Question 4.7: For a given earth orbit, the elements are h = 80,000 km²/s, ......
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 4.5, find the state vectors r and v in the geocentric equatorial frame.
Step 1:
\{{\bf r}\}_{\bar{x}}=\frac{h^{2}}{\mu}\frac{1}{1+e\cos\theta}\left\{\begin{array}{c}{{\cos\theta}}\\ {{\sin\theta}}\\ {{0}}\end{array}\right\} = \frac{80,000^{2}}{398,600} \frac{1}{1 + 1.4 \cos 30°} \left\{\begin{array}{c}{{\cos30°}}\\ {{\sin30°}}\\ {{0}}\end{array}\right\} =\left\{\begin{array}{c}{{6285.0}}\\ {{3628.6}}\\ {{0}}\end{array}\right\} km
Step 2:
\{{\bf v}\}_{\bar{x}}=\frac{\mu}{h} \left\{\begin{array}{c}{{ – \sin\theta}}\\ {{ e + \cos\theta}}\\ {{0}}\end{array}\right\} = \frac{398,600}{80,000} \left\{\begin{array}{c}{{ – \sin 30°}}\\ {{ 1.4 + \cos 30°}}\\ {{0}}\end{array}\right\} =\left\{\begin{array}{c}{{- 2.4913 }}\\ {{11.290}}\\ {{0}}\end{array}\right\} km / s
Step 3:
[ Q] _{X \bar{x}}= \begin{bmatrix} cos ω & sin ω & 0 \\ -sin ω & cos ω & 0 \\ 0 & 0& 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos i & sin i \\ 0 & -sin i & cos i \end{bmatrix} \begin{bmatrix} cos Ω & sin Ω & 0 \\ -sin Ω & cos Ω & 0 \\ 0 & 0& 1 \end{bmatrix}
= \begin{bmatrix} cos 60° & sin 60° & 0 \\ -sin 60° & cos 60° & 0 \\ 0 & 0& 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos 30° & sin 30° \\ 0 & -sin 30° & cos 30° \end{bmatrix} \begin{bmatrix} cos 40° & sin 40° & 0 \\ -sin 40° & cos 40° & 0 \\ 0 & 0& 1 \end{bmatrix}
= \begin{bmatrix} -0.099068 & 0.89593 & 0.43301 \\ – 0.94175 & – 0.22496 & 0.25 \\ 0.32139 & – 0.38302 & 0.86603 \end{bmatrix}
This is the direct cosine matrix for XYZ → \bar{x} \bar{y} \bar{z} . The transformation matrix for \bar{x} \bar{y} \bar{z} → XYZ is the transpose ,
[ Q]_{\bar{x} X} = \begin{bmatrix} – 0.099068 & -0.94175 & 0.32139 \\ 0.89593 & – 0.22496 & -0.38302 \\ 0.43301 & 0.25 & 0.86603 \end{bmatrix}
Step 4:
The geocentric equatorial position vector is
\{{\bf{r}}\}_{X}=[Q]_{\bar{x} X} \{{\bf{r}}\}_{\overline{{{x}}}}= \begin{bmatrix}- 0.099068 & -0.94175 & 0.32139 \\ 0.89593 & – 0.22496 & -0.38302 \\ 0.43301 & 0.25 & 0.86603 \end{bmatrix} \left\{{\begin{array}{c}{6285.0}\\ {3628.6}\\ {0}\end{array}}\right\} = \left\{{\begin{array}{c}{- 4040}\\ {4815}\\ {3629}\end{array}}\right\} (km) (a)
whereas the geocentric equatorial velocity vector is
\{{\bf{v}}\}_{X}=[Q]_{\bar{x} X} \{{\bf{v}}\}_{\overline{{{x}}}}= \begin{bmatrix}- 0.099068 & -0.94175 & 0.32139 \\ 0.89593 & – 0.22496 & -0.38302 \\ 0.43301 & 0.25 & 0.86603 \end{bmatrix} \left\{{\begin{array}{c}{- 2.4913}\\ {11.290}\\ {0}\end{array}}\right\} = \left\{{\begin{array}{c}{- 10.39}\\ {-4.772}\\ {1.744}\end{array}}\right\} (km/s)
The state vectors r and v are shown in Figure 4.17. By holding all 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 r_{\bar{x}} from Eqn (4.37). Each of these is projected into the geocentric equatorial frame as in Eqn (a), using repeatedly the same transformation matrix [Q]_{ \bar{x} X} . By connecting the end points of all the position vectors r_{X} , we trace out the trajectory illustrated in Figure 4.17.
[\mathbf{Q}]=[\mathbf{R}_{3}(\gamma)][\mathbf{R}_{1}(\beta)][\mathbf{R}_{3}(\alpha)]\quad(0\leq\alpha\lt 360^{\circ}\quad0\leq\beta\leq180^{\circ}\quad0\leq\gamma\lt 360^{\circ}) (4.37)
