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## Q. 7.1

Spacecraft A is in an elliptical earth orbit having the following parameters:

h=52 059  km²/s , e=0.025724 , i=60°, Ω=40°, ω =30°, θ=40°            (a)

Spacecraft B is likewise in an orbit with these parameters:

h = 52 362  km²/s, e = 0.0072696, i = 50°, Ω= 40°, ω = 120°, θ = 40°            (b)

Calculate the velocity $v_{rel})_{xyz}$ and acceleration $a_{rel})_{xyz}$ of spacecraft B relative to spacecraft A, measured along the xyz axes of the co-moving coordinate system of spacecraft A, as defined in Figure 7.1. ## Verified Solution

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)

$a_{A}=-\mu \frac{r_{A}}{\left\|r_{A}\right\|^{3} }= 0.00035876\hat{I}− 0.0051989\hat{J}− 0.0072975\hat{K}(km/s^{2})$   (e)

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

$h_{A}=r_{A}\times v_{A}=\left|\begin{matrix} \hat{I} & \hat{J} & \hat{K} \\ -266.74 & 3865.4 & 5425.7 \\ -6.4842 &-3.6201 &2.4159 \end{matrix} \right|$ $=28980\hat{I}-34537{I}+26030 \hat{K}(Km/s^{2})$

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)

$\Omega =\frac{ 28 980\hat{I}-34537\hat{J}+26030\hat{K}}{6667.1^{2}}$

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

$\dot{\Omega }=-\frac{2(844.41)}{6667.1^{2}}(0.00065196\hat{I} − 0.00077698\hat{J}+ 0.00058559\hat{K})$

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

$0.93594 \hat{ I }-5.2409 \hat{ J }-5.5016 K =(-6.4842 \hat{ I }-3.6201 \hat{ J }+2.4159 \hat{ K })$ $+\left|\begin{matrix} \hat{I} & \hat{J} & \hat{K} \\ 0.00065196 & −0.00077698 & 0.00058559 \\ -5623.3 & −6844.8 & −3633.7 \end{matrix} \right| +v_{rel}$

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

$0.0073377\hat{I} + 0.0037117\hat{J} − 0.0022325\hat{K}$

$= 0.00035876\hat{I} + 0.0051989 \hat{J} − 0.0072975\hat{K}$

$+\left|\begin{matrix} \hat{I} & \hat{J} & \hat{K} \\-2.4770(10^{-8}) & 2.9520(10^{-8}) & −2.2248(10^{-8}) \\ −5623.3 & −6844.8 & −3633.7 \end{matrix} \right|$

$+(0.00065196 \hat{I} -0.00077698 \hat{J} +0.00058559 \hat{K})$

$\times \left|\begin{matrix} \hat{I} & \hat{J} & \hat{K} \\0.00065196 & -0.00077698 & 0.00058559 \\ -5623.3 & -6844.8 & -3633.7 \end{matrix} \right|$

$+2 \left|\begin{matrix} \hat{I} & \hat{J} & \hat{K} \\0.00065196 & -0.00077698 & 0.00058559 \\ 0.58865 & -0.69692 & 0.91414 \end{matrix} \right| +a_{rel}$

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

$\left[Q\right]_{Xx}= \left[\begin{matrix}−0.040008 &0.57977 &0.81380 \\ −0.82977& −0.47302& 0.29620\\ 0.55667& −0.66341& 0.5000 \end{matrix} \right]$

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:

$r_{rel})_{xyz}=\left|\begin{matrix} −0.040008 & 0.57977 & 0.81380 \\ −0.82977 & −0.47302 & 0.29620 \\ 0.5566 7& −0.66341 & 0.5000 \end{matrix}\right| \left\{\begin{matrix} −5623.3 \\ −6844.8 \\ −3633.7 \end{matrix} \right\} =\left\{\begin{matrix} −6700.5 \\ 6827.4 \\ −406.22 \end{matrix} \right\} (km)$

$v_{rel})_{xyz}=\left|\begin{matrix} −0.040008 & 0.57977 & 0.81380 \\ −0.82977 & −0.47302 & 0.29620 \\ 0.5566 7& −0.66341 & 0.5000 \end{matrix}\right| \left\{\begin{matrix} 0.58865 \\ -0.69692 \\ 0.91414 \end{matrix} \right\}=\left\{\begin{matrix} 0.31632 \\ 0.11199 \\ 1.2471 \end{matrix} \right\}(Km/s)$

$a_{rel})_{xyz}=\left|\begin{matrix} −0.040008 & 0.57977 & 0.81380 \\ −0.82977 & −0.47302 & 0.29620 \\ 0.5566 7& −0.66341 & 0.5000 \end{matrix}\right| \left\{\begin{matrix} 0.00043984 \\ -0.00038019 \\ 0.000017988 \end{matrix} \right\}=\left\{\begin{matrix} -0.00022338 \\ -0.00017980 \\ 0.00050607 \end{matrix} \right\}( Km/s^{2})$ 