Question bank — Kepler's equation M = E − e·sin E — derivation, eccentric anomaly

False. $M$ ek kalpnik uniformly-moving body ka angle hai, $M=n(t-t_p)$; Sun se real angle true anomaly $\nu$ hai. Dono sirf perihelion aur aphelion par milte hain.

True. Perihelion par $E=0$, isliye $\sin E=0$ aur $M=E-e\sin E=0$; aur $\nu =0$ bhi same perihelion axis ke along point karta hai. Teeno turning points par agree karte hain.

False. $E$ ko ellipse ke centre $O$ se radius $a$ ke auxiliary circle par measure kiya jaata hai. Centre se measure karna hi woh clean $x=a\cos E,y=b\sin E$ squash exact rakhta hai.

True. Correction directly $e$ ke proportional hai, isliye bada $e$ matlab planet ki speed zyada vary karti hai, aur uniform clock $M$ aur geometric angle $E$ ke beech ka gap barhta hai.

True. $e=0$ se $e\sin E$ term khatam ho jaata hai, isliye $M=E$; motion uniform hai aur mean, eccentric, aur true anomalies sab milte hain — dekho ellipse geometry.

False. $e>1$ ke liye geometry hyperbolic form $M=e\sinh F-F$ use karti hai hyperbolic sine aur eccentric anomaly $F$ ke saath; ordinary $\sin$ aur circle sirf ellipses ($0\le e<1$) par apply hote hain.

True. Wahan $\cos E=-1$ se $r=a(1+e)$ milta hai, maximum distance. $E=0$ par, $\cos E=1$ se $r=a(1-e)$ milta hai, minimum (perihelion).

False. $M$ ke equal increments matlab equal time aur equal swept area (Kepler's 2nd law), lekin $\nu$ perihelion ke paas fast jump karta hai aur aphelion ke paas slow — dekho Kepler's Second Law — equal areas in equal times.

False. Yeh equation areas/arcs se derive hui hai jahan angle radians mein construction se aata hai. $E$ term ek raw angle-as-arc hai, sirf $\sin$-only quantity nahi, isliye degrees isko tod dete hain. Sirf radians kaam karte hain.

Rearranged form mein abhi bhi $E$ dono sides par hai — yeh ek fixed-point/transcendental equation hai, closed formula nahi. Tumhe iterate karna hoga (fixed-point ya Newton–Raphson).

$y$-coordinate $b/a$ se squash hota hai: woh $y=b\sin E$ hai, $a\sin E$ nahi. $a\sin E$ use karna auxiliary circle par point $Q$ describe karta hai, ellipse par planet $P$ ko nahi.

Focus $S$, centre $O$ se $ae$ doori par hai, $a$ nahi. Woh $ae$ base ($O$ se $S$ tak) hi exactly $\frac{1}{2}{a}^{2}e\sin E$ triangle area generate karta hai, aur isliye $e\sin E$ term aata hai.

Sign galat hai. Correct $E$-form $r=a(1-e\cos E)$ hai minus ke saath. Check karo: $E=0$ par perihelion $a(1-e)$ milna chahiye; plus wala version galat tarike se $a(1+e)$ deta hai, jo aphelion value hai.

Small-to-moderate $e$ ke liye, natural seed $E_0=M$ hai, kyunki jab $e$ chhota ho toh $E\approx M$. $0$ se start karna iterations waste karta hai; $M$ sabse fast converge karta hai.

Equal areas in equal times ka matlab equal arc length in equal times nahi hota. Perihelion ke paas radius chhoti hoti hai, isliye planet ko same area sweep karne ke liye faster move karna padta hai; speed constant nahi hai.

Auxiliary circle ellipse ko radius $a$ (semi-major axis) ke saath circumscribe karta hai. Us radius-$a$ circle ko $b/a$ se squash karne par ellipse banta hai.

Kyunki ellipse exactly ek circle hai jo centre ke baare mein vertically $b/a$ se squash hua hai, isliye centre se measure karna clean parametrization $x=a\cos E,y=b\sin E$ rakhta hai. Focus offset $ae$ phir alag handle hota hai aur $e\sin E$ term ban jaata hai.

Auxiliary circle par tum triangle $OSQ$ (corners centre $O$, Sun $S$, circle-point $Q$; base $OS=ae$, height $a\sin E$) ko circular sector se subtract karte ho, jisse $\frac{1}{2}{a}^{2}e\sin E$ area milta hai. Ellipse area paane ke liye pure circular figure ko vertical squash factor $b/a$ se multiply karna us term ko $-e\sin E$ ke roop mein carry karta hai — planet ke Sun ke paas fast chalne ka physical correction.

$E$ linearly aur $\sin E$ ke andar dono jagah appear karta hai, ek algebraic aur ek transcendental term ko mix karta hai — elementary functions ka koi finite combination unhe unmix nahi kar sakta, isliye hume numerically solve karna padta hai.

Kyunki $\nu$ time ke saath non-uniformly change karta hai, lekin $M=n(t-t_p)$ ek perfectly linear clock hai. Phir Kepler's equation easy clock $M$ ko geometric angle $E$ se bridge karta hai, jisse $\nu$ aur $r$ follow karte hain.

Bada $e$ matlab $-e\sin E$ correction bada hota hai aur motion zyada non-uniform hoti hai, isliye $E$, $M$ se aur door ho jaata hai; starting guess $E_0=M$ worse hoti hai aur iterations zyada slowly converge karti hain.

Mean motion $n$ set karta hai ki $M$ kitna fast tick karta hai; Kepler's third law ke zariye, $n=2\pi /T=\sqrt{\mu /a^3}$ jahan $\mu =GM$ gravitational parameter hai, isliye orbit ka size $a$ aur central mass akele clock rate fix karte hain. Yeh $M=n(t-t_p)$ ko physical time se link karta hai.

Kyunki ellipse par har $y$-coordinate auxiliary circle par corresponding $y$ ka $b/a$ times hota hai, aur area us same vertical factor se scale hoti hai. Yeh messy elliptical region ko ek easy circular sector minus triangle mein convert kar deta hai.

Yeh $M=E$ mein collapse ho jaata hai, uniform circular motion describe karta hai; ellipse circle ban jaati hai aur teeno anomalies milti hain.

Perihelion distance $a(1-e)$ $0$ ki taraf shrink karta hai jabki aphelion $a(1+e)$ $2a$ ki taraf grow karta hai; orbit extremely elongated ho jaata hai aur perihelion aur aphelion ke beech speed contrast extreme ho jaata hai.

Kyunki $\sin 0=\sin \pi =0$, correction term dono perihelion aur aphelion turning points par vanish ho jaata hai, isliye clock aur geometry momentarily agree karte hain.

Pehle $M$ ko $[0,2\pi )$ mein modulo $2\pi$ reduce karo, kyunki geometry har revolution ke baad repeat hoti hai; phir us principal range mein $E$ solve karo.

$M=n(t-t_p)$ negative ho jaata hai, odd symmetry $\sin (-E)=-\sin E$ se negative $E$ deta hai; planet simply perihelion se door jaane ki jagah uski taraf approach kar raha hota hai.

Nahi — $e=1$ ek parabolic escape trajectory hai, bound ellipse nahi, isliye circular auxiliary construction aur $M=E-e\sin E$ ab apply nahi hote; ek alag parabolic (Barker) treatment chahiye hota hai, aur $e>1$ hyperbolic form use karta hai.

$0\le e<1$ ke liye derivative $1-e\cos E$ hamesha strictly positive hota hai (kabhi zero nahi), isliye iteration well-behaved hai; seed $E_0=M$ ke saath yeh reliably converge karta hai, sirf $e$ ke $1$ ke paas jaane par slow hota hai.