Question 10.15: The uniform, monolithic 10,000 kg slab, having the dimension...
The uniform, monolithic 10,000 kg slab, having the dimensions shown in Figure 10.31, is in a circular LEO. Determine the orientation of the satellite in its orbit for gravity gradient stabilization, and compute the periods of the pitch and yaw/roll oscillations in terms of the orbital period T.
According to Figure 9.9c, the principal moments of inertia around the xyz axes through the center of mass are
A=\frac{10,000}{12}\left(1^{2}+9^{2}\right)=68,333 kg \cdot m ^{2}
B=\frac{10,000}{12}\left(3^{2}+9^{2}\right)=75,000 kg \cdot m ^{2}
C=\frac{10,000}{12}\left(3^{2}+1^{2}\right)=8333.3 kg \cdot m ^{2}
Let us first determine whether we can stabilize this object as a minor axis spinner. In that case,
I_{\text {pitch }}=C=8333.3 kg \cdot m ^{2} \quad I_{ yaw }=A=68,333 kg \cdot m ^{2} \quad I_{ roll }=B=75,000 kg \cdot m ^{2}Since I_{\text {roll }}>I_{ yaw }, the satellite would be stable in pitch. To check yaw/roll stability, we first compute
k_{Y}=\frac{I_{ pitch }-I_{ roll }}{I_{ yaw }}=-0.97561 \quad k_{R}=\frac{I_{ pitch }-I_{ yaw }}{I_{ roll }}=-0.8000We see that k_{Y} k_{R}>0, which is one of the two requirements. The other one is found in Equation 10.187, but in this case
3 k_{R}+k_{Y} k_{R}+1>4 \sqrt{k_{Y} k_{R}} (10.187)
1+3 k_{R}+k_{Y} k_{R}-4 \sqrt{k_{Y} k_{R}}=-4.1533<0so 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
k_{Y}=\frac{I_{ pitch }-I_{ roll }}{I_{ yaw }}=0.8000 \quad k_{R}=\frac{I_{ pitch }-I_{ yaw }}{I_{ roll }}=0.97561we have
k_{Y} k_{R}=0.7805>0
1+3 k_{R}+k_{Y} k_{R}-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 10.32 , 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 10.177, the frequency of the pitch oscillation is
\omega_{f \text { 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_{\text {pitch }}=\frac{2 \pi}{\omega_{f \text { pitch }}}=0.6455 \frac{2 \pi}{n}=0.6455 TFor the yaw/roll frequencies, we use Equation 10.190,
\left.\omega_{f \text { yaw/roll }}\right)_{1}=n \sqrt{\frac{1}{2}\left(b+\sqrt{b^{2}-4 c}\right)}where
b=1+3 k_{R}+k_{Y} k_{R}=4.7073 \text { and } c=4 k_{Y} k_{R}=3.122Thus,
\left.\omega_{f \text { yaw } / \text { roll }}\right)_{1}=2.3015 nLikewise,
\left.\omega_{f \text { yaw } / \text { roll }}\right)_{2}=\sqrt{\frac{1}{2}\left(b-\sqrt{b^{2}-4 c}\right)}=1.977 nFrom these we obtain
T_{\text {yaw } / \text { roll }_{1}}=0.5058 T T_{\text {yaw } / \text { roll }_{2}}=0.4345 T
Finally, observe that
\frac{I_{ roll }-I_{ yaw }}{I_{ pitch }}=0.8so that we are far from the pitch resonance condition that exists if the orbit has a small eccentricity.
