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I'm working on a theoretical problem that I came up with; Given the extremely low atmosphere in low Earth orbit (300-500 km), would an arrow that had orbital velocity keep flying 'true' due to the center of mass being ahead of the center of pressure, or would the gravity gradient forces take over so the arrow aligns with zenith/nadir?

My issue is that I can't find any simple method to calculate the gravity gradient forces.

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  • $\begingroup$ There's a third possibility, which is that the arrow remains in a basically fixed orientation, if both the gravity gradient and the atmospheric effects are negligible. (One or the other has to dominate eventually, I suppose, but possibly not over the time scale you care about.) $\endgroup$ – Russell Borogove May 6 '15 at 16:41
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Assume the arrow is 1m long, and the reference altitude is 400km.

Acceleration due to gravity is µ / r2; Earth's µ (standard gravitational parameter) is 398,600.442 km3/s2, and radius is 6371 km. Figuring for altitude of 400km and 400.001km (these values added to the planet's radius), the difference in acceleration is about 0.000002568 m/s2 -- that's 2.6 micrometers per second squared.

So that's how fast the opposite ends of the arrow want to escape from each other when it's in the zenith-nadir orientation. At any LEO altitude the gradient is similar.

(I suppose calculus can be applied to get the gradient more directly.)

If the acceleration due to drag force at 400km is more than that, it should stabilize in the "airstream"; if it's less than that, it should stay radial. I believe the drag equations will require you to estimate a mass for the arrow as well; the gravity gradient acceleration is independent of the mass, so I worked in acceleration units rather than force units.

In practice, solar pressure and solar-heating-evaporation of surface layers would probably push the thing around, too; I don't know how to estimate those forces.

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