From the orbital elements in (a) and (b) we can use Algorithm 4.2 to find the position and velocity of the spacecraft relative to the geocentric equatorial reference frame. Omitting those calculations, we find, for spacecraft A,

r_{A}=-266.74\hat{I} + 3865.4 \hat{J}+ 5425.7\hat{K}(Km)                          (a)

v_{A}=-6.4842\hat{I}-3.6201\hat{J}+ 2.4159( km/s)                         (b)

and for spacecraft B,

r_{B}=−5890.0\hat{I}− 2979.4\hat{J}+ 1792.0\hat{K} (km)                        (c)

v_{B}=0.93594\hat{I}− 5.2409\hat{J}- 5.5016\hat{K}(Km/s)                      (d)

According to Equation 2.15, the accelerations of the two spacecraft are

  \ddot{r}= – \frac{\mu }{r^{3} }r                             (2.15)

a_{B}=-\mu \frac{r_{B}}{\left\|r_{B}\right\|^{3} }=  0.0073377\hat{I}+0.0037117\hat{J}− 0.0022325\hat{K}(km/s^{2})  (f)

The unit vector \hat{i} along the x axis of spacecraft A’s rigid, co-moving frame of reference is

\hat{i}=\frac{r_{A}} {r_{A}}=-0.040008\hat{I}+0.57977\hat{J}+0.81380\hat{K}                                      (g)

Since the z axis is in the direction of h_{A}, and

we obtain

\hat{K}=\frac{h_{A}}{h_{A}}=0.55667\hat{I} -0.66341 \hat{J}+0.5000\hat{K}                       (h)

Finally,\hat{j} = \hat{K}\times \hat{i} ,so that

\hat{j}=−0.82977\hat{I} – 0.47302\hat{J}+0.29620\hat{K}                              (i)

The angular velocity Ω of the xyz frame attached to spacecraft A is given by Equation 7.1,

\Omega =\frac{r_{0}\times v_{0}}{r^{2}_{0} }                                         (7.1)

= 0.00065196\hat{I}− 0.00077698\hat{J}+ 0.00058559\hat{K} (rad/s)        (j)

We find the angular acceleration \dot{\Omega } using Equation 7.5,

\dot{\Omega } =-\frac{2(r_{0}\cdot v_{0})}{r^{2}_{0} } \Omega                                    (7.5)

= −2.4763(10^{-8})\hat{I}+ 2.9512(10^{-8})\hat{J} − 2.2242(10^{-8})\hat{K} (rad/s^{2})                 (k)

According to Equation 1.38, the relative velocity relation is

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

v_{B}=v_{A}+\Omega \times r_{rel}+v_{rel}            (I)

where  r_{rel}  and   v_{rel} are the position and velocity of B as measured relative to the moving xyz frame attached to A. From (a) and (b), we have

r_{rel} = r_{B} − r_{A} = −5623.3\hat{I}− 6844.8\hat{J}− 3633.7 \hat{K} (km)          (m)

Substituting this, together with (b), (d) and (j) into (l), we get

Solving for v_{rel} yields

v_{rel}= 0.58865\hat{I} − 0.69692\hat{J} + 0.91414 \hat{K} (km/s)           (n)

The relative acceleration formula, Equation 1.42, is

a=a_{O}+\dot{\Omega }\times r_{rel}+\Omega \times (\Omega \times r_{rel})+2\Omega \times v_{rel}+a_{rel}     (1.42)

a_{B}=a_{A}+\dot{\Omega }\times r_{rel}+\Omega \times(\Omega \times r_{rel})+2\Omega \times v_{rel}+a_{rel}  (o)

Substituting (e), (f), (j), (k), (m), and (n) into (o), we get

Carrying out the cross products, combining terms and solving for a_{rel} yields

a_{rel}=0.00043984 \hat{I} − 0.00038019\hat{j} + 0.000017988\hat{K} (km/s^{2})    (p)

From (g), (h), and (i), we see that the orthogonal transformation matrix [Q]_{Xx} from the inertial XYZ frame into the co-moving xyz frame is

To get the components of r_{rel} , v_{rel} ,  and   a_{rel}, along the axes of the co-moving xyz frame of spacecraft A, we multiply each of their expressions as components in the XYZ frame [(m), (n) and (p), respectively] by [Q]_{Xx }as follows: