We use the given data in Algorithm 4.1 to obtain the state vectors of the two spacecraft in the geocentric equatorial frame

Space station:

r_{0}= 1622.39 \hat{I}+ 5305.10\hat{J} + 3717.44\hat{K} (km)

 

v_{0}= −7.29977\hat{I}+ 0.492357\hat{J} + 2.48318\hat{K} (km/s)

Spacecraft:

r= 1612.75\hat{I}+ 5310.19 \hat{J} + 3750.33\hat{K} (km)

 

v= −7.35321\hat{I}+ 0.463856 \hat{J} + 2.46920\hat{K} (km/s)

The space station reference frame unit vectors (at this instant) are, by definition:

\hat{i}=\frac{r_{0}}{\left\|r_{0}\right\| } = 0.242945\hat{I}+ 0.794415 \hat{J} + 0.556670\hat{K}

 

\hat{j}=\frac{v_{0}}{\left\|v_{0}\right\| } = −0.944799\hat{I}+ 0.063725 \hat{J} + 0.321394\hat{K}

 

\hat{k}=\hat{i}\times \hat{j} = 0.219846\hat{I}− 0.604023 \hat{J} + 0.766044\hat{K}

Therefore, the transformation matrix from the geocentric equatorial frame into space station frame is (at this instant)

[Q]_{Xx}=\left[\begin{matrix} 0.242945& 0.794415& 0.556670 \\ −0.944799 & 0.063725 & 0.321394 \\0.219846 & −0.604023 & 0.766044 \end{matrix} \right]

The position vector of the spacecraft relative to the space station (in the geocentric equatorial frame) is

\delta r=r-r_{0}= −9.63980\hat{I}+5.08240 \hat{J} + 32.8821\hat{K}(km)

The relative velocity is given by the formula (Equation 1.38)

v=v_{O}+\Omega \times r_{rel}+v_{rel}       (1.38)

\delta v=v-v_{0}-\Omega_{\text{ space station}}\times \delta r

where \Omega_{\text{ space station}}=n\hat{k} and n, the mean motion of the space station, is

n=\frac{v_{0}}{r_{0}}=\frac{7.72627}{6678}=0.00115697   rad/s         (a)

Thus

\delta v= −7.35321\hat{I}+0.463856 \hat{J} + 2.46920\hat{K}-(-7.29977\hat{I}+0.492357 \hat{J}+2.48318 \hat{K} )

so that

In space station coordinates, the relative position vector δr_{0} at the beginning of the rendezvous maneuver is

Likewise, the relative velocity δ{v_{0}}^{-} just before launch into the rendezvous trajectory is

 

=\left\{\begin{matrix} −0.02000 \\ 0.02000\\−0.005000 \end{matrix} \right\} (km/s)

The Clohessy–Wiltshire matrices, for t = t_{f} = 8 hr = 28800 s and n = 0.00115697 rad/s [from (a)], are

[\Phi _{rr}]=\left[\begin{matrix} 4−3\cos nt& 0 &0 \\ 6( \sin nt − nt)&1 &0 \\ 0& 0& \cos nt \end{matrix} \right] =\left[\begin{matrix} 4.98383& 0 &0 \\ −194.257 &1.000 &0 \\ 0& 0& −0.327942 \end{matrix} \right]

 

[\Phi _{rv}]=\left[\begin{matrix} \frac{1}{2} \sin nt & \frac{2}{n} (1 − \cos nt) &0 \\ \frac{2}{n} (\cos nt − 1) & \frac{1}{n} (4 \sin nt − 3nt) &0 \\ 0& 0&\frac{1}{n} \sin nt \end{matrix} \right]

 

=\left[\begin{matrix}816.525 & 2295.54 &0 \\ −2295.54 & −83 133.9 &0 \\ 0& 0& 816.525 \end{matrix} \right]

 

[\Phi _{vr}]=\left[\begin{matrix}3n \sin nt & 0 &0 \\6n(\cos nt − 1)&0&0\\0&0&−n \sin nt\end{matrix}\right] =\left[\begin{matrix}0.00327897 & 0 &0 \\−0.00921837 & 0&0 \\ 0& 0& −0.00109299 \end{matrix} \right]

 

[\Phi _{vv}]=\left[\begin{matrix} \cos nt& 2 \sin nt &0 \\ −2 \sin nt & 4 \cos nt − 3&0 \\ 0& 0& \cos nt \end{matrix} \right]

 

=\left[\begin{matrix}−0.327942&1.88940 &0 \\ −1.88940 &−4.31177 &0 \\ 0& 0&−0.327942 \end{matrix} \right]

From Equation 7.46 and (b) we find \delta v^{+}_{0}:

\left\{ \delta v^{+}_{0}= \right\} =-\left[\Phi _{rv} (t_{f})\right]^{-1}\left[\Phi _{rr} (t_{f})\right] \left\{\delta r_{0}\right\}     (7.46)

\left\{\begin{matrix}\delta u^{+}_{0} \\\delta v^{+}_{0}\\ \delta w^{+}_{0}\end{matrix} \right\} =- \left[\begin{matrix}816.525& 2295.54&0 \\−2295.54 &−83 133.9 &0 \\ 0& 0& 816.525\end{matrix} \right] ^{-1}

 

\times \left[\begin{matrix} 4.98383 &0 &0 \\ −194.257 &1.000 & 0 \\ 0 &0 &−0.327942 \end{matrix} \right] \left\{\begin{matrix} 20 \\ 20 \\20 \end{matrix} \right\}

 

 

=\left\{\begin{matrix}0.00936084 \\ −0.0467514\\0.00803263 \end{matrix} \right\} (km/s)      (c)

From Equation 7.42b, evaluated at t = t_{f} , we have

\left\{\delta v(t)\right\}=\left[\Phi _{vr}(t)\right]\left\{\delta r_{0}\right\}+ \left[\Phi _{vv}(t)\right]\left\{\delta v_{0}\right\}    (7.24b)

\left\{ \delta v_{f} \right\}=\left[\Phi _{vr}\left(t_{f}\right) \right]\left\{\delta r_{0}\right\}+\left[\Phi _{vv}\left(t_{f}\right) \right] \left\{\delta v_{0}^{+}\right\}

Substituting (b) and (c),

\left\{\begin{matrix} \delta u^{-}_{f} \\\delta v^{-}_{f} \\\delta w^{-}_{f} \end{matrix} \right\} =\left[\begin{matrix} 0.00327897 &0 &0 \\ −0.00921837& 0& 0 \\ 0& 0& −0.00109299 \end{matrix} \right] \left\{\begin{matrix} 20 \\ 20 \\ 20 \end{matrix} \right\}

 

 

\left\{\begin{matrix} \delta u^{-}_{f} \\\delta v^{-}_{f} \\\delta w^{-}_{f} \end{matrix} \right\} =\left\{\begin{matrix}−0.0258223 \\ −0.000472444 \\\ −0.0222449 \end{matrix} \right\} (Km/s)         (d)

Delta-v at the beginning of the rendezvous maneuver is found as

\left\{\Delta v_{0}\right\}=\left\{\delta v^{+}_{0}\right\}-\left\{\delta v^{-}_{0} \right\} =\left\{\begin{matrix} 0.00936084 \\−0.0467514 \\ 0.00803263 \end{matrix} \right\}-\left\{\begin{matrix} −0.02\\ 0.02 \\−0.005 \end{matrix} \right\} =\left\{\begin{matrix} 0.0293608 \\ −0.0667514 \\0.0130326 \end{matrix} \right\}

Delta-v at the conclusion of the maneuver is

\left\{\Delta v_{f}\right\}=\left\{\delta v^{+}_{f}\right\}-\left\{\delta v^{-}_{f} \right\} =\left\{\begin{matrix} 0 \\ 0 \\0 \end{matrix} \right\}- \left\{\begin{matrix} −0.0258223\\ −0.000472444 \\ −0.0222449 \end{matrix} \right\}=\left\{\begin{matrix} 0.0258223 \\ 0.000472444 \\ 0.0222449 \end{matrix} \right\}(km/s)

The total delta-v requirement is

\Delta v_{total}= \left\|\Delta v_{0}\right\| +\left\|\Delta v_{f}\right\| =0.0740787 + 0.03559465 = 0.109673   km/s = 109.7 m/s

From Equation 7.42a, we have, for 0 < t < t_{f} ,

\left\{\delta r(t)\right\}=\left[\Phi _{rr}(t)\right]\left\{\delta r_{0}\right\}+ \left[\Phi _{rv}(t)\right]\left\{\delta v_{0}\right\}  (7.42a)

\left\{ \begin{matrix} \delta x(t)\\ \delta y(t) \\ \delta z(t) \end{matrix} \right\}=\left[\begin{matrix} 4 − 3 \cos nt & 0 & 0 \\ 6(\sin nt − nt) & 1 &0 \\ 0 & 0 & \cos nt \end{matrix} \right]\left\{\begin{matrix} 20 \\ 20 \\ 20 \end{matrix} \right\}

 

Substituting n from (a), we obtain the relative position vector as a function of time. It is plotted in Figure 7.7.