Question 5.11: A tracking station is located at Φ = 40° north latitude at a...
A tracking station is located at \phi = 40° north latitude at an altitude of H = 1 km. Three observations of an earth satellite yield the values for the topocentric right ascension and declination listed in the following table, which also shows the local sidereal time θ of the observation site.
Use the Gauss Algorithm 5.5 to estimate the state vector at the second observation time. Recall that µ = 398 600 km^{3}/s^{2}.
Data for Example 5.11
| Observation | Time
(seconds) |
Right ascension,α
(degrees) |
Declination,δ
(degrees) |
Local sidereal time,θ (degrees) |
| 1 | 0 | 43.537 | −8.7833 | 44.506 |
| 2 | 118.10 | 54.420 | −12.074 | 45.000 |
| 3 | 237.58 | 64.318 | −15.105 | 45.499 |
Recalling that the equatorial radius of the earth is R_{e} = 6378 km and the flattening factor is f = 0.003353, we substitute \phi = 40°, H = 1 km and the given values of θ into Equation 5.56 to obtain the inertial position vector of the tracking station at each of the three observation times:
R=\left[\frac{R_{e}}{\sqrt{1-(2f-f^{2})\sin^{2} \phi } } +H\right] \cos \phi (\cos \theta \hat{I}+\sin \theta \hat{J} )+\left[\frac{R_{e}}{\sqrt{1-(2f-f^{2})\sin^{2} \phi } } +H\right]\sin \phi \hat{K} (5.56)
R_{1} = 3489.8 \hat{ I}+ 3430.2\hat{J}+ 4078.5\hat{K} (km)
R_{2} = 3460.1\hat{ I} + 3460.1\hat{J} + 4078.5\hat{K} (km)
R_{3} = 3429.9\hat{ I}+ 3490.1\hat{J} + 4078.5\hat{K} (km)
Using Equation 5.57 we compute the direction cosine vectors at each of the three observation times from the right ascension and declination data:
\hat{\varrho }=\cos \delta \cos \alpha \hat{I} +\cos \delta \sin \alpha \hat{J}+\sin \delta \hat{K} (5.57)
\hat{\varrho}_{1} = \cos(−8.7833°) \cos 43.537° \hat{ I} + \cos(−8.7833°) \sin 43.537° \hat{J} + \sin(−8.7833°)\hat{K}
= 0.71643 \hat{ I}+ 0.68074\hat{J} − 0.15270 \hat{K}
\hat{\varrho} _{2} = \cos(−12.074°) \cos 54.420° \hat{ I} + \cos(−12.074° ) \sin 54.420° \hat{J} + \sin(−12.074° )\hat{K}
= 0.56897 + 0.79531\hat{J} − 0.20917 \hat{K}
\hat{\varrho}_{ 3} = \cos(−15.105°) \cos 64.318° + \cos(−15.105°) \sin 64.318° \hat{J} + \sin(−15.105°)\hat{K}
= 0.41841\hat{ I} + 0.87007\hat{J} − 0.26059\hat{K}
We can now proceed with Algorithm 5.5.
Step 1:
\tau _{1} = 0 − 118.10 = −118.10 s
\tau _{3} = 237.58 − 118.10 = 119.47 s
\tau = 119.47 − (−118.1) = 237.58 s
Step 2:
p_{1} = \hat{\varrho} _{2} ×\hat{\varrho} _{3} = −0.025258\hat{ I} + 0.060753\hat{ J} + 0.16229\hat{K}
p_{2} = \hat{\varrho}_{1} × \hat{\varrho}_{ 3} = −0.044538\hat{ I} + 0.12281\hat{ J} + 0.33853\hat{K}
p_{3} = \hat{\varrho}_{1} × \hat{\varrho}_{2} = −0.020950 \hat{ I} + 0.062977\hat{ J} + 0.18246\hat{K}
Step 3:
D_{0} = \hat{\varrho}_{1} · p_{1} = −0.0015198
Step 4:
D_{11}= R_{1} · p_{1} = 782.15 km D_{12} = R_{1} · p_{2}= 1646.5 km D_{13} = R_{1} · p_{3} = 887.10 km
D_{21}= R_{2}· p_{1} = 784.72 km D_{22} = R_{2} · p_{2} = 1651.5 km D_{23} = R_{2} · p3 = 889.60 km
D_{31}= R_{3} · p_{1} = 787.31 km D_{32} = R_{3} p_{2} = 1656.6 km D_{33} = R_{3} · p_{3} = 892.13 km
Step 5:
+\left.1656.6[237.58^{2}− (−118.10)^{2}]\frac{(−118.10)}{237.58} \right\}
= 7.6667 × 10^{9} km ·s^{2}
Step 6:
E = R_{2} · \hat{\varrho}_{2} = 3875.8 km
R_{2}^{2} = R_{2} · R_{2} = 4.058 × 10^{7} km^{2}
Step 7:
a = −[(−6.6858)^{2} + 2(−6.6858)(3875.8) + 4.058 × 10^{7}] = −4.0528 × 10^{7} km^{2}
b = −2(389 600)(7.6667 × 10^{9})(−6.6858 + 3875.8) = −2.3597 × 10^{19} km^{5}
c = −(398 600)^{2}(7.6667 × 10^{9})^{2} = −9.3387 × 10^{30} km^{8}
Step 8:
F(x) = x^{8} − 4.0528 × 10^{7}x^{6} − 2.3597 × 10^{19}x^{3} − 9.3387 × 10^{30} = 0
The graph of F(x) in Figure 5.15 shows that it changes sign near x = 9000 km. Let us use that as the starting value in Newton’s method for finding the roots of F(x). For the case at hand, Equation 5.119 is
x_{i+1}=x_{i}-\frac{x_{i}^{8}+ax_{i}^{6}bx_{i}^{3}+c}{8x_{i}^{7}+6ax_{i}^{5}+3bx_{i}^{2}} (5.119)
x_{i+1}=x_{i}-\frac{x_{i}^{8}-4.0528 × 10^{7}x_{i}^{6}− 2.3622 × 10^{19}x_{i}^{3}-9.3186 × 10^{30}}{x_{i}^{7}− 2.4317 × 10^{8}x_{i}^{5}− 7.0866 × 10^{19}x_{i}^{2}}Stepping through Newton’s iterative procedure yields
x_{0} = 9000
x_{1} = 9000 − (−276.93) = 9276.9
x_{2} = 9276.9 − 34.526 = 9242.4
x_{3} = 9242.4 − 0.63428 = 9241.8
x_{4} = 9241.8 − 0.00021048 = 9241.8
Thus, after four steps we converge to
r_{2} = 9241.8 km
The other roots are either negative or complex and are therefore physically unacceptable.
Step 9:
= 3639.1 km
\varrho _{2}=−6.6858 +\frac{398 600 · 7.6667 × 10^{9} }{9241.8^{3}}= 3864.8 km
=4156.9 km
Step 10:
r_{1} = (3489.8\hat{ I} + 3430.2\hat{ J} + 4078.5 \hat{K } ) + 3639.1(0.71643 \hat{ I} + 0.68074\hat{ J} − 0.15270\hat{K } )
= 6096.9\hat{ I} + 5907.5 \hat{ J} + 3522.9\hat{K } (km)
r_{2} = (3460.1\hat{ I} + 3460.1\hat{ J} + 4078.5 \hat{K } ) + 3864.8(0.56897 \hat{ I} + 0.79531\hat{ J} − 0.20917 \hat{K } )
= 5659.1 \hat{ I} + 6533.8\hat{ J} + 3270.1\hat{K } (km)
r_{3} = (3429.9\hat{ I} + 3490.1\hat{ J} + 4078.5\hat{K } ) + 4156.9(0.41841 \hat{ I} + 0.87007\hat{ J} − 0.26059 \hat{K } )
= 5169.1 \hat{ I} + 7107.0 \hat{ J}+ 2995.3 \hat{K } (km)
Step 11:
f_{3}\approx 1-\frac{1}{2}\frac{398 600}{9241.8^{3}}(119.47)^{2}=0.99640
g_{1}\approx−118.10 −\frac{1}{2}\frac{398 600}{9241.8^{3}}(-118.10)^{3}= −117.97
g_{3}\approx119.47−\frac{1}{2}\frac{398 600}{9241.8^{3}}(119.47)^{3}= 119.33
Step 12:
= −3.9080\hat{ I} + 5.0573 \hat{ J} − 2.2222\hat{K } (km/s)
In summary, the state vector at time t2 is, approximately,
\underline{r_{2} = 5659.1\hat{I } + 6533.8\hat{J } + 3270.1\hat{K } (km)}
\underline{v_{2} = −3.9080\hat{ I} + 5.0573\hat{J } − 2.2222\hat { K} (km/s)}
