Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya: Worked examples

3.2.37 · D3 · Physics › Orbital Mechanics & Astrodynamics › Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

Throughout: $μ = G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ aur $R_{\oplus} = 6371$ km. Neeche har symbol parent topic mein aur Kepler's Laws of Planetary Motion, Two-Body Problem and the Vis-Viva Equation, J2 Perturbation and Nodal Precession mein build kiya gaya tha.

The scenario matrix

Calculate karne se pehle, poori space of things dekho jo ek problem mein ho sakti hain. Har row ek case class hai; aakhri column us worked example ka naam deta hai jo uspe land karta hai.

#	Case class	Yahan kya tricky hai	Example
A	Low & circular (LEO) — chhota $r$	badi speed, chhota period	Ex 1
B	High & circular (GEO) — bada $r$	period pehle fixed, backwards solve karo	Ex 2
C	Sign / direction of inclination ( $cos i < 0$ )	retrograde $i > 9 0^{\circ}$ , SSO	Ex 3
D	Eccentric orbit, both ends (Molniya)	apogee vs perigee, do radii, ek $a$	Ex 4
E	Energy comparison across altitudes	kaun zyada $Δ v$ maangta hai? $E$ ka sign	Ex 5
F	Limiting / degenerate input ( $e \to 0$ , $e \to 1$ , $r \to R_{\oplus}$ )	formulas edges par kya karte hain	Ex 6
G	Real-world word problem	English → sahi tool mein translate karo	Ex 7
H	Exam twist (sidereal vs solar, wrong-frame trap)	deliberate trap	Ex 8

Is matrix ko ek checklist ki tarah padho: Ex 8 tak har cell fill ho jaati hai.

Case A — Low & circular

Worked example Ex 1 (Cell A):

$h = 550$ km par Starlink shell Orbital speed aur period nikalo.

Forecast: Padhne se pehle guess karo — kya speed $7 km/s$ se upar hai ya neeche? Kya period 90 min ke kareeb hai ya 100 min ke?

Step 1 — Orbital radius build karo. $r = R_{\oplus} + h = 6371 + 550 = 6921 km = 6.921 \times 1 0^{6} m$ . Yeh step kyun? Har formula Earth ke centre se distance use karta hai, surface ke upar ki altitude nahi. $R_{\oplus}$ bhoolna #1 slip hai.

Step 2 — Speed with $v = μ / r$ . $v = \frac{3.986 \times 1 0 ^{14}}{6.921 \times 1 0 ^{6}} = 5.759 \times 1 0^{7} \approx 7589 m/s \approx 7.59 km/s .$ Yeh step kyun? Gravity ek circle ke liye exact centripetal pull supply karta hai, jisne $v = μ / r$ force kiya. Satellite ka koi mass nahi aata — ek pankh aur ek bus orbit share karte hain.

Step 3 — Period with $T = 2 π r^{3} / μ$ . $T = 2 π \frac{( 6.921 \times 1 0 ^{6} ) ^{3}}{3.986 \times 1 0 ^{14}} \approx 5731 s \approx 95.5 min .$ Yeh step kyun? Ek lap = circumference $2 π r$ jo speed $v$ par travel ki; algebra use karke Kepler's 3rd milti hai.

Verify: Cross-check $T = 2 π r / v = 2 π (6.921 \times 1 0^{6}) /7589 \approx 5731$ s. ✓ Do independent routes agree karte hain. Units: $(m^{3} / s^{2}) / m = m/s$ . ✓ Speed $7.66 km/s$ (ISS at 420 km) se thodi neeche baith ti hai — zyada oocha orbit, thoda slower, jaisa hona chahiye.

Case B — High & circular (solve backwards)

Worked example Ex 2 (Cell B): Galileo-style MEO satellite, period 14 h

Ek navigation satellite ka period $T = 14 h = 50400 s$ hai. Uski altitude aur speed nikalo.

Forecast: GPS se zyada lamba period (~12 h) — kya uski altitude GPS ki 20 200 km se upar hogi ya neeche?

Step 1 — $r$ paane ke liye Kepler invert karo. $r = (\frac{μ T ^{2}}{4 π ^{2}})^{1/3} = (\frac{3.986 \times 1 0 ^{14} \cdot ( 50400 ) ^{2}}{39.478})^{1/3} .$ Yeh step kyun? Yahan period given hai, altitude nahi. Cell A mein altitude pehle aaya tha; yahan hum wahi law ulta chalate hain. Cube karna phir cube-root lena exactly $T^{2} \propto r^{3}$ ka undo hai.

Step 2 — Calculate karo. Numerator $= 3.986 \times 1 0^{14} \times 2.540 \times 1 0^{9} = 1.0125 \times 1 0^{24}$ ; $39.478$ se divide karo $\Rightarrow 2.565 \times 1 0^{22}$ ; cube root $\Rightarrow r \approx 2.948 \times 1 0^{7} m = 29480 km$ . Yeh step kyun? Hum arithmetic usi order mein karte hain jisme formula group karta hai — period square karo, $μ /4 π^{2}$ se scale karo, phir cube root lo — kyunki cube root aakhir mein lene se har intermediate number sane size mein rehta hai aur final step par units ( $m^{3} \to m$ ) sanity-check karna aasaan hota hai.

Step 3 — Altitude aur speed. $h = 29480 - 6371 = 23109 km$ → firmly MEO. $v = μ / r = 3.986 \times 1 0^{14} /2.948 \times 1 0^{7} \approx 3.68 km/s$ .

Verify: $r$ ko aage feed karo: $T = 2 π r^{3} / μ \approx 50400$ s. ✓ GPS altitude se upar jaisa forecast kiya (lamba period ⇒ bada orbit). Speed 3.68 km/s < LEO ka 7.6 — zyada oocha = slower, confirmed.

Case C — Sign of inclination (retrograde)

Definition Do orbit-plane words jo pehle chahiye

$Ω$ (right ascension of the ascending node, RAAN) woh compass-direction hai, equator plane mein measure ki gayi, jahan orbit equator ko northward jaate hue cross karti hai. Orbit ki tilted hoop ko equatorial disc mein ghuste hue picture karo: $Ω$ woh jagah hai jahan woh stab-point baith ta hai. Yeh orbit plane ki Earth ke axis ke around swivel fix karta hai.
$\dot{Ω}$ (padho " $Ω$ -dot") uski rate of change per day hai — woh crossing point equator ke around kitni tezi se drift karta hai. Sign convention: $\dot{Ω} > 0$ matlab node eastward drift karta hai (Earth ke spin jaise sense mein); $\dot{Ω} < 0$ matlab westward. Sun-synchronism ke liye hum chahte hain ki plane Earth ke yearly rate se eastward swing kare, toh hum $\dot{Ω} > 0$ demand karte hain.
$p$ (semi-latus rectum) orbit ka "width scale" hai: $p = a (1 - e^{2})$ . Near-circle ke liye $e \approx 0$ , toh $p \approx a$ . Yeh $J_{2}$ formula mein appear karta hai kyunki bulge ka tug orbit ki actual shape par depend karta hai, sirf uski size par nahi.

Worked example Ex 3 (Cell C): 700-km orbit ko Sun-synchronous banane ke liye kaunsa inclination chahiye?

$\dot{Ω} = - \frac{3}{2} J_{2} (\frac{R _{\oplus}}{p})^{2} n cos i$ use karo. Near-circular orbit ke liye $e \approx 0$ toh $p \approx a = R_{\oplus} + h$ . $\dot{Ω} = + 0.985 6^{\circ} / day$ (eastward, Sun ko track karne ke liye) require karo.

Forecast: Hume $\dot{Ω}$ positive chahiye. Formula ke leading minus sign ko dekho — kya $cos i$ positive aayega ya negative? Negative $cos i$ angle ke baare mein kya kehta hai?

Step 1 — Numbers set up karo. $a = 6371 + 700 = 7071 km = 7.071 \times 1 0^{6}$ m, aur kyunki $e \approx 0$ hum $p \approx a$ lete hain. Mean motion $n = μ / a^{3}$ : $n = \frac{3.986 \times 1 0 ^{14}}{( 7.071 \times 1 0 ^{6} ) ^{3}} = 1.128 \times 1 0^{- 6} = 1.062 \times 1 0^{- 3} rad/s .$ Yeh step kyun? $n$ orbit ki angular rate hai; precession is scale karti hai ki satellite kitni tezi se lap karta hai, toh hume yeh pehle chahiye.

Step 2 — Target rate ko rad/s mein convert karo. $0.985 6^{\circ} / day = 0.9856 \cdot \frac{π}{180} \div 86400 = 1.991 \times 1 0^{- 7} rad/s$ . Yeh step kyun? Formula ke har term SI radians aur seconds mein hai; silently degrees/day mix karna nonsense dega.

Step 3 — $cos i$ solve karo. $cos i = \frac{Ω ˙}{- \frac{3}{2} J _{2} ( R _{\oplus} / p ) ^{2} n} .$ Denominator (sign ke bina): $\frac{3}{2} (1.0826 \times 1 0^{- 3}) (6371/7071)^{2} (1.062 \times 1 0^{- 3}) = 1.400 \times 1 0^{- 6}$ . Minus ke saath: $cos i = \frac{1.991 \times 1 0 ^{- 7}}{- 1.400 \times 1 0 ^{- 6}} = - 0.1422$ . $i = arccos (- 0.1422) \approx 98. 2^{\circ} .$ Yeh step kyun? Required $\dot{Ω}$ positive hai lekin physics mein built-in minus hai, toh $cos i$ negative hona hi chahiye. Negative cosine matlab $i > 9 0^{\circ}$ — orbit straight-up-over-the-pole se aage tilt hoti hai ek retrograde (thoda backwards) path mein. Woh negative sign hi poora SSO signature hai.

Verify: $i = 98.1 8^{\circ}$ , $cos i = - 0.1422$ back plug karo: $\dot{Ω} = - 1.400 \times 1 0^{- 6} \times (- 0.1422) = 1.991 \times 1 0^{- 7}$ rad/s $= 0.985 6^{\circ} / day$ . ✓ Aur $9 8^{\circ}$ parent note mein quoted real SSO band se match karta hai. J2 Perturbation and Nodal Precession dekho.

Case D — Eccentric orbit, dono ends

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

Upar ka figure derivation hai: Earth green focus (origin) par baitha hai, ellipse ke centre par nahi. Yellow perigee point nearest end hai, red apogee point farthest, aur white dashed line poora long axis hai jiska half-length $a$ hai. Dekho ki Earth ka focus ellipse ke centre se kitna offset hai — woh offset exactly wajah hai ki do radii itni unequal hain. Hum Step 1 mein seedha is picture se numbers padhte hain.

Worked example Ex 4 (Cell D): Molniya apogee & perigee speeds

Molniya ka period = half a sidereal day, toh $a = 26560$ km, aur perigee altitude $h_{p} = 600$ km. Dono radii aur apogee aur perigee par speeds nikalo.

Forecast: Kepler's 2nd law (equal areas) kehta hai satellite kahaan sabse slow hoga? Kaunsa end "loiter" karega? Calculate karne se pehle speed ratio guess karo.

Step 1 — Perigee radius, phir ellipse identity se apogee. $r_{p} = 6371 + 600 = 6971$ km. Figure mein perigee (yellow) aur apogee (red) white dashed long axis ke do ends par baithe hain; poora axis $2 a$ length ka hai, toh do extreme radii, green focus se measure ki gayi, mila kar usme add honi chahiye: $2 a = r_{p} + r_{a}$ . $r_{a} = 2 a - r_{p} = 2 (26560) - 6971 = 46149 km .$ Yeh step kyun? Orbit ki size akele $a$ se fix hoti hai; $r_{p}$ aur $r_{a}$ bas woh do ends hain jo woh long axis share karte hain. Yeh ek geometry fact hai jo tum figure mein literally dekh sakte ho, extra physics nahi.

Step 2 — Vis-viva se speeds. Radius $r$ par kisi bhi point ke liye, energy conservation deta hai (dekho Two-Body Problem and the Vis-Viva Equation): $v = μ (\frac{2}{r} - \frac{1}{a}) .$ Perigee par ( $r_{p} = 6.971 \times 1 0^{6}$ m, $a = 2.656 \times 1 0^{7}$ m): $v_{p} = 3.986 \times 1 0^{14} (\frac{2}{6.971 \times 1 0 ^{6}} - \frac{1}{2.656 \times 1 0 ^{7}}) \approx 9.85 km/s .$ Apogee par ( $r_{a} = 4.6149 \times 1 0^{7}$ m): $v_{a} = 3.986 \times 1 0^{14} (\frac{2}{4.6149 \times 1 0 ^{7}} - \frac{1}{2.656 \times 1 0 ^{7}}) \approx 1.49 km/s .$ Yeh step kyun? Ellipse par speed constant nahi hoti, toh $v = μ / r$ (sirf circles) use nahi ho sakta. Vis-viva shape ko account karne ke liye $1/ a$ term subtract karta hai — eccentric case ke liye sahi tool.

Verify (Kepler's 2nd law): areal speed constant hai, toh $r_{p} v_{p}$ ko $r_{a} v_{a}$ equal karna chahiye: $6.971 \times 1 0^{6} \times 9850 \approx 6.87 \times 1 0^{10}$ ; $4.6149 \times 1 0^{7} \times 1489 \approx 6.87 \times 1 0^{10}$ . ✓ Match. Apogee speed perigee ki ~ $1/6.6$ hai — satellite north ke upar (figure ka red end) crawl karta hai, exactly woh loitering jo parent note ne promise ki thi.

Case E — Energy comparison

Worked example Ex 5 (Cell E): Kya LEO→GEO raise karne mein energy lagti hai? Per kg kitni?

Specific orbital energy (energy per kg) of a circular orbit $ε = - \frac{μ}{2 r}$ hai. $r_{1} = 6791$ km (ISS, 420 km) ko $r_{2} = 42164$ km (GEO) se compare karo.

Forecast: Dono energies negative hain (bound orbits). Kaun kam negative hai — yaani zero ke kareeb, yaani zyada energy? Predict karo kaun orbit "richer" hai.

Step 1 — Har specific energy compute karo. $ε_{1} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 6.791 \times 1 0 ^{6} )} = - 2.934 \times 1 0^{7} J/kg,$ $ε_{2} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 4.2164 \times 1 0 ^{7} )} = - 4.727 \times 1 0^{6} J/kg .$ Yeh step kyun? Energy "kitna mushkil hai pahuunchna" fix karta hai. Signed formula use karne se sign truth bani rehti hai: bound orbits negative hain, aur zyada oocha orbit = kam negative = zyada energy.

Step 2 — Jo energy add karni hai. $Δ ε = ε_{2} - ε_{1} = (- 4.727 \times 1 0^{6}) - (- 2.934 \times 1 0^{7}) = + 2.461 \times 1 0^{7} J/kg .$ Yeh step kyun? Change positive hai, toh climb karne ke liye energy spend karni padegi — parent ki "high = expensive to reach" confirm karta hai.

Verify: GEO energy LEO se kam negative hai ( $- 4.7 \times 1 0^{6} > - 2.9 \times 1 0^{7}$ ), toh GEO higher-energy orbit hai — forecast se match. Order of magnitude: ~ $25$ MJ/kg extra, yaani wajah hai ki GEO ka Hohmann transfer hefty $Δ v$ budget carry karta hai. Hohmann Transfer and Delta-v Budgets dekho.

Case F — Limiting / degenerate inputs

Numbers se pehle yeh figure padho: horizontal axis orbit radius hai, vertical axis speed. Blue curve circular speed $μ / r$ hai aur red curve escape speed $2 μ / r$ ; dono downward slope karti hain — bahar farther hamesha slower hota hai, koi exception nahi. Teen marked dots woh teen edge cases hain jo hum neeche test karte hain: yellow dot lowest possible circular orbit hai (surface grazing), green dot usi radius par escape speed hai, aur white dot far right par GEO hai. Notice karo ki red curve hamesha blue se exactly $2$ upar hai — woh constant gap Case F3 hai.

Worked example Ex 6 (Cell F): Formulas edges par kya karte hain

Teen extreme inputs test karo. Koi naye numbers memorize nahi karne — bas check karo tools sahi behave karte hain.

Forecast: Jaise eccentricity $e \to 0$ kya vis-viva $v = μ / r$ mein collapse hoga? Jaise $r \to R_{\oplus}$ (surface skim) kya speed blow up karegi ya finite value par settle karegi?

Case F1 — $e \to 0$ (ellipse circle ban jaata hai). Circle ke liye $r = a$ , toh vis-viva deta hai $v = μ (\frac{2}{a} - \frac{1}{a}) = \frac{μ}{a} .$ Yeh step kyun? Eccentric formula ko circular mein reduce hona chahiye, warna unme se ek galat hai. Hota hai — $2/ r$ aur $1/ a$ terms exactly $μ / r$ chodh jaate hain. Dono tools consistent hain. (Yahi wajah hai ki figure mein blue aur red sirf do hi curves hain — har ellipse unke beech rehti hai.)

Case F2 — $r \to R_{\oplus}$ (grazing orbit, air ignore karte hue). $r = 6.371 \times 1 0^{6}$ m set karo — figure ke far left mein yellow dot: $v = 3.986 \times 1 0^{14} /6.371 \times 1 0^{6} \approx 7.91 km/s .$ Yeh step kyun? Yeh maximum possible circular speed hai — blue curve yahan peak karti hai aur koi orbit neeche nahi ho sakta. Yeh finite hai, infinite nahi: formula ~7.9 km/s par cap karta hai (first cosmic velocity). Real satellites yahan nahi baith sakte kyunki Atmospheric Drag and Orbital Decay unhe minutes mein mita dega.

Case F3 — $e \to 1$ (parabolic escape limit). $e = 1$ par, $a \to \infty$ , toh $ε = - μ /2 a \to 0$ aur vis-viva → $v = 2 μ / r$ (escape speed) — red curve. $r = R_{\oplus}$ par (green dot): $v_{esc} = \frac{2 μ}{R _{\oplus}} = \frac{2 ( 3.986 \times 1 0 ^{14} )}{6.371 \times 1 0 ^{6}} = 2 \times 7.91 \approx 11.19 km/s .$ Yeh step kyun? Eccentricity knob, apni limit tak push kiya ( $a \to \infty$ se $1/ a$ term vanish hota hai), tumhe free mein escape-velocity formula de deta hai — dikhata hai ki ellipse family smoothly ek open (unbound) trajectory ban jaati hai. Figure mein $2$ vertical gap exactly yahi limit hai: escape hamesha usi radius par circular se $2$ factor upar hoti hai.

Verify: F1 algebraically $μ / a$ reduce hoti hai (checked). F2: $7.91$ km/s textbook first cosmic velocity hai. ✓ F3: $v_{esc} / v_{c i r c} = 2 \approx 1.414$ , aur $7.91 \times 1.414 = 11.19$ km/s = known Earth escape speed. ✓ Teeno edges sahi behave karti hain — koi blow-up nahi, koi contradiction nahi.

Case G — Real-world word problem

Worked example Ex 7 (Cell G): "Ek weather agency chahti hai ki ek satellite hamesha ek hi equatorial city ke upar hover kare." Kaun sa orbit, altitude, aur speed?

Forecast: Kaunsi named family? Period kya equal hona chahiye? Sidereal ya solar day?

Step 1 — English ko constraints mein translate karo. "Ek hi spot ke upar hover karo" ⇒ sky mein motionless dikhna ⇒ geostationary: circular ( $e = 0$ ), equatorial ( $i = 0$ ), aur period = ek sidereal day $= 86164$ s. Yeh step kyun? Word problems phrasing mein chupi numeric requirement nikaalne se jeete hain. "Hamesha same spot" code hai period-match-Earth-rotation ke liye.

Step 2 — Altitude ke liye Kepler invert karo. $r = (\frac{3.986 \times 1 0 ^{14} \cdot ( 86164 ) ^{2}}{39.478})^{1/3} \approx 4.216 \times 1 0^{7} m = 42164 km .$ Yeh step kyun? Same inversion tool jaisa Ex 2 mein, ab us sidereal-day period se driven jo humne words se nikala. $h = 42164 - 6371 = 35793 km$ (standard precision tak ≈35 786 km).

Step 3 — Speed. $v = μ / r = 3.986 \times 1 0^{14} /4.216 \times 1 0^{7} \approx 3.07 km/s$ . Yeh step kyun? Jab orbit circular hai aur $r$ pata hai, circular-speed tool seedha cruise speed deta hai.

Verify: $T = 2 π r^{3} / μ$ is $r$ ke saath $86164$ s return karta hai. ✓ 3.07 km/s parent ke quoted GEO speed se match karta hai. Answer: GEO, ~35 800 km, ~3.07 km/s.

Case H — Exam twist (deliberate trap)

Worked example Ex 8 (Cell H): "GEO ke liye 24-hour day use karo." Kitna galat hai yeh?

Ek student GEO compute karta hai solar day $86400$ s use karke sidereal $86164$ s ki jagah. Unhe jo altitude milti hai aur woh correct value se kitni dur hai, nikalo.

Forecast: Lamba assumed period ⇒ bada ya chhota radius? "GEO" altitude roughly kitne km off hogi?

Step 1 — Wrong (solar) period se radius. $r_{wrong} = (\frac{μ T _{solar}^{2}}{4 π ^{2}})^{1/3} = (\frac{3.986 \times 1 0 ^{14} \cdot ( 86400 ) ^{2}}{39.478})^{1/3} \approx 4.2241 \times 1 0^{7} m .$ Yeh step kyun? Yeh wahi Kepler inversion hai jo Ex 2 aur Ex 7 mein use hua, lekin trap value $86400$ s feed ki gayi hai, taaki hum measure kar sakein ki wrong frame kitna damage karta hai.

Step 2 — Correct (sidereal) period se radius. $r_{correct} = (\frac{3.986 \times 1 0 ^{14} \cdot ( 86164 ) ^{2}}{39.478})^{1/3} \approx 4.2164 \times 1 0^{7} m .$ Yeh step kyun? Hum already jaante hain yeh true GEO radius hai (Ex 7); dono calculate karne se hum subtract karke seedha gap dekh sakte hain.

Step 3 — Error. $Δ r = r_{wrong} - r_{correct} \approx (4.2241 - 4.2164) \times 1 0^{7} = 0.0077 \times 1 0^{7} m \approx 77 km too high .$ Yeh step kyun? $T^{2} \propto r^{3}$ matlab $0.27%$ period error sirf $\frac{2}{3} \times 0.27% \approx 0.18%$ radius error mein bada hota hai — lekin $0.18%$ of 42 000 km phir bhi ~77 km hai, jo matter karne ke liye kaafi hai.

Verify: $Δ r \approx 77$ km — percentage mein chhota lekin mission-fatal: 77 km zyada oocha park kiya satellite ka period ek sidereal day se lamba hai, toh uska "fixed" spot sky mein eastward drift karta rehta hai aur woh ab geostationary nahi raha. Fix (parent note se): hamesha sidereal day $86164$ s use karo, kyunki GEO ko Earth ke rotation ke saath stars ke relative sync karna hota hai, Sun ke nahi.

Recall Self-test — answers cover karo

Kaun sa cell $μ / r$ ki jagah vis-viva maangta hai? ::: Eccentric case D (aur F3) — ellipse par speed $r$ ke saath vary karti hai. SSO $i > 9 0^{\circ}$ kyun force karta hai? ::: Precession formula mein leading minus hai, toh positive required $\dot{Ω}$ ke liye $cos i < 0$ chahiye. $Ω$ ka matlab kya hai aur $\dot{Ω} > 0$ kya signify karta hai? ::: $Ω$ orbit plane ki equator-crossing (ascending-node) direction hai; $\dot{Ω} > 0$ matlab woh node eastward drift karta hai. $J_{2}$ kya represent karta hai? ::: Earth ke gravity field mein equatorial-bulge term ki strength, ~ $1.08 \times 1 0^{- 3}$ . Surface par grazing circular speed aur escape speed? ::: 7.91 km/s aur 11.19 km/s (ratio $2$ ). GEO ke liye kaun sa day length, aur galat use karne ka penalty? ::: Sidereal 86 164 s; 86 400 s use karne se altitude ~77 km zyada ho jaati hai.