A tracking station is located at Φ = 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

Question

A tracking station is located at [latex]\phi = 40°[/latex] 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 [latex] µ = 398 600 km^{3}/s^{2}.[/latex]

Accepted Answer

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 [latex]R_{e}[/latex] = 6378 km and the flattening factor is f = 0.003353, we substitute [latex]\phi = 40°[/latex], 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:

[latex]R=\left[\frac{R_{e}}{\sqrt{1-(2f-f^{2})\sin^{2} \phi } } +H ight] \cos \phi (\cos heta \hat{I}+\sin heta \hat{J} )[/latex]

[latex]+\left[\frac{R_{e}}{\sqrt{1-(2f-f^{2})\sin^{2} \phi } } +H ight]\sin \phi \hat{K} [/latex] (5.56)
[latex]R_{1} = 3489.8 \hat{ I}+ 3430.2\hat{J}+ 4078.5\hat{K} (km)[/latex]
[latex]R_{2} = 3460.1\hat{ I} + 3460.1\hat{J} + 4078.5\hat{K} (km)[/latex]
[latex]R_{3} = 3429.9\hat{ I}+ 3490.1\hat{J} + 4078.5\hat{K} (km)[/latex]
Using Equation 5.57 we compute the direction cosine vectors at each of the three observation times from the right ascension and declination data:

[latex]\hat{\varrho }=\cos \delta \cos \alpha \hat{I} +\cos \delta \sin \alpha \hat{J}+\sin \delta \hat{K} [/latex] (5.57)
[latex]\hat{\varrho}_{1} = \cos(−8.7833°) \cos 43.537° \hat{ I} + \cos(−8.7833°) \sin 43.537° \hat{J} + \sin(−8.7833°)\hat{K}[/latex]
[latex]= 0.71643 \hat{ I}+ 0.68074\hat{J} − 0.15270 \hat{K}[/latex]
[latex]\hat{\varrho} _{2} = \cos(−12.074°) \cos 54.420° \hat{ I} + \cos(−12.074° ) \sin 54.420° \hat{J} + \sin(−12.074° )\hat{K}[/latex]
[latex]= 0.56897 + 0.79531\hat{J} − 0.20917 \hat{K}[/latex]
[latex]\hat{\varrho}_{ 3} = \cos(−15.105°) \cos 64.318° + \cos(−15.105°) \sin 64.318° \hat{J} + \sin(−15.105°)\hat{K}[/latex]
[latex]= 0.41841\hat{ I} + 0.87007\hat{J} − 0.26059\hat{K}[/latex]
We can now proceed with Algorithm 5.5.
Step 1:
[latex] au _{1} = 0 − 118.10 = −118.10 s[/latex]
[latex] au _{3} = 237.58 − 118.10 = 119.47 s[/latex]
[latex] au = 119.47 − (−118.1) = 237.58 s[/latex]
Step 2:
[latex]p_{1} = \hat{\varrho} _{2} ×\hat{\varrho} _{3} = −0.025258\hat{ I} + 0.060753\hat{ J} + 0.16229\hat{K}[/latex]
[latex]p_{2} = \hat{\varrho}_{1} × \hat{\varrho}_{ 3} = −0.044538\hat{ I} + 0.12281\hat{ J} + 0.33853\hat{K}[/latex]
[latex]p_{3} = \hat{\varrho}_{1} × \hat{\varrho}_{2} = −0.020950 \hat{ I} + 0.062977\hat{ J} + 0.18246\hat{K}[/latex]
Step 3:
[latex]D_{0} = \hat{\varrho}_{1} · p_{1} = −0.0015198[/latex]
Step 4:
[latex]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[/latex]
[latex]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[/latex]
[latex]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[/latex]
Step 5:

[latex]A=\frac{1}{−0.0015198}\left[−1646.5\frac{119.47}{237.58} + 1651.5 + 1656.6\frac{(−118.10)}{237.58} ight]= −6.6858 km[/latex]

[latex]B=\frac{1}{6(−0.0015198)}\left\{1646.5(119.47^{2}− 237.58^{2})\frac{119.47}{237.58} ight.[/latex]

[latex]+\left.1656.6[237.58^{2}− (−118.10)^{2}]\frac{(−118.10)}{237.58} ight\}[/latex]

[latex]= 7.6667 × 10^{9} km ·s^{2}[/latex]

Step 6:
[latex]E = R_{2} · \hat{\varrho}_{2} = 3875.8 km[/latex]

[latex]R_{2}^{2} = R_{2} · R_{2} = 4.058 × 10^{7} km^{2}[/latex]
Step 7:
[latex]a = −[(−6.6858)^{2} + 2(−6.6858)(3875.8) + 4.058 × 10^{7}] = −4.0528 × 10^{7} km^{2}[/latex]
[latex]b = −2(389 600)(7.6667 × 10^{9})(−6.6858 + 3875.8) = −2.3597 × 10^{19} km^{5}[/latex]
[latex]c = −(398 600)^{2}(7.6667 × 10^{9})^{2} = −9.3387 × 10^{30} km^{8}[/latex]
Step 8:
[latex]F(x) = x^{8} − 4.0528 × 10^{7}x^{6} − 2.3597 × 10^{19}x^{3} − 9.3387 × 10^{30} = 0[/latex]
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

[latex]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}} [/latex] (5.119)

[latex]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}}[/latex]

Stepping through Newton’s iterative procedure yields
[latex]x_{0} = 9000[/latex]
[latex]x_{1} = 9000 − (−276.93) = 9276.9[/latex]
[latex]x_{2} = 9276.9 − 34.526 = 9242.4[/latex]
[latex]x_{3} = 9242.4 − 0.63428 = 9241.8[/latex]
[latex]x_{4} = 9241.8 − 0.00021048 = 9241.8[/latex]
Thus, after four steps we converge to
[latex]r_{2} = 9241.8 km[/latex]
The other roots are either negative or complex and are therefore physically unacceptable.
Step 9:

[latex]\varrho _{1}=\frac{1}{-0.0015198}[/latex]

[latex] imes \left\{\frac{ \begin{matrix} 6[787.31\frac{(-118.10)}{119.47}+784.72\frac{237.58}{119.47} ]9241.8^{3} \ +398600· 787.31[237.58^{2}− (−118.10)^{2}]\frac{−118.10}{119.47} \end{matrix} }{6 · 9241.8^{3}+398600(237.58^{2}-119.47^{2})^{2}}− 782.15 ight\} [/latex]

= 3639.1 km

[latex]\varrho _{2}=−6.6858 +\frac{398 600 · 7.6667 × 10^{9} }{9241.8^{3}}= 3864.8 km[/latex]

[latex]\varrho _{3}=\frac{1}{-0.0015198} imes \left[ \frac{\begin{matrix} 6(887.10\frac{119.47}{−118.10}− 889.60 \frac{237.58}{−118.10})9241.8^{3} \ + 398 600 · 887.10(237.58^{2} − 119.47^{2})\frac{ 119.47}{−118.10} \end{matrix} }{6 · 9241.8^{3} + 398 600(237.58^{2} − 119.47^{2})}− 892.13 ight][/latex]

=4156.9 km

Step 10:
[latex]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 } )[/latex]
[latex]= 6096.9\hat{ I} + 5907.5 \hat{ J} + 3522.9\hat{K } (km)[/latex]
[latex]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 } )[/latex]
[latex]= 5659.1 \hat{ I} + 6533.8\hat{ J} + 3270.1\hat{K } (km)[/latex]
[latex]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 } )[/latex]
[latex]= 5169.1 \hat{ I} + 7107.0 \hat{ J}+ 2995.3 \hat{K } (km)[/latex]
Step 11:

[latex]f_{1}\approx 1-\frac{1}{2}\frac{398 600}{9241.8^{3}}(-118.10)^{2}= 0.99648[/latex]

[latex]f_{3}\approx 1-\frac{1}{2}\frac{398 600}{9241.8^{3}}(119.47)^{2}=0.99640[/latex]

[latex]g_{1}\approx−118.10 −\frac{1}{2}\frac{398 600}{9241.8^{3}}(-118.10)^{3}= −117.97[/latex]

[latex]g_{3}\approx119.47−\frac{1}{2}\frac{398 600}{9241.8^{3}}(119.47)^{3}= 119.33[/latex]

Step 12:

[latex]V_{2}=\frac{ \begin{matrix} −0.99640(6096.9\hat{ I} + 5907.5\hat{ J} + 3522.9\hat{K })+0.99648(5169.1 \hat{ I} +7107.0\hat{ J} \ +2995.3\hat{K }) \end{matrix} }{0.99648 · 119.33 − 0.99640(−117.97)}[/latex]

[latex]= −3.9080\hat{ I} + 5.0573 \hat{ J} − 2.2222\hat{K } (km/s)[/latex]

In summary, the state vector at time t2 is, approximately,
[latex]\underline{r_{2} = 5659.1\hat{I } + 6533.8\hat{J } + 3270.1\hat{K } (km)}[/latex]
[latex]\underline{v_{2} = −3.9080\hat{ I} + 5.0573\hat{J } − 2.2222\hat { K} (km/s)}[/latex]

Question 5.11: A tracking station is located at Φ = 40° north latitude at a...

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