The cosmological expansion of the universe tends to occur in the space between galaxies or even clusters of galaxies. The distinction is whether the systems are gravitationally bound or not. The following is a crude attempt to define "gravitationally bound" based in just Newtonian mechanics.

The gravitational interaction force for two masses is GM1M2/r2 and Hubble’s expansion rate is Hr. Here G = 6.674×10−11, H = 2.27 x 10-18, and r is the separation of the masses (all in SI units).

Assume the two isolated masses are in roughly circular orbits separated by distance r. The masses are at distances rM2/(M1+M2) and rM1/(M1+M2), respectively, from their common center of mass (which is also taken as an origin, r=0). The energy of the system is -(1/2)GM1M2/r. We take the expansion of the universe to have been caused by a radial impulse of force applied to all objects in proportion to their mass, M, and to their distance, r, from the center of mass. This impulse is scaled so the resulting radial velocity follows Hubble’s expansion, vr = Hr. Since the expansion velocities are at right angles to the orbital velocities, the energies simply add. The new orbital radius will be given by
-(1/2)GM1M2/(r+$\Delta$r) = -(1/2)GM1M2/r + (1/2)M1(Hr1)2 + (1/2)M2(Hr2)2
This results in an expression for the increase, $\Delta$r, in orbital radius: $\Delta$r = H2r4/(M1+M2)/G

Since the expansion velocity is Hr, it takes $\Delta$r/Hr = Hr3/(M1+M2)/G seconds for this change in radius to happen. Somewhat arbitrarily, if this is shorter than the age of the universe, 1/H, then the bodies are taken as gravitationally bound.
The resulting inequality is (M1+M2)/r3 > H2/G = 7.72E-26 kg/m3
This has units of mass-density which makes some sort of sense. It's a much bigger number than the current mass density of the universe.

This does seem similar to the conclusion reached via a different route in another question: Gravitationally bound systems in an expanding universe


If we have a system in a circular orbit with radius $r$ and orbital frequency $\omega$, and if the secular trend is slow compared to $\omega$, we have for the secular trend

$$\frac{\dot{r}}{r} = \omega^{-2} \frac{d}{dt}\left(\frac{\ddot{a}}{a}\right).$$

See Cooperstock, http://arxiv.org/abs/astro-ph/9803097v1 for a calculation leading to an equivalent result. This holds regardless of the matter fields. In the case of a vacuum-dominated cosmology, the time dependence of $r$ seems to integrate to something like

$$r = \left[\text{const.}-(\text{const.})h^{-1}e^{-ht}\right]^{-1/3},$$

where $h=\sqrt{\Lambda/3}$ is currently roughly the same as the Hubble constant. So really there aren't big variations in the final amount of expansion for systems that meet the criteria for the above treatment to be a good approximation. The tightness of their binding makes little difference.

So since the above treatment says that everything is bound, we're interested in the conditions under which that treatment fails, so that the system will come apart. I don't know about highly elliptical orbits, but if we continue to assume a circular orbit, then the only other assumption that can fail is that the secular trend is slow compared to $\omega$. In the present universe, the value of $(d/dt)(\ddot{a}/a)$ is on the order of $H^3$, so this condition is just that $\omega \gg H$. By Kepler's law of periods, this is about the same as the criterion you proposed. However, in the future, as the universe becomes more vacuum dominated, $(d/dt)(\ddot{a}/a)$ will no longer be on the order of $H^{-3}$, and will in fact exponentially approach zero. So in the future the criterion you propose for binding will be wrong. Eventually, arbitrarily loosely bound systems will still be bound against the effects of cosmological expansion, and the rate at which this future change occurs is much faster than what we would get by applying your criterion and taking $H$ to be on the order of the inverse age of the universe.


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