Angular Momentum, Orbital Planarity, and Kepler's Second Law

1 Where GPSORB03 fits in the derivation chain

GPSORB01 derived the ideal Earth-satellite relative equation of motion,

|------μ---| |¨r = − -3r | -------r---|

(1)

where $r$ is the satellite position relative to Earth’s center, $r = ∥r∥$ , and $μ$ is Earth’s gravitational parameter. GPSORB02 then explained why this three-dimensional second-order equation requires six independent scalar constants to specify one unique orbit.

The next task is to expose some of those constants physically. The first conserved quantity to derive is angular momentum. It is especially important because it proves that the ideal orbit is planar before any conic-section geometry is introduced.

The chain developed here is

|---------| |--------| | μ- | |---------------------| |------------------| |dA- h-| ¨r = − r3r| = ⇒ -h-=-r-×-v-=-constant-- =⇒ fixed--orbital-plane-- =⇒ |dt = 2 . ----------- ----------

That last expression is Kepler’s second law in differential form.

2 Angular momentum: physical and specific forms

For a particle of mass $m$ moving with position $r$ and velocity $v$ , the angular momentum about the origin is

|------------| L--=-r-×-mv.--

(2)

Because the satellite mass is constant, orbital mechanics usually divides by $m$ and works with the specific angular momentum,

|----------------| | L | |h = -- = r × v.| -----m------------

(3)

The adjective “specific” means “per unit mass.” The units are therefore

m2- [h] = s .

The magnitude of a cross product is

|------------| -h-=-rv-sin-γ-|

(4)

where $γ$ is the angle between $r$ and $v$ . If $vt$ denotes the velocity component perpendicular to $r$ , then

|--------| -h-=-rvt.-

(5)

Thus angular momentum measures the transverse, or sideways, part of the orbital motion. Purely radial velocity contributes no angular momentum about Earth’s center.

Figure 1. Central gravity is radial, while the specific angular momentum is perpendicular to the plane containing position and velocity.

3 Why central gravity produces zero torque

The rotational analogue of Newton’s second law is obtained from torque. The gravitational force on a satellite of mass $m$ is

F = − μm-r. g r3

(6)

The torque about Earth’s center is

τ = r × Fg.

(7)

Substituting Eq. (6),

( μm ) μm τ = r × − -r3-r = − -r3-(r × r).

(8)

But any vector crossed with itself vanishes,

r × r = 0,

|------| τ--=-0.-

(9)

This is the geometric core of the result. Gravity points along the radius vector, so it has no moment arm about Earth’s center. A radial force can change the magnitude and direction of velocity, but it cannot exert a torque about the force center.

4 Direct derivation of conserved angular momentum

Start with the definition

h = r × v.

Differentiate with respect to time. The cross product obeys a product rule,

dh- = dr-× v + r × dv-. dt dt dt

(10)

Since $dr∕dt = v$ and $dv ∕dt = a$ ,

dh- dt = v × v + r × a.

(11)

The first cross product is zero,

v × v = 0.

Using the two-body acceleration from Eq. (1),

dh ( μ ) μ --- = r × − -3 r = − -3 (r × r) = 0. dt r r

(12)

Therefore

|--------| |dh- | |dt = 0 | ---------

(13)

and hence

|--------------------| -h-=-constant-vector.-

(14)

The word vector matters. Conservation of $h$ means both its magnitude and its direction remain fixed in the ideal two-body problem.

Equation (13) is equivalent to the zero-torque result because

dL- dt = τ, L = mh.

For a constant satellite mass, zero torque implies constant $h$ .

5 A subtle but powerful point: the inverse-square form was not required

The conservation proof did not actually use the $1 ∕r2$ magnitude of Newtonian gravity. It used only the fact that the acceleration is parallel or antiparallel to $r$ .

For any central-force acceleration of the form

a = f(r)r,

(15)

we obtain

r × a = f(r)r × r = 0,

|----------------------------------------------| h = r × v is conserved for every central force.| ------------------------------------------------

(16)

Newton’s inverse-square law is special for additional reasons: it leads to closed conic trajectories and to another conserved vector, the eccentricity vector. But planarity and constant areal velocity arise more generally from central-force symmetry.

6 Conservation of angular momentum proves orbital planarity

By definition,

h = r × v.

(17)

A cross product is perpendicular to both input vectors, so

|------------------------| -h-⋅ r-=-0,----h-⋅ v-=-0.|

(18)

At every instant, the position and velocity therefore lie in the plane perpendicular to $h$ .

Because Eq. (14) says that $h$ is a constant vector, the orientation of this plane does not change with time. The plane also passes through the force center because $r$ is measured from Earth’s center. Thus

|----------------------------------------------------------------------------| |ideal two -body motion is con fined to one fixed plane through Earth’s center. | -----------------------------------------------------------------------------

(19)

Figure 2. Position and velocity remain in the fixed orbital plane perpendicular to the conserved angular-momentum vector.

This is a major reduction in the problem. We began with a three-dimensional vector differential equation, yet conservation of angular momentum tells us that once the orbital plane is known, all subsequent Keplerian motion may be analyzed in two dimensions inside that plane.

The radial special case

r × v = 0,

then $h = 0$ . Position and velocity are parallel, so the motion is purely radial. This is a degenerate case rather than an ordinary orbit with a well-defined orbital plane. In normal satellite motion, $h > 0$ .

7 The angular-momentum vector in Cartesian coordinates

Let

⌊ ⌋ ⌊ ⌋ x vx r = ⌈ y⌉ , v = ⌈vy⌉ . z vz

Then

| | ||i j k|| h = r × v = ||x y z|| . |vx vy vz|

(20)

Expanding the determinant,

|----⌊----------⌋--| | yvz − zvy | |h = ⌈ zvx − xvz⌉ .| | xvy − yvx | --------------------

(21)

Its magnitude is

|----∘---------------| |h = h2x + h2y + h2z.| ---------------------|

(22)

These equations are the practical bridge from a Cartesian navigation state to orbital geometry. Given $(r,v)$ at one epoch, computing $h$ immediately gives the orbital-plane normal.

8 A second derivation using polar coordinates

Because the motion is planar, introduce polar basis vectors $er$ and $e𝜃$ inside the orbital plane. The position is

r = rer.

(23)

The polar basis rotates with angle $𝜃$ , so

der-= 𝜃˙e 𝜃, de-𝜃= − ˙𝜃er. dt dt

(24)

Differentiate Eq. (23):

d- der- ˙ v = dt(rer) = ˙rer + r dt = r˙er + r𝜃e 𝜃.

(25)

The velocity therefore has a radial component

vr = ˙r

and a transverse component

vt = r˙𝜃.

Figure 3. Polar decomposition of orbital velocity into radial and transverse components.

Now compute the specific angular momentum:

h = r × v.

(26)

Substituting the polar forms of position and velocity gives

h = (re ) × (r˙e + r𝜃˙e ). r r 𝜃

(27)

Expanding the cross product,

2 ˙ h = r˙r(er × er) + r 𝜃(er × e𝜃).

(28)

Since

er × er = 0, er × e𝜃 = k,

where $k$ is the unit normal to the orbital plane,

|----------| -h-=-r2𝜃˙k.-|

(29)

Taking magnitudes,

|--------| -h-=-r2 ˙𝜃.

(30)

This is one of the most useful identities in orbital mechanics.

9 The same result from the transverse equation of motion

Differentiating the polar velocity gives the standard polar acceleration,

|------------------------------| a = (¨r − r𝜃˙2)er + (r¨𝜃 + 2˙r𝜃˙)e𝜃. --------------------------------

(31)

Central gravity has no transverse component, so

r¨𝜃 + 2r˙𝜃˙= 0.

(32)

Multiply by $r$ :

r2¨𝜃 + 2rr˙𝜃˙= 0.

(33)

But the left-hand side is exactly

( ) d- r2 ˙𝜃 = 2rr˙𝜃˙+ r2¨𝜃. dt

(34)

Therefore

|---(---)------| |-d 2 ˙ | |dt r 𝜃 = 0,| ---------------

(35)

which gives again

|--------------------| -r2 ˙𝜃-=-h-=-constant.|

The vector proof and the polar-coordinate proof are the same physics expressed in two different mathematical languages. The vector proof emphasizes zero torque and plane orientation; the polar proof emphasizes the coupling between orbital radius and angular rate.

10 Deriving Kepler’s second law from angular momentum

Consider the radius vector moving from $r$ to $r + dr$ during a small time $dt$ . The small swept area is approximately the area of a triangle,

dA = 1-∥r × dr∥. 2

(36)

Because

dr = v dt,

(37)

we obtain

1 1 dA = -∥r × v ∥dt = -h dt. 2 2

(38)

Divide by $dt$ :

|---------| |dA- h- | -dt-=--2.-|

(39)

Since $h$ is constant,

|----------------| |dA- = constant. | --dt-------------|

(40)

This is Kepler’s second law: the radius vector sweeps out equal areas in equal times.

The same result follows immediately from Eq. (30). A polar area element is

dA = 1r2d𝜃. 2

(41)

Hence

dA- 1- 2d𝜃- 1-2 ˙ h- dt = 2 r dt = 2r 𝜃 = 2 .

(42)

Figure 4. Equal time intervals sweep equal areas; the satellite therefore moves faster near periapsis and slower near apoapsis.

11 Why orbital speed changes around an ellipse

Equation (5) gives

|--------| |v = h-.| --t---r--|

(43)

For a fixed $h$ , the transverse speed is larger when $r$ is smaller and smaller when $r$ is larger.

At periapsis and apoapsis, $r$ is at a local minimum or maximum, so

˙r = 0.

(44)

The velocity is therefore purely transverse at those two points. Thus

|----------------------| |vp = -h , va = -h. | ------rp-----------ra--|

(45)

Dividing,

---------- |vp ra | |--= --.| -va---rp--

(46)

This result follows from angular momentum alone. Energy conservation will later supply the absolute values of the speeds and lead to the vis-viva equation.

12 Angular rate is strongly coupled to orbital radius

From Eq. (30),

|--------| |˙𝜃 = -h .| -----r2--|

(47)

The angular rate therefore varies as $2 1∕r$ . This is stronger than the $1∕r$ dependence of transverse speed because the same angular change corresponds to a larger arc length at larger radius.

This relationship is central to later anomaly variables. True anomaly does not increase uniformly with time on an eccentric orbit. The satellite advances rapidly in angle near periapsis and slowly near apoapsis. The mean anomaly introduced later is useful precisely because it advances linearly in the ideal Kepler problem.

13 The direction of angular momentum encodes the orbital plane

Let the reference $z$ -axis be represented by the unit vector

⌊ ⌋ 0 k = ⌈ 0⌉ . 1

The inclination $i$ is the angle between $h$ and the positive $z$ -axis. Since $k$ is a unit vector, the dot product gives

k ⋅ h = hz = hcos i.

(48)

Therefore

|----------| | h | |cosi = -z-| ---------h--

(49)

and

|---------(---)--| i = cos−1 hz- .| | h | ------------------

(50)

The line where the orbital plane intersects the reference equatorial plane is the line of nodes. A vector toward the ascending-node direction is

|-----------| n-=--k-×-h.-|

(51)

Figure 5. The angular-momentum direction determines orbital inclination, while the node vector lies along the intersection of the orbital and reference planes.

Thus a single conserved vector already provides two of the geometric ingredients needed for the classical orbital elements:

its direction determines the orientation of the orbital plane;
its angle from the reference pole determines inclination;
together with the reference pole, it determines the node line.

The right ascension of the ascending node will be derived carefully when the full classical element set is introduced.

14 Numerical scale for a nominal GPS orbit

Consider a nearly circular GPS-like orbit with radius

7 r ≈ 26,560 km = 2.6560 × 10 m

(52)

and use

μ ≈ 3.986005 × 1014 m3∕s2.

(53)

For a circular orbit, radial force balance gives

v2- μ- r = r2,

(54)

∘ -- v = μ. r

(55)

Numerically,

|--------------------| v-≈--3.874-×-103-m/s.--

(56)

Because the velocity is perpendicular to the radius in a circular orbit,

h = rv,

(57)

|----------------------| h ≈ 1.029 × 1011 m2 ∕s. ------------------------

(58)

The areal velocity is therefore

|----------------------------| |dA- h- 10 2 | |dt = 2 ≈ 5.14 × 10 m ∕s.| ------------------------------

(59)

The orbital angular rate is

𝜃˙= h- ≈ 1.459 × 10−4 rad/s. r2

(60)

Hence the period is approximately

T = 2π-≈ 4.31 × 104 s ≈ 11.97 h. ˙𝜃

(61)

This provides a useful scale check for GPS orbital mechanics.

15 What angular momentum does and does not determine

Conservation of $h$ is extremely powerful, but it does not by itself determine the entire orbit. The three components of $h$ specify the plane orientation and the amount of transverse motion, yet additional information is required to determine the size and shape of the conic and the satellite’s phase on it.

A useful summary is:

Direction of $h$ : orientation of the orbital plane.
Magnitude $h$ : strength of transverse motion and constant areal rate.
Specific energy: orbit size and bound versus unbound character.
Eccentricity vector: orbit shape and periapsis direction.
Phase variable: location of the satellite on the conic at an epoch.

This is why GPSORB02 emphasized that the conserved vectors are constrained and should not be counted naively as independent scalar constants.

16 What happens in the real GPS problem

The ideal result

dh- dt = 0

assumes a perfectly central force. Real satellite dynamics contain perturbing accelerations,

-μ a = − r3r + apert.

(62)

Then

dh --- = r × a = r × apert, dt

(63)

|--------------| |˙h = r × apert.| ---------------

(64)

Any perturbing acceleration with a component that produces nonzero torque about Earth’s center can change the angular-momentum vector.

For GPS satellites, important departures from the ideal central model include Earth’s nonspherical gravity field, third-body gravity, solar radiation pressure, and other smaller effects. The most important conceptual consequence for this article is that the orbital plane is no longer perfectly fixed. Its orientation changes slowly, which later appears in orbital-element rates such as nodal precession and inclination variation.

This provides a physical bridge to the extra rate parameters seen in GPS broadcast ephemerides. The broadcast model does not discard Keplerian geometry; it augments that geometry so a compact parameter set can track a real perturbed satellite over the intended fit interval.

17 Connection to the next derivation

Angular momentum has reduced the three-dimensional problem to a two-dimensional one. The next step is to determine the shape of the trajectory inside that plane.

The radial equation in polar coordinates is obtained from Eq. (31):

¨r − r𝜃˙2 = − μ-. r2

(65)

Using

˙ -h 𝜃 = r2 ,

this becomes

|-----2--------| r¨− h--= − μ-.| -----r3-----r2--

(66)

The term

h2 -3- r

encodes the angular-motion contribution to the radial dynamics.

In the next stages of the series, energy conservation and the eccentricity vector will turn this radial dynamics into the conic equation

r = -----p---- , 1 + e cosν

and will establish

h2 p = -μ-.

Those results will connect the conserved quantities derived from Newton’s law to semimajor axis, eccentricity, periapsis, and ultimately the six orbital elements used as the foundation of GPS ephemeris models.

Key results

The principal results of this article are

˙ h = r × v, h = 0,

h ⋅ r = 0, h ⋅ v = 0,

h = rvt, h = r2𝜃˙,

dA h (hz ) --- = --, i = cos−1 --- , dt 2 h

n = k × h.

Together they establish the most important geometric consequence of a central gravitational force:

zero central torque = ⇒ constant angular momentum = ⇒ fixed orbital plane,

fixed orbital plane with constant h = ⇒ constant areal velocity.

References

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

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

[3] H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann, 2020.

[4] R. H. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, Revised ed., AIAA, 1999.

[5] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2001.

"Angular Momentum, Orbital Planarity, and Kepler's Second Law" is owned by bloftin.
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