For a given earth orbit, the elements are h=80 000 km²/s, e = 1.4, i = 30°, Ω = 40°, ω = 60° and θ = 30°. Using Algorithm 4.2 find the state vectors r and v in the geocentric equatorial frame.

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Accepted Answer

Step 1:

[latex]\{ r \}_{\bar{x}}=\frac{h^{2}}{\mu} \frac{1}{1+e \cos heta}\left\{\begin{array}{c}\cos heta \\sin heta \0\end{array} ight\}=\frac{80 000^{2}}{398 600} \frac{1}{1+1.4 \cos 30^{\circ}}\left\{\begin{array}{c}\cos 30^{\circ} \\sin 30^{\circ} \0\end{array} ight\}=\left\{\begin{array}{c}6285.0 \3628.6 \0\end{array} ight\} km[/latex]

Step 2:

[latex]\{ v \}_{\bar{x}}=\frac{\mu}{h}\left\{\begin{array}{c}-\sin heta \e+\cos heta \0\end{array} ight\}=\frac{398 600}{80 000}\left\{\begin{array}{c}-\sin 30^{\circ} \1.4+\cos 30^{\circ} \0\end{array} ight\}=\left\{\begin{array}{c}-2.4913 \11.290 \0\end{array} ight\} km / s[/latex]

Step 3:

[latex][ Q ]_{X \bar{x}}=\left[\begin{array}{ccc}\cos \omega & \sin \omega & 0 \-\sin \omega & \cos \omega & 0 \0 & 0 & 1\end{array} ight]\left[\begin{array}{ccc}1 & 0 & 0 \0 & \cos i & \sin i \0 & -\sin i & \cos i\end{array} ight]\left[\begin{array}{ccc}\cos \Omega & \sin \Omega & 0 \-\sin \Omega & \cos \Omega & 0 \0 & 0 & 1\end{array} ight][/latex]
[latex]=\left[\begin{array}{ccc}\cos 60^{\circ} & \sin 60^{\circ} & 0 \-\sin 60^{\circ} & \cos 60^{\circ} & 0 \0 & 0 & 1\end{array} ight]\left[\begin{array}{ccc}1 & 0 & 0 \0 & \cos 30^{\circ} & \sin 30^{\circ} \0 & -\sin 30^{\circ} & \cos 30^{\circ}\end{array} ight]\left[\begin{array}{ccc}\cos 40^{\circ} & \sin 40^{\circ} & 0 \-\sin 40^{\circ} & \cos 40^{\circ} & 0 \0 & 0 & 1\end{array} ight][/latex]
[latex]=\left[\begin{array}{lrl}-0.099068 & 0.89593 & 0.43301 \-0.94175 & -0.22496 & 0.25 \0.32139 & -0.38302 & 0.86603\end{array} ight][/latex]

This is the transformation matrix for [latex]X Y Z ightarrow \bar{x} \bar{y} \bar{z}[/latex]. The transformation matrix for [latex]\bar{x} \bar{y} \bar{z} ightarrow X Y Z[/latex] is the transpose,

[latex][ Q ]_{\bar{x} X}=\left[\begin{array}{ccr}-0.099068 & -0.94175 & 0.32139 \0.89593 & -0.22496 & -0.38302 \0.43301 & 0.25 & 0.86603\end{array} ight][/latex]

Step 4:
The geocentric equatorial position vector is
[latex]\{ r \}_{X}=[ Q ]_{\bar{x} X}\{ r \}_{\bar{x}}[/latex]

[latex]=\left[\begin{array}{ccc}-0.099068 & -0.94175 & 0.32139 \0.89593 & -0.22496 & -0.38302 \0.43301 & 0.25 & 0.86603\end{array} ight]\left\{\begin{array}{c}6285.0 \3628.6 \0\end{array} ight\}=\underline{\left\{\begin{array}{r}-4040 \4815 \3629\end{array} ight\}( km )} [/latex] (a)

whereas the geocentric equatorial velocity vector is

[latex]\{ v \}_{X}=[ Q ]_{\bar{x} X}\{ v \}_{\bar{x}}[/latex]

[latex]=\left[\begin{array}{ccc}-0.099068 & -0.94175 & 0.32139 \0.89593 & -0.22496 & -0.38302 \0.43301 & 0.25 & 0.86603\end{array} ight]\left\{\begin{array}{c}-2.4913 \11.290 \0\end{array} ight\}=\underline{\left\{\begin{array}{c}-10.39 \-4.772 \1.744\end{array} ight\}( km / s )} [/latex]

The state vectors r and v are shown in Figure 4.16. By holding all of the orbital parameters except the true anomaly fixed and allowing θ to take on a range of values, we generate a sequence of position vectors [latex]r _{\bar{x}}[/latex] from Equations 4.37. Each of these is projected into the geocentric equatorial frame as in (a), using repeatedly the same transformation matrix [latex][ Q ]_{\bar{x} X}[/latex]. By connecting the end points of all of the position vectors [latex]r _{X}[/latex], we trace out the trajectory illustrated in Figure 4.16.

[latex]\{ r \}_{\bar{x}}=\frac{h^{2}}{\mu} \frac{1}{1+e \cos heta}\left\{\begin{array}{c}\cos heta \\sin heta \0\end{array} ight\}[/latex] (4.37)

Question 4.5: For a given earth orbit, the elements are h=80 000 km²/s, e ...

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