That’s the thing, I think Pluto is the closest example we have in a planetary situation. Some binary star pairs are closer in mass ratio, I think. Realistically speaking, it would be pretty improbable for two objects with precisely the same mass to end up orbiting each other. The center of mass of the system would be slightly closer to the slightly more massive object, so that one would be called the ‘planet’ and the other would be the ‘moon’.
Yay, my 200th post. 
The bigger the primary the smaller the chance of a decent sized moon-moon (i.e. tertiary). Proto-moon-moons would get sucked into the primary or flung away before they had the chance to become locked to a moon (i.e. secondary). The more spatial bodies involved in the orbital dance the more complex the mechanics required to keep them all stable. Still, it’s a Big™ universe so somewhere, there are probably even ?quaternaries.
Wouldn’t that be the cube of the distance?
Nope, F = Gm1m2/r^2
KP is correct, if i remember my celestial mechanics correctly. The situation is basically this: A two-body system “wants” to have both bodies locked to each other, which is to say both objects always having the same side facing each other. Most planetary systems are such that the satellites already are tidally locked to their primary, though I believe only one primary (Pluto) is locked to its secondary (Charon). The Moon is locked to the Earth, and is taking energy from Earth’s rotation to lock it. That energy is going into the Moon’s orbit, lengthening its period of revolution. Eventually, the Moon will take enough energy out of the Earth’s orbit that the day will equal a month and the Moon will stop moving (well, there are other effects, but that’s the classical picture). Because the Moon formed at a time and place where its period of revolution was longer than the day, that’s what happens.
For Phobos, the situation is different. Its period of revolution is * shorter * than a martian day. Going through the math shows that its orbital energy is going into speeding up the martian day, which makes its orbit smaller. Eventually it will crash into Mars. Deimos is outside the critical distance, and will evolve outward like the Moon.
To the point, then: A hypothetical satellite of the Moon would evolve relative to the Moon the same way. Either it would be orbiting the Moon faster than the lunar day (28 days) and would eventually crash into the Moon, or would be orbiting more slowly and evolve outwards. Once you get that far from the Moon, you’re dealing with a 3-body problem, not a 2-body one, and the Moon’s satellites’ orbit is unstable (not on immediate timescales necessarily, but eventually). The gravitational effect of the hypothetical third party is very small, so the Earth and Moon don’t notice its influence.
For the binary asteroids, there is no third body involved, so orbits can theoretically be stable.
As your own link says, it’s tidal forces that drop with the cube of the distance. The Actual force only drops with the square. Hence the tidal force of the sun on earth is smaller than the tidal force of the moon, even though the gravitational force of the sun is greater. Tidal forces might be why moons don’t have moons.
that’s odd
Eh… no.
Charon is approx. half the diameter pf pluto. All else being equal, that would make pluto 8 times (2^3) more massive than Charon - so “half the size” is not really correct.
Dan Abarbanel
Thaumaturge, the link got screwed up cause vB interpreted an internal string as a smiley. To prevent that, check “Disable Smilies” just before submitting (problems like that should show up on preview).
I confess I have forgotten my physics (how often do I have to compute the attraction of two bodies other than near my girlfriend? 
). But the site linked to above says
Is this site wrong? If so, are tides (water) subject to a different equation than other parts of the same celestial body like rocks and gases?
See also Wikipedia on gravity and tidal force.