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

3.2.8 · D2 · Physics › Orbital Mechanics & Astrodynamics › Orbital elements (Keplerian) — semi-major axis a, eccentrici

Kisi bhi symbol se pehle, yahan puri cast hai jo hum milenge, seedhe shabdon mein:

Step 1 — Ek satellite, ek arrow, ek pull

KYA. Earth ke centre ko origin pe rakho. Satellite ko kahin draw karo, aur origin se satellite tak arrow $r$ draw karo. Iska length $r$ hai — satellite kitna dur hai.

KYU. Har orbital idea isi ek arrow ke baare mein ek statement hai — time ke saath iska length kitna hai aur kis direction mein point karta hai. Agar hum $r$ ka ek rule kisi angle ke function ke roop mein dhundh sakein, toh hamein orbit mil gayi.

PICTURE. Blue arrow $r$ hai. Gravity (red) hamesha origin ki taraf wapas point karti hai — yahi "central force" ka matlab hai: pull exactly $- r$ ke saath align hoti hai.

Gravitational acceleration hai $\ddot{r} = - \frac{μ}{r ^{3}} r .$

Ise term by term padhte hain, wahan se jahan har symbol baitha hai:

$\ddot{r}$ — velocity arrow kitni tez change ho raha hai (acceleration).
minus sign — pull ko andar ki taraf point karta hai, $r$ ke opposite.
$μ$ — pull ki strength.
$1/ r^{3}$ times $r$ — yeh equals karta hai $\frac{1}{r ^{2}} \cdot \frac{r}{r}$ : ek $1/ r^{2}$ strength along unit direction $r / r$ ke. Toh force distance ke inverse square ke saath weak hoti jaati hai.

Yahi poora two-body gravity hai. Baaki sab consequence hai.

Step 2 — Plane kabhi tilt nahi hota: angular momentum $h$

KYA. Arrow define karo $h = r \times v .$ Yeh " $\times$ " cross product hai: yeh ek aisa arrow produce karta hai jo $r$ aur $v$ dono ke perpendicular hota hai, jiska length un donon se bane parallelogram ka area ke barabar hota hai.

YEH TOOL KYU. Hum prove karna chahte hain ki motion ek flat plane mein rehti hai. Cross product exactly woh tool hai jo yeh sawaal answer karta hai: "in do arrows ke plane ke perpendicular kaunsa ek direction hai?" Agar woh perpendicular arrow kabhi move nahi karta, toh plane kabhi tilt nahi hota.

PICTURE. $r$ (blue) aur $v$ (green) ek plane mein lete hain; $h$ (yellow) isse seedha bahar, ek flagpole ki tarah khada hai.

Dekho yeh constant kaise rehta hai. Differentiate karo: $\dot{h} = \dot{r} \times v + r \times \dot{v} = = 0 v \times v + r \times (- \frac{μ}{r ^{3}} r) = 0.$

$v \times v = 0$ — koi bhi arrow apne aap ke saath cross karo toh zero milta hai.
$r \times r = 0$ — same reason, toh gravity term bhi vanish ho jaata hai.

Toh $h$ frozen hai. Iska constant length $h$ aur constant direction ka matlab hai:

orbit ek fixed plane mein rehti hai (woh plane jiske perpendicular $h$ hai),
aur $h = r v_{⊥}$ conserved hai (yeh $\vec h$ hai aur yeh hi Kepler ka second law hai).

Ab se hum usi single plane ke andar kaam karenge. 3D tilt angles $i, Ω$ bas yeh batate hain ki yeh plane space mein kaise baitha hai — ek alag kahani jo perifocal frame mein handle ki jaati hai.

Step 3 — Ek doosra frozen arrow closest approach ki taraf point karta hai: $e$

KYA. Eccentricity vector banao $e = \frac{v \times h}{μ} - \frac{r}{r} .$

KYU. Hamein pehle se ek constant arrow mil chuka hai, $h$ , jo plane batata hai. Ellipse ko us plane ke andar pin karne ke liye hume ek constant arrow chahiye jo plane mein ho, kisi landmark ki taraf point kare. Nikalta hai $e$ exactly wahi hai: woh time mein frozen hai aur focus se perigee (closest point) ki taraf aim karta hai. Iska length shape number $e$ ban jaayega.

PICTURE. Yellow arrow $e$ orbital plane mein baitha hai aur ellipse ke near end ki taraf point karta hai. $e$ se $r$ tak ka angle hamara true anomaly $ν$ hai.

$\frac{v \times h}{μ}$ — ek vector jo velocity aur frozen $h$ ko combine karta hai.
$\frac{r}{r}$ — woh unit arrow jo seedha satellite ki taraf bahar point karta hai.
Unka difference ek constant nikalta hai, aur iska length woh number hai jise hum $e$ kehte hain.

$e$ constant hai yeh ek chhoti (thodi grind wali) derivative check hai, Eccentricity vector note mein dikhaya gaya hai; yahan hum ise use karte hain.

Step 4 — $e$ ko $r$ mein dot karo: angle $ν$ enter karta hai

KYA. $e$ aur $r$ ka dot product lo.

YEH TOOL KYU. Dot product $a \cdot b = ∣ a ∣∣ b ∣ cos (angle between them)$ exactly woh tool hai jo do arrows mein se angle nikaalta hai. Hamein ek constant landmark $e$ aur moving arrow $r$ mila hai; inhe dot karna "satellite kahan hai" ko "usne perigee se kitna angle $ν$ sweep kiya hai" mein convert karta hai.

PICTURE. Fixed yellow $e$ aur moving blue $r$ ke beech ka shaded angle $ν$ hai.

Dot product do tareekon se calculate karo.

Geometric tarika (dot product ki definition se): $e \cdot r = e r cos ν .$

$e$ — $e$ ki length.
$r$ — $r$ ki length.
$cos ν$ — kyunki $ν$ unke beech ka angle hai.

Algebraic tarika (Step 3 se $e$ ki definition plug in karo aur vector identity $r \cdot (v \times h) = (r \times v) \cdot h = h \cdot h = h^{2}$ use karo): $e \cdot r = \frac{h ^{2}}{μ} - r .$

$h^{2} / μ$ — frozen $h$ se bana ek pure constant.
$- r$ — kyunki $\frac{r}{r} \cdot r = r$ .

Step 5 — Dono expressions ko equal karo: orbit equation appear hoti hai

KYA. Left sides identical hain, toh right sides match karne chahiye: $e r cos ν = \frac{h ^{2}}{μ} - r .$

KYU. Yeh akeli line pehle se hi orbit hai — yeh distance $r$ ko angle $ν$ se link karti hai. Hum bas ise theek karte hain.

$r$ ke liye solve karo: dono $r$ terms ko saath laao, $r + e r cos ν = \frac{h ^{2}}{μ} ⟹ r (1 + e cos ν) = \frac{h ^{2}}{μ},$ $r = \frac{h ^{2} / μ}{1 + e cos ν} .$

Upar ke constant ko semi-latus rectum naam do: $p \equiv \frac{h ^{2}}{μ}, so r = \frac{p}{1 + e c o s ν} .$

PICTURE. Yeh ek conic section hai. Dekho $r$ kaise sweep karta hai jab $ν$ ghoomta hai: yeh sabse chhota hota hai jab $cos ν = + 1$ (perigee) aur sabse bada jab $cos ν = - 1$ (apogee).

Term by term:

$p$ — "size scaffold": yeh literally $r$ hai jab $ν = 9 0^{\circ}$ (tab $cos ν = 0$ ).
$e cos ν$ — "squash": yeh denominator ko apogee ke paas badhata hai ( $r$ bada karta hai) aur perigee ke paas ghataata hai ( $r$ chhota karta hai).
$r$ — woh distance jo hum chahte the, ab ek angle ka clean function.

Step 6 — Perigee, apogee padhna, aur $a$ se connect karna

KYA. Do special angles plug in karo.

$ν = 0$ par ( $cos ν = + 1$ ): $r_{p} = \frac{p}{1 + e}$ — perigee, closest point. $ν = 18 0^{\circ}$ par ( $cos ν = - 1$ ): $r_{a} = \frac{p}{1 - e}$ — apogee, farthest point.

KYU. Yeh hume $e$ ka physical meaning dete hain aur hume $p$ ko zyada familiar semi-major axis $a$ se swap karne dete hain.

Dono apside distances ko add karke half karo ( $a$ ko unke average ke roop mein define karta hai yeh): $a = \frac{r _{p} + r _{a}}{2} = \frac{p}{1 - e ^{2}} ⟹ p = a (1 - e^{2}) .$

Wapas substitute karne se parent note ki headline form milti hai: $r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν} .$

PICTURE. $r_{p}$ (perigee, red) aur $r_{a}$ (apogee, green) marked ke saath ellipse, aur $a$ long axis ki half-length ke roop mein dikhaya gaya.

$e$ ab ek squash ratio ke roop mein padhta hai: $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}}$ .
$a$ average distance / long axis ki half hai. (Dekho Vis-viva equation ke liye ki $a$ energy kaise fix karta hai.)

Step 7 — $e$ ke har case: circle, ellipse, parabola, hyperbola

KYA. Single formula $r = \frac{p}{1 + e cos ν}$ chup-chaap saare conic shapes contain karta hai, bas $e$ change karke. Hume har ek check karna hoga taaki koi reader ko surprise na mile.

KYU. Denominator $1 + e cos ν$ hamesha positive ho sakta hai, ya zero reach kar sakta hai, ya negative bhi ho sakta hai — aur wahi shape switch karta hai.

PICTURE. Ek hi focus share karte hue chaar orbits, har ek $e$ ki value ke liye.

Cases walk karo:

Worked example

$e$ ke chaar regimes

$e = 0$ (circle). Denominator $= 1$ hamesha, toh $r = p$ constant. Koi near/far point nahi — perigee aur apogee coincide karte hain. $ν$ phir bhi $0 \to 36 0^{\circ}$ run karta hai lekin distance kabhi nahi badalti.
$0 < e < 1$ (ellipse). Denominator positive rehta hai (kyunki $∣ e cos ν ∣ < 1$ ). $r$ , $r_{p}$ aur $r_{a}$ ke beech oscillate karta hai. Closed orbit — parent note mein har satellite.
$e = 1$ (parabola). Jaise $ν \to 18 0^{\circ}$ , denominator $1 + cos ν \to 0$ , toh $r \to \infty$ . Path khul jaata hai: exactly zero speed bachne ke saath escape. Yahan $a \to \infty$ , toh hume $p = h^{2} / μ$ use karna hoga, na ki $a (1 - e^{2})$ — woh form break ho jaata hai (gives $0$ ).
$e > 1$ (hyperbola). Denominator ek finite angle $ν^{*}$ par zero hit karta hai jahan $cos ν^{*} = - 1/ e$ . Uske aaage satellite infinity ki taraf ud chuka hai; sirf $∣ ν ∣ < ν^{*}$ hi physical hai. Ek flyby.

Ek-picture summary

Upar ka sab kuch, compress karke: do frozen arrows ( $h$ plane se bahar, $e$ plane mein perigee ki taraf), moving arrow $r$ angle $ν$ par $e$ se, aur resulting conic jiska radius $r = p / (1 + e cos ν)$ follow karta hai.

Recall Feynman: poora walkthrough seedhe shabdon mein

Shuru karo kuch bhi nahi se bas ek satellite se jo seedha Earth ke centre ki taraf khich raha hai. Pehli trick: position arrow ko velocity arrow ke saath cross karo — isse ek flagpole arrow $h$ banta hai jo kabhi move nahi karta, jo prove karta hai ki poori orbit ek flat, non-tilting plane mein rehti hai. Doosri trick: velocity aur $h$ ko bilkul sahi recipe mein mix karo aur outward unit arrow subtract karo — tumhe ek aur frozen arrow milta hai, $e$ , jo hamesha closest approach ki taraf point karta hai. Ab clever part: us landmark arrow ko position arrow mein "dot" karo. Dot do alag tareekon se karna — ek baar geometrically (jo $cos ν$ ugalta hai) aur ek baar algebraically (jo ek constant ugalta hai) — aur unhe equal set karna hi orbit hai. Saaf karo aur tumhe milta hai $r = p / (1 + e cos ν)$ : distance ek angle ka simple function ke roop mein. Upar ka constant $p$ size hai; $e cos ν$ term squash hai. $e$ ko $0$ se upar dial karo aur wahi formula circle, ellipse, parabola, phir hyperbola ban jaata hai — tumhe kabhi nayi equation ki zaroorat nahi padi, bas ek naye number ki.

Recall Quick self-test

Kaunsa tool do arrows ko ek angle mein convert karta hai, aur hume Step 4 mein iske kyon zaroorat padi? ::: Dot product, kyunki $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ fixed $e$ aur moving $r$ ke beech angle $ν$ isolate karta hai. Orbit ek fixed plane mein kyun rehti hai? ::: Kyunki $h = r \times v$ ka zero time-derivative hai, toh iska direction frozen hai aur motion iske perpendicular rehti hai. $p$ physically kya hai? ::: Radius $r$ jab $ν = 9 0^{\circ}$ (semi-latus rectum), $h^{2} / μ$ ke barabar. Orbit equation ka kaunsa form $e = 1$ survive karta hai? ::: $r = p / (1 + e cos ν)$ ke saath $p = h^{2} / μ$ ; $a (1 - e^{2})$ form fail ho jaata hai kyunki $a \to \infty$ .

Related: Kepler's equation and mean anomaly ( $ν$ ko ek clock mein badalna), State vectors to orbital elements ( $r, v$ se ulta jaana), Orbital perturbations (J2) (jo slowly $e$ aur $h$ ko un-freeze karta hai).