Visual walkthrough — Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

3.2.37 · D2 · Physics › Orbital Mechanics & Astrodynamics › Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

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Step 1 — "In orbit" ka matlab kya hota hai: hamesha falling

KYA. Ek satellite "wahan upar nahi hai jahan gravity nahi hai". Gravity wahan bhi almost utni hi strong hai jitni ground par. Ek satellite ek aisa object hai jo itne tez sideways throw kiya gaya hai ki jaise woh girta hai, ground neeche curve ho jaati hai bilkul utni hi tez se. Woh Earth ko miss karta rehta hai. Yahi orbit hai.

YAHAN SE KYUN shuru karein. Neeche diya har equation aslmein sirf yeh hai: "girna exactly curving away se match karta hai". Agar aap is picture par believe karte hain, toh algebra sirf bookkeeping hai.

PICTURE. Red ball seedha neeche drop ki jaati hai — land karti hai. Ise sideways zyada se zyada tez throw karo (black arcs) aur yeh dur jaake land karti hai. Itna tez throw karo ki fall curve kabhi ground se na mile: woh circle karti hai.

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Step 2 — Drawing ko naam dena: $r$, $v$, aur $m$

KYA. Koi bhi formula se pehle, hum drawing ko label karte hain. Earth ke centre ko ek dot ke roop mein draw karo. Satellite ek point hai jo circle ke around ja raha hai.

• $r$ = Earth ke centre se satellite tak ki distance (ground se nahi!). Circle ka radius.
• $v$ = satellite ki circle ke saath speed (kitne metres of arc per second).
• $m$ = satellite ka mass (woh kitni cheez se bana hai).
• $M$ = Earth ka mass. $G$ = gravitational constant, woh fixed number jo universe mein kahin bhi gravity kitni strong hai yeh set karta hai.

YEH CHAAR KYUN. Gravity dono masses aur distance par depend karti hai. Circle mein motion speed aur radius par depend karta hai. Yeh exactly woh knobs hain.

PICTURE. Red satellite, centre se $r$ ka radius arrow, aur $v$ ka velocity arrow tangent draw kiya gaya (circle ko touch karta hai, travel ki direction mein point karta hai).

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Step 3 — Do forces, ek balance

KYA. Do physical facts yahan milte hain.

Fact A — gravity andar ki taraf kheenchti hai. Newton ka law of gravitation kehta hai ki Earth aur satellite ke beech ka pull hai $F_{\text{grav}}=\frac{GMm}{r^2}$

• Upar $GMm$: bada Earth ($M$), bada satellite ($m$), ya zyada strong gravity ($G$) → zyada strong pull.
• Neeche $r^2$: distance double karo aur pull quarter ho jaata hai. Yeh famous inverse-square law hai — pull ek sphere ke upar spread hota hai jiska area $r^2$ ke saath badhta hai.

Fact B — circle mein jaane ke liye ek inward force chahiye. Jo bhi cheez circle mein move karti hai woh constantly direction badal rahi hai, aur direction badalna ek acceleration hai jo centre ki taraf point karta hai. Ise supply karne ke liye jo force chahiye woh hai $F_{\text{centripetal}}=\frac{m{v}^2}{r}$

• Upar $v^2$: tez matlab direction zyada tezi se whip karta hai, zyada force chahiye.
• Neeche $r$: tighter (smaller) circle aapko zyada tez turn karti hai, zyada force chahiye.

DONO KO EQUAL KYUN SET KAREIN. Step 1 mein humne kaha tha ki girna exactly curving se match karta hai. Force language mein yeh hai: gravity jo inward force supply karti hai woh exactly woh inward force hai jo circle demand karta hai. Koi leftover nahi, koi shortfall nahi. Toh $\underset{\text{what gravity gives}}{\underbrace{\frac{GMm}{r^2}}}=\underset{\text{what the circle needs}}{\underbrace{\frac{m{v}^2}{r}}}$

PICTURE. Ek red arrow (gravity, inward) exactly dashed black arrow (centripetal requirement, inward) ke upar baitha hai. Same length = balance.

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Step 4 — Speed ke liye solve karna: $v=\sqrt{\mu /r}$

KYA. Ab balance equation par pure algebra. Pehle $m$ cancel karo (woh dono sides par hai — satellite ka khud ka mass matter nahi karta, ek bowling ball aur ek pea bilkul same orbit karte hain): $\frac{GM}{r^2}=\frac{v^2}{r}$ Dono sides ko $r$ se multiply karo taaki $r$ ki ek power clear ho: $\frac{GM}{r}=v^2⟹v=\sqrt{\frac{GM}{r}}$

Hum $GM$ ko ek single symbol $\mu$ (the "standard gravitational parameter") mein bundle karte hain kyunki $G$ aur $M$ hamesha saath travel karte hain aur $\mu$ ko dono akele se kahin zyada precisely measure kiya jaata hai: ${\mu }_{\oplus }\approx 3.986\times 10^{14}{\text{m}}^3/{\text{s}}^2$.

$\boxed{v=\sqrt{\frac{\mu }{r}}}$

Square root KYUN, aur bada $r$ ek chota $v$ KYUN deta hai? $r$ root ke andar bottom par hai. $r$ badhao → fraction chota hota hai → uska root chota hota hai → $v$ chota hota hai. Physically: dur bahar, gravity kamzor hoti hai (Fact A mein woh $r^2$ denominator), toh sirf ek lazy, slow circle hi balance mein rehne ke liye chahiye. Yeh highway intuition "farther must be faster" ko kill kar deta hai.

PICTURE. $v$ ka $r$ ke against curve: ek steep drop jo flat hoti hai. Red dot LEO mark karta hai (fast), black dot GEO mark karta hai (slow).

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Step 5 — Speed se period tak: ek lap kitna lamba hota hai?

KYA. Period $T$ woh time hai jisme ek full trip around hoti hai. Ek trip circumference $2\pi r$ cover karti hai (circle ka perimeter) speed $v$ par. Time = distance ÷ speed: $T=\frac{2\pi r}{v}$

• $2\pi r$: circular path ki total length.
• $v$ se divide karna: metres of path ÷ metres per second = seconds.

Ab $v$ jo humne abhi nikala hai substitute karo, $v=\sqrt{\mu /r}$: $T=\frac{2\pi r}{\sqrt{\mu /r}}=2\pi r\sqrt{\frac{r}{\mu }}=2\pi \sqrt{\frac{r^3}{\mu }}$

Beech ka step: $\sqrt{\mu /r}$ se divide karna $\sqrt{r/\mu }$ se multiply karna hai; phir $r\cdot \sqrt{r}=\sqrt{r^2}\cdot \sqrt{r}=\sqrt{r^3}$.

$\boxed{T=2\pi \sqrt{\frac{r^3}{\mu }}}$

$r^3$ ki KYUN parwah karein. Dono sides square karo: $T^2=\frac{4{\pi }^2}{\mu }{r}^3$, yani $T^2\propto r^3$ — yeh exactly Kepler's 3rd law hai, jo humne ab ek rock on a string se build kiya hai. (Stretched ellipses ke liye $r$ ki jagah semi-major axis use karo; dekho Two-Body Problem and the Vis-Viva Equation.)

PICTURE. Period radius ke saath steeply climb karti hai; horizontal line ek sidereal day mark karti hai; jahaan yeh curve se milti hai woh GEO radius hai jo hum dhundh rahe hain.

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Step 6 — Trick: $T$ = ek sidereal day demand karo

KYA. "Geostationary" ka matlab hai satellite equator par ek spot ke upar still hang karta hua dikh raha hai. Yeh tabhi hoga jab woh exactly utna hi fast ghume jitna Earth spin karti hai. Toh hum $T$ ko Earth ke rotation period ke barabar set karte hain.

Lekin kaun sa rotation period? 24-hour solar day nahi. Yeh woh time hai jab tak Sun wapas overhead aata hai, aur yeh thoda lamba hai kyunki Earth apni Sun ke around orbit mein bhi creep karti hai, toh use Sun ko face karne ke liye thoda extra spin karna hota hai. Fixed stars ke relative actual spin sidereal day hai: $T_{\text{sid}}=86164\text{s}(\approx 23\text{ h }56\text{ m }4\text{ s})$

SIDEREAL KYUN. Satellite ko koi fark nahi padta Sun kahan hai; use Earth ki actual spin in space se match karna hai. $86164$ s ki jagah $86400$ s use karne par altitude ~50 km overshoot ho jaata hai.

PICTURE. Do clocks: ek solar day (Sun wapas overhead aata hai) ek sidereal day (ek fixed star wapas overhead aata hai) se thoda lamba hai, kyunki uss din Earth ki Sun ke around motion ki wajah se.

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Step 7 — Kepler ko invert karo aur number nikalo

KYA. Hamare paas $T$ hai aur hum $r$ chahte hain, toh period formula flip karo. Shuru karo $T=2\pi \sqrt{\frac{r^3}{\mu }}.$ Dono sides square karo: $T^2=4{\pi }^2\frac{r^3}{\mu }$. $r^3$ ke liye solve karo: $r^3=\frac{\mu T^2}{4{\pi }^2}$. Cube root lo: $\boxed{r={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}}$

$\mu =3.986\times 10^{14}$ aur $T=86164$ s plug in karo: $r={\left(\frac{3.986\times 10^{14}\cdot (86164{)}^2}{4{\pi }^2}\right)}^{1/3}\approx 4.216\times 10^7\text{m}=42164\text{km}.$

Finally ground ke upar altitude mein convert karne ke liye Earth ka radius subtract karo (Step 2 ki warning!): $h=r-R_{\oplus }=42164-6371\approx 35793\mathbf{km}(\approx 35786\text{standard precision tak}).$

Cube root KYUN. $T$ depend karta hai $r^3$ par; cube ko undo karne ke liye cube root lo. Yahi wajah hai ki exponent $1/3$ hai.

PICTURE. Orbits ki number line: LEO low, MEO middle mein (GPS), aur GEO par single red tick, $35786$ km, jahan period ek sidereal day hit karta hai.

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Step 8 — Edge aur degenerate cases (kabhi skip mat karo)

KYA & KYUN — har woh scenario walk karo jo reader hit kar sakta hai:

• $e=0,i=0$ (true GEO). Perfect circle in the equatorial plane. Ground track ek single point mein collapse ho jaata hai — satellite motionless hang karta hai. Yeh woh case hai jo humne solve kiya.
• $e=0$ lekin $i\ne 0$ (tilted geosynchronous). Same period, same $r$, lekin tilt isse har din north–south drift karta hai, sky mein figure-8 analemma trace karta hai. Same altitude, geostationary nahi. Dekho Satellite Ground Tracks.
• $i=0$ lekin $e\ne 0$ (eccentric geosynchronous). Same average $a$, lekin woh speed up aur slow down karta hai (Kepler's 2nd law), daily east–west slide karta hai — phir se ek figure-8, fixed point nahi.
• $r\to R_{\oplus }$ (degenerate low limit). Jaise $r$ Earth ki surface tak shrink hota hai, $T\to 2\pi \sqrt{R_{\oplus }^3/\mu }\approx 84$ min — kisi bhi circular orbit ki sabse fast speed, aur wahan $v$ sabse badi hai. Neeche tum planet ke andar ho: koi orbit nahi.
• $r\to \infty$ (degenerate high limit). $v\to 0$ aur $T\to \infty$: infinitely dur, infinitely slow, infinitely lamba. GEO physics mein koi special jagah nahi hai — woh sirf woh ek $r$ hai jahan $T$ ek din ke barabar hota hai.
• ~2000 km se neeche: atmospheric drag har pass mein energy chura leta hai; ek "circular" orbit slowly spiral down karti hai. Hamaara frictionless balance (Step 3) ek idealisation hai jo sirf sensible atmosphere ke upar hold karta hai.

PICTURE. Teen sky-tracks side by side: ek still red dot (true GEO), ek figure-8 (tilted), aur ek east–west smear (eccentric) — sab same altitude par.

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Ek-picture summary

Upar sab kuch, compressed: do forces balance karo → $v$ ke liye solve karo → $T$ mein convert karo → ek sidereal day demand karo → $r$ nikalane ke liye invert karo → $R_{\oplus }$ subtract karo.

Recall Feynman retelling — plain words mein wapas bolo

Ek satellite bas woh cheez hai jo itni tez sideways throw ki gayi hai ki woh ground miss karta rehta hai. Circle mein jaane ke liye centre ki taraf constant pull chahiye, aur gravity exactly wahi pull hai — toh maine "jo pull gravity deti hai" ko "jo pull circle needs" ke barabar set kiya. Satellite ka khud ka weight cancel ho jaata hai (ek pea aur ek boulder same orbit karte hain), baaki reh jaata hai speed = square-root of (Earth ka gravity number ÷ distance). Kyunki distance bottom par hai, higher orbits slower hoti hain, faster nahi. Ek lap circle ki perimeter ko us speed se divide karne mein lagta hai, jo tidy ho ke $T=2\pi \sqrt{r^3/\mu }$ ban jaata hai — Kepler's law, ek rock on a string se born. Ek spot ke upar still hang karne ke liye, main ek lap exactly ek Earth-spin mein laata hun — sidereal day of 86 164 s, 24-hour solar day nahi. Main period formula flip karta hun (cube root, kyunki time distance cubed par depend karta tha), 42 164 km ka radius milta hai, Earth ka khud ka radius subtract karo, aur magic altitude nikal aata hai: approximately 35 786 km. Aur agar orbit tilted ya stretched hai, same altitude lekin woh figure-8 trace karta hai instead of still khada rehne ke.

Recall Quick self-check

Bade $r$ ke liye $v$ smaller kyun hota hai? ::: $r$ $\sqrt{\mu /r}$ ke denominator ke andar hai, toh $r$ badhane se $v$ chota hota hai; dur bahar, gravity kamzor hai aur sirf ek slow circle hi use balance karne ke liye chahiye. GEO ke liye solar day ki jagah sidereal kyun? ::: Satellite ko Earth ki true spin relative to the stars se match karna hai ($86164$ s); solar day lamba hota hai kyunki Earth bhi Sun ke orbit mein hai. $R_{\oplus }$ subtract karna kya karta hai? ::: Centre-distance $r$ ko altitude-above-ground $h$ mein convert karta hai. Same altitude par agar $i\ne 0$ ho toh kya hota hai? ::: Phir bhi geosynchronous hai, lekin woh figure-8 analemma trace karta hai — geostationary nahi.