Question 12.2: The Intelsat II series of communication satellites were depl...
The Intelsat II series of communication satellites were deployed in geostationary-equatorial orbit (GEO) in the late 1960s. Intelsat II was an oblate cylinder with a diameter of 1.42 \mathrm{~m} and height of 0.67 \mathrm{~m}. Before firing the apogee rocket engine for insertion into GEO, the Intelsat II was spun along its (symmetric) 3 axis at 12.57 \mathrm{rad} / \mathrm{s} (about 120 \mathrm{rpm} ). The mass of the Intelsat II was 162 \mathrm{~kg} before the apogee engine was fired. Let us assume that a pure spin about the 3 axis was not achieved and that at time t=0 the angular velocity is \boldsymbol{\omega}(0)=0.3419 \mathbf{u}_{1}-0.1974 \mathbf{u}_{2}+12.5638 \mathbf{u}_{3} \mathrm{rad} / \mathrm{s} along the 123 principal body axes as shown in Figure 12.14.
a) Compute the angular momentum vector \mathbf{H} in body coordinates.
b) Determine the nutation angle \theta and angle \gamma.
c) Compute the coning period and compare it with the spin period.
d) Compute the precession period and compare it with the spin period.
a) We know that angular momentum is \mathbf{H}=\mathbf{I} \boldsymbol{\omega} and we are given the angular velocity vector \boldsymbol{\omega} in 123 (or body) coordinates. Therefore, we need the inertia matrix of the Intelsat II. Assuming a homogeneous cylinder, the moment of inertia about the symmetric 3 axis is
I_{3}=\frac{1}{2} m R^{2}=40.8321 \mathrm{~kg}-\mathrm{m}^{2}
where the cylinder radius is R=(1 / 2)(1.42 \mathrm{~m})=0.71 \mathrm{~m} and mass is m=162 \mathrm{~kg}. The moments of inertia about the transverse 1 and 2 axes are
I_{1}=I_{2}=\frac{1}{12} m\left(3 R^{2}+h^{2}\right)=26.4762 \mathrm{~kg}-\mathrm{m}^{2}
where height h=0.67 \mathrm{~m}. Because I_{3}>I_{1}, the Intelsat II is an oblate (“flat”) cylinder.
Angular momentum is
\mathbf{H}=\mathbf{I} \boldsymbol{\omega}=\left[\begin{array}{ccc} 26.4762 & 0 & 0 \\ 0 & 26.4762 & 0 \\ 0 & 0 & 40.8321 \end{array}\right]\left[\begin{array}{c} 0.3419 \\ -0.1974 \\ 12.5638 \end{array}\right]=\left[\begin{array}{c} 9.052 \\ -5.226 \\ 513.006 \end{array}\right] \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}
Or, we can express angular momentum as \mathbf{H}=9.052 \mathbf{u}_{1}-5.226 \mathbf{u}_{2}+513.006 \mathbf{u}_{3} \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}.
b) The nutation angle is determined from the body-axis components of \mathbf{H}; we may use either Eq. (12.57) or (12.59).
\theta=\tan^{-1}\left({\frac{H_{12}}{H_{3}}}\right) (12.57)
\theta=\tan^{-1}\left({\frac{I_{1}\omega_{12}}{I_{3}n}}\right) (12.59)
Thus, we need to compute either the 1-2 projection H_{12} or \omega_{12}. The projection of \omega onto the 1-2 plane is
\omega_{12}=\sqrt{\omega_{1}^{2}+\omega_{2}^{2}}=0.3948 \mathrm{rad} / \mathrm{s}
Using Eq. (12.59), the nutation angle is
\theta=\tan ^{-1}\left(\frac{I_{1} \omega_{12}}{I_{3} \omega_{3}}\right)=1.17^{\circ}
The angle between \boldsymbol{\omega} and the 3 axis is computed using Eq. (12.56)
\gamma=\tan ^{-1}\left(\frac{\omega_{12}}{\omega_{3}}\right)=1.80^{\circ}
Because the Intelsat II is an oblate satellite, \theta<\gamma and the angular momentum vector \mathbf{H} is between the 3 axis and the angular velocity vector \boldsymbol{\omega} as shown in Figure 12.10.
c) We use Eq. (12.43) to compute the coning angular velocity
\lambda=\frac{I_{3}-I_{2}}{I_{1}} n
where I_{1}=I_{2}=26.4762 \mathrm{~kg}-\mathrm{m}^{2}, I_{3}=40.8321 \mathrm{~kg}-\mathrm{m}^{2}, and n=\omega_{3}=12.5638 \mathrm{rad} / \mathrm{s} (i.e., the constant spin component along the 3 axis). Using these values, we obtain \lambda=6.8123 \mathrm{rad} / \mathrm{s}. The period of this coning frequency is
\tau_{\text {coning }}=\frac{2 \pi}{\lambda}=0.922 \mathrm{~s}
Therefore, the \boldsymbol{\omega} vector completes one cycle of coning motion in 0.922 \mathrm{~s}. The “spin period” is
\tau_{\text {spin }}=\frac{2 \pi}{\omega}=0.500 \mathrm{~s}
where \omega=\|\boldsymbol{\omega}\|=\sqrt{\omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2}}=12.57 \mathrm{rad} / \mathrm{s}. The spin period is the time required for the satellite to make one complete revolution (recall that the satellite makes two revolutions per second, or 120 \mathrm{rpm} ). Therefore, the ratio of the coning-to-spin period is 0.922 / 0.5=1.84; in other words, the Intelsat II makes 1.84 revolutions during one coning cycle.
d) The precession rate is computed using Eq. (12.69)
\dot{\psi}=\frac{I_{3} \dot{\phi}}{\left(I_{1}-I_{3}\right) \cos \theta}
where the Euler angle spin rate is \dot{\phi}=-\lambda=-6.8123 \mathrm{rad} / \mathrm{s}. Using this value and the moments of inertia and \cos \theta=0.99979, we obtain
\dot{\psi}=19.3802 \mathrm{rad} / \mathrm{s}
The precession period is
\tau_{\text {precess }}=\frac{2 \pi}{\dot{\psi}}=0.324 \mathrm{~s}
As expected, the precession rate is greater than the coning angular velocity; it is also greater than the angular velocity magnitude. The ratio of the precession wobble-to-spin period is 0.324 / 0.5=0.65 and the Intelsat II makes 0.65 revolutions during one “wobble” precession cycle.
As previously mentioned in this section, satellites are often “spin stabilized” before firing an onboard rocket. The Intelsat II was spin stabilized in order to provide “gyroscopic resistance” to any disturbance torques encountered during the second Hohmann-transfer burn.
