True anomaly from eccentric anomaly: Worked examples

3.2.16 · D3 · Physics › Orbital Mechanics & Astrodynamics › True anomaly from eccentric anomaly

Yeh note parent conversion note ka practice-drill child hai. Parent ne formulas banaye. Yahan hum $E \to ν$ conversion mein aane wale har case ko dhundhte hain aur har ek ko zero se karte hain. Agar aapne derivation nahi dekhi, pehle parent padhein; agar aapko identity engine chahiye, Half-angle trigonometric identities khula rakho.

Neeche har line par teen quantities aati hain, toh pehle unhe fix kar lete hain.

Definition Is page par use hone wale symbols

==Eccentricity $e$ == — ek single number, $0 \leq e < 1$ , jo batata hai ki ellipse kitna squashed hai. $e = 0$ ek perfect circle hai; jaise-jaise $e \to 1$ , ellipse lamba aur patla hota jaata hai. Dekho Orbit geometry — semi-major axis and eccentricity.
==Semi-major axis $a$ == — ellipse ki lambi width ka aadha (yaani "size"). Is page par hum $a = 1$ "orbit radius units" mein lete hain taaki lengths padhne mein aasaan ho; end mein apne real $a$ se multiply kar lena.
==Focus offset $c = a e$ == — ellipse ke center se focus (jahan mass baitha hai) tak ki distance. Yahi offset hi poori wajah hai ki $E$ (center se angle) aur $ν$ (focus se angle) alag kyun hote hain. Jab $e = 0$ , $c = 0$ : focus aur center ek jagah milte hain.
$E$ (eccentric anomaly, center se) aur $ν$ (true anomaly, focus se) woh do angles hain jo parent note ne banaye; hum inhe yahan convert karte hain.

Woh do tools jo hum baar baar use karenge:

Pehle yeh samjho: "quadrant" yahan ka matlab kya hai? Focus par centered ek ghadi ka face imagine karo. $ν = 0$ perihelion ki taraf point karta hai (3 o'clock). Jaise body orbit karti hai, $ν$ counter-clockwise sweep karta hai: $0$ – $9 0^{\circ}$ pehla quadrant hai (upar-daayein), $90$ – $18 0^{\circ}$ doosra (upar-baayein), $180$ – $27 0^{\circ}$ teesra (neeche-baayein), $270$ – $36 0^{\circ}$ chautha (neeche-daayein). "Quadrant sahi karna" ka matlab bas itna hai: kya hamara formula body ko inme se sahi wale mein land karta hai?

Neeche ki figure woh map hai jise hum har descending example mein refer karenge. Har quadrant cell par seedha uska naam aur angle range likha hai ("Q I: 0–90°", etc.), isliye label text — color nahi — batata hai ki kaunsa cell kaunsa hai. Dhyan do ki Q I aur Q IV red focus dot ki daayein (perihelion) taraf hain, jabki Q II aur Q III baayein taraf. $ν = 12 0^{\circ}$ par bani body Q II position dikhati hai — perihelion direction se $ν$ ka arc trace karo us tak. Jab bhi koi worked example puche "kaunsa quadrant?", is picture par matching labelled cell dhundho.

Figure — True anomaly from eccentric anomaly

Scenario matrix

Is topic ka har problem inme se kisi ek cell mein aata hai. Neeche ke worked examples mein har ek par uska cell tag laga hai.

#	Case class	Kya tricky banata hai	Example
A	$e = 0$ (circle, degenerate)	focus = center, sab angles equal	Ex 1
B	$E$ quadrant I mein ( $0 < E < 9 0^{\circ}$ )	seedha, ascending, perihelion ke paas	Ex 2
C	$E$ quadrant II mein ( $9 0^{\circ} < E < 18 0^{\circ}$ )	abhi bhi ascending, $ν > E$	Ex 3
D	$E = 18 0^{\circ}$ exactly (aphelion)	$tan (E /2) \to \infty$ , woh trap	Ex 4
E	$E$ quadrant III mein ( $18 0^{\circ} < E < 27 0^{\circ}$ )	descending, $sin ν < 0$ , sign flip	Ex 5
F	High eccentricity $e \to 1$	stretch factor bahut badh jaata hai	Ex 6
G	Reverse: $ν$ diya, $E$ nikalo	formula invert karo, real-world	Ex 7
H	Word problem: radius → time link	Kepler chain se connect hota hai	Ex 8
I	$E$ quadrant IV mein ( $27 0^{\circ} < E < 36 0^{\circ}$ ) / exam twist	$arccos$ -only trap exposed	Ex 9

Jab bhi koi length aaye, hum $a = 1$ (in "orbit radius units") lete hain — scaling trivial hai.

Worked Examples

Worked example Example 1 — Cell A: circle (

$e = 0$ ) Statement. Ek body circular orbit par ( $e = 0$ ) hai aur uski eccentric anomaly $E = 4 0^{\circ}$ hai. $ν$ aur $r$ nikalo ( $a = 1$ lo).

Forecast: circle par Sun bilkul center mein baitha hai — toh dono angles mein se kaunse equal hone chahiye? Padhne se pehle andaza lagao.

Stretch factor. $\frac{1 + 0}{1 - 0} = 1 = 1$ . Yeh step kyun? Poori $E \to ν$ distortion focus ke off-center hone ki wajah se aati hai — offset $c = a e$ (upar define kiya). Jab $e = 0$ , $c = 0$ : koi offset nahi, koi distortion nahi.
Convert. $tan \frac{ν}{2} = 1 \cdot tan 2 0^{\circ} \Rightarrow \frac{ν}{2} = 2 0^{\circ} \Rightarrow ν = 4 0^{\circ}$ . Yeh step kyun? Factor of 1 matlab equation degenerate hokar $ν = E$ ban jaati hai.
Radius. $r = 1 \cdot (1 - 0 \cdot cos 4 0^{\circ}) = 1$ . Yeh step kyun? Constant radius circle ki definition hai — yeh ek accha self-consistency check hai.

Verify: $ν = E = 4 0^{\circ}$ aur $r = a = 1$ . Bilkul wahi circle sanity anchor jo parent note mein tha (wahan $ν = E$ mean anomaly ke bhi barabar hoti hai, lekin yahan us baat ki zaroorat nahi). ✓

Worked example Example 2 — Cell B: quadrant I, perihelion ke paas

Statement. $e = 0.5$ , $E = 6 0^{\circ}$ . $ν$ , $cos ν$ , $sin ν$ , $r$ nikalo.

Forecast: body helper circle ke ek-tihaayi raaste par hai, perihelion side mein. Kya expect karte ho — $ν > E$ ya $ν < E$ ?

Stretch factor. $\frac{1.5}{0.5} = 3 \approx 1.7320$ . Yeh step kyun? $e = 0.5$ focus ko perihelion ki taraf kaafi dur kheenchta hai (offset $c = a e = 0.5$ ), isliye perihelion ke paas ke angles magnify hote hain.
Half-angle convert. $tan \frac{ν}{2} = 3 tan 3 0^{\circ} = 3 \cdot \frac{1}{3} = 1 \Rightarrow \frac{ν}{2} = 4 5^{\circ} \Rightarrow ν = 9 0^{\circ}$ . Yeh step kyun? $tan 3 0^{\circ} = 1/ 3$ conveniently $3$ ko cancel kar deta hai — ek clean landmark case.
Component cross-check. $cos ν = \frac{cos 6 0 ^{\circ} - 0.5}{1 - 0.5 cos 6 0 ^{\circ}} = \frac{0.5 - 0.5}{1 - 0.25} = 0$ , aur $sin ν = \frac{1 - 0.25 sin 6 0 ^{\circ}}{0.75} = \frac{0.8660 \cdot 0.8660}{0.75} = + 1$ . Yeh step kyun? $(cos ν, sin ν) = (0, + 1)$ $ν = 9 0^{\circ}$ ko quadrant I/II boundary par unambiguously pin karta hai.
Radius. $r = 1 (1 - 0.5 cos 6 0^{\circ}) = 1 - 0.25 = 0.75$ . Yeh step kyun? Direct formula $r = a (1 - e cos E)$ focus-to-body distance deta hai bina $ν$ ki zaroorat ke — same $E$ ka ek sasta, singularity-free by-product.

Verify: $ν = 9 0^{\circ} > E = 6 0^{\circ}$ — perihelion-side stretching confirmed, bilkul forecast ki tarah. ✓

Worked example Example 3 — Cell C: quadrant II, abhi bhi ascending

Statement. $e = 0.6$ , $E = 12 0^{\circ}$ . $ν$ nikalo.

Forecast: minor axis ke paas par abhi aphelion nahi aaya. Ascending hai, toh $sin ν$ positive rehna chahiye. $ν$ bada hoga ya $E$ se chhota?

Stretch factor. $\frac{1.6}{0.4} = 4 = 2$ . Yeh step kyun? Factor $(1 + e) / (1 - e)$ focus offset $c = a e = 0.6$ ko encode karta hai; pehle isko compute karne se pata chalta hai ki trig touch karne se pehle yeh orbit angles kitna magnify karti hai.
Half-angle. $tan \frac{ν}{2} = 2 tan 6 0^{\circ} = 23 \approx 3.4641 \Rightarrow \frac{ν}{2} = arctan (3.4641) = 73.89 8^{\circ} \Rightarrow ν = 147.8 0^{\circ}$ . Yeh step kyun? $tan (E /2) = tan 6 0^{\circ} = 3$ positive hai aur $E /2 < 9 0^{\circ}$ , isliye $ν /2$ ek plain first-quadrant arctan hai — koi sign problem nahi.
Component cross-check. $cos ν = \frac{cos 12 0 ^{\circ} - 0.6}{1 - 0.6 cos 12 0 ^{\circ}} = \frac{- 0.5 - 0.6}{1 + 0.3} = \frac{- 1.1}{1.3} = - 0.8462$ ; $sin ν = \frac{0.8 \cdot 0.8660}{1.3} = + 0.5329$ . Yeh step kyun? $cos ν < 0$ , $sin ν > 0$ ⇒ quadrant II ⇒ $ν = 18 0^{\circ} - 32. 2^{\circ} = 147. 8^{\circ}$ . Match ho gaya.

Verify: $arccos (- 0.8462) = 147.8 0^{\circ}$ aur $sin > 0$ ise quadrant II mein rakha. ✓ Abhi bhi $ν > E$ , distortion aphelion ke paas aate-aate fade ho rahi hai.

Worked example Example 4 — Cell D: aphelion,

$E = 18 0^{\circ}$ (woh $\infty$ trap) Statement. $e = 0.6$ , $E = 18 0^{\circ}$ . $ν$ aur $r$ nikalo.

Forecast: body sabse door waale end par hai. $tan (E /2) = tan 9 0^{\circ}$ kya hoga? Kya formula toot jaayega?

Singularity spot karo. $tan \frac{E}{2} = tan 9 0^{\circ} = + \infty$ . Yeh step kyun? Yahi woh jagah hai jahan students panic karte hain. Half-angle form mein yahan ek removable blow-up hai.
Limit lo. $tan \frac{ν}{2} = 2 \cdot (+ \infty) = + \infty \Rightarrow \frac{ν}{2} = 9 0^{\circ} \Rightarrow ν = 18 0^{\circ}$ . Yeh step kyun? Dono sides $+ \infty$ par saath-saath jaati hain, aur $arctan (+ \infty) = 9 0^{\circ}$ . Toh $ν = E = 18 0^{\circ}$ aphelion par exactly — woh ek jagah perihelion ke alaawa jahan dono angles milte hain.
Component sanity (koi infinities nahi). $cos ν = \frac{- 1 - 0.6}{1 - 0.6 ( - 1 )} = \frac{- 1.6}{1.6} = - 1$ , $sin ν = \frac{0.8 \cdot 0}{1.6} = 0$ . Toh $(cos ν, sin ν) = (- 1, 0) \Rightarrow ν = 18 0^{\circ}$ . Yeh step kyun? Component form mein koi singularity nahi hai, isliye yeh limit confirm karne ka safe tarika hai.
Radius (aphelion distance). $r = 1 (1 - 0.6 cos 18 0^{\circ}) = 1 + 0.6 = 1.6 = a (1 + e)$ . Yeh step kyun? Aphelion par $cos E = - 1$ , toh $r = a (1 + e)$ sabse badi possible distance hai — ek memorable landmark value jo poore orbit ki size scale ko anchor karti hai.

Verify: aphelion radius $a (1 + e) = 1.6$ ✓, aur $ν = 18 0^{\circ}$ dono methods se match karta hai. " $\infty$ " harmless nikla. ✓

Worked example Example 5 — Cell E: quadrant III, descending sign flip

Statement. $e = 0.6$ , $E = 24 0^{\circ}$ . $ν$ nikalo.

Forecast: body ab perihelion ki taraf wapas aa rahi hai. Uska $sin ν$ negative hona chahiye. Kya ek $arccos$ -only method yeh sahi dega? (Nahi — yahi toh point hai.)

Half-angle, honestly. $E /2 = 12 0^{\circ}$ , $tan 12 0^{\circ} = - 3$ . Toh $tan \frac{ν}{2} = 2 (- 3) = - 23 \approx - 3.4641$ . Yeh step kyun? $E /2 = 12 0^{\circ}$ quadrant II mein hai jahan tangent negative hota hai — sign real kaam kar raha hai.
Carefully invert karo. $\frac{ν}{2} = arctan (- 3.4641)$ . Kyunki $E /2 = 12 0^{\circ}$ $(9 0^{\circ}, 18 0^{\circ})$ mein hai, humein $ν /2$ usi half-turn mein chahiye: $ν /2 = 18 0^{\circ} - 73.89 8^{\circ} = 106.10 2^{\circ} \Rightarrow ν = 212.2 0^{\circ}$ . Yeh step kyun? $arctan$ $- 73.89 8^{\circ}$ return karta hai (ek quadrant-IV value). Kyunki $E /2$ quadrant II mein tha aur $tan (ν /2)$ monotonically $tan (E /2)$ ko follow karta hai, $ν /2$ ko correct branch mein rakhne ke liye $18 0^{\circ}$ add karo.
Component cross-check (asli safety net). $cos ν = \frac{cos 24 0 ^{\circ} - 0.6}{1 - 0.6 cos 24 0 ^{\circ}} = \frac{- 0.5 - 0.6}{1 - 0.6 ( - 0.5 )} = \frac{- 1.1}{1.3} = - 0.8462$ ; $sin ν = \frac{0.8 sin 24 0 ^{\circ}}{1.3} = \frac{0.8 ( - 0.8660 )}{1.3} = - 0.5329$ . Yeh step kyun? $cos < 0$ , $sin < 0$ ⇒ quadrant III (map figure par labelled Q III cell). $atan2 (- 0.5329, - 0.8462) = - 147.8 0^{\circ} \equiv 212.2 0^{\circ}$ . Match ho gaya.

Verify: $36 0^{\circ} - 212.2 0^{\circ} = 147.8 0^{\circ}$ — Example 3 ka mirror image, bilkul waise jaise symmetry demand karti hai (same $∣ E - 18 0^{\circ} ∣$ ). ✓ Descending half ko sign chahiye; sirf $arccos$ use karne se galat $147. 8^{\circ}$ aata.

Worked example Example 6 — Cell F: high eccentricity,

$e = 0.9$ Statement. Ek comet jisme $e = 0.9$ hai, uski $E = 3 0^{\circ}$ hai. $ν$ aur $r$ nikalo.

Forecast: focus almost edge par hone se, perihelion ke paas ke angles bahut zyada magnify hote hain. Expect karo $ν ≫ E$ .

Stretch factor. $\frac{1.9}{0.1} = 19 \approx 4.3589$ . Yeh step kyun? Jaise $e \to 1$ , offset $c = a e \to a$ (focus almost ellipse edge par) aur factor $(1 + e) / (1 - e) \to \infty$ : extreme perihelion whip.
Half-angle. $tan \frac{ν}{2} = 4.3589 tan 1 5^{\circ} = 4.3589 \cdot 0.26795 = 1.1680 \Rightarrow \frac{ν}{2} = 49.4 4^{\circ} \Rightarrow ν = 98.8 8^{\circ}$ . Yeh step kyun? Sirf $3 0^{\circ}$ ke helper angle se $9 9^{\circ}$ ke paas ka real angle aata hai — comet Sun ke paas lunge karta hua lagta hai.
Radius. $r = 1 (1 - 0.9 cos 3 0^{\circ}) = 1 - 0.9 (0.86603) = 1 - 0.77942 = 0.22058$ . Yeh step kyun? Bahut chhota $r$ (perihelion $a (1 - e) = 0.1$ se compare karo) confirm karta hai ki comet gravitational well mein deep hai, jahan speed aur angular rate sabse zyada hai.

Verify: component check $cos ν = \frac{0.86603 - 0.9}{1 - 0.9 ( 0.86603 )} = \frac{- 0.03397}{0.22058} = - 0.1540$ ; $arccos (- 0.1540) = 98.8 6^{\circ}$ (rounding) aur $sin ν > 0$ . ✓ $ν ≫ E$ forecast ki tarah. ✓

Worked example Example 7 — Cell G: reverse —

$ν$ diya, $E$ nikalo Statement. Ek satellite ka $e = 0.4$ aur true anomaly $ν = 12 0^{\circ}$ hai. Uski eccentric anomaly $E$ nikalo.

Forecast: hum formula ulta chalate hain. Kyunki $ν$ (focus se) perihelion side par $E$ (center se) se aage hota hai, yahan $E < ν$ expect karo.

Stretch invert karo. $tan \frac{E}{2} = \frac{1 - e}{1 + e} tan \frac{ν}{2}$ . Yeh step kyun? $tan \frac{ν}{2} = k tan \frac{E}{2}$ ko $tan \frac{E}{2}$ ke liye solve karna bas $k$ se divide karna hai, yaani root ke andar fraction ulta karna.
Numbers. $\frac{0.6}{1.4} = 0.42857 = 0.65465$ ; $tan 6 0^{\circ} = 1.73205$ . Toh $tan \frac{E}{2} = 0.65465 \cdot 1.73205 = 1.13389 \Rightarrow \frac{E}{2} = 48.59 6^{\circ} \Rightarrow E = 97.1 9^{\circ}$ . Yeh step kyun? Dono $ν /2$ aur $E /2$ quadrant I mein hain, isliye plain arctan branch sahi hai — koi sign patching nahi chahiye.
Forward-check. $E = 97.1 9^{\circ}$ ko aage plug karo: $tan \frac{ν}{2} = 1.4/0.6 tan 48.59 6^{\circ} = 1.52753 \cdot 1.13389 = 1.73206 \Rightarrow ν /2 = 60.0 0^{\circ} \Rightarrow ν = 12 0^{\circ}$ . Yeh step kyun? Round-tripping inversion confirm karta hai.

Verify: $E = 97.1 9^{\circ} < ν = 12 0^{\circ}$ ✓ — focus-offset lead, bilkul forecast ki tarah. (Aage $E$ ko Kepler's equation mein feed karoge, Kepler's Equation, mean anomaly aur time pane ke liye.)

Worked example Example 8 — Cell H: word problem, radius → time link

Statement. Ek orbit ka semi-major axis $a = 10, 000$ km aur $e = 0.25$ hai. Ek certain instant par body ki eccentric anomaly $E = 10 8^{\circ}$ hai. (a) Woh focus se kitni door hai? (b) Focus par baitha observer kaunsa true anomaly measure karta hai? (c) Perihelion ke baad se orbital period ka kitna fraction guzra hai? (Kepler's Equation se $M = E - e sin E$ aur Mean anomaly and time se $M = \frac{2 π}{T} (t - t_{p})$ use karo.)

Forecast: $E = 10 8^{\circ}$ par hum minor axis se thoda aage hain, toh $r$ thoda $a$ se zyada hona chahiye, $ν$ thoda $E$ se zyada, aur — kyunki yahan $M < E$ — period ka $10 8^{\circ} /36 0^{\circ}$ se thoda kam guzra hona chahiye.

$E$ se Radius. $r = a (1 - e cos E) = 10000 (1 - 0.25 cos 10 8^{\circ})$ . $cos 10 8^{\circ} = - 0.30902$ , toh $r = 10000 (1 + 0.077254) = 10772.5$ km. Yeh step kyun? $r = a (1 - e cos E)$ direct, singularity-free distance formula hai — ise sirf $E$ chahiye, jo humhare paas already hai, isliye yeh sabse sasta quantity hai pehle nail karne ke liye.
True anomaly. $\frac{1.25}{0.75} = 1.66667 = 1.29099$ ; $tan 5 4^{\circ} = 1.37638$ . $tan \frac{ν}{2} = 1.29099 \cdot 1.37638 = 1.77690 \Rightarrow ν /2 = 60.63 8^{\circ} \Rightarrow ν = 121.2 8^{\circ}$ . Yeh step kyun? Standard forward conversion; $E /2 = 5 4^{\circ}$ quadrant I mein hai toh koi sign fixing nahi.
$ν$ se Radius (independent route). $r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν}$ use karo. $cos ν = cos 121.2 8^{\circ} = - 0.51840$ . Denominator $1 + 0.25 (- 0.51840) = 0.87040$ . Numerator $10000 (1 - 0.0625) = 9375$ . $r = 9375/0.87040 = 10770.3$ km. Yeh step kyun? Do alag anomalies se do alag formulas ek hi physical distance dene chahiye — ek strong consistency test (chhota gap rounding ka hai).
Mean anomaly se Time. $M = E - e sin E$ mein $E$ radians mein: $E = 10 8^{\circ} = 1.88496$ rad, $sin 10 8^{\circ} = 0.95106$ . $M = 1.88496 - 0.25 (0.95106) = 1.88496 - 0.23776 = 1.64720$ rad $= 94.3 8^{\circ}$ . Elapsed fraction $= M / (2 π) = 1.64720/6.28319 = 0.26216$ . Yeh step kyun? $M$ time mein linear angle hai, isliye $M / (2 π) = (t - t_{p}) / T$ directly elapsed period ka fraction hai — yahi $M \to E \to ν$ chain ka poora point hai, reporting ke liye ulta chalaya.

Verify: dono radius routes $r \approx 10.77 \times 1 0^{3}$ km dete hain ✓; $ν = 121.2 8^{\circ} > E = 10 8^{\circ}$ ✓; elapsed fraction $0.2622 < 108/360 = 0.30$ ✓ (forecast ki tarah, $M < E$ ). Units: $r$ ke liye km, time ke liye dimensionless fraction. ✓

Worked example Example 9 — Cell I: quadrant IV,

$arccos$ -only trap expose karo Statement. Ek student compute karta hai, $e = 0.3$ , $E = 30 0^{\circ}$ ke liye: " $cos ν = \frac{cos 30 0 ^{\circ} - 0.3}{1 - 0.3 cos 30 0 ^{\circ}} = \frac{0.5 - 0.3}{1 - 0.15} = \frac{0.2}{0.85} = 0.23529$ , toh $ν = arccos (0.23529) = 76.3 9^{\circ}$ ." Kya yeh sahi hai? Agar nahi, toh sahi $ν$ batao.

Forecast: $E = 30 0^{\circ}$ fourth quadrant mein hai — body perihelion ki taraf descend kar rahi hai. Descending body ka $ν$ $(18 0^{\circ}, 36 0^{\circ})$ mein hona chahiye. Kya $76.3 9^{\circ}$ sahi ho sakta hai?

$sin ν$ ka asli sign nikalo. $sin ν = \frac{1 - 0.09 sin 30 0 ^{\circ}}{1 - 0.15} = \frac{0.95394 ( - 0.86603 )}{0.85} = - 0.97186$ . Yeh step kyun? $arccos$ sirf $[0, 18 0^{\circ}]$ return karta hai, isliye akela woh orbit ke upper half ko lower half se distinguish nahi kar sakta. $sin ν$ ka sign woh ek extra fact hai jo $ν$ ko $36 0^{\circ} - ν$ se distinguish karta hai; yahi woh information hai jo student ne throw away ki.
Quadrant correct karo. $cos ν = + 0.23529 (> 0)$ aur $sin ν = - 0.97186 (< 0)$ ⇒ quadrant IV (map figure par labelled Q IV cell). Toh $ν = 36 0^{\circ} - 76.3 9^{\circ} = 283.6 1^{\circ}$ . Yeh step kyun? Positive cosine, negative sine unambiguously quadrant IV hai; student ka $76.3 9^{\circ}$ uska quadrant-I mirror hai — orbit ka galat half.
Half-angle confirmation. $E /2 = 15 0^{\circ}$ , $tan 15 0^{\circ} = - 0.57735$ . $tan \frac{ν}{2} = 1.3/0.7 (- 0.57735) = 1.36277 (- 0.57735) = - 0.78678$ . Kyunki $E /2 \in (9 0^{\circ}, 18 0^{\circ})$ , $ν /2 = 18 0^{\circ} - 38.1 9^{\circ} = 141.8 1^{\circ} \Rightarrow ν = 283.6 1^{\circ}$ lo. Yeh step kyun? Half-angle form, correct branch ke saath padha, component method se agree karna chahiye — aur karta hai, toh hum error do independent tareekon se pakad chuke hain.

Verify: $atan2 (- 0.97186, 0.23529) = - 76.3 9^{\circ} \equiv 283.6 1^{\circ}$ ✓. Student ka $76.3 9^{\circ}$ galat quadrant mein tha; sahi answer $ν = 283.6 1^{\circ}$ hai. ✓

Recall Matrix par quick self-test

$e = 0.6, E = 24 0^{\circ}$ kaun sa cell hai, aur uska quadrant fix karne wala ek fact kya hai? ::: Cell E (quadrant III); $sin ν < 0$ $ν$ ko lower half mein force karta hai, $ν = 212.2 0^{\circ}$ deta hai. $E = 18 0^{\circ}$ par half-angle form $tan (ν /2) = \infty$ deta hai — $ν$ kya hai? ::: $ν = 18 0^{\circ}$ exactly (dono angles aphelion par milte hain); $(cos ν, sin ν) = (- 1, 0)$ se confirm karo. $ν$ diya ho toh $tan (E /2)$ kaise nikalte hain? ::: $tan \frac{E}{2} = \frac{1 - e}{1 + e} tan \frac{ν}{2}$ (root ke andar fraction ulta karo). Offset $c = a e$ physically kya represent karta hai? ::: Center-to-focus distance; yahi akela $E$ -vs- $ν$ difference ki wajah hai, aur $e = 0$ hone par yeh khatam ho jaata hai.