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In other words, if a group of stars born within the same nebula, e.g. the Pleiades, are initially gravitationally bound - what causes them to finally get separated and proceed individually? Is it the influence of "external" mass? Or, are they now and we're they always, even at birth, having the escape velocity? (If the latter is true then I think they should not be considered truly bound from the beginning...)

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  • $\begingroup$ From en.wikipedia.org/wiki/Pleiades#Composition The cluster contains over 1,000 statistically confirmed members, a figure that excludes an unresolved likely further number of binary stars.[...] The total mass contained in the cluster is estimated to be about 800 solar masses and is dominated by fainter and redder stars. [...] The cluster contains many brown dwarfs [...] They may constitute up to 25% of the total population of the cluster, although they contribute less than 2% of the total mass. $\endgroup$
    – PM 2Ring
    Commented May 9, 2021 at 9:13

2 Answers 2

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The boundedness or otherwise of clusters remains to be established in most cases. The vast majority of clusters become unbound and disperse at a much younger age than the Pleiades. Or they may be born unbound as you suggest.

The stars in a cluster have a distribution of velocities and there will always be a tail of high speed stars that will be able to escape the gravitational potential in an "evaporation process", even in a cluster which is globally bound (i.e. the sum of kinetic and gravitational potential energies is negative).

It turns out that this isn't usually sufficient to explain the disappearance of clusters. A more effective disruptive mechanism are the Galactic tides. Stars that stray beyond a tidal radius can be stripped from the cluster. The reduction in the cluster's mass then reduces the tidal radius allowing more stars to escape in what eventually would become a runaway process. The tidal radius for a cluster near the Sun is about $5 M^{1/3}$ light years, where $M$ is the cluster mass expressed in units of the solar mass. For the Pleiades, $M \sim 1000 M_{\odot}$, so stars that stray beyond 50 light years can be lost.

In some cases, the evaporation process causes a cluster core collapse. The escaping stars take away kinetic energy and the remaining stars become more tightly bound. This is what has happened in most of the massive and long-lived globular clusters. Less massive open clusters like the Pleiades are much more susceptible to tidal dissipation.

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I’d like to add that globular clusters are usually much larger than open/irregular ones. Often we are talking about many thousands or even millions of stars in a globular cluster. That’s a lot of mass and, therefore, most of stars in such clusters are well bound by gravitational force. Of course when a cluster passes near a large mass (e.g. a galactic center or a massive dust cloud) stars are gravitationally attracted and some may live the cluster, actually sometimes a lot may leave forming a “leaving” group in the form of a huge tail of stars. Such tails (tail-like collections) of stars may be very large. Simply put, it all comes down to the magnitude of gravitational pull exerted on them. Large pulls are called tidal shocks and tidal waves as they are complicated, and the kinematics is not as easy as one may think it is. Anyway, large globular clusters under the force of gravity (there’s a lot of force there) are very tight in the center and can even form massive black holes in the center. Stars that are far away from the core of the cluster are more likely to leave under the influence of external gravitational forces at some point in time as the clusters rotate around the center of the galaxy and things change over time.

Open clusters are much smaller and are not so tightly bound by gravitation force. So, it is no surprise that stars in such clusters are much more susceptible to massive gravitational sources that can pull them out of their clusters.

Escape velocities are more likely to be attained later on (somewhat more or less haphazard) or, to be more precise, during the lifespan of the cluster, rather than in the very beginning, especially for large clusters. Smaller stars (in terms of mass) are of course more likely to leave their clusters, which is basic physics and it needs no further comment in my opinion.

The brightest globular cluster 47 Tucanae (too-kAH-nee) in the night sky very probably contains a huge black hole near the center which “sucks” in nearby objects and stars. Those won’t escape. enter image description here So, it's all about gravitational forces in our Galaxy that causes stars to get separated from their clusters.

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    $\begingroup$ Please remove the sentence about black holes sucking. They exert no more gravitational influence than any other mass. $\endgroup$
    – ProfRob
    Commented Sep 16, 2017 at 8:14
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    $\begingroup$ I you don't like my phrasing you can go ahead and vote down. The points I made are crystal clear. $\endgroup$
    – user18491
    Commented Sep 16, 2017 at 18:16
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    $\begingroup$ ok, so I look at Kizilstan et al (2017). Figs 1 and 2 show that the presence of a 2000 solar mass BH in 47 Tuc would reduce the density of the cluster core and accelerate the surrounding stars to higher speeds. arxiv.org/abs/1702.02149 $\endgroup$
    – ProfRob
    Commented Sep 16, 2017 at 18:48
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    $\begingroup$ In here I don't want to quote from the suggested arxiv.org/abs/1702.02149; I'm just gonna say this: Folks you should, please, read carefully. One more simple explanation: A massive black hole won't repel or expel close-by stars. Slightly further (small) stars are often get accelerated and get expelled by the presence of large masses and thus leave the cluster. Larger stars usually remain in the cluster. I repeat that globular clusters are more stable and have larger lifespans. The kinematics is complicated but there's no need to turn everything upside down. $\endgroup$
    – user18491
    Commented Sep 16, 2017 at 19:44
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    $\begingroup$ But I have to agree with a commenter that "sucks", although quoted, might mislead other readers. $\endgroup$
    – Alchimista
    Commented Oct 19, 2017 at 20:14

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