Solving Kepler's equation — Newton-Raphson iteration: Question bank

3.2.15 · D5 · Physics › Orbital Mechanics & Astrodynamics › Solving Kepler's equation — Newton-Raphson iteration

Ye teen angles hain jinhe tumhe poore waqt seedha rakhna hai:

$M$ = mean anomaly — ek fake angle jo time ke saath bilkul uniformly badhta hai.
$E$ = eccentric anomaly — ellipse ke centre par geometric angle, auxiliary circle par padha jaata hai.
$ν$ = true anomaly — body ka asli angle jo focus se dekha jaata hai.

Agar inme se koi bhi shaky lage toh Eccentric Anomaly and the Auxiliary Circle aur True Anomaly and the Orbit Equation dekho.

True or false — justify karo

Kepler's equation

M = E - e sin E

ka har

M

ke liye ek unique solution

E

hota hai.

True — kyunki

f (E) = E - e sin E - M

ka derivative

1 - e cos E > 0

hai sabhi

E

ke liye (kyunki

e < 1

), isliye

f

strictly increasing hai aur zero ko exactly ek baar cross karti hai.

Ek circular orbit (

e = 0

) mein mean aur eccentric anomalies identical hote hain.

True —

e = 0

ke saath equation collapse hokar

M = E

ban jaata hai, matlab body genuinely uniformly move karti hai, toh iterate karne ki koi zaroorat nahi.

M

E

se instantly compute kar sakte ho, lekin

E

M

se nahi.

True —

E \to M

jaana sirf plug-and-evaluate hai, lekin

M \to E

jaane ke liye ek transcendental equation solve karni padti hai jahan

E

akele bhi aata hai aur

sin E

ke andar bhi, jise algebraically isolate nahi kar sakte.

M = E - e sin E

sahi answer deta hai chahe degrees mein kaam karo ya radians mein.

False — ye sirf radians mein exact hai, kyunki

sin E

(ek pure number) ko seedha

E

(ek angle) mein add kiya jaata hai, aur ye balance tabhi hold karta hai jab angle radians mein ho.

0 < E < π

ke liye eccentric anomaly

E

hamesha mean anomaly

M

se bada hota hai.

True — perihelion se aphelion ki taraf jaate waqt

sin E > 0

hota hai, isliye

M = E - e sin E < E

; physically body focus ke paas uniform clock-hand se aage nikal jaati hai.

Correction term

- e sin E

perihelion aur aphelion dono par vanish ho jaata hai.

True — perihelion par

E = 0

aur aphelion par

E = π

hai, aur

sin 0 = sin π = 0

, isliye

M = E

exactly dono apsides par hota hai.

Newton-Raphson hamesha correct root par converge karta hai, seed

E_{0}

chahe kuch bhi ho.

False — bahut high

e

ke liye ek bura seed overshoot kar sakta hai ya bahut dheere converge karta hai kyunki

f^{'} = 1 - e cos E

perihelion ke paas tiny ho jaata hai, jisse early steps violent ho jaate hain; seed choice genuinely matter karti hai.

Distance

r = a (1 - e cos E)

sabse bada hota hai jab

E = π

ho.

True —

cos π = - 1

deta hai

r = a (1 + e)

, jo aphelion distance hai, jo indeed maximum hai kyunki

cos E

wahaan minimised hota hai.

Eccentricity ko double karne se hamesha Newton iterations ki sankhya double ho jaati hai.

False — iteration count non-linearly badhta hai aur dono

e

aur

M

ki position par depend karta hai; near-parabolic (

e \to 1

) orbits mein difficulty kisi bhi linear rule se kahin zyada tezi se badhti hai.

True anomaly

ν

aur eccentric anomaly

E

hamesha orbit ke same half mein hote hain.

True — dono perihelion se same rotational sense mein measure kiye jaate hain, isliye ye saath mein

0

π

, aur

2 π

cross karte hain aur kabhi opposite half-planes mein nahi rehte.

Error dhundho

f (E) = E - e sin E - M

ka derivative

f^{'} = 1 - e sin E

hai."

Galat — tumne Kepler's equation se

sin

copy kar liya differentiate karne ki jagah;

\frac{d}{d E} (- e sin E) = - e cos E

, isliye correct slope hai

f^{'} = 1 - e cos E

"Kyunki

f^{'}

yahan constant hai, Newton-Raphson sirf ek single division hai."

Galat —

f^{'} = 1 - e cos E

E

par depend karta hai, isliye ise har guess

E_{k}

par re-evaluate karna padta hai; ise constant treat karna update ko poori tarah tod deta hai.

"Main

E_{0} = M = 35 0^{\circ}

se seed karunga aur degrees mein iterate karunga."

Do errors hain — equation radians mein honi chahiye (pehle convert karo), aur haalaanki

E_{0} = M

yahan ek accha seed hai, degree ki galti har

sin E

term ko nonsense bana degi.

"Kyunki mere guess par

f < 0

hai, true root chota hona chahiye, isliye main

E

ghataata hoon."

Ulta hai —

f

increasing hai, isliye

f < 0

matlab guess root se neeche hai; Newton ka

E_{k + 1} = E_{k} - f / f^{'}

automatically ek positive amount add karta hai aur

E

ko upar move karta hai.

"Maine

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

arctan

se solve kiya, toh

ν

ho gaya."

Incomplete — plain

arctan

sirf

(- \frac{π}{2}, \frac{π}{2})

mein value return karta hai aur quadrant drop kar deta hai, isliye tumhe half-angle sign (ya atan2) use karna padega taaki

ν

E

ke same half-plane mein rakha ja sake.

"Mera galat

f^{'}

phir bhi sahi answer tak converge ho gaya, toh formula theek hoga."

Misleading — ek galat-lekin-positive slope phir bhi correct root ki taraf limp kar sakta hai, lekin tum quadratic convergence kho dete ho jo Newton ko itna useful banati hai; limit ki correctness broken speed ko hide kar leti hai.

M /2 π

orbit ka angle fraction hai jo perihelion se sweep hua hai."

Galat — Kepler's 2nd law ke hisaab se

M /2 π

period ka (aur isliye swept area ka) fraction hai jo beeta hai, geometric angle ka fraction nahi; angle non-uniformly sweep karta hai.

Why questions

Hum fictitious mean anomaly

M

kyun invent karte hain?

Kyunki time aur geometric angle ek ellipse par proportional nahi hote;

M

ko linearly time ke saath badhne ke liye define kiya jaata hai (

M = n (t - t_{p})

), jo hume ek clean uniform handle deta hai jise baad mein real angle mein convert kar sakte hain.

Newton-Raphson ko plain fixed-point iteration

E_{k + 1} = M + e sin E_{k}

ke upar kyun prefer kiya jaata hai?

Newton quadratically converge karta hai (roughly har step mein correct digits double ho jaate hain), jabki fixed-point sirf linearly converge karta hai aur

e \to 1

par crawl ho jaata hai; Newton aim karne ke liye slope

f^{'}

use karta hai, sirf repeat nahi karta.

Seed

E_{0} = M + e sin M

E_{0} = M

se kyun better kaam karta hai?

Ye pehle se

- e sin E

term ke liye ek first-order correction apply kar leta hai, initial guess ko root ke paas land karta hai toh fewer iterations chahiye hoti hain — ye essentially ek manual correction step pehle se built in hai.

Kepler's equation ke root ki uniqueness guaranteed kyun hai?

Kyunki

e < 1

f^{'} = 1 - e cos E

har jagah strictly positive hota hai, isliye

f

kabhi nahi mudi; ek strictly monotonic function zero ko zyada se zyada ek baar cross kar sakti hai.

- e sin E

term physically kyun exist karta hai?

Ye focus ke ellipse ke centre se

c = a e

offset hone ko correct karta hai; focus se measure kiya gaya equal-areas law exactly yahi centre-to-focus triangle term produce karta hai.

High eccentricity iteration ko "violent" kyun banata hai?

Perihelion ke paas

cos E \approx 1

hota hai, isliye

f^{'} = 1 - e cos E \approx 1 - e

tiny ho jaata hai; ek small slope se divide karne par enormous first steps aate hain jo root ko overshoot kar sakte hain.

ν

ko half-angle formula se kyun nikalna padta hai, aur simply

E

ke barabar kyun nahi set kar sakte?

E

centre se measure hota hai aur

ν

focus se, isliye ye sirf apsides par barabar hote hain; half-angle tangent relation ek geometric viewpoint se doosre mein convert karta hai.

Edge cases

Jab

e = 0

ho toh Kepler's equation ka kya hota hai?

M = E

ban jaata hai: orbit circular hai, motion uniform hai, aur koi iteration required nahi.

Jab

e \to 1

(near-parabolic) ho toh kya hota hai?

f^{'} \to 1 - cos E

perihelion ke paas zero approach kar sakta hai, isliye Newton wahaan unstable ho jaata hai; ek robust seed jaise

E_{0} = π

apside ke paas, ya ek specialised near-parabolic formulation, ki zaroorat hai.

Jab

M = 0

ho toh

E

kya hai?

E = 0

— perihelion, jahan

M = E

exactly hai kyunki

sin 0 = 0

, isliye koi correction nahi aur koi iteration nahi chahiye.

Jab

M = π

ho toh

E

kya hai?

E = π

— aphelion, phir

M = E

exactly kyunki

sin π = 0

; ye do symmetric apsides hi woh jagahein hain jahan

M

aur

E

e > 0

ke liye, coincide karte hain.

π

se thoda aage

M

ke liye (jaise

M = 3.2

), kya correction

- e sin E

positive hai ya negative?

Return leg par

π < E < 2 π

deta hai

sin E < 0

, isliye

- e sin E > 0

aur ab

M > E

— body focus se door hote waqt uniform clock-hand se peeche reh jaati hai.

Agar tumhara computed

E

ek small negative

M

ke liye negative aata hai, kya ye valid hai?

Haan — Kepler's equation odd-symmetric hai (

M \to - M

E \to - E

milta hai), isliye negative anomalies sirf orbit ke pre-perihelion half ko describe karte hain aur bilkul legitimate hain.

Agar

f (E_{0})

seed par hi tumhari tolerance se neeche hai toh kya?

Tab tum

E_{0}

ko zero iterations ke saath accept kar lete ho — seed root par hi land kar gaya, jo small

e

ke liye apside ke paas common hota hai.

Recall Ek-line self-test

Upar sab kuch cover karo aur aloud jawab do: Hum $M = E - e sin E$ ko algebraically kyun nahi solve kar sakte, aur kaunsi ek property guarantee karti hai ki Newton ek sahi $E$ dhundh lega? Answer ::: $E$ akele bhi aata hai aur $sin E$ ke andar bhi isliye ise isolate nahi kar sakte (transcendental); aur $f^{'} = 1 - e cos E > 0$ sabhi $E$ ke liye, exactly ek root ke saath strictly increasing $f$ guarantee karta hai.