the relative angle $\psi (t)={\theta }_2(t)-{\theta }_1(t)$, the target's lead over the chaser. WHY: the phasing geometry is entirely captured by $\psi$; a launch window recurs when $\psi$ returns to the same value. Each body's angle grows linearly: ${\theta }_i(t)={\theta }_i(0)+n_{i}t$ (constant mean motion on a circle). So $\psi (t)=\psi (0)+(n_2-n_1)t.$ WHAT: the geometry repeats when $\psi$ has advanced by a full $2\pi$: $(n_2-n_1)T_{\text{syn}}=\pm 2\pi .$ WHY the $\pm$: whichever body is faster, the magnitude of one relative revolution is $2\pi$. Solving: $T_{\text{syn}}=\frac{2\pi }{|n_1-n_2|}.■$ This ties directly to Mean Motion and Orbital Period — mean motion is exactly the constant slope $n_i$ used here.

(a) $\frac{r_1+r_2}{2r_2}=\frac{48842}{84328}=0.57920$; ${}^{3/2}=0.44078$; $\varphi =\pi (1-0.44078)=1.7569$ rad $=100.66^{\circ }$. (Matches parent Example 2.) (b) $a_t=24421$ km; $t_t=\pi \sqrt{24421^3/\mu }=\pi \sqrt{1.4566\times 10^{13}/3.986\times 10^5}$. $=\pi \sqrt{3.6543\times 10^7}=\pi \times 6045.1=18991$ s $=5.275$ h. (c) $n_1=1.1500\times 10^{-3}$ rad/s (from $6678^3$); $n_2=7.2921\times 10^{-5}$ rad/s (from $42164^3$). $|n_1-n_2|=1.0771\times 10^{-3}$; $T_{\text{syn}}=2\pi /1.0771\times 10^{-3}=5833$ s $=1.620$ h. (d) Miss the window ⇒ wait one synodic period $\approx 1.62$ h for the same phasing to return (ignoring the separate ground-track/plane condition). ✅ WHY $T_{\text{syn}}$ is short here: the two mean motions are very different (LEO fast, GEO slow), so the relative drift is large ⇒ the geometry recycles quickly — the opposite of the ISS case in L4.1.

Let $x=\frac{r_2}{r_1}>1$. Then $\frac{r_1+r_2}{2r_2}=\frac{1+x}{2x}=\frac{1}{2}\left(1+\frac{1}{x}\right).$ As $x$ ranges over $(1,\infty )$, the quantity $\frac{1}{2}(1+1/x)$ decreases monotonically from $\frac{1}{2}(1+1)=1$ (at $x=1$) down to $\frac{1}{2}(1+0)=\frac{1}{2}$ (as $x\to \infty$). So the ratio lives in $(\frac{1}{2},1)$. Raising to $3/2$ (an increasing function on positive reals) keeps the ordering, giving the bracket value in $(2^{-3/2},1)=(0.3536,1)$. Therefore $\varphi =\pi (1-\text{bracket})\in \pi (0,1-2^{-3/2})=(0,116.4^{\circ }).■$ Interpretation: confirms L3.3 — outer targets always need a positive lead, but never more than $\approx 116^{\circ }$. The whole outer-transfer world is squeezed into that band, which the figure above illustrates.

Can you state, from memory, the three master formulas and what each ignores or keeps $\mu$? $\varphi$ (no $\mu$) ::: $\varphi =\pi [1-((r_1+r_2)/2r_2{)}^{3/2}]$ — pure geometry. $t_t$ (keeps $\mu$) ::: $t_t=\pi \sqrt{a_t^3/\mu }$, $a_t=(r_1+r_2)/2$. $T_{\text{syn}}$ (keeps $\mu$ via $n$) ::: $2\pi /|n_1-n_2|$, $n_i=\sqrt{\mu /r_i^3}$.