1 Where GPSORB01 fits in the derivation chain

GPSORB00 established the long path

(1)

The purpose of the present article is to make the first arrow rigorous.

Kaplan and Hegarty write the ideal satellite equation of motion in the familiar form

(2)

with r = ∥r∥. Equation (2) contains a remarkable amount of physics in one line. It assumes a particular force law, a particular choice of coordinates, and an idealization of Earth and the satellite. Before extracting orbital elements from it, we should understand exactly where every factor comes from.

2 The physical two-body system

Consider an Earth of mass M and a satellite of mass m. We begin in a Newtonian inertial frame with fixed, nonrotating axes. Let

RM(t)	= position of Earth’s center of mass,	(3)
Rm(t)	= position of the satellite’s center of mass.	(4)

Define the relative position from Earth to satellite by

(5)

and its magnitude and unit vector by

(6)

Figure 1 shows the geometry.

A key point is already visible: in the exact two-body problem, Earth is not perfectly fixed. The satellite pulls on Earth with the same force magnitude that Earth exerts on the satellite. The Earth-centered orbit equation will therefore emerge most cleanly from relative motion, not from pretending at the outset that Earth cannot move.

3 Newton’s law of universal gravitation

Newton’s universal gravitation law gives the magnitude of the attractive force between two point masses separated by distance r:

(7)

where G is the universal gravitational constant.

The force on the satellite points toward Earth, which is opposite the direction of r. Therefore

(8)

Using Eq. (6),

(9)

so

(10)

This is the source of the r³ in the vector equation. The magnitude still follows an inverse-square law. Multiplying by the radial unit vector contributes one additional factor of 1∕r:

(11)

There is no inverse-cube force law hidden in Eq. (2).

Figure 2 emphasizes the distinction between the vector expression and the physical inverse-square magnitude.

4 Newton’s third law and the force on Earth

Newton’s third law gives an equal and opposite force on Earth:

(12)

Applying Newton’s second law separately to both bodies gives

(13)

and

(14)

Canceling the corresponding masses yields

Rm	= −r,	(15)
RM	= +r.	(16)

These are the accelerations of the two bodies with respect to the original inertial frame.

5 The exact relative-motion equation

Differentiate the definition of the relative vector, Eq. (5), twice:

(17)

Substitute Eqs. (15) and (16):

r	= −r −r,	(18)
	= −r.	(19)

Define the exact two-body gravitational parameter

(20)

Then the exact relative-motion equation is

(21)

This result is worth pausing over. Equation (21) is not obtained by assuming Earth is immovable. It follows from subtracting the two inertial equations of motion. The coordinate r measures the separation between the bodies, and the parameter contains the sum of their masses.

5.1 Why satellite texts usually write μ ≈ GM

For an artificial satellite,

(22)

Hence

(23)

and therefore

(24)

The relative equation becomes

(25)

This is the form used throughout basic satellite dynamics and is the physical origin of Kaplan’s equation.

The approximation is extraordinarily good for a navigation satellite. Even a spacecraft mass of several tonnes is negligible compared with Earth’s mass. The spacecraft’s own mass cancels from the satellite acceleration equation and contributes negligibly to M + m in the exact relative equation.

6 Center-of-mass motion: why the two-body problem separates

Define the system center of mass

(26)

Differentiate twice:

(27)

Using Eqs. (13) and (14), the internal gravitational forces cancel:

(28)

Therefore

(29)

The barycenter moves at constant velocity in an isolated two-body system.

The full motion therefore separates naturally into two pieces:

1.

uniform translation of the center of mass; and

2.

relative orbital motion governed by Eq. (21).

For orbital mechanics we normally discard the dynamically uninteresting uniform translation and study the relative coordinate r.

7 Why a spherical Earth may be treated as a point mass

Newton’s inverse-square force law in Eqs. (8)–(10) was written as though all of Earth’s mass were concentrated at one point. Why is this allowed for a satellite outside Earth?

Newton’s shell theorem states that a spherically symmetric mass distribution attracts an exterior point mass exactly as if the entire mass were concentrated at the center. Thus for an ideal spherical Earth, the external gravitational acceleration is

(30)

The real Earth is not perfectly spherical. It is oblate and has a nonuniform mass distribution. Those departures generate the J₂ and higher spherical-harmonic corrections introduced later in GPSORB10. Equation (25) is therefore the central-body baseline model, not the final high-fidelity GPS force model.

8 A second derivation from gravitational potential

The same equation can be obtained from potential energy. For a satellite in the point-mass Earth approximation, the gravitational potential energy is

(31)

Divide by satellite mass to obtain the specific gravitational potential

(32)

The gravitational acceleration is the negative gradient:

(33)

In Cartesian coordinates,

(34)

The derivative of 1∕r with respect to x is

	= (x2 + y2 + z2)−1∕2,	(35)
	= −(x2 + y2 + z2)−3∕2(2x),	(36)
	= −.	(37)

Similarly,

(38)

Thus

(39)

Using Eq. (32),

(40)

Finally,

(41)

The force-law derivation and the potential-gradient derivation are mathematically equivalent. The potential viewpoint becomes especially valuable when nonspherical gravity is added, because additional gravitational structure is naturally represented by additional terms in the potential.

9 Dimensional check

A useful engineering habit is to verify units before trusting a dynamical equation. The gravitational parameter has dimensions

(42)

The factor r∕r³ has dimensions

(43)

Therefore

(44)

which is correctly an acceleration.

10 The role of the reference frame

Equation (25) is a Newtonian inertial-frame equation, or equivalently an equation for the relative coordinate expressed in axes that are not rotating with Earth. This condition matters.

Suppose a vector is represented in an inertial frame I and a frame E rotating with angular velocity Ω_E∕I. The transport theorem gives

(45)

Differentiate again:

I =	E + 2ΩE∕I ×E	(46)
	+ × r + ΩE∕I ×.	(47)

For nearly constant Earth rotation rate,

(48)

If the inertial acceleration is central gravity,

(49)

then the Earth-fixed coordinate acceleration satisfies

(50)

The additional terms are the Coriolis and centrifugal accelerations associated with the rotating coordinates.

This is why one must not casually label the components x,y,z as ECEF and still use Eq. (25) unchanged. The force is the same physical gravity, but the coordinate acceleration is frame dependent.

GPS broadcast ephemeris eventually produces Earth-fixed coordinates, so the inertial-to-rotating-frame distinction will return in GPSORB09 and GPSORB15.

11 Cartesian component equations

Let the nonrotating Cartesian coordinates of the relative vector be

(51)

Then Eq. (25) becomes

(52)

Thus

	ẍ = − ,	(53)
	ÿ = − ,	(54)
	z = − .	(55)

The equations are coupled and nonlinear. The x acceleration, for example, depends not only on x but also on y and z through r.

12 From one second-order vector equation to six first-order equations

Numerical propagators and navigation filters usually work with first-order state equations. Define velocity

(56)

and the six-component state

(57)

Then

(58)

In scalar form,

	= vx,	(59)
	= vy,	(60)
	= vz,	(61)
	= −,	(62)
	= −,	(63)
	= −.	(64)

A single trajectory is selected by specifying

(65)

This requires six scalar initial values. GPSORB02 will examine carefully why those six degrees of freedom can instead be represented by six constants such as the classical orbital elements.

13 A first circular-orbit sanity check

The full solution of Eq. (25) is deferred until later articles, but a circular orbit provides an immediate physical check.

For uniform circular motion of radius r, the required inward centripetal acceleration is

(66)

Gravity provides the inward acceleration

(67)

Equating them,

(68)

so

(69)

Because v = 2πr∕T,

(70)

and therefore

(71)

This is the circular-orbit form of Kepler’s third law and previews the mean-motion relation that will later become

(72)

14 GPS-scale numerical check

Take a representative GPS orbital radius or semimajor axis

(73)

and the legacy GPS user-model gravitational parameter

(74)

The central gravitational acceleration magnitude is

|ag|	= ,	(75)
	≈ 0.565 m∕s2.	(76)

This is much smaller than surface gravity because a GPS satellite is several Earth radii from Earth’s center.

For a circular orbit at the same radius, Eq. (69) gives

(77)

and Eq. (71) gives

(78)

These familiar GPS-scale numbers emerge directly from the same equation of motion that will later generate the orbital-element and broadcast-ephemeris equations.

15 What is hidden inside the phrase “two-body problem”?

Equation (25) is powerful precisely because it removes many physical effects. The assumptions should be explicit.

These are not defects. They are deliberate model reductions. The central two-body solution provides the zeroth-order orbit around which higher-fidelity effects can be understood as perturbations.

16 A hierarchy of orbit models

It is useful to see where Eq. (25) sits in a realistic navigation force model. One may write schematically

(79)

The first term is

(80)

Every later term describes a departure from the ideal two-body model. The GPS control segment may use a much higher-fidelity force model to estimate and predict the spacecraft trajectory, while the broadcast navigation message provides users with a compact fitted representation suitable for receiver computation.

The present article therefore supplies the physical baseline from which “perturbation” has meaning: a perturbation is a correction to an already understood central two-body motion.

17 Why the equation is nonlinear but highly structured

The acceleration is nonlinear because

(81)

Nevertheless, the force has two special properties:

1.

it is central, because a is always parallel or antiparallel to r;

2.

it is conservative, because it derives from the scalar potential −μ∕r.

Those two properties create the conserved quantities that make the two-body problem analytically solvable. In later articles they will lead to

(82)

conservation of specific mechanical energy,

(83)

and the eccentricity vector. Together these reveal that the trajectory is a conic section and allow the Cartesian state to be replaced by orbital elements.

We intentionally stop short of deriving those integrals here. GPSORB02 will first clarify what “six constants” means mathematically, and GPSORB03–GPSORB04 will derive the angular-momentum, energy, and eccentricity invariants in detail.

18 A useful linearization preview: the gravity-gradient matrix

Although the exact two-body dynamics are nonlinear, navigation and estimation often need a local linear approximation. Let

(84)

For a small position perturbation δr,

(85)

The Jacobian is

(86)

This is the point-mass gravity-gradient or tidal matrix. It is not needed yet to derive the orbit, but it is a direct descendant of Eq. (25) and later appears naturally in variational equations, covariance propagation, and inertial-navigation error dynamics.

18.1 Derivation of the Jacobian

Write the ith component as

(87)

Then

	= −μ,	(88)
	= −μ,	(89)
	= μ.	(90)

Collecting the components gives Eq. (86).

This preview reinforces an important theme of the series: once the physics is written carefully in vector form, many later navigation equations are obtained by systematic calculus rather than by memorizing isolated formulas.

19 From GPSORB01 to GPSORB02

We can now state precisely what Kaplan’s compact equation means:

(91)

is the relative, central, Newtonian acceleration of an ideal two-body Earth-satellite system, normally written with μ ≈ GM_E and expressed in nonrotating axes. The factor r⁻³r is simply the inverse-square magnitude multiplied by the radial direction. The equation may be written as three coupled second-order scalar equations or as six first-order state equations.

The next question is no longer “where does the force law come from?” It is:

Why does solving this three-dimensional second-order system require six independent constants, and how can those six constants be reorganized into quantities with direct orbital meaning?

That is the subject of GPSORB02.

References

References