Exercises — Orbital perturbations — J2 effect (oblateness), derivation of nodal precession

$\dot{\Omega }$ mein sirf inclination-dependent piece $-\cos i$ hai, aur $n,J_2,(R_E/p{)}^2$ sab positive hain. Toh $\dot{\Omega }$ ka sign $\cos i$ ka ulta sign hai — figure mein solid drift curve (labelled $\dot{\Omega }$) follow karo.

• $i=30°$ par: $\cos 30°=+0.866>0$, toh $-\cos i<0$ → $\dot{\Omega }<0$ → westward (node regresses).
• $i=120°$ par: $\cos 120°=-0.5<0$, toh $-\cos i>0$ → $\dot{\Omega }>0$ → eastward (node advances).

Switch $i=90°$ par hota hai jahaan $\cos i=0$ aur zero nodal precession hai (curve horizontal axis ko cross karti hai, dotted vertical line se marked).

• Sun-synchronous node $\Omega$ ko exploit karta hai (poora plane ghoomta hai): Sun-synchronous orbits.
• Molniya argument of perigee $\omega$ ko exploit karta hai (ellipse apne plane mein rotate karta hai), use critical inclination $i=63.4°$ par freeze karta hai.

$n$ pehle kyun? Har precession rate $n$ ke saath linearly scale karta hai, isliye yeh base timescale hai. $n=\sqrt{\frac{\mu }{a^3}}=\sqrt{\frac{3.986\times 10^{14}}{(7.0\times 10^6{)}^3}}.$ Denominator: $(7.0\times 10^6{)}^3=3.43\times 10^{20}$. Ratio $=1.1621\times 10^{-6}$. Square root: $n=1.0780\times 10^{-3}\text{rad/s}.$ Period $T=2\pi /n=6.2832/1.0780\times 10^{-3}=5828\text{s}=97.1\text{min}$. Yeh seedha Keplerian $T=2\pi \sqrt{a^3/\mu }$ se jud jaata hai.

L2.1 se $n=1.0780\times 10^{-3}$ rad/s use karo. Kyunki $e=0$, $p=a$ hai, toh ${\left(\frac{R_E}{p}\right)}^2={\left(\frac{6.378\times 10^6}{7.0\times 10^6}\right)}^2=(0.9111{)}^2=\mathrm{0.8302.}$ $\cos 55°=0.5736$. Phir $\dot{\Omega }=-\frac{3}{2}(1.0780\times 10^{-3})(1.0826\times 10^{-3})(0.8302)(0.5736).$ Step by step multiply karo: $\frac{3}{2}\times 1.0780\times 10^{-3}=1.617\times 10^{-3}$; $\times 1.0826\times 10^{-3}=1.7506\times 10^{-6}$; $\times 0.8302=1.4535\times 10^{-6}$; $\times 0.5736=8.337\times 10^{-7}$. $\dot{\Omega }=-8.337\times 10^{-7}\text{rad/s}.$ Convert karo: $\times 86400\times (180/\pi )=\times 4.9536\times 10^6$: $\dot{\Omega }\approx -4.13°/\text{day}(\text{westward}).$

Dono same $a$ share karte hain, isliye same $n$, aur same $i$, toh fark sirf $(R_E/p{)}^2$ factor mein hai, kyunki $p=a(1-e^2)$. $\frac{{\dot{\Omega }}_B}{{\dot{\Omega }}_A}=\frac{(R_E/p_B{)}^2}{(R_E/p_A{)}^2}=\frac{p_A^2}{p_B^2}=\frac{a^2}{[a(1-e^2){]}^2}=\frac{1}{(1-e^2{)}^2}.$ $e=0.3$ ke saath: $1-e^2=0.91$, $(0.91{)}^2=0.8281$. $\frac{{\dot{\Omega }}_B}{{\dot{\Omega }}_A}=\frac{1}{0.8281}=\mathrm{1.2076.}$ Eccentric orbit ~21% tez precess karta hai. Yahi hai "$p$ use karo, $a$ nahi" waali warning: eccentricity ek small correction nahi hai.

$\dot{\omega }=0$ kyun set karein? Vanishing apsidal rate ka matlab hai perigee frozen hai — yahi Molniya design ka poora point hai. $\dot{\omega }\propto (5{\cos }^{2}i-1)$, toh bracket ko zero set karo: $5{\cos }^{2}i-1=0\Rightarrow {\cos }^{2}i=\frac{1}{5}\Rightarrow \cos i=\pm \sqrt{0.2}=\pm \mathrm{0.4472.}$ $i=\arccos (0.4472)=63.43°,i=\arccos (-0.4472)=116.57°.$ Dono critical inclination hain. Molniya prograde wala use karta hai, $63.4°$.

Step 1 — mean motion. $n=\sqrt{\mu /a^3}=\sqrt{3.986\times 10^{14}/(7.2\times 10^6{)}^3}$. Cube: $(7.2\times 10^6{)}^3=3.7325\times 10^{20}$; ratio $=1.0679\times 10^{-6}$; root $n=1.0334\times 10^{-3}$ rad/s.

Step 2 — bulge factor. $e=0\Rightarrow p=a$, toh $(R_E/p{)}^2=(6.378/7.2{)}^2=(0.8858{)}^2=0.7847$.

Step 3 — $\cos i$ isolate karo. Humein $i$ chahiye, lekin $i$ $\cos i$ ke andar baitha hai, jo known constants $-\frac{3}{2}n{J}_2(R_E/p{)}^2$ se multiply hai. Toh hum dono sides ko us poore constant block se divide karte hain taaki $\cos i$ akela right par reh jaaye — yeh ek legal move hai kyunki block ek fixed nonzero number hai. Note karo block negative hai (leading $-\frac{3}{2}$), aur yeh sign matter karega: $\cos i=\frac{\dot{\Omega }}{-\frac{3}{2}n{J}_2(R_E/p{)}^2}.$ Denominator: $-\frac{3}{2}(1.0334\times 10^{-3})(1.0826\times 10^{-3})(0.7847)=-1.3167\times 10^{-6}$. $\cos i=\frac{+1.991\times 10^{-7}}{-1.3167\times 10^{-6}}=-\mathrm{0.1512.}$ Ek positive target $\dot{\Omega }$ ko ek negative constant block se divide karne par negative $\cos i$ milta hai — sign flip forced hai, aur yeh batata hai ki orbit retrograde honi chahiye.

Step 4 — invert karo. $\cos i$ negative hai, toh $i$ $90°$ ke baad hona chahiye: $i=\arccos (-0.1512)=98.70°$. Retrograde ($i>90°$) inclination exactly yahi hai jo $\cos i$ ko negative banata hai, $-\cos i$ ko positive flip karta hai taaki node eastward drift karke Sun ko chase kare. Chart mein marked crossing point dekho.

Rate construction se $0.9856°/\text{day}$ hai. $2\times 365.25=730.5$ days mein: $\Delta \Omega =0.9856\times 730.5=719.98°\approx 720°.$ Yeh exactly do poore chakkar hain ($2\times 360°$) — ek per year — jo ek Sun-synchronous orbit ki defining property hai: plane Sun ke saath kadam milaata hai, isliye har equator crossing par local solar time fixed rehti hai.

Mean motion. $a^3=(2.656\times 10^7{)}^3=1.8735\times 10^{22}$; $\mu /a^3=2.1275\times 10^{-8}$; $n=1.4586\times 10^{-4}$ rad/s. (Yeh $T\approx 11.97$ h deta hai — half-sidereal-day Molniya period.)

Semi-latus rectum. $p=a(1-e^2)=2.656\times 10^7(1-0.5476)=2.656\times 10^7\times 0.4524=1.2016\times 10^7$ m. $(R_E/p{)}^2=(6.378\times 10^6/1.2016\times 10^7{)}^2=(0.5308{)}^2=0.2818$.

(a) Node. $\cos 63.43°=0.4472$. $\dot{\Omega }=-\frac{3}{2}(1.4586\times 10^{-4})(1.0826\times 10^{-3})(0.2818)(0.4472).$ Chain: $\frac{3}{2}\times 1.4586\times 10^{-4}=2.1879\times 10^{-4}$; $\times 1.0826\times 10^{-3}=2.369\times 10^{-7}$; $\times 0.2818=6.676\times 10^{-8}$; $\times 0.4472=2.985\times 10^{-8}$ rad/s. $\dot{\Omega }=-2.985\times 10^{-8}\text{rad/s}\times 4.9536\times 10^6=-0.148°/\text{day}.$

(b) Perigee. $5{\cos }^{2}i-1=5(0.4472{)}^2-1=5(0.2)-1=0$. Toh $\dot{\omega }=0$ — perigee frozen hai, exactly Molniya design goal. Apogee northern hemisphere ke upar parked rehta hai.

Observation ko rad/s mein convert karo. $-3.0°/\text{day}÷4.9536\times 10^6=-6.056\times 10^{-7}$ rad/s.

Mean motion. $a^3=(7.5\times 10^6{)}^3=4.2188\times 10^{20}$; $\mu /a^3=9.4479\times 10^{-7}$; $n=9.7200\times 10^{-4}$ rad/s.

Bulge factor ($e=0$, $p=a$): $(R_E/p{)}^2=(6.378/7.5{)}^2=(0.8504{)}^2=0.7232$.

$\cos i$ isolate karo. L4.1 jaisa hi isolation move: observed $\dot{\Omega }$ ko constant block $-\frac{3}{2}n{J}_2(R_E/p{)}^2$ se divide karo taaki $\cos i$ se sab kuch strip ho jaaye. Block banao: $-\frac{3}{2}n{J}_2(R_E/p{)}^2:\frac{3}{2}\times 9.7200\times 10^{-4}=1.458\times 10^{-3};\times 1.0826\times 10^{-3}=1.5784\times 10^{-6};\times 0.7232=1.1415\times 10^{-6}.$ Leading minus ke saath block $-1.1415\times 10^{-6}$ hai. Ab divide karo: $\cos i=\frac{\dot{\Omega }}{-\frac{3}{2}n{J}_2(R_E/p{)}^2}=\frac{-6.056\times 10^{-7}}{-1.1415\times 10^{-6}}=\mathrm{0.5305.}$ Yahan numerator aur block dono negative hain, toh ratio positive aata hai — hum prograde regime mein hain ($i<90°$).

Invert karo. $\cos i=0.5305\Rightarrow i=\arccos (0.5305)=57.96°\approx 58.0°$. Kyunki observed $\dot{\Omega }<0$ (westward) $\cos i>0$ force karta hai, hum prograde root $i\approx 58.0°$ lete hain. (Retrograde alias $i=122.0°$ $\cos i<0$ dega isliye $\dot{\Omega }>0$, jo westward observation se contradict karta hai.)