Converting between orbital elements and state vectors (r, v): Worked examples

First, every symbol earned

We will use these on the very first line of the scenario matrix, so let us build them all now — no symbol appears before this list.

Definition The symbols we will use

$r$ — position vector: an arrow from Earth's centre to the spacecraft, measured in km.
$v$ — velocity vector: an arrow whose length is the speed and whose direction is the motion, in km/s.
$r = ∣ r ∣$ , $v = ∣ v ∣$ — the plain lengths of those arrows (no direction, just size).
$μ = GM$ — the "gravitational strength" of the central body. For Earth $μ = 398600 km^{3} / s^{2}$ .
$h = r \times v$ — the cross product: an arrow perpendicular to both $r$ and $v$ , so it sticks straight out of the orbit plane. Its length $h$ measures how fast area is swept. Its third component $h_{z}$ is the "how much does it lean north" part.
$ε = \frac{v ^{2}}{2} - \frac{μ}{r}$ — energy per kilogram. Negative = bound (ellipse), zero = parabola, positive = escaping (hyperbola).
$\overset{z}{^} = (0, 0, 1)$ — the "north" reference direction of the fixed frame; the reference plane is the flat $x y$ -plane (Earth's equator).
$i$ — inclination: the tilt angle between $h$ and $\overset{z}{^}$ , i.e. how steeply the orbit plane leans away from the equator. It runs from $0^{\circ}$ (flat, in the equator) to $18 0^{\circ}$ (flat but going the "wrong way," retrograde). Found from $cos i = h_{z} / h$ .
$n = \overset{z}{^} \times h$ — the node vector: an arrow lying in the equator, pointing at the ascending node (the spot where the orbit crosses the equator heading north). Its length is $n = ∣ n ∣$ , and its components are $n_{x}, n_{y}$ . If the orbit lies flat in the equator, $h ∥ \overset{z}{^}$ and $n$ collapses to zero.
$e$ — the eccentricity vector: an arrow pointing from Earth toward periapsis (the closest point of the orbit). Its length $e = ∣ e ∣$ is the eccentricity (shape). Its third component is $e_{z}$ . Formula: $e = \frac{1}{μ} [(v^{2} - μ / r) r - (r \cdot v) v]$ .
$Ω$ — RAAN: the angle in the equator from the $x$ -axis round to the node vector $n$ .
$ω$ — argument of periapsis: the angle from the node $n$ round to periapsis $e$ , measured in the orbit plane.
$ν$ — true anomaly: the angle from periapsis $e$ round to the current position $r$ .
$r \cdot v$ — the dot product: a single number that is positive when $r$ and $v$ point "with" each other (satellite climbing away, $r$ growing) and negative when "against" ( $r$ shrinking, falling inward). It is our tie-breaker for $ν$ .
$p$ — semi-latus rectum: the orbit's "width scale," $p = a (1 - e^{2}) = h^{2} / μ$ . It is the radius $r$ when $ν = 9 0^{\circ}$ , and unlike $a$ it stays finite even for a parabola.
$u = ω + ν$ — argument of latitude: the angle from the ascending node straight to the satellite. Used when $e = 0$ kills $ω$ .
$ϖ = Ω + ω$ — true longitude of periapsis: periapsis direction measured from the fixed $x$ -axis. Used when $i = 0$ kills $Ω$ .
$ℓ$ — true longitude: the angle of $r$ from the $x$ -axis. Used when the orbit is both circular and equatorial.

Definition The rotation matrix

$\tilde{R}$ (perifocal → inertial) To turn a perifocal vector into an inertial one we apply the single combined matrix. We use the compact shorthand $c θ = cos θ$ and $s θ = sin θ$ ; when $θ$ is one of our elements this reads $c O = cos Ω$ , $s O = sin Ω$ , $c w = cos ω$ , $s w = sin ω$ , and $c i = cos i$ , $s i = sin i$ — that is, $c i$ and $s i$ are just this same shorthand applied to the inclination $i$ .

c\Omega\,c\omega - s\Omega\,s\omega\,ci & -c\Omega\,s\omega - s\Omega\,c\omega\,ci & s\Omega\,si\\ s\Omega\,c\omega + c\Omega\,s\omega\,ci & -s\Omega\,s\omega + c\Omega\,c\omega\,ci & -c\Omega\,si\\ s\omega\,si & c\omega\,si & ci \end{bmatrix}$$ Then $\vec r=\tilde R\,\vec r_{pqw}$ and $\vec v=\tilde R\,\vec v_{pqw}$. Being a rotation, $\tilde R$ never changes a vector's length — a fact we use to sanity-check Ex 6.

The scenario matrix

Now that every symbol is defined, we list what can vary so we make sure nothing is skipped. Each row is a "cell" of behaviour; the worked examples are tagged with the cell(s) they hit.

#	Cell (the thing being tested)	Why it is dangerous	Covered by
C1	True anomaly $ν$ in the second half ( $r \cdot v < 0$ )	$cos^{- 1}$ hides it; must flip with the sign of $r \cdot v$	Ex 1
C2	RAAN $Ω$ with $n_{y} < 0$ (node past $18 0^{\circ}$ )	same $cos^{- 1}$ trap on $Ω$	Ex 2
C3	Argument of periapsis $ω$ with $e_{z} < 0$	same trap on $ω$	Ex 2
C4	Degenerate: circular non-equatorial orbit ( $e = 0$ , $i \neq = 0$ ) → $ω$ undefined	division by $e = 0$ ; use argument of latitude $u$	Ex 3
C5	Degenerate: equatorial orbit ( $i = 0$ ) → $Ω$ undefined	node vector $n \to 0$	Ex 4
C6	Limiting: hyperbola ( $e > 1$ , $ε > 0$ , $a < 0$ )	sign of $a$ flips; must not panic	Ex 5
C7	Elements → State with $ν$ off periapsis (full rotation)	the $\tilde{R}$ matrix must be applied correctly	Ex 6
C8	Real-world word problem (ISS-like circular orbit)	translating English into vectors	Ex 7
C9	Exam twist: parabola limit ( $e \to 1$ , $ε = 0$ )	$a \to \infty$ ; must use $p$ , not $a$	Ex 8

Ex 1 — Second-half true anomaly (cell C1)

Forecast: guess first — is the satellite heading toward periapsis or away from it? If $r$ and $v$ point "with" each other (positive dot product), it is climbing away; if "against" each other, falling in. Jot your guess.

Compute $r$ and $v$ . $r = 604 5^{2} + 349 0^{2} + 250 0^{2} = 7414.0$ km, $v = 3.45 7^{2} + 6.61 8^{2} + 2.53 3^{2} = 7.884$ km/s. Why this step? We need the plain lengths before any ratio; the formulas for $e$ and $ν$ use $r$ and $v$ directly.
Angular momentum. $h = r \times v = (- 25384.4, 6669.5, - 52070.7)$ , $h = 58311.7$ . Why this step? $h$ fixes the plane and lets us build $e$ next.
Eccentricity vector. Using $e = \frac{1}{μ} [(v^{2} - μ / r) r - (r \cdot v) v]$ , first compute the dot product carefully: $r \cdot v = (- 6045) (- 3.457) + (- 3490) (6.618) + (2500) (2.533) = 20897.9 - 23096.8 + 6332.5 = + 4133.6$ . Then $e = (- 0.1090, - 0.1244, 0.02386)$ , so $e = 0.1655 \approx 0.166$ . Why this step? $e$ points at periapsis; $ν$ is the angle from $e$ to $r$ . Note the dot product came out positive — hold that thought for step 5.
True anomaly (raw). $cos ν = \frac{e \cdot r}{e r} = 0.6295 \Rightarrow ν = 51. 0^{\circ}$ or $309. 0^{\circ}$ . Why this step? $cos^{- 1}$ cannot tell $51. 0^{\circ}$ from $309. 0^{\circ}$ — both have the same cosine. We must break the tie.
Quadrant fix. $r \cdot v = + 4133.6 > 0$ , meaning the satellite is moving away from periapsis, so $ν < 18 0^{\circ}$ . Take $ν = 51. 0^{\circ}$ . Why this step? Positive $r \cdot v$ = $r$ is growing = we are on the outbound half. This is exactly the sign rule from the parent recipe.

Verify: the sign was positive, so we kept the first-half value — matching the physical meaning "climbing away from periapsis." Had it been negative we would have flipped to $309. 0^{\circ}$ . ✓ (units: $cos ν$ is dimensionless, a ratio of $km^{2} \cdot$ km $^{- 1} \cdot$ km $^{- 1}$ — clean.)

Ex 2 — RAAN and $ω$ in the far quadrants (cells C2, C3)

Forecast: with $h_{z} < 0$ (from Ex 1, $h_{z} = - 52070.7$ ), the orbit's "north pole" points below the equator. Guess: is $i$ less than or greater than $9 0^{\circ}$ ?

Inclination. $cos i = \frac{h _{z}}{h} = \frac{- 52070.7}{58311.7} = - 0.8930 \Rightarrow i = 153. 2^{\circ}$ . Why this step? $i$ (defined above) is the tilt of $h$ from $\overset{z}{^}$ ; a negative $h_{z}$ means $h$ points below the equator → retrograde ( $i > 9 0^{\circ}$ ). No quadrant fix needed: $i$ lives only in $[0, 18 0^{\circ}]$ .
Node vector. $n = \overset{z}{^} \times h = (- h_{y}, h_{x}, 0) = (- 6669.5, - 25384.4, 0)$ , $n = 26247.9$ . Why this step? $n$ (defined above) points along the line where the orbit crosses the reference plane going up — the ascending node, the anchor for $Ω$ .
RAAN (raw). $cos Ω = \frac{n _{x}}{n} = \frac{- 6669.5}{26247.9} = - 0.2541 \Rightarrow Ω = 104. 7^{\circ}$ or $255. 3^{\circ}$ . Why this step? Again $cos^{- 1}$ is two-valued.
RAAN fix. $n_{y} = - 25384.4 < 0 \Rightarrow Ω = 36 0^{\circ} - 104. 7^{\circ} = 255. 3^{\circ}$ . Why this step? The parent rule: if $n_{y} < 0$ the node is in the lower half-plane, so $Ω$ is past $18 0^{\circ}$ .
Argument of periapsis (raw + fix). $cos ω = \frac{n \cdot e}{n e}$ . With $e = (- 0.1090, - 0.1244, 0.02386)$ (from Ex 1) and $n \cdot e = (- 6669.5) (- 0.1090) + (- 25384.4) (- 0.1244) = 727.0 + 3157.8 = 3884.8$ , $cos ω = \frac{3884.8}{26247.9 \times 0.1655} = 0.8944 \Rightarrow ω = 26. 6^{\circ}$ or $333. 4^{\circ}$ . Since $e_{z} = + 0.02386 > 0$ , keep $ω = 26. 6^{\circ}$ . Why this step? $e_{z} > 0$ means periapsis is above the equator, so $ω < 18 0^{\circ}$ .

Verify: all three angles land in their proper ranges ( $i \in [0, 180]$ , $Ω \in [0, 360)$ , $ω \in [0, 360)$ ). The retrograde read ( $i = 153. 2^{\circ}$ ) is consistent with $h_{z} < 0$ . ✓

Ex 3 — Circular non-equatorial orbit: $ω$ undefined, use $u$ (cell C4)

Forecast: if the orbit is a perfect circle but still tilted, where is periapsis — and can we still measure "how far along the orbit" the satellite is? Guess before reading.

Radius and angular momentum. $r = 495 0^{2} + 495 0^{2} = 7000$ km. $h = r \times v = (4950 \cdot 5.405 - 0, 0 - 4950 \cdot 5.405, 4950 \cdot 3.822 + 4950 \cdot 3.822) = (26754.8, - 26754.8, 37837.8)$ , $h = 52822$ . Why this step? $h_{z} \neq = 0$ but $h_{x}, h_{y} \neq = 0$ too, so the plane is tilted — this is not equatorial. Good, that's the point of the cell.
Eccentricity. $v = 3.82 2^{2} + 3.82 2^{2} + 5.40 5^{2} = 7.546$ , $v^{2} - μ / r = 56.94 - 56.94 = 0$ , and $r \cdot v = 4950 (- 3.822) + 4950 (3.822) + 0 = 0$ . Both terms vanish, so $e = 0$ , $e = 0$ . Why this step? A zero eccentricity vector has no direction — there is no periapsis to point at.
The singularity in $ω$ . $cos ω = \frac{n \cdot e}{n e}$ has $e = 0$ in the denominator → $0/0$ , undefined. But the node vector survives: $n = \overset{z}{^} \times h = (26754.8, 26754.8, 0)$ , $n = 37837.8 \neq = 0$ . Why this step? Because $i \neq = 0$ the node is well-defined; only $ω$ (which needs periapsis) breaks. This is the pure $e = 0$ case, unpolluted by an equatorial collapse.
The fix — argument of latitude $u$ . Measure the angle straight from the node $n$ to the satellite $r$ : $cos u = \frac{n \cdot r}{n r} = \frac{( 26754.8 ) ( 4950 ) + ( 26754.8 ) ( 4950 )}{37837.8 \times 7000} = \frac{264872520}{264864600} = 1.000 \Rightarrow u = 0^{\circ}$ . Why this step? $u = ω + ν$ folds the undefined $ω$ and the position $ν$ into a single angle we can measure — from the surviving node. Here $u = 0^{\circ}$ means the satellite sits right at the ascending node. See Orbital singularities and equinoctial elements.

Verify: $e = 0$ ✓ and $r = a = 7000$ km (circular). Vis-viva check: $v = μ / r = 398600/7000 = 7.546$ km/s ✓. Since $r = (4950, 4950, 0)$ lies exactly along $n = (26754.8, 26754.8, 0)$ , they are parallel, so $u = 0^{\circ}$ ✓.

Ex 4 — Equatorial orbit: $Ω$ undefined (cell C5)

Forecast: if the orbit never leaves the equator, there is no "crossing point" going up. What happens to the ascending node?

Angular momentum. $h = (0, 0, 8000 \times 6.5) = (0, 0, 52000)$ , purely $+ \overset{z}{^}$ . Why this step? $h ∥ \overset{z}{^}$ means $i = 0$ (the plane is the reference plane).
Inclination. $cos i = h_{z} / h = 52000/52000 = 1 \Rightarrow i = 0^{\circ}$ . Why this step? Confirms the equatorial case.
Node vector vanishes. $n = \overset{z}{^} \times h = \overset{z}{^} \times (52000 \overset{z}{^}) = 0$ , $n = 0$ . Why this step? The node is defined as where the orbit rises through the plane — but it never leaves the plane, so there is no such point. $Ω = cos^{- 1} (n_{x} / n)$ is $0/0$ .
Eccentricity survives. $v = 6.5$ , $v^{2} - μ / r = 42.25 - 49.825 = - 7.575$ , $r \cdot v = 0$ , so $e = \frac{- 7.575}{398600} (8000, 0, 0) = (- 0.1520, 0, 0)$ , $e = 0.1520$ . Periapsis lies in the $- x$ direction, so the satellite is currently at apoapsis. Why this step? Shows eccentricity is fine even when $Ω$ is not — only the node-dependent element breaks.
The fix — true longitude of periapsis. Replace $Ω$ and $ω$ by $ϖ = Ω + ω$ (defined above), here just the angle of $e$ from the $x$ -axis: $ϖ = 18 0^{\circ}$ . Why this step? With no node to measure from, we measure periapsis directly from the fixed $x$ -axis.

Verify: $i = 0$ ✓, $n = 0$ ✓ (equatorial), $e = 0.152 \neq = 0$ so the shape is a genuine ellipse. Energy $ε = \frac{6. 5 ^{2}}{2} - \frac{398600}{8000} = 21.125 - 49.825 = - 28.700$ , so $a = - μ /2 ε = 6944$ km. Then apoapsis distance $a (1 + e) = 6944 (1.152) = 8000$ km $= r$ — the satellite is indeed at apoapsis. ✓

Ex 5 — Hyperbolic flyby: $a < 0$ (cell C6)

Forecast: with a whopping 12 km/s at 6000 km, is this bound or escaping? Guess the sign of $ε$ .

Energy. $v = 12$ , $ε = \frac{144}{2} - \frac{398600}{6000} = 72 - 66.433 = + 5.567$ . Why this step? Positive energy = unbound = hyperbola. The formula $a = - μ /2 ε$ still applies.
Semi-major axis. $a = - \frac{398600}{2 \times 5.567} = - 35802$ km. Why this step? A negative $a$ is correct and normal for a hyperbola — it is not an error to be "fixed."
Eccentricity. $r \cdot v = 0$ , $v^{2} - μ / r = 144 - 66.433 = 77.567$ , $e = \frac{77.567}{398600} (6000, 0, 0) = (1.1676, 0, 0)$ , $e = 1.168 > 1$ . Why this step? $e > 1$ is the algebraic signature of a hyperbola, consistent with $ε > 0$ .

Verify: cross-check via $p = h^{2} / μ$ and $e = 1 + 2 ε h^{2} / μ^{2}$ : $h = 6000 \times 12 = 72000$ , $e = 1 + \frac{2 ( 5.567 ) ( 7200 0 ^{2} )}{39860 0 ^{2}} = 1.168$ ✓. Both routes agree, and $a (1 - e^{2}) = p$ holds even with $a < 0$ . ✓

Ex 6 — Elements → State off periapsis (cell C7)

Forecast: at $ν = 9 0^{\circ}$ the satellite is a quarter-orbit past periapsis. Guess whether $r$ is bigger or smaller than the periapsis distance $a (1 - e) = 7200$ km.

Figure — Converting between orbital elements and state vectors (r, v)

Semi-latus rectum & radius. $p = a (1 - e^{2}) = 8000 (0.99) = 7920$ ; $r = \frac{p}{1 + e cos 9 0 ^{\circ}} = \frac{7920}{1 + 0} = 7920$ km. Why this step? The orbit equation gives the distance for any $ν$ ; at $9 0^{\circ}$ , $cos ν = 0$ so $r = p$ exactly (this is precisely why $p$ is called the orbit's "width scale"). In the figure: this radius is the length of the yellow arrow.
Perifocal position — read it off the figure. $r_{pq w} = r (cos ν, sin ν, 0) = 7920 (cos 9 0^{\circ}, sin 9 0^{\circ}, 0) = 7920 (0, 1, 0) = (0, 7920, 0)$ km. Why this step? In the perifocal frame periapsis sits on the $\overset{p}{^}$ axis (green dot in the figure). Because $ν = 9 0^{\circ}$ , the yellow arrow $r_{pq w}$ rotates a quarter-turn off $\overset{p}{^}$ and points straight up the $\overset{q}{^}$ axis — that is why only the middle ( $q$ ) component is non-zero. This is how $(a, e, ν)$ became the concrete vector $(0, 7920, 0)$ .
Rotate to inertial. Apply $\tilde{R}$ (defined above, with $c i = cos i$ , $s i = sin i$ ) using $c Ω = cos 4 0^{\circ} = 0.766$ , $s Ω = sin 4 0^{\circ} = 0.643$ , $c ω = cos 6 0^{\circ} = 0.5$ , $s ω = sin 6 0^{\circ} = 0.866$ , $c i = cos 3 0^{\circ} = 0.866$ , $s i = sin 3 0^{\circ} = 0.5$ . Since $r_{pq w} = (0, 7920, 0)$ has only a $q$ -component, it selects the middle column of $\tilde{R}$ : $r_{x} = 7920 (- c Ω s ω - s Ω c ω c i) = 7920 (- 0.766 \cdot 0.866 - 0.643 \cdot 0.5 \cdot 0.866) = - 7459$ km; $r_{y} = 7920 (- s Ω s ω + c Ω c ω c i) = 7920 (- 0.643 \cdot 0.866 + 0.766 \cdot 0.5 \cdot 0.866) = - 1783$ km; $r_{z} = 7920 (c ω s i) = 7920 (0.5 \cdot 0.5) = 1980$ km. Why this step? We transform from perifocal to inertial, which uses the negative-angle rotations baked into $\tilde{R}$ . In the figure: the red arrow is this tilted-and-swivelled result — same length as the yellow arrow, new direction.

Verify: length preserved by rotation: $∣ r ∣ = 745 9^{2} + 178 3^{2} + 198 0^{2} = 7920$ km $= r$ ✓ (rotations never stretch). Units all km. ✓

Ex 7 — Real-world: ISS-like circular orbit (cell C8)

Forecast: roughly how fast does the ISS go — closer to 3 km/s or 8 km/s? Guess.

Circular speed via vis-viva. For $e = 0$ , $r = a$ , so $v = μ / r = 398600/6798 = 7.657$ km/s. Why this step? Vis-viva $v^{2} = μ (2/ r - 1/ a)$ reduces to $μ / r$ when $a = r$ .
Period. $T = 2 π a^{3} / μ = 2 π 679 8^{3} /398600 = 5578$ s $= 92.97$ min. Why this step? Kepler's third law gives the time to sweep the whole ellipse; see Kepler's equation and time-of-flight.
A consistent velocity. For a circular equatorial orbit with $r$ along $+ x$ , the velocity is perpendicular and in-plane: $v = (0, 7.657, 0)$ km/s. Why this step? Circular ⇒ velocity purely tangential (no radial part), and equatorial ⇒ no $z$ -component.

Verify: $92.97$ min matches the real ISS period ( $\approx 93$ min) ✓. And $r \cdot v = 0$ confirms zero radial speed, exactly what "circular" demands. ✓

Ex 8 — Exam twist: the parabolic limit (cell C9)

Forecast: what is $ε$ for a parabola? And what does $a = - μ /2 ε$ do when $ε = 0$ ?

Escape speed. $v_{esc} = 2 μ / r = 2 \times 398600/10000 = 8.929$ km/s. Why this step? Parabola = exactly zero total energy = just barely escaping.
Energy. $ε = \frac{v _{esc}^{2}}{2} - \frac{μ}{r} = \frac{79.72}{2} - 39.86 = 39.86 - 39.86 = 0$ . Why this step? Confirms the parabola's defining property $ε = 0$ .
Why $a$ breaks. $a = - μ / (2 \times 0) = - \infty$ . A parabola has no finite semi-major axis — its two "ends" run off to infinity. Why this step? Shows why textbooks parametrise open orbits by $p$ (or perigee $r_{p}$ ), never $a$ .
Use $p$ instead. With $r \cdot v = 0$ (at perigee), $h = r v = 10000 \times 8.929 = 89290$ , $p = h^{2} / μ = 8929 0^{2} /398600 = 20000$ km. And indeed $r_{p} = p / (1 + e) = 20000/2 = 10000$ km $= r$ ✓. Why this step? $p$ stays finite and fully describes the parabola; perigee radius $= p /2$ for $e = 1$ .

Verify: $ε = 0$ exactly ✓; $r_{p} = 10000$ km recovers the given position ✓. A parabola is the clean boundary between the ellipses ( $ε < 0$ , Ex 3–4) and hyperbolas ( $ε > 0$ , Ex 5). ✓

Recall Which companion sign fixes which angle?

For $Ω$ , check $n_{y}$ ::: if $n_{y} < 0$ then $Ω = 36 0^{\circ} - Ω_{raw}$ . For $ω$ , check $e_{z}$ ::: if $e_{z} < 0$ then $ω = 36 0^{\circ} - ω_{raw}$ . For $ν$ , check $r \cdot v$ ::: if $r \cdot v < 0$ then $ν = 36 0^{\circ} - ν_{raw}$ .

Flashcards

When is semi-major axis

a

negative?

For hyperbolas, where

ε > 0

, since

a = - μ /2 ε

What replaces

ω

for a circular non-equatorial orbit?

The argument of latitude

u = ω + ν

, measured from the ascending node to the satellite.

Why is

a

useless for a parabola?

Because

ε = 0

makes

a = - μ / (2 ε) \to \infty

; use the semi-latus rectum

p = h^{2} / μ

instead.

Which companion sign tells you

ν

is past

18 0^{\circ}

A negative

r \cdot v

(radius shrinking, falling toward periapsis).