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

3.2.8 · D1 · Physics › Orbital Mechanics & Astrodynamics › Orbital elements (Keplerian) — semi-major axis a, eccentrici

Is page pe assume kiya gaya hai ki aapne kuch nahi dekha. Hum har letter, har triangle, har arrow jo parent note use karta hai, us order mein build karte hain jo har ek ko pehle wale ke upar khara rehne deta hai. Upar se neeche padho.

1. Centre se doori: $r$

Ek dhage ka picture banao jo beech ke ek dot (Earth) se ek loop (satellite) ke dot tak jata ho. Us dhage ki lambai $r$ hai. Jaise satellite loop ke around ghoomta hai, dhaga lamba aur chota hota hai — isliye $r$ ek number hai jo time ke saath badalta hai.

Figure 1. Ek blue orbit loop jisme Earth yellow dot hai focus par. White dashed line woh dhaga hai jiska length $r$ hai — yeh pink satellite tak stretch karta hai aur satellite ke move karne ke saath badalta hai.

2. Speed: $v$ , aur hum ise square kyun karte hain

Parent note hamesha $v^{2}$ kyun likhta hai, $v$ nahi? Kyunki motion ki energy — kinetic energy — speed ke square ke saath badhti hai. Yeh squaring koi trick nahi hai; yeh physics se naturally nikalta hai.

3. Gravitational parameter: $μ$

4. Vectors: $r$ aur $v$ — chhote arrows

Plain letter $r$ ek single number hai (ek length). Arrow letter $r = (x, y, z)$ teen numbers hai — ek har axis ke liye. Unke beech ka relation: $r = ∣ r ∣ = x^{2} + y^{2} + z^{2} (arrow ki length) .$

Figure 2. Yellow $r$ arrow origin (Earth) se satellite tak point karta hai; uske blue dotted shadows $x$ aur $y$ components dikhate hain. Pink $v$ arrow, satellite par attached, travel ki direction mein point karta hai. Plain $r$ sirf yellow arrow ki length hai.

5. Ellipse aur uske do size numbers: $a$ aur $e$

Ek ellipse ek squashed circle hai. Koi bhi ellipse do numbers se completely describe hoti hai.

Figure 3. Ek blue ellipse jisme Earth (yellow) ek focus par hai. Pink double-arrow $a$ ko centre se edge tak long axis ke along mark karta hai. Long axis par do white dots perigee (Earth ke sabse kareeb) aur apogee (sabse dur) hain.

Long axis par do special points ke apne naam hain, dono parent dwara constantly use hote hain:

6. Focus par measure kiye gaye angles: $ν$ (true anomaly)

Satellite ellipse par kahi hai. Hume ek angle chahiye yeh batane ke liye ki kahan.

Yeh single angle woh ek element hai jo time ke saath badalta hai (baaki sab fixed ellipse describe karte hain).

Har conic case mein $ν$ ki range

$ν$ ka allowed span sab orbits ke liye same nahi hai — §5 se shape ladder ise control karta hai, kyunki $r$ positive aur finite rehna chahiye.

Definition

$ν$ kahan point kar sakta hai — charo cases $r = \frac{p}{1 + e cos ν}$ dekho. Radius blow up hota hai ( $r \to \infty$ ) exactly jab denominator zero hit karta hai, yani jab $cos ν = - 1/ e$ .

Circle ( $e = 0$ ): denominator hamesha $1$ hai; $ν$ puri $0^{\circ}$ se $36 0^{\circ}$ freely run karta hai (lekin starting point ambiguous hai — §7 dekho).
Ellipse ( $0 < e < 1$ ): $- 1/ e < - 1$ , toh $cos ν = - 1/ e$ impossible hai; $ν$ puri $0^{\circ}$ – $36 0^{\circ}$ run karta hai, closed loop, koi restriction nahi.
Parabola ( $e = 1$ ): $cos ν = - 1$ at $ν = 18 0^{\circ}$ , toh $r \to \infty$ wahan. $ν$ open interval $- 18 0^{\circ} < ν < 18 0^{\circ}$ tak restricted hai; endpoints $\pm 18 0^{\circ}$ point at infinity hain aur kabhi actually reach nahi hote.
Hyperbola ( $e > 1$ ): $- 1/ e$ between $- 1$ aur $0$ hai, toh do asymptote angles $ν_{\infty} = \pm arccos (- 1/ e)$ hain. $ν$ $- ν_{\infty} < ν < ν_{\infty}$ tak restricted hai; usse aage formula negative $r$ deta hai (physically meaningless — woh branch "doosra" hyperbola hai).

Orbit equation kahan se aata hai

Parent assert karta hai $r = \frac{a ( 1 - e ^{2} )}{1 + e c o s ν} = \frac{p}{1 + e c o s ν} .$ Aao hum $(1 - e^{2})$ factor ko trust karne ki jagah earn karte hain. Ek ellipse ko uske do focus points $F$ (Earth) aur $F^{'}$ (empty focus) se define kiya ja sakta hai: ellipse par har point ke liye, do focus-distances same constant mein add hoti hain, $2 a$ : $r + r^{'} = 2 a, jahan r^{'} = empty focus tak doori .$

Step 1 — do triangles set up karo. Do foci $2 c$ se separated hain, jahan $c = a e$ centre-to-focus doori hai (yeh $e$ ki definition hai: $e = c / a$ ). Satellite ko true anomaly $ν$ par rakho. Triangle draw karo sides $r$ , $r^{'}$ aur base $2 c$ ke saath. Earth ke focus par $r$ aur long axis ke beech ka angle $ν$ hai.

Step 2 — us triangle par law of cosines. Side $r^{'}$ angle $(18 0^{\circ} - ν)$ ke opposite satisfy karta hai $r^{'2} = r^{2} + (2 c)^{2} - 2 r (2 c) cos (18 0^{\circ} - ν) = r^{2} + 4 c^{2} + 4 r c cos ν .$

Step 3 — $r^{'} = 2 a - r$ use karo. Square karo: $r^{'2} = 4 a^{2} - 4 a r + r^{2}$ . $r^{'2}$ ke liye do expressions equal set karo, aur $r^{2}$ cancel ho jata hai: $4 a^{2} - 4 a r = 4 c^{2} + 4 r c cos ν .$

Step 4 — $r$ ke liye solve karo. $4$ se divide karo aur $r$ terms gather karo: $a^{2} - c^{2} = r (a + c cos ν) \Rightarrow r = \frac{a ^{2} - c ^{2}}{a + c c o s ν} .$ Ab $c = a e$ substitute karo: numerator $a^{2} - a^{2} e^{2} = a^{2} (1 - e^{2})$ , denominator $a (1 + e cos ν)$ . $a$ ka ek factor cancel ho jata hai: $r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν} = \frac{p}{1 + e cos ν} p \equiv a (1 - e^{2}) .$ Toh mysterious $(1 - e^{2})$ aur kuch nahi, bas $a^{2} - c^{2}$ over $a$ hai — "focus kitna off-centre baith hai" ka direct fingerprint. Isliye $p = a (1 - e^{2})$ hai.

7. Teen tilt angles: $i$ , $Ω$ , $ω$

Ab tak ellipse ek page par flat draw ki gayi hai. Reality mein yeh 3D mein ek angle par hang karta hai. Teen aur angles ise orient karte hain. Orientation describe karne ke liye humein pehle reference directions ka ek fixed set chahiye.

Figure 4. White oval equatorial plane hai; yellow arrow $X$ vernal equinox $♈$ ki taraf point karta hai aur blue arrow $Z$ North ki taraf. Pink oval tilted orbit plane hai; woh jitna lean karta hai woh angle inclination $i$ hai, aur $X$ se pink ascending node (marked) tak ka angle RAAN $Ω$ hai.

Common mistake Degenerate orbits — jab

$Ω$ ya $ω$ ka matlab khatam ho jata hai Yeh do angles tabhi defined hain jab geometry jinhe woh point karte hain exist kare. Do cases unhe break karte hain:

Equatorial orbit ( $i = 0^{\circ}$ ya $i = 18 0^{\circ}$ ): orbit plane aur equator same plane hain, toh woh kabhi cross nahi karte — koi line of nodes nahi aur koi ascending node nahi. Tab $Ω$ undefined hai (swing measure karne ke liye kuch nahi). Fix: engineers longitude of perigee $ϖ = Ω + ω$ use karte hain, ya bas $Ω = 0$ set kar dete hain.
Circular orbit ( $e = 0$ ): circle mein koi perigee nahi hota (har point equidistant hota hai), toh koi closest point nahi jisse measure karein — $ω$ undefined hai. Fix: argument of latitude $u = ω + ν$ (node → satellite) use karo, jo perfectly well-defined rehta hai.
Circular aur equatorial ( $e = 0, i = 0$ ): dono ek saath fail hote hain; sirf true longitude $ℓ = Ω + ω + ν$ bachta hai. Yeh bugs nahi hain — yeh honest gaps hain jahan chosen coordinates ke paas kuch pakadne ko nahi hai.

8. Do conserved companions jinhe parent use karta hai: $h$ aur $ε$

Energy size $a$ ko kyun fix karti hai — poori derivation

Parent $ε = - μ /2 a$ box karta hai lekin kyun nahi dikhata. Yeh raha, scratch se, sirf conservation use karke.

Step 1 — do apsides par $ε$ likho. Perigee aur apogee par, $r ⊥ v$ , toh $h = r v$ , giving $v = h / r$ . $ε$ mein substitute karo: $ε = \frac{h ^{2}}{2 r _{p}^{2}} - \frac{μ}{r _{p}} aur ε = \frac{h ^{2}}{2 r _{a}^{2}} - \frac{μ}{r _{a}} .$ Dono same $ε$ equal karte hain kyunki energy conserved hai.

Step 2 — equal set karo aur $h^{2}$ isolate karo. $\frac{h ^{2}}{2} (\frac{1}{r _{p}^{2}} - \frac{1}{r _{a}^{2}}) = μ (\frac{1}{r _{p}} - \frac{1}{r _{a}}) .$ Left side ko factor karo $\frac{1}{r _{p}^{2}} - \frac{1}{r _{a}^{2}} = (\frac{1}{r _{p}} - \frac{1}{r _{a}}) (\frac{1}{r _{p}} + \frac{1}{r _{a}})$ use karke, phir common factor $(\frac{1}{r _{p}} - \frac{1}{r _{a}})$ cancel karo: $\frac{h ^{2}}{2} (\frac{1}{r _{p}} + \frac{1}{r _{a}}) = μ \Rightarrow h^{2} = \frac{2 μ}{\frac{1}{r _{p}} + \frac{1}{r _{a}}} = \frac{2 μ r _{p} r _{a}}{r _{p} + r _{a}} .$

Step 3 — geometry ke saath $h^{2}$ evaluate karo. $r_{p} + r_{a} = 2 a$ aur $r_{p} r_{a} = a (1 - e) \cdot a (1 + e) = a^{2} (1 - e^{2})$ ke saath: $h^{2} = \frac{2 μ a ^{2} ( 1 - e ^{2} )}{2 a} = μ a (1 - e^{2}) .$

Step 4 — perigee par $ε$ mein back substitute karo. $ε = \frac{h ^{2}}{2 r _{p}^{2}} - \frac{μ}{r _{p}}$ use karo $r_{p} = a (1 - e)$ aur $h^{2} = μ a (1 - e^{2})$ ke saath: $ε = \frac{μ a ( 1 - e ^{2} )}{2 a ^{2} ( 1 - e ) ^{2}} - \frac{μ}{a ( 1 - e )} .$ Pehle term mein, $1 - e^{2} = (1 - e) (1 + e)$ likho; $(1 - e)$ ka ek factor aur $a$ ka ek factor cancel ho jata hai: $ε = \frac{μ ( 1 + e )}{2 a ( 1 - e )} - \frac{μ}{a ( 1 - e )} .$

Step 5 — common factor $\frac{μ}{a ( 1 - e )}$ par combine karo. $ε = \frac{μ}{a ( 1 - e )} (\frac{1 + e}{2} - 1) = \frac{μ}{a ( 1 - e )} \cdot \frac{1 + e - 2}{2} = \frac{μ}{a ( 1 - e )} \cdot \frac{e - 1}{2} .$

Step 6 — final cancellation. Note karo $e - 1 = - (1 - e)$ , toh denominator mein $(1 - e)$ numerator wale ko cancel kar deta hai: $ε = \frac{μ}{a ( 1 - e )} \cdot \frac{- ( 1 - e )}{2} = - \frac{μ}{2 a} .$ $e$ ka har trace gayab ho gaya — energy sirf $a$ par depend karti hai. Exactly isliye energy jaanna orbit ka size lock kar deta hai, chahe uski shape koi bhi ho ya satellite kahan bhi baitha ho. $ε = v^{2} /2 - μ / r = - μ /2 a$ rearrange karne par vis-viva equation $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ wapas milta hai jo §2 mein promise ki gayi thi.

Six Keplerian elements — ek consolidated list

Formula Complete set (yeh chhe yaad karo)

Symbol	Naam	Kaam
$a$	semi-major axis	ellipse ki size
$e$	eccentricity	ellipse ki shape (squash)
$i$	inclination	plane ka equator se tilt
$Ω$	RAAN	tilted plane ka $Z$ ke around swing
$ω$	argument of perigee	plane ke andar perigee kahan point karta hai
$ν$	true anomaly	satellite abhi kahan hai (sirf yahi time-varying hai)
Pehle paanch ideal two-body gravity mein constant hain; sirf $ν$ move karta hai.

Foundations topic ko kaise feed karte hain

Legend: $r$ = focus se doori; $v$ = speed; $μ$ = gravity parameter $GM$ ; $ε$ = specific energy; $a, e$ = ellipse size aur shape; $r, v$ = position aur velocity arrows; $h$ = angular-momentum vector $r \times v$ ; $ν$ = true anomaly; ECI = fixed reference frame; $i, Ω, ω$ = teen tilt angles; KEP = chhe Keplerian elements jo full 3D position dete hain.

Equipment checklist

Plain letter

r

ka kya matlab hai vs arrow letter

r

r

ek single number hai (ek doori);

r

ek arrow hai — teen numbers jo doori aur direction dete hain.

G

kya hai aur yeh

μ

se kaise alag hai?

G

universal gravitational constant hai (har jagah same);

μ = GM

is body ki mass fold in karta hai uski specific pull dene ke liye.

Vis-viva equation ko words mein state karo.

v^{2} = μ (2/ r - 1/ a)

— kisi bhi point par speed-squared sirf tumhari doori

r

aur orbit size

a

par depend karta hai.

Ek orbit ko exactly chhe numbers ki zaroorat kyun hai?

Kyunki position (3) plus velocity (3) chhe hai, aur woh chhe saare future motion fix kar dete hain.

Chhe Keplerian elements aur unke kaam name karo.

a

size,

e

shape,

i

tilt,

Ω

swing,

ω

perigee direction,

ν

current position.

Altitude ko

r

mein kaise convert karte ho?

Earth ka radius add karo:

r = altitude + R_{E}

a

aur

e

kya describe karte hain, aur

e = 1

ka kya matlab hai?

a

size hai,

e

squash hai;

e = 1

parabolic escape boundary hai bound aur unbound orbits ke beech.

Ellipse vs hyperbola ke liye

ν

kis range mein run karta hai?

Ellipse: full