True anomaly from eccentric anomaly: Question bank

3.2.16 · D5 · Physics › Orbital Mechanics & Astrodynamics › True anomaly from eccentric anomaly

Child of True anomaly from eccentric anomaly. Yeh ek concept-trap bank hai: koi bhari arithmetic nahi, bas woh ideas jo log miss kar lete hain. Har item ek one-line reveal hai — pehle answer cover karo, apni reasoning commit karo, phir check karo.

Shuru karne se pehle, players ko seedha rakhna (parent ki geometry dekho):

$a$ — semi-major axis (lambe diameter ka aadha); $b = a 1 - e^{2}$ — semi-minor axis (chhote diameter ka aadha). Auxiliary circle ki radius $a$ hai, aur ellipse woh circle hai jo vertically $b / a$ factor se squash hua hai.
$ν$ (true anomaly) — angle focus $F$ par, asli "Sun se dekha gaya" angle.
$E$ (eccentric anomaly) — angle center $O$ par, radius $a$ ke auxiliary circle par.
$e$ (eccentricity) — ellipse kitna squash hai; $c = a e$ center se focus ka offset hai.

Master relations jinhe aap baar baar use karoge (valid for $e < 1$ ): $cos ν = \frac{c o s E - e}{1 - e c o s E}, sin ν = \frac{1 - e ^{2} s i n E}{1 - e c o s E}, tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} .$

True or false — justify

ν = E

circular orbit (

e = 0

) ke liye hamesha hold karta hai.

True.

e = 0

ke saath stretch factor

(1 + e) / (1 - e) = 1

aur focus center par hi baithta hai (

c = a e = 0

), isliye dono angles ek hi point se measure hote hain →

ν = E

(aur

= M

bhi).

Kisi bhi elliptical orbit (

0 < e < 1

) ke liye,

ν \geq E

poore orbit mein hold karta hai.

False. Perihelion wale half mein (

0 < E < π

) stretch

ν > E

banata hai; aphelion-return wale half mein (

π < E < 2 π

) effect ulta ho jaata hai aur

ν < E

. Woh exactly

E = 0, π, 2 π

par equal aur cross hote hain.

Perihelion aur aphelion par,

ν = E

True.

E = 0

(perihelion) par dono angles

0

hain;

E = π

(aphelion) par dono

π

hain. Yeh sirf do points hain jahan

e > 0

ke liye teeno anomalies ek saath milti hain.

Formula

tan (ν /2) = (1 + e) / (1 - e) tan (E /2)

ko plain

arctan

ke saath apply karke har baar sahi

ν

milega.

False.

arctan

sirf

(- 90°, 90°)

return karta hai, isliye

E = π

ke paas jahan

ν /2 \to 90°

hai, aapko track karna hoga ki

E /2

ne kitne half-turns complete kiye hain; safest yeh hai ki

sin ν, cos ν

compute karo aur atan2 use karo.

r = a (1 - e cos E)

perihelion par sabse chhota hota hai.

True.

cos E

sabse bada (

= 1

)

E = 0

par hota hai,

1 - e cos E

ko sabse chhota

= 1 - e

banata hai, isliye

r_{m i n} = a (1 - e)

— perihelion distance.

Radius formula

r = a (1 - e cos E)

aur orbit ka

r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν}

alag curves describe karte hain.

False. Yeh same ellipse par same point hain, sirf do alag anomalies se likhe gaye hain; ek mein

cos ν

relation substitute karne par doosra recover hota hai. Orbital radius equation dekho.

Agar

sin ν > 0

hai to body orbit ke outbound (perihelion→aphelion) half par hai.

True.

sin ν = 1 - e^{2} sin E / (1 - e cos E)

ka sign

sin E

ka hi sign hai;

sin E > 0

ka matlab hai

E \in (0, π)

, ascending half jahan body focus se door jaati hai.

cos ν

akela

ν

ka quadrant determine kar sakta hai.

False.

cos ν

ν

aur

- ν

mein fark nahi kar sakta (dono ka cosine same hai). Quadrant fix karne ke liye

sin ν

ka sign chahiye — yaani

sin E

ka sign.

Spot the error

"Center coordinates mein body ki position

x = a sin E, y = b cos E

hai."

Wrong axes.

E

perihelion se measure hota hai jo major (

x

) axis par hai, isliye

x = a cos E

aur

y = b sin E = a 1 - e^{2} sin E

(yaad raho

b = a 1 - e^{2}

semi-minor axis hai).

ν

pane ke liye hum

(a cos E, b sin E)

ka angle seedha padhte hain."

Yeh angle center par hai, focus par nahi. Pehle origin ko

F

par shift karna hoga

x

ke saath

c = a e

subtract karke:

x_{F} = a (cos E - e)

ν

sirf focus par define hai.

b = a (1 - e^{2})

, isliye squash factor

1 - e^{2}

hai."

Square root miss ho gayi. Semi-minor axis

b = a 1 - e^{2}

hai; circle ka vertical squash

b / a = 1 - e^{2}

hai,

1 - e^{2}

nahi.

"Kyunki

cos ν = \frac{cos E - e}{1 - e cos E}

E = 90°

par

cos ν = \frac{- e}{1} = - e

milta hai, isliye

ν = 90° +

kuch chhota."

Value

- e

sahi hai lekin conclusion fuzzy hai:

cos ν = - e < 0

ν

ko second quadrant (

> 90°

) mein rakhta hai, aur

sin ν > 0

isse confirm karta hai.

e = 0.6

ke liye

ν \approx 126.9°

hai,

90°

se kaafi aage.

"Stretch factor

(1 + e) / (1 - e)

ν

E

se aphelion ke paas bada karta hai."

Ulta hai. Perihelion ke paas (chhota

E

) tiny

tan (E /2)

amplify hota hai, isliye

ν > E

wahan hota hai. Aphelion ke paas ulta hota hai; single factor dono encode karta hai kyunki

tan (E /2)

E = π

approach karte waqt unbounded grow karta hai.

e = 1

ke liye formula phir bhi finite

ν

deta hai."

e = 1

par orbit ek parabola hai,

(1 + e) / (1 - e) \to \infty

, aur eccentric-anomaly picture (bounded ellipse/circle) bilkul toot jaata hai — is formula ki jagah parabolic/hyperbolic analogue chahiye.

M

E

, aur

ν

same angle ke sirf alag units mein hain."

Nahi — sirf

M

time ke saath linearly scale karta hai;

E

aur

ν

genuinely alag geometric angles hain (alag points se measure hote hain), nonlinearly related. Same units, alag definitions.

Why questions

ν

focus par kyun measure hota hai, center par kyun nahi?

Kyunki physical force (gravity) focus se act karta hai, aur radius

r

jo orbit aur energy mein appear karta hai woh focus-to-body distance hai;

ν

us focus frame mein natural polar angle hai. Position and velocity in the perifocal frame dekho.

Helper angle

E

introduce karne ki zaroorat hi kyun hai,

ν

se directly kaam kyun nahi karte?

Kyunki

E

Kepler's Equation (

M = E - e sin E

) ke through time se cleanly connect hota hai, jiska

ν

mein koi simple closed form nahi hai.

E

time ($M$) aur geometry (

ν

) ke beech bridge hai.

E

mein ek chhota step perihelion ke paas

ν

mein ek bada step kyun produce karta hai?

Perihelion ke paas body focus ke paas hoti hai, isliye same arc length nearby focus se dekhe jaane par bahut bada angle subtend karti hai —

(1 + e) / (1 - e)

factor exactly yahi focus-offset magnification hai.

Code mein

cos ν

expression ki jagah half-angle

tan

form prefer kyun karein?

tan (ν /2)

tan (E /2)

poore orbit mein monotonic hai, isliye koi quadrant ambiguity nahi, jabki

arccos (cos ν)

ν

aur

- ν

ko collapse kar deta hai aur sirf

[0, π]

return karta hai.

Half-angle identity

tan^{2} (θ /2) = \frac{1 - c o s θ}{1 + c o s θ}

derivation mein kyun aati hai?

1 \pm cos ν

banana messy rationals ko

(1 \mp e) (1 \pm cos E)

mein factor kar deta hai; unhe divide karne par

(1 - e cos E)

denominators cancel ho jaate hain, aur identity bacha hua

\frac{1 - c o s}{1 + c o s}

ratio ko clean

tan^{2}

mein convert kar deti hai. Half-angle trigonometric identities dekho.

Practice mein

r = a (1 - e cos E)

aksar

ν

ke saath kyun compute kiya jaata hai?

Yeh same Step-3 algebra (

r = x_{F}^{2} + y_{F}^{2}

) se free mein nikal aata hai, aur body ko place karne ke liye almost hamesha dono angle aur distance chahiye — inhe saath compute karna redundant work bachata hai.

1 - e^{2}

factor sirf

sin ν

mein kyun aata hai,

cos ν

mein kyun nahi?

Yeh vertical squash

y = b sin E = a 1 - e^{2} sin E

se aata hai, jo sirf

y

-component ko affect karta hai; focus tak

x

-shift (ek horizontal operation) mein aisa koi factor nahi hota, isliye

cos ν

root-free rehta hai.

Edge cases

E = 0

(perihelion) par, har formula kya deta hai?

cos ν = 1, sin ν = 0 \Rightarrow ν = 0

, aur

r = a (1 - e)

— closest approach. Teeno anomalies zero ke barabar hain.

E = 180°

(aphelion) par,

tan (E /2)

ka kya hota hai aur

ν

phir bhi well-defined kyun hai?

tan (E /2) = tan 90°

blow up karta hai, lekin

tan (ν /2)

bhi blowup hona chahiye: dono half-angles saath

90°

hit karte hain,

ν = 180°

dete hain. Infinity se bachne ke liye atan2 ke saath

sin ν = 0, cos ν = - 1

use karo. Yahan

r = a (1 + e)

, sabse door point.

E

just below

360°

ke liye

ν

kya hai (body orbit complete karne wali hai)?

sin E < 0

isliye

sin ν < 0

ν

360°

se thoda kam hai (equivalently ek chhota negative angle), descending side se perihelion approach kar raha hai. Plain

arctan

half-angle ise galat jagah rakhega — atan2 zaroori hai.

e \to 1^{-}

(near-parabolic ellipse) ke saath, aphelion distance aur stretch factor ka kya hota hai?

r_{m a x} = a (1 + e) \to 2 a

fixed

a

ke liye finite rehta hai, lekin stretch

(1 + e) / (1 - e) \to \infty

, isliye perihelion ke door tiny

E

changes

ν

ko enormously swing karte hain — orbit apna almost sara angle-range perihelion ke paas spend karta hai.

e = 0

aur kisi bhi

E

wale degenerate "orbit" ke liye, Kepler's equation trivial kyun hai?

e = 0

ke saath,

M = E - 0 = E

aur

ν = E

, isliye time directly aur linearly true angle par map karta hai — circle mein solve karne wali koi non-uniform-angle problem hi nahi hai.

Negative

ν

(ya

ν > 180°

) physically kya matlab hai, aur kya yeh valid hai?

Bilkul valid — iska matlab hai body descending half par hai (

sin ν < 0

E \in (π, 2 π)

), major axis ke neeche perihelion ki taraf wapas ja rahi hai. Anomalies mod

360°

wrap karti hain.

Recall One-line self-test

Upar ke har answer ko cover karo; agar aap kyun justify kar sako ki har true/false hold karta hai aur naam le sako ki kaunsa sign ya quadrant ambiguous wale ko fix karta hai, to concept aapka hai.