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So I'm looking at the following problem;

Arrow type configuration in orbit. Each end of the arrow has a mass and a cross sectional area (we can assume the head and tail are connected by a negligibly thin beam). The arrow start facing directly away from the Earth (Zenith), but it has normal orbital properties (speed = ~7.7km/s). Drag in orbit is considered not negligible.

Does the arrow align it's self with the velocity vector? My intuition is that it does, but I've tried to do some calculations on this and I get the arrow swinging back and forth from Zenith to Nadir, with increasing angles (I expected the angles to slowly decrease).

Any advice on what I'm doing wrong here would be great.

My method in detail;

  1. Calculate the drag area of the tail (this is reduced if the head is in front of the tail, but the head can never completely cover the tail).
  2. Calculate the drag area of the head (the tail never goes in front of the head from my starting configuration, but since the angle changes the wetted area of the head changes)
  3. Calculate the linear speed of the head and tail based on the current rate of rotation about the CoM.
  4. Calculate the drag on both head and tail (velocity = orbital velocity +/- linear speed based on rate of rotation) This is done using D = 0.5pV^2CdA
  5. Calculate the resultant moment about the CoM. M = F*r, Since the two forces will always be in opposition one moment - the other
  6. Calculate the induced angular acceleration from a = M/I (moment/moment of inertia)
  7. Calculate the angular velocity from V = U + at
  8. Calculate the Angle from S = Ut + 0.5at^2

I do all this with a time step model on excel. Unfortunately this gives me increases maximum angle of rotation (it swings back further than it originated). Any advice/comments welcome.

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  • $\begingroup$ Use a better programing environment and a better quadrature (numeric integration; you are using one of the simplest possible ways to go about that and it has some bad failure modes). You might also consider Computational Science as a better home for this. Let me know if you want it migrated. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 3, 2015 at 17:53
  • $\begingroup$ Hi, I was hoping for a simple analysis at first (then I was planning on doing a more in depth method). It could be moved to computational science, but I expect my issue to be my approach as opposed to my implementation (if you think otherwise please migrate) $\endgroup$
    – ThePlanMan
    Commented Jul 3, 2015 at 18:02
  • $\begingroup$ Well, look at the leap-frog method which is the simplest integration that will do decent orbits. Don't ask me how to implement it in a spreadsheet, however, I learned "proper" programming early enough to never get any good at spreadsheets. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 3, 2015 at 19:19

1 Answer 1

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I would do a hand analysis at first. You have two torques on the arrow. One is the drag torque, which is maximum when the arrow is transverse to the orbital velocity, as in your initial condition. The second is gravity gradient torque, which will be maximum when the arrow is horizontal. Compute each of these for your arrow. If one is much greater than the other, that tells you which direction the arrow will be close to. You have to assume there is enough rotational damping due to drag to bring the arrow to rest. If there were no damping, it would oscillate around the equilibrium angle. Since you are computing the drag forces, there will be the damping you nee.

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  • $\begingroup$ Thanks. I'm not to concerned about the gravity gradient forces at the moment. How would I include some damping to this system? $\endgroup$
    – ThePlanMan
    Commented Jul 3, 2015 at 23:18
  • $\begingroup$ If you don't include gravity gradient, then there is nothing to prevent the arrow from flying straight. The damping comes from transverse drag. If the arrow is spinning, there will be drag to slow the rotation down. That may be in your step 4. $\endgroup$
    – Ross Millikan
    Commented Jul 3, 2015 at 23:36
  • $\begingroup$ At the moment I'm looking at the system of forces that move the arrow from nadir aligned to flying straight (azimuth aligned if we're in equatorial orbit). The problem I'm having at the moment is I can identify the damping in the system. The way I see it the drag will just cause oscillation, no damping. $\endgroup$
    – ThePlanMan
    Commented Jul 3, 2015 at 23:55
  • $\begingroup$ Gravity gradient torque is zero when the arrow is flying horizontal as well as vertical. It's maximized when the arrow is flying at about 45 degrees to vertical or horizontal. $\endgroup$
    – David Hammen
    Commented Jul 4, 2015 at 1:21

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