Specific Mechanical Energy, the Eccentricity Vector, and the Conic-Orbit Equation

1 Where GPSORB04 fits in the derivation chain

The starting point remains the ideal two-body equation derived in GPSORB01,

μ- ¨r = − r3r,

(1)

and the conserved specific angular momentum developed in GPSORB03,

h = r × v, ˙h = 0.

(2)

Because $h$ is constant, the motion is confined to a fixed plane perpendicular to $h$ . What remains is to determine the shape of the trajectory within that plane.

The development in this article is

p (r,v ) − → ℰ ,h −→ e −→ p, e,ν − → r = ----------. 1 + ecos ν

The central idea is that the orbit equation is not imposed geometrically. It is extracted from Newton’s differential equation through its conserved quantities.

Figure 1. The Cartesian state determines conserved quantities, which in turn determine the conic orbit and its classification.

2 Specific mechanical energy

For a satellite of mass $m$ , the ordinary mechanical energy is

1 2 μm E = -mv − ----. 2 r

(3)

Orbital mechanics usually divides by the satellite mass and works with specific mechanical energy,

|------------| | v2 μ | |ℰ = ---− -- . ------2----r-

(4)

The units are

2 [ℰ] = m--= -J-. s2 kg

The first term is specific kinetic energy and the second is gravitational potential energy per unit mass, with zero potential chosen at infinite distance.

2.1 Derivation directly from the equation of motion

Take the dot product of Eq. (1) with the velocity $v = ˙r$ :

μ v ⋅ ¨r = −-3v ⋅ r. r

(5)

The left-hand side is

( 2 ) v ⋅ ˙v = d- v-- . dt 2

(6)

For the right-hand side, begin with

√---- r = r ⋅ r.

(7)

Differentiation gives

˙r = r-⋅ v-. r

(8)

Therefore

( ) -d 1- r˙ r-⋅ v- dt r = − r2 = − r3 .

(9)

Multiplying by $μ$ ,

( ) -d μ- = − -μ r ⋅ v. dt r r3

(10)

Equation (5) therefore becomes

( ) d v2 d ( μ) dt -2- = dt r- .

(11)

Move both terms to the same side:

( 2 ) d- v--− μ- = 0. dt 2 r

(12)

Hence

|-----------------------| | 2 | |ℰ = v-− μ-= constant . ------2----r-------------

(13)

This is the energy first integral of the Kepler problem.

2.2 Physical interpretation

As a satellite falls inward, $r$ decreases and the potential term $− μ∕r$ becomes more negative. Since the sum $ℰ$ remains constant, the kinetic term $v2∕2$ must increase. As the satellite climbs outward, the reverse occurs.

Figure 2. In an elliptical orbit the satellite moves fastest at periapsis and slowest at apoapsis while total specific mechanical energy remains constant.

This energy exchange is another way to understand why the orbital speed is not generally constant even though no dissipative force is present.

3 Why angular momentum alone is not enough

The conserved vector $h$ determines the orbital plane and the amount of transverse motion, but it does not by itself determine the orientation of the ellipse within that plane. Two ellipses can lie in the same plane and have the same angular-momentum magnitude while pointing their periapsides in different directions.

A second conserved vector is therefore needed. The inverse-square gravitational field has a special additional invariant known as the eccentricity vector. In classical mechanics it is closely related to the Laplace-Runge-Lenz vector.

4 The eccentricity vector

Define

|--------------| | v × h | |e = ------− ^r| -------μ--------

(14)

where

r ^r = -. r

(15)

The eccentricity vector is dimensionless. Its magnitude is the orbital eccentricity,

|--------| |e = ∥e∥.| ----------

(16)

For a noncircular Keplerian orbit, $e$ points from the attracting focus toward periapsis.

Figure 3. The eccentricity vector points toward periapsis, and the true anomaly is measured from that direction to the position vector.

5 Proving that the eccentricity vector is constant

The constancy of $e$ is special to the inverse-square gravitational law. Differentiate Eq. (14):

˙ ˙e = v˙×-h- + v-×-h-− ˙^r. μ μ

(17)

Because $˙ h = 0$ ,

e˙= a ×-h-− ˙^r. μ

(18)

The two-body acceleration is

a = − μ-^r. r2

(19)

Hence

a × h 1 ------= − -2 [^r × (r × v)]. μ r

(20)

Use the vector triple-product identity

A × (B × C ) = B (A ⋅ C ) − C (A ⋅ B ).

(21)

With $A = ^r$ , $B = r$ , and $C = v$ ,

^r × (r × v) = r(^r ⋅ v) − v(^r ⋅ r).

(22)

Now

^r ⋅ v = ˙r, ^r ⋅ r = r,

(23)

^r × (r × v ) = r˙r^r − rv = − r(v − r˙^r).

(24)

Substituting into Eq. (20),

a-×--h v-−--˙r^r μ = r .

(25)

But differentiating $^r = r∕r$ gives

|-----------| |˙ v-−-r˙^r| -^r =----r----.

(26)

Therefore the two terms in Eq. (18) cancel exactly:

|------| -˙e =-0.-

(27)

Thus the eccentricity vector is constant in both magnitude and direction for the ideal Kepler problem.

6 The geometric meaning of the eccentricity vector

Because $e$ is fixed in the orbital plane, it provides a natural reference direction. Define the true anomaly $ν$ as the angle from $e$ to the instantaneous position vector $r$ . Then

e ⋅^r = ecos ν.

(28)

This apparently simple dot product is the key to deriving the conic equation.

First use Eq. (14):

(v-×-h)-⋅^r- e ⋅^r = μ − 1.

(29)

The scalar triple product may be cyclically permuted:

(v × h) ⋅^r = h ⋅ (^r × v).

(30)

Since

h = r × v = r(^r × v ),

(31)

we have

h- ^r × v = r.

(32)

Therefore

2 (v × h ) ⋅^r = h-. r

(33)

Substitution into Eq. (29) gives

h2- e cosν = μr − 1.

(34)

Rearrange:

h2 1 + e cosν = --. μr

(35)

Hence

|----------------| | h2∕μ | |r = ---------- .| -----1-+-e-cosν--

(36)

Define the semilatus rectum

|------2-| |p = h--.| ------μ--|

(37)

The orbit equation becomes

|----------------| | -----p---- | |r = 1 + e cosν .| -----------------

(38)

This is the polar equation of a conic section with the attracting body at one focus.

7 Why this is a major result

Equation (38) was obtained entirely from the Newtonian equation of motion. No ellipse was assumed. The derivation shows that inverse-square gravity forces the trajectory to be a conic section.

The constants have direct physical meanings:

Quantity	Physical role
$h$	fixes the orbital plane and transverse angular momentum per unit mass
$h$	sets the semilatus rectum through $p = h2∕μ$
$e$	fixes the periapsis direction in the orbital plane
$e$	determines the conic shape
$ν$	locates the satellite around the conic relative to periapsis

8 Conic classification from eccentricity

The same orbit equation describes all nondegenerate Keplerian conics:

0 ≤ e < 1 : ellipse, e = 1 : parabola, e > 1 : hyperbola.

(39)

The special case $e = 0$ is a circle. In that case the eccentricity vector vanishes and there is no unique periapsis direction.

Figure 4. The same focus-centered orbit equation produces elliptical, parabolic, and hyperbolic trajectories as eccentricity crosses unity.

The conic classification emerges even more deeply from energy, as shown next.

9 The relation among energy, angular momentum, and eccentricity

Square the magnitude of Eq. (14):

∥∥ ∥∥2 e2 = ∥v-×--h − ^r∥ . ∥ μ ∥

(40)

Expanding,

2 ∥v-×--h∥2 2(v-×-h-) ⋅^r- e = μ2 + 1 − μ .

(41)

Because $v$ lies in the orbital plane and $h$ is normal to it,

v ⋅ h = 0,

(42)

∥v × h∥2 = v2h2.

(43)

From Eq. (33),

h2 (v × h ) ⋅^r = --. r

(44)

Thus

2 v2h2- 2h2- e = μ2 + 1 − μr .

(45)

Factor $2h2 ∕μ2$ :

( ) 2 2h2- v2- μ- e = 1 + μ2 2 − r .

(46)

Recognizing the specific mechanical energy,

|------------2-| e2 = 1 + 2ℰ-h-.| | μ2 | ----------------

(47)

This equation is one of the most useful bridges between orbital dynamics and conic geometry.

Because $2 h > 0$ and $2 μ > 0$ , the sign of $ℰ$ determines whether $e$ is less than, equal to, or greater than one:

ℰ < 0 ⇐ ⇒ e < 1, ℰ = 0 ⇐ ⇒ e = 1, ℰ > 0 ⇐ ⇒ e > 1.

(48)

Therefore negative specific orbital energy corresponds to a bound ellipse, zero energy to the parabolic escape boundary, and positive energy to an unbound hyperbola.

10 Periapsis and apoapsis from the conic equation

Periapsis occurs at $ν = 0$ , so $cosν = 1$ . Equation (38) gives

|-----------| r = --p--. | -p---1-+-e---

(49)

For an ellipse, apoapsis occurs at $ν = π$ , so $cos ν = − 1$ :

|-------p----| |ra = -----. | ------1-−-e--|

(50)

The semimajor axis of an ellipse is half the sum of the apsidal distances,

r + r a = -p----a. 2

(51)

Substituting Eqs. (49) and (50),

( ) 1- --p-- --p-- a = 2 1 + e + 1 − e .

(52)

Combining the fractions,

p a = -----2 . 1 − e

(53)

Hence

|--------------| |p = a(1 − e2).| ---------------

(54)

This relation applies to ellipses and, with signed semimajor-axis conventions, extends naturally to hyperbolic conics.

11 Energy and semimajor axis

Substitute

h2 = μp

(55)

into Eq. (47):

e2 = 1 + 2ℰp-. μ

(56)

Rearrange:

2 2ℰp 1 − e = − ----. μ

(57)

For an ellipse, Eq. (54) gives

p 1 − e2 = --. a

(58)

Therefore

p-= − 2ℰp-. a μ

(59)

Cancel the nonzero $p$ :

|----------| |ℰ = − -μ-.| -------2a--|

(60)

This is a remarkably compact result: for every ideal elliptical orbit having the same semimajor axis, the total specific mechanical energy is the same, regardless of eccentricity.

12 The vis-viva equation

Combine the energy definition

v2 μ ℰ = ---− -- 2 r

(61)

with Eq. (60):

2 v--− μ-= − -μ-. 2 r 2a

(62)

Multiply by two and solve for $2 v$ :

|------------------| | ( ) | |v2 = μ 2-− 1- .| ----------r---a----

(63)

This is the vis-viva equation. It gives the speed at any orbital radius without first solving the full time history of the trajectory.

At periapsis,

∘ ------------ ( 2 1) vp = μ --− -- , rp a

(64)

and at apoapsis,

∘ ------------ ( 2 1) va = μ --− -- . ra a

(65)

For an ellipse, $rp < ra$ , so $vp > va$ , consistent with both energy conservation and Kepler’s second law.

13 Radial and transverse velocity components

GPSORB03 established

2 h = r ˙ν

(66)

when the polar angle is measured in the orbital plane from a fixed apsidal direction. Therefore the transverse speed is

v𝜃 = rν˙= h. r

(67)

Using Eq. (38),

v = μ-(1 + ecosν ). 𝜃 h

(68)

Differentiate the orbit equation with respect to $ν$ :

dr-= ---pesinν----. dν (1 + ecosν )2

(69)

Since

dr- ˙r = dνν˙

(70)

and

˙ν = -h , r2

(71)

substitution gives

|------------| | μ- | |˙r = he sin ν.| --------------

(72)

Thus the in-plane velocity may be written as

|----------------------------------| |v = μ-[esinν er + (1 + ecosν )e𝜃].| -----h------------------------------

(73)

This form is useful later when converting between orbital elements and Cartesian state vectors.

14 A numerical GPS-like example

Consider an idealized medium-Earth orbit with

a = 26,560 km, e = 0.010.

(74)

Use

μ = 3.986004418 × 1014 m3 ∕s2.

(75)

The periapsis and apoapsis radii are

rp = a (1 − e) = 26,294.4 km,

(76)

ra = a (1 + e) = 26,825.6 km.

(77)

The specific mechanical energy is

μ ℰ = − ---≈ − 7.504 × 106 m2 ∕s2. 2a

(78)

The semilatus rectum is

p = a(1 − e2) ≈ 26,557.34 km.

(79)

The specific angular-momentum magnitude is

h = √ μp-≈ 1.029 × 1011 m2 ∕s.

(80)

Using vis-viva,

vp ≈ 3.913 km ∕s, va ≈ 3.835 km ∕s.

(81)

Even for a small eccentricity of only $0.01$ , the speed changes measurably over the orbit.

15 From a Cartesian state directly to the conic

Given one inertial state $(r,v)$ , the ideal conic can be reconstructed without numerically integrating the trajectory first.

Compute

h = r × v,

(82)

v2- μ- ℰ = 2 − r,

(83)

and

e = v-×-h-− ^r. μ

(84)

Then

h2 h = ∥h ∥, e = ∥e ∥, p = --. μ

(85)

For a bound noncircular orbit,

a = − -μ-. 2ℰ

(86)

The true anomaly can be found robustly with an atan2 construction. One convenient form is

cos ν = e-⋅ r, er

(87)

with the sign of $sinν$ obtained from the radial velocity or from a triple product involving $h$ . This avoids quadrant ambiguity that appears if only an inverse cosine is used.

16 Connection to the GPS broadcast ephemeris

The legacy GPS navigation message does not transmit an instantaneous Cartesian state vector. Instead, among its broadcast ephemeris parameters it provides the eccentricity $e$ and the square root of semimajor axis $-- √ A$ , together with angular elements, rates, an epoch, and harmonic correction coefficients. The official interface specification describes these parameters as Keplerian in appearance while noting that their values are obtained by fitting a propagated satellite trajectory over a finite interval.

The results derived in this article explain why $e$ and $a$ are such natural coordinates for the broadcast model:

μ ℰ = − ---, p = a(1 − e2), 2a

(88)

so $a$ controls the orbital energy and characteristic size while $e$ controls the departure from circularity and the apsidal geometry. Later articles will show how $√-- A$ is converted to $A$ , then to mean motion, mean anomaly, eccentric anomaly, and finally satellite position.

17 Important limiting cases and cautions

17.1 Circular orbit

For $e = 0$ ,

e = 0,

(89)

so the eccentricity-vector direction is undefined. The orbit is still perfectly well defined, but periapsis has no unique direction because every point on a circle is geometrically equivalent.

17.2 Radial motion

If $h = 0$ , then the motion is purely radial and the standard nondegenerate conic-element description breaks down. This is a singular limiting case, not the normal satellite-orbit situation.

17.3 Perturbed motion

For a real GPS satellite, Earth’s nonspherical gravity, third-body forces, solar radiation pressure, and other perturbations cause the ideal invariants to evolve slowly. In that setting one can still define instantaneous or osculating values of $ℰ$ , $h$ , and $e$ , but they are no longer exactly constant.

This distinction matters later when interpreting broadcast ephemeris parameters: the equations look Keplerian, but the transmitted coefficients represent a finite-interval fit to a perturbed orbit rather than a set of eternal two-body constants.

18 Summary of the derivation

Starting from Newton’s inverse-square two-body equation,

μ ¨r = − r3r,

(90)

we obtained a second scalar first integral,

|------2-----------------| |ℰ = v-− μ-= constant. | ------2----r-------------|

(91)

Together with the conserved angular momentum,

h--=-r ×-v,| ------------

(92)

the inverse-square law admits the conserved eccentricity vector,

|---------------| | v × h | e = ------− ^r. | -------μ---------

(93)

Its dot product with $^r$ yields

|----------------------------| | -----p---- h2- | |r = 1 + e cosν, p = μ .| -----------------------------

(94)

For an ellipse,

|----------------------------| |p = a(1 − e2), ℰ = − -μ-.| -------------------------2a---

(95)

Finally,

|------(-------)-| |v2 = μ 2-− 1- | ---------r----a---

(96)

provides the vis-viva speed relation.

The next step is to use $h$ , $e$ , and the inertial reference axes to construct the classical orbital elements $(a,e,i,Ω, ω,ν )$ explicitly from a Cartesian state vector.

References

[1] E. D. Kaplan and C. J. Hegarty, eds., Understanding GPS/GNSS: Principles and Applications, 3rd ed., Artech House, 2017.

[2] R. R. Bate, D. D. Mueller, and J. E. White, Fundamentals of Astrodynamics, Dover Publications, 1971.

[3] D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.

[4] H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Elsevier, 2020.

[5] Global Positioning Systems Directorate, IS-GPS-200N: Navstar GPS Space Segment/Navigation User Interfaces, 1 August 2022. Available from GPS.gov.

"Specific Mechanical Energy, the Eccentricity Vector, and the Conic-Orbit Equation" is owned by bloftin.
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