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I Shuttle Orientation While Orbiting

  1. Mar 7, 2016 #1
    An interesting thing that caught my attention before is how the space shuttle was oriented when it was in orbit (feels kinda sad to say "was" instead of "is"...). Maybe I'm not remembering right, or maybe the source was wrong, but I recall an astronaut mentioning that the tail end of the shuttle was pointed towards Earth, because in this posture, the shuttle was more stable from rotations under gravity's influence. He then mentioned something about a gravitational torque, which clearly must've been a Newtonian term.

    That... Seems very odd, but regardless of my inability to remember the details, it does have me wondering about what happens to, say, the space shuttle when it's orbiting.

    Now, in the context of GR, orbits are the result of geodesics - lines that are locally "straight," in that the derivative of a transposed vector is always zero along a geodesic.

    That would seem to imply, at least in the point-approximation case, that if the shuttle's nose was aligned with the velocity vector (with the Earth's center as the center of the coordinate system, of course), then it would stay aligned with the velocity vector through the entire orbit. Likewise if the tail, right wingtip, or any other arbitrary point was lined up with the velocity vector, it would stay that way through the entire orbit (the sidereal rotation period of the craft would be equal to the sidereal rotation period of its orbit).

    But perhaps that's a naive over-simplification? The back of my mind says that the fact that an orbit slightly closer to the Earth has a slightly smaller circumference means that there might be some rotational effect, since whichever end of the craft is closer to the Earth is travelling a smaller path. I really can't think of how this would alter the results of a point-approximation, though.

    Placing this in relativity instead of classic because of the appeal to geodesics. Theoretically, Newtonian should give the same result(?), but the only argument I can think of is the geodesic one.
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  3. Mar 8, 2016 #2


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    Consider the tidal forces on the shuttle. There are two main force components. In the radial direction, the tidal force is a tension. In Newtonian theory, this happens because the closer object is pulled more strongly. In the other two directions at right angle to the radial direction, the force is compressive. This happens in Newtonian theory because the force is directed towards the center of the Earth.

    If messing around with the forces is too annoying, you can find the same result from potential theory. Calculate the total energy of the shuttle pointing in the orbital direction, compare it to the total energy if it's angled. I believe Goldstein has a section on tidal torques in "Classical Mechanics" using this approach.

    Hopefully if you think about it, you'll see which configuration is more stable, and has the lowest energy, and why. I'd assume that the shuttle a) attempted to keep a fixed orientation relative to the Earth (to keep the antenae aligned if nothing else) and b) of the fixed-Earth configurations, used the most stable one. But I don't know for sure.

    If you like science fiction, try reading the short story "Neutron Star" by Larry Niven. Though I won't guarantee he got everything right (the gravitational blueshift effects described in the wiki summary seem to be off, for instance), the main points about the tidal forces are right.
  4. Mar 8, 2016 #3


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    You are confusing geodesics in spacetime with paths in space. The orientation throughout the orbit depends on the initial angular velocity.

    A point has no orientation, so a point-approximation doesn't give any results on this.
  5. Mar 8, 2016 #4
    Ah, I knew about tidal forces, I don't know why I didn't connect the dots.* So we have two points, the center of volume (where the center of mass would be if the density was uniform) and the true center of mass. The tidal forces will operate with the center of volume as the center, then? Which means that the effect would be to try to orient the craft so that the center of mass, center of volume, and Earth's center are all lined up?

    *Actually, I remember now why I didn't connect the dots; in Newton's theory, the gravitational force is directly proportional to the object's mass which the force is being applied to (hence the equal accelerations for all objects - at least when m1 >> m2), and torque is proportional to the "lever-arm" distance (of course being the distance in-between the force vector and the mass being acted on at the force vector's closest point). I guess I figured those two might act in such a way as to cancel their effects and not create a net torque, which I think would be true in a uniform density case, but apparently isn't true in a more realistic non-uniform density case.

    And actually... I think the proportionality to the affected object's mass, ignoring the lever-arm distances for torque (making them all equal), would cancel out so as to make it so the tidal forces have a net zero torque. But then add the lever-arm term in, and suddenly the torques across the vehicle are unbalanced, since although the different masses no longer make a difference, the different distances to the center of the object, do.

    I think I'll toy around with some spread of points and see if the math confirms or disproves this.
  6. Mar 8, 2016 #5


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    I think you've got the basic idea. A potential based approach would just compute the integral of -GMm/r for different orientations of the shuttle, but you can also do this by considering the tidal stretching forces (units of force/meter, the tidal force is proportional to the displacement from the center of mass) of -2GMm/r^3 for the stretching forces in the radial direction and +GMm/r^3 for the compressive forces in the non-radial direction.

    More sophisticated analysis are possible as well.
  7. Mar 9, 2016 #6

    Popular posture of space shuttle is upside down for protection against sunshine by floor tiles. I assume that the standing position you mentioned has larger inertial momentum for gravity torque and more stable in rotation caused by mechanical jobs like load bay door open-close or robot arm operation.
  8. Mar 12, 2016 #7


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    Doesn't help much if you move with ~8km/s relative to the ground stations, and have to switch between different ground stations frequently. But at least you know which side of the shuttle will point towards Earth and which side will not.

    @MattRob: there is no buoyancy involved, the center of volume is irrelevant (also, what would be included in the volume?). You can model the space shuttle as a lengthy brick. Tidal forces make an orientation along the tension axis the lowest-energy state.
  9. Mar 15, 2016 #8
    Made a simple c++ console code to get the sum of the torques on the points. It's a bit messy, but short. http://pastebin.com/waU60cZu

    Not sure how accurate it is, though.

    It probably makes life a lot easier for station-keeping since you don't have to have some function of time for your orientation - you know that relative to the local horizon, your orientation will always remain thus.

    I was thinking the center of volume might matter because of the lever-arm term in the torque (the r in the r x F), or other more complex arguments to do with the tidal forces. In any case, the mental picture I had had a 90* position being stable in an exact case, but unstable to any perturbation, like a pencil balancing on its tip. The code I wrote, though - if it's right - seems to say that that's not even the case, since there are torques when the inputted angle is 90*.

    Still, though, for symmetry reasons I keep thinking the center of volume and the center of mass are important - that there must be some at least semi-intuitive way to picture this. Does the tension axis run along a line on which lies both the center of mass and volume?
  10. Mar 15, 2016 #9


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    The center of volume does not matter. Torque is also coming from the mass, not from any volume.
    There is an unstable equilibrium point in horizontal orientation, not necessarily exactly at 0 degrees relative to some arbitrary shuttle axis.
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