Help with my tensor tension; how to derive and calculate this rigid body gravity gradient torque?

Question

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!

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Screen Shot

SMAD first edition

Answer 1

Let $\theta$ be the angle between the direction of the rod and the direction toward the Earth, measured counter-clockwise. Then in the system of coordinates you used to write $I$ as $$\begin{bmatrix} \frac{1}{12}ml^2 & 0\\ 0 & 0\\ \end{bmatrix}, $$ $\mathbf{R_C}$ has the form $$ \left(\begin{array}{c} R_C\cos\theta \\ R_C\sin\theta\end{array}\right). $$ Then $I\mathbf{R_C}$ is obtained by matrix multiplication: $$ I\mathbf{R_C} = \left(\begin{array}{c} \frac{1}{12}ml^2R_C\cos\theta \\ 0\end{array}\right), $$ and the cross product is $$ \mathbf{R_C}\times I\mathbf{R_C} = - R_C\sin\theta\cdot \frac{1}{12}ml^2R_C\cos\theta = -\frac{1}{24}ml^2R_C^2\sin 2\theta. $$ (Strictly speaking, the cross product is a three-dimensional vector, but if we restrict ourselves to a plane, then this vector is always perpendicular to this plane, so we can view it as a scalar.)

And the final result is $$ L_G = -\frac{GM_Eml^2}{8R_C^3}\sin 2\theta. $$