I used a instead of alpha.
The quadrupole energy contribution equation can be written as quadrupole energy = – a*f(theta) with f(theta) = 2cos2(theta) – sin2(theta). The monopole contribution to the energy is independent of a, so we consider the quadrupole contribution.
* If a > 0: For the lowest energy, we look at the orientation in which the energy is most negative. This is the case if the dumb-bell is oriented such that the angle dependent function is most positive. This is the case when theta=0° (dumb-bell parallell to z axis). Intuitively, we can understand this by saying that the distance between the rings is larger than the radius (because of a > 0). Therefore, due to the gravitational attraction between the dumb-bell mass and the ring, the dumb-bell will “try” to orient itself parallel to the z axis.
* If a < 0: The energy is most negative if the angle dependent function is most negative. This is the case when theta= 90° (dumb-bell in xy plane). Intuitively, this can be understood, because a < 0 so the radius of the rings is larger than de distance between them. This means the masses of the dumb-bell “feel” the attraction of a larger mass parallell to the xy plane. The dumb-bell will “try” to orient itself parallell to the xy plane.
* If a = 0: The quadrupole energy contribution is zero in this case. The dipole contribution is zero in general, so the only contribution to the gravitational energy is the monopole term. This term is only dependent of the masses, the distance between rings and their radius. Therefore, if none of those properties change (especially those of the dumb-bell), the energy will be the same, regardless of the orientation of the dumb-bell. Intuitively, this is the “middle” between the two previous situations.
I think you come to the right conclusion for a=0, but you did not take higher order terms into account (octopoles and so on). I don’t think these are sensitive to the orientation, but I’m just saying they exist as well.