Yes, that’s what I mean. I think the rotation of the moon should be considered separately, but it’s not. The rotation of the moon induces a bulge on all sides–the moon then is slightly oblate. Without that bulge, the reasoning–that the centrifugal force increases with distance from the Earth–fails, as the centrifugal force is the same at all points of the moon then.
That’s why I don’t really like the centrifugal force arguments for tides. What if there were no centrifugal force? What if the moon were free-falling to the Earth? The Earth’s gravity would induce a tide then too–of a magnitude the same as it does now.
Well, the Earth is free-falling towards the Earth. And, as I point out, the centrifugal force is the same across the moon, so it doesn’t contribute at all to the differential. Plus, we don’t have to use the c word, as the BA doesn’t in his explanation.
As to jezzaOZ’s question about the symmetry of the principal semidiurnal tidal components M2 and S2, the intuition is correct. There is a lack of symmetry, but it gets complicated by the fact that the Earth/moon/Sun system is tilted–so that there are not only components like O1 (the lunar diurnal component), K1 (the lunisolar diurnal component), and P1 (the solar diurnal component) which together express the effect of the moon’s and Sun’s declination, but there are also M1 and S1. M1 and S1 are just a lot smaller, as you might expect, since the gravity fall-off at those distances can be approximated linearly (that is, the increase on one side of the Earth is approx. the same as the decrease on the other side).
Why are M2 and S2 symmetrical? Well, that’s the definition. They’re sorta like the coefficients in the second degree component of a power spectrum–it’s the second degree component function that is symmetrical, not the tide itself.
Of course, that’s not the complete story.