Intuition The one core ideaA planet moves on a squashed circle (an ellipse ) with the Sun sitting off-center at a special point called the focus . To convert the easy-to-compute "helper angle" measured from the middle into the real angle you'd actually see from the Sun, you only need to know how the circle was squashed and how far off-center the Sun sits — that is the entire content of the parent topic .
This page builds every symbol the parent note uses, from absolute zero, in an order where each piece rests on the one before it. (The two angles named in that one-idea sentence — the "helper angle" and the "real angle" — get their formal symbols in §4; we deliberately avoid naming them with letters until then.) If a word or letter ever shows up in the parent and you are not 100% sure what it pictures , it is defined here.
An ellipse is a circle that has been squashed in one direction. Take a perfect circle and press it flat toward its horizontal axis: what you get is an oval. That oval is the path (the orbit ) the planet travels on.
Look at the figure. The long way across (left–right, red) is the major axis . The short way across (up–down, blue) is the minor axis . Half of each of these has a name:
Definition Semi-major axisa b
==a longest width of the ellipse. Picture the red arrow from the exact middle to the far right edge.
==b shortest width. Picture the blue arrow from the middle straight up to the top edge.
"Semi" just means "half". These two numbers completely fix the ellipse's size and how squashed it is. See Orbit geometry — semi-major axis and eccentricity .
Why the topic needs it: the parent's very first formula writes the body's height as b sin E b / a how much the circle was flattened, and without b
Before any angle, cos sin where the origin is, which way x Every formula on the parent page silently uses the convention below — here we make it loud.
Definition The perifocal convention (used everywhere in this topic)
The ==x from the center toward perihelion (the near end of the orbit, defined in §4).
The ==y 90° counter-clockwise from the x
Angles are measured counter-clockwise (CCW) starting from the positive x positive direction; turning clockwise gives negative angles.
Two origins get used: the center O (§1) and the focus F (§2b). A subscript tells you which — coordinates ( x O , y O ) ( x F , y F )
This is the perifocal frame . Fix this picture once and every cos sin
Why the topic needs it: a bare "cos E E + x ν
O F
==Center O ( x O , y O )
==Focus F shifted away from the center toward perihelion. The Sun (the heavy mass) sits here , not at the center. This is the origin of ( x F , y F )
The gap between the center and the focus has its own name:
c c O F + x
Why the topic needs it: the angle you actually observe is measured at the focus , but the easy helper angle is measured at the center . The whole conversion is really just "account for the shift c
e e 0 1 how squashed the ellipse is:
e = a c = semi-major axis focus offset .
e = 0 perfect circle .e 1 long thin cigar .
e c = e a as a fraction of a . If e = 0.6 a = 10 c = 6 c = a e
The three key lengths are tied together by one Pythagorean-flavoured relation:
Why the topic needs it: the term 1 − e 2 not magic — it is just the squash factor written using e
The word anomaly in astronomy just means "an angle that tells you where the body is on its orbit." There are three of them, and the entire topic is about converting between two of them. The key trap: an angle is meaningless until you say from which point and from which starting direction you measure it — which is exactly why §2 came first.
Definition Perihelion — the starting linePerihelion = the point on the orbit closest to the focus (the Sun). It sits at the near end of the major axis, on the + x
Definition The three anomalies
==True anomaly ν real angle from perihelion to the body, measured at the focus F This is the answer you want. (This is the "real angle" promised in the intro.)
==Eccentric anomaly E helper angle measured at the center O
==Mean anomaly M fake angle that grows perfectly evenly with the clock. It has no direct geometric picture on the orbit — it is a stand-in for time . See Mean anomaly and time .
Common mistake Confusing "measured at the focus" with "measured at the center"Why it feels the same: both are angles from the perihelion direction. Why it is not: you are standing at different points . From the center you see angle E different angle ν
Here is the picture that defines E
Definition Auxiliary circle andE a O auxiliary circle . Now take the planet at point P straight up (vertically) until you hit the circle at point P ′ P ′ measured at the center O and CCW, is the ==eccentric anomaly E
Because P ′ a ( x O , y O ) = ( a cos E , a sin E )
Intuition Why go through a circle at all?Angles on a circle are dead simple: a point at angle E ( cos E , sin E ) b / a ( x O , y O ) = ( a cos E , b sin E ) E
The true anomaly ν at the focus , so we must re-express the planet's position with F F ( c , 0 ) = ( a e , 0 ) slide the origin right by c :
Definition Focus-relative coordinatesx F , y F x F = x O − c = a cos E − a e = a ( cos E − e ) , y F = y O = a 1 − e 2 sin E .
==x F right of the focus (positive x
==y F above the focus (positive y
These are the very symbols the parent's shift-to-focus step produces. The figure below shows both arrows from F
r r focus F P r = x F 2 + y F 2 . r = a ( 1 − e cos E ) Orbital radius equation .
Why the topic needs it: to read off the true anomaly you use cos ν = x F / r sin ν = y F / r must have x F y F r
The formulas use three trig ideas. Here is each, anchored to a picture in your head.
cos sin 1 θ CCW from the positive x sits at:
cos θ horizontal position (how far right/left),sin θ vertical position (how far up/down).
That is why ( a cos E , a sin E ) a a
tan tan θ = c o s θ s i n θ = horizontal vertical = adjacent opposite . tan θ tan
Definition The half-angle identity — the workhorsetan 2 2 θ = 1 + c o s θ 1 − c o s θ , and also tan 2 θ = 1 + c o s θ s i n θ . Why the topic leans on it: the raw conversion gives cos ν sin ν tan ( ν /2 ) = 1 − e 1 + e tan ( E /2 ) Half-angle trigonometric identities .
Common mistake Branch and sign: taking the square root and gettingν tan 2 ( θ /2 ) ± which sign? For a full orbit E ν 0 360° together , so E /2 ν /2 [ 0° , 180° ) tan positive root: tan ( ν /2 ) = + 1 − e 1 + e tan ( E /2 ) tan ( θ /2 ) monotonic over this range, each E ν Recovering ν compute ν /2 = atan2 ( 1 + e sin 2 E , 1 − e cos 2 E ) atan2 ( sin ν , cos ν )
Common mistake Why not just takearccos cos ν Why it tempts you: you get a clean closed form for cos ν Why it breaks: arccos [ 0° , 180° ] sin ν < 0 tan atan2) is monotonic across the whole orbit — no ambiguity. This is the single most important reason the topic prefers the half-angle formula.
Intuition How to read this mapEach box is a foundation built above. Arrows mean "feeds into". Follow them top-to-bottom: shape facts (a b c e r ν
Eccentricity e equals c over a
Squash factor root of 1 minus e squared
Auxiliary circle point a cosE and a sinE
Ellipse point squashed height
Coordinate convention x to perihelion CCW
Shift origin to focus xF yF
Radius r and angle nu at focus
Clean tan nu over 2 formula
Test yourself — say the answer out loud before revealing.
What does the semi-major axis a Half the longest width of the ellipse — from center to the far end of the long axis.
What is b a e b = a 1 − e 2 (the vertical squash factor times
a ).
State the coordinate convention this topic uses. Origin at center or focus; + x + x
Define eccentricity e e = c / a
Where does the mass (Sun) actually sit — center or focus? At the focus F c = a e + x
Where is perihelion? The orbit point closest to the focus, on the + x
Where is the true anomaly ν From the perihelion direction, CCW, measured at the focus .
Where is the eccentric anomaly E From the perihelion direction, CCW, measured at the center , using the auxiliary circle.
What are the focus-relative coordinates x F , y F x F = a ( cos E − e ) ,
y F = a 1 − e 2 sin E .
How do you get the ellipse point from the auxiliary-circle point? Take ( a cos E , a sin E ) b / a ( a cos E , b sin E )
Write the half-angle identity for tan 2 ( θ /2 ) tan 2 ( θ /2 ) = 1 + cos θ 1 − cos θ
Which square-root sign do you keep converting E ν The positive root, because E /2 ν /2 [ 0° , 180° ) tan ( θ /2 ) atan2 to recover ν
Why avoid arccos ( cos ν ) It only returns [ 0 , 180° ]
What does r Focus-to-planet distance
r = x F 2 + y F 2 ;
r = a ( 1 − e cos E ) .
