Part of Kepler's Equation solving chain: M → E → ν M \to E \to \nu M → E → ν
An orbiting body moves on an ellipse with the central mass at one focus . There are three different "angles" we use to track it:
True anomaly ν \nu ν real geometric angle from focus to the body. This is what you actually want.Eccentric anomaly E E E helper angle measured from the center of the ellipse, on an imaginary circle that hugs the orbit. Easy to relate to time via Kepler's equation.Mean anomaly M M M fake angle that grows linearly with time.
We solve Kepler's equation M = E − e sin E M = E - e\sin E M = E − e sin E E E E E → ν E \to \nu E → ν only that last conversion.
Definition The three anomalies geometricallyPlace the ellipse with semi-major axis a a a e e e O O O F F F P P P
==True anomaly ν \nu ν ∠ ( perihelion , F , P ) \angle (\text{perihelion}, F, P) ∠ ( perihelion , F , P ) focus .
==Eccentric anomaly E E E ∠ ( perihelion , O , P ′ ) \angle (\text{perihelion}, O, P') ∠ ( perihelion , O , P ′ ) P ′ P' P ′ P P P vertically up onto the auxiliary circle of radius a a a center .
We set up coordinates with origin at the center O O O x x x
( 1 + e ) / ( 1 − e ) \sqrt{(1+e)/(1-e)} ( 1 + e ) / ( 1 − e ) E ≈ 0 E\approx 0 E ≈ 0 close to the focus, so a tiny step of E E E big true-angle ν \nu ν E ≈ π E\approx\pi E ≈ π E E E ν \nu ν e = 0 e=0 e = 0 1 1 1 ν = E = M \nu = E = M ν = E = M
Start from Step 3 results.
1 + cos ν = 1 + cos E − e 1 − e cos E = ( 1 − e cos E ) + ( cos E − e ) 1 − e cos E = ( 1 − e ) ( 1 + cos E ) 1 − e cos E 1 + \cos\nu = 1 + \frac{\cos E - e}{1 - e\cos E} = \frac{(1-e\cos E)+(\cos E - e)}{1-e\cos E} = \frac{(1-e)(1+\cos E)}{1-e\cos E} 1 + cos ν = 1 + 1 − e c o s E c o s E − e = 1 − e c o s E ( 1 − e c o s E ) + ( c o s E − e ) = 1 − e c o s E ( 1 − e ) ( 1 + c o s E ) 1 − cos ν = ( 1 − e cos E ) − ( cos E − e ) 1 − e cos E = ( 1 + e ) ( 1 − cos E ) 1 − e cos E 1 - \cos\nu = \frac{(1-e\cos E)-(\cos E - e)}{1-e\cos E} = \frac{(1+e)(1-\cos E)}{1-e\cos E} 1 − cos ν = 1 − e c o s E ( 1 − e c o s E ) − ( c o s E − e ) = 1 − e c o s E ( 1 + e ) ( 1 − c o s E ) 1 − cos ν 1 + cos ν = ( 1 + e ) ( 1 − cos E ) ( 1 − e ) ( 1 + cos E ) . \frac{1-\cos\nu}{1+\cos\nu} = \frac{(1+e)(1-\cos E)}{(1-e)(1+\cos E)}. 1 + c o s ν 1 − c o s ν = ( 1 − e ) ( 1 + c o s E ) ( 1 + e ) ( 1 − c o s E ) . tan 2 ( θ / 2 ) = 1 − cos θ 1 + cos θ \tan^2(\theta/2) = \dfrac{1-\cos\theta}{1+\cos\theta} tan 2 ( θ /2 ) = 1 + cos θ 1 − cos θ both sides:
tan 2 ν 2 = 1 + e 1 − e tan 2 E 2 ⇒ tan ν 2 = 1 + e 1 − e tan E 2 . \tan^2\frac{\nu}{2} = \frac{1+e}{1-e}\,\tan^2\frac{E}{2}\ \Rightarrow\ \tan\frac{\nu}{2} = \sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}. tan 2 2 ν = 1 − e 1 + e tan 2 2 E ⇒ tan 2 ν = 1 − e 1 + e tan 2 E . Why the half-angle? The 1 − cos 1 + cos \tfrac{1-\cos}{1+\cos} 1 + c o s 1 − c o s
Worked example Example 1 — Circular check (e = 0 e=0 e = 0 e = 0 e=0 e = 0 E = 50 ∘ E = 50^\circ E = 5 0 ∘ tan ( ν / 2 ) = 1 / 1 tan ( 25 ∘ ) = tan 25 ∘ ⇒ ν = 50 ∘ \tan(\nu/2) = \sqrt{1/1}\tan(25^\circ) = \tan25^\circ \Rightarrow \nu = 50^\circ tan ( ν /2 ) = 1/1 tan ( 2 5 ∘ ) = tan 2 5 ∘ ⇒ ν = 5 0 ∘ Why this step? With no eccentricity the ellipse is a circle, focus = center, so all three angles coincide — a sanity anchor.
Worked example Example 2 — Eccentric orbit,e = 0.6 e=0.6 e = 0.6 E = 90 ∘ E=90^\circ E = 9 0 ∘ tan ( E / 2 ) = tan 45 ∘ = 1 \tan(E/2)=\tan45^\circ = 1 tan ( E /2 ) = tan 4 5 ∘ = 1 = ( 1 + 0.6 ) / ( 1 − 0.6 ) = 1.6 / 0.4 = 4 = 2 =\sqrt{(1+0.6)/(1-0.6)} = \sqrt{1.6/0.4} = \sqrt{4} = 2 = ( 1 + 0.6 ) / ( 1 − 0.6 ) = 1.6/0.4 = 4 = 2 tan ( ν / 2 ) = 2 ⋅ 1 = 2 ⇒ ν / 2 = arctan 2 = 63.43 ∘ ⇒ ν = 126.87 ∘ \tan(\nu/2) = 2\cdot 1 = 2 \Rightarrow \nu/2 = \arctan 2 = 63.43^\circ \Rightarrow \nu = 126.87^\circ tan ( ν /2 ) = 2 ⋅ 1 = 2 ⇒ ν /2 = arctan 2 = 63.4 3 ∘ ⇒ ν = 126.8 7 ∘ Why ν > E \nu > E ν > E At E = 90 ∘ E=90^\circ E = 9 0 ∘ offset focus , it has already swept more than 90 ∘ 90^\circ 9 0 ∘ r = a ( 1 − e cos E ) = a ( 1 − 0 ) = a r = a(1-e\cos E)=a(1-0)=a r = a ( 1 − e cos E ) = a ( 1 − 0 ) = a
Worked example Example 3 — Verify via cosineSame numbers, e = 0.6 e=0.6 e = 0.6 E = 90 ∘ E=90^\circ E = 9 0 ∘ cos ν = cos 90 ∘ − 0.6 1 − 0.6 cos 90 ∘ = − 0.6 1 = − 0.6 ⇒ ν = 126.87 ∘ . \cos\nu = \dfrac{\cos90^\circ - 0.6}{1 - 0.6\cos90^\circ} = \dfrac{-0.6}{1} = -0.6 \Rightarrow \nu = 126.87^\circ. cos ν = 1 − 0.6 cos 9 0 ∘ cos 9 0 ∘ − 0.6 = 1 − 0.6 = − 0.6 ⇒ ν = 126.8 7 ∘ . Why cross-check? Cosine alone is quadrant-ambiguous, but combined with the sign of sin ν = 1 − 0.36 ⋅ 1 / 1 = + 0.8 > 0 \sin\nu = \sqrt{1-0.36}\cdot 1 / 1 = +0.8 >0 sin ν = 1 − 0.36 ⋅ 1/1 = + 0.8 > 0 ν \nu ν
Common mistake Using onlycos ν = cos E − e 1 − e cos E \cos\nu = \frac{\cos E - e}{1-e\cos E} cos ν = 1 − e c o s E c o s E − e arccos \arccos arccos Why it feels right: it's a clean closed form. Why it's wrong: arccos \arccos arccos [ 0 , π ] [0,\pi] [ 0 , π ] sin ν < 0 \sin\nu<0 sin ν < 0 E ∈ ( π , 2 π ) E\in(\pi,2\pi) E ∈ ( π , 2 π ) ν \nu ν Fix: use the half-angle tan \tan tan or use atan2(sin ν, cos ν).
Common mistake Squashing the wrong coordinateSome write x = a sin E x = a\sin E x = a sin E Why it feels right: confusing which axis is the major axis. Fix: perihelion is along the major axis (x x x E E E x = a cos E x = a\cos E x = a cos E y = b sin E y = b\sin E y = b sin E
Common mistake Forgetting to shift to the focusComputing ν \nu ν eccentric angle, not the true one. Fix: subtract c = a e c = ae c = a e x x x
Recall Feynman: explain to a 12-year-oldImagine a planet running around an oval track with the Sun sitting off-center at a special spot called the focus. To find where the planet is, scientists use a trick: they draw a perfect circle around the oval and a helper angle E E E middle of the oval. But the angle you'd actually see looking from the Sun is different — call it ν \nu ν tan ( ν / 2 ) = 1 + e 1 − e tan ( E / 2 ) \tan(\nu/2) = \sqrt{\frac{1+e}{1-e}}\tan(E/2) tan ( ν /2 ) = 1 − e 1 + e tan ( E /2 ) e = 0 e=0 e = 0
"True is Eccentric, plus a Perihelion kick."
Subtract the focus offset (cos E − e \cos E - e cos E − e
Stretch toward perihelion with 1 + e 1 − e \sqrt{\tfrac{1+e}{1-e}} 1 − e 1 + e
"One-plus-e over one-minus-e , root it, times tan-half-E."
What is the radius formula in terms of eccentric anomaly? r = a ( 1 − e cos E ) r = a(1 - e\cos E) r = a ( 1 − e cos E ) Give the quadrant-safe formula relating true and eccentric anomaly. tan ν 2 = 1 + e 1 − e tan E 2 \tan\frac{\nu}{2} = \sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2} tan 2 ν = 1 − e 1 + e tan 2 E Express cos ν \cos\nu cos ν E E E e e e cos ν = cos E − e 1 − e cos E \cos\nu = \dfrac{\cos E - e}{1 - e\cos E} cos ν = 1 − e cos E cos E − e Express sin ν \sin\nu sin ν E E E e e e sin ν = 1 − e 2 sin E 1 − e cos E \sin\nu = \dfrac{\sqrt{1-e^2}\,\sin E}{1 - e\cos E} sin ν = 1 − e cos E 1 − e 2 sin E Why is the half-angle tan \tan tan arccos ( cos ν ) \arccos(\cos\nu) arccos ( cos ν ) tan \tan tan of half-angle is monotonic over a full orbit, so it has no quadrant ambiguity;
arccos \arccos arccos only returns
[ 0 , π ] [0,\pi] [ 0 , π ] .
In center coordinates, what are the body's x , y x,y x , y x = a cos E , y = a 1 − e 2 sin E x = a\cos E,\ y = a\sqrt{1-e^2}\sin E x = a cos E , y = a 1 − e 2 sin E (circle squashed vertically by
b / a b/a b / a ).
What is the focus offset from the center? c = a e c = ae c = a e toward perihelion.
For e = 0 e=0 e = 0 ν \nu ν E E E M M M They are all equal — the orbit is a circle and focus = center.
Why is ν > E \nu > E ν > E The focus is offset toward perihelion, so the same
E E E -step sweeps a larger true angle when the body is close to the focus.
Kepler's Equation — gives E E E M M M Mean anomaly and time — defines M = n ( t − t 0 ) M = n(t-t_0) M = n ( t − t 0 ) Orbit geometry — semi-major axis and eccentricity — source of b = a 1 − e 2 b=a\sqrt{1-e^2} b = a 1 − e 2 c = a e c=ae c = a e Orbital radius equation — r = a ( 1 − e cos E ) = a ( 1 − e 2 ) 1 + e cos ν r = a(1-e\cos E) = \frac{a(1-e^2)}{1+e\cos\nu} r = a ( 1 − e cos E ) = 1 + e c o s ν a ( 1 − e 2 ) Position and velocity in the perifocal frame — next step using ν \nu ν Half-angle trigonometric identities — the algebraic engine here.
Aux circle point a cosE, a sinE
cos nu and sin nu formulas
Intuition Hinglish mein samjhoDekho, ek planet Sun ke around ellipse (oval) orbit me ghoomta hai, aur Sun center pe nahi, ek focus pe baitha hota hai. Position track karne ke liye hum teen angles use karte hain: true anomaly ν \nu ν eccentric anomaly E E E center se, ek imaginary circle pe), aur mean anomaly M M M E E E E E E ν \nu ν
Trick simple hai: circle ka point ( a cos E , a sin E ) (a\cos E, a\sin E) ( a cos E , a sin E ) focus pe shift karo (focus center se a e ae a e tan ( ν / 2 ) = 1 + e 1 − e tan ( E / 2 ) \tan(\nu/2) = \sqrt{\frac{1+e}{1-e}}\tan(E/2) tan ( ν /2 ) = 1 − e 1 + e tan ( E /2 ) r = a ( 1 − e cos E ) r = a(1 - e\cos E) r = a ( 1 − e cos E )
Yeh half-angle wala form kyun? Kyunki agar tum sirf cos ν \cos\nu cos ν arccos \arccos arccos 0 0 0 180 ∘ 180^\circ 18 0 ∘ tan \tan tan E E E ν \nu ν ( 1 + e ) / ( 1 − e ) \sqrt{(1+e)/(1-e)} ( 1 + e ) / ( 1 − e ) e = 0 e=0 e = 0
