This answer deals with a particular class of systems, where these assumptions hold:
- Gravity is the only important force between bodies (this would also work for any other purely attractive inverse square law).
- Collisions (or encounters) in the sense of processes modelled by the Fokker-Planck equation (rather than the collisionless Boltzmann equation) occur. For e.g. a stellar system, this means that stars may gravitationally scatter off of each other.
- Inelastic collisions (direct hits between stars, also strong tidal interactions) occur rarely enough to be unimportant.
These assumptions are valid for stellar system like a globular cluster, or most galaxies. I'll touch on this validity a bit more below.
One more concept to introduce before I actually get to an answer. In stellar dynamics, "relaxation" refers to the process by which the orbits of individual stars are perturbed, which eventually tends to drive the system to a configuration with a dense stellar core and a diffuse "halo" of stars surrounding it. The characteristic timescale for this to occur is $t_{\rm rh}$:
$$t_{\rm rh} = \frac{0.17 N}{\ln(\lambda N)}\sqrt{\frac{r_{\rm h}^3}{GM}}$$
$N$ is the number of particles in the system, $\lambda$ is determined empirically and is typically taken to be $0.1$, $r_{\rm h}$ is the radius enclosing half the mass of the system, $M$ is the total mass of the system and $G$ is of course the gravitational constant.
Now the actual answer. As stars orbit in a stellar system, they interact and exchange energy. A particular star may happen to gradually gain energy through multiple interactions until it has positive energy (i.e. kinetic energy that exceeds the potential binding it to the system), it can then escape the system and go off "to infinity". The remaining particles are then less bound to the system, and gradually more and more particles can escape via the same process. Eventually only two particles (stars) remain, and are in a Keplerian orbit about each other.
A crude way to estimate the timescale for evaporation is to assume that stars with speeds exceeding the escape speed are removed on the timescale $t_{\rm rh}$. Assuming a Maxwellian speed distribution, the fraction of stars with speeds exceeding the escape speed is $\gamma=7.38\times10^{-3}$. The loss rate is then:
$$\frac{{\rm d}N}{{\rm d}t}=-\frac{\gamma N}{t_{\rm rh}}$$
The natural definition for the evaporation timescale is therefore $t_{\rm evap}=t_{\rm rh}/\gamma$, so the system evaporates in about $140$ relaxation timescales. A more detailed calculation gives a value closer to $t_{\rm evap}\approx 300 t_{\rm rh}$.
Regarding the rarity of inelastic encounters, the timescale for collisions in a typical globular cluster is $t_{\rm coll}\approx4\times10^{3}t_{\rm rh}$ (even longer in a galaxy), so generally these collisions can safely be neglected in the context of evaporation, but in some cases it can be as short as $t_{\rm coll}\approx20t_{\rm rh}$.
My reference for all of the above is Binney & Tremaine's Galactic Dynamics text (2 ed.). Chapter 7 goes through the evaporation of stellar systems, and many related processes, probably in more detail than you ever wanted to see. If you're unsatisfied with the above I suggest reading that, as I don't really want to produce a rehash of an entire book chapter here. If anything above requires clarification, though, please let me know!
