Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true ano: Foundations

This page assumes you have seen nothing. We build every letter, every triangle, every arrow the parent note leans on, in the order that lets each one stand on the one before it. Read top to bottom.

1. Distance from a centre: $r$

Picture a piece of string from a dot in the middle (Earth) to a dot on a loop (the satellite). The length of that string is $r$ . As the satellite moves around the loop, the string gets longer and shorter — so $r$ is a number that changes with time.

Figure 1. A blue orbit loop with Earth as the yellow dot at the focus. The white dashed line is the string whose length is $r$ — it stretches to the pink satellite and changes as the satellite moves.

2. Speed: $v$ , and why we square it

Why does the parent note always write $v^{2}$ and not $v$ ? Because energy of motion — kinetic energy — grows with the square of speed. That squaring is not a trick; it falls out of the physics.

3. The gravitational parameter: $μ$

4. Vectors: $r$ and $v$ — the little arrows

A plain letter $r$ is a single number (a length). An arrow letter $r = (x, y, z)$ is three numbers — one for each axis. The relationship between them: $r = ∣ r ∣ = x^{2} + y^{2} + z^{2} (the length of the arrow) .$

Figure 2. The yellow $r$ arrow points from the origin (Earth) to the satellite; its blue dotted shadows show the $x$ and $y$ components. The pink $v$ arrow, attached at the satellite, points along the direction of travel. Plain $r$ is just the length of the yellow arrow.

5. The ellipse and its two size numbers: $a$ and $e$

An ellipse is a squashed circle. Two numbers describe any ellipse completely.

Figure 3. A blue ellipse with Earth (yellow) at one focus. The pink double-arrow marks $a$ from centre to edge along the long axis. The two white dots on the long axis are perigee (closest to Earth) and apogee (farthest).

The two special points on the long axis have their own names, both used constantly by the parent:

6. Angles measured at the focus: $ν$ (true anomaly)

The satellite is somewhere on the ellipse. We need one angle to say where.

This single angle is the only element that changes with time (all others describe the fixed ellipse).

The range of $ν$ in every conic case

The allowed span of $ν$ is not the same for all orbits — the shape ladder from §5 controls it, because $r$ must stay positive and finite.

Definition Where

$ν$ is allowed to point — all four cases Look at $r = \frac{p}{1 + e cos ν}$ . The radius blows up ( $r \to \infty$ ) exactly when the denominator hits zero, i.e. when $cos ν = - 1/ e$ .

Circle ( $e = 0$ ): denominator is always $1$ ; $ν$ runs the full $0^{\circ}$ to $36 0^{\circ}$ freely (but the starting point is ambiguous — see §7).
Ellipse ( $0 < e < 1$ ): $- 1/ e < - 1$ , so $cos ν = - 1/ e$ is impossible; $ν$ runs the full $0^{\circ}$ – $36 0^{\circ}$ , closed loop, no restriction.
Parabola ( $e = 1$ ): $cos ν = - 1$ at $ν = 18 0^{\circ}$ , so $r \to \infty$ there. $ν$ is restricted to the open interval $- 18 0^{\circ} < ν < 18 0^{\circ}$ ; the endpoints $\pm 18 0^{\circ}$ are the point at infinity and are never actually reached.
Hyperbola ( $e > 1$ ): $- 1/ e$ lies between $- 1$ and $0$ , so there are two asymptote angles $ν_{\infty} = \pm arccos (- 1/ e)$ . $ν$ is restricted to $- ν_{\infty} < ν < ν_{\infty}$ ; beyond those the formula gives negative $r$ (physically meaningless — that branch is the "other" hyperbola).

Where the orbit equation comes from

The parent asserts $r = \frac{a ( 1 - e ^{2} )}{1 + e c o s ν} = \frac{p}{1 + e c o s ν} .$ Let us earn the factor $(1 - e^{2})$ instead of trusting it. An ellipse can be defined by its two focus points $F$ (Earth) and $F^{'}$ (the empty focus): for every point on the ellipse, the two focus-distances add to the same constant, $2 a$ : $r + r^{'} = 2 a, where r^{'} = distance to the empty focus .$

Step 1 — set up the two triangles. The two foci are separated by $2 c$ , where $c = a e$ is the centre-to-focus distance (that is the definition of $e$ : $e = c / a$ ). Place the satellite at true anomaly $ν$ . Draw the triangle with sides $r$ , $r^{'}$ and base $2 c$ . The angle at Earth's focus between $r$ and the long axis is $ν$ .

Step 2 — law of cosines on that triangle. The side $r^{'}$ opposite the angle $(18 0^{\circ} - ν)$ satisfies $r^{'2} = r^{2} + (2 c)^{2} - 2 r (2 c) cos (18 0^{\circ} - ν) = r^{2} + 4 c^{2} + 4 r c cos ν .$

Step 3 — use $r^{'} = 2 a - r$ . Square it: $r^{'2} = 4 a^{2} - 4 a r + r^{2}$ . Set the two expressions for $r^{'2}$ equal, and the $r^{2}$ cancels: $4 a^{2} - 4 a r = 4 c^{2} + 4 r c cos ν .$

Step 4 — solve for $r$ . Divide by $4$ and gather the $r$ terms: $a^{2} - c^{2} = r (a + c cos ν) \Rightarrow r = \frac{a ^{2} - c ^{2}}{a + c c o s ν} .$ Now substitute $c = a e$ : numerator $a^{2} - a^{2} e^{2} = a^{2} (1 - e^{2})$ , denominator $a (1 + e cos ν)$ . One factor of $a$ cancels: $r = \frac{a ( 1 - e ^{2} )}{1 + e cos ν} = \frac{p}{1 + e cos ν} p \equiv a (1 - e^{2}) .$ So the mysterious $(1 - e^{2})$ is nothing but $a^{2} - c^{2}$ over $a$ — a direct fingerprint of "how far the focus sits off-centre." That is why $p = a (1 - e^{2})$ .

7. Three tilt angles: $i$ , $Ω$ , $ω$

So far the ellipse has been drawn flat on a page. In reality it hangs at an angle in 3D. Three more angles orient it. To describe orientation we first need a fixed set of reference directions.

Figure 4. The white oval is the equatorial plane; the yellow arrow $X$ points to the vernal equinox $♈$ and the blue arrow $Z$ points North. The pink oval is the tilted orbit plane; the angle it leans is the inclination $i$ , and the angle from $X$ round to the pink ascending node (marked) is the RAAN $Ω$ .

Common mistake Degenerate orbits — when

$Ω$ or $ω$ stop making sense These two angles are defined only when the geometry they point at exists. Two cases break them:

Equatorial orbit ( $i = 0^{\circ}$ or $i = 18 0^{\circ}$ ): the orbit plane and the equator are the same plane, so they never cross — there is no line of nodes and no ascending node. Then $Ω$ is undefined (there is nothing to measure the swing to). Fix: engineers use the longitude of perigee $ϖ = Ω + ω$ instead, or just set $Ω = 0$ .
Circular orbit ( $e = 0$ ): a circle has no perigee (every point is equidistant), so there is no closest point to measure from — $ω$ is undefined. Fix: use the argument of latitude $u = ω + ν$ (node → satellite), which stays perfectly well-defined.
Circular and equatorial ( $e = 0, i = 0$ ): both fail at once; only the true longitude $ℓ = Ω + ω + ν$ survives. These aren't bugs — they are honest gaps where the chosen coordinates have nothing to grab onto.

8. Two conserved companions the parent leans on: $h$ and $ε$

Why energy fixes the size $a$ — the full derivation

The parent boxes $ε = - μ /2 a$ but does not show why. Here it is, from scratch, using only conservation.

Step 1 — write $ε$ at the two apsides. At perigee and apogee, $r ⊥ v$ , so $h = r v$ , giving $v = h / r$ . Substitute into $ε$ : $ε = \frac{h ^{2}}{2 r _{p}^{2}} - \frac{μ}{r _{p}} and ε = \frac{h ^{2}}{2 r _{a}^{2}} - \frac{μ}{r _{a}} .$ Both equal the same $ε$ because energy is conserved.

Step 2 — set them equal and isolate $h^{2}$ . $\frac{h ^{2}}{2} (\frac{1}{r _{p}^{2}} - \frac{1}{r _{a}^{2}}) = μ (\frac{1}{r _{p}} - \frac{1}{r _{a}}) .$ Factor the left side using $\frac{1}{r _{p}^{2}} - \frac{1}{r _{a}^{2}} = (\frac{1}{r _{p}} - \frac{1}{r _{a}}) (\frac{1}{r _{p}} + \frac{1}{r _{a}})$ , then cancel the common factor $(\frac{1}{r _{p}} - \frac{1}{r _{a}})$ : $\frac{h ^{2}}{2} (\frac{1}{r _{p}} + \frac{1}{r _{a}}) = μ \Rightarrow h^{2} = \frac{2 μ}{\frac{1}{r _{p}} + \frac{1}{r _{a}}} = \frac{2 μ r _{p} r _{a}}{r _{p} + r _{a}} .$

Step 3 — evaluate $h^{2}$ with the geometry. With $r_{p} + r_{a} = 2 a$ and $r_{p} r_{a} = a (1 - e) \cdot a (1 + e) = a^{2} (1 - e^{2})$ : $h^{2} = \frac{2 μ a ^{2} ( 1 - e ^{2} )}{2 a} = μ a (1 - e^{2}) .$

Step 4 — substitute back into $ε$ at perigee. Use $ε = \frac{h ^{2}}{2 r _{p}^{2}} - \frac{μ}{r _{p}}$ with $r_{p} = a (1 - e)$ and $h^{2} = μ a (1 - e^{2})$ : $ε = \frac{μ a ( 1 - e ^{2} )}{2 a ^{2} ( 1 - e ) ^{2}} - \frac{μ}{a ( 1 - e )} .$ In the first term, write $1 - e^{2} = (1 - e) (1 + e)$ ; one factor of $(1 - e)$ and one of $a$ cancel: $ε = \frac{μ ( 1 + e )}{2 a ( 1 - e )} - \frac{μ}{a ( 1 - e )} .$

Step 5 — combine over the common factor $\frac{μ}{a ( 1 - e )}$ . $ε = \frac{μ}{a ( 1 - e )} (\frac{1 + e}{2} - 1) = \frac{μ}{a ( 1 - e )} \cdot \frac{1 + e - 2}{2} = \frac{μ}{a ( 1 - e )} \cdot \frac{e - 1}{2} .$

Step 6 — the final cancellation. Note $e - 1 = - (1 - e)$ , so the $(1 - e)$ in the denominator cancels the one in the numerator: $ε = \frac{μ}{a ( 1 - e )} \cdot \frac{- ( 1 - e )}{2} = - \frac{μ}{2 a} .$ Every trace of $e$ has vanished — energy depends only on $a$ . That is exactly why knowing the energy locks the size of the orbit, no matter its shape or where the satellite sits. Rearranging $ε = v^{2} /2 - μ / r = - μ /2 a$ gives back the vis-viva equation $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ promised in §2.

The six Keplerian elements — one consolidated list

Formula The complete set (memorise these six)

Symbol	Name	Job
$a$	semi-major axis	size of the ellipse
$e$	eccentricity	shape (squash) of the ellipse
$i$	inclination	tilt of the plane off the equator
$Ω$	RAAN	swing of the tilted plane around $Z$
$ω$	argument of perigee	where perigee points within the plane
$ν$	true anomaly	where the satellite is right now (the only time-varying one)
The first five are constant under ideal two-body gravity; only $ν$ moves.

How the foundations feed the topic

Legend: $r$ = distance from focus; $v$ = speed; $μ$ = gravity parameter $GM$ ; $ε$ = specific energy; $a, e$ = ellipse size and shape; $r, v$ = position and velocity arrows; $h$ = angular-momentum vector $r \times v$ ; $ν$ = true anomaly; ECI = the fixed reference frame; $i, Ω, ω$ = the three tilt angles; KEP = the six Keplerian elements that give full 3D position.

Equipment checklist

What does a plain letter

r

mean vs an arrow letter

r

r

is a single number (a distance);

r

is an arrow — three numbers giving distance and direction.

What is

G

and how does it differ from

μ

G

is the universal gravitational constant (same everywhere);

μ = GM

folds in this body's mass to give its specific pull.

State the vis-viva equation in words.

v^{2} = μ (2/ r - 1/ a)

— speed-squared at any point depends only on your distance

r

and the orbit size

a

Why does an orbit need exactly six numbers?

Because position (3) plus velocity (3) is six, and those six fix all future motion.

Name the six Keplerian elements and their jobs.

a

size,

e

shape,

i

tilt,

Ω

swing,

ω

perigee direction,

ν

current position.

How do you turn altitude into

r

Add Earth's radius:

r = altitude + R_{E}

What do

a

and

e

describe, and what does

e = 1

mean?

a

is size,

e

is squash;

e = 1

is the parabolic escape boundary between bound and unbound orbits.

Over what range does

ν

run for an ellipse vs a hyperbola?

Ellipse: full