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

3.2.14 · Physics › Orbital Mechanics & Astrodynamics

YEH ZAROORI KYU HAI?

Ek circle ke liye position trivial hai: angle $= ω t$ . Lekin ek ellipse ke liye, planet equal time mein equal areas sweep karti hai, equal angles nahi. Jo angle aap actually chahte ho (the true anomaly $ν$ , Sun se measure kiya hua) woh time ke saath non-uniformly badalta hai. Isse suljhaane ke liye hume teen angles ki ek chain chahiye.

ISKO KAISE BANATE HAIN: auxiliary circle

Ellipse banao (semi-major $a$ , semi-minor $b$ , eccentricity $e$ ). Uske around ellipse centre $O$ par radius $a$ ka ek circle banao. Sun focus $S$ par hai, $O$ se $a e$ door.

Ellipse par planet ke point $P$ ke liye, $P$ ke through ek vertical line girao; woh circle par point $Q$ par milti hai. Angle $∠ QO X$ (centre se, $X$ = perihelion direction) hi eccentric anomaly $E$ hai.

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

Ellipse ki magical property: woh ek circle hai jo $y$ direction mein $b / a$ factor se squash hua hai. Toh agar circle par $Q = (a cos E, a sin E)$ hai, toh planet ki position hai $x = a cos E, y = b sin E .$

DERIVATION 1 — Area se Kepler's equation (first principles)

Kepler's 2nd law: focus se radius constant rate se area sweep karta hai. Toh elapsed orbital period ka fraction, perihelion ke baad swept ellipse area ke fraction ke barabar hota hai:

$\frac{t - t _{p}}{T} = \frac{Area swept S X P}{total ellipse area π ab} .$

Left side ko $2 π$ se multiply karo, yeh by definition $M$ hai: $M = \frac{2\pi}{T}(t-t_p) = \frac{2\pi \cdot (\text{Area } SXP)}{\pi a b}. \tag{1}$

Step: Area $S X P$ compute karo. Squash use karo. Ellipse par har $y$ -coordinate, circle ke corresponding $y$ ka $b / a$ times hota hai. Area bhi usi tarah scale hoti hai, toh ellipse par koi bhi region $= \frac{b}{a} \times$ (circle par uska image region).

Yeh step kyun? Yeh ek messy elliptical area ko circular-sector area mein convert kar deta hai, jo simple geometry se compute ho sakta hai.

$Area S X P_{ellipse} = \frac{b}{a} Area S X Q_{circle} .$

Step: Circle par Area $S X Q$ = (circular sector $O X Q$ ) − (triangle $O S Q$ ).

Kyun? $S X Q$ woh sector hai jo perihelion se centre ke around sweep hua, minus woh wedge jo focus offset se cut off hota hai.

Sector $O X Q$ : radius $a$ ke circle ka ek slice, angle $E$ : area $= \frac{1}{2} a^{2} E$ .
Triangle $O S Q$ : base $O S = a e$ axis ke along, height $= y_{Q} = a sin E$ : $area = \frac{1}{2} (a e) (a sin E) = \frac{1}{2} a^{2} e sin E .$

Toh $Area S X Q_{circle} = \frac{1}{2} a^{2} E - \frac{1}{2} a^{2} e sin E = \frac{1}{2} a^{2} (E - e sin E) .$

Step: Ellipse mein convert karo aur (1) mein plug karo. $Area S X P = \frac{b}{a} \cdot \frac{1}{2} a^{2} (E - e sin E) = \frac{1}{2} ab (E - e sin E) .$

$M = \frac{2 π}{π ab} \cdot \frac{1}{2} ab (E - e sin E) .$

DERIVATION 2 — $E$ ke terms mein radius aur true anomaly

Focus $S = (a e, 0)$ se planet $P = (a cos E, b sin E)$ tak: $r^{2} = (a cos E - a e)^{2} + (b sin E)^{2} .$ $b^{2} = a^{2} (1 - e^{2})$ use karo: $r^{2} = a^{2} (cos E - e)^{2} + a^{2} (1 - e^{2}) sin^{2} E .$ Expand karo aur $cos^{2} E + sin^{2} E = 1$ use karo: $r^{2} = a^{2} [cos^{2} E - 2 e cos E + e^{2} + sin^{2} E - e^{2} sin^{2} E] = a^{2} (1 - e cos E)^{2} .$ $r = a (1 - e cos E)$ Yeh useful kyun hai? Jab ek baar $E$ solve kar lo, toh distance $r$ turant mil jaati hai, aur components $x = a cos E, y = b sin E$ se poori position milti hai. True anomaly is se milta hai: $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} .$

ACTUALLY SOLVE KAISE KARTE HAIN (yeh transcendental hai!)

$M = E - e sin E$ ko algebraically $E (M)$ ke liye invert nahi kar sakte. Hum iterate karte hain (Newton–Raphson): $E_{k + 1} = E_{k} - \frac{E _{k} - e s i n E _{k} - M}{1 - e c o s E _{k}}, E_{0} = M .$

Common mistakes (Steel-manned)

Quick recall

Recall Cloze self-test

$M === E - e sin E ==$
$M === n (t - t_{p}) ==$ , with $n === μ / a^{3} ==$
$r === a (1 - e cos E) ==$
Position: $x === a cos E ==, y === b sin E ==$
$E$ ko centre se auxiliary circle par, radians mein measure kiya jaata hai.

Recall Feynman: 12 saal ke bacche ko explain karo

Socho ek runner ek oval track par daud raha hai, lekin ek focus par ek magnet hai jo use paas hone par sprint karta hai aur door hone par crawl. Predict karna mushkil hai woh kahan hai! Toh hum cheat karte hain: hum imagine karte hain ek robot jo ek perfect circle par steady pace se daud raha hai, aur ek clock jo bas evenly tick karta hai — woh hai "mean" angle $M$ . Phir hum ek clever shadow trick use karte hain: runner ki oval ko ek bade circle ke andar rakhte hain aur unka "shadow angle" $E$ beech se draw karte hain. Kepler's equation, $M = E - e sin E$ , woh secret decoder ring hai jo steady robot ki clock ko real runner ke shadow position mein convert karti hai. $e sin E$ wala part bas "kitna sprint kiya ya crawl kiya" ki correction hai.

Flashcards

Mean anomaly

M

physically kya hai?

Ek fictitious body ka angle jo constant angular speed se chalta hai; ek rescaled clock

M = n (t - t_{p})

, real planet ki position nahi.

Kepler's equation batao.

M = E - e sin E

(angles radians mein).

Eccentric anomaly

E

kahan se aur kis point se measure hota hai?

Ellipse ke centre se, radius

a

ke auxiliary circle par jo ellipse ko circumscribe karta hai.

Kepler's equation ko

E

ke liye algebraically kyun solve nahi kar sakte?

Yeh transcendental hai —

E

linearly bhi aata hai aur

sin E

ke andar bhi; numerically solve karo (Newton/fixed-point).

E

ke liye Newton iteration do.

E_{k + 1} = E_{k} - \frac{E _{k} - e sin E _{k} - M}{1 - e cos E _{k}}

, start

E_{0} = M

E

ke terms mein radius?

r = a (1 - e cos E)

E

ke terms mein planet ke

x, y

kya hain?

x = a cos E, y = b sin E

(centre origin), ellipse = circle squashed by

b / a

use karke.

M

E

, aur

ν

kahan coincide karte hain?

Perihelion (

= 0

) aur aphelion (

= π

) par, jahan

sin E = 0

Mean motion

n

kya hai?

n = 2 π / T = μ / a^{3}

E

ko true anomaly

ν

mein convert karo.

tan (ν /2) = \frac{1 + e}{1 - e} tan (E /2)

e sin E

term geometrically kyun aata hai?

Yeh triangle

O S Q

ka area hai (focus offset

a e

times height

a sin E

) jo circular sector mein se subtract hota hai jab swept area compute karte hain.

Connections

True anomaly ν and the orbit equation r = a(1−e²)/(1+e cos ν)
Kepler's Second Law — equal areas in equal times
Kepler's Third Law — T² ∝ a³ and mean motion n
Eccentricity and ellipse geometry (a, b, ae, b²=a²(1−e²))
Numerical root-finding — Newton–Raphson
Time of flight and orbit propagation
Hyperbolic Kepler equation M = e·sinh F − F