1 Why GPSORB07 belongs between time propagation and frame rotation

GPSORB06 ended with a complete phase-propagation chain for an ideal ellipse,

(1)

At that stage the semimajor axis a, eccentricity e, and true anomaly ν determine where the satellite lies on the conic, but we have not yet written its Cartesian position and velocity in the orbital plane.

GPSORB08 will rotate an orbital-plane state into an Earth-centered inertial frame using Ω, i, and ω. Therefore the missing bridge is

(2)

before any three-dimensional orientation transformation is applied.

This separation is conceptually useful. The elements a, e, and ν describe the instantaneous geometry and motion within the orbital plane. The elements Ω, i, and ω describe how that plane is oriented relative to an external reference frame. The PQW frame isolates the first problem from the second.

2 The perifocal frame

The perifocal frame is an orthonormal right-handed frame attached to the Keplerian orbit. Its basis vectors are

(3)

They are defined geometrically:

• P points from the occupied focus toward periapsis.
• Q lies in the orbital plane, 90^∘ ahead of P in the direction of increasing true anomaly.
• W = h is normal to the orbital plane and points along the specific angular-momentum vector.

Thus

(4)

In this frame, true anomaly ν is especially simple: it is the angle measured from the positive P axis to the radius vector. Therefore the orbital geometry itself supplies the polar coordinate system.

3 The three scalar quantities needed first

Before writing the Cartesian state, compute three familiar orbital quantities.

The semilatus rectum is

(5)

From GPSORB03 and GPSORB04,

(6)

so the magnitude of the specific angular momentum is

(7)

Finally, the conic-orbit equation is

(8)

Equations (5)–(8) already contain nearly everything required for the orbital-plane state. The remaining work is to express the local radial and transverse directions in the fixed PQW basis and then derive the corresponding velocity components.

4 Radial and transverse unit vectors

Because ν is measured from the P axis, the radial unit vector is

(9)

The transverse unit vector points 90^∘ ahead in the direction of increasing ν,

(10)

These satisfy

(11)

and

(12)

Differentiating Eq. (9) with respect to ν gives

(13)

so by the chain rule

(14)

Similarly,

(15)

These moving-basis derivatives are the essential kinematic fact behind the orbital velocity formula.

5 Position in the PQW frame

The position vector is simply

(16)

Substituting Eq. (9),

(17)

Using Eq. (8), this can also be written

(18)

This is the first half of the perifocal state. The third component vanishes because the PQ plane is the orbital plane.

6 Velocity in polar form

Differentiate

(19)

The product rule gives

(20)

Using Eq. (14),

(21)

The first term changes the orbital radius. The second term sweeps the radius vector around the focus. A circular orbit has zero radial rate, but an eccentric orbit generally has both terms.

7 Deriving the radial speed

Start with the conic equation,

(22)

Differentiate with respect to ν:

(23)

From angular-momentum conservation,

(24)

so

(25)

The chain rule gives

(26)

Substitute Eqs. (23) and (25):

(27)

But Eq. (8) implies

(28)

Therefore

(29)

Using p = h²∕μ,

(30)

This formula immediately gives several physical checks. At periapsis, ν = 0, so the radial rate vanishes. At apoapsis, ν = π, so the radial rate vanishes again. Between periapsis and apoapsis, 0 < ν < π and the radial rate is positive: the satellite moves outward. On the return half of the orbit, sin ν < 0 and the satellite moves inward.

8 Deriving the transverse speed

The transverse speed is the second scalar in Eq. (21),

(31)

Using the angular-momentum relation from Eq. (25),

(32)

Now substitute Eq. (8):

(33)

Again using p = h²∕μ,

(34)

Unlike the radial component, the transverse component does not vanish at periapsis or apoapsis. Indeed, at periapsis it is largest because the orbital radius is smallest while the angular momentum is fixed.

9 Assembling the compact perifocal velocity vector

Insert Eqs. (30) and (34) into Eq. (21):

(35)

Now substitute the two unit vectors from Eqs. (9) and (10). The P component becomes

(36)

The two mixed terms cancel, leaving

(37)

The Q component becomes

(38)

Use sin ²ν + cos ²ν = 1 to obtain

(39)

Therefore

(40)

Because h = ,

(41)

This is the standard perifocal velocity formula. The derivation shows exactly where each term comes from: the − sin ν term emerges from cancellation between radial and transverse motion, while the e + cos ν term contains both the circular part and the eccentricity correction.

10 The whole state reconstruction in one chain

For implementation, the minimum sequence is

(42)

(43)

(44)

followed by Eqs. (17) and (40). No node angle, inclination, or argument of periapsis is needed yet because the result is still expressed in the orbit’s own PQW coordinates.

11 Check 1: the cross product must reproduce angular momentum

A strong verification is

(45)

Using Eqs. (17) and (40), only the W component survives:

(46)

Collect terms:

(47)

Thus

(48)

Since r(1 + e cos ν) = p,

(49)

Therefore

(50)

The reconstructed state has exactly the angular momentum used to build it.

12 Check 2: recovering the vis-viva equation

Square Eq. (41):

(51)

Expand the bracket:

(52)

Hence

(53)

From r = p∕(1 + e cos ν),

(54)

Using p = a(1 − e²), one can verify that

(55)

Therefore Eq. (53) becomes

(56)

which is the vis-viva equation derived in GPSORB04. The vector reconstruction and the scalar energy relation are therefore consistent.

13 Flight-path angle

The velocity vector is not generally perpendicular to the radius vector. Define the flight-path angle γ as the angle from the local transverse direction toward the velocity vector. Then

(57)

Substituting Eqs. (30) and (34) gives

(58)

For a circular orbit, e = 0 and therefore γ = 0 everywhere: velocity is exactly transverse. For an eccentric ellipse, γ = 0 at periapsis and apoapsis but is nonzero elsewhere. This is the precise mathematical version of the statement that the velocity vector is tangent to the orbital trajectory but is not generally perpendicular to the radius vector.

14 An equivalent reconstruction using eccentric anomaly

GPSORB06 derived the focus-centered ellipse coordinates

(59)

(60)

Therefore the perifocal position can be written directly as

(61)

Kepler’s equation gives

(62)

where

(63)

Because

(64)

we can rewrite the eccentric-anomaly rate as

(65)

Differentiate Eq. (61):

(66)

Substitution gives

(67)

Equations (61) and (67) are especially convenient when Kepler’s equation has just been solved numerically for E. The true-anomaly formulas and eccentric-anomaly formulas must produce the same state to numerical precision.

15 Worked GPS-like example

Use the same idealized orbit employed in GPSORB05 and GPSORB08,

(68)

with

(69)

First compute the semilatus rectum:

(70)

Then

(71)

The orbital radius is

(72)

The position vector is therefore

(73)

The velocity scale is

(74)

Thus

(75)

The radial and transverse components are

(76)

(77)

Thus most of the velocity is transverse, but the nonzero positive radial component shows that the satellite is moving outward at ν = 70^∘.

The total speed is

(78)

The specific mechanical energy computed from the reconstructed state is

(79)

which agrees with

(80)

For an independent anomaly check, the corresponding eccentric anomaly is

(81)

Using Eqs. (61) and (67) reproduces the same position and velocity components shown above.

16 Important limiting cases

16.1 Circular orbit

If e = 0, then

(82)

The velocity becomes

(83)

which is exactly perpendicular to the radius vector.

The caveat is geometric: in a perfectly circular orbit, periapsis is undefined, so the P axis itself is not unique. One may choose a convenient in-plane reference direction and use argument of latitude instead. The state remains perfectly well defined even though the classical periapsis-based coordinates become singular.

16.2 Periapsis and apoapsis

At periapsis, ν = 0, so

(84)

At apoapsis, ν = π, so

(85)

The negative Q component at apoapsis is not a contradiction: at ν = π, the local transverse direction points in the negative P-frame Q sense associated with the instantaneous geometry of the position angle. The vector remains tangent to the orbit and consistent with increasing ν.

17 A compact implementation recipe

Given a, e, ν, and μ for a nonsingular Keplerian ellipse:

1.

Compute p = a(1 − e²).

2.

Compute h = .

3.

Compute r = p∕(1 + e cos ν).

4.

Form r^PQW = [r cos ν, r sin ν, 0]^T .

5.

Form v^PQW = [− sin ν, e + cos ν, 0]^T .

6.

Verify ∥r × v∥ = h and v² = μ(2∕r − 1∕a).

If the propagator naturally produces E first, Eqs. (61) and (67) provide an equally valid route. Comparing the E-based and ν-based states is an excellent software unit test.

18 Bridge to GPSORB08

GPSORB07 has deliberately stopped before applying the three orientation elements. The state currently has the form

(86)

These vectors describe the full instantaneous motion in the orbital plane, but they do not yet tell us how that plane sits in inertial space.

GPSORB08 supplies the missing transformation,

(87)

where

(88)

Thus the clean conceptual sequence is

(89)

GPSORB06 supplied the clock, GPSORB07 supplies the orbital-plane state, and GPSORB08 supplies the three-dimensional orientation.

References

References