Hi Thomas,
On the subject of orbital precession (i.e. nodal regression) you asked:
> So, what would account for such a huge precession of the plane of orbit?
The earth's equatorial bulge.
> The most significant torque (or force acting in a non-center-directed way)
> on the ISS would seem to be gravitational tugging by the moon & sun (which
> I'd think would largely balance out, over the course of an orbit),
> atmospheric drag, and the solar wind.
The earth's non-spherical shape overwhelms all of these for LEO
satellites in non-polar orbits.
> Do more compact satellites (such as Hubble, for example) have a similar
> orbital precession?
Yes. Nodal regression is a function of altitude and orbital inclination.
The lower the altitude and the lower the inclination, the greater the
nodal regression rate per day (the maximum possible is about 9 degrees/day).
The rate of change of the right ascension of ascending node (RAAN) due
to earth's J2 zonal coefficient is given by:
delta Omega = -1.5 * n * J2 * (Re/a)^2 * cos i * (1-e^2)^-2
where n is mean motion in deg/day
Re is earth's equatorial radius
a is the semi-major axis in km
e is eccentricity
i is orbital inclination
J2 = .00108263
delta Omega is in deg/day
Since n is a function only of a, the formula can be simplified. After
substituting all the constants, you get:
delta Omega = -2.06474E+14 * a^(-3.5) * cos i * (1-e^2)^-2
For a circular orbit, the last term is unity. So, try plugging in
51.6 degrees for the inclination, and an altitude of 380 km (a=6758 km):
delta Omega = -2.06574E+14 * 3.94127E-14 * cos 51.6 = -5.06 deg/day
(the negative sign indicating westward movement of the RAAN).
--Rob
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