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

3.2.14 · D4 · Physics › Orbital Mechanics & Astrodynamics › Kepler's equation M = E − e·sin E — derivation, eccentric an

Saare angles radians mein hain, jab tak problem kuch aur na kahe. Kuch recurring facts jinpar hum rely karte hain: $M = E - e sin E, M = n (t - t_{p}), n = \frac{2 π}{T} = \frac{μ}{a ^{3}}, r = a (1 - e cos E) .$

Definition True-anomaly half-angle bridge

Parent note mein bataya gaya hai, aur D2 mein derive kiya gaya hai, woh identity jo geometric angle $E$ ko Sun se dekhe jaane wale physical angle $ν$ se jorti hai: $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} .$ Yeh kahan se aata hai (sketch): $x = a cos E$ , $y = b sin E$ use karo, focus $(a e, 0)$ par hai, toh milta hai $cos ν = \frac{cos E - e}{1 - e cos E}$ aur $sin ν = \frac{1 - e ^{2} sin E}{1 - e cos E}$ ; inhe half-angle identity $tan (ν /2) = \frac{sin ν}{1 + cos ν}$ mein daalo aur simplify karo, toh boxed form mil jaata hai. Hum ise ek given tool ki tarah use karte hain yahan. Dekho True anomaly ν and the orbit equation r = a(1−e²)/(1+e cos ν). Half-angle form kyun? Kyunki $tan (ν /2)$ ek poore revolution $ν \in (- π, π)$ par monotonic hota hai, isliye yeh woh quadrant ambiguity door kar deta hai jo $tan ν$ mein hoti hai — $E$ aur $ν$ hamesha same half-turn mein hote hain, jo hume branch cleanly choose karne deta hai.

Level 1 — Recognition

Ye check karte hain ki tum har symbol ko naam de aur padh sako.

L1.1

$M$ , $E$ , aur $ν$ physically kya hain, aur inhe kis point se measure kiya jaata hai — yeh ek-ek sentence mein batao.

Recall Solution

$M$ (mean anomaly): ek fictitious body ka angle jo constant angular speed se move karta hai. Yeh koi real geometric angle nahi hai — yeh ek rescaled clock hai, $M = n (t - t_{p})$ , jahan $t_{p}$ perihelion passage ka time hai.
$E$ (eccentric anomaly): ek geometric helper angle jo ellipse ke centre $O$ se measure hota hai, radius $a$ ke auxiliary circle par.
$ν$ (true anomaly): planet ki real angular position, focus $S$ (Sun) se measured.

L1.2

Kuch solve kiye bina batao: orbit par kin do points par $M = E$ exactly hota hai? Kyun?

Recall Solution

Perihelion ( $E = 0$ ) aur aphelion ( $E = π$ ) par. Wahan $sin E = 0$ hota hai, isliye correction term $e sin E$ disappear ho jaata hai aur $M = E - 0 = E$ . Physically, yeh woh do turning points hain jahan fictitious uniform body aur geometric angle momentarily line up kar jaate hain.

L1.3

Ek orbit mein $e = 0$ hai. Kepler's equation kya ban jaata hai, aur woh motion ke baare mein kya kehta hai?

Recall Solution

$e = 0$ ke saath: $M = E - 0 \cdot sin E = E$ . Ellipse ek circle hai, $r = a (1 - 0) = a$ (constant radius), aur planet uniform angular speed se move karta hai — teeno anomalies ek saath coincide karti hain.

Level 2 — Application

Formulas mein plug karo, dono directions mein.

L2.1 (easy direction: $E \to M \to t$ )

Diya hai $E = π /3$ , $e = 0.4$ , period $T = 200$ days. $M$ aur perihelion ke baad ka time $t - t_{p}$ nikalo.

Recall Solution

$M = E - e sin E = \frac{π}{3} - 0.4 sin \frac{π}{3} = 1.047198 - 0.4 (0.866025) = 1.047198 - 0.346410 = 0.700788 rad .$ Phir $t - t_{p} = \frac{M}{2 π} T = \frac{0.700788}{6.283185} \times 200 = 22.306$ days.

L2.2 (radius from $E$ )

Usi orbit ke liye ( $e = 0.4$ ) semi-major axis $a = 1.5 \times 1 0^{8}$ km aur $E = π /3$ ke saath, distance $r$ nikalo.

Recall Solution

$r = a (1 - e cos E) = 1.5 \times 1 0^{8} (1 - 0.4 cos \frac{π}{3}) = 1.5 \times 1 0^{8} (1 - 0.4 \times 0.5) = 1.5 \times 1 0^{8} (0.8) = 1.2 \times 1 0^{8} km .$

L2.3 (hard direction: one Newton step)

$e = 0.3$ , $M = 0.8$ rad ke liye, $E_{0} = M = 0.8$ se shuru karke ek Newton–Raphson step karo. $E_{1}$ report karo.

Recall Solution

Newton step: $E_{1} = E_{0} - \frac{E _{0} - e sin E _{0} - M}{1 - e cos E _{0}}$ . (Neeche saare trig values 6 decimal places tak carry kiye gaye hain; last digit mein koi choti rounding isi se aati hai.)

$f (E_{0}) = 0.8 - 0.3 sin (0.8) - 0.8 = - 0.3 \times 0.717356 = - 0.215207.$
$f^{'} (E_{0}) = 1 - 0.3 cos (0.8) = 1 - 0.3 \times 0.696707 = 0.790988.$
$E_{1} = 0.8 - \frac{- 0.215207}{0.790988} = 0.8 + 0.272073 = 1.072073.$

Level 3 — Analysis

Ab explain karo kyun, aur signs aur quadrants ko probe karo.

L3.1 (all quadrants — correction ka sign)

$e = 0.5$ ke liye, $E = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ (ek each quadrant mein) ke liye $M$ compute karo. Kin quadrants mein $M < E$ hai, aur kin mein $M > E$ ? Physically explain karo.

Figure ko kaise padhe. Horizontal axis eccentric anomaly $E$ hai (ek poora revolution, $0$ se $2 π$ tak); vertical axis mean anomaly $M$ hai. Dashed slate line reference $M = E$ hai (woh jo bina correction ke milta). Coral curve actual $M = E - 0.5 sin E$ hai. Jahan coral curve dashed line ke neeche dips karti hai, $M < E$ hai (lavender shading, outbound half $0 < E < π$ ); jahan upar jaati hai, $M > E$ hai (mint shading, inbound half $π < E < 2 π$ ). Chaar butter dots woh sample points hain jo tum compute karte ho.

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

Recall Solution

$M = E - 0.5 sin E$ .

$E = \frac{π}{4} = 0.785398$ : $sin E = + 0.707107$ , $M = 0.785398 - 0.353553 = 0.431845$ . $M < E$ .
$E = \frac{3 π}{4} = 2.356194$ : $sin E = + 0.707107$ , $M = 2.356194 - 0.353553 = 2.002641$ . $M < E$ .
$E = \frac{5 π}{4} = 3.926991$ : $sin E = - 0.707107$ , $M = 3.926991 + 0.353553 = 4.280544$ . $M > E$ .
$E = \frac{7 π}{4} = 5.497787$ : $sin E = - 0.707107$ , $M = 5.497787 + 0.353553 = 5.851340$ . $M > E$ .

Pattern: jahan $sin E > 0$ hai (upper half, quadrants I–II, perihelion se outbound leg), $M < E$ ; jahan $sin E < 0$ hai (lower half, quadrants III–IV, perihelion ki taraf inbound), $M > E$ . Physically: bahar jaate waqt planet perihelion se sprint kar chuka hota hai, isliye real geometry ( $E$ ) uniform clock ( $M$ ) se aage ho jaati hai; wapas aate waqt woh lag karta hai aur clock pakad leta hai. Figure mein coral curve left half par dashed $M = E$ line ke neeche hoti hai (lavender) aur right half par upar (mint).

L3.2 (monotonicity — Newton kabhi zero se divide kyun nahi karta)

Dikhao ki $\frac{d M}{d E} = 1 - e cos E > 0$ sabhi $E$ ke liye hota hai jab $0 \leq e < 1$ . Yeh kyun guarantee karta hai ki har $M$ ke liye ek unique $E$ hai, aur Newton ka denominator safe kyun hai?

Recall Solution

$M = E - e sin E$ ko $E$ ke saath differentiate karo: $\frac{d M}{d E} = 1 - e cos E$ . Kyunki $- 1 \leq cos E \leq 1$ aur $0 \leq e < 1$ hai, toh $e cos E \leq e < 1$ , isliye $1 - e cos E \geq 1 - e > 0$ . Yeh har jagah strictly positive hai.

Uniqueness: $M (E)$ strictly increasing hai, isliye one-to-one hai — har $M$ exactly ek $E$ par map hota hai.
Newton safety: Newton step mein denominator $f^{'} (E) = 1 - e cos E$ hai, exactly yahi quantity, jo $1 - e > 0$ se neeche bounded hai. Yeh kabhi zero nahi ho sakti, isliye koi division-by-zero blowup nahi hoga.

L3.3 (limiting behaviour — high eccentricity)

Jaise $e \to 1^{-}$ , perihelion ( $E = 0$ ) ke paas $d M / d E$ ka kya hota hai? Iska kya matlab hai ki wahan clock ke saath $E$ kitni tezi se badhta hai?

Recall Solution

$E = 0$ par: $d M / d E = 1 - e cos 0 = 1 - e$ . Jaise $e \to 1^{-}$ , yeh $\to 0^{+}$ ho jaata hai. Kyunki $d E / d M = 1/ (1 - e cos E)$ hai, perihelion ke paas $d E / d M = 1/ (1 - e) \to \infty$ ho jaata hai. Toh clock $M$ mein ek choti si change $E$ mein bahut badi change produce karta hai: near-parabolic orbits mein planet perihelion se bahut tezi se guzarta hai. Yahi wajah hai ki high- $e$ cases numerically stiff hote hain aur $E_{0} = M$ se better seed ki zaroorat hoti hai.

Level 4 — Synthesis

Kai relations ko ek saath chain mein jodo.

L4.1 ( $t \to M \to E \to r \to ν$ , full propagation)

Ek orbit mein $e = 0.2$ , $a = 2$ AU, $T = 1000$ days hai. $t - t_{p} = 100$ days par, (a) $M$ , (b) $E$ (Newton se 5 dp tak), (c) $r$ , (d) true anomaly $ν$ nikalo.

Recall Solution

(a) Clock $\to$ mean anomaly. $n = \frac{2 π}{T} = \frac{6.283185}{1000} = 0.00628319$ rad/day. $M = n (t - t_{p}) = 0.00628319 \times 100 = 0.628319 rad .$ (b) Newton se Kepler solve karo, $E_{0} = M = 0.628319$ , $e = 0.2$ . Stopping rule: ruko jab $∣ E_{k + 1} - E_{k} ∣ < 1 0^{- 5}$ , yaani 5th decimal settle ho jaye.

$f_{0} = 0.628319 - 0.2 sin (0.628319) - 0.628319 = - 0.2 \times 0.587785 = - 0.117557$ ; $f_{0}^{'} = 1 - 0.2 cos (0.628319) = 1 - 0.2 \times 0.809017 = 0.838197$ . $E_{1} = 0.628319 + 0.140250 = 0.768569$ . (step size $0.140250$ )
$f_{1} = 0.768569 - 0.2 sin (0.768569) - 0.628319 = 0.768569 - 0.138931 - 0.628319 = 0.001319$ ; $f_{1}^{'} = 1 - 0.2 cos (0.768569) = 0.857157$ . $E_{2} = 0.768569 - 0.001539 = 0.767030$ . (step size $0.001539$ )
$f_{2} = 0.767030 - 0.2 sin (0.767030) - 0.628319 = 0.767030 - 0.138710 - 0.628319 = 0.000001$ ; $f_{2}^{'} = 0.857005$ . $E_{3} = 0.767030 - 0.0000012 = 0.767029$ . (step size $\approx 1.2 \times 1 0^{- 6} < 1 0^{- 5}$ → ruko.)
Converged: $E \approx 0.767029$ rad.

(c) Radius. $r = a (1 - e cos E) = 2 (1 - 0.2 cos 0.767029) = 2 (1 - 0.2 \times 0.719963) = 2 (0.856007) = 1.712014 AU .$

(d) True anomaly. $tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} = \frac{1.2}{0.8} tan (0.383515) = 1.224745 \times 0.403553 = 0.494292.$ $\frac{ν}{2} = arctan (0.494292) = 0.457546$ , toh $ν = 0.915093 rad \approx 52.4 3^{\circ} .$ Quadrant check: yahan $E = 0.767$ pehle half-turn mein hai ( $0 < E < π$ ), aur half-angle form guarantee karta hai ki $ν$ bhi usi half-turn mein hoga, isliye $ν /2 \in (0, π /2)$ aur principal $arctan$ branch sahi hai — koi $π$ add karne ki zaroorat nahi. (Rule of thumb: agar $E \in (π, 2 π)$ hai, toh doubling ke baad $ν$ mein $2 π$ add karo, ya equivalently $ν /2$ ko matching half mein lo taaki $ν$ aur $E$ ek hi revolution share karein.)

L4.2 (period unknown — Kepler's 3rd law combine karna)

Ek planet ek star orbit karta hai jiska mass $μ = 1.327 \times 1 0^{20} m^{3} / s^{2}$ (Sun) deta hai, $a = 1.496 \times 1 0^{11}$ m ke saath. $E = 1.2$ rad aur $e = 0.3$ par, perihelion ke baad kitne days beete hain?

Recall Solution

Kepler's 3rd law se mean motion: $n = \frac{μ}{a ^{3}}$ . $a^{3} = (1.496 \times 1 0^{11})^{3} = 3.348071 \times 1 0^{33} m^{3}, \frac{μ}{a ^{3}} = \frac{1.327 \times 1 0 ^{20}}{3.348071 \times 1 0 ^{33}} = 3.963779 \times 1 0^{- 14} s^{- 2} .$ $n = 3.963779 \times 1 0^{- 14} = 1.990924 \times 1 0^{- 7} rad/s .$ Mean anomaly: $M = E - e sin E = 1.2 - 0.3 sin (1.2) = 1.2 - 0.3 \times 0.932039 = 1.2 - 0.279612 = 0.920388$ rad. Time: $t - t_{p} = \frac{M}{n} = \frac{0.920388}{1.990924 \times 1 0 ^{- 7}} = 4.622882 \times 1 0^{6} s .$ Days mein: $\frac{4.622882 \times 1 0 ^{6}}{86400} = 53.506$ days.

Level 5 — Mastery

Edge cases, stiff numerics, aur reverse engineering.

L5.1 (high eccentricity — bad seed vs good seed)

$e = 0.9$ , $M = 0.3$ rad ke liye Kepler's equation iterate karo. (a) $E_{0} = M$ se do fixed-point iterations $E_{k + 1} = M + e sin E_{k}$ karo. (b) $E_{0} = π$ se do Newton iterations karo. True root $E \approx 1.16227$ ki taraf kaun faster converge karta hai?

Recall Solution

(a) Fixed point, $E_{0} = 0.3$ :

$E_{1} = 0.3 + 0.9 sin (0.3) = 0.3 + 0.9 \times 0.295520 = 0.565968.$
$E_{2} = 0.3 + 0.9 sin (0.565968) = 0.3 + 0.9 \times 0.535896 = 0.782306.$ Abhi bhi $1.16227$ se bahut door hai — fixed point rengta hai kyunki $e$ bada hai (iska convergence factor $\approx e cos E$ hai, $0.9$ ke paas).

(b) Newton, $E_{0} = π = 3.141593$ :

$f_{0} = 3.141593 - 0.9 sin (3.141593) - 0.3 = 3.141593 - 0.9 \times 2.65 \times 1 0^{- 6} - 0.3 = 2.841590$ ; $f_{0}^{'} = 1 - 0.9 cos (3.141593) = 1 + 0.9 = 1.900000$ . $E_{1} = 3.141593 - 1.495574 = 1.646019.$
$f_{1} = 1.646019 - 0.9 sin (1.646019) - 0.3 = 1.646019 - 0.9 \times 0.997191 - 0.3 = 0.448547$ ; $f_{1}^{'} = 1 - 0.9 cos (1.646019) = 1 - 0.9 \times (- 0.074937) = 1.067443$ . $E_{2} = 1.646019 - 0.420210 = 1.225809.$ Do Newton steps ke baad root ke $0.06$ ke andar aa gaye, jabki fixed-point abhi bhi $0.78$ ke paas atka hua hai. Newton clearly jeet jaata hai large $e$ ke liye; parent ka default seed $E_{0} = M$ yahan kharaab hai, aur $E_{0} = π$ (ya $M + e sign$ ) ek common robust choice hai.

L5.2 (degenerate case — $e$ just below 1, perihelion speed)

$e = 0.99$ ke liye, perihelion ( $E = 0$ ) aur aphelion ( $E = π$ ) par $d E / d M$ compare karo. Perihelion par eccentric anomaly clock ke saath kitne guna faster respond karta hai?

Recall Solution

$\frac{d E}{d M} = \frac{1}{1 - e cos E}$ .

Perihelion $E = 0$ : $\frac{1}{1 - 0.99 \times 1} = \frac{1}{0.01} = 100.$
Aphelion $E = π$ : $\frac{1}{1 - 0.99 \times ( - 1 )} = \frac{1}{1.99} = 0.502513.$ Ratio $= \frac{100}{0.502513} = 199.0$ . Toh $E$ (aur isliye planet) clock ke liye perihelion par aphelion ki tulna mein roughly 199 guna zyada responsively sweep karta hai — near-parabolic sprint-and-crawl ki ek vivid picture.

L5.3 (reverse — observation se $e$ nikalo)

$E = π /2$ par ek planet $r = 0.85 a$ distance par measure kiya gaya. Eccentricity $e$ nikalo, phir us instant par mean anomaly $M$ nikalo.

Recall Solution

$r = a (1 - e cos E)$ use karo $E = π /2$ ke saath, toh $cos E = 0$ : $0.85 a = a (1 - e \times 0) = a \Rightarrow contradiction agar literally liya jaye!$ Yahi lesson hai: $E = π /2$ par radius $a (1 - e cos (π /2)) = a (1 - 0) = a$ hota hai har eccentricity ke liye, isliye $E = π /2$ par $r$ mein $e$ ke baare mein koi information nahi hai. Diya gaya datum $r = 0.85 a$ at $E = π /2$ isliye impossible hai — ek badiya trap. Corrected version: maan lo ki $r = 0.85 a$ $E = π /3$ par hota hai. Tab $0.85 = 1 - e cos \frac{π}{3} = 1 - 0.5 e \Rightarrow e = \frac{1 - 0.85}{0.5} = 0.30.$ Phir $M = E - e sin E = \frac{π}{3} - 0.30 sin \frac{π}{3} = 1.047198 - 0.30 \times 0.866025 = 1.047198 - 0.259808 = 0.787390$ rad.

L5.4 (hyperbolic case ki taraf bridge)

Ek unbound trajectory mein ellipse ki jagah hyperbola aata hai aur equation ban jaata hai $M = e sinh F - F$ (dekho Hyperbolic Kepler equation M = e·sinh F − F). $e = 1.5$ , $F = 1.0$ diya hai, $M$ nikalo. Elliptic case ke saath sign structure ka contrast karo.

Recall Solution

$F$ kya hai? Hyperbolic (escape) orbit ke liye, $e > 1$ hota hai aur koi auxiliary circle nahi hota; uski jagah geometry hyperbolic eccentric anomaly $F$ se capture hoti hai, jo ek real parameter hai hyperbolic functions ( $cosh F, sinh F$ ) ke through define hota hai, bilkul waise jaise $E$ ellipse ke liye $cos E, sin E$ ke through define tha. Yeh same "geometric helper" ka role play karta hai, lekin cycle karne ki jagah unbounded grow karta hai. $M = e sinh F - F = 1.5 sinh (1.0) - 1.0 = 1.5 \times 1.175201 - 1.0 = 1.762802 - 1.0 = 0.762802.$ Contrast (complete):

Elliptic: $M = E - e sin E$ . $sin E$ term bounded hai aur oscillate karta hai; leading sign $E$ par hai (plus) aur correction minus ke saath.
Hyperbolic: $M = e sinh F - F$ . $sinh F$ term unbounded hai aur exponentially grow karta hai; $e$ -term ab plus sign carry karta hai aur dominate karta hai, jabki akela $F$ subtract hota hai.
Same idea: dono mein, $M$ abhi bhi uniform clock hai $M = n (t - t_{p})$ (hyperbola ke liye $n = μ /∣ a ∣^{3}$ ke saath), aur dono mein, $M \to E$ (ya $M \to F$ ) solve karna transcendental hai aur Newton iteration ki zaroorat hai. Sirf geometry alag hai — circle vs. hyperbola, $sin$ vs. $sinh$ . Yahi wajah hai ki dono equations ko saath padha jaata hai.

L5.4 trap (steel-manned) aur neeche wala L5 trap alag cautions hain.

Recall Ladder ka ek-line recap

Anomalies ko pehchano ::: L1 Dono directions mein plug karo ( $E \to M$ easy, $M \to E$ ko Newton chahiye) ::: L2 $e sin E$ ka sign hemisphere se flip hota hai; $d M / d E = 1 - e cos E > 0$ ::: L3 Full chain $t \to M \to E \to r \to ν$ , $n = μ / a^{3}$ ke saath ::: L4 High- $e$ stiffness, degenerate $cos E = 0$ , hyperbolic bridge ::: L5