Solving Kepler's equation — Newton-Raphson iteration: Exercises

3.2.15 · D4 · Physics › Orbital Mechanics & Astrodynamics › Solving Kepler's equation — Newton-Raphson iteration

Cast se pehle, teen quantities jinpar neeche ke formulas depend karte hain:

Teen characters jo tum baar baar use karoge:

Newton-Raphson recipe jo hum baar baar use karte hain:

$E_{k + 1} = E_{k} - \frac{E _{k} - e s i n E _{k} - M}{1 - e c o s E _{k}}, f (E) = E - e sin E - M, f^{'} (E) = 1 - e cos E .$

Neeche ki picture dikhati hai ki ek Newton step actually kya karta hai — hum ise Level 2 mein build karenge, isliye abhi ek baar dekho aur Exercise 2.2 ke baad wapas aao:

Figure — Solving Kepler's equation — Newton-Raphson iteration

Blue curve root function $f (E)$ hai; red dot hamara starting guess (seed) hai; dashed green line woh tangent hai jiska slope $f^{'}$ hai. Newton us tangent par slide karke wahan pahunchta hai jahan woh zero cross karti hai (green square), true root (yellow dot) ke kaafi kareeb land karke.

Level 1 — Recognition

Exercise 1.1

Words mein batao har symbol ka kya matlab hai aur uski units kya hain: $M$ , $E$ , $e$ , $n$ , $a$ , $t_{p}$ , $μ$ .

Recall Solution

$M$ — mean anomaly: ek fictional angle jo time ke saath uniformly badhta hai. Units: radians.
$E$ — eccentric anomaly: ellipse ke centre par geometric angle, body ke upar auxiliary circle ke point tak. Units: radians.
$e$ — eccentricity: dimensionless shape number, ellipse ke liye $0 \leq e < 1$ .
$n$ — mean motion: average angular rate, $n = μ / a^{3}$ . Units: rad/s.
$a$ — semi-major axis: ellipse ke long diameter ka aadha. Units: metres (ya koi bhi length).
$t_{p}$ — time of perihelion passage: woh clock reading jab body focus ke sabse paas hoti hai. Units: seconds.
$μ$ — gravitational parameter $= G M_{central}$ : central body ki gravity ki strength. Units: $km^{3} / s^{2}$ .

Exercise 1.2

Inme se Kepler's equation kaun si hai? (a) $E = M - e sin M$ (b) $M = E - e sin E$ (c) $M = E - e cos E$ .

Recall Solution

(b) $M = E - e sin E$ . Option (a) mein $M$ aur $E$ ke roles swap hain aur $sin M$ use hota hai — ye actually first-order seed hai, exact equation nahi. Option (c) mein $cos$ use hota hai; $cos$ derivative $f^{'} = 1 - e cos E$ mein aata hai, Kepler's equation mein nahi.

Exercise 1.3

Root function $f (E) = E - e sin E - M$ diya gaya hai, iska derivative $f^{'} (E)$ likho.

Recall Solution

$M, e$ ko constant maanke term by term differentiate karo: $\frac{d}{d E} E = 1, \frac{d}{d E} (- e sin E) = - e cos E, \frac{d}{d E} (- M) = 0.$ $f^{'} (E) = 1 - e cos E .$ Kyunki $e < 1$ hai, $f^{'} (E) \geq 1 - e > 0$ har jagah — $f$ strictly increasing hai, isliye root unique hai.

Level 2 — Application

Exercise 2.1

$M = 0.8$ rad, $e = 0.3$ ke liye, first-order seed $E_{0} = M + e sin M$ compute karo.

Recall Solution

$sin (0.8) = 0.71736$ . $E_{0} = 0.8 + 0.3 (0.71736) = 0.8 + 0.21521 = 1.01521 rad .$ Seed $- e sin E$ shift ke liye pre-correct karta hai, isliye humein answer ke paas se shuru karta hai.

Exercise 2.2

$M = 0.8$ , $e = 0.3$ ke liye $E_{0} = 1.01521$ se EK Newton-Raphson step karo. (Apna kaam page ke top par di gayi figure se match karo.)

Recall Solution

$f (E_{0}) = 1.01521 - 0.3 sin (1.01521) - 0.8 = 1.01521 - 0.25548 - 0.8 = - 0.04027.$ $f^{'} (E_{0}) = 1 - 0.3 cos (1.01521) = 1 - 0.15853 = 0.84147.$ $E_{1} = 1.01521 - \frac{- 0.04027}{0.84147} = 1.01521 + 0.04786 = 1.06307 rad .$ $f < 0$ ne bataya ki guess bahut chhoti thi, isliye Newton ne $E$ ko upar push kiya — bilkul wahi "tangent par slide karke zero tak" jo upar ki figure dikhati hai. Ye parent note ke Worked Example 1 se match karta hai.

Exercise 2.3

$E = 1.0627$ rad, $e = 0.3$ , $a = 1$ AU diya gaya hai, focus se distance $r$ find karo.

Recall Solution

$r = a (1 - e cos E) = 1 \cdot (1 - 0.3 cos (1.0627)) = 1 - 0.3 (0.48679) = 1 - 0.14604 = 0.85396 AU .$ Kyunki $r < a$ hai, body orbit ke near half mein hai — $E < \frac{π}{2}$ ke saath consistent.

Level 3 — Analysis

Exercise 3.1

$M = 0.1$ rad, $e = 0.9$ ke liye, seed $E_{0} = M + e sin M$ se pehla Newton step karo aur $f^{'}$ ke terms mein kyun step itna bada hai explain karo.

Recall Solution

Seed: $sin (0.1) = 0.09983$ , toh $E_{0} = 0.1 + 0.9 (0.09983) = 0.18985$ . Newton ka step size $- f / f^{'}$ hai. Yahan $f (E_{0}) = 0.18985 - 0.9 sin (0.18985) - 0.1 = 0.18985 - 0.16989 - 0.1 = - 0.08004,$ $f^{'} (E_{0}) = 1 - 0.9 cos (0.18985) = 1 - 0.9 (0.98204) = 1 - 0.88384 = 0.11616,$ ek chhota denominator. Phir $E_{1} = 0.18985 - \frac{- 0.08004}{0.11616} = 0.18985 + 0.68908 = 0.87893 rad .$ Kyunki $e$ 1 ke kareeb hai, $f^{'} = 1 - e cos E$ near zero tak dip karta hai — curve $f (E)$ wahan almost flat hai, isliye ek chhoti si vertical error ek badi horizontal correction mein translate ho jaati hai. Ye exactly woh reason hai kyun high- $e$ orbits numerically mushkil hoti hain: corrections "violent" hote hain, aur iteration settle hone mein kai steps leta hai. Figure mein flat region dekho.

Exercise 3.2

Algebraically dikhao ki $r = a (1 - e cos E)$ perihelion distance $a (1 - e)$ aur aphelion distance $a (1 + e)$ deta hai, aur identify karo ki kaun si value of $E$ inhe produce karta hai.

Recall Solution

$cos E$ ka range $[- 1, 1]$ hai.

Perihelion (sabse paas): $r$ minimum tab jab $cos E$ maximum ho, yaani $cos E = 1 \Rightarrow E = 0$ . Phir $r = a (1 - e \cdot 1) = a (1 - e)$ . ✓
Aphelion (sabse door): $r$ maximum tab jab $cos E = - 1 \Rightarrow E = π$ . Phir $r = a (1 - e (- 1)) = a (1 + e)$ . ✓

Inke beech $r$ smoothly vary karta hai, ellipse ki near/far geometry se match karta hua.

Exercise 3.3

Kepler's equation $M = E - e sin E$ ko fixed-point iteration $E_{k + 1} = M + e sin E_{k}$ se bhi solve kiya ja sakta hai. Is simpler scheme ke converge hone ki $e$ par condition batao, discuss karo ki iterate kahan se shuru hona chahiye, aur batao kyun large $e$ ke liye Newton prefer kiya jaata hai.

Recall Solution

Fixed-point map $g (E) = M + e sin E$ hai, jiska $g^{'} (E) = e cos E$ hai. Banach/contraction test kehta hai ki iteration converge hoti hai agar map ek contraction ho us interval par jahan hum iterate kar rahe hain, yaani $∣ g^{'} (E) ∣ \leq L < 1$ wahan. Kyunki har $E$ ke liye $∣ g^{'} (E) ∣ = ∣ e cos E ∣ \leq e$ hai — bound $L = e$ poori real line par hold karta hai, sirf locally nahi — isliye $g$ ek global contraction hai jab bhi $e < 1$ . Ye purely local result se stronger hai: koi bhi starting guess $E_{0}$ ek bounded region mein map hoti hai aur iterates ek Cauchy sequence banate hain jo unique fixed point par converge karti hai.

Convergence ka region: kyunki contraction bound uniform hai, $e < 1$ ke liye koi bhi $E_{0}$ kaam karta hai; ek natural aur safe choice $E_{0} = M$ hai (ya $E_{0} = M + e sin M$ ).
Rate: error roughly har step mein factor $e$ se shrink hoti hai. $e \to 1$ ke liye ye barely shrink hoti hai (har baar $0.9$ se multiply karo → bahut slow).

Newton-Raphson iske bajaye root ke paas quadratically converge karta hai (error har step mein square hoti hai), isliye woh fixed-point ko badi $e$ hone par badly beat karta hai — $f^{'}$ aur ek decent seed ki zaroorat ke cost par. Fixed-Point Iteration for Kepler dekho.

Level 4 — Synthesis

Exercise 4.1

Ek satellite ka $a = 7000$ km, $μ = 398600 km^{3} / s^{2}$ , $e = 0.1$ hai. Mean motion $n$ aur orbital period $T$ find karo.

Recall Solution

$n = \frac{μ}{a ^{3}} = \frac{398600}{700 0 ^{3}} = \frac{398600}{3.43 \times 1 0 ^{11}} = 1.16210 \times 1 0^{- 6} = 1.07801 \times 1 0^{- 3} rad/s .$ $T = \frac{2 π}{n} = \frac{6.28319}{1.07801 \times 1 0 ^{- 3}} = 5828.7 s \approx 97.1 min .$ Ek sensible low-Earth-orbit period. Mean Motion and Orbital Period dekho.

Exercise 4.2

Usi satellite ke liye, $t - t_{p} = 600$ s hai. $M$ find karo, phir $E$ ke liye solve karo ( $E_{0} = M + e sin M$ se do Newton steps), phir distance $r$ find karo. $n = 1.07801 \times 1 0^{- 3}$ rad/s, $e = 0.1$ , $a = 7000$ km use karo.

Recall Solution

Step 1 — mean anomaly: $M = n (t - t_{p}) = 1.07801 \times 1 0^{- 3} \times 600 = 0.64681 rad .$ Step 2 — seed: $sin (0.64681) = 0.60266$ , toh $E_{0} = 0.64681 + 0.1 (0.60266) = 0.70708 rad .$ Newton step 1: $f = 0.70708 - 0.1 sin (0.70708) - 0.64681 = 0.70708 - 0.06497 - 0.64681 = - 0.00470,$ $f^{'} = 1 - 0.1 cos (0.70708) = 1 - 0.07603 = 0.92397,$ $E_{1} = 0.70708 - \frac{- 0.00470}{0.92397} = 0.70708 + 0.00509 = 0.71217 rad .$ Newton step 2: $f = 0.71217 - 0.1 sin (0.71217) - 0.64681 = 0.71217 - 0.06537 - 0.64681 = - 0.00001,$ already machine-tiny hai, toh $E \approx 0.71218$ rad. Step 3 — distance: $r = a (1 - e cos E) = 7000 (1 - 0.1 cos (0.71218)) = 7000 (1 - 0.1 (0.75648)) = 7000 (0.92435) = 6470.5 km .$

Exercise 4.3

Ex 4.2 se $E = 0.71218$ rad aur $e = 0.1$ use karke, true anomaly $ν$ find karo (upar symbol list mein define kiya gaya actual focus-measured angle).

Recall Solution

$tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} = \frac{1.1}{0.9} tan (0.35609) .$ $1.1/0.9 = 1.22222 = 1.10554$ , aur $tan (0.35609) = 0.37193$ , toh $tan \frac{ν}{2} = 1.10554 \times 0.37193 = 0.41118 \Rightarrow \frac{ν}{2} = arctan (0.41118) = 0.38984,$ $ν = 0.77968 rad .$ Note karo $ν > E > M$ : body dono geometric aur uniform angles se aage nikal gayi hai, kyunki woh orbit ke fast (near-focus) part par hai. Kyunki $E$ pehle half-plane $(0, π)$ mein hai, $ν$ bhi hai, toh plain $arctan$ yahan safe hai — True Anomaly and the Orbit Equation dekho.

Level 5 — Mastery

Exercise 5.1

Prove karo ki $e < 1$ ke liye Kepler's equation $M = E - e sin E$ ka exactly ek solution $E$ hota hai, aur Newton-Raphson physical range mein kisi bhi point se start karke stuck nahi ho sakta (iska denominator kabhi zero nahi hota).

Recall Solution

Ek target $M \in [0, 2 π]$ fix karo aur physical interval $E \in [0, 2 π]$ par kaam karo (baki saare cases is par reduce ho jaate hain $2 π$ ka poora number $M$ aur $E$ dono mein add karke, kyunki $sin$ ka period $2 π$ hai).

Existence aur uniqueness. Define karo $f (E) = E - e sin E - M$ , $[0, 2 π]$ par continuous. Iska derivative $f^{'} (E) = 1 - e cos E$ hai, aur kyunki $∣ cos E ∣ \leq 1$ aur $e < 1$ hai, $f^{'} (E) \geq 1 - e > 0 for all E .$ Toh $f$ $[0, 2 π]$ par strictly increasing hai. Do endpoints check karo: $f (0) = 0 - 0 - M = - M \leq 0, f (2 π) = 2 π - 0 - M = 2 π - M \geq 0,$ $0 \leq M \leq 2 π$ use karke. Ek continuous, strictly increasing function jo left end par $\leq 0$ aur right end par $\geq 0$ hai, $[0, 2 π]$ par zero exactly ek baar cross karti hai — ye Intermediate Value Theorem hai jo physical domain ke andar apply ho rahi hai, aur strict monotonicity ek second crossing forbid karti hai. Hence ek unique $E$ .

Newton kabhi stuck nahi hota. Newton ka update $f^{'} (E_{k}) = 1 - e cos E_{k}$ se divide karta hai. Kyunki $∣ cos E_{k} ∣ \leq 1$ aur $e < 1$ hai, hamesha $f^{'} (E_{k}) \geq 1 - e > 0$ hota hai, toh denominator strictly zero se door bounded hai aur kabhi vanish nahi ho sakta — update kisi bhi $E_{0}$ ke liye well-defined hai. Kyunki $f$ monotone hai aur koi stationary points nahi hain, iterates trapped nahi ho sakte, aur Newton-Raphson guaranteed hai ki single root par home in kare. Isse proof complete hota hai. Newton-Raphson Method dekho.

Exercise 5.2

Ek comet ka $e = 0.967$ hai (Halley-jaisa). $M = 0.2$ rad par, $E_{0} = M + e sin M$ se Newton-Raphson perform karo jab tak $∣ f ∣ < 1 0^{- 6}$ na ho jaaye. Steps ki sankhya aur final $E$ report karo. $e = 0.3$ case ke comparison mein speed par comment karo.

Recall Solution

Seed: $sin (0.2) = 0.19867$ , $E_{0} = 0.2 + 0.967 (0.19867) = 0.39211$ .

$k$	$E_{k}$	$f (E_{k})$	$f^{'} (E_{k})$
0	0.39211	$- 0.17646$	$0.10787$
1	$0.39211 + 0.17646/0.10787 = 2.02790$	$0.96906$	$1.43046$
2	$2.02790 - 0.96906/1.43046 = 1.35544$	$0.20915$	$1.20441$
3	$1.35544 - 0.20915/1.20441 = 1.18178$	$0.02010$	$1.36135$
4	$1.18178 - 0.02010/1.36135 = 1.16702$	$0.00013$	$1.37262$
5	$1.16702 - 0.00013/1.37262 = 1.16693$	$\sim 4 \times 1 0^{- 8}$ ✓	—

Answer: lagbhag 5–6 steps, $E \approx 1.16693$ rad (chhaatha step $∣ f ∣$ ko $1 0^{- 6}$ se neeche polish karta hai). Seed par bahut chhota $f^{'} \approx 0.108$ ne pehle step ko wildly root se past $2.03$ tak throw kar diya, phir agli steps mein wapas aaya. Ex 2.2 mein $e = 0.3$ se contrast karo, jo essentially 2 steps mein converge ho gaya tha. High eccentricity = root ke paas flat $f$ = slow, dramatic convergence, exactly jaisa L3 par predict kiya gaya tha.

Exercise 5.3

Poore time-to-position pipeline ko quantities ke flow ke roop mein sketch karo aur har arrow par use hone wala tool naam do.

Recall Solution

$t \to M$ : trivial multiplication by mean motion $n$ (linear, instant).
$M \to E$ : hard step — transcendental hai, iteration chahiye (Newton-Raphson).
$E \to ν$ : half-angle tangent identity, quadrant dhyan rakho.
$E \to r$ : direct plug-in $r = a (1 - e cos E)$ .

Recall Jaane se pehle ek-line self-test

Poore pipeline mein genuinely hard arrow kaun sa hai, aur kaun sa method use karta hai? $M \to E$ arrow (solving $M = E - e sin E$ ) ::: ye transcendental hai, isliye hum Newton-Raphson use karte hain: $E_{k + 1} = E_{k} - \frac{E _{k} - e sin E _{k} - M}{1 - e cos E _{k}}$ .