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

3.2.14 · D2 · Physics › Orbital Mechanics & Astrodynamics › Kepler's equation M = E − e·sin E — derivation, eccentric an

Hum bilkul zero se shuru karte hain: do shapes, ek squash, aur ek scoop of area.

Step 1 — Ellipse banao aur uski bones ko naam do

KYA. Ek ellipse ek oval hota hai. Uski ek lambi half-width aur ek choti half-width hoti hai. Hum unhe naam dete hain:

$a$ = semi-major axis = sabse lambe diameter ka aadha (lambi half-width).
$b$ = semi-minor axis = sabse chote diameter ka aadha (choti half-width).
$O$ = centre, oval ka bilkul beech ka point.

YEH PEHLE KYUN. Har cheez aage chalke $a$ aur $b$ ke against measure hoti hai. Jab tak do half-widths compare karne ke liye na hon, tab tak "oval kitna squashed hai" — yeh bolna possible nahi.

PICTURE. Figure dekho: lambi green arrow $a$ hai, choti yellow arrow $b$ hai, dono centre $O$ (white dot) se shuru hoti hain. Real ellipse ke liye $a > b$ hamesha hota hai — yahi cheez isse oval banati hai, circle nahi.

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

Step 2 — Sun ko focus par rakho, centre se $a e$ door

KYA. Sun centre par nahi hai. Woh ek special point par baitha hai jise focus $S$ kehte hain, jo long axis ke saath centre se exactly $a e$ shift hua hai.

KYUN. Kepler's First Law kehta hai ki planets orbit karte hain Sun ko focus par rakhke. Yeh off-centre shift $a e$ hi pura reason hai ki motion lopsided kyun lagti hai — $S$ ke paas planet close hota hai, $S$ se door hone par door hota hai. Length $a e$ ko yaad rakho: yeh $e sin E$ term ke roop mein wapas aayega.

PICTURE. Red dot Sun $S$ hai. $O$ se $S$ tak ka red segment length $a e$ ka hai. Notice karo ki $a e$ badha jaata hai jab $e$ badha — rounder orbit ( $e$ chota) mein Sun almost centre par hota hai; stretched orbit mein Sun zyada off-centre hota hai.

Step 3 — Ellipse ke around radius $a$ ka ek circle banao (auxiliary circle)

KYA. $O$ ko centre rakhke $a$ radius ka ek pura circle banao. Iske left/right edges ellipse ke tips ko just touch karte hain; top/bottom ellipse ke upar bulge karta hai. Yahi auxiliary circle hai.

YEH CIRCLE HI KYUN, koi aur nahi. Kyunki ek ellipse literally ek dabaya hua circle hota hai. Agar tum auxiliary circle ka har point ka height $b / a$ factor se daba do, tum exactly ellipse par pahunch jaate ho. Hum apni geometry friendly circle par karenge, phir answer ko squash karenge.

PICTURE. Blue circle = auxiliary circle. Grey ellipse uske andar baitha hai, left aur right tips par touch karta hua. Do vertical green arrows squash dikhate hain: ek tall circle-height $a sin E$ ek choti ellipse-height $b sin E$ ban jaati hai.

Step 4 — Eccentric anomaly $E$ define karo (helper angle)

KYA. Ellipse par planet ki position $P$ pick karo. $P$ se seedha upar (ya neeche) jao jab tak auxiliary circle par point $Q$ na mile. Centre $O$ se, perihelion direction $X$ se line $O Q$ tak ka angle, eccentric anomaly $E$ kehlaata hai.

CENTRE SE KYUN, Sun se nahi. Squash factor $b / a$ sirf centre $O$ ke baare mein exact hai. Wahan $E$ measure karo toh coordinates clean rehte hain: $Q = (a cos E, a sin E), P = (a cos E, b sin E) .$ Dono ka same $x = a cos E$ hai ( $Q$ se $P$ tak sirf vertically gaye); sirf height squash hui.

PICTURE. $Q$ (blue dot) circle par, $P$ (yellow dot) directly neeche ellipse par. Angle $E$ $O$ par blue wedge hai. Dashed vertical line $Q$ ko $P$ se jodti hai — same $x$ , chota $y$ .

Step 5 — Kepler's Second Law: area hi clock hai

KYA. Kepler's Second Law kehta hai ki Sun se planet tak ki line equal areas in equal times sweep karti hai. Toh perihelion se ab tak sweep kiya gaya orbit ke total area ka fraction = elapsed period ka fraction.

ANGLE NAHI, AREA KYUN. Planet ki speed badlti hai, isliye Sun se angle steadily nahi badha. Lekin prateek second mein sweep kiya gaya area ek fixed constant hai. Area honest clock hai. Hum "time" ko "area" mein convert karte hain, aur area geometrically calculate kar sakte hain.

PICTURE. Sun ke paas aur door se do equal-area red slivers. Sun ke paas planet ek wide-but-short wedge quickly cover karta hai; door se ek thin-but-long wedge slowly. Same area, same time.

Step 6 — Swept area ko circle par wapas squash karo

KYA. Swept area $S X P$ ellipse par hai aur awkward hai. Squash undo karo: har ellipse region, circle par matching region ke $\frac{b}{a}$ times ke barabar hai.

KYUN. Hum jaante hain circle par areas kaise measure karte hain (sectors aur triangles). Toh $S X P$ ko circle par push karo, wahan measure karo, phir wapas aane ke liye $b / a$ se multiply karo.

PICTURE. Yellow ellipse region $S X P$ aur uska taller circle twin $S X Q$ directly uske upar. Yeh exactly ek number se related hain: $Area (S X P) = \frac{b}{a} Area (S X Q)$ .

Step 7 — Circle region ko ek sector minus ek triangle mein kato

KYA. Circle par, region $S X Q$ (Sun se, perihelion ke saath, $Q$ tak) seedha split hota hai: $Area (S X Q) = centre se fan sector O X Q - Sun offset se kata wedge triangle O S Q .$

YEH DO PIECE KYUN. Fan $O X Q$ ek plain circular slice hai — easy. Lekin hum woh region chahte the jo Sun $S$ se measure ho, centre $O$ se nahi. Sun $a e$ se offset hai (Step 2), isliye hum woh chota triangle $O S Q$ subtract karte hain jo offset create karta hai. Yahi subtraction hai jahan $a e$ — aur isliye $e sin E$ — enter karta hai.

PICTURE. Blue fan = sector $O X Q$ opening angle $E$ ke saath. Red triangle = $O S Q$ , base $O S = a e$ axis ke saath aur height $a sin E$ ( $Q$ ki height). Jo area chahiye woh fan minus triangle hai.

Step 8 — Assemble karo: Kepler's equation appear hoti hai

KYA. Circle area ko $b / a$ se multiply karo (Step 6) ellipse area paane ke liye, phir isse area-fraction / mean-anomaly relation mein daalo (Step 5).

KYUN. Yahi woh moment hai jab sab pieces click karte hain: geometry ka $\frac{1}{2} a^{2}$ total area $π ab$ ke saath cancel ho jaata hai, aur naked $(E - e sin E)$ khada reh jaata hai.

$Area (S X P) = \frac{b}{a} \cdot \frac{1}{2} a^{2} (E - e sin E) = \frac{1}{2} ab (E - e sin E)$

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

PICTURE. Ek cancellation cartoon: numerator mein $ab$ denominator ke $ab$ ko kill karta hai, $2 π / (π) \cdot \frac{1}{2} = 1$ , boxed result bacha rehta hai.

Step 8b — Usi picture se distance $r$ nikalo

KYA. Hamare paas pehle se planet ke coordinates $P = (a cos E, b sin E)$ aur Sun $S = (a e, 0)$ par hain. Distance $r$ (Step 2 mein define kiya) Pythagorean theorem se segment $S P$ ki length hi hai.

YAHAN KYUN. $r$ neeche edge-case examples mein aur summary mein use hoti hai, isliye hum ise quote karne ki bajaye ab earn karte hain. Isme sirf ek line algebra lagti hai.

$r^{2} = (a cos E - a e)^{2} + (b sin E)^{2}$

$b^{2} = a^{2} (1 - e^{2})$ (Step 1) substitute karo, expand karo, aur $cos^{2} E + sin^{2} E = 1$ use karo:

$r^{2} = a^{2} (cos E - e)^{2} + a^{2} (1 - e^{2}) sin^{2} E = a^{2} (1 - e cos E)^{2}$

Step 9 — Edge cases (koi scenario kabhi mat chodo)

KYA AUR KYUN. Ek derivation jis par trust kar sako use apni extremes survive karni chahiye. Turning points aur degenerate circle check karo.

PICTURE. Teen snapshots: perihelion ( $E = 0$ ), aphelion ( $E = π$ ), aur ek perfect circle ( $e = 0$ ).

Ek-picture summary

Sab kuch ek canvas par: ellipse, uska auxiliary circle, planet $P$ aur uska shadow $Q$ , Sun offset $a e$ , swept area, aur woh sector-minus-triangle jo $E - e sin E$ ban jaata hai.

Recall Feynman retelling — walkthrough plain words mein

Ek oval par planet Sun ke paas speed up karta hai aur door slow ho jaata hai, isliye "3 baje woh kahan hai?" ka koi aasaan jawab nahi. Hamara trick: planet ka real angle bhool jao aur area ko clock banao, kyunki area steady rate se sweep hota hai. Woh swept area measure karne ke liye hum cheat karte hain — flat oval ko fair circle mein wapas blow up karte hain apni heights un-squash karke (ek factor $b / a$ ). Circle par, perihelion se hamare helper point tak ka area sirf ek pizza slice hota hai (sector, worth $\frac{1}{2} a^{2} E$ ) jisme ek triangular bite remove ki gayi hai (kyunki Sun $a e$ off-centre baitha hai, woh bite $\frac{1}{2} a^{2} e sin E$ hai). Slice minus bite hai $\frac{1}{2} a^{2} (E - e sin E)$ . Ise wapas squash karo, poore oval ke area se divide karo, $2 π$ se scale karo, aur clean bookkeeping exactly $M = E - e sin E$ chhod jaati hai. $E$ pizza-slice angle hai; $- e sin E$ off-centre bite hai. Tips par bite zero hoti hai (clock = geometry); ek circle ke liye koi bite hi nahi. Aur Sun se distance $r$ sirf planet tak line ki length hai, jo same coordinates se hume $r = a (1 - e cos E)$ de deta hai.

Recall Cloze self-test

Area clock hai kyunki Kepler's Second Law (equal areas in equal times).
Hum geometry auxiliary circle (radius $a$ ) par karte hain phir == $b / a$ == se squash karke ellipse par aate hain.
Sector $O X Q$ ka area == $\frac{1}{2} a^{2} E$ == hai aur $E$ radians mein chahiye.
Triangle $O S Q$ ki base == $a e$ aur height $a sin E$ hai, area $\frac{1}{2} a^{2} e sin E$ ==.
$r$ Sun-to-planet distance hai, jo == $a (1 - e cos E)$ == ke barabar hai.
Ek orbit mein dono $E$ aur $M$ == $0 \leq \cdot < 2 π$ == se guzarte hain.
Result: $M === E - e sin E ==$ aur $r === a (1 - e cos E) ==$ .
Related: $M \to E$ solve karne ke liye Newton–Raphson chahiye; parabola/hyperbola cousin hai hyperbolic Kepler equation; $E \to ν$ convert karne ke liye true anomaly use hoti hai.