True anomaly from eccentric anomaly: Worked examples

This is the practice-drill child of the parent conversion note. The parent built the formulas. Here we hunt down every case the conversion $E \to ν$ can throw at you and work each one from zero. If you have not seen the derivation, read the parent first; if you want the identity engine, keep Half-angle trigonometric identities open.

Three quantities appear on every line below, so let us pin them down first.

Definition The symbols used on this page

==Eccentricity $e$ == — a single number, $0 \leq e < 1$ , that says how squashed the ellipse is. $e = 0$ is a perfect circle; as $e \to 1$ the ellipse gets long and thin. See Orbit geometry — semi-major axis and eccentricity.
==Semi-major axis $a$ == — half the long width of the ellipse (the "size"). On this page we set $a = 1$ in "orbit radius units" so lengths are easy to read; multiply back by your real $a$ at the end.
==Focus offset $c = a e$ == — the distance from the ellipse's center to the focus (where the mass sits). This offset is the entire reason $E$ (angle from center) and $ν$ (angle from focus) differ. When $e = 0$ , $c = 0$ : focus and center coincide.
$E$ (eccentric anomaly, from the center) and $ν$ (true anomaly, from the focus) are the two angles the parent note built; we convert between them here.

The two tools we will use again and again:

Before anything: what does "quadrant" even mean here? Picture a clock face centred on the focus. $ν = 0$ points at perihelion (3 o'clock). As the body orbits, $ν$ sweeps counter-clockwise: $0$ – $9 0^{\circ}$ is the first quadrant (upper right), $90$ – $18 0^{\circ}$ the second (upper left), $180$ – $27 0^{\circ}$ the third (lower left), $270$ – $36 0^{\circ}$ the fourth (lower right). "Getting the quadrant right" just means: does our formula land the body in the correct one of these four?

The figure below is the map we will refer back to in every descending example. Each quadrant cell is labelled directly on the picture with its name and angle range ("Q I: 0–90°", etc.), so the label text — not the colour — tells you which cell is which. Note how Q I and Q IV sit on the right (perihelion) side of the red focus dot, while Q II and Q III sit on the left. The body drawn at $ν = 12 0^{\circ}$ shows a Q II position — trace the arc marked $ν$ from the perihelion direction round to it. Whenever a worked example asks "which quadrant?", find the matching labelled cell on this picture.

Figure — True anomaly from eccentric anomaly

The scenario matrix

Every problem this topic can hand you falls into one of these cells. The worked examples below are each tagged with the cell they cover.

#	Case class	What makes it tricky	Example
A	$e = 0$ (circle, degenerate)	focus = center, all angles equal	Ex 1
B	$E$ in quadrant I ( $0 < E < 9 0^{\circ}$ )	plain, ascending, near perihelion	Ex 2
C	$E$ in quadrant II ( $9 0^{\circ} < E < 18 0^{\circ}$ )	still ascending, $ν > E$	Ex 3
D	$E = 18 0^{\circ}$ exactly (aphelion)	$tan (E /2) \to \infty$ , the trap	Ex 4
E	$E$ in quadrant III ( $18 0^{\circ} < E < 27 0^{\circ}$ )	descending, $sin ν < 0$ , sign flip	Ex 5
F	High eccentricity $e \to 1$	stretch factor blows up	Ex 6
G	Reverse: given $ν$ , find $E$	invert the formula, real-world	Ex 7
H	Word problem: radius → time link	connects to Kepler chain	Ex 8
I	$E$ in quadrant IV ( $27 0^{\circ} < E < 36 0^{\circ}$ ) / exam twist	$arccos$ -only trap exposed	Ex 9

We use $a = 1$ (in "orbit radius units") whenever a length appears — scaling is trivial.

Worked Examples

Worked example Example 1 — Cell A: the circle (

$e = 0$ ) Statement. A body on a circular orbit ( $e = 0$ ) has eccentric anomaly $E = 4 0^{\circ}$ . Find $ν$ and $r$ (take $a = 1$ ).

Forecast: on a circle the Sun sits dead center — so which of the angles should be equal? Guess before reading.

Stretch factor. $\frac{1 + 0}{1 - 0} = 1 = 1$ . Why this step? The whole $E \to ν$ distortion comes from the focus being off-center by the offset $c = a e$ (defined above). With $e = 0$ , $c = 0$ : no offset, no distortion.
Convert. $tan \frac{ν}{2} = 1 \cdot tan 2 0^{\circ} \Rightarrow \frac{ν}{2} = 2 0^{\circ} \Rightarrow ν = 4 0^{\circ}$ . Why this step? Factor of 1 means the equation degenerates to $ν = E$ .
Radius. $r = 1 \cdot (1 - 0 \cdot cos 4 0^{\circ}) = 1$ . Why this step? Constant radius is the definition of a circle — good self-consistency.

Verify: $ν = E = 4 0^{\circ}$ and $r = a = 1$ . Exactly the circle sanity anchor from the parent note (there $ν = E$ also equals the mean anomaly, but we do not need that here). ✓

Worked example Example 2 — Cell B: quadrant I, near perihelion

Statement. $e = 0.5$ , $E = 6 0^{\circ}$ . Find $ν$ , $cos ν$ , $sin ν$ , $r$ .

Forecast: the body is one-third of the way round the helper circle, on the perihelion side. Do you expect $ν > E$ or $ν < E$ ?

Stretch factor. $\frac{1.5}{0.5} = 3 \approx 1.7320$ . Why this step? $e = 0.5$ pulls the focus a long way toward perihelion (offset $c = a e = 0.5$ ), so near-perihelion angles get magnified.
Half-angle convert. $tan \frac{ν}{2} = 3 tan 3 0^{\circ} = 3 \cdot \frac{1}{3} = 1 \Rightarrow \frac{ν}{2} = 4 5^{\circ} \Rightarrow ν = 9 0^{\circ}$ . Why this step? $tan 3 0^{\circ} = 1/ 3$ conveniently cancels the $3$ — a clean landmark case.
Component cross-check. $cos ν = \frac{cos 6 0 ^{\circ} - 0.5}{1 - 0.5 cos 6 0 ^{\circ}} = \frac{0.5 - 0.5}{1 - 0.25} = 0$ , and $sin ν = \frac{1 - 0.25 sin 6 0 ^{\circ}}{0.75} = \frac{0.8660 \cdot 0.8660}{0.75} = + 1$ . Why this step? $(cos ν, sin ν) = (0, + 1)$ pins $ν = 9 0^{\circ}$ unambiguously in quadrant I/II boundary.
Radius. $r = 1 (1 - 0.5 cos 6 0^{\circ}) = 1 - 0.25 = 0.75$ . Why this step? The direct formula $r = a (1 - e cos E)$ gives the focus-to-body distance without needing $ν$ — a cheap, singularity-free by-product of the same $E$ .

Verify: $ν = 9 0^{\circ} > E = 6 0^{\circ}$ — perihelion-side stretching confirmed, exactly as forecast. ✓

Worked example Example 3 — Cell C: quadrant II, still ascending

Statement. $e = 0.6$ , $E = 12 0^{\circ}$ . Find $ν$ .

Forecast: past the minor axis but not yet at aphelion. Ascending, so $sin ν$ should stay positive. $ν$ bigger or smaller than $E$ ?

Stretch factor. $\frac{1.6}{0.4} = 4 = 2$ . Why this step? The factor $(1 + e) / (1 - e)$ is what encodes the focus offset $c = a e = 0.6$ ; computing it first tells us how strongly this orbit magnifies angles before we touch the trig.
Half-angle. $tan \frac{ν}{2} = 2 tan 6 0^{\circ} = 23 \approx 3.4641 \Rightarrow \frac{ν}{2} = arctan (3.4641) = 73.89 8^{\circ} \Rightarrow ν = 147.8 0^{\circ}$ . Why this step? $tan (E /2) = tan 6 0^{\circ} = 3$ is positive and $E /2 < 9 0^{\circ}$ , so $ν /2$ is a plain first-quadrant arctan — no sign worries.
Component cross-check. $cos ν = \frac{cos 12 0 ^{\circ} - 0.6}{1 - 0.6 cos 12 0 ^{\circ}} = \frac{- 0.5 - 0.6}{1 + 0.3} = \frac{- 1.1}{1.3} = - 0.8462$ ; $sin ν = \frac{0.8 \cdot 0.8660}{1.3} = + 0.5329$ . Why this step? $cos ν < 0$ , $sin ν > 0$ ⇒ quadrant II ⇒ $ν = 18 0^{\circ} - 32. 2^{\circ} = 147. 8^{\circ}$ . Matches.

Verify: $arccos (- 0.8462) = 147.8 0^{\circ}$ and $sin > 0$ keeps it in quadrant II. ✓ Still $ν > E$ , distortion fading as we near aphelion.

Worked example Example 4 — Cell D: aphelion,

$E = 18 0^{\circ}$ (the $\infty$ trap) Statement. $e = 0.6$ , $E = 18 0^{\circ}$ . Find $ν$ and $r$ .

Forecast: the body is at the far end. What is $tan (E /2) = tan 9 0^{\circ}$ ? Does the formula break?

Spot the singularity. $tan \frac{E}{2} = tan 9 0^{\circ} = + \infty$ . Why this step? This is the classic place students panic. The half-angle form has a removable blow-up here.
Take the limit. $tan \frac{ν}{2} = 2 \cdot (+ \infty) = + \infty \Rightarrow \frac{ν}{2} = 9 0^{\circ} \Rightarrow ν = 18 0^{\circ}$ . Why this step? Both sides run to $+ \infty$ together, and $arctan (+ \infty) = 9 0^{\circ}$ . So $ν = E = 18 0^{\circ}$ exactly at aphelion — the one place besides perihelion where the two angles coincide.
Component sanity (no infinities). $cos ν = \frac{- 1 - 0.6}{1 - 0.6 ( - 1 )} = \frac{- 1.6}{1.6} = - 1$ , $sin ν = \frac{0.8 \cdot 0}{1.6} = 0$ . So $(cos ν, sin ν) = (- 1, 0) \Rightarrow ν = 18 0^{\circ}$ . Why this step? The component form has no singularity here, so it's the safe way to confirm the limit.
Radius (aphelion distance). $r = 1 (1 - 0.6 cos 18 0^{\circ}) = 1 + 0.6 = 1.6 = a (1 + e)$ . Why this step? At aphelion $cos E = - 1$ , so $r = a (1 + e)$ is the largest possible distance — a memorable landmark value that anchors the whole orbit's size scale.

Verify: aphelion radius $a (1 + e) = 1.6$ ✓, and $ν = 18 0^{\circ}$ matches both methods. The " $\infty$ " was harmless. ✓

Worked example Example 5 — Cell E: quadrant III, the descending sign flip

Statement. $e = 0.6$ , $E = 24 0^{\circ}$ . Find $ν$ .

Forecast: the body is now on the way back down toward perihelion. Its $sin ν$ must be negative. Will an $arccos$ -only method get this right? (No — that's the point.)

Half-angle, honestly. $E /2 = 12 0^{\circ}$ , $tan 12 0^{\circ} = - 3$ . So $tan \frac{ν}{2} = 2 (- 3) = - 23 \approx - 3.4641$ . Why this step? $E /2 = 12 0^{\circ}$ is in quadrant II where tangent is negative — the sign is doing real work.
Invert carefully. $\frac{ν}{2} = arctan (- 3.4641)$ . Since $E /2 = 12 0^{\circ}$ lies in $(9 0^{\circ}, 18 0^{\circ})$ , we need $ν /2$ in the same half-turn: $ν /2 = 18 0^{\circ} - 73.89 8^{\circ} = 106.10 2^{\circ} \Rightarrow ν = 212.2 0^{\circ}$ . Why this step? $arctan$ returns $- 73.89 8^{\circ}$ (a quadrant-IV value). Because $E /2$ was in quadrant II and $tan (ν /2)$ tracks $tan (E /2)$ monotonically, add $18 0^{\circ}$ to keep $ν /2$ in the correct branch.
Component cross-check (the real safety net). $cos ν = \frac{cos 24 0 ^{\circ} - 0.6}{1 - 0.6 cos 24 0 ^{\circ}} = \frac{- 0.5 - 0.6}{1 - 0.6 ( - 0.5 )} = \frac{- 1.1}{1.3} = - 0.8462$ ; $sin ν = \frac{0.8 sin 24 0 ^{\circ}}{1.3} = \frac{0.8 ( - 0.8660 )}{1.3} = - 0.5329$ . Why this step? $cos < 0$ , $sin < 0$ ⇒ quadrant III (the Q III cell labelled on the map figure). $atan2 (- 0.5329, - 0.8462) = - 147.8 0^{\circ} \equiv 212.2 0^{\circ}$ . Matches.

Verify: $36 0^{\circ} - 212.2 0^{\circ} = 147.8 0^{\circ}$ — mirror image of Example 3, exactly as symmetry demands (same $∣ E - 18 0^{\circ} ∣$ ). ✓ The descending half needs the sign; an $arccos$ -only answer would have wrongly reported $147. 8^{\circ}$ .

Worked example Example 6 — Cell F: high eccentricity,

$e = 0.9$ Statement. A comet with $e = 0.9$ has $E = 3 0^{\circ}$ . Find $ν$ and $r$ .

Forecast: with the focus almost at the edge, near-perihelion angles get hugely magnified. Expect $ν ≫ E$ .

Stretch factor. $\frac{1.9}{0.1} = 19 \approx 4.3589$ . Why this step? As $e \to 1$ the offset $c = a e \to a$ (focus almost at the ellipse edge) and the factor $(1 + e) / (1 - e) \to \infty$ : extreme perihelion whip.
Half-angle. $tan \frac{ν}{2} = 4.3589 tan 1 5^{\circ} = 4.3589 \cdot 0.26795 = 1.1680 \Rightarrow \frac{ν}{2} = 49.4 4^{\circ} \Rightarrow ν = 98.8 8^{\circ}$ . Why this step? A helper angle of only $3 0^{\circ}$ already gives a real angle near $9 9^{\circ}$ — the comet appears to lunge around the Sun near perihelion.
Radius. $r = 1 (1 - 0.9 cos 3 0^{\circ}) = 1 - 0.9 (0.86603) = 1 - 0.77942 = 0.22058$ . Why this step? Very small $r$ (compare perihelion $a (1 - e) = 0.1$ ) confirms the comet is deep in the well, where speed and angular rate are highest.

Verify: component check $cos ν = \frac{0.86603 - 0.9}{1 - 0.9 ( 0.86603 )} = \frac{- 0.03397}{0.22058} = - 0.1540$ ; $arccos (- 0.1540) = 98.8 6^{\circ}$ (rounding) with $sin ν > 0$ . ✓ $ν ≫ E$ as forecast.

Worked example Example 7 — Cell G: reverse — given

$ν$ , find $E$ Statement. A satellite has $e = 0.4$ and true anomaly $ν = 12 0^{\circ}$ . Find its eccentric anomaly $E$ .

Forecast: we run the formula backward. Since $ν$ (from focus) leads $E$ (from center) on the perihelion side, expect $E < ν$ here.

Invert the stretch. $tan \frac{E}{2} = \frac{1 - e}{1 + e} tan \frac{ν}{2}$ . Why this step? Solving $tan \frac{ν}{2} = k tan \frac{E}{2}$ for $tan \frac{E}{2}$ just divides by $k$ , i.e. flips the fraction inside the root.
Numbers. $\frac{0.6}{1.4} = 0.42857 = 0.65465$ ; $tan 6 0^{\circ} = 1.73205$ . So $tan \frac{E}{2} = 0.65465 \cdot 1.73205 = 1.13389 \Rightarrow \frac{E}{2} = 48.59 6^{\circ} \Rightarrow E = 97.1 9^{\circ}$ . Why this step? Both $ν /2$ and $E /2$ sit in quadrant I, so the plain arctan branch is correct — no sign patching needed.
Forward-check. Plug $E = 97.1 9^{\circ}$ forward: $tan \frac{ν}{2} = 1.4/0.6 tan 48.59 6^{\circ} = 1.52753 \cdot 1.13389 = 1.73206 \Rightarrow ν /2 = 60.0 0^{\circ} \Rightarrow ν = 12 0^{\circ}$ . Why this step? Round-tripping confirms the inversion.

Verify: $E = 97.1 9^{\circ} < ν = 12 0^{\circ}$ ✓ — the focus-offset lead, exactly as forecast. (Next you'd feed $E$ into Kepler's equation, Kepler's Equation, to get mean anomaly and time.)

Worked example Example 8 — Cell H: word problem, radius → time link

Statement. An orbit has semi-major axis $a = 10, 000$ km and $e = 0.25$ . At a certain instant the body's eccentric anomaly is $E = 10 8^{\circ}$ . (a) How far is it from the focus? (b) What true anomaly does an observer at the focus measure? (c) What fraction of the orbital period has elapsed since perihelion? (Use $M = E - e sin E$ from Kepler's Equation and the time link $M = \frac{2 π}{T} (t - t_{p})$ from Mean anomaly and time.)

Forecast: at $E = 10 8^{\circ}$ we're just past the minor axis, so $r$ should be a bit more than $a$ , $ν$ a bit more than $E$ , and — since $M < E$ here — a little less than $10 8^{\circ} /36 0^{\circ}$ of the period should have passed.

Radius from $E$ . $r = a (1 - e cos E) = 10000 (1 - 0.25 cos 10 8^{\circ})$ . $cos 10 8^{\circ} = - 0.30902$ , so $r = 10000 (1 + 0.077254) = 10772.5$ km. Why this step? $r = a (1 - e cos E)$ is the direct, singularity-free distance formula — it needs only $E$ , which we already have, so it is the cheapest quantity to nail down first.
True anomaly. $\frac{1.25}{0.75} = 1.66667 = 1.29099$ ; $tan 5 4^{\circ} = 1.37638$ . $tan \frac{ν}{2} = 1.29099 \cdot 1.37638 = 1.77690 \Rightarrow ν /2 = 60.63 8^{\circ} \Rightarrow ν = 121.2 8^{\circ}$ . Why this step? Standard forward conversion; $E /2 = 5 4^{\circ}$ in quadrant I means no sign fixing.
Radius from $ν$ (independent route). Use $r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν}$ . $cos ν = cos 121.2 8^{\circ} = - 0.51840$ . Denominator $1 + 0.25 (- 0.51840) = 0.87040$ . Numerator $10000 (1 - 0.0625) = 9375$ . $r = 9375/0.87040 = 10770.3$ km. Why this step? Two different formulas from two different anomalies must give the same physical distance — a strong consistency test (tiny gap is rounding).
Time via mean anomaly. $M = E - e sin E$ with $E$ in radians: $E = 10 8^{\circ} = 1.88496$ rad, $sin 10 8^{\circ} = 0.95106$ . $M = 1.88496 - 0.25 (0.95106) = 1.88496 - 0.23776 = 1.64720$ rad $= 94.3 8^{\circ}$ . Elapsed fraction $= M / (2 π) = 1.64720/6.28319 = 0.26216$ . Why this step? $M$ is the linear-in-time angle, so $M / (2 π) = (t - t_{p}) / T$ is directly the fraction of the period elapsed — this is the whole point of the $M \to E \to ν$ chain, run in reverse for reporting.

Verify: both radius routes give $r \approx 10.77 \times 1 0^{3}$ km ✓; $ν = 121.2 8^{\circ} > E = 10 8^{\circ}$ ✓; elapsed fraction $0.2622 < 108/360 = 0.30$ ✓ (as forecast, $M < E$ ). Units: km for $r$ , dimensionless fraction for time. ✓

Worked example Example 9 — Cell I: quadrant IV, expose the

$arccos$ -only trap Statement. A student computes, for $e = 0.3$ , $E = 30 0^{\circ}$ : " $cos ν = \frac{cos 30 0 ^{\circ} - 0.3}{1 - 0.3 cos 30 0 ^{\circ}} = \frac{0.5 - 0.3}{1 - 0.15} = \frac{0.2}{0.85} = 0.23529$ , so $ν = arccos (0.23529) = 76.3 9^{\circ}$ ." Is this right? If not, give the correct $ν$ .

Forecast: $E = 30 0^{\circ}$ is in the fourth quadrant — the body is descending toward perihelion. A descending body must have $ν$ in $(18 0^{\circ}, 36 0^{\circ})$ . Can $76.3 9^{\circ}$ be right?

Find the true sign of $sin ν$ . $sin ν = \frac{1 - 0.09 sin 30 0 ^{\circ}}{1 - 0.15} = \frac{0.95394 ( - 0.86603 )}{0.85} = - 0.97186$ . Why this step? $arccos$ only ever returns $[0, 18 0^{\circ}]$ , so by itself it cannot tell the upper half of the orbit from the lower. The sign of $sin ν$ is the one extra fact that distinguishes $ν$ from $36 0^{\circ} - ν$ ; that is exactly the information the student threw away.
Correct the quadrant. $cos ν = + 0.23529 (> 0)$ and $sin ν = - 0.97186 (< 0)$ ⇒ quadrant IV (the Q IV cell labelled on the map figure). So $ν = 36 0^{\circ} - 76.3 9^{\circ} = 283.6 1^{\circ}$ . Why this step? Positive cosine, negative sine is unambiguously quadrant IV; the student's $76.3 9^{\circ}$ is its quadrant-I mirror — the wrong half of the orbit.
Half-angle confirmation. $E /2 = 15 0^{\circ}$ , $tan 15 0^{\circ} = - 0.57735$ . $tan \frac{ν}{2} = 1.3/0.7 (- 0.57735) = 1.36277 (- 0.57735) = - 0.78678$ . Since $E /2 \in (9 0^{\circ}, 18 0^{\circ})$ , take $ν /2 = 18 0^{\circ} - 38.1 9^{\circ} = 141.8 1^{\circ} \Rightarrow ν = 283.6 1^{\circ}$ . Why this step? The half-angle form, read with the correct branch, must agree with the component method — and it does, so we have caught the error two independent ways.

Verify: $atan2 (- 0.97186, 0.23529) = - 76.3 9^{\circ} \equiv 283.6 1^{\circ}$ ✓. The student's $76.3 9^{\circ}$ was off by being in the wrong quadrant; the correct answer is $ν = 283.6 1^{\circ}$ . ✓

Recall Quick self-test on the matrix

Which cell is $e = 0.6, E = 24 0^{\circ}$ , and what one fact fixes its quadrant? ::: Cell E (quadrant III); $sin ν < 0$ forces $ν$ into the lower half, giving $ν = 212.2 0^{\circ}$ . At $E = 18 0^{\circ}$ the half-angle form gives $tan (ν /2) = \infty$ — what is $ν$ ? ::: $ν = 18 0^{\circ}$ exactly (both angles meet at aphelion); confirm with $(cos ν, sin ν) = (- 1, 0)$ . Given $ν$ , how do you get $tan (E /2)$ ? ::: $tan \frac{E}{2} = \frac{1 - e}{1 + e} tan \frac{ν}{2}$ (flip the fraction under the root). What does the offset $c = a e$ physically represent? ::: The center-to-focus distance; it is the sole cause of the $E$ -vs- $ν$ difference, and vanishes when $e = 0$ .