CLASSICAL ORBITAL ELEMENTS

Five independent quantities called "orbital elements" are sufficient to completely describe the size , shape and orientation of an orbit . A sixth element is required to pinpoint the position of the satellite along the orbit at a particular time . The classical set of six orbital elements are defined with the help of Figure 2 .3- 1 as follows :

1 . a, semi-major axis-a constant defining the size of the conic orbit.

2 . e, eccentricity-a constant defining the shape of the conic orbit . 3 . i, inclination-the angle between the

K

unit vector and the angular momentum vector,

h.

4. Q longitude of the ascending node-the angle , in the fundamental plane , between the

I

unit vector and the point where the satellite crosses through the fundamental plane in a northerly direction (ascending node) measured counterclockwise when viewed from the north side of the fundamental plane .

5 . w, argument of periapsis-the angle , in the plane of the satellite's orbit, between the ascending node and the periapsis point, measured in the direction of the satellite's motion.

6. T, time of periapsis passage-the time when the satellite was at periapsis.

The above definitions are valid whether we are describing the orbit of an earth satellite in the geocentric-equatorial system or the orbit of a planet in the heliocentric-ecliptic system. Only the definition of the unit vectors and the fundamental plane would be different.

It is common when referring to earth satellites to use the term

"argument of perigee" for w. Similarly, the term "argument of perihelion" is used for sun-centered orbits. In the remainder of this chapter we shall tacitly assume that we are describing the orbit of an earth satellite in the geocentric-equatorial system using IJK unit vectors.

The list of six orbital elements defined above is by no means exhaustive . Frequently the semi-latus rectum, p, is substituted for a in the above list. Obviously, if you know a and e you can compute p.

Instead of argument of periapsis, the following is sometimes used:

n longitude of periapsis-the angle from

I

to periapsis measured

Sec. 2.3 C LASS I C A L O R B ITA L E L E M E N TS 59

h

vernal

periapsis direction ^_

\

Figure 2.3-1 Orbital elements

60 O R B I T D ET E RM I N AT I O N F R OM O BS E RVAT IO N S Ch . 2

eastward to the ascending node (if it eXists) and then in the orbital plane to periapsis. If both n and W are defined then

(2 .3-1) If there is no periapsis (circular orbit), then b oth W and

n

^are

undefined.

Any of the following may be substituted for time of periapsis passage and would suffice to locate the satellite at to:

va true anomaly at epoch

-

the angle , in the plane of the satellite's orbit, between periapsis and the position of the satellite at a particular time , to' called the "epoch."

ua argument of latitude at epoch

-

the angle , in the plane of the orbit, between the ascending node (if it exists) and the radius vector to the satellite at time to' If w and V

0

are both defined then

Uo = w + ^Vo"

^{(2 .3-2)}

If there is no ascending node (equatorial orbit), then both w and

U

o are undefined.

£

a true longitude at epoch

-

the angle between I and ro (the radius vector to the satellite at t

J

measured eastward to the ascending node (if it exists) and then in the orbital plane to roo If n, w and va are all both of which are always defined .

Two other terms frequently used to describe orbital motion are

"direct" and "retrograde ." Direct means easterly. This is the direction in which the sun, earth, and most of the planets and their moons rotate on their axes and the direction in which all of the planets revolve around the sun. Retrograde is the opposite of direct. From Figure 2.3-1 you can see that inclinations between 00 and 900 imply direct orb its and inclinations between 900 and 1800 are retrograde.

Sec. 2 .4 D ETE R M I N I N G O R B IT A L E l E M E NTS 6 1 2.4 Let us assume that a radar site on the earth is able to provide us with DETERMINING THE ORBITAL ELEMENTS FROM r AND

v

the vectors r and v representing the po sition and velo city of a satellite relative to the geocentric-equatorial reference frame at a particular time ,

to.

How do we find the six orbital elements which describe the motion of the satellite? The first step is to form the three vectors, h, ⁿ and e.

2.4. 1 Three Fundamental Vectors-h, n and e. We have already encountered the angular momentum vector, h.

Thus

h = r x v

I J ^K

h = rl rJ rK = hII + hl+ hKK . vI vJ vK

(2 .4-1 )

(2 .4-2)

An important thing to remember is that h is a vector perpendicular to the plane of the orbit.

The node vector, n, is defined as

(2 .4-3) Thus

From the definition of a vector cro ss product , n must be perpendicular to both K and h. To be perpendicular to ^K, n would have to lie in the equatorial plane . To be perpendicular to h, ⁿ would have to lie in the orbital plane. Therefore , n must lie in both the equatorial and orbital planes, or in their intersection which is called the "line of

62 O R B I T D ET E R M I N AT I O N F R OM OBSE RVAT I O N S Ch . 2

nodes." Specifically, n is a vector pointing along the line of nodes in the direction of the ascending node. The magnitude of n is of no consequence to us. We are only interested in its direction.

The vector, e, is obtained from

(2.4·5)

and is derived in Chapter 1 (equation ( l .S - 1 1 )). The vector e points from the center of the earth (focus of the orbit) toward perigee with a magnitude exactly equal to the eccentricity of the orbit.

All three vectors, h, n and e are illustrated in Figure 2.3-1 . Study this figure carefully. An understanding of it is essential to what follows.

2.4.2 Solving for the Orbital Elements. Now that we have h, n and e

we can proceed rather easily to obtain the orbital elements. The parameter, p, and eccentricity follow directly from h and e while all the remaining orbital elements are simply angles between two vectors whose components are now known . If we know how to find the angle between two vectors the problem is solved . In general, the cosine of the angle, ex, between two vectors A and B is found by dotting the two vectors together and dividing by the product of their magnitudes. Since

then

A ^• B = AB

cos a

cos a

= A . B

AB (see Figure (2.4-1) (2.4-6)

Of course , being able to evaluate the cosine of an angle does not mean that you know the angle . You still have to decide whether the angle is smaller or greater than 1 800 . The answer to this quadrant resolution problem must come from other information in the problem as we shall see .

We can outline the method of finding the orbital elements as follows:

Sec. 2.4 D E T E R M I N I N G O R B I TA L E L E M E NTS

2. e ⁼

I

^e

l

3 . Since

i

is the angle between K and h,

(Inclination is always less than 1 8 0° .) 4. Since n is the angle between I and n,

(If ⁿ

J

> 0 then n is less than 1 80° .) 5 . Since w is the angle between n and e,

(If e K > O then w is less than 1 80° .) 6. Since Vo is the angle between e and f,

coS P

o = � er

(If f • v > O then Po is less than 180° .) 7 . Since Uo is the angle between n and f,

cos

U

o

⁼!!...:.L_nr ^{. (If}

r� O

^then U is less than 1 80

�

⁾

63

(2.4-7)

(2 .4-8)

(2.4-9)

(2.4-1 0)

(2.4- 1 1)

64 O R B I T D ET E R M I N AT I O N F R OM O BS E RVAT I O N S

x

Figure 2.4-1 Angle between vectors

8 . Q = _o

n +

^w

+

^V ₀ ⁼

n +

^u ₀

Ch . 2

All of the quadrant checks in parentheses make physical sense. If they don't make sense to you, look at the geometry of Figure 2.3-1 and study it until they do . The quadrant check for Vo is nothing more than a method of determining whether the satellite is between periapsis and apoapsis (where flight-path angle is always positive) or between apoapsis and periapsis (where ¢ is always negative). With this hint see if you can fathom the logic of checking ^{r • v.}

The three orbits illustrated in the following example problem should help you visualize what we have been talking about up to now.

Sec. VI D E T E R M I N I N G O R B I T A L E lE M E NTS 65

Find the orbital elements of the three orbits in the following illustrations.

Figure 2.4-2 Orbit 1

ORBIT 1 (retrograde equatorial)

p = 1 .5 DU

e == .2

n = undefined

v =

^o

2700

U o

= undefined

Q

^o

= 31 50

66 ORBIT DETERMINATION f ROM OBSERV AT IONS

Ch . 2

Figure ^2.4-3 Orbit 2

ORBIT ² (polar)

w ==

'\800

p == ,\ .5 DU

e == .2

Sec. 2.4

p :::: 1.5 DU

e :::: O

D E T E R M I N I N G O R B ITAL E L E M E NTS

Figure 2 .4-4 Orbit 3

ORBIT 3 (direct circular)

w :::: undefined

v 0 ::::

undefined

u ^o

^::::

2700

Q

^o

^::::

4200

67

68 O R B I T D ET E R M I N AT I O N F RO M O BS E RVAT IO N S Ch . 2

EXAMPLE PROBLEM. A radar tracks a meteoroid and from the tracking data the following inertial position and velocity vectors are found (expressed in the geocentric-equatorial coordinate system).

r =

²¹

DU

^EEl

v = 1J DU (TU

^{EEl EEl}

Determine the 6 orbital elements for the observed meteroid.

Find p using equation (2.4- 1 )

Then from equation ( 1 .6- 1 )

h2 .

p

= - = 4DU = 13,775.74n.ml. _f.l

^{. EEl}

Find eccentricity, e, using equation (2.4-5) e

= ^.1. _f.l ^{[ (}

^{v2 -}.I:!:....) r ^{r - (r} .

v)v 1 = 11 e = l el = 1

Since h ^=;i:.

0

^{and e =} 1 the path of the meteoroid is parabolic with respect to the earth.

Find inclination, i. Using equation (2.4-7)

i =

^COS

^-1

^(�) _h

= 0°

Therefore the meteoroid is traveling in the equatorial plane . Find longitude of ascending node ,

n.

From equation (2.4-8)

Sec. 2 .4 D ETE R M I N I N G O R B I T A L E L E M E NTS

.n

=

^COS

^-1

(.!!...:L) _n

69

However, since the meteoroid is located in the equatorial plane there is no ascending node because its trajectory does not cross the equatorial plane and therefore , for this case .n, is undefined.

Find argument of periapsis, W.

From equation (2.4-9)

n · e W

= ^cos-1

^(--ne ⁾

Again, since there is no ascending node w is also undefined . In lieu of the w the longitude of periapsis, II, can be determined in this case.

Find longitude of periapsis, II.

Since the orbital plane is coincident with the equatorial plane

(i =

^{(fJ) II}

is measured from the I-axis to periapsis which is colocated with the e

vector. Therefore from the definition of the dot product

1

e • I 0

II

= ^{cos -}

⁽

--

e ⁾

= 0

it is determined that perigee is located along the I axis.

Find true anomaily, va.

From equation (2.4-1 0)

1

e · r 0 v == ^a

cos-

₍^--er ₎

= 0

and it is found that the meteoroid is presently at periapsis.

7 0 O R B I T D E T E R M I N AT I O N F RO M O B S E RVAT I O N S

Find true longitude of epoch,

Qo'

From equation (2.3-3)

Q ^o =

^IT

+

^v

⁰ = 0°

Ch . 2

EXAMPLE PROBLEM. The following inertial position and velocity vectors are expressed in a geocentric equatorial coordinate system for an observed space object :

r = 3v'3 I + L J _{4 4} ^DU

EB

Determine the 6 orbital elements for the observed object.

Find p.

First find

h

by equation (2^.4^-1 )

h=r xv=-1 _{8 8} ^{DU2 /TU}

EB EB

In document Fundamentals of Astrodynamics Bate Mueller and White 0486600610 (Page 73-85)