Converting between orbital elements and state vectors (r, v): Question bank

Definition Symbols and tools you'll need on THIS page (so you don't jump away)

$μ = GM$ — the gravitational parameter: the central body's mass $M$ times Newton's constant $G$ , bundled into one number (km³/s²). It sets the strength of gravity's pull; every speed, energy, and period formula below carries it.
$ε = \frac{v ^{2}}{2} - \frac{μ}{r}$ — specific energy (energy per kg): kinetic minus gravitational potential. Negative for bound orbits, zero for parabolic escape, positive for hyperbolic.
$h = r \times v$ — specific angular momentum: perpendicular to the orbit plane.
$e$ — eccentricity vector: points to periapsis, magnitude $= e$ .
$n = \overset{z}{^} \times h$ — node vector: points to the ascending node.
Rotation matrices (the recipe's step-B tools). $R_{3} (θ)$ turns a vector about the $\overset{z}{^}$ axis by angle $θ$ ; $R_{1} (θ)$ turns about the $\overset{x}{^}$ axis. Look at figure s01 for what "about that axis" and "positive angle" mean.

A positive rotation is counter-clockwise when you look down the axis toward the origin (right-hand rule). The recipe uses the negative-angle versions because it undoes the frame build-up (see the "Spot the error" note below and figure s01).

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

True or false — justify

Six numbers, one ellipse. Decide, then defend.

State vectors

(r, v)

and elements

(a, e, i, Ω, ω, ν)

carry exactly the same information.

True in general — both are 6 numbers describing the same orbit, and conversion loses no physics. Caveat: at the circular (

e = 0

) or equatorial (

i = 0

) singularities some angles (

ω

Ω

) become undefined, so the classical element set is momentarily incomplete even though

(r, v)

stays perfectly valid.

The eccentricity vector

e

always points from the focus toward apoapsis.

False — it points toward periapsis, the closest point. At periapsis

r

lies along

e

, so

cos ν = e \cdot r / (er) = 1

and

ν = 0

Because

h = r \times v

is constant, the orbit plane never changes in the two-body problem.

True — constant

h

means a fixed perpendicular direction, so both

r

and

v

stay in the one plane

h

is normal to; that's exactly why

h

fixes

i

and

Ω

Specific angular momentum

h

points along the orbit's motion.

False —

h

is the cross product

r \times v

, so it is perpendicular to the plane of motion (it points "out of" the orbit), not along the path.

A larger semi-major axis

a

always means a faster satellite.

False — by vis-viva, larger

a

means lower total speed on average (higher, slower orbits); energy

ε = - μ / (2 a)

rises toward zero as

a

grows, meaning less kinetic energy for a given

r

r \cdot v = 0

at some instant, the body is at either periapsis or apoapsis.

True —

r \cdot v

is the rate of change of distance; zero means

r

is momentarily neither growing nor shrinking, which happens only at the two turning points of the ellipse.

The node vector

n = \overset{z}{^} \times h

points toward periapsis.

False —

n

points toward the ascending node (where the orbit crosses the reference plane going up). Periapsis is fixed by

e

, and the angle between

n

and

e

is the argument of periapsis

ω

For a hyperbolic trajectory the semi-major axis

a

comes out negative.

True — hyperbolas have positive specific energy

ε > 0

, and

a = - μ / (2 ε)

then forces

a < 0

; the formula handles all conics without a special case.

Two orbits with identical

(a, e, i, Ω, ω)

but different

ν

are different orbits.

False — same orbit, different positions along it;

ν

is the only element that changes with time, so it names the point, not the path.

Spot the error

Each line contains a plausible-sounding mistake. Name it and fix it. (See figure s01 for the rotation-axis convention.)

"To go from perifocal to inertial, rotate by

R_{3} (+ ω) R_{1} (+ i) R_{3} (+ Ω)

Wrong signs and order. Building the perifocal frame from inertial is a swivel

Ω

, tilt

i

, spin

ω

; going the other way is the inverse, and the inverse of a rotation product reverses the order and negates each angle. So the correct transform is

R_{3} (- Ω) R_{1} (- i) R_{3} (- ω)

"I used

ν = cos^{- 1} (\frac{e \cdot r}{er}) = 4 8^{\circ}

, done."

Missing the quadrant check —

cos^{- 1}

only returns

[0, π]

. If

r \cdot v < 0

(moving toward periapsis) the true value is

36 0^{\circ} - 4 8^{\circ}

. More robust: compute

ν = atan2 (r \cdot v / (scaled), e \cdot r)

style two-argument arctangent, which returns the full

[0, 2 π)

in one step with no sign patch. See figure s03.

e = 0.02

so the argument of periapsis is

ω = 0

by definition."

False reasoning — small

e

does not fix

ω

to zero;

ω

is still the well-defined angle from the node to periapsis. Only when

e = 0

exactly does

ω

become undefined.

cos i = h_{z} / h

, and since cosine can be negative I must check the sign of

i

separately."

No quadrant check needed for

i

— inclination lives in

[0, π]

, exactly the range

cos^{- 1}

returns, so

cos i = h_{z} / h

gives

i

uniquely with no ambiguity.

"For

v_{pq w}

I wrote

μ / p (- sin ν, cos ν, 0)

Dropped the

e

— the tangential component is

e + cos ν

, not

cos ν

. Forgetting it makes velocity wrong everywhere except the special points, and violates energy conservation.

a = μ / (2 ε)

gives the orbit size from energy."

Sign slip — the correct relation is

a = - μ / (2 ε)

. Why the minus: for a bound orbit the specific energy is negative (

ε < 0

, the body is trapped in the potential well), so

a = - μ / (2 ε)

flips that into a positive size. A plus sign would give a nonsensical negative

a

for an ordinary ellipse.

"The RAAN

Ω

is measured from periapsis."

Wrong reference. Why from $\overset{x}{^}$ :

Ω

describes how the orbit plane is swivelled, which is set entirely by where the plane crosses the equator — the ascending node. That crossing is measured in the reference plane from the fixed inertial direction

\overset{x}{^}

(the reference/vernal direction). Periapsis lies inside the plane and is measured separately by

ω

(from the node) and

ν

(from periapsis).

Why questions

The parent recipe does things — here's why they can't be skipped. (Step numbers refer to the one-glance recipe above.)

Why do steps 6, 7, 8 need a companion-sign check but step 5 (inclination) does not?

Because

i \in [0, π]

matches

cos^{- 1}

's output range exactly, but

Ω, ω, ν

span the full

[0, 2 π)

, so a second quantity's sign is needed to know which half of the circle the angle lives in. See figure s03.

Why is

r \cdot v

the right sign-flag for the true anomaly

ν

Because

r \cdot v > 0

means the body is moving away from the focus, i.e. climbing from periapsis toward apoapsis, so

ν \in (0, π)

; a negative value means descending, so

ν \in (π, 2 π)

Why does converting elements → state split into "build in PQW, then rotate"?

In the perifocal frame periapsis sits on one axis, so

r

and

v

have dead-simple

cos ν / sin ν

forms; all the messy orientation (

Ω, i, ω

) is then delivered by a single rotation matrix afterward.

Why does the eccentricity vector deserve to be derived rather than memorised?

Its derivation shows it is a genuine constant of motion (crossing the two-body equation with

h

integrates to a fixed vector), which is why it can reliably encode both

e

and the periapsis direction

ω

Why does

h

fix two elements (

i

and

Ω

) while

ε

fixes only one (

a

h

is a vector — its direction pins the plane's orientation (giving both tilt

i

and swivel

Ω

), while

ε

is a single scalar and can only set one size number,

a

Why does the vis-viva relation let a single speed measurement (with

r

) reveal

a

Because

v^{2} = μ (2/ r - 1/ a)

ties speed, position, and size together; knowing

v

and

r

leaves only

a

unknown — see Two-body problem and the vis-viva equation.

Why must the same rotation matrix

\tilde{R}

be applied to both

r

and

v

They are two vectors expressed in the same perifocal frame; changing frames is one geometric re-orientation, so it acts identically on every vector living in that frame.

Edge cases

The scenarios where naive formulas quietly break. (Figure s04 shows the two singular geometries.)

Circular orbit (

e = 0

): which element becomes undefined, and how is it patched?

ω

becomes undefined because there is no periapsis to measure from (

e = 0

). Patch: use the argument of latitude

u = ω + ν

, the angle from the node to the body. See Orbital singularities and equinoctial elements.

Equatorial orbit (

i = 0

): what fails, and why?

n = \overset{z}{^} \times h = 0

because

h

is parallel to

\overset{z}{^}

, so there is no ascending node and

Ω

is undefined; the fix is to track the true longitude instead of

Ω

and

ω

separately.

Both

e = 0

and

i = 0

(circular equatorial): how many elements collapse?

Both

ω

and

Ω

vanish, and only

a

plus a single angle (the true longitude

ℓ = Ω + ω + ν

) remain meaningful; equinoctial elements are built precisely to avoid these dual singularities.

Parabolic escape (

e = 1

): what happens to

a

Energy

ε = 0

, so

a = - μ / (2 ε)

blows up (infinite) — a parabola has no finite semi-major axis, and one instead uses the semi-latus rectum

p = h^{2} / μ

to describe it.

At exact apoapsis,

r \cdot v = 0

— how does the recipe pick

ν = π

over

ν = 0

The dot product is zero at both turning points, so the sign rule is inconclusive; you resolve it from

cos ν = e \cdot r / (er)

, which is

- 1

at apoapsis (giving

ν = π

) and

+ 1

at periapsis.

Radial (straight-line) trajectory,

r ∥ v

: why does the whole scheme break?

Then

h = r \times v = 0

, so the orbit plane,

i

Ω

, and

p = h^{2} / μ

are all undefined — a degenerate "orbit" with no angular momentum that the element set simply cannot represent.

Retrograde orbit (

i

between

9 0^{\circ}

and

18 0^{\circ}

): does

cos i = h_{z} / h

still work?

Yes —

h_{z} < 0

makes the cosine negative and

cos^{- 1}

returns an angle above

9 0^{\circ}

, correctly flagging retrograde motion with no extra sign fix (that's why

i

needs no quadrant check).

Nearly-zero eccentricity where numerical noise makes

e \cdot r / (er)

slightly exceed

1

: what's the safe habit?

Clamp the argument to

[- 1, 1]

before calling

cos^{- 1}

; floating-point round-off can push it past the valid range and crash an otherwise correct computation.

Recall One-sentence summary of every trap

The recurring lessons: $h$ points out of the plane (not along), $e$ points to periapsis, semi-major axis uses a minus sign ( $a = - μ /2 ε$ ), $cos^{- 1}$ needs a companion sign except for $i$ , transforming PQW→ECI uses negative angles, and whenever $e = 0$ or $i = 0$ some element dissolves and must be swapped for $u$ , longitude, or equinoctial elements.