Orbit Determination
3.8 Orbit Determination from Three Position Vectors
Comparing the inertial velocities computed in Examples 3.4 and 3.5, we see that the largest component difference is less than 0.5 m/s.
The derivation of the Gibbs method presented here follows Bate et al. [2; pp. 109–115].
The basis of the Gibbs method is that the three position vectors (r1,r2,r3) must lie in the same plane. Therefore, the third position vectorr3must be a linear combination of the two previous vectorsr1andr2. This linear dependence may be written as
c1r1+ c2r2+ c3r3=0 (3.61) where c1, c2, and c3are constants. Next, take the cross product of Eq. (3.61) with each of the three position vectors to produce three vector equations:
0 r r r r r
r1× 1+ 2 2× 1+ 33× 1=
1 c c
c
0 3 62a
0 r r r r r
r1× 2+ 2 2× 2+ 33× 2=
1 c c
c 0 3 62b
0 r r r r r
r1× 3+ 2 2× 3+ 3 3× 3=
1 c c
c
0 3 62c
Of course, the cross productri×ri=0. Rearranging Eq. (3.62a)–(3.62c) and noting that ri×rj=−rj×ri, we obtain
c2r1×r2= c3r3×r1 (3.63a) c1r1×r2= c3r2×r3 (3.63b) c1r3×r1= c2r2×r3 (3.63c) The next step involves the projection of each position vector ri onto the periapsis direction. Figure 3.4 shows that the dot product of position vectorr and eccentricity vectore is
r e = recosθ (3.64)
We can also obtain the result recosθ from the trajectory equation:
r= p
1 + e cosθ or r+ re cosθ = p (3.65) Therefore,r e = recosθ = p−r. Taking the dot product of each term in Eq. (3.61) with the eccentricity vectore and substituting ri e = p−ri, we obtain
c1 p−r1 + c2 p−r2 + c3 p−r3 = 0 (3.66) Multiplying all terms in Eq. (3.66) by the vector productr3×r1 yields
r3×r1 c1 p−r1 + r3×r1 c2 p−r2 + r3×r1 c3 p−r3 =0 (3.67) Finally, we make use of equation set (3.63) by substituting Eq. (3.63c) for the term c1r3×r1and Eq. (3.63a) for the term c3r3×r1. After these substitutions, Eq. (3.67) only involves one unknown constant c2
r2×r3 c2 p−r1 + r3×r1 c2 p−r2 + r1×r2 c2 p−r3 =0 (3.68)
Or, after factoring out the single constant c2from Eq. (3.68), we obtain
c2 r2×r3 p−r1 + r3×r1 p−r2 + r1×r2 p−r3 =0 (3.69) Constant c2can be ignored in Eq. (3.69) because the bracketed term must equal the zero vector. Finally, we can group all terms involving parameter p on the left-hand side and all terms involving positions rion the right-hand side to yield
p r2×r3 + r3×r1 + r1×r2 = r1 r2×r3 + r2 r3×r1 + r3 r1×r2 (3.70) Equation (3.70) allows us to compute the parameter p from the three position vectors.
Let us define the auxiliary vectorsD and N as the left- and right-hand sides of Eq. (3.70):
D = r2×r3 + r3×r1 + r1×r2 (3.71) N = r1 r2×r3 + r2 r3×r1 + r3 r1×r2 (3.72) Therefore, Eq. (3.70) becomes pD = N and the parameter is p = N/D where N = N and D = D .
The next steps involve determining expressions for the perifocal-frame unit vectorsP, Q, and W. Note that vectors N and D both involve the two cross products (r1×r2and r2×r3) that are in the direction of the angular momentum vectorh, and one cross prod-uct (r3×r1) that is opposite ofh. Therefore, both N and D are in the direction of vector h and the perifocalW axis can be defined as
W =N
N (3.73)
The periapsis direction is the unit vector along the eccentricity vector,P = e/e, and the Q axis is defined by the cross product of unit vectors W and P
Q = W × P =N × e
Ne (3.74)
Or,
NeQ = N × e (3.75)
Substituting Eq. (3.72) for vectorN in Eq. (3.75), we obtain
NeQ = r1 r2×r3 ×e + r2 r3×r1 ×e + r3 r1×r2 ×e (3.76) Each right-hand side term in Eq. (3.76) is a vector triple product:
a × b × c = a c b− b c a Using the vector triple product, Eq. (3.76) becomes
NeQ = r1 r2 e r3−r1 r3 e r2+ r2 r3 e r1−r2 r1 e r3+ r3 r1 e r2−r3 r2 e r1 (3.77) Recalling Eqs. (3.64) and (3.65), the six dot products in Eq. (3.77) are ri e = p−ri; Eq. (3.77) becomes
NeQ = r1 p−r2 r3−r1 p−r3 r2+ r2 p−r3 r1−r2 p−r1 r3+ r3 p−r1 r2−r3 p−r2 r1
(3.78)
After some algebra, Eq. (3.78) becomes
NeQ = p r2−r3 r1+ r3−r1 r2+ r1−r2 r3 (3.79) Let us define the right-hand side bracketed vector in Eq. (3.79) as auxiliary vectorS
S = r2−r3 r1+ r3−r1 r2+ r1−r2 r3 (3.80) Hence, unit vectorQ = S/S. The magnitude of each side of Eq. (3.79) is
Ne= pS (3.81)
Substituting p = N/D into Eq. (3.81), we obtain e=S
D (3.82)
Now we have obtained the orbit’s size and shape (parameter p and eccentricity e) in terms of the magnitudes of the three auxiliary vectorsN, D, and S. However, we can use these three vectors to determine the ECI velocity vector corresponding to the middle observation. To show this, we recall an intermediate step in our derivation of the trajec-tory equation in Chapter 2. Repeat Eq. (2.40) with the middle observation position and velocity vectorsr2andv2
v2×h = μ r2
r2 +e (3.83)
Next, cross Eq. (3.83) with angular momentumh h × v2×h = μ h × r2
r2
+h × e (3.84)
The left-hand side is a vector triple product and the result is h2v2becauseh and v2are orthogonal. Next, we substituteh = hW and e = eP into Eq. (3.84) to yield
h2v2=μ hW × r2
r2
+ hW × eP (3.85)
SubstitutingQ = W × P and solving Eq. (3.85) for velocity yields v2=μ
h
W × r2
r2
+ eQ (3.86)
The final substitutions involve h = pμ = Nμ/D, e = S/D, and unit vectors W = D/D andQ = S/S. Applying these substitutions Eq. (3.86) becomes
v2= 1 r2
μ
NDD × r2 + μ
NDS (3.87)
Equation (3.87) is our final result: the ECI velocity vector of the middle observation as a function of vectorsr2,N, D, and S. Of course, N, D, and S are determined by the three ECI position vectorsr1,r2, andr3.
The Gibbs method fails if the three position vectors are not coplanar. We may use the following test:
ε =r1 r2×r3
r1 r2×r3
(3.88)
Scalar quantityε is the cosine of the angle between position vector r1and the cross-product resultr2×r3. Becauser2×r3 is along the angular momentum vectorh (which is orthogonal tor1), the scalarε should be exactly zero if all three vectors are coplanar.
However, measurement errors and/or sensor noise will lead to three position vectors that are not exactly coplanar. A reasonable test is to ensure that ε < cos88o(or ε < 0 0349) for coplanar position vectors.
The Gibbs method has numerical instability if the position vectors are too closely spaced.
Vallado [1; p. 459] recommends a 1 angular separation as a lower threshold. We can easily compute the angular separation using the dot product and inverse cosine function:
θ12= cos−1 r1 r2
r1r2
and θ23= cos−1 r2 r3
r2r3
(3.89) We may think of the separation angles θ12 and θ23 as the changes in true anomaly between two successive position vectors.
We may now summarize the Gibbs method as follows:
1) Computeε using Eq. (3.88) and the three position vectors r1,r2, andr3. If ε < 0 0349, then the three position vectors are essentially coplanar and the Gibbs method can be used.
2) Using Eq. (3.89), compute the angular separations between position vectors. If either θ12orθ23is less than 1 , the Gibbs method may produce inaccurate results.
3) Using the three position vectorsr1,r2, andr3, compute the auxiliary vectorsD, N, and S using Eqs. (3.71), (3.72), and (3.80).
4) Compute the ECI velocity vector of the middle observation,v2, using Eq. (3.87).
5) Use the techniques of Section 3.4 to compute the orbital elements from the state vec-tor (r2,v2).
The following example illustrates the Gibbs method for orbit determination.
Example 3.6 A ground station makes LOS measurements of a satellite at three obser-vation times and determines the following three ECI position vectors:
r1=
−11,052 902
−12,938 738 8, 505 244
km r2=
−10,378 257
−15,955 205 14, 212 351
km r3=
−9,336 222
−17,747 079 18, 337 068
km
The ground-station operators suspect that this satellite is in a Molniya orbit. Compute the satellite’s state vector for the middle observation and determine if the satellite is indeed in a Molniya orbit.
The Gibbs method relies on the magnitudes of each position vector and the three cross products. The three magnitudes are
r1= 19, 024 110 km, r2= 23, 754 320 km, and r3= 27, 173 000 km The three cross products are
r1×r2=
−48,186,974 358 68,818, 114 713 42,069, 769 035
km2
r2×r3=
−40,343,963 066 57,617, 140 253 35,222, 410 926
km2
r3×r1=
86,315, 281 358
−123,270,969 423
−75,357,794 605 km2
First, we must use Eq. (3.88) to check the coplanar condition:
ε =r1 r2×r3
r1 r2×r3
=−8 54 10−9
The magnitude of theε parameter is extremely small and hence the three position vec-tors are coplanar. The next step is to determine the angular spacing between the position vectors. Using Eq. (3.89), we obtain
θ12= cos−1 r1 r2
r1r2 = cos−1 442,029, 587 85
451,904, 790 21 = 12 00 θ23= cos−1 r2 r3
r2r3
= cos−1 640,664, 841 65
645,476, 124 86 = 7 00
The angular separations are greater than 1 and therefore the Gibbs method should provide sufficient accuracy. Next, we compute the auxiliary vectorsD, N, and S using Eqs. (3.71), (3.72), and (3.80):
D = r2×r3 + r3×r1 + r1×r2 =
−2,215,656 066 3,164, 285 542 1,934, 385 356
km2
N = r1 r2×r3 + r2 r3×r1 + r3 r1×r2 =
−26,531,841,547 5 37, 891,405, 572 8 23, 163,695, 418 9
km3
S = r2−r3 r1+ r3−r1 r2+ r1−r2 r3=
−2,622,648 754
−1,836,395 621
−3 881
km2
Finally, use Eq. (3.87) to compute the ECI velocity vector corresponding to the middle observation:
v2= 1 r2
μ
NDD × r2 + μ NDS =
0 7610
−1 8108 3 8337
km/s
Hence, the state vector for the middle observation is
r2=
−10,378 257
−15,955 205 14, 212 351
km, v2=
0 7610
−1 8108 3 8337
km/s
We can now compute the six orbital elements from this state vector using the methods of Section 3.4. However, the Gibbs method allows us to take a few short cuts. For exam-ple, parameter p can be computed from magnitudes N and D
p=N
D= 11, 974 710 km Eccentricity is the ratio of S and D; see Eq. (3.82)
e=S
D= 0 7411 Semimajor axis can be computed from p and e
a= p
1−e2= 26,565 km
Referring back to Section 3.4, we see that these values of semimajor axis a and eccen-tricity e match the values for a Molniya orbit. To complete the orbit-determination proc-ess, we can use Eq. (3.8) to compute inclination:
cosi =K h h
where the angular momentum ish = r2×v2. Carrying out the calculations, we find that cosi = 0.447759, or i = 63.4 which is the inclination of a Molniya orbit.
The last critical element is the argument of perigee as computed by Eq. (3.14) cosω = n e
n e
We must compute the ascending node vector using Eq. (3.11) n = K × h
and the eccentricity vector using Eq. (3.6) and state vector (r2,v2) e =1
μ v22−μ
r2 r2− r2 v2 v2
The reader can carry out these vector manipulations to find that
n =
−50,603 214
−35,432 759 0
and e =
−0 1903 0 2718
−0 6627
Hence, cosω = −1 45 10−6 ≈0. The argument of perigee is either +90 or –90 . Because theK component of the eccentricity vector e is negative, the perigee is south of the
equatorial plane and the argument of perigee isω = –90 . We have now determined that the satellite is indeed in a Molniya orbit (the reader should note that the longitude of the ascending nodeΩ is not needed to define a specific Molniya orbit unless we are given the epoch time t2and requirements for the orbit’s apogee to pass over a particular geographic longitude).