Question 4.12: An earth satellite has the following orbital parameters: rp ......
An earth satellite has the following orbital parameters:
\begin{array}{ll}r_{ p }=6700 km & \text { Perigee } \\r_{ a }=10 000 km & \text { Apogee } \\\theta_0=230^{\circ} & \text { True anomaly } \\\Omega_0=270^{\circ} & \text { Right ascension of the ascending node } \\i_0=60^{\circ} & \text { Inclination } \\\omega_0=45^{\circ} & \text { Argument of perigee }\end{array}
Calculate the right ascension (longitude east of x^{\prime}) and declination (latitude) relative to the rotating earth 45 min later.
Step-by-Step
First, we compute the semimajor axis a, eccentricity e, the angular momentum h, the semimajor axis a, and the period T. For the semimajor axis, we recall that
a=\cfrac{r_{ p }+r_{ a }}{2}=\cfrac{6700+10,000}{2}=8350 kmFrom Eqn (2.84) we get
e=\cfrac{r_{ a }-r_{ p }}{r_{ a }+r_{ p }}=\cfrac{10,000-6700}{10,000+6700}=0.19760Equation (2.50) yields
r_p=\cfrac{h^2}{\mu} \cfrac{1}{1+e} (2.50)
h=\sqrt{\mu r_{ p }(1+e)}=\sqrt{398,600 \times 6700 \times(1+0.19760)}=56,554 km ^2 / sFinally, we obtain the period from Eqn (2.83),
T=\cfrac{2 \pi}{\sqrt{\mu}} a^{3 / 2}=\cfrac{2 \pi}{\sqrt{398,600}} 8350^{3 / 2}=7593.5 sNow we can proceed with Algorithm 4.6.
Step 1:
\begin{aligned}& \dot{\Omega}=-\left[\frac{3}{2} \frac{\sqrt{\mu} J_2 R_{\text {earth }}^2}{\left(1-e^2\right) a^{7 / 2}}\right] \cos i=-\left[\frac{3}{2} \frac{\sqrt{398,600} \times 0.0010836 \times 6378^2}{\left(1-0.19760^2\right) 8350^{7 / 2}}\right] \cos 60^{\circ}=-2.3394 \times 10^{-7} {}^{\circ} / s \\\\& \dot{\omega}=\dot{\Omega} \frac{5 / 2 \sin ^2 i-2}{\cos i}=-2.3394 \times 10^{-5}\left(\frac{5 / 2 \sin ^2 60^{\circ}-2}{\cos 60^{\circ}}\right)=5.8484 \times 10^{-6} {}^{\circ} / s\end{aligned}
Step 2:
a.
b.
M=E-e \sin E=-2.1059-0.19760 \sin (-2.1059)=-1.9360 radc.
t_0=\cfrac{M}{2 \pi} T=\cfrac{-1.9360}{2 \pi} \cdot 7593.5=-2339.7 s \quad(2339.7 s \text { until perigee })Step 3: t=t_0+45 \text{min} =-2339.7+45 \times 60=360.33 s \quad(360.33 s \text { after perigee })
a.
\begin{aligned}M & =2 \pi \cfrac{t}{T}=2 \pi \cfrac{360.33}{7593.5}=0.29815 \text{rad} \\\\E-0.19760 \sin E & =0.29815 \overbrace{\Rightarrow}^{\text {Algorithm 3.1 }} E=0.36952 rad \\\\\theta & =2 \tan ^{-1}\left(\tan \cfrac{E}{2} \sqrt{\cfrac{1+e}{1-e}}\right)=2 \tan ^{-1}\left(\tan \cfrac{0.36952}{2} \sqrt{\cfrac{1+0.19760}{1-0.19760}}\right)=25.723^{\circ}\end{aligned}
b.
\begin{aligned}& \Omega=\Omega_0+\dot{\Omega} \Delta t=270^{\circ}+\left(-2.3394 \times 10^{-5 \%} / s \right)(2700 s )=269.94 \\& \omega=\omega_0+\dot{\omega} \Delta t=45^{\circ}+\left(5.8484 \times 10^{-6} / s \right)(2700 s )=45.016^{\circ}\end{aligned}c.
\{ r \}_X \stackrel{\text { Algorithm } 4.5}{\widehat{=}}\left\{\begin{array}{c}3212.6 \\-2250.5 \\5568.6\end{array}\right\}( km )d.
\begin{aligned}&\theta=\omega_E \Delta t=\frac{360^{\circ}\left(1+\frac{1}{365.26}\right)}{24 \times 3600 s } \times 2700 s =11.281^{\circ}\\\\&\begin{aligned}& {\left[ R _3(\theta)\right]=\left[\begin{array}{ccc}\cos 11.281^{\circ} & \sin 11.281^{\circ} & 0 \\-\sin 11.281^{\circ} & \cos 11.281^{\circ} & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0.98068 & 0.19562 & 0 \\-0.19562 & 0.98068 & 0 \\0 & 0 & 1\end{array}\right]} \\\\& \{ r \}_{X^{\prime}}=\left[ R _3(\theta)\right]\{ r \}_X=\left[\begin{array}{ccc}0.98068 & 0.19562 & 0 \\-0.19562 & 0.98068 & 0 \\0 & 0 & 1\end{array}\right]\left\{\begin{array}{c}3212.6 \\-2250.5 \\5568.6\end{array}\right\}=\left\{\begin{array}{c}2710.3 \\-2835.4 \\5568.6\end{array}\right\}( km )\end{aligned}\end{aligned}
e.
r =2710.3 I ^{\prime}-2835.4 \hat{\jmath}^{\prime}+5568.6 \hat{ K }^{\prime}\overbrace{\Rightarrow }^{\text{Algorithm 4:1}} \boxed{\alpha=313.7^{\circ} \quad \delta=54.84^{\circ}}Related Answered Questions
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