Orbit Determination
3.4 Transforming Cartesian Coordinates to Orbital Elements
equatorial plane fromI to position vector r0. True longitude is always defined because the vectorsI and r0always exist.
large inclination (i = 63.4 ) tells us that the orbital plane is tilted away from the equatorial (IJ) plane. Because the argument of perigee ω is –90 (or 270 ), the perigee of the Molniya orbit is in the southern hemisphere. Finally, because the true anomaly at epoch isθ0= 180 , we know that the satellite is at apogee; becauseω = –90 , the satellite’s apogee is at its
“highest point” (geographic latitude) on its passage through the northern hemisphere.
Now consider the state vector (inIJK coordinates) that corresponds to the satellite’s position at epoch t0(apogee in this case):
r0=
−15,865 13,312 41,357
km, v0=
−0 9601
−1 1443 0
km/s
It is very difficult to visualize or ascertain the characteristics of the Molniya orbit directly from the state vector (r0,v0). Therefore, computing the six orbital elements from the six-dimensional state vector (r0,v0) is crucial to the orbit-determination process.
Before we proceed, it is important to stress that the state vector (r0,v0) expressed in a body-centered Cartesian frame and the six orbital elements (a, e, i,Ω, ω, θ0) both com-pletely define the two-body orbit. In other words, there is a unique mapping between the six-element state vector (r,v) and the six classical orbital elements. Furthermore, all six coordinates of the state vector (r,v) will change as the satellite moves from epoch t0to t1 (unless time t1is k periods after epoch t0). In contrast, the five orbital elements (a, e, i,Ω, ω) will remain constant for two-body motion and only the true anomaly θ will change as the satellite moves in its orbit.
Let us develop a systematic process for determining the classical orbital elements (a, e, i,Ω, ω, θ0) from the state vector (r0,v0) at epoch t0. We will follow the processes presented in Vallado [1; pp. 95–100] and Bate et al. [2; pp. 61–63]. The first step is to determine total energy from the magnitudes of the position and velocity vectorsr0andv0
ξ =v20 2−μ
r0
(3.4) where r0= r0 and v0= v0 . Recall from Chapter 2 that semimajor axis is solely a func-tion of energy:
a=−μ
2ξ (3.5)
Next, we compute the eccentricity vector e directly from state vector (r0,v0) by using Eq. (2.57):
e =1
μ v20−μ
r0 r0− r0 v0 v0 (3.6)
We derived this expression for the eccentricity vector in Chapter 2 when we developed the trajectory equation (2.45). The vectore points from the center of the gravitational body to periapsis (see Figure 3.4) and its magnitude is the eccentricity of the orbit:
e= e (3.7)
Figure 3.4 shows that inclination is the angle between vectorsK and h and therefore it can be computed from the dot (or scalar) product:
cosi =K h
h (3.8)
where the angular momentum vectorh is computed from r0andv0
h = r0×v0 (3.9)
Equation (3.8) is derived from the dot-product definitionA B = AB cos α where α is the angle between vectorsA and B and A and B are their respective magnitudes. Note that the unity magnitude of vectorK is not explicitly shown in Eq. (3.8). Because the range for inclination is 0 ≤ i ≤ 180 , we may simply apply the inverse cosine operation to Eq. (3.8) without an additional quadrant check.
Longitude of the ascending nodeΩ is the angle between I and ascending node vector n (see Figure 3.4), and therefore its cosine is
cosΩ =I n
n (3.10)
(again, note that I = 1). The node vector n is the cross product of vectors K and h (see Figure 3.4):
n = K × h (3.11)
We cannot determineΩ by simply taking the inverse cosine of Eq. (3.10) because the resulting angle will always be in the first or second quadrant (i.e., between 0 and 180 ).
Longitude of ascending nodeΩ ranges from 0 to 360 ; therefore, we need a quadrant check. From Figure 3.4, we see that a unit vector in the direction of node vector n may be expressed in terms ofI and J components:
n
n = cosΩI + sinΩJ (3.12)
Note that dotting all terms in Eq. (3.12) with vectorI yields Eq. (3.10) because I I = 1 and I J = 0. Similarly, we can dot all terms in Eq. (3.12) with vector J to yield
sinΩ =J n
n (3.13)
We must use Eqs. (3.10) and (3.13) together to determine the proper quadrant forΩ. For computer applications (such as MATLAB), theatan2 function with input arguments sinΩ and cos Ω will place angle Ω in the proper quadrant. When using a hand-held calculator, the signs of sinΩ and cos Ω will determine the proper quadrant. For example, if sinΩ = −0 5 and cosΩ = −0 866025, we know that the longitude of the ascending node mustbe in the third quadrant, orΩ = 210 . This simple example should serve as a strong warning to the reader: do not simply press the inverse-cosine (or inverse-sine) button on a calculator and assume that you have determined the correct angle!
Figure 3.4 shows that the argument of periapsisω is the angle between n and e and therefore its cosine is
cosω = n e
n e (3.14)
For reasons previously mentioned, we cannot determineω by simply taking the inverse cosine of Eq. (3.14). The quadrant check forω depends on the sign of eZ, which is theK component of eccentricity vectore. If eZ> 0 (as shown in Figure 3.4), the vectore is in the northern hemisphere (“above” the equatorial plane) and argument of periapsis ranges from 0 to 180 . In this case, the“calculator” inverse-cosine operation of Eq. (3.14) pro-duces the correct answer. If eZ< 0 (periapsis is below the equatorial plane), the argument of periapsis is between 180 and 360 . In this case, the correct argument of periapsis is
If eZ< 0 ω = 360o−cos−1 n e
n e (3.15)
The sixth orbital element is the true anomaly at epoch t0that corresponds to the state vector (r0,v0). The cosine ofθ0is determined by the dot product of the eccentricity vector e and the satellite’s position vector r0(see Figure 3.4)
cosθ0=e r0
er0 (3.16)
True anomaly ranges from 0 to 360 ; hence, we cannot rely solely on the inverse cosine function. The easiest quadrant check utilizes the sign of the dot productr0 v0. Recall from Chapter 2 thatr0 v0= r0r0where r0is the radial velocity component. Therefore, ifr0 v0> 0 the radius is increasing (r0> 0) and the satellite is moving toward apoapsis and true anomaly is between 0 and 180 . On the other hand, ifr0 v0< 0 the radius is decreasing and the satellite is moving toward periapsis and θ0is between 180 and 360 . In this case, we must use
If r0 v0< 0 θ0= 360o−cos−1 e r0
er0
(3.17) It is worth repeating that orbital elementsΩ, ω, and θ0can range from 0 to 360 and therefore an inverse cosine operation alone will not correctly determine the orbital ele-ment if the angle is between 180 and 360 . The correct quadrant must be determined by applying the quadrant checks denoted by Eqs. (3.13), (3.15), and (3.17). When determin-ing inclination, one may simply take the inverse cosine of Eq. (3.8) because inclination is always between 0 and 180 .
It is easy to recognize that a computer program will readily perform these steps.
MATLAB is well suited to carry out the vector manipulations (cross product, dot prod-uct, and vector norm) and the inverse-angle calculations with theatan2 function. The reader is encouraged to develop a general-purpose M-file that computes the classical orbital elements given the state vector (r,v) as an input (see Problem 3.18). The following example illustrates this orbit-determination process.
Example 3.1 A tracking station determines the following state vector for an Earth-orbiting satellite in ECI coordinates:
r0=
9, 031 5
−5,316 9
−1,647 2
km, v0=
−2 8640 5 1112
−5 0805 km/s
Determine the classical orbital elements at this epoch. Is this satellite in a Mol-niya orbit?
We will useμ = 3.986(105) km3/s2for Earth-satellite example problems in this chapter.
The reader can consult Appendix A for a more precise numerical value if desired. First, we determine the magnitude of the position and velocity vectors:
r0= 9, 031 52+ −5,316 9 2+ −1,647 22= 10,609 km v0= −2 8640 2+ 5 11122+ −5 0805 2= 7 7549 km/s Energy is
ξ =v20 2−μ
r0
=– 7 5027 km2/s2=− μ 2a Therefore, semimajor axis is a = 26, 563 6 km
Using Eq. (3.6), the eccentricity vector is e =1
μ v20−μ
r0 r0− r0 v0 v0
The dot productr0 v0 is
r0 v0= 9,031 5 –2 8640 + – 5,316 9 5 1112 + – 1,647 2 – 5 0805
=– 44,673 4 km2/s The eccentricity vector is
e =
0 1903 0 2718
−0 6627
Eccentricity is the vector norm, e = e = 0 19032+ 0 27182+ −0 66272= 0 7411 Computing inclination requires the angular momentum vector:
h = r0×v0=
I J K
9, 031 5 −5,316.9 −1,647.2
−2 8640 5 1112 −5 0805
=
35,432 50,602 30,934
km2/s
Using Eq. (3.8), the cosine of inclination is cosi =K h
h =30,934
69,086= 0 4478
where the dot productK h = hZ. Taking the inverse cosine, we obtain i = 63 4 Computing the longitude of the ascending node (Ω) and argument of perigee (ω) requires the ascending node vector:
n = K × h =
I J K
0 0 1
35,432 50,602 30,934
=
−50,602 35,432
0
Equations (3.10) and (3.13) allow us to compute the cosine and sine ofΩ cosΩ = I n
n =−50,602
61,774 =−0 8192 sinΩ = J n
n =35,432
61,774= 0 5736
Because cosΩ < 0 and sinΩ > 0, the ascending node is in the second quadrant. The lon-gitude of ascending node is Ω = 145
Using Eq. (3.14), the cosine ofω is cosω = n e
n e= 0 8570
61,774 0 7411 = 10– 5
The inverse cosine of 10–5 is 90 . However, we must check theK component of the eccentricity vector e in order to determine whether ω is +90 or –90 . Because eZ=−0 6627 < 0, the perigee is in the southern hemisphere and Eq. (3.15) shows that the argument of perigee is
ω = 360o−90o= 270 or−90
Finally, we compute the cosine of true anomaly at epoch using Eq. (3.16) cosθ0=e r0
er0
= 0 1903 9,031 5 + 0 2718 −5,316 9 + −0 6627 −1,647 2 0 7411 10,609
=1,365
7,863= 0 1736
Becauser0 v0< 0 (see computations fore), the satellite is approaching perigee and we must use Eq. (3.17) to compute true anomaly:
θ0= 360o−cos−1 e r0
er0
= 360 −80 = 280
If we compare these calculated orbital elements to the description of the Molniya orbit at the beginning of this section, we see that the orbit in this example is indeed a Molniya orbit. The critical orbital features for a Molniya orbit are a = 26,564 km (i.e., the period is about 12 h), e = 0.7411, i = 63.4 , andω = –90 (perigee in southern hemisphere). The argument of perigee must be–90 so that the Molniya apogee is at the highest possible northern latitude for good observation of Russia. Because the apogee altitude is very large (~39,874 km) and the perigee altitude is low (~500 km), the apogee speed is relatively slow (~1.5 km/s) and the perigee speed is very fast (~10 km/s). Therefore, a satellite in a Molniya orbit will spend most of its time near apogee where it can send and receive signals with ground stations in Russia.