Question 10.14: The uniform, monolithic 10 000 kg slab, having the dimension...

According to Figure 9.9(c), the principal moments of inertia around the xyz axes through the center of mass are
A =\frac{10000}{12}(1^{2} + 9^{2}) = 68 333   kg·m^{3}
B=\frac{10000}{12}(3^{2} + 9^{2})=75 000  kg·m^{3}
C=\frac{10000}{12}(3^{2} + 1^{2})=8333.3  kg·m^{3}
Let us first determine whether we can stabilize this object as a minor axis spinner. In that case,

Since I_{roll} > I_{yaw}, the satellite would be stable in pitch. To check yaw/roll stability, we first compute

We see that k_{Y} k_{R} > 0, which is one of the two requirements. The other one is found in Equation 10.181, but in this case

3k_{R} + k_{Y} k_{R}+1>4\sqrt{k_{Y} k_{R}}                                       (10.181)
1 + 3k_{R} + k_{Y} k_{R}−4\sqrt{k_{Y} k_{R}}=−4.1533 < 0
so that condition is not met. Hence, the object cannot be gravity-gradient stabilized as a minor axis spinner.
As a major axis spinner, we must have

Then I_{roll} > I_{yaw}, so the pitch stability condition is satisfied. Furthermore, since

we have
k_{Y} k_{R} = 0.7805 > 0
1 + 3kR + kY kR − 4\sqrt{k_{Y} k_{R}}= 1.1735 > 0
which means the two criteria for stability in the yaw and roll modes are met. The satellite should therefore be orbited as shown in Figure 10.31, with its minor axis aligned with the radial from the earth’s center, the plane abcd lying in the orbital plane, and the body x axis aligned with the local horizon.

According to Equation 10.171, the frequency of the pitch oscillation is
\omega_{f_{pitch}}=n\sqrt{3\frac{I_{roll}-I_{yaw}}{I_{pitch}} }=n\sqrt{3\frac{68 333 − 8333.3}{75 000}}= 1.5492  n
where n is the mean motion. Hence, the period of this oscillation, in terms of that of the orbit, is
T_{pitch}=\frac{2\pi }{\omega_{f_{pitch}}}=0.6455\frac{2\pi }{n}=\underline{0.6455 T}
For the yaw/roll frequencies, we use Equation 10.184,
\omega_{f_{yaw/roll}})_{1,2}=n\sqrt{\frac{1}{2}(b\pm \sqrt{b^{2}-4c)}}                                 (10.184)

\omega_{f_{yaw/roll}})_{1}=n\sqrt{\frac{1}{2}(b+\sqrt{b^{2}-4c)}}
where
b = 1 + 3k_{R} + k_{Y} k_{R} = 4.7073   and   c = 4k_{Y} k_{R} = 3.122
Thus,
\omega_{f_{yaw/roll}})_{1}=2.03015  n
Likewise,
\omega_{f_{yaw/roll}})_{2}=n\sqrt{\frac{1}{2}(b-\sqrt{b^{2}-4c)}}= 1.977  n
From these we obtain
\begin{matrix} T_{yaw/roll_{1}} = \underline{0.5058 T} & T_{yaw/roll_{2}} =\underline{ 0.4345 T} \end{matrix}
Finally, observe that
\frac {I_{roll} – I_{yaw}}{I_{pitch}}= 0.8
so that we are far from the pitch resonance condition that exists if the orbit has a small eccentricity.