Question 12.6: Suppose GEOSAT (shown in Figure 12.24) has principal moments...
Suppose GEOSAT (shown in Figure 12.24) has principal moments of inertia I_{3}=100 \mathrm{~kg}-\mathrm{m}^{2} (long minor axis) and I_{1}=I_{2}=2,600 \mathrm{~kg}-\mathrm{m}^{2} (transverse major axes). If GEOSAT operated in an 800-km altitude circular orbit, compute the frequency and period of the pitch libration motion caused by the gravity-gradient torque.
Equation (12.130) provides the libration frequency for the pitching motion
\Omega_{p}=\omega_{0} \sqrt{\frac{3\left(I_{1}-I_{3}\right)}{I_{2}}}
We need the orbital angular velocity \omega_{0}. The orbital radius is r_{0}=R_{E}+800 \mathrm{~km}=7,178 \mathrm{~km} (recall that the Earth’s equatorial radius is R_{E}=6,378 \mathrm{~km} ). The orbital angular velocity is
\omega_{0}=\sqrt{\frac{\mu}{r_{0}^{3}}}=0.001038 \mathrm{rad} / \mathrm{s}
Using this value for \omega_{0} and the moments of inertia, the libration frequency is
\Omega_{p}=\omega_{0} \sqrt{\frac{3\left(I_{1}-I_{3}\right)}{I_{2}}}=0.001763 \mathrm{rad} / \mathrm{s}
The period of the pitch libration motion is \tau_{\mathrm{lib}}=2 \pi / \Omega_{p}=3,563.5 \mathrm{~s}=59.4 \mathrm{~min}.
Note that GEOSAT’s orbital period is 2 \pi / \omega_{0}=6,053.2 \mathrm{~s}=100.9 \mathrm{~min}.