True anomaly from eccentric anomaly: Question bank

Child of True anomaly from eccentric anomaly. This is a concept-trap bank: no heavy arithmetic, just the ideas that catch people. Each item is a one-line reveal — cover the answer, commit to a reason, then check.

Before you start, keep the players straight (see the parent's geometry):

$a$ — semi-major axis (half the long diameter); $b = a 1 - e^{2}$ — semi-minor axis (half the short diameter). The auxiliary circle has radius $a$ , and the ellipse is that circle squashed vertically by the factor $b / a$ .
$ν$ (true anomaly) — angle at the focus $F$ , the real "as-seen-from-the-Sun" angle.
$E$ (eccentric anomaly) — angle at the center $O$ , on the auxiliary circle of radius $a$ .
$e$ (eccentricity) — how squashed the ellipse is; $c = a e$ is the focus offset from center.

The master relations you will keep leaning on (valid for $e < 1$ ): $cos ν = \frac{c o s E - e}{1 - e c o s E}, sin ν = \frac{1 - e ^{2} s i n E}{1 - e c o s E}, tan \frac{ν}{2} = \frac{1 + e}{1 - e} tan \frac{E}{2} .$

True or false — justify

ν = E

always holds for a circular orbit (

e = 0

True. With

e = 0

the stretch factor

(1 + e) / (1 - e) = 1

and the focus sits at the center (

c = a e = 0

), so both angles are measured from the same point →

ν = E

(and

= M

too).

For any elliptical orbit (

0 < e < 1

ν \geq E

throughout the whole orbit.

False. On the perihelion half (

0 < E < π

) the stretch makes

ν > E

; on the aphelion-return half (

π < E < 2 π

) the effect reverses and

ν < E

. They cross and are equal exactly at

E = 0, π, 2 π

At perihelion and at aphelion,

ν = E

True. At

E = 0

(perihelion) both angles are

0

; at

E = π

(aphelion) both are

π

. These are the only two points where the three anomalies all coincide for

e > 0

The formula

tan (ν /2) = (1 + e) / (1 - e) tan (E /2)

can be applied with a plain

arctan

and give the right

ν

every time.

False.

arctan

returns only

(- 90°, 90°)

, so near

E = π

where

ν /2 \to 90°

you must track how many half-turns

E /2

has made; safest is to compute

sin ν, cos ν

and use atan2.

r = a (1 - e cos E)

is smallest at perihelion.

True.

cos E

is largest (

= 1

) at

E = 0

, making

1 - e cos E

smallest

= 1 - e

, so

r_{m i n} = a (1 - e)

— the perihelion distance.

The radius formula

r = a (1 - e cos E)

and the orbit's

r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν}

describe different curves.

False. They are the same point on the same ellipse written with two different anomalies; substituting the

cos ν

relation into one recovers the other. See Orbital radius equation.

sin ν > 0

then the body is on the outbound (perihelion→aphelion) half of the orbit.

True.

sin ν = 1 - e^{2} sin E / (1 - e cos E)

has the sign of

sin E

;

sin E > 0

means

E \in (0, π)

, the ascending half where the body climbs away from the focus.

cos ν

alone determines the quadrant of

ν

False.

cos ν

can't tell

ν

from

- ν

(they share a cosine). You need

sin ν

's sign — i.e. the sign of

sin E

— to fix the quadrant.

Spot the error

"Body position in center coordinates is

x = a sin E, y = b cos E

Wrong axes.

E

is measured from perihelion which lies on the major (

x

) axis, so

x = a cos E

and

y = b sin E = a 1 - e^{2} sin E

(recall

b = a 1 - e^{2}

is the semi-minor axis).

"To get

ν

we read the angle of

(a cos E, b sin E)

directly."

That gives an angle at the center, not the focus. You must first shift the origin to

F

by subtracting

c = a e

along

x

x_{F} = a (cos E - e)

ν

is only defined at the focus.

b = a (1 - e^{2})

, so the squash factor is

1 - e^{2}

Off by a square root. The semi-minor axis is

b = a 1 - e^{2}

; the vertical squash of the circle is

b / a = 1 - e^{2}

, not

1 - e^{2}

"Since

cos ν = \frac{cos E - e}{1 - e cos E}

, at

E = 90°

we get

cos ν = \frac{- e}{1} = - e

, so

ν = 90° +

something small."

The value

- e

is right but the conclusion is fuzzy:

cos ν = - e < 0

puts

ν

in the second quadrant (

> 90°

), and

sin ν > 0

confirms it. For

e = 0.6

that's

ν \approx 126.9°

, well past

90°

"The stretch factor

(1 + e) / (1 - e)

makes

ν

bigger than

E

near aphelion."

Backwards. Near perihelion (small

E

) the tiny

tan (E /2)

is amplified, so

ν > E

there. Near aphelion the reverse holds; the single factor still encodes both because

tan (E /2)

grows without bound approaching

E = π

"For

e = 1

the formula still gives a finite

ν

e = 1

the orbit is a parabola,

(1 + e) / (1 - e) \to \infty

, and the eccentric-anomaly picture (a bounded ellipse/circle) breaks down entirely — you need the parabolic/hyperbolic analogue, not this formula.

M

E

, and

ν

are just the same angle in different units."

No — only

M

scales linearly with time;

E

and

ν

are genuinely different geometric angles (measured from different points), related nonlinearly. Equal units, different definitions.

Why questions

Why is

ν

measured at the focus and not the center?

Because the physical force (gravity) acts from the focus, and the radius

r

that appears in the orbit and energy is the focus-to-body distance;

ν

is the natural polar angle in that focus frame. See Position and velocity in the perifocal frame.

Why introduce the helper angle

E

at all, instead of working with

ν

directly?

Because

E

connects cleanly to time through Kepler's Equation (

M = E - e sin E

), which has no simple closed form in

ν

E

is the bridge between time ($M$) and geometry (

ν

Why does a small step in

E

near perihelion produce a large step in

ν

Near perihelion the body is close to the focus, so the same arc length subtends a much larger angle as seen from the nearby focus — the

(1 + e) / (1 - e)

factor is exactly this focus-offset magnification.

Why prefer the half-angle

tan

form over the

cos ν

expression in code?

tan (ν /2)

tan (E /2)

is monotonic across a full orbit, so there's no quadrant ambiguity, whereas

arccos (cos ν)

collapses

ν

and

- ν

and only returns

[0, π]

Why does the half-angle identity

tan^{2} (θ /2) = \frac{1 - c o s θ}{1 + c o s θ}

show up in the derivation?

Forming

1 \pm cos ν

factors the messy rationals into

(1 \mp e) (1 \pm cos E)

; dividing them makes the

(1 - e cos E)

denominators cancel, and the identity turns the leftover

\frac{1 - c o s}{1 + c o s}

ratio into a clean

tan^{2}

. See Half-angle trigonometric identities.

Why is

r = a (1 - e cos E)

often computed alongside

ν

in practice?

It falls out of the same Step-3 algebra (

r = x_{F}^{2} + y_{F}^{2}

) for free, and you almost always need both the angle and the distance to place the body — computing them together avoids redundant work.

Why does the

1 - e^{2}

factor appear only in

sin ν

and not in

cos ν

It comes from the vertical squash

y = b sin E = a 1 - e^{2} sin E

, which only affects the

y

-component; the

x

-shift to the focus (a horizontal operation) carries no such factor, so

cos ν

stays root-free.

Edge cases

E = 0

(perihelion), what does every formula give?

cos ν = 1, sin ν = 0 \Rightarrow ν = 0

, and

r = a (1 - e)

— the closest approach. All three anomalies equal zero.

E = 180°

(aphelion), what happens to

tan (E /2)

and how is

ν

still well-defined?

tan (E /2) = tan 90°

blows up, but so should

tan (ν /2)

: both half-angles hit

90°

together, giving

ν = 180°

. Use

sin ν = 0, cos ν = - 1

(via atan2) to sidestep the infinity. Here

r = a (1 + e)

, the farthest point.

What is

ν

for

E

just below

360°

(body about to complete an orbit)?

sin E < 0

sin ν < 0

ν

is just below

360°

(equivalently a small negative angle), approaching perihelion from the descending side. The plain

arctan

half-angle would misplace it — atan2 is required.

e \to 1^{-}

(near-parabolic ellipse), what happens to the aphelion distance and the stretch factor?

r_{m a x} = a (1 + e) \to 2 a

stays finite for fixed

a

, but the stretch

(1 + e) / (1 - e) \to \infty

, so away from perihelion tiny

E

changes swing

ν

enormously — the orbit spends nearly all its angle-range near perihelion.

For a degenerate "orbit" with

e = 0

and any

E

, why is Kepler's equation trivial?

With

e = 0

M = E - 0 = E

and

ν = E

, so time maps linearly and directly onto the true angle — the circle has no non-uniform-angle problem to solve.

What does a negative

ν

(or

ν > 180°

) physically mean, and is it valid?

Perfectly valid — it means the body is on the descending half (

sin ν < 0

E \in (π, 2 π)

), below the major axis heading back toward perihelion. Anomalies wrap mod

360°

Recall One-line self-test

Cover every answer above; if you can justify why each true/false holds and name which sign or quadrant fixes the ambiguous ones, you own the concept.