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7 Einstein's Modification of the Orbit Equation

7 Einstein's Modification of the Orbit Equation

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4.8 Universality of Newton’s Law



93



increases u by a very small constant quantity. The third term,

˛ 2

e cos 2.Â

6p2



Â0 /



(4.75)



is very small and periodic. However, the term

˛

e sin.Â

p2



Â0 /



(4.76)



is periodic, and steadily increasing in amplitude as  increases. So this is bound to

have some effect with increased time. Considering only the observable effects,

uD



1

Œ1 C e cos.Â

p



Â0 / C



˛e

 sin.Â

p2



Â0 /



(4.77)



Let k D ˛Â=p and neglect ˛ 2 , then Eq. (4.77) can be written as

uD



1

f1 C eŒcos.Â

p



Â0 / C k sin.Â



Â0 /g



1

Œ1Ce cos. Â0 kÂ/

p



(4.78)



Â0 C k is the angular coordinate of the perihelion. So the planet is moving in an

ellipse with a moving line of apsides, which is a slowly rotating ellipse. The angular

change of the line of apsides is ! D 2 ˛=p per period. Substituting values for

the planets we have

Mercury: ! D 43:03 arcsec per century.

Venus: ! D 8:63 arcsec per century.

Earth: ! D 3:84 arcsec per century.

For Mercury this small effect was observable and remained unexplained until

Einstein’s relativity. It was fudged in Newcomb’s planetary theories and called an

empirical term (Danby 1962, pp. 66–67).



4.8 Universality of Newton’s Law

Newton’s law follows from Kepler’s first two laws of planetary motion; therefore,

any two bodies traveling around each other according to Kepler’s first two laws

are subject to Newton’s law. When two stars are observed moving around each

other, these are called visual binaries. In other cases of binaries, the two stars have

different spectra, continuously shifting spectra, or eclipse each other. For visual

binaries, when the fainter star is plotted with respect to the brighter, the orbits are

ellipses, and the law of areas is followed. However, a projection of the true orbit is

being plotted, so the brighter star is not at the focus. It is unlikely in any case that

the orbital plane is perpendicular to the line of sight.



94



4 Central Force Motion



An ellipse always projects into another ellipse. Since the apparent orbit as plotted

is an ellipse, the true orbit must be an ellipse. The law of areas of the apparent

orbit will hold for the true orbit, because the law of areas depends on ratios, which

are not affected by projections. For the true orbit, the theory of central forces and

conservative fields applies, and an ellipse with the brighter star at the focus can be

found. It can be shown that for visual binaries. elliptical motion requires the center

of attraction to be at the focus of the ellipse.

So Newton’s law is the only plausible law governing Keplerian motion within and

exterior to the solar system. Also, this law has explained deviations from Keplerian

motion, when the relativity effect is included, and has led to the correction of

Kepler’s third law (Danby 1962, pp. 73–76).



References

McCuskey, S.W.: Introduction to Celestial Mechanics. Addison-Wesley, Reading (1963)

Danby, J.: Fundamentals of Celestial Mechanics. The Macmillan Company, New York (1962)



Chapter 5



The Two-Body Problem



5.1 Introduction

Assume that the masses are spherically symmetrical and homogeneous in concentric

layers. So they attract one another as if the mass were concentrated at spherical

centers, i.e. gravitationally they act like two mass particles separated by the distance

between the centers. The two masses are assumed to be isolated from other masses,

so the only force acting is the inverse square force of their mutual attractions along

the line joining the centers. In astronomical applications, the distance between

centers is large, compared to the diameters of the spheres. This is not true for

artificial satellites.

The dynamics of the motion of two masses presents two problems for celestial

mechanics and astrodynamics:

1. Given the position and velocity, or three positions, of a mass as a function of time,

find the elements of the orbit. This is the computation of orbits to be considered

in Chap. 6.

2. Given the orbital elements, or parameters defining the orbital motion, find

the position in space of the mass at a given time. This we will take up now (the

first problem is by far the more difficult), with the first step being defining the

classical orbital elements.



5.2 Classical Orbital Elements

Assume that a mass m is rotating counterclockwise in orbit when viewed from the

planet’s north pole. If the orbital plane and the fundamental plane of some reference

frame intersect, then we define two points of interest on the line of intersection,



© Springer-Verlag Berlin Heidelberg 2016

P. Gurfil, P.K. Seidelmann, Celestial Mechanics and Astrodynamics: Theory

and Practice, Astrophysics and Space Science Library 436,

DOI 10.1007/978-3-662-50370-6_5



95



96



5 The Two-Body Problem



e(eccentricity vector)







uˆ A



Periapsis



f

i



Orbital plane



r



Descending

Node

‫܁‬



ω

Ω



Reference plane







i

‫܀‬

Ascending

Node



Reference frame



xˆ (vernal equinox)



ˆl (line of nodes)



Fig. 5.1 Definitions of the right ascension of the ascending node,

and the inclination, i. Also shown is the true anomaly, f



, the argument of periapsis, !,



as shown in Fig. 5.1: The first is the ascending node, denoted by . This point

marks the location on the line of intersection when moving eastward; the second is

the descending node, denoted by . This point marks the location on the line of

intersection when moving westward. The line connecting to is called the line

of nodes (LON); we will use the notation Ol to denote a unit vector that lies along

the LON.

We now define three angles that determine the orientation of the orbital plane

with respect to the reference frame: , the right ascension of the ascending node

(RAAN), also referred to as the longitude of the ascending node, an angle measured

from the vernal equinox (see Sect. 3.1) to the LON; !, the argument of periapsis,

which is an angle measured from the LON to the eccentricity vector (see Sect. 4.6.1);

and i, the inclination, an angle measured from the z axis of the reference frame, zO ,

to the vector normal to the orbital plane, uO A . These angles are shown in Fig. 5.1. It

is convenient to express the position vector in a perifocal coordinate system. This

coordinate system is centered at the attraction center. The fundamental plane is the

orbital plane. The unit vector PO is directed from the center to the periapsis (recall

O is normal to the fundamental plane, positive in the direction of the

Sect. 4.6.1), R

O is pointed toward the point where the true

orbital angular momentum vector, and Q

ı

anomaly is 90 , thus completing the right-hand Cartesian triad. Using the definition

of the true anomaly (Sect. 4.6.1), we can write the position vector in the perifocal



5.2 Classical Orbital Elements



97



frame as

rp D r Œcos f ; sin f ; 0T



(5.1)



where, as before, r D p=.1 C e cos f /.

We can transform from the orbital frame to the reference frame using three

consecutive clockwise rotations: a rotation about uO A by 0 Ä ! Ä 2 , mapping

the eccentricity vector, eO , onto the LON, Ol; a rotation about Ol by 0 Ä i Ä , mapping

uO A onto zO ; and a rotation about zO by 0 Ä Ä 2 , mapping Ol onto xO .

The composite rotation, transforming any vector in the orbital frame into the

inertial frame is given by

2



c c! s s! ci

T D 4 s c! C c s ! ci

s! si



3

c s! s c! ci s si

s s! C c c! ci c si 5

c! si

ci



(5.2)



where we used the compact notation cx D cos x, sx D sin x.

Transforming into inertial reference coordinates using Eqs. (5.1) and (5.2), we

obtain the position vector

3

2

cf C! c

ci sf C! s

p

4 ci c sf C! C cf C! s 5

rD

1 C e cos f

si sf C!



(5.3)



The true anomaly f depends on time and on the epoch of observation, T. Thus, the

inertial position and velocity depend on time t and the classical orbital elements,

given by

fa; e; i;



; !; Tg



(5.4)



5.2.1 Osculating Orbital Elements

In the two-body problem, the orbital elements are constant; f is time-varying.

However, in the presence of perturbations and/or thrust forces, the orbital elements

may become time-varying, and are referred to as osculating orbital elements.

In general, elliptical motion constitutes a correct approximation to the real

motion observed in the solar system. Thus, for example if, starting from an instant

t0 , all the perturbing forces were neglected, the movement of a body would become

exactly elliptical. It would represent the real movement quite well for a certain time,

even though strictly speaking, it would not be identical with the real movement as

regards position and velocity, except at the instant t0 . The elements of an ellipse that

would be followed by a body after a specific time t are thus said to be osculating, or



98



5 The Two-Body Problem



instantaneous, if starting from this instant, all the forces with the exception of the

central force were to disappear. The elements of such an unperturbed orbit can be

defined at any instant; they correspond to the elliptical orbit followed by a moving

body, which would have at the given instant the same position and velocity as the

real body. As in fact the real orbit is simply tangential to the osculating orbit, at an

instant tCıt the osculating orbit will be different, with different osculating elements.

It follows that the osculating elements in perturbed motion are no longer constant,

but are functions of time.

Osculating elements can be used to describe the perturbed motion of a body.

They possess the advantage of having a precise and simple geometrical significance

while having small variations.

The coordinates and velocity components of perturbed motion at an instant t are

those which would be obtained at this instant t, assuming that the orbit is elliptical,

from elements equal to the osculating elements at the same time t.



5.2.2 Nonsingular Orbital Elements

While the angles ; i; ! may become degenerate is some cases (for instance, !

is undefined for circular orbits; both ! and are undefined for equatorial orbits),

the position and velocity vectors are always well-defined. However, occasionally

alternative orbital elements are used to alleviate these deficiencies. These alternative elements are collectively referred to as nonsingular orbital elements, see

Sect. 10.12. A thorough survey of these elements was performed by Hintz (2008).



5.3 Motion of the Center of Mass

Let an origin, O, define an inertial system with Newton’s laws of motion. The

positions of two masses are given by vectors r1 and r2 , and R is the vector to the

center of mass of the pair, C. r is the position vector of m2 relative to m1 , as shown

in Fig. 5.2.

From Newton’s law of gravitation, the force on m1 due to m2 is r12 k2 m1 m2 uO r and

that on m2 due to m1 is r12 k2 m1 m2 uO r . uO r is a unit vector in the direction of r, and

k2 is the constant of gravitation. The reason and significance of the notation k2 will

become apparent later. The equations of motion are

m1 rR 1 D

m2 rR 2 D



k 2 m1 m2

r

r3

k 2 m1 m2

r

r3



(5.5)

(5.6)



5.4 Relative Motion



99

1



Fig. 5.2 Motion of the center

of mass



1



2

2



Adding Eqs. (5.5) and (5.6) and integrating twice, we have

m1 r 1 C m2 r 2 D c 1 t C c 2



(5.7)



where c1 and c2 are vector constants. The left side of Eq. (5.7) is MR, by the

definition of the center of mass, with M D m1 C m2 . Thus,

RD



c2

c1 t

C

M

M



(5.8)



so the center of mass moves uniformly in a straight line in space.

This is in agreement with previous results and what one would expect, since

there is no external force acting on this system. This result is applicable to double

star observations, but of little interest otherwise (McCuskey 1963, pp. 32–33).



5.4 Relative Motion

The motions of m1 and m2 relative to the center of mass can be derived as follows.

Let r1 D R C r01 and r2 D R C r02 , where r01 and r02 denote position vectors to m1

and m2 from the center of mass, C, respectively, as shown in Fig. 5.3.

R D 0, from definitions of r0 and r0 , m1 rR 1 D m1 rR 0 and

Then r D r02 r01 . Since R

1

2

1

0

m2 rR2 D m2 rR 2 . Thus Eqs. (5.5) and (5.6) become

m1 rR 01 D

m2 rR 02 D



k2 m1 m2 .r02

r3

k2 m1 m2 .r02

r3



r01 /

r01 /



(5.9)

(5.10)



100



5 The Two-Body Problem

1



Fig. 5.3 Motion of two

masses relative to the center

of mass



′1



1



′2



2

2



and m1 r01 C m2 r02 D 0 due to the center of mass definition. So r02 can be eliminated

from Eq. (5.9) and r01 can be eliminated from Eq. (5.10). Thus,

m1 rR 01 D

m2 rR 02 D



Ä

m1

k 2 m1 m2 1 C

m2

Ä

m2

k 2 m1 m2 1 C

m1



r01

r3



(5.11)



r02

r3



(5.12)



Since

rD



M 0

M 0

r1 D

r

m2

m1 2



(5.13)



we may write by dividing though by m1 or m2 and replacing the sum of the masses

by M, and then substituting from above,

rR 01 D



k2 M 0

r D

r3 1



k2



rR 02



k2 M 0

r D

r3 2



2



D



k



Â

Â



m32

M2

m31

M2



Ã

Ã



r01

r103



(5.14)



r02

r203



(5.15)



The accelerations of m1 and m2 relative to the center of mass are given. They are the

same as Eqs. (5.5) and (5.6), with m1 and m2 , respectively, replaced by the adjusted

effective mass.

From Eqs. (5.14) and (5.15) and constants c1 and c2 of Eq. (5.8), the positions

of m1 and m2 can be determined for any time. However, the constants cannot be

determined, because they are with respect to an origin fixed in space. So the solution

must be for one mass with respect to the other.



5.5 The Integral of Areas



101



Consider m1 as the origin of the two-body system. Then, from the first parts of

Eqs. (5.14) and (5.15)

rR D



k2 M

r

r3



(5.16)



where r is the relative radius vector. This is the acceleration of m2 around m1 . In a

planetary system, m1 is the Sun and m2 is the planet. In the case of a satellite and a

planet, m1 is the planet and m2 is the satellite.

For computations, the equations of relative motion can be expressed in Cartesian

O so that r D xOi C yOj C zk.

O Hence, Eq. (5.16) is

form, with unit vectors Oi; Oj; k,

xR D



k2 Mx.x2 C y2 C z2 /



3=2



yR D



k2 My.x2 C y2 C z2 /



3=2



zR D k2 Mz.x2 C y2 C z2 /



3=2



(5.17)



The differential equations, Eqs. (5.5) and (5.6), in vector form, are three secondorder equations. Each solution introduces two constants of integration, which would

be the initial conditions. So there are twelve constants in the original system.

If we ignore the motion of the center of gravity, the number of constants reduces

to six. So, the solution of Eqs. (5.17) will result in six constants. These six constants

can be determined, if we know the 3 position coordinates and 3 velocity components

at any instant. So to determine an orbit, six pieces of information are required; they

can be three observations of two angles each, or two observations of two angles and

a distance, each. An orbit is thus defined by six values: position and velocity or six

parameters (McCuskey 1963, pp. 33–35).



5.5 The Integral of Areas

The motion of m2 around m1 is a central force motion, so the areal velocity is

constant. That is

P D 1 .r

A

2



v/ D



1

huO A

2



(5.18)



where uO A is a unit vector with constant direction perpendicular to the orbital plane

defined by r and v. The components of areal velocity in Cartesian coordinates are

1

.yPz

2

1

.zPx

2

1

.xPy

2



1

c1

2

1

xPz/ D c2

2

1

yPx/ D c3

2

zPy/ D



(5.19)



102



5 The Two-Body Problem



where c1 ; c2 ; c3 are constants related to h by

q



c1 2 C c2 2 C c23 D h



(5.20)



From the initial coordinate and velocity components of m2 , the constants

c1 ; c2 ; c3 can be determined. When known, they must be related to the elements

of the orbit, which were defined in Sect. 5.2 (see also Fig. 5.1). In terms of ; i the

unit vector is

uO A D sin i sin



Oi



Oj C cos i kO



(5.21)



Oj C 1 h cos i kO

2



(5.22)



sin i cos



The areal velocity is

P D 1 h sin i sin

A

2



Oi



1

h sin i cos

2



Comparing Eqs. (5.19) and (5.20) we have

c1 D h sin i sin

c2 D h sin i cos

c3 D h cos i

q

h D c21 C c22 C c23



(5.23)



From the initial conditions, c1 ; c2 ; c3 are determined, and Eqs. (5.23) determine

and i. These elements orient the orbital plane with respect to a Cartesian coordinate

system (McCuskey 1963, pp. 35–36).



5.6 Elements of the Orbit from Position and Velocity

The orientation of the orbital plane is established by the constants c1 ; c2 ; c3 . The

size, shape, and orientation in the orbital plane must be determined. It will be a conic

section with a central force mass, M, at the conic focus. k2 D G is the gravitational

constant, m1 is at the focus, and m2 is the moving mass.

We seek the elements: a, the semimajor of an ellipse or semitransverse axis of a

hyperbola, respectively; q, the distance from the focus to the vertex of the parabola;

e, the eccentricity; !, the argument of periapsis, which is the angle in the orbital

plane between the line of nodes and the eccentricity vector, as explained in Sect. 5.2;

and T, the time of periapsis passage. The longitude of periapsis is !Q D C !. As

defined in Sect. 4.6.1, the true anomaly, f , is the angle in the orbital plane between

the eccentricity vector and the m2 position vector, as shown in Fig. 5.1.



5.6 Elements of the Orbit from Position and Velocity



103



The equation of the conic for the m2 motion is

rD



p

1 C e cos f



(5.24)



The coordinates, x0 ; y0 ; z0 , and velocity components, xP 0 ; yP 0 ; zP0 , are for m2 at time

t D 0. Then Eqs. (5.19) and (5.23) are



which determine

q



are r0 D



x20



C



y20



c1 D y0 zP0



z0 yP0 D h sin i sin



c2 D z0 xP0



x0 zP0 D



c3 D x0 yP0



y0 xP0 D h cos i



h sin i cos

(5.25)



; i, and h, as previously

indicated. The initial distance and speed

q



C z20 and v0 D



xP 20 C yP 20 C zP20 . Then, for an ellipse,



2

1

D

a

r0



v02

k2 M



(5.26)



2

r0



(5.27)



For a hyperbola,

v2

1

D 20

a

k M

For a parabola,

qD



h2

2k2 M



(5.28)



We can determine the eccentricity from the values of h by

e2 D 1



h2

k2 Ma



(5.29)



where there is a minus sign for an ellipse and a plus sign for a hyperbola. The angle

! can be calculated as follows. Set the argument of latitude as u D f C !, then

r cos u D x cos

r sin u D . x sin



C y sin

C y cos / cos i C z sin i



(5.30)



Once and i are known, and from initial conditions r0 is known, Eq. (5.30) yields

u0 at time t0 . From Eq. (5.24),

e cos f D



p

r



1



(5.31)



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