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

3.2.8 · D5 · Physics › Orbital Mechanics & Astrodynamics › Orbital elements (Keplerian) — semi-major axis a, eccentrici

Kisi bhi trap se pehle, hum har symbol aur reference frame ko lock down kar lete hain, taaki is page par koi cheez undefined na rahe.

Definition Chhe orbital elements (zero se define kiye hue)

Ek tilti hui hula-hoop jiski surface par ek bead slide kar rahi hai ki picture banao. Chhe numbers hoop ko aur bead ko pin down karti hain:

$a$ = semi-major axis = hoop ke sabse lambe diameter ka aadha, ek distance. Ye orbit ki size set karta hai.
$e$ = eccentricity = ek dimensionless number jo batata hai ki hoop kitni squashed hai. $e = 0$ ek perfect circle hai, $0 < e < 1$ ek ellipse hai, $e = 1$ ek parabola hai, $e > 1$ ek hyperbola hai. Ye shape set karta hai.
$i$ = inclination = hoop flat equator plane se kitna tila hua hai, ek angle $[0^{\circ}, 18 0^{\circ}]$ mein.
$Ω$ = RAAN (Right Ascension of the Ascending Node) = tilti hui hoop kis taraf ghumi hui hai (agle callout mein precisely define kiya gaya hai).
$ω$ = argument of perigee = hoop ke apne plane ke andar, sabse close point (perigee) kahaan hai — ascending node se perigee tak measure kiya gaya.
$ν$ = true anomaly = bead abhi kahaan hai, perigee se satellite tak ka angle focus par measure kiya gaya, motion ki direction mein. Ye akela element hai jo time ke saath change hota hai.

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

Definition Is page par use kiye gaye supporting symbols

Focus = woh point jahan Earth ka center hai; har $r$ (distance) aur $ν$ (angle) yahin se measure kiya jaata hai.
Apsides = perigee (sabse close point, $ν = 0^{\circ}$ ) aur apogee (sabse door point, $ν = 18 0^{\circ}$ ).
$μ$ = Earth ka gravitational parameter: Newton's constant times Earth ki mass, $μ = GM$ , units km³/s². Gravity ki strength ka ek single number. Earth ke liye $μ \approx 3.986 \times 1 0^{5}$ km³/s².
$v$ = velocity vector (ek arrow: direction + speed); $v = ∣ v ∣$ uski length hai, scalar speed. Vector $v$ $h = r \times v$ mein appear karta hai; scalar $v$ energy $ε = \frac{v ^{2}}{2} - \frac{μ}{r}$ mein appear karta hai jahan sirf magnitude matter karta hai.
$ε$ = specific orbital energy = orbital energy per unit mass, $ε = \frac{v ^{2}}{2} - \frac{μ}{r}$ . Ye motion ka ek constant hai aur $- μ /2 a$ ke barabar hai.
$T$ = orbital period, $T = 2 π a^{3} / μ$ .
$p$ = semi-latus rectum, $p = a (1 - e^{2})$ — perigee ke ek quarter-turn baad orbit ki radius; ye $r = p / (1 + e cos ν)$ mein appear karta hai.
$h$ = angular-momentum vector, $h = r \times v$ . Ye orbit plane ke perpendicular point karta hai (right-hand rule: ungliyan uss taraf curl karti hain jis taraf satellite move karta hai, thumb $h$ deta hai), isliye uski direction hi plane ka tilt hai.
$M$ = mean anomaly, ek kalpnik angle jo time ke saath uniformly $2 π / T$ ki rate se badhta hai; $ν$ ka "clock hand" version (dekho Kepler's equation and mean anomaly).
$R_{z} (θ)$ , $R_{x} (θ)$ = active rotation matrices jo ek vector ko angle $θ$ se $Z$ ya $X$ axis ke baare mein physically rotate karti hain (right-hand rule: thumb axis ke saath, ungliyan positive turn direction deti hain).

True or false — justify

TF1. Ek circular orbit mein bhi argument of perigee

ω

well-defined hota hai.

False. Circle mein

e = 0

hota hai, isliye koi perigee nahi hota — koi "closest point" nahi jahan se measure karein — aur

ω

undefined hai; engineers ki jagah argument of latitude

u = ω + ν

use karte hain.

TF2. Same

a

wale do orbits ka period hamesha same hota hai.

True. Period sirf

a

par depend karta hai via

T = 2 π a^{3} / μ

; eccentricity, tilt, aur orientation enter nahi karte, isliye equal

a

wale ek nearly circular aur ek bahut elongated orbit ek hi period share karte hain.

TF3.

a

ko fixed rakhte hue eccentricity badhane se total orbital energy badhti hai.

False. Energy

ε = - μ /2 a

hai, jo sirf

a

par depend karti hai.

e

change karne se ellipse ki shape badal jaati hai lekin energy — aur isliye

a

— change nahi hoti (dekho Vis-viva equation).

TF4.

i = 9 0^{\circ}

wala orbit dono poles ke upar se guzarta hai.

True. Inclination equator plane se plane ka tilt hai;

9 0^{\circ}

par plane

Z

(spin) axis ko contain karti hai, isliye ground track North aur South poles ke upar sweep karta hai.

TF5. RAAN

Ω

exactly equatorial orbit ke liye meaningful hai.

False.

i = 0

hone par plane kabhi equator nahi chodti, isliye koi ascending node nahi hota jishe point kiya ja sake;

Ω

undefined ho jaata hai (ek degenerate/singular case).

TF6. True anomaly

ν

ek orbit mein constant rate se badhti hai.

False. Kepler ke 2nd law ke anusaar satellite equal times mein equal areas sweep karta hai, isliye

ν

perigee par tezi se aur apogee par dheere chalti hai; uniform angle mean anomaly

M

hai (dekho Kepler's equation and mean anomaly).

TF7. Semi-major axis

a

apogee aur perigee radii ke average ke barabar hota hai.

True. Kyunki

r_{a} = a (1 + e)

aur

r_{p} = a (1 - e)

hain, add karne par

r_{a} + r_{p} = 2 a

milta hai, isliye

a = (r_{a} + r_{p}) /2

— focus se radii ke, altitudes ke nahi.

TF8. Retrograde orbit ki inclination negative hoti hai.

False. Inclination

[0^{\circ}, 18 0^{\circ}]

par define ki gayi hai; retrograde motion

i > 9 0^{\circ}

se encode hoti hai (

h

vector equator ke neeche jhukta hai), kabhi negative angle se nahi.

TF9. Agar ek instant par

(r, v)

pata ho toh saare chhe elements recover ho sakte hain.

True. Chhe state numbers exactly wahi information rakhte hain jo chhe elements rakhte hain; conversion

h = r \times v

se plane compute karta hai, eccentricity vector se

e

aur perigee direction, aur energy se

a

— ye State vectors to orbital elements dwara hota hai, jo standard state→elements recipe hai (dekho bhi Angular momentum vector h, Eccentricity vector).

Spot the error

SE1. "Perigee altitude

400

km hai, isliye

r_{p} = 400

km."

Error ye hai ki focus Earth ke center par hai ye bhool gaye:

r_{p} = R_{E} + altitude = 6378 + 400 = 6778

km. Altitude surface se measure hoti hai, radius center se.

SE2. "Orbit bahut eccentric hai (

e = 0.9

), isliye zaroor bahut badi hogi."

Eccentricity shape set karti hai, size nahi; ek chhoti orbit aur ek badi orbit dono ka

e = 0.9

ho sakta hai. Sirf

a

(aur isliye

r_{a}, r_{p}

) size scale set karta hai.

SE3. "

ω

aur

ν

dono ascending node se measure kiye jaate hain."

Sirf

ω

ascending node se shuru hota hai (perigee tak).

ν

perigee se shuru hota hai (satellite tak), isliye unka sum

u = ω + ν

node-se-satellite angle hai.

SE4. "Vis-viva kehta hai

v^{2} = μ (2/ r - 1/ a)

, isliye apogee par

v

sabse bada hoga kyunki

r

bada hai."

Ulta hai: bada

r

2/ r

ko chhhota banata hai, isliye

v

(scalar speed) apogee par sabse chhhoti hoti hai. Speed perigee par peak karti hai jahan

r

sabse chhhota hota hai.

SE5. "Inclination orbit plane aur

Z

-axis ke beech ka angle hai."

h

(plane ke perpendicular) aur

Z

ke beech ka angle hai, jo orbit plane aur equatorial plane ke beech ke angle ke barabar hai — plane aur axis ke beech nahi.

SE6. "Perifocal position ko ECI mein convert karne ke liye pehle

i

se rotate karo, phir

ω

se, phir

Ω

se."

Sahi order

R_{z} (Ω) R_{x} (i) R_{z} (ω)

hai jo

r_{P Q W}

par act karta hai. Ye active rotations hain (wo vector ko move karte hain, axes ko fixed rakhte hain), aur matrix form mein sabse right wali matrix vector ko pehle touch karti hai — isliye right-to-left padhne par hum perigee ko plane ke andar rakhte hain (

ω

), phir plane ko node line ke baare mein tilt karte hain (

i

), phir node ko uski longitude par swing karte hain (

Ω

). Pehle

i

karna perigee rakhne se pehle tilt kar dega, alag aur galat orientation milegi (dekho Perifocal coordinate frame, jo P–Q–W axes define karta hai jahan se ye shuru hota hai).

SE7. "Escape trajectory ke liye

ε = - μ /2 a

positive

a

ke saath abhi bhi hold karta hai."

Parabola ke liye

ε = 0

(isliye

a \to \infty

), aur hyperbola ke liye

ε > 0

a < 0

force karta hai. Formula hold karta hai lekin

a

ab positive "size" nahi raha.

Why questions

WQ1. Exactly chhe orbital elements kyun hote hain, paanch ya saat kyun nahi?

Kyunki poora state 3 position + 3 velocity = chhe independent numbers hai; elements inhi chhe degrees of freedom ki ek repackaging hain (dekho Two-body problem, jo dikhata hai ki motion inhi chhe se fix hoti hai).

WQ2. Ideal two-body motion mein paanch elements constant aur sirf

ν

kyun change hota hai?

Ellipse ki shape aur orientation conserved quantities se fix hoti hai (energy

a

fix karta hai;

h

plane fix karta hai; Eccentricity vector — perigee ki taraf point karne wala fixed arrow —

e

aur

ω

fix karta hai); sirf fixed curve par satellite ki position chalti hai, aur woh

ν

hai.

WQ3. Elements ke saath kyun bothered karein, sirf

(r, v)

track kyun na karein?

Elements unchanging geometry (5 constants) ko ek moving number se alag karte hain, orbit ki shape, tilt, aur orientation ko instantly readable banate hain — chhe wiggling coordinates mein dube hone ki jagah.

WQ4.

J_{2}

is "constant" picture ko kyun matter karta hai?

J_{2}

sabse bada zonal harmonic hai — ek dimensionless number (

J_{2} \approx 1.083 \times 1 0^{- 3}

Earth ke liye) jo measure karta hai ki Earth ka equatorial bulge gravity ko perfect

1/ r^{2}

se kitna hata deta hai. Woh extra pull slowly

Ω

aur

ω

ko drift karta hai, isliye paanch "constants" slowly varying ho jaate hain (dekho Orbital perturbations (J2)).

WQ5. Semi-latus rectum

p

ν = 9 0^{\circ}

par radius kyun hai?

Orbit equation

r = p / (1 + e cos ν)

cos 9 0^{\circ} = 0

deta hai, isliye

r = p

exactly perigee ke ek quarter-orbit baad —

p

woh "sideways" radius hai.

WQ6.

ν

ko motion ki direction mein kyun measure karna chahiye?

Warna sign ambiguous ho jaata hai aur

r = p / (1 + e cos ν)

apogee se guzarte hue satellite ko sahi track nahi karta; motion direction fix karta hai ki

ν

apogee ki taraf badhti hai ya perigee ki taraf waapas.

Edge cases

EC1. Elements ka kya hota hai jab

e \to 0

Perigee vanish ho jaata hai, isliye

ω

(aur uske saath

ν

ka reference) undefined ho jaata hai; fix argument of latitude

u = ω + ν

hai, jo well-defined rehta hai.

EC2.

i \to 0

hone par

Ω

aur

ω

ka kya hota hai?

Node line disappear ho jaati hai (plane kabhi equator nahi chhodti), isliye dono apna reference kho dete hain; fix true longitude

ℓ = Ω + ω + ν

hai, jo seedha

X

-axis se measure ki jaati hai.

EC3.

e = 1

physically kya matlab hai, aur

a

ka kya hota hai?

Ye ek parabolic escape trajectory hai jisme total energy bilkul zero hoti hai

ε = 0

; kyunki

ε = - μ /2 a

, ye

a \to \infty

force karta hai — "ellipse" khul jaata hai aur kabhi close nahi hota.

EC4. Ek orbit jiski

h

- Z

ki taraf point karti hai, uski inclination kya hogi?

i = 18 0^{\circ}

: plane equatorial hai lekin motion poori tarah retrograde hai,

i = 0^{\circ}

par prograde equatorial orbit ka extreme opposite.

EC5. Agar perigee altitude apogee altitude ke barabar ho, toh

e

aur

ω

kya hain?

Orbit ek circle hai, isliye

e = 0

exactly hai, aur

ω

undefined hai kyunki koi unique closest point nahi hai.

EC6. Hyperbolic flyby (

e > 1

) ke liye kya apogee hota hai?

Nahi — trajectory open hai, isliye koi farthest point nahi hai; sirf perigee (closest approach,

ν = 0

) exist karta hai, aur

ν

asymptote angles se bounded hai jahan

r \to \infty

Recall Har trap ki ek-line summary

Size = $a$ (energy, period); shape = $e$ ; tilt = $i$ ; swing = $Ω$ ; perigee-kahan-point-karta-hai = $ω$ ; satellite-abhi-kahan-hai = $ν$ . Degenerate cases ( $e = 0$ , $i = 0$ , $e \geq 1$ ) exactly wahan hain jahan "where" references toot jaate hain aur $u$ , $ℓ$ , ya open-orbit thinking ki zaroorat padti hai.