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!

U. Colorado Boulder Kinetics: Studying Spacecraft Motion Coursera

Screen Shot


SMAD first edition

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    $\begingroup$ possibly helpful? I see a $\sin(2\theta)$ in Eq. A.16 here but I've run out of steam... $\endgroup$ – uhoh Jan 22 '19 at 2:45
  • $\begingroup$ @Paul It is from intuition. I know it's got to be there from symmetry. $\endgroup$ – uhoh Jan 22 '19 at 3:02
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    $\begingroup$ Hint: recall that $sin(2\theta)=2sin(\theta)cos(\theta)$. Everything else should be straight forward from that linked appendix. $\endgroup$ – Paul Jan 22 '19 at 3:02
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    $\begingroup$ Multiplying a rank 2 tensor to a vector is no different from multiplying a matrix to a vector. The trigonometric terms come from the euler rotations in the formula for the torque. Remember: attitude is everything! $\endgroup$ – Paul Jan 22 '19 at 3:10
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Paul Jan 22 '19 at 3:50

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 $$ L_G = -\frac{GM_Eml^2}{8R_C^3}\sin 2\theta. $$

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    $\begingroup$ That's excellent! I probably shot myself in the foot by trying to force a 3D problem into 2D, rather than just leave some zeros. Thank you very much! $\endgroup$ – uhoh Jan 22 '19 at 9:20
  • $\begingroup$ Can it really be true that the instantaneous angular acceleration for the thin rod is independent of length (to first order), and just $\ddot{\theta} = \dot{\omega} = -(3GM_E/2R_C^3)\sin(2\theta)$? That maxes out at about 0.4 degrees/minute^2 at 45 degrees; that's pretty cool! $\endgroup$ – uhoh Jan 22 '19 at 9:51
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    $\begingroup$ @uhoh You're welcome. And yes, your formula looks correct. $\endgroup$ – Litho Jan 22 '19 at 10:10
  • $\begingroup$ sweet! I've used it here now. $\endgroup$ – uhoh Jan 22 '19 at 10:13

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