- #1
- 38
- 0
If a globular cluster is a dense(relatively speaking) collection of stars, why don't the stars attract each other gravitationally? Why don't they all collide to form a mega-star?
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Why aren't stars in a star cluster attracted to each other?
- Thread starter texasblitzem
- Start date
Answers and Replies
- #2
- 172
- 1
If a globular cluster is a dense(relatively speaking) collection of stars, why don't the stars attract each other gravitationally? Why don't they all collide to form a mega-star?
They do attract each other gravitationally. They don't all crash into one another for the same reason the Earth doesn't crash into the sun.
- #3
- 309
- 0
If they all collided and merged it wouldn't result in a mega star. It would be a black hole that would be created.
- #4
- 24
- 0
They do attract each other gravitationally. They don't all crash into one another for the same reason the Earth doesn't crash into the sun.
It is not that I'm a expert on astronomy. But isn't that the same as saying the cluster is rotating?
- #5
- 56
- 0
I'm not an expert either, but I was playing with someone's orbital simulation software once where you could define any number of planets and stars and set their initial velocities, etc., and you then see how they all interact. I noticed that it is very difficult to get interstellar objects to collide. They have to hit practically dead-on. Otherwise they just whip around each other.
Obviously, if that were to happen in real life there would be severe damage to both bodies in terms of atmospheric and other disturbances. But literal collisions are hard to do even when done on purpose.
Obviously, if that were to happen in real life there would be severe damage to both bodies in terms of atmospheric and other disturbances. But literal collisions are hard to do even when done on purpose.
- #6
- 65
- 0
^ this is true
Even when 2 galaxies merge, there are only a few of the billons of stars that actually collide, otherwise they just swing around each other.
Even when 2 galaxies merge, there are only a few of the billons of stars that actually collide, otherwise they just swing around each other.
- #7
Drakkith
Staff Emeritus
Science Advisor
- 21,138
- 4,980
It is not that I'm a expert on astronomy. But isn't that the same as saying the cluster is rotating?
Of course. The stars in the cluster are moving around each other, all orbiting the barycenter of the cluster.
- #8
- 2,102
- 468
The stars in the cluster are moving around each other, all orbiting the barycenter of the cluster.
That doesn't mean that the cluster is rotating because the orbital angular momentums doesn't have the same direction. Star clusters usually have no total angular momentum. In addition the trajectory of the stars doesn't have to be orbits.
- #9
Dotini
Gold Member
- 621
- 231
That doesn't mean that the cluster is rotating because the orbital angular momentums doesn't have the same direction. Star clusters usually have no total angular momentum. In addition the trajectory of the stars doesn't have to be orbits.
Is it being suggesting that stars in a globular cluster might be oscillating like bees in a hive?
Respectfully,
Steve
- #10
- 2,102
- 468
Is it being suggesting that stars in a globular cluster might be oscillating like bees in a hive?
That's a nice analogy but the motion of the stars is determined by the common gravitational field and not by aerodynamics.
- #11
Drakkith
Staff Emeritus
Science Advisor
- 21,138
- 4,980
That doesn't mean that the cluster is rotating because the orbital angular momentums doesn't have the same direction. Star clusters usually have no total angular momentum. In addition the trajectory of the stars doesn't have to be orbits.
I'm not saying every star will be moving in the same direction, only that they will be moving around in orbits. If the trajectory isn't in an orbital then the star will escape the cluster. Unless I'm misunderstanding the way orbit is being used here.
- #12
- 2,102
- 468
If the trajectory isn't in an orbital then the star will escape the cluster.
How about a chaotic path due to many close flybys?
- #13
- 984
- 174
As was pointed out earlier, cluster stars seldom collide with each other. Let's try to quantify this result.
Collision rate = (number density of targets) * (collision cross section) * (velocity)
For stars with radius r separated by distance a and traversing their separation distance in time T, the rate becomes
(1/T) * (r/a)2
One has to be careful of gravitational effects, since the stars will be pulled together before they collide. This changes the cross section from pi*r2 to
pi*r*(r + 2(G*M)/(v2))
for mass M and velocity at infinity v. Even then, it's easy to show that collisions are rare.
Collision rate = (number density of targets) * (collision cross section) * (velocity)
For stars with radius r separated by distance a and traversing their separation distance in time T, the rate becomes
(1/T) * (r/a)2
One has to be careful of gravitational effects, since the stars will be pulled together before they collide. This changes the cross section from pi*r2 to
pi*r*(r + 2(G*M)/(v2))
for mass M and velocity at infinity v. Even then, it's easy to show that collisions are rare.
- #14
Drakkith
Staff Emeritus
Science Advisor
- 21,138
- 4,980
How about a chaotic path due to many close flybys?
Isn't that an orbit still? It's still bound to the cluster.
- #15
- 984
- 174
One can estimate the likelihood of close flybys per orbit.
The average orbit speed ~ sqrt(G*M/A)
M = total mass, A = total size
For N stars, they are related to individual mass m and separation a as
M ~ N*m
A ~ N1/3*a
To be deflected significantly, a star must pass within about G*m/v2 of another star, and that is about A*(m/M) or A/N.
A star's mean free path, its travel distance between such deflections, is
1/((stars' number density) * (deflection cross section))
Working it out, 1/((N/A3) * (A/N)2) or N*A.
So a star will make on average about N orbits between sizable deflections.
The average orbit speed ~ sqrt(G*M/A)
M = total mass, A = total size
For N stars, they are related to individual mass m and separation a as
M ~ N*m
A ~ N1/3*a
To be deflected significantly, a star must pass within about G*m/v2 of another star, and that is about A*(m/M) or A/N.
A star's mean free path, its travel distance between such deflections, is
1/((stars' number density) * (deflection cross section))
Working it out, 1/((N/A3) * (A/N)2) or N*A.
So a star will make on average about N orbits between sizable deflections.
- #16
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
- 12,796
- 1,669
Maybe they're not the right type.
- #17
Drakkith
Staff Emeritus
Science Advisor
- 21,138
- 4,980
Maybe they're not the right type.
*Groan* Terrible joke!
- #18
- 172
- 1
How about a chaotic path due to many close flybys?
You're using a different definition of the word orbit than I'm used to.
- #19
DaveC426913
Gold Member
- 19,298
- 2,785
You're using a different definition of the word orbit than I'm used to.
It's a good question.
I am not sure if orbits are really stable in a cluster where the orbits have a wide distribution of planes.
For any given star, as long as it does not manage to acquire enough velocity to actually escape the cluster (and it's entirely possible some do) it could happily drift around on an highly unstable path.
- #20
Dotini
Gold Member
- 621
- 231
I am not sure if orbits are really stable in a cluster where the orbits have a wide distribution of planes.
For any given star, as long as it does not manage to acquire enough velocity to actually escape the cluster (and it's entirely possible some do) it could happily drift around on an highly unstable path.
Yup, maybe! I found this in wiki:
"The results of N-body simulations have shown that the stars can follow unusual paths through the cluster, often forming loops and often falling more directly toward the core than would a single star orbiting a central mass. In addition, due to interactions with other stars that result in an increase in velocity, some of the stars gain sufficient energy to escape the cluster."
http://en.wikipedia.org/wiki/Globular_cluster
Respectfully submitted,
Steve
- #21
- 2,102
- 468
Isn't that an orbit still? It's still bound to the cluster.
It is not periodic.
- #22
- 2,102
- 468
You're using a different definition of the word orbit than I'm used to.
My definition for orbit is a periodic path around a point in space. What is yours?
- #23
- 2,102
- 468
So a star will make on average about N orbits between sizable deflections.
On average. But in the core of the cluster close encounters are much more common. In case of a core collapse the number of such events increases dramatically (compared to the average) so that formations of double stars lead to a significant increase of the "temperature" of the whole cluster.
- #24
- 172
- 1
My definition for orbit is a periodic path around a point in space. What is yours?
Are hyperbolic orbits not orbits?
You mentioned chaos...
How about a chaotic path due to many close flybys?
If those chaotic paths are closed, are they not orbits?
- #25
- 2,102
- 468
Are hyperbolic orbits not orbits?
To my understanding of "orbit" they are not. But I checked that and found some sources that uses this word also for open trajectories. Do you have a source for an official definition of this word?
If those chaotic paths are closed, are they not orbits?
A closed chaotic path is very unlikely.
Share:
