Question 4.11: Given the following state vector of a satellite in geocentri......
Given the following state vector of a satellite in geocentric equatorial coordinates,
\begin{aligned}& r =-3670 \hat{ I }-3870 \hat{J }+4400 \hat{ K } km \\& v =4.7 \hat{ I }-7.4 \hat{J }+1 \hat{ K } km / s\end{aligned}find the state vector after 4 days (96 h) of coasting flight, assuming that there are no perturbations other than the influence of the earth’s oblateness on \Omega \text { and } \omega \text {. }
Four days is a long enough time interval that we need to take into consideration not only the change in true anomaly but also the regression of the ascending node and the advance of perigee. First, we must determine the orbital elements at the initial time using Algorithm 4.2, which yields
\begin{aligned}h & =58,930 km ^2 / s \\i & =39.687^{\circ} \\e & =0.42607 \text { (The orbit is an ellipse) } \\\Omega_0 & =130.32^{\circ} \\\omega_0 & =42.373^{\circ} \\\theta_0 & =52.404^{\circ}\end{aligned}We use Eqn (2.71) to determine the semimajor axis,
a=\cfrac{h^2}{\mu} \cfrac{1}{1-e^2}=\cfrac{58,930^2}{398,600} \cfrac{1}{1-0.4261^2}=10,640 kmso that, according to Eqn (2.83), the period is
T=\cfrac{2 \pi}{\sqrt{\mu}} a^{\frac{3}{2}}=10,928 sFrom this we obtain the mean motion
n=\cfrac{2 \pi}{T}=0.00057495 rad / sThe initial value E_0 of eccentric anomaly is found from the true anomaly \theta_0 using Eqn (3.13a),
\tan \cfrac{E}{2}=\sqrt{\cfrac{1-e}{1+e}} \tan \cfrac{\theta}{2} (3.13a)
\tan \cfrac{E_0}{2}=\sqrt{\cfrac{1-e}{1+e}} \tan \cfrac{\theta_0}{2}=\sqrt{\cfrac{1-0.42607}{1+0.42607}} \tan \cfrac{52.404^{\circ}}{2} \Rightarrow E_0=0.60520 radWith E_0, we use Kepler’s equation to calculate the time t_0 since perigee at the initial epoch,
n t_0=E_0-e \sin E_0 \Rightarrow 0.00057495 t_0=0.60520-0.42607 \sin 0.60520 \Rightarrow t_1=631.00 sNow we advance the time to t_f, that of the final epoch, given as 96 h later. That is, \Delta t=345,600 s, so that
t_{ f }=t_1+\Delta t=631.00+345,600=346,230 sThe number of periods n_P since passing perigee in the first orbit is
n_{ P }=\cfrac{t_{ f }}{T}=\cfrac{346,230}{10,928}=31.682From this we see that the final epoch occurs in the 32nd orbit, whereas t_0 was in orbit 1. Time since passing perigee in the 32nd orbit, which we will denote t_{32}, is
t_{32}=(31.682-31) T \Rightarrow t_{32}=7455.7 sThe mean anomaly corresponding to that time in the 32nd orbit is
M_{32}=n t_{32}=0.00057495 \times 7455.7=4.2866 radKepler’s equation yields the eccentric anomaly:
\begin{aligned}& E_{32}-e \sin E_{32}=M_{32} \\& E_{32}-0.42607 \sin E_{32}=4.2866 \\& \therefore E_{32}=3.9721 rad \text { (Algorithm 3.1) }\end{aligned}The true anomaly follows in the usual way,
\tan \cfrac{\theta_{32}}{2}=\sqrt{\cfrac{1+e}{1-e}} \tan \cfrac{E_{32}}{2} \Rightarrow \theta_{32}=211.25^{\circ}At this point, we use the newly found true anomaly to calculate the state vector of the satellite in perifocal coordinates. Thus, from Eqn (4.43)
r =\bar{x} \hat{ p }+\bar{y} \hat{ q }=\cfrac{h^2}{\mu} \cfrac{1}{1+e \cos \theta}(\cos \theta \hat{ p }+\sin \theta \hat{ q }) (4.43)
r =r \cos \theta_{32} \hat{ p }+r \sin \theta_{32} \hat{ q }=-11,714 \hat{ p }-7108.8 \hat{ q }( km )or, in matrix notation,
\{ r \}_{\bar{x}}=\left\{\begin{array}{c}-11,714 \\-7108.8 \\0\end{array}\right\}( km )Likewise, from Eqn (4.43),
\text{v} =-\cfrac{\mu}{h} \sin \theta_{32} \hat{ p }+\cfrac{\mu}{h}\left(e+\cos \theta_{32}\right) \hat{ q }=3.5093 \hat{ p }-2.9007 \hat{ q }( km / s )or
\{ \text{v} \}_{\bar{x}}=\left\{\begin{array}{c}3.5093 \\-2.9007 \\0\end{array}\right\}( km / s )Before we can project r and v into the geocentric equatorial frame, we must update the right ascension of the node and the argument of perigee. The regression rate of the ascending node is
or
\dot{\Omega}=-2.2067 \times 10^{-5 } {}^{\circ}/ sTherefore, right ascension at epoch in the 32nd orbit is
\Omega_{32}=\Omega_0+\dot{\Omega} \Delta t=130.32+\left(-2.2067 \times 10^{-5}\right) \times 345,600=122.70^{\circ}Likewise, the perigee advance rate is
\dot{\omega}=-\left[\cfrac{3}{2} \cfrac{\sqrt{\mu} J_2 R^2}{\left(1-e^2\right)^2 a^{\frac{7}{2}}}\right]\left(\cfrac{5}{2} \sin ^2 i-2\right)=4.9072 \times 10^{-7} \text{rad / s} =2.8116 \times 10^{-5} {}^{ \circ} / swhich means the argument of perigee at epoch in the 32nd orbit is
\omega_{32}=\omega_0+\omega \Delta t=42.373+2.8116 \times 10^{-5} \times 345,600=52.090^{\circ}Substituting the updated values of Ω and ω, together with the inclination i, into Eqn (4.47) yields the updated transformation matrix from geocentric equatorial to the perifocal frame,
[ Q ]_{X \bar{\text{x}}}=\left[ R _3(\omega)\right]\left[ R _1(i)\right]\left[ R _3(\Omega)\right] (4.47)
or
[ Q ]_{X \bar{\text{x}}}=\left[\begin{array}{ccc}-0.84285 & 0.18910 & 0.50383 \\0.028276 & -0.91937 & 0.39237 \\0.53741 & 0.34495 & 0.76955\end{array}\right]For the inverse transformation, from perifocal to geocentric equatorial, we need the transpose of this matrix,
[ Q ]_{\bar{\text{x}} X}=\left[\begin{array}{ccc}-0.84285 & 0.18910 & 0.50383 \\0.028276 & -0.91937 & 0.39237 \\0.53741 & 0.34495 & 0.76955\end{array}\right]^T=\left[\begin{array}{ccc}-0.84285 & 0.028276 & 0.53741 \\0.18910 & -0.91937 & 0.34495 \\0.50383 & 0.39237 & 0.76955\end{array}\right]Thus, according to Eqn (4.51), the final state vector in the geocentric equatorial frame is
\{ r \}_X=[ Q ]_{\bar{\text{x}} X}\{ r \}_{\bar{\text{x}}} \quad\{ v \}_X=[ Q ]_{\bar{x} X}\{ v \}_{\bar{\text{x}}} (4.51)
or, in vector notation,
The two orbits are plotted in Figure 4.21.
