3.2.16
Part of Kepler's Equation solving chain: M → E → ν
Ek orbiting body ek ellipse par chalti hai jisme central mass ek focus par hoti hai. Usse track karne ke liye hum teen alag-alag "angles" use karte hain:
True anomaly ν asli geometric angle. Yahi aapko actually chahiye hota hai.Eccentric anomaly E helper angle jo ellipse ke center se measure hota hai, ek khayal ki circle par jo orbit ko hug karti hai. Isse Kepler's equation ke through time se relate karna aasaan hai.Mean anomaly M nakli angle jo time ke saath linearly badhta hai.
Hum Kepler's equation M = E − e sin E E E → ν sirf usi last conversion ke baare mein hai.
Definition Teen anomalies geometricallyEllipse ko semi-major axis a e O F P
==True anomaly ν ∠ ( perihelion , F , P ) focus par measure kiya gaya.
==Eccentric anomaly E ∠ ( perihelion , O , P ′ ) P ′ P vertically upar radius a kiya gaya hai, center par measure kiya gaya.
Hum coordinates set up karte hain origin center O x
( 1 + e ) / ( 1 − e ) E ≈ 0 karib hoti hai, isliye E bada true-angle ν E ≈ π E ν e = 0 1 ν = E = M
Step 3 ke results se shuru karo.
1 + cos ν = 1 + 1 − e c o s E c o s E − e = 1 − e c o s E ( 1 − e c o s E ) + ( c o s E − e ) = 1 − e c o s E ( 1 − e ) ( 1 + c o s E ) 1 − cos ν = 1 − e c o s E ( 1 − e c o s E ) − ( c o s E − e ) = 1 − e c o s E ( 1 + e ) ( 1 − c o s E ) 1 + c o s ν 1 − c o s ν = ( 1 − e ) ( 1 + c o s E ) ( 1 + e ) ( 1 − c o s E ) . Dono sides par tan 2 ( θ /2 ) = 1 + cos θ 1 − cos θ tan 2 2 ν = 1 − e 1 + e tan 2 2 E ⇒ tan 2 ν = 1 − e 1 + e tan 2 E . Half-angle kyun? 1 + c o s 1 − c o s
Worked example Example 1 — Circular check (e = 0 e = 0 E = 5 0 ∘ tan ( ν /2 ) = 1/1 tan ( 2 5 ∘ ) = tan 2 5 ∘ ⇒ ν = 5 0 ∘ Yeh step kyun? Jab koi eccentricity nahi hoti toh ellipse hi circle hai, focus = center, isliye teeno angles ek jaisi hoti hain — ek sanity anchor.
Worked example Example 2 — Eccentric orbit,e = 0.6 E = 9 0 ∘ tan ( E /2 ) = tan 4 5 ∘ = 1 = ( 1 + 0.6 ) / ( 1 − 0.6 ) = 1.6/0.4 = 4 = 2 tan ( ν /2 ) = 2 ⋅ 1 = 2 ⇒ ν /2 = arctan 2 = 63.4 3 ∘ ⇒ ν = 126.8 7 ∘ ν > E E = 9 0 ∘ offset focus se dekha jaaye toh usne pehle hi 9 0 ∘ r = a ( 1 − e cos E ) = a ( 1 − 0 ) = a
Worked example Example 3 — Cosine se verify karoSame numbers, e = 0.6 E = 9 0 ∘ cos ν = 1 − 0.6 cos 9 0 ∘ cos 9 0 ∘ − 0.6 = 1 − 0.6 = − 0.6 ⇒ ν = 126.8 7 ∘ . Cross-check kyun? Cosine akele quadrant-ambiguous hota hai, lekin sin ν = 1 − 0.36 ⋅ 1/1 = + 0.8 > 0 ν
cos ν = 1 − e c o s E c o s E − e arccos Kyun sahi lagta hai: yeh ek clean closed form hai. Kyun galat hai: arccos [ 0 , π ] sin ν < 0 E ∈ ( π , 2 π ) ν Fix: half-angle tan ya atan2(sin ν, cos ν) use karo.
Common mistake Galat coordinate ko squash karnaKuch log x = a sin E Kyun sahi lagta hai: confuse ho jaate hain ki major axis kaun sa hai. Fix: perihelion major axis (x E x = a cos E y = b sin E
Common mistake Focus par shift karna bhool jaanaCenter coordinates se ν eccentric angle deta hai, true angle nahi. Fix: pehle x c = a e
Recall Feynman: explain to a 12-year-oldSocho ek planet ek oval track par daud raha hai aur Sun off-center ek special jagah baith ke hai jise focus kehte hain. Kahan hai planet, yeh jaanne ke liye scientists ek trick use karte hain: woh oval ke around ek perfect circle banate hain aur oval ke middle se ek helper angle E aap actually dekhenge Sun se dekhne par woh alag hoti hai — ise ν tan ( ν /2 ) = 1 − e 1 + e tan ( E /2 ) e = 0
"True is Eccentric, plus a Perihelion kick."
Focus offset subtract karo (cos E − e
Perihelion ki taraf 1 − e 1 + e
"One-plus-e over one-minus-e , root it, times tan-half-E."
Eccentric anomaly ke terms mein radius formula kya hai? r = a ( 1 − e cos E )
True aur eccentric anomaly ko relate karne wala quadrant-safe formula do. cos ν E e cos ν = 1 − e cos E cos E − e
sin ν E e Half-angle tan arccos ( cos ν ) Half-angle ka tan arccos [ 0 , π ]
Center coordinates mein body ke x , y x = a cos E , y = a 1 − e 2 sin E (circle ko vertically
b / a se squash kiya gaya).
Center se focus ka offset kya hai? c = a e
e = 0 ν E M Teeno equal hain — orbit ek circle hai aur focus = center.
Perihelion ke paas ν > E Focus perihelion ki taraf offset hai, isliye wahi E
Kepler's Equation — M E Mean anomaly and time — M = n ( t − t 0 ) Orbit geometry — semi-major axis and eccentricity — b = a 1 − e 2 c = a e Orbital radius equation — r = a ( 1 − e cos E ) = 1 + e c o s ν a ( 1 − e 2 ) Position and velocity in the perifocal frame — ν Half-angle trigonometric identities — yahan algebraic engine.
Aux circle point a cosE, a sinE
cos nu and sin nu formulas
