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## The galactic center

With regards to the STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER

Based on the following article:
http://iopscience.iop.org/article/10...37X/692/2/1075

It is stated at the end of chapter 6:

"The late-type star S111 is marginally unbound to the MBH, a result of its large radial velocity (−740 km s−1) at r = 1farcs48 which brings its total velocity up to a value ≈1σ above the local escape velocity."

If I understand it correctly:
Regards S111
Based on its High velocity and radius from the MBH, it should be at escape velocity.
However, it doesn't escape from the MBH.

So the question is:
Why it does not escape from the MBH, although its velocity is above the escape velocity?

2. "Marginally unbound" means that it is escaping, because it has a velocity greater than escape velocity. It's passing by on a parabolic orbit.

Grant Hutchison

3. Originally Posted by grant hutchison
"Marginally unbound" means that it is escaping, because it has a velocity greater than escape velocity. It's passing by on a parabolic orbit.
Or it might be in the process of being ejected from an initially closed orbit after interaction with other stars.

Or, apparently, it might have excess velocity relative to the black hole, but not relative to the whole galactic centre cluster:
Hence we see that S111 might be not bound to Sgr A*. This is in agreement with the findings by Gillessen et al. (2008) who conclude that S111’s orbit around Sgr A* might be hyperbolic. It is however possible that S111 is still bound to the GC star cluster if it (a) it follows a highly eccentric orbit and (b) we happen to observe it close to its pericentre. In this case, the stellar mass enclosed by the star’s orbit can be (together with Sgr A*) sufficient to bind the star to the cluster.

Trippe et al. 2008
(1MB pdf)
Grant Hutchison

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Originally Posted by grant hutchison
Or, apparently, it might have excess velocity relative to the black hole, but not relative to the whole galactic centre cluster:
Grant Hutchison
Thanks

Let's assume that it has excess velocity relative to the black hole, but not relative to the whole galactic centre cluster.

What does it mean?

1. Does it mean that S111 rotates around some virtual object which represents an equivalent mass of the whole galactic cluster?
2. Does it mean that the whole galactic center has a significant more mass then the black hole in order to hold that excess velocity?
Last edited by Dave Lee; 2016-May-07 at 04:22 AM.

5. Originally Posted by Dave Lee
1. Does it mean that S111 rotates around some virtual object which represents an equivalent mass of the whole galactic cluster?
It orbits around the centre of gravity of the galactic centre cluster. It would follow a non-elliptical orbit, because as it moved further outwards it would "see" more of the mass of the cluster. It would be a similar situation to a star orbiting within a globular cluster. But the CofG of the cluster is likely to be within the black hole, given how massive it is.

Originally Posted by Dave Lee
2. Does it mean that the whole galactic center has a significant more mass then the black hole in order to hold that excess velocity?
Well, that's a given, really - [black hole plus stars] masses more than just [black hole]. The question is whether the additional mass is enough to hold S111 in orbit, or if it will escape.

Grant Hutchison

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Originally Posted by grant hutchison
It orbits around the centre of gravity of the galactic centre cluster. It would follow a non-elliptical orbit, because as it moved further outwards it would "see" more of the mass of the cluster. It would be a similar situation to a star orbiting within a globular cluster. But the CofG of the cluster is likely to be within the black hole, given how massive it is.

Grant Hutchison
Thanks

With regards to the center of gravity of the galactic center cluster:
If I understand you correctly, it includes MSB + Mass/star in the galactic center which it can "see" (Inside the orbit cycle).

However:
1. How can we calculate that center of Gravity? Is it based on Newton law?
2. If S111 rotates around the center of gravity of the galactic center (which it can "see"), then could it be that any other star in the galactic center should also rotates around its own galactic center? So it should be SMB + Mass/stars inside its uniqe orbit cycle - (which it can "see")

7. Originally Posted by Dave Lee
With regards to the center of gravity of the galactic center cluster:
If I understand you correctly, it includes MSB + Mass/star in the galactic center which it can "see" (Inside the orbit cycle).
I'm not really sure what you mean by "see".

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Originally Posted by Noclevername
I'm not really sure what you mean by "see".
Inside its orbit cycle.
As stated by Grant Hutchison:
"It would follow a non-elliptical orbit, because as it moved further outwards it would "see" more of the mass of the cluster."
Last edited by Dave Lee; 2016-May-07 at 03:05 PM.

9. Originally Posted by Dave Lee
Inside its the orbit cycle.
No, AFAIK the only part that needs to be inside the orbit is the center of gravity. Most of the core stars are not inside its orbit.

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Originally Posted by Noclevername
No, AFAIK the only part that needs to be inside the orbit is the center of gravity. Most of the core stars are not inside its orbit.
What do you mean by: "Most of the core stars are not inside its orbit."

Do you agree that some of the core stars could be inside its orbit?
For example:
Let's assume that there is a star which is called S-one
This star orbits the MSB + 10 other stars.
So, can we claim that the S-one central gravity is based on the MSB + those 10 stars?

11. Originally Posted by Dave Lee
What do you mean by: "Most of the core stars are not inside its orbit."

Do you agree that some of the core stars could be inside its orbit?
Some could, sure.

Let's assume that there is a star which is called S-one
This star orbits the MSB + 10 other stars.
The orbit of S111 passes through the galactic core. There will possibly be many stars inside its orbital plane, and definitely millions of stars outside its orbital plane.

So, can we claim that the S-one central gravity is based on the MSB + those 10 stars?
Its "central gravity"? I don't know what you mean by that. It orbits the center of gravity of the entire core.

12. You're wondering about a gnat within a swarm of gnats. The center of gnat-mass depends at every moment where all the gnats are right then. And a moment later the center moves.

Because all the gnats are moving.

Why does this bug you?

13. I think i see your problem Dave Lee;

AFAIK indeed nothing outside its orbit is contributing to the gravitational force that acts on it. Because of it's velocity it will move outward, thus automatically the further out the more core stars will come inside its orbit, and thus contribute to the gravitational force that acts on it (however it is also moving away from the black hole, this one alone exerts slowly less).

I think that's why it would non-elliptical orbit; because the gravitational force acting on it is not constant. ( not accounting flyby's with other core stars ofcourse)

Am I right?

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You are basically trying to apply a static solution method to a dynamic problem. The force the star feels at any given moment is simply the sum of the forces due to all the bodies around it. For some simple situations you can treat this like a static centre of gravity around which the body orbits - but in this example this is unlikely to be the case. For this case the simple two body solution is only very approximate.

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Originally Posted by 01101001
The center of gnat-mass depends at every moment where all the gnats are right then. And a moment later the center moves.

Because all the gnats are moving.
However, moment is a relative time frame.

Based on all stars orbits/velocities it might takes days and even years in order to see significant change in their locations.
Therefore, the center of that "gnat-mass" is quite stable in a short range of time (Days - for sure).
So, when we claim "moments" it means days or even years.

Originally Posted by Shaula
The force the star feels at any given moment is simply the sum of the forces due to all the bodies around it.
I really like this simple explanation.

So, any star has its own center which at any given moment is simply the sum of the forces due to all the bodies around it.

16. I'd say the sum of the forces due to all the bodies inside its orbit.

Check out wiki on galaxy rotation curves;

https://en.m.wikipedia.org/wiki/Galaxy_rotation_curve

Notice how the curve quickly drops near the galaxy center? While there is an incredible amount of mass around, it's the inside that counts as they say

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Originally Posted by AFJ
I'd say the sum of the forces due to all the bodies inside its orbit.
That is true for spherically symmetric systems thanks to the shell theorem. However for any system deviating from that there is a residual effect. This is generally small asyour example shows - however when you are dealing with a busy, complex area like the galactic centre these residual effects can be quite significant especially when making statements about the stability or otherwise of an orbit. Which is why it is dangerous to completely discount them in this sort of situation.

18. Originally Posted by Shaula
That is true for spherically symmetric systems thanks to the shell theorem. However for any system deviating from that there is a residual effect. This is generally small asyour example shows - however when you are dealing with a busy, complex area like the galactic centre these residual effects can be quite significant especially when making statements about the stability or otherwise of an orbit. Which is why it is dangerous to completely discount them in this sort of situation.
Ah yes i see, it is a scale thing. Just for the sake of simplicity of the concept of mass inside / mass outside an orbit I theoretically discarded these effects, close star flyby's etc.

I think that is what the question is about to be honest, general shell theorem so to say. Accounting for residual side-effects might confuse things unnecesarely.

19. Originally Posted by Shaula
That is true for spherically symmetric systems thanks to the shell theorem. However for any system deviating from that there is a residual effect. This is generally small asyour example shows - however when you are dealing with a busy, complex area like the galactic centre these residual effects can be quite significant especially when making statements about the stability or otherwise of an orbit. Which is why it is dangerous to completely discount them in this sort of situation.
Even more specific than that, it's only true for objects in circular orbits around the center of mass of a spherically symmetric system. Roughly the case for a star in the galactic core, but generally not a valid simplification. At any given moment, in a spherically symmetric system, the total gravitation from masses outside a sphere around the center of the system with the current distance from the center of the system cancels out, leaving only the gravitation of masses within that sphere. But if the outer shell is not spherically symmetric, the gravitation does not cancel out: the net gravitation of a ring is toward the ring, not toward the point in the center of the ring. (This is why Niven's Ringworld was gravitationally unstable.)

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Originally Posted by cjameshuff
Even more specific than that, it's only true for objects in circular orbits around the center of mass of a spherically symmetric system. Roughly the case for a star in the galactic core, but generally not a valid simplification. At any given moment, in a spherically symmetric system, the total gravitation from masses outside a sphere around the center of the system with the current distance from the center of the system cancels out, leaving only the gravitation of masses within that sphere. But if the outer shell is not spherically symmetric, the gravitation does not cancel out: the net gravitation of a ring is toward the ring, not toward the point in the center of the ring. (This is why Niven's Ringworld was gravitationally unstable.)
Thanks

So if I understand it correctly - with regards to spiral galaxy;
If a star is located between the galactic center and the inward side of the spiral ring, "the net gravitation of a ring is toward the ring", therefore it should "move" toward the ring.
Is it correct?

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Originally Posted by Dave Lee
Thanks

So if I understand it correctly - with regards to spiral galaxy;
If a star is located between the galactic center and the inward side of the spiral ring, "the net gravitation of a ring is toward the ring", therefore it should "move" toward the ring.
Is it correct?
No. Because now you are ignoring the galactic centre. Cjameshuff was specifically talking about a non-symmetric ring - so the net gravitational pull of this ring is towards the ring - but the galaxy is not a ring. The disk, being not perfectly symmetric, exerts a net outwards component - however said component is basically 'what is left' after you remove the symmetric parts that cancel and even then is small compared to the inwards force exerted by the galactic centre.

Edit: This is basically why I gave the answer I did before. What you are doing here is trying to break a complex dynamic system down into a series of static approximations. You are taking an interacting mass of stars and saying "this bit is like a simple central object, this bit is like a ring, this bit is like...". Essentially you are treating each component as a perturbation of the first order central mass solution. This approach can give approximate answers but you rapidly get into an ever expanding series of extra components added to make it work better. In general there are not simple two body like answers for many body systems.
Last edited by Shaula; 2016-May-08 at 05:53 AM.

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Originally Posted by Shaula
No. Because now you are ignoring the galactic centre. Cjameshuff was specifically talking about a non-symmetric ring - so the net gravitational pull of this ring is towards the ring - but the galaxy is not a ring. The disk, being not perfectly symmetric, exerts a net outwards component - however said component is basically 'what is left' after you remove the symmetric parts that cancel and even then is small compared to the inwards force exerted by the galactic centre.
.

Thanks

However, let me introduce the following picture of sombrero-galaxy.

For me, it looks like a galactic center surrounded by a ring of dust.
If you claim that it is not a ring - then it is not a ring.

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Originally Posted by Dave Lee
Thanks

However, let me introduce the following picture of sombrero-galaxy.

For me, it looks like a galactic center surrounded by a ring of dust.
If you claim that it is not a ring - then it is not a ring.
I did not say that it was not a ring, nor did I say that a disk was not ring-like. What I said was that your statement:
If a star is located between the galactic center and the inward side of the spiral ring, "the net gravitation of a ring is toward the ring", therefore it should "move" toward the ring.
was not correct because the galactic core effects were larger than any effects due to asymmetry of the ring. So the star won't move towards the ring - it will orbit the centre. There may be a small effect on its orbit due to said ring if the orbit is non-circular and/or the ring asymmetric. There will also be effects due to all sorts of other things because in the crowded area near the galactic centre several of the simplifying assumptions used to do this kind of decomposition into independent two body solutions are questionable.

24. Originally Posted by cjameshuff
Even more specific than that, it's only true for objects in circular orbits around the center of mass of a spherically symmetric system. Roughly the case for a star in the galactic core, but generally not a valid simplification. At any given moment, in a spherically symmetric system, the total gravitation from masses outside a sphere around the center of the system with the current distance from the center of the system cancels out, leaving only the gravitation of masses within that sphere. But if the outer shell is not spherically symmetric, the gravitation does not cancel out: the net gravitation of a ring is toward the ring, not toward the point in the center of the ring. (This is why Niven's Ringworld was gravitationally unstable.)
As a small side note; could a Dyson Sphere ( solid shell of matter around a star version) be thought of as many rings joined together, eg gravitationally unstable?

25. Originally Posted by AFJ
As a small side note; could a Dyson Sphere ( solid shell of matter around a star version) be thought of as many rings joined together, eg gravitationally unstable?
The most realistic Dyson sphere concepts consist of independently orbiting objects. However, for one composed of multiple rings, a large number of rings approaching a spherically symmetric distribution of mass would reduce the instability as far as the interactions between the shell and the central mass go. The interactions between the rings would be a different matter...they would approach relatively closely to each other and the mass distribution they see would unavoidably be noticeably non-uniform, I think. However, they might be stabilized against each other.

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Originally Posted by Shaula
I did not say that it was not a ring, nor did I say that a disk was not ring-like. What I said was that your statement:

was not correct because the galactic core effects were larger than any effects due to asymmetry of the ring. So the star won't move towards the ring - it will orbit the centre. There may be a small effect on its orbit due to said ring if the orbit is non-circular and/or the ring asymmetric. There will also be effects due to all sorts of other things because in the crowded area near the galactic centre several of the simplifying assumptions used to do this kind of decomposition into independent two body solutions are questionable.
So, with regards the sombrero-galaxy, let me ask the following:

1. Do you agree that there is a ring in sombrero-galaxy?
2. Is this ring symmetric or asymmetric?
3. Assuming that the ring is symmetric, does it mean that if a star is located between the galactic center and the inward side of that ring, then due to "the net gravitation of a ring is toward the ring", therefore the star should "move" toward the ring?
Last edited by Dave Lee; 2016-May-08 at 01:36 PM.

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Originally Posted by Dave Lee
So, with regards the sombrero-galaxy, let me ask the following:

1. Do you agree that there is a ring in sombrero-galaxy?
I agree that there is a dusk lane in the Sombrero Galaxy. I don't agree that it necessarily represents a significant density enhancement over the rest of the disk.

Originally Posted by Dave Lee
2. Is this ring symmetric or asymmetric?
From what we can see essentially symmetric. But nothing is every going to be perfectly symmetric

Originally Posted by Dave Lee
3. Assuming that the ring is symmetric, does it mean that if a star is located between the galactic center and the inward side of that ring, then due to "the net gravitation of a ring is toward the ring", therefore the star should "move" toward the ring?
It depends on the orbit of the star, as cjameshuff said. For a star in an elliptical orbit, if the dust lane represents a significant over density (which it may well not do) then compared to a star in the same orbit in an identical galaxy without a ring there will be an extra component to the forces on it due to the ring. If the star is in a circular orbit then there will not be.

However, as I said, the gravity from the galactic core will be orders of magnitude greater than the residual effect from a ring unless there is a very large and very asymmetric one. So the star will not move towards the ring - it will move in an orbit around the centre that is modified by effects due to the ring.

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Originally Posted by Shaula
I agree that there is a dusk lane in the Sombrero Galaxy. I don't agree that it necessarily represents a significant density enhancement over the rest of the disk.

From what we can see essentially symmetric. But nothing is every going to be perfectly symmetric

It depends on the orbit of the star, as cjameshuff said. For a star in an elliptical orbit, if the dust lane represents a significant over density (which it may well not do) then compared to a star in the same orbit in an identical galaxy without a ring there will be an extra component to the forces on it due to the ring. If the star is in a circular orbit then there will not be.

However, as I said, the gravity from the galactic core will be orders of magnitude greater than the residual effect from a ring unless there is a very large and very asymmetric one. So the star will not move towards the ring - it will move in an orbit around the centre that is modified by effects due to the ring.
Thanks

That's clear.

29. Originally Posted by Dave Lee
So, with regards the sombrero-galaxy, let me ask the following:

1. Do you agree that there is a ring in sombrero-galaxy?
2. Is this ring symmetric or asymmetric?
3. Assuming that the ring is symmetric, does it mean that if a star is located between the galactic center and the inward side of that ring, then due to "the net gravitation of a ring is toward the ring", therefore the star should "move" toward the ring?
The net gravitation of a ring is toward the ring. A spiral galaxy is not a ring. You have to consider all the gravity sources, you can't just take one group of stars, compute the orbital motion due to them, and then move on to another group of stars.

The gravitational influence on a star in a galaxy is the sum of the gravitational influences of every other star, black hole, gas and dust cloud, dark matter particle, etc in that galaxy. These influences are in all different directions, and for some particular special cases, such as a circular orbit around the center of a spherically symmetric distribution of mass, you can simplify things by ignoring the portions that will cancel out, but outside of those special cases those simplifications are invalid. You can approximate the mass distribution by decomposing it into multiple simpler mass distributions, such as a small ellipsoid for the core, a big one for the dark matter halo, and a concentric series of rings for the disk, but you can't pick just one of these and compute the motion of stars.

30. Originally Posted by Shaula
It depends on the orbit of the star, as cjameshuff said. For a star in an elliptical orbit, if the dust lane represents a significant over density (which it may well not do) then compared to a star in the same orbit in an identical galaxy without a ring there will be an extra component to the forces on it due to the ring. If the star is in a circular orbit then there will not be.
I don't understand this. Why doesn't it, in a circular orbit? Isn't the net effect outward, towards the ring? Regardless of where the star is (except at the center)?
However, as I said, the gravity from the galactic core will be orders of magnitude greater than the residual effect from a ring unless there is a very large and very asymmetric one. So the star will not move towards the ring - it will move in an orbit around the centre that is modified by effects due to the ring.

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