Exercises — True anomaly from eccentric anomaly

• $\nu$ (true anomaly) focus par measure hota hai — yeh woh real angle hai jo tum Sun se dekhoge.
• $E$ (eccentric anomaly) ellipse ke center par, auxiliary circle par measure hota hai.
• $M$ (mean anomaly) geometric angle nahi hai; yeh time ke saath linearly badhta hai (dekho Mean anomaly and time).

$e=0$ ke saath: $r=a(1-0)=a,\cos \nu =\frac{\cos E}{1}=\cos E,\sin \nu =\frac{1\cdot \sin E}{1}=\sin E,\tan \frac{\nu }{2}=\sqrt{\frac{1}{1}}\tan \frac{E}{2}=\tan \frac{E}{2}.$ Har equation kehti hai $\nu =E$. Ek circle ka focus center par hi hota hai, isliye dono angles ek ho jaate hain. Yeh tumhara sanity anchor hai baaki saari problems ke liye.

• $E=0$: $\cos 0=1$, to $r=a(1-e)=a-ae=a-c$ — sabse chhhoti distance (perihelion).
• $E=\pi$: $\cos \pi =-1$, to $r=a(1+e)=a+ae=a+c$ — sabse badi distance (aphelion). To perihelion, $a$ se $c$ andar hai aur aphelion, $a$ se $c$ bahar hai: focus offset $c=ae$ literally half-difference $(r_{\text{apo}}-r_{\text{peri}})/2=ae=c$ hai. Body perihelion par sabse paas hoti hai, aphelion par sabse door.

Figure s01 (neeche): $e=0.6$ ellipse black mein draw ki gayi hai jisme center $O$ (black dot) aur focus $F$ (ek red dot — $F$ key object hai, isliye sirf yahi accent red mein draw kiya gaya hai). $F$ se perihelion point aur aphelion point tak do red arrows hain, labeled $r=a(1-e)$ aur $r=a(1+e)$; ye red arrows woh "accent-red features" hain jinhe text point karta hai. Black center dot aur red focus dot ke beech ka gap focus offset $c=ae$ hai.

$\tan (E/2)=\tan 45^{\circ }=1$. Stretch factor: $\sqrt{\frac{1+0.6}{1-0.6}}=\sqrt{\frac{1.6}{0.4}}=\sqrt{4}=2.$ To $\tan (\nu /2)=2\cdot 1=2$, jisse $\nu /2=\arctan 2=63.43^{\circ }$, isliye $\nu =126.87^{\circ }.$ Note karo $\nu >E$: offset focus se dekha jaaye to body halfway mark se pehle hi nikal chuki hai — perihelion-side stretching in action.

$\cos \nu =\frac{\cos 90^{\circ }-0.6}{1-0.6\cos 90^{\circ }}=\frac{-0.6}{1}=-\mathrm{0.6.}$ $\sin \nu =\frac{\sqrt{1-0.36}\sin 90^{\circ }}{1}=\sqrt{0.64}=+\mathrm{0.8.}$ $\cos \nu <0$ aur $\sin \nu >0$ ⟹ second quadrant. $\arctan \frac{0.8}{-0.6}$ ko Q2 mein resolve karne par $\nu =126.87^{\circ }$ ✓ — L2.1 jaisa hi.

$r=a(1-e\cos E)=2(1-0.6\cos 90^{\circ })=2(1-0)=2\text{ AU}.$ $E=90^{\circ }$ par $\cos E$ term zero ho jaata hai, isliye $r=a$ hota hai chahe $e$ kuch bhi ho — ek handy checkpoint.

$E$ ke liye half-angle formula invert karo: $\tan \frac{E}{2}=\sqrt{\frac{1-e}{1+e}}\tan \frac{\nu }{2}.$ $\tan (\nu /2)=\tan 60^{\circ }=1.7320508$. Factor $\sqrt{0.7/1.3}=\sqrt{0.5384615}=0.7337994$. $\tan \frac{E}{2}=0.7337994\times 1.7320508=\mathrm{1.2709803.}$ $E/2=\arctan 1.2709803=51.798^{\circ }$, to $E=103.597^{\circ }.$ Sanity: yahan $E<\nu$ hai (reverse stretch), jo perihelion-approaching side par body ke saath consistent hai.

L2.1 ke saath symmetry se, yeh axis ke neeche mirror image hai. Signed components use karo: $\cos \nu =\frac{\cos 270^{\circ }-0.6}{1-0.6\cos 270^{\circ }}=\frac{0-0.6}{1-0}=-0.6,$ $\sin \nu =\frac{\sqrt{0.64}\sin 270^{\circ }}{1}=\frac{0.8\cdot (-1)}{1}=-\mathrm{0.8.}$ $\cos \nu <0$, $\sin \nu <0$ ⟹ third quadrant: $\nu =233.13^{\circ }$ (equivalently $-126.87^{\circ }$). Ab half-angle check. Hum $E=270^{\circ }$ lete hain, to $E/2=135^{\circ }$ aur $\tan (E/2)=\tan 135^{\circ }=-1$, jisse $\tan (\nu /2)=2\times (-1)=-2$ milta hai. Equation $\tan (\nu /2)=-2$ ke do candidate half-angles hain: principal $\arctan (-2)=-63.43^{\circ }$ aur shifted $-63.43^{\circ }+180^{\circ }=116.57^{\circ }$. Kyun hume $116.57^{\circ }$ choose karna hai na ki $-63.43^{\circ }$: half-angle $\nu /2$ ko usi half-turn mein rehna chahiye jisme $E/2$ hai. Yahan $E/2=135^{\circ }$, $(90^{\circ },180^{\circ })$ mein hai, isliye $\nu /2$ bhi $(90^{\circ },180^{\circ })$ mein hona chahiye — yeh $116.57^{\circ }$ branch force karta hai, negative wala nahi. Double karne par $\nu =233.13^{\circ }$ ✓ milta hai, jo signed-component answer se match karta hai. Yahi woh jagah hai jahan naive $\arccos$ fail hota hai — woh $126.87^{\circ }$ return karta, sign kho deta.

$E=10^{\circ }$ par: $\tan (E/2)=\tan 5^{\circ }=0.0874886$. $\tan (\nu /2)=2\times 0.0874886=0.1749773$, to $\nu /2=9.923^{\circ }$, $\nu =19.846^{\circ }$. Gap $\nu -E=+9.85^{\circ }$. $E=170^{\circ }$ par: $\tan (E/2)=\tan 85^{\circ }=11.4300523$. $\tan (\nu /2)=2\times 11.4300523=22.8601$, to $\nu /2=87.494^{\circ }$, $\nu =174.988^{\circ }$. Gap $\nu -E=+4.99^{\circ }$. Perihelion ke paas (chhhota $E$) body focus ke paas hoti hai, isliye ek chhhota $E$-step ek bada $\nu$ sweep karta hai — stretching strong hai, bada gap. Aphelion ke paas (bada $E$) body door hoti hai, same $E$-step ek chhhota $\nu$ sweep karta hai — chhhota gap. $\sqrt{(1+e)/(1-e)}$ factor is focus-offset distortion ko encode karta hai.

Figure s02 (neeche): horizontal axis $E$ hai $0^{\circ }$ se $180^{\circ }$ tak; vertical axis gap $\nu -E$ degrees mein hai. Ek red curve (key object hone ki wajah se sirf yahi red element hai) $e=0.6$ ke liye gap hai; yeh left edge par perihelion se steeply uthti hai aur right edge par aphelion ki taraf gently wapas aati hai. Do black dots un $E=10^{\circ }$ aur $E=170^{\circ }$ points ko mark karte hain jo tumne abhi compute kiye — yeh woh "black dots" hain jinhe text refer karta hai.

$\nu -E$, $E=0$ par $0$ hai aur $E=\pi$ par phir $0$ (dono angles saath $0$ aur $\pi$ hit karte hain). Dono ends par $0$ hona aur beech mein positive rehna, matlab yeh kahi beech mein peak karta hai — yeh figure s02 mein red curve ka crest hai. $E=90^{\circ }$ check karo: $\nu =126.87^{\circ }$ (L2.1), gap $=36.87^{\circ }$. $E=60^{\circ }$ check karo: $\tan (30^{\circ })=0.5773503$, $\tan (\nu /2)=1.1547005$, $\nu /2=49.107^{\circ }$, $\nu =98.21^{\circ }$, gap $=38.21^{\circ }$. To maximum thoda $90^{\circ }$ se pehle baithta hai (is $e$ ke liye $E\approx 60^{\circ }$ ke aaspaas), perihelion-leaning side par — consistent hai kyunki body abhi bhi kaafi paas hai tab strongest stretching hoti hai.

Yeh Mean anomaly and time → Kepler's Equation → is note ko chain karta hai. $E=73.90^{\circ }$ ke saath: $\tan (E/2)=\tan 36.95^{\circ }=0.7522$. Stretch factor $\sqrt{1.5/0.5}=\sqrt{3}=1.7320508$. $\tan \frac{\nu }{2}=1.7320508\times 0.7522=\mathrm{1.3028.}$ $\nu /2=\arctan 1.3028=52.49^{\circ }$, to $\nu =104.98^{\circ }.$ Expected hai ki $M(45^{\circ })<E(73.9^{\circ })<\nu (105^{\circ })$: har successive angle perihelion side ki taraf "stretched" hai. Outgoing perihelion side par yeh ordering $M<E<\nu$ ek great final sanity check hai.

Direct: $r=a(1-e\cos E)=4(1-0.5\cos 73.90^{\circ })=4(1-0.5\times 0.27731)=4\times 0.86134=3.4454\text{ AU}$. Cross-check via true anomaly: $\cos \nu =\cos 104.98^{\circ }=-0.25834$. $r=\frac{4(1-0.25)}{1+0.5\times (-0.25834)}=\frac{3.0}{0.87083}=3.4454\text{ AU}.✓$ Dono routes agree karte hain — dono radius formulas same physics hain, bas alag kapde mein.

Step 1 — square karo aur half-angle identity se rewrite karo. Di gayi relation ko square karo: ${\tan }^2\frac{E}{2}=\frac{1-e}{1+e}{\tan }^2\frac{\nu }{2}.$ Ab Half-angle trigonometric identities, ${\tan }^2\frac{\theta }{2}=\frac{1-\cos \theta }{1+\cos \theta }$, dono sides par apply karo: $\frac{1-\cos E}{1+\cos E}=\frac{1-e}{1+e}\cdot \frac{1-\cos \nu }{1+\cos \nu }.$ Step 2 — clutter kam karne ke liye unknowns ka naam rakho. $u=\cos E$ (jo chahiye) aur $w=\cos \nu$ (known) lo. Equation hai: $\frac{1-u}{1+u}=\frac{(1-e)(1-w)}{(1+e)(1+w)}.$ Step 3 — cross-multiply karo. Dono denominators clear karo: $(1-u)(1+e)(1+w)=(1+u)(1-e)(1-w).$ Step 4 — har side expand karo. Pehle $e,w$ brackets multiply out karo: $(1+e)(1+w)=1+e+w+ew,(1-e)(1-w)=1-e-w+ew.$ To equation yeh ban jaati hai: $(1-u)(1+e+w+ew)=(1+u)(1-e-w+ew).$ Step 5 — $u$ terms ek side, constants doosri side. Expand karo: $\underset{\text{const}}{\underbrace{(1+e+w+ew)}}-u(1+e+w+ew)=\underset{\text{const}}{\underbrace{(1-e-w+ew)}}+u(1-e-w+ew).$ Saare $u$-terms right, saare constants left move karo: $(1+e+w+ew)-(1-e-w+ew)=u[(1-e-w+ew)+(1+e+w+ew)].$ Step 6 — har bracket simplify karo. Left side par $1$'s aur $ew$'s cancel hote hain, $2e+2w$ bachta hai. Right side par $e,w$ terms cancel hote hain, $2+2ew$ bachta hai: $2e+2w=u(2+2ew).$ Step 7 — 2 se divide karo aur $u$ solve karo. Factor $2$ cancel karo: $e+w=u(1+ew)\Rightarrow u=\frac{w+e}{1+ew}.$ Restoring names: $\boxed{\cos E=\frac{\cos \nu +e}{1+e\cos \nu }}.$ Sign flip ka "kyun": forward $\cos \nu =\frac{\cos E-e}{1-e\cos E}$ ke compare mein, inverse mein $-e$ ki jagah $+e$ hai. Center-view se focus-view jaane par perihelion kick subtract hota hai; wapas jaane par add hota hai — offset direction reverse hota hai, aur yahi Steps 5–6 ki algebra ne produce kiya jab $ew$ terms cancel hue.

$e\to 1^{-}$ hone par, $1-e\to 0^{+}$ to factor $\sqrt{\frac{1+e}{1-e}}\to +\infty$. Tab kisi bhi fixed $E>0$ ke liye $\tan \frac{\nu }{2}\to \infty$, yaani $\nu \to 180^{\circ }$. Physically: near-parabolic orbit enormously elongated hoti hai, focus extremely offset hota hai, isliye perihelion passage ka almost poora hissa $\nu$ ko rapidly $180^{\circ }$ ki taraf sweep karta hai jabki $E$ barely move karta hai. Eccentric-anomaly picture (auxiliary circle) degenerate ho jaati hai jaise ellipse unbounded stretch hoti hai — yahi wajah hai ki parabolic orbits mein ek alag parameter use hota hai (ek alag note).

$\sin \nu =\frac{\sqrt{1-e^2}\sin E}{1-e\cos E}$ se: prefactor $\frac{\sqrt{1-e^2}}{1-e\cos E}$ strictly positive hai (numerator $>0$ $e<1$ ke liye; denominator $1-e\cos E\ge 1-e>0$). To $\sin \nu$ exactly $\sin E$ ka sign carry karta hai — woh kabhi disagree nahi karte. Yahi wajah hai ki upper/lower half preserve hoti hai aur half-angle formula consistent rehta hai. Numeric at $E=120^{\circ }$, $e=0.4$: ratio $=\frac{\sqrt{1-0.16}}{1-0.4\cos 120^{\circ }}=\frac{\sqrt{0.84}}{1-0.4\times (-0.5)}=\frac{0.916515}{1.2}=0.763763$.

• Main $r=a(1-e\cos E)$ aur dono $\cos \nu ,\sin \nu$ formulas memory se bol sakta hoon.
• Main full orbit ke liye atan2 / half-angle use karta hoon, bare arccos nahi.
• Main $\nu \to E$ invert karte waqt stretch factor flip karta hoon.
• Main $\cos E(\nu )$ inverse mein $e$ ka sign flip karta hoon.
• Main $M\to E\to \nu \to r$ chain early rounding ke bina complete kar sakta hoon, last step tak 5+ significant figures rakhta hoon.