Exercises — Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true ano

Hum kya test kar rahe hain: $e$ ka matlab definition se seedha padhna. $e$ shape number hai. Jab $e=0$ hota hai toh conic ek circle hota hai. Orbit equation $r=\frac{a(1-e^2)}{1+e\cos \nu }$ mein $e=0$ daalne par har $\nu$ ke liye $r=a$ ho jaata hai — radius kabhi nahi badlata. Isliye $r_p=r_a=a$. Ek circle: apogee equals perigee equals $a$.

Time ke saath badalta hai: ==true anomaly $\nu$== — yeh "satellite abhi kahan hai?" wala angle hai. Constant: $a,e,i,\Omega ,\omega$. Yeh ellipse ki unchanging geometry (size, shape, tilt, swing, in-plane pointing) fix karte hain. Sirf moving point ko moving number chahiye.

$i$ orbital plane ka equator se tilt hai.

• $i=90^{\circ }$ → plane seedha poles ke through khada ho jaata hai → polar orbit.
• $i=0^{\circ }$ → plane equatorial plane mein flat leta hai → equatorial orbit. ($i>90^{\circ }$ retrograde hoga — Earth ke spin ke against jaana.)

Step 1 — altitude → radius. Kyun: elements Earth ke centre (focus) se measure hote hain, isliye $R_E$ jodo. $r_p=6378+500=6878\text{km}$, $r_a=6378+2500=8878\text{km}$. Step 2 — semi-major axis. Poora major axis $r_p+r_a=2a$ hai, toh $a=\frac{r_p+r_a}{2}=\frac{6878+8878}{2}=7878\text{km}.$ Step 3 — eccentricity. $r_a-r_p=2ae$ aur $r_a+r_p=2a$ se, dono divide karne par $2a$ cancel ho jaata hai: $e=\frac{r_a-r_p}{r_a+r_p}=\frac{2000}{15756}=\mathrm{0.1269.}$

Kyun vis-viva: yeh speed ko position se link karta hai sirf $a$ use karke, kyunki energy $\epsilon =-\mu /2a$ constant hai. $v^2=\mu \left(\frac{2}{r}-\frac{1}{a}\right)\text{at}r=r_p=6878.$ $v_p^2=3.986\times 10^5\left(\frac{2}{6878}-\frac{1}{7878}\right)=3.986\times 10^5(2.9078\times 10^{-4}-1.2694\times 10^{-4}).$ $v_p^2=3.986\times 10^5(1.6384\times 10^{-4})=65.31\Rightarrow v_p=8.08\text{km/s}.$ Sensible hai: perigee sabse tez point hai, circular LEO speed se thoda upar.

Step 1 — semi-latus rectum. $p=a(1-e^2)=7878(1-0.1269^2)=7878(1-0.016103)=7751.1\text{km}$. Step 2 — orbit equation. Kyun yeh formula: $r=\frac{p}{1+e\cos \nu }$ ellipse ka focus-centred polar form hai. $r=\frac{7751.1}{1+0.1269\cos 60^{\circ }}=\frac{7751.1}{1+0.1269(0.5)}=\frac{7751.1}{1.06345}=7288.9\text{km}.$

$p=a(1-e^2)=8000(1-0.04)=7680\text{km}$. $r=\frac{p}{1+e\cos \nu }$ use karo:

• $\nu =0^{\circ }$: $\cos =1$, $r=\frac{7680}{1.20}=6400\text{km}$ (perigee $=a(1-e)=6400$ ✓).
• $\nu =90^{\circ }$: $\cos =0$, $r=7680\text{km}$ ($=p$).
• $\nu =180^{\circ }$: $\cos =-1$, $r=\frac{7680}{0.80}=9600\text{km}$ (apogee $=a(1+e)=9600$ ✓).
• $\nu =270^{\circ }$: $\cos =0$, $r=7680\text{km}$ ($=p$). Symmetry: $\cos \nu =\cos (360^{\circ }-\nu )$, isliye $r(90^{\circ })=r(270^{\circ })$. Ellipse major axis ke baare mein mirror-symmetric hai; perigee se ek quarter-turn dono taraf wale do points ek hi distance par hain.

RAAN. South→north equator crossing hi ascending node hai. Uska right ascension (equatorial plane mein $X$/♈ se angle) $\Omega =70^{\circ }$ hai. Argument of perigee. $\omega$ ascending node se perigee tak ka angle hai, orbital plane mein, motion ke direction mein — diya gaya hai $\omega =50^{\circ }$. True anomaly. $\nu$ perigee se satellite tak measure hota hai. Satellite perigee se $50^{\circ }$ pehle hai, yaani usse $50^{\circ }$ aur jaana hai, toh woh $\nu =360^{\circ }-50^{\circ }=310^{\circ }$ par hai (equivalently $-50^{\circ }$). Argument of latitude. $u=\omega +\nu =50^{\circ }+310^{\circ }=360^{\circ }\equiv 0^{\circ }$. Iska matlab node→satellite $0^{\circ }$ hai: satellite abhi ascending node par hi hai — "usne abhi equator cross kiya" ke consistent hai. ✓

Pehle energy. $\epsilon =-\mu /2a$ sirf $a$ par depend karta hai — dono ke liye same. Equal energy, alag shapes. Perigee speed. $r_p=a(1-e)$, toh orbit 2 ka perigee deeper hai (chhota $r_p$), aur vis-viva kehta hai chhota $r$ → bada $v$. $r_{p1}=7500(0.95)=7125\text{km}$, $r_{p2}=7500(0.65)=4875\text{km}$. $v_p^2=\mu \left(\frac{2}{r_p}-\frac{1}{a}\right).$ $v_{p1}^2=3.986\times 10^5(2/7125-1/7500)=3.986\times 10^5(2.8070\times 10^{-4}-1.3333\times 10^{-4})=58.74$, $v_{p1}=7.664\text{km/s}$. $v_{p2}^2=3.986\times 10^5(2/4875-1/7500)=3.986\times 10^5(4.1026\times 10^{-4}-1.3333\times 10^{-4})=110.4$, $v_{p2}=10.51\text{km/s}$. Ratio $v_{p2}/v_{p1}=1.371$. Orbit 2 perigee par ~37% tez hai — phir bhi dono ki total energy equal hai. Eccentric orbit sirf ek tez perigee aur slow apogee ke beech speed trade karta hai.

Step 1 — radii. $r_p=6378+600=6978\text{km}$, $r_a=6378+35786=42164\text{km}$. Step 2 — $a$ aur $e$. $a=\frac{r_p+r_a}{2}=\frac{6978+42164}{2}=24571\text{km},e=\frac{r_a-r_p}{r_a+r_p}=\frac{35186}{49142}=\mathrm{0.7160.}$ Step 3 — period. Kyun yeh formula: Kepler's third law, $T=2\pi \sqrt{a^3/\mu }$, gravity ko ek loop mein centripetal need ke saath equate karne se aata hai; yeh sirf $a$ par depend karta hai. $T=2\pi \sqrt{\frac{24571^3}{3.986\times 10^5}}=2\pi \sqrt{\frac{1.4835\times 10^{13}}{3.986\times 10^5}}=2\pi \sqrt{3.7218\times 10^7}=2\pi (6100.7)=38331\text{s}\approx 10.65\text{h}.$ Step 4 — apogee speed. vis-viva at $r=r_a$ par: $v_a^2=\mu \left(\frac{2}{r_a}-\frac{1}{a}\right)=3.986\times 10^5\left(\frac{2}{42164}-\frac{1}{24571}\right)=3.986\times 10^5(4.7433\times 10^{-5}-4.0698\times 10^{-5}).$ $v_a^2=3.986\times 10^5(6.735\times 10^{-6})=2.6847\Rightarrow v_a=1.639\text{km/s}.$ Apogee par slow — exactly jaisa Kepler's 2nd law demand karta hai.

Step 1 — $\nu =120^{\circ }$ par radius. $p=a(1-e^2)=24571(1-0.7160^2)=24571(1-0.51266)=24571(0.48734)=11973\text{km}$. $r=\frac{p}{1+e\cos 120^{\circ }}=\frac{11973}{1+0.7160(-0.5)}=\frac{11973}{0.6420}=18650\text{km}.$ Step 2 — perifocal components. Perifocal (P–Q–W) frame mein P perigee ki taraf point karta hai, Q $90^{\circ }$ aage hai: ${\vec{r}}_{PQW}=r\left(\begin{array}{c}\cos \nu \\ \sin \nu \\ 0\end{array}\right)=18650\left(\begin{array}{c}\cos 120^{\circ }\\ \sin 120^{\circ }\\ 0\end{array}\right)=18650\left(\begin{array}{c}-0.5\\ 0.8660\\ 0\end{array}\right)=\left(\begin{array}{c}-9325\\ 16151\\ 0\end{array}\right)\text{km}.$ Step 3 — teen rotations (words mein). ECI tak pahunchne ke liye ${\vec{r}}_{ECI}=R_z(\Omega )R_x(i)R_z(\omega ){\vec{r}}_{PQW}$ apply karo:

1. $R_z(\omega )$ — $W/Z$ ke baare mein spin karo taaki perigee node se sahi in-plane angle par ho.
2. $R_x(i)$ — line of nodes ke baare mein plane ko inclination $i$ se tilt karo.
3. $R_z(\Omega )$ — tilted plane ko $Z$ ke baare mein swing karo taaki node right ascension $\Omega$ par land kare.

Kepler's third law invert karo. $T=2\pi \sqrt{a^3/\mu }$ se, $a={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}={\left(\frac{3.986\times 10^5\times 43082^2}{4{\pi }^2}\right)}^{1/3}.$ Numerator: $3.986\times 10^5\times 1.85606\times 10^9=7.3982\times 10^{14}$. $4{\pi }^2=39.478$ se divide karo: $1.87401\times 10^{13}$. Cube root: $a=2.6607\times 10^4=26607\text{km}$. Circular altitude ($e=0$ → $r=a$): altitude $=a-R_E=26607-6378=20229\text{km}$. (Yeh essentially GPS orbit hai.)

Energy aur $a$. Specific energy $\epsilon =-\mu /2a$ hai. $\epsilon =0$ set karne par $-\mu /2a=0$ force hota hai, jo sirf $a\to \infty$ par hold ho sakta hai — ellipse ek parabola mein stretch ho jaata hai. Yahi precisely $e=1$ boundary hai. ✓ Escape speed. $\epsilon =0$ par: $\frac{v^2}{2}-\frac{\mu }{r}=0\Rightarrow v^2=\frac{2\mu }{r}\Rightarrow v_{esc}=\sqrt{\frac{2\mu }{r}}$. (Vis-viva ke saath same, jab $1/a\to 0$.) $r=6978$ par numbers: $v_{esc}=\sqrt{\frac{2\times 3.986\times 10^5}{6978}}=\sqrt{114.24}=10.69\text{km/s}.$ GTO perigee speed (L4·Q1) tha $v_p=\sqrt{\mu (2/r_p-1/a)}$ with $a=24571$: $v_p^2=3.986\times 10^5(2/6978-1/24571)=3.986\times 10^5(2.8661\times 10^{-4}-4.0698\times 10^{-5})=89.63$, $v_p=9.467\text{km/s}$. Comparison: $v_p=9.467<10.69=v_{esc}$. GTO bound hai (elliptic, $e<1$) — yeh escape nahi karta, $a$ finite ke consistent hai.

(a) Perigee khatam ho jaata hai. $\nu$ perigee se measure hota hai. Lekin jab $e=0$ hota hai toh orbit ka har point same radius $r=a$ par hota hai — koi unique closest point nahi hota, isliye "perigee" undefined hai, aur isliye $\nu$ ka koi anchor nahi. Yeh ek degenerate coordinate hai. (b) Radius survive karta hai. $r=\frac{a(1-e^2)}{1+e\cos \nu }\to a$ sab $\nu$ ke liye jab $e\to 0$. Geometry (radius $a$ ka ek circle) completely well-defined hai; sirf labelling angle toota. (c) Fix — argument of latitude. Ascending node abhi bhi exist karta hai (plane abhi bhi equator cross karta hai), isliye hum satellite ka angle node se measure kar sakte hain: $u=\omega +\nu$. Kyunki $\omega$ (node→perigee) aur $\nu$ (perigee→satellite) dono individually meaning kho dete hain lekin unka sum (node→satellite) nahi kho ta, mission software near-circular orbits ke liye directly $u$ use karta hai. Isliye $u$ ko $e\to 0$ ke liye "non-singular" element kaha jaata hai.

$r$ kab blow up hota hai? $r\to \infty$ ke liye denominator $1+e\cos \nu \to 0$ chahiye, yaani $\cos {\nu }_{\infty }=-\frac{1}{e}=-\frac{1}{1.2}=-\mathrm{0.8333.}$ Solve karo, dono signs. ${\nu }_{\infty }=\pm \arccos (-0.8333)=\pm 146.44^{\circ }$.

• $+146.44^{\circ }$ outbound asymptote hai (infinity ki taraf ja raha hai).
• $-146.44^{\circ }$ (equivalently $213.56^{\circ }$) inbound asymptote hai (infinity se aa raha hai). Sign check: bound ellipse ke liye $e<1$, $-1/e<-1$, toh $\cos \nu =-1/e$ ka koi solution nahi — $r$ kabhi infinity tak nahi pahunchta, orbit close ho jaata hai. Sirf $e\ge 1$ ke liye asymptote exist karta hai. ✓ Physical meaning: $r\to \infty$ woh direction mark karta hai jis se spacecraft infinite distance se aata hai / jaata hai — hyperbolic asymptotes. Dono ke beech ka angle, $2(180^{\circ }-146.44^{\circ })=67.1^{\circ }$... zyada precisely turn angle $\delta =2\arcsin (1/e)$ involve karta hai; asymptotes khud perigee se $\nu =\pm 146.44^{\circ }$ par hain.

• L1–L2 clean mile ::: tum padh aur plug kar sakte ho. Solid foundation.
• L3 mila ::: tum signs handle karte ho aur energy + geometry combine karte ho.
• L4 mila ::: tum poora elements↔state pipeline run kar sakte ho.
• L5 mila ::: tum limits aur edge cases ke baare mein reason karte ho — mastery.