Exercises — Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

WHAT: teen numbers ko ek family se match karo. Altitude $\approx 35786$ km ke saath $e=0,i=0$ yeh GEO (geostationary) ka fingerprint hai. WHY it matters: is radius par period ek sidereal day ke barabar hoti hai, aur kyunki yeh circular ($e=0$) aur equatorial ($i=0$) hai, satellite equator ke ek point ke upar bilkul stationary dikhta hai. Ground dish ko kabhi hilana nahin padta. Answer: Geostationary orbit; yeh sky mein fixed dikhta hai.

WHY this reasoning: speed hai $v=\sqrt{\mu /r}$ — $r$ ki decreasing function. Bada $r$ ⇒ chhota $v$. A chhote $r$ par hai, isliye A faster hai. Period $a$ ke saath badhti hai ($T\propto a^{3/2}$), isliye jo zyada upar hai, B ki period zyada lambi hai. Answer: A faster, B ki period lambi.

WHAT: altitude ko radius mein convert karo, $v=\sqrt{\mu /r}$ mein daalo. $r=(6371+550)\text{km}=6921\text{km}=6.921\times 10^6$ m. WHY this tool: yeh ek circular orbit hai, isliye gravity = centripetal exactly $v=\sqrt{\mu /r}$ deta hai; ellipse machinery ki zaroorat nahin. $v=\sqrt{\frac{3.986\times 10^{14}}{6.921\times 10^6}}=\sqrt{5.759\times 10^7}\approx 7589\text{m/s}\approx 7.59\text{km/s}.$

WHAT: $T=2\pi \sqrt{r^3/\mu }$ use karo (circle ⇒ $a=r$). $r^3=(6.921\times 10^6{)}^3=3.316\times 10^{20}{\text{m}}^3$. $T=2\pi \sqrt{\frac{3.316\times 10^{20}}{3.986\times 10^{14}}}=2\pi \sqrt{8.319\times 10^5}=2\pi (912.1)\approx 5731\text{s}\approx 95.5\text{min}.$ WHY: period ek circumference $2\pi r$ hai jo speed se divide hoti hai — algebra usse Kepler ke 3rd law mein collapse kar deta hai.

WHAT: period → radius jaane ke liye Kepler ke 3rd law ko invert karo. $r={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}.$ WHY invert: humein woh time pata hai jo hum chahte hain (ek Earth-spin) aur distance solve kar rahe hain. $\mu T^2=3.986\times 10^{14}\times (86164{)}^2=3.986\times 10^{14}\times 7.4242\times 10^9=2.9593\times 10^{24}$. $4{\pi }^2=39.478$ se divide karo: $7.4959\times 10^{22}$. Cube root: $r\approx 4.216\times 10^7$ m $=42164$ km. $h=42164-6371\approx 35793\text{km}$ (book precision mein ≈35 786 km).

WHAT: L2.3 jaisa hi inversion, new $T$ ke saath. $\mu T^2=3.986\times 10^{14}\times (43082{)}^2=3.986\times 10^{14}\times 1.8561\times 10^9=7.398\times 10^{23}$. $/4{\pi }^2=1.874\times 10^{22}$; cube root $r\approx 2.656\times 10^7$ m $=26560$ km. $h=26560-6371\approx 20190\text{km}$. WHY the check: $20190$ km, 2000 aur 35 786 km ke beech hai → MEO ✓. Dhyan do ki period ko half karne se radius half nahin hoti: $r\propto T^{2/3}$, isliye radius period se kahin zyada gently shrink hoti hai.

Yeh figure woh geometry dikhata hai jo hum use kar rahe hain: perigee (Earth ke paas) aur apogee (door) long axis ke opposite ends par hain, Earth ek focus par hai, aur do radii $r_p$ aur $r_a$ jo hum neeche relate karte hain woh usi focus se drawn hain. $r_a=2a-r_p$ seedha picture se padho — poori long axis $2a$ hai, aur chhota piece $r_p$ hatane par lamba piece $r_a$ bachta hai.

WHAT: $a=\frac{1}{2}(r_p+r_a)$ use karo, yaani $r_a=2a-r_p$. $r_p=6371+600=6971$ km. $2a=2\times 26560=53120$ km. $r_a=53120-6971=46149$ km $\Rightarrow h_a=46149-6371\approx 39780\text{km}$. WHY this works: semi-major axis by definition do extreme radii ka average hota hai — yeh ek ellipse ki geometry hai, koi forces nahin chahiye. Bonus mein, eccentricity yahan $e=\frac{r_a-r_p}{r_a+r_p}=\frac{46149-6971}{53120}\approx 0.74$ hai — ek strongly stretched ellipse, exactly Molniya ka signature.

WHAT: bracket ko zero set karo. ${\cos }^{2}i=1/5\Rightarrow \cos i=\pm 1/\sqrt{5}=\pm 0.4472$. Root 1 (positive cosine, prograde): $i=\arccos (+0.4472)\approx 63.43^{\circ }$. Root 2 (negative cosine, retrograde): $i=\arccos (-0.4472)\approx 116.57^{\circ }$. WHY this tool ($\arccos$): hume cosine pata hai aur angle wapas chahiye — $\arccos$ exactly "kaunse angle ka cosine yeh hai?" wala undo-button hai. Kyunki equation mein sirf ${\cos }^{2}i$ hai, dono ek prograde tilt aur uska retrograde mirror ($180^{\circ }-63.43^{\circ }=116.57^{\circ }$) isko satisfy karte hain. WHAT IT MEANS: dono inclinations perigee ko freeze karti hain — physics genuinely dono allow karti hai. Real Molniya orbits prograde $63.43^{\circ }$ choose karte hain kyunki launch cheaper hota hai (Earth ke eastward spin ka faayda milta hai) aur apogee northern hemisphere ke upar park hota hai; retrograde $116.57^{\circ }$ twin ke liye extra launch energy chahiye aur koi benefit nahin. Yeh $J_2$ critical inclination hai jo ellipse ki orientation ko freeze karti hai.

WHAT: har factor ka sign track karo. $J_2>0$, $(R_{\oplus }/p{)}^2>0$ (ek real length ka square), aur $n=\sqrt{\mu /a^3}>0$. Isliye overall prefactor $-\frac{3}{2}{J}_2(\cdots )n$ negative hai. Humein $\dot{\Omega }>0$ (positive/eastward) chahiye. Ek negative prefactor times $\cos i$ positive aana chahiye, isliye $\cos i$ negative hona chahiye. WHY it matters: $\cos i<0$ ka matlab hai $i>90^{\circ }$ — ek retrograde orbit (typically ~98°). Woh "thoda backwards, near-polar" tilt SSO ka signature hai. Dhyan do ki point Sun angle hai, pole ke upar fly karna nahin. Answer: retrograde, $i>90^{\circ }$.

WHAT: har $a$ (metres mein) ko $\epsilon =-\mu /(2a)$ mein daalo. LEO: $a=6.771\times 10^6$ m ⇒ $\epsilon =-\frac{3.986\times 10^{14}}{2(6.771\times 10^6)}=-2.943\times 10^7\text{J/kg}$. GEO: $a=4.2164\times 10^7$ m ⇒ $\epsilon =-\frac{3.986\times 10^{14}}{2(4.2164\times 10^7)}=-4.727\times 10^6\text{J/kg}$. WHY GEO costs more: GEO ka $\epsilon$ kam negative hai (zero ke zyada paas), yaani zyada energy. Satellite ko LEO se GEO tak raise karne ka matlab energy add karna hai. Actual burns ke liye Hohmann Transfer and Delta-v Budgets dekho. WHY negative at all: ek bound orbit ki total energy escape threshold (jo $\epsilon =0$ hai) se neeche hoti hai. Negative ka sirf matlab hai "gravitationally trapped." Answer: GEO zyada energy maangta hai; energy negative hai kyunki orbit bound hai.

WHAT (step 1 — period): 15 orbits ek sidereal day mein fit hote hain, isliye $T=\frac{86164}{15}=5744.3\text{s}.$ WHY 15 into a sidereal day: ground track tab repeat hota hai jab satellite ki orbits aur Earth ka spin star frame mein re-align ho jaate hain — woh alignment period sidereal day hai. WHAT (step 2 — radius): Kepler ko invert karo. $\mu T^2=3.986\times 10^{14}\times (5744.3{)}^2=3.986\times 10^{14}\times 3.2997\times 10^7=1.3153\times 10^{22}$. $/4{\pi }^2=3.3318\times 10^{20}$; cube root $a\approx 6.936\times 10^6$ m $=6936$ km. WHAT (step 3 — altitude): $h=6936-6371\approx 565\text{km}$. WHY it lands in LEO: 565 km, 160–2000 km ke andar hai, exactly wahan jahan imaging satellites rehte hain — resolution ke liye kaafi deep, worst drag se bachne ke liye kaafi high. Satellite Ground Tracks bhi dekho.

WHY vis-viva: orbit elliptical hai, isliye speed position ke saath change hoti hai — plain $\sqrt{\mu /r}$ (sirf circles ke liye) ab kaam nahin karta. Vis-viva kisi bhi radius par speed deta hai. Metres mein kaam karo. Perigee: $\frac{2}{r_p}-\frac{1}{a}=\frac{2}{6.971\times 10^6}-\frac{1}{2.656\times 10^7}=2.869\times 10^{-7}-3.765\times 10^{-8}=2.493\times 10^{-7}$. $v_p=\sqrt{3.986\times 10^{14}\times 2.493\times 10^{-7}}=\sqrt{9.936\times 10^7}\approx 9968\text{m/s}\approx 9.97\text{km/s}.$ Apogee: $\frac{2}{r_a}-\frac{1}{a}=\frac{2}{4.6149\times 10^7}-3.765\times 10^{-8}=4.334\times 10^{-8}-3.765\times 10^{-8}=5.69\times 10^{-9}$. $v_a=\sqrt{3.986\times 10^{14}\times 5.69\times 10^{-9}}=\sqrt{2.268\times 10^6}\approx 1506\text{m/s}\approx 1.51\text{km/s}.$ WHAT IT MEANS: $v_p/v_a\approx 6.6$. Satellite perigee se screaming speed se guzarta hai aur apogee par dawdle karta hai — woh northern hemisphere ke upar dawdling hi poora Molniya ka point hai, aur yeh Kepler ke 2nd law se match karta hai (equal time mein equal areas sweep ⇒ door par slow). Sanity: $r_p{v}_p=6.971\times 10^6\times 9968=6.95\times 10^{10}$ aur $r_a{v}_a=4.6149\times 10^7\times 1506=6.95\times 10^{10}$ ✓ angular momentum conserved.