Tensors make me tense.
Imagine a long thin rod in a circular orbit. The gravity gradient will produce a net torque on the rod whenever it is not oriented parallel, or perpendicular to the radius vector (pointing up/down or pointing forward/backward). Let's keep the problem 2D and ignore out-of-plane orientations.
The Coursera video 1: Gravity Gradient Torque Development from the course by University of Colorado Boulder Kinetics: Studying Spacecraft Motion taught by Hanspeter Schaub includes the following:
$$ L_G = \frac{3 GM_E}{R_C^5} \mathbf{R_C} \times [I] \mathbf{R_C},$$
which comes from a first order expansion of the local gravity gradient.
He explains that at this point (about 16:30
) that you need to stop and think about coordinate systems and frames, and that's when I start feeling like this guy lower your volume first!
For an infinitely thin rod in 2D, I am guessing that the moment of inertia tensor is just
$$ I = \begin{bmatrix} \frac{1}{12}ml^2 & 0\\ 0 & 0\\ \end{bmatrix}. $$
Now what? I need to get a $\sin(2\theta)$ somehow so that the torque will be zero at both 0 and 90 degrees. What sort of tensor magic multiplication can get me there?
SMAD first edition gives:
$$\frac{3 \mu}{R_0^3} \mathbf{u_e} \times (\mathbf{I} \cdot \mathbf{u_e})$$
where $\mathbf{u_e}$ is the unit vector towards the nadir. It seems to be essentially the same, but with the magical tension-inducing tensor math written slightly differently. Dot product between two vectors gives a scalar, but here I don't know what to do.
In each case $\mathbf{R_C}$ or $\mathbf{R_0}$ are from the center of the Earth to the rod's center of mass.
Question: How would I derive that expression for torque in terms of an angle $\theta$ that the rod makes with respect to the nadir, such that the torque has a $\sin(2\theta)$ term, using my simple 2D moment of inertia?
Please, no small angle approximations!
SMAD first edition