From Eq. 2.13 from Orbital Mechanics for Engineering Students (Curtis, 2005) the equation of relative motion between two bodies orbiting around a center of mass can be written as:

$$\ddot{\mathbf{r}} = \frac{\mu}{r^3} \mathbf{r}$$



How would you obtain the same for a dumbbell system. A dumbbell system consisting of point masses at the ends of a rigid massless tether rotating around its center of mass, in in-plane motion about the Earth.

It would be very helpful to see the equation of motion derived for a rigid body mentioned in the question involving translational motion and rotational motion.

  • $\begingroup$ This question is difficult for me to understand, what is it you are really asking? How do you know this equation? Where did it come from? Also, "What are equations of motion (in celestial dynamics)?" is probably a separate question, but you'd probably have to show some independent work of some sort. Try to narrow down to one specific and clear question only. You can ask many questions; break them up a bit. But maybe wait for one answer before asking the follow-up. $\endgroup$ – uhoh Mar 17 '18 at 13:27
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    $\begingroup$ @Saiboogu This question is about orbital mechanics and is certainly on-topic here. There is no theoretical physics, just basic engineering (forces, torques, dynamics, etc.). If you think about it, standard Gravity gradient stabilization of spacecraft is already a non-rotating, asymmetric "dumbbell". An answer will contain some insight into spacecraft dynamics that could be applied elsewhere, perhaps for artificial gravity or tethers or something else. $\endgroup$ – uhoh Mar 18 '18 at 23:35
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    $\begingroup$ Well, I guess you just convinced me to cast the deciding vote on reopening. It seems better fleshed out from the original. $\endgroup$ – Saiboogu Mar 19 '18 at 1:12
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    $\begingroup$ @uhoh Thank yo very much for re-writing and helping with the question being presented in a better way and thank you for the explanation. :) $\endgroup$ – C J Spitzer Mar 19 '18 at 5:46
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    $\begingroup$ @uhoh It is very much okay. I agree that the initial problem should be simplistic to solve. :) $\endgroup$ – C J Spitzer Mar 19 '18 at 12:44

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