Exercises — Launch window — phasing with target orbit

$\varphi$ woh angular lead of the target over the chaser hai departure burn ke instant par, jo guarantee karta hai ki rendezvous transfer ke arrival point par hoga.

• $\varphi >0$: target ko chaser se aage hona chahiye (motion ki direction mein).
• $\varphi <0$: target ko chaser ke peeche hona chahiye.
• $\varphi =0$: launch ke waqt dono ko same angular position par hona chahiye. Yeh central body (Earth ke centre) par measure hota hai, radians mein.

Gravitational parameter ==$\mu$== $\varphi$ se cancel ho jaata hai (angle pure geometry hai — radii ka ratio). Yeh transfer time $t_t=\pi \sqrt{a_t^3/\mu }$ aur synodic period $T_{\text{syn}}=2\pi /|n_1-n_2|$ se cancel nahi hota, kyunki woh measure karte hain ki cheezein kitni der leti hain, aur time gravity ki strength se set hota hai.

Exactly $180^{\circ }=\pi$ radians. Hohmann transfer aadha ellipse hai — perigee se apogee tak — aur perigee aur apogee major axis ke opposite ends par hote hain, yaani focus (Earth ka centre) ke across $180^{\circ }$ alag.

WHAT: ratio compute karo, $3/2$ power tak raise karo, 1 se subtract karo, $\pi$ se multiply karo. $\frac{r_1+r_2}{2r_2}=\frac{7000+8000}{2\cdot 8000}=\frac{15000}{16000}=0.9375$ $0.9375^{3/2}=0.9077$ $\varphi =\pi (1-0.9077)=\pi (0.0923)=0.2900\text{rad}=16.6^{\circ }$ WHY positive: target orbit thodi hi zyada high hai, isliye transfer mein woh just under $180^{\circ }$ ($\Delta {\theta }_2=163.4^{\circ }$) cover karta hai, jo ek chhota positive lead chhod deta hai. ✅

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

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. $\Delta {\theta }_2=n_2{t}_t=8.823\times 10^{-4}\times 3232=2.851\text{rad}=163.4^{\circ }$ Phir $\varphi =180^{\circ }-163.4^{\circ }=16.6^{\circ }$. ✅ L2.1 se match karta hai — closed formula aur direct time computation agree karte hain, kyunki $\mu$ unke beech cancel ho gaya.

$\frac{r_1+r_2}{2r_2}=\frac{42164+6678}{2\cdot 6678}=\frac{48842}{13356}=3.6570$ $3.6570^{3/2}=6.9926$ $\varphi =\pi (1-6.9926)=\pi (-5.9926)=-18.83\text{rad}$ Reduce mod $2\pi$: $-18.83+6\pi =-18.83+18.850=+0.020$ rad $\approx +1.1^{\circ }$... lekin physically meaningful statement raw value hai. WHY negative aur bada: inner LEO target fast hai — lambe transfer mein woh kai baar lap kar leta hai ($\Delta {\theta }_2\approx 6.99\pi \approx 1259^{\circ }$). Isliye $\varphi$ ek bada negative number hai: target bahut "peeche" hona chahiye (equivalently, kaafi baar wrap around karta hai).

Honest interpretation: inner target ke liye tum integer multiples of $2\pi$ add karte ho taki ise $[0,2\pi )$ mein rakh sako. $\varphi \mathrm{mod}2\pi$ lo: $-18.83+6\pi =0.0192$ rad $=1.1^{\circ }$. Toh launch par LEO target essentially chaser ki angular position par hona chahiye (zara sa aage). Un-reduced $\varphi$ ka sign phir bhi sahi encode karta hai ki "target $\pi$ se zyada cover karta hai." ✅

Jab $r_2\to r_1$: $\frac{r_1+r_2}{2r_2}\to \frac{2r_1}{2r_1}=1$, aur $1^{3/2}=1$, isliye $\varphi \to \pi (1-1)=0.$ Matlab: agar dono orbits (almost) same hain, to transfer (almost) exactly aadha period leta hai, jis mein target (almost) exactly $180^{\circ }$ cover karta hai — wahi $180^{\circ }$ jo chaser cover karta hai. Toh lead karne ki zaroorat nahi: dono ko same angular position par start karna chahiye. Yeh exactly near-co-orbital ISS case hai, jahan phasing razor-thin hoti hai — dekho Rendezvous and Phasing Orbits.

Jab $r_2\to \infty$, $\frac{r_1+r_2}{2r_2}\to \frac{r_2}{2r_2}=\frac{1}{2}$, aur $(1/2{)}^{3/2}=0.3536$. $\varphi \to \pi (1-0.3536)=\pi (0.6464)=2.031\text{rad}=116.4^{\circ }.$ Toh yeh $\pi$ tak nahi pahunchta — yeh $116^{\circ }$ ke paas saturate ho jaata hai. WHY: ek infinitely far, infinitely slow target ke liye bhi, ratio $1/2$ par floor ho jaata hai (perigee $r_1$ negligible hai isliye $a_t\approx r_2/2$). Target bahut lambe transfer mein thoda sa aage badhta hai, isliye lead maximum $\approx 116^{\circ }$ par cap hoti hai, poore $180^{\circ }$ par nahi. GEO example ($100.6^{\circ }$) is ceiling ki taraf ja raha tha.

(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 faster hai, kyunki lower hai). $T_{\text{syn}}=\frac{2\pi }{5.01\times 10^{-6}}=1.254\times 10^6\text{s}=14.5\text{days}.$ (c) Dono orbits sirf 20 km differ karti hain, isliye unke $n$ sirf $0.4\%$ differ karte hain. Relative drift bahut chhoti hai, isliye ek correct phasing sirf har ~2 hafte mein repeat hoti hai. Itna intezaar karna unacceptable hai, isliye real missions mein kaafi phasing orbits use hote hain jo phase ko jaldi nudge karte hain. Yahi Synodic Period bol raha hai. ✅

$\frac{r_1+r_2}{2r_2}=\frac{6778+6798}{2\cdot 6798}=\frac{13576}{13596}=0.99853$ $0.99853^{3/2}=0.99779$ $\varphi =\pi (1-0.99779)=\pi (0.00221)=0.00693\text{rad}=0.397^{\circ }.$ Haan — L3.2 se consistent hai: almost equal radii almost-zero lead dete hain. Target ko sirf $\approx 0.4^{\circ }$ aage rehna chahiye. L4.1 ke 14.5-day synodic period ke saath milake, yeh dikhata hai ki poori window yahan kitni restrictive hai. ✅

(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: ~3-ghante ke transfer mein, GPS target sirf $\Delta {\theta }_2=180^{\circ }-90.8^{\circ }=89.2^{\circ }$ move karta hai, isliye use $\approx 91^{\circ }$ aage se start karna chahiye. LEO ($5.7^{\circ }$) aur GEO ($100.6^{\circ }$) parent examples ke beech mein neatly baitha hai. ✅

WHAT: relative angle $\psi (t)={\theta }_2(t)-{\theta }_1(t)$ define karo, jo target ka chaser par lead hai. WHY: phasing geometry poori tarah $\psi$ se capture hoti hai; launch window tab repeat hoti hai jab $\psi$ wahi value par wapas aa jaaye. Har body ka angle linearly badhta hai: ${\theta }_i(t)={\theta }_i(0)+n_{i}t$ (circle par constant mean motion). Toh $\psi (t)=\psi (0)+(n_2-n_1)t.$ WHAT: geometry tab repeat hoti hai jab $\psi$ ek poora $2\pi$ advance kar le: $(n_2-n_1)T_{\text{syn}}=\pm 2\pi .$ WHY $\pm$: chahe kaun sa body faster ho, ek relative revolution ki magnitude $2\pi$ hai. Solve karo: $T_{\text{syn}}=\frac{2\pi }{|n_1-n_2|}.■$ Yeh directly Mean Motion and Orbital Period se jodta hai — mean motion exactly woh constant slope $n_i$ hai jo yahan use hua.

(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 $=100.66^{\circ }$. (Parent Example 2 se match karta hai.) (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 ($6678^3$ se); $n_2=7.2921\times 10^{-5}$ rad/s ($42164^3$ se). $|n_1-n_2|=1.0771\times 10^{-3}$; $T_{\text{syn}}=2\pi /1.0771\times 10^{-3}=5833$ s $=1.620$ h. (d) Window miss karo ⇒ same phasing ke wapas aane ke liye ek synodic period $\approx 1.62$ h wait karo (ground-track/plane condition ko alag rakhte hue). ✅ WHY $T_{\text{syn}}$ yahan chhota hai: dono mean motions bahut alag hain (LEO fast, GEO slow), isliye relative drift badi hai ⇒ geometry jaldi recycle hoti hai — L4.1 ke ISS case ke bilkul opposite.

$x=\frac{r_2}{r_1}>1$ lene do. Phir $\frac{r_1+r_2}{2r_2}=\frac{1+x}{2x}=\frac{1}{2}\left(1+\frac{1}{x}\right).$ Jab $x$ $(1,\infty )$ mein range karta hai, quantity $\frac{1}{2}(1+1/x)$ monotonically decrease karti hai $\frac{1}{2}(1+1)=1$ (at $x=1$) se lekar $\frac{1}{2}(1+0)=\frac{1}{2}$ tak (jab $x\to \infty$). Toh ratio $(\frac{1}{2},1)$ mein rehta hai. $3/2$ tak raise karne par (positive reals par increasing function) ordering maintain rehti hai, aur bracket value $(2^{-3/2},1)=(0.3536,1)$ mein aa jaati hai. Isliye $\varphi =\pi (1-\text{bracket})\in \pi (0,1-2^{-3/2})=(0,116.4^{\circ }).■$ Interpretation: L3.3 confirm hota hai — outer targets ko hamesha positive lead chahiye, lekin kabhi bhi $\approx 116^{\circ }$ se zyada nahi. Outer-transfer ki poori duniya us band mein squeezed hai, jo upar ki figure mein illustrated hai.

Kya tum memory se teen master formulas bata sakte ho aur kaun sa $\mu$ ignore karta hai ya rakhta hai? $\varphi$ (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 $n$) ::: $2\pi /|n_1-n_2|$, $n_i=\sqrt{\mu /r_i^3}$.