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Dave Lee
2016-May-06, 08:34 PM
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/10.1088/0004-637X/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?

grant hutchison
2016-May-06, 08:57 PM
"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

grant hutchison
2016-May-06, 09:32 PM
"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 (https://arxiv.org/pdf/0810.1040.pdf) (1MB pdf)
Grant Hutchison

Dave Lee
2016-May-07, 04:10 AM
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?

grant hutchison
2016-May-07, 08:38 AM
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.


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

Dave Lee
2016-May-07, 02:32 PM
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")

Noclevername
2016-May-07, 02:53 PM
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".

Dave Lee
2016-May-07, 03:01 PM
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."

Noclevername
2016-May-07, 03:05 PM
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.

Dave Lee
2016-May-07, 03:18 PM
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?

Noclevername
2016-May-07, 03:27 PM
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.

01101001
2016-May-07, 03:30 PM
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?

AFJ
2016-May-07, 03:45 PM
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?

Shaula
2016-May-07, 03:46 PM
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.

Dave Lee
2016-May-07, 05:22 PM
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.


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.

AFJ
2016-May-07, 05:32 PM
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 :D

Shaula
2016-May-07, 05:47 PM
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.

AFJ
2016-May-07, 05:54 PM
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.

cjameshuff
2016-May-07, 06:18 PM
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.)

Dave Lee
2016-May-08, 05:19 AM
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?

Shaula
2016-May-08, 05:49 AM
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.

Dave Lee
2016-May-08, 08:12 AM
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

I do appriciate your answer.
However, let me introduce the following picture of sombrero-galaxy.

http://pics-about-space.com/sombrero-galaxy-dust-ring?p=1#img6000360227297524069

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.

Shaula
2016-May-08, 08:23 AM
Thanks

I do appriciate your answer.
However, let me introduce the following picture of sombrero-galaxy.

http://pics-about-space.com/sombrero-galaxy-dust-ring?p=1#img6000360227297524069

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.

AFJ
2016-May-08, 08:28 AM
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?

cjameshuff
2016-May-08, 11:17 AM
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.

Dave Lee
2016-May-08, 01:33 PM
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.

Sorry, I'm not sure that I fully understand your answer.
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?

Shaula
2016-May-08, 01:44 PM
Sorry, I'm not sure that I fully understand your answer.
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.


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


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.

Dave Lee
2016-May-08, 01:52 PM
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.

cjameshuff
2016-May-08, 02:09 PM
Sorry, I'm not sure that I fully understand your answer.
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.

grapes
2016-May-08, 02:25 PM
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.

AFJ
2016-May-08, 03:07 PM
Probably assuming the orbital plane of the star is exactly lined up with the galactic plane / ring plane?

cjameshuff
2016-May-08, 03:31 PM
Probably assuming the orbital plane of the star is exactly lined up with the galactic plane / ring plane?

Even in that case, the gravitation of a ring will oppose the gravitation of the core for objects in circular orbits between them. Circular orbits are still possible, but their velocities will be lower than you would estimate by only considering the mass "enclosed by the orbit".

AFJ
2016-May-08, 04:12 PM
Thinking about that;

Woulnd't a (perfect) ring rule out an elliptical orbit? Say it's either a perfect circular orbit OR non-elliptical orbit?

cjameshuff
2016-May-08, 04:37 PM
Thinking about that;

Woulnd't a (perfect) ring rule out an elliptical orbit? Say it's either a perfect circular orbit OR non-elliptical orbit?

Anything that's not a spherically symmetric mass distribution entirely enclosed by the orbit rules out perfect elliptical orbits, and once you include tides and relativistic effects, you don't even get them in that case. Realistically, there are only approximately-elliptical orbits.

Fiery Phoenix
2016-May-08, 06:52 PM
Anything that's not a spherically symmetric mass distribution entirely enclosed by the orbit rules out perfect elliptical orbits, and once you include tides and relativistic effects, you don't even get them in that case. Realistically, there are only approximately-elliptical orbits.
I think it was Ken G who once said that the concept of circular and elliptical orbits is nothing but abstract models we employ to convey our understanding of phenomena.

chornedsnorkack
2016-May-08, 08:36 PM
The orbit is also elliptical for a body orbiting inside a constant density.

Noclevername
2016-May-08, 10:11 PM
I think it was Ken G who once said that the concept of circular and elliptical orbits is nothing but abstract models we employ to convey our understanding of phenomena.

Yes, but he says that about everything. So much that it's hard to filter out the useful applications of that from the nonuseful.

Dave Lee
2016-May-09, 01:51 PM
The net gravitation of a ring is toward the ring.

That is clear.
However, why is it so important to verify the orbit shape of a star?
If the net gravitation is towards the ring, then what is the difference if the star has an elliptical or circular orbit?
In any case it should obey to the gravitation power.

cjameshuff
2016-May-10, 01:01 AM
That is clear.
However, why is it so important to verify the orbit shape of a star?
If the net gravitation is towards the ring, then what is the difference if the star has an elliptical or circular orbit?
In any case it should obey to the gravitation power.

The words I bolded, which you then ignored, represent an important detail that you keep missing. That's why I bolded them. The gravitation of the disk of a galaxy, which can be roughly approximated as a ring or series of rings, is only one component of the overall gravitation influencing a star, and isn't even the most significant component.

All other stars, every atom of gas and dust, and every particle of dark matter in a galaxy contributes to the total gravitational force on a star. Many of these influences cancel out, and the overall gravitation depends on the position of the star within the galaxy. That position changes in different ways for circular and non-circular orbits: for a radially symmetric distribution, circular orbits are possible because distance from the center/axis is constant, but they may be highly unstable. Certain specific distributions of mass produce elliptical orbits. Real-world systems never exactly match these distributions, but star systems are close enough for most purposes. When you are dealing with spiral galaxies, the resulting orbit only somewhat resembles an ellipse.

Dave Lee
2016-May-10, 04:07 AM
Thanks


The words I bolded, which you then ignored, represent an important detail that you keep missing. That's why I bolded them. The gravitation of the disk of a galaxy, which can be roughly approximated as a ring or series of rings, is only one component of the overall gravitation influencing a star, and isn't even the most significant component.

All other stars, every atom of gas and dust, and every particle of dark matter in a galaxy contributes to the total gravitational force on a star. Many of these influences cancel out, and the overall gravitation depends on the position of the star within the galaxy. That position changes in different ways for circular and non-circular orbits: for a radially symmetric distribution, circular orbits are possible because distance from the center/axis is constant, but they may be highly unstable. Certain specific distributions of mass produce elliptical orbits. Real-world systems never exactly match these distributions, but star systems are close enough for most purposes. When you are dealing with spiral galaxies, the resulting orbit only somewhat resembles an ellipse.

Let's consider that we are dealing with a pure symmetric disc ring with Millions or even billions Sun mass. (Similar to Sombrero- Galaxy).
Now, if a star which orbits the galactic center penetrates the gravity force of that ring:
So, if the star is in circular orbit (around the galactic center) at the same plane as the ring plane – Then, in this case I assume that it should move towards the ring and actually join all the other stars in that ring.
However, what might be the impact of the following:
1. The star is in circular orbit, but with a shift from the ring plane.
2. The star is in elliptical orbit.
Does it mean that the star might move back to the galactic center, or be ejected out of the ring?

chornedsnorkack
2016-May-10, 09:10 PM
Let's consider that we are dealing with a pure symmetric disc ring with Millions or even billions Sun mass. (Similar to Sombrero- Galaxy).
Now, if a star which orbits the galactic center penetrates the gravity force of that ring:
So, if the star is in circular orbit (around the galactic center) at the same plane as the ring plane – Then, in this case I assume that it should move towards the ring and actually join all the other stars in that ring.
Suppose a light test particle is moving inside a spherically symmetric, empty shell.
There is no gravity inside a spherical shell. Therefore the test particle cannot follow a circular orbit, or any other curved orbit - if it moves, it does so on a straight line, whether a chord or diametre.
When the test particle passes the shell, it does so at a nonzero speed. It therefore cannot "join" the shell - it passes through the shell. Outside the shell it is attracted to the shell, and follows an inward curving trajectory (unless it is decelerating at a radial direction) , whether an arc of a hyperbole, a parabole or an ellipse.
If it was an arc of ellipse, it will intersect the shell again, and again follow a straight chord inside the shell.

cjameshuff
2016-May-11, 12:36 AM
Thanks



Let's consider that we are dealing with a pure symmetric disc ring with Millions or even billions Sun mass. (Similar to Sombrero- Galaxy).
Now, if a star which orbits the galactic center penetrates the gravity force of that ring:
So, if the star is in circular orbit (around the galactic center) at the same plane as the ring plane – Then, in this case I assume that it should move towards the ring and actually join all the other stars in that ring.
However, what might be the impact of the following:
1. The star is in circular orbit, but with a shift from the ring plane.
2. The star is in elliptical orbit.
Does it mean that the star might move back to the galactic center, or be ejected out of the ring?

You have once again taken one subset of the system and jumped off to conclusions based on that subset. Gravity doesn't work that way. Stars don't switch back and forth between orbiting object A, object B, or some empty point C, they orbit within the gravitational field of the galaxy as a whole. Add up all the gravitational influences, then you can talk about the object's motion.

For radially symmetric mass distributions, all points at distance R from the axis have the same magnitude of gravitation, with the direction being toward the axis. Circular orbits are always possible with such distributions. However, the magnitude of the gravitation will change with different distances, and it will change in different ways with different mass distributions, and orbits that cross a range of distances will have a shape that depends on the way the gravitation changes with distance. In the more complex case of the sun's orbit through the Milky Way, it's a complex shape that forms a rosette figure viewed from above, while bobbing up and down through the disk. For certain rough approximations you can consider it to be an ellipse, but that is never anything other than an approximation, it is not some underlying fundamental principle.

You keep asking variations of the same question reworded in innumerable different ways, and then going on with the same mistaken reasoning you had before. You seem to be hunting for some specific answer (apparently something along the lines of "objects follow elliptical orbits around virtual centers of mass"). Someone might eventually give you the answer you're looking for, or something close enough to be interpreted as it, but that's not going to make it the right answer, and getting it isn't going to get you any closer to understanding gravity.

chornedsnorkack
2016-May-11, 10:49 AM
In the more complex case of the sun's orbit through the Milky Way, it's a complex shape that forms a rosette figure viewed from above, while bobbing up and down through the disk. For certain rough approximations you can consider it to be an ellipse, but that is never anything other than an approximation, it is not some underlying fundamental principle.

That's also description of, for example, orbit of Moon.

Dave Lee
2016-May-11, 02:27 PM
Sorry for asking too many questions about gravity power of ring.
Actually, when I have started this thread I had no idea about this issue.
I have got the first message about the ring – in pg. 19.
As stated:

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.)
From that moment, I'm fascinating form the power of ring.
AJF have asked a question about circular/elliptical orbit in a ring – in pg. 33.
As stated:

Thinking about that;
Wouldn't a (perfect) ring rule out an elliptical orbit? Say it's either a perfect circular orbit OR non-elliptical orbit?

That was the base for my last question.

However, this question was quite hypothetical as I have asked about some theoretical and unrealistic ring.
Never the less, while I was waiting for an answer, I have found the following information about ring galaxy:
https://en.wikipedia.org/wiki/Ring_galaxy#/media/File:Hoag%27s_object.jpg
It almost seems to me as spiral galaxy – and it's Huge:
"the surrounding ring has an inner 28″ diameter of 75±3 kly (24.8±1.1 kpc) and an outer 45″ diameter of121±4 kly (39.9±1.7 kpc), which is slightly larger than the Milky Way Galaxy"
So, now it might not be just hypothetical question.
In any case, I do not want to promote any sort of idea and I do not want to upset anyone. Hence, I have no more questions about the ring.
However, I might have few more questions about stars in the galactic center.

chornedsnorkack
2016-May-11, 06:14 PM
Sorry for asking too many questions about gravity power of ring.
Actually, when I have started this thread I had no idea about this issue.
I have got the first message about the ring – in pg. 19.

From that moment, I'm fascinating form the power of ring.
AJF have asked a question about circular/elliptical orbit in a ring – in pg. 33.

However, this question was quite hypothetical as I have asked about some theoretical and unrealistic ring.
Hence, I have no more questions about the ring.
However, I might have few more questions about stars in the galactic center.
However, would you like some more exposition about gravity and orbits of various mass distributions?
A point mass attracts test masses to it, inwards to centre - with force decreasing outwards as inverse square.
An empty spherical shell has the same gravity for everything outside the shell as a point mass at the centre. And inside the shell, there is no gravity.
A sphere of uniform density has the same gravity for everything outside the sphere as a spherical shell or a point mass. And inside the sphere, the gravity increases outwards.
Inside a ring, the gravity is towards the ring, outwards from the centre.
So, do you want an exposition about what then happens to orbits?

Dave Lee
2016-May-11, 07:08 PM
However, would you like some more exposition about gravity and orbits of various mass distributions?
A point mass attracts test masses to it, inwards to centre - with force decreasing outwards as inverse square.
An empty spherical shell has the same gravity for everything outside the shell as a point mass at the centre. And inside the shell, there is no gravity.
A sphere of uniform density has the same gravity for everything outside the sphere as a spherical shell or a point mass. And inside the sphere, the gravity increases outwards.
Inside a ring, the gravity is towards the ring, outwards from the centre.
So, do you want an exposition about what then happens to orbits?

Thanks

At this phase, I have no more questions about the ring.

Dave Lee
2016-May-11, 07:27 PM
Based on the following stars orbit diagram around the galactic center:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f16
It seems that most of the stars has ellipse shape orbit.
Only for S13 it looks like a circular orbit.
So, is there any possibility that some of those orbits are also circular?
Could it be that as we don't face on their orbit plane, we see it as ellipse but in reality it is circular?

mkline55
2016-May-11, 07:57 PM
So, do you want an exposition about what then happens to orbits?

Dave Lee may not want an exposition, but I'm curious about the orbital path. I don't know if this is the right approach, but here's where I'd start. Take a cross-section of the ring and analyze it from there. Draw a circle to represent the cross-section. Put a dot in the center to represent the center of gravity. But remember that this is part of a ring, so the center of gravity will actually be offset either toward the center or toward the edge. Consider that at the outer edge of the ring all of the ring's mass is inward, and at the inner edge, some of the mass is inwards and some outwards. So move the dot appropriately. Since the ring is symetrical, if you then expand the cross-section into a ring again, the dot becomes a circle. Something orbiting the ring fairly close to the ring's surface and in a plane that is not exactly perpendicular to the ring's plane, might describe a path that circles the ring and simultaneously procedes around the ring.

Cougar
2016-May-11, 08:34 PM
- An empty spherical shell has the same gravity for everything outside the shell as a point mass at the centre. And inside the shell, there is no gravity.

- Inside a ring, the gravity is towards the ring, outwards from the centre.


These two points seem to be mutually contradictory. How do you (and cjameshuff) reconcile this?

Hornblower
2016-May-11, 10:31 PM
Based on the following stars orbit diagram around the galactic center:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f16
It seems that most of the stars has ellipse shape orbit.
Only for S13 it looks like a circular orbit.
So, is there any possibility that some of those orbits are also circular?
Could it be that as we don't face on their orbit plane, we see it as ellipse but in reality it is circular?
S13 has an eccentricity of about 0.4, so it is distinctly elliptical, but it appears to have its major axis tipped toward us so that it projects as nearly circular. All of the others are more eccentric, so none of them will be circular.

Hornblower
2016-May-11, 10:33 PM
These two points seem to be mutually contradictory. How do you (and cjameshuff) reconcile this?
There is no contradiction. A spherically uniform shell is the one form that gives total cancellation inside. Others, including a ring, do not fully cancel.

pzkpfw
2016-May-11, 11:19 PM
I may have missed something, but I had the impression two meanings of "in" were in use. In one, "in the ring" meaning among the stuff that makes up the ring, and in another usage, "in the ring" meaning located, say, halfway between the stuff of the ring and the centre.
(i.e. positionally, like the difference between being inside the gold that makes up a wedding ring, and being in the bone of the finger the ring is on.)

?

Dave Lee
2016-May-12, 03:57 AM
S13 has an eccentricity of about 0.4, so it is distinctly elliptical, but it appears to have its major axis tipped toward us so that it projects as nearly circular. All of the others are more eccentric, so none of them will be circular.

What about the rotation plane for each star?
Does this diagram represent the orbits as we see it from Earth?

If so, that is a severe mistake.

Those stars are moving in a 3D. (It is not a disc).
Therefore, it is quite logical that each one of them should have its own rotation plane.

Actually, if we look at circular orbit from the top, it will be circular.
However, if we look at the same orbit with some offset from the top, we should see it as elliptical orbit.
If the offset is close to 90 degree, then it might look like the path of S27, S14 and even S6.

Hence, there is good chance that the orbits of all/most of the stars are circulars, but we see it from our location as ellipticals

So, we must verify the rotation plane for each star. Based on that, we can draw the correct shape of each orbit.
However, the main question will be - How can we estimate the exact rotation plane for each star?

chornedsnorkack
2016-May-12, 05:07 AM
If we look at a planar ellipse at any angle, it is still seen as an ellipse.
If we look at a circle at an angle, it is seen as an ellipse, whose central body is at the centre of the ellipse - not at a focus.
If we look at an ellipse whose major axis is towards us, at an angle, then it can be seen as a circle. But the central body is then not at the centre of the circle.
In a general case of an ellipse seen at an angle to its plane and its axes at an angle to us, it is still an ellipse as said, but with central body somewhere off its axes.

Dave Lee
2016-May-12, 05:33 AM
O.K.

If I understand it correctly, you claim that we don't need to verify the orbit plane due to the location of the central body.
So, as the location of the MBH doesn't meet a circular path, then we take it for granted that it is ellipse.
Hence, we didn't even try to verify the correct orbit plane for each star.

Is it correct?

Shaula
2016-May-12, 07:10 AM
Hence, there is good chance that the orbits of all/most of the stars are circulars, but we see it from our location as ellipticals
That simply does not follow. You have gone from "this is a possible effect" to assessing it as a probable cause for what is seen. In order to have made that call I would have expected you to have analysed the stellar distribution in the area and worked out whether what we see in terms of horizontal and vertical densities of stars is consistent with the required distribution of circular orbits to reproduce what we see. You have basically said something akin to "it is possible that at night you could mistake a starling for a blackbird - therefore there is a good chance that all blackbirds are actually starlings". You also ignore that the converse (that any observed circular orbit is actually an ellipse) is just as valid a leap to make.

Dave Lee
2016-May-12, 07:13 AM
Please try to give direct answer:

Did we verify the correct orbit plane for each star?
Yes or No?

Shaula
2016-May-12, 07:23 AM
Please try to give direct answer:

Did we try to verify the correct orbit plane for each star?
Yes or No?
Irrelevant to the point that you are making a series of unjustified logical leaps.

If you need an answer to your question you should really refer to the paper you took the orbital plots from. In it they are pretty clearly using a 6DOF orbital fit model into which they put both positional and velocity estimates - they even discuss how they can get to those 6DOF from the available observations. Given the way orbits work then this implicitly determines the orbital plane. They then verified this using a Markov chain model to check they were not looking at a degenerate system. So your answer is Yes. Possibly Yes^2.

Dave Lee
2016-May-12, 07:33 AM
Irrelevant to the point that you are making a series of unjustified logical leaps.

If you need an answer to your question you should really refer to the paper you took the orbital plots from. In it they are pretty clearly using a 6DOF orbital fit model into which they put both positional and velocity estimates - they even discuss how they can get to those 6DOF from the available observations. Given the way orbits work then this implicitly determines the orbital plane. They then verified this using a Markov chain model to check they were not looking at a degenerate system. So your answer is Yes. Possibly Yes^2.

Thanks

So the answer is Yes.
You even specify that they could determines the orbital plane:
"Given the way orbits work then this implicitly determines the orbital plane."

So, let me ask the following:

1. What does the diagram represent? Is it what we see from our location, or does it represnt the real orbit plane of each star?
2. Do you mean that for each star they have verified the correct plane and based on that they have set the diagram? So they actually set a 3D rotation on 2D while each orbit represents a central top view.
3. Why it is not specify in the article what is the shift between each orbit plane to our location view?

Dave Lee
2016-May-12, 09:54 AM
If you need an answer to your question you should really refer to the paper you took the orbital plots from.
I have tried to verify the orbital plane in that article. The word "plane" is located in only 5 places in the whole article:
The first one is related for the MBH motion:
"Clearly, the motion of Sgr A* perpendicular to the galactic plane is very small as expected."
All the others (4 times) are related to figure 19.
"Figure 19. Orientation of the orbital planes of those S-stars for which we were able to determine orbits."
However, in figure 19 we only see few of the total stars.
I was quite surprised to discover that most of those stars orbits around some virtual center – which is not the MBH?
How could it be? Why they do not orbit around the MSB (which is located at 0)?
In that figure they also discuss about S5 and S31:
"The orbital plane of S5 is also consistent with the disk given its distance of 18◦. However, the lower brightness (mK = 15.2) of the star and the higher eccentricity (e > 0.8) of the orbit make it unlikely that S5 is a true disk member. The next closest star to the disk beyond the six disk stars and S5 is S31 with an angular distance of 27◦. We also note that the orbital solutions for S96 and S97 derived from marginal accelerations are consistent with the disk hypothesis. Therefore, we are confident in these orbits, too."
Based on this information, does it mean that they have verified the correct orbit plane for S5 and S13?.
Did they adjust it in the following diagram?
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f16
In any case, what about orbit planes of all the other stars in this diagram?
For example – S1, S2, S6 S13 and all the others?
How do we know for sure that this diagram represents the real orbit plane for each star?

01101001
2016-May-12, 09:54 AM
UCLA Galactic Center Group: Animations (http://www.galacticcenter.astro.ucla.edu/animations.html)


The movie shows a 3-dimensional visualization of the stellar orbits in the Galactic center based on data obtained by the W. M. Keck Telescopes between 1995 and 2012. Stars with the best determined orbits are shown with full ellipses and trails behind each star span ~15-20 years.

LaurieAG
2016-May-12, 11:42 AM
Dave Lee may not want an exposition, but I'm curious about the orbital path. I don't know if this is the right approach, but here's where I'd start. Take a cross-section of the ring and analyze it from there. Draw a circle to represent the cross-section. Put a dot in the center to represent the center of gravity. But remember that this is part of a ring, so the center of gravity will actually be offset either toward the center or toward the edge. Consider that at the outer edge of the ring all of the ring's mass is inward, and at the inner edge, some of the mass is inwards and some outwards. So move the dot appropriately. Since the ring is symetrical, if you then expand the cross-section into a ring again, the dot becomes a circle. Something orbiting the ring fairly close to the ring's surface and in a plane that is not exactly perpendicular to the ring's plane, might describe a path that circles the ring and simultaneously procedes around the ring.

If you just look at the optical parts of rotating sources (pure SR) you can use the constant speed of light and basic geometry to give you a solid platform for an observation model based on rotating sources (for 11-16 year cycles as well as billions of year cycles).

The image below shows the photon paths present, at the time of an observation, of the photons emitted and in transit between 2 identical sources that are rotating around a common center of mass and observers, at various angles to the plane of rotation, who are stationary with respect to the center of mass of the rotating sources. The distance between the observer and the sources cycle start locations 1,0 and 3,0 always equals \frac {2 \pi r c}{v} (where v is the angular velocity and r the radius of rotation) after one complete rotation regardless of the angle of the observer to the plane of rotation of the sources. The observer may or may not be rotating around the common center of mass of the rotating sources to be considered stationary with respect to that center of mass. The scale shown in the images below is for v = 0.99c.

http://home.onthenet.com.au/~laurieag/PoolfrogsMisc_files/Rotations%20shift%20three.jpg
Here's how to plot a 45 degree angle path over 4 quarters of rotation. You know the start and end points after each quarter, join them together and push them along to the observer.
http://home.onthenet.com.au/~laurieag/PoolfrogsMisc_files/image019.jpg

chornedsnorkack
2016-May-12, 11:58 AM
O.K.

If I understand it correctly, you claim that we don't need to verify the orbit plane due to the location of the central body.
So, as the location of the MBH doesn't meet a circular path, then we take it for granted that it is ellipse.
Hence, we didn't even try to verify the correct orbit plane for each star.

Is it correct?
We may not so much as verify the correct orbital plane as deduce its inclination.
And in addition to the line of orbit in sky, we also can follow the orbital speed.

Swift
2016-May-12, 12:59 PM
At three pages I consider this now an extended discussion, and so the thread is moved from Q&A to Astronomy. Carry on...

01101001
2016-May-12, 02:05 PM
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.


I stand corrected. This professional science guy says: bees.

ESO: Unprecedented 16-Year Long Study Tracks Stars Orbiting Milky Way Black Hole (https://www.eso.org/public/usa/news/eso0846/)


For the first time the number of known stellar orbits is now large enough to look for common properties among them. "The stars in the innermost region are in random orbits, like a swarm of bees," says Gillessen. "However, further out, six of the 28 stars orbit the black hole in a disc. In this respect the new study has also confirmed explicitly earlier work in which the disc had been found, but only in a statistical sense. Ordered motion outside the central light-month, randomly oriented orbits inside – that's how the dynamics of the young stars in the Galactic Centre are best described."

I always was weak in entomological etymology.

Anyone fear this 6-star disc will result in a trip down another garden path of other stars being inside or outside of the disc?

Don't get stung!

Dave Lee
2016-May-12, 03:31 PM
I stand corrected. This professional science guy says: bees.
ESO: Unprecedented 16-Year Long Study Tracks Stars Orbiting Milky Way Black Hole (https://www.eso.org/public/usa/news/eso0846/)
I always was weak in entomological etymology.
Anyone fear this 6-star disc will result in a trip down another garden path of other stars being inside or outside of the disc?
Don't get stung!

With regards to that article, it is stated:

"One particular star, known as S2, orbits the Milky Way's centre so fast that it completed one full revolution within the 16-year period of the study. Observing one complete orbit of S2 has been a crucial contribution to the high accuracy reached and to understanding this region. Yet the mystery still remains as to how these young stars came to be in the orbits they are observed to be in today. They are much too young to have migrated far, but it seems even more improbable that they formed in their current orbits where the tidal forces of the black hole act. Excitingly, future observations are already being planned to test several theoretical models that try to solve this riddle."

Hence, let's focus on S2 as it is quite important star.
In the other article the word "S2" repeats 88 times. (Quite significant with related to all the other stars.)
It is also stated:
"Our data set covers the pericenter passages of several stars. Particularly important to our analysis is the one of the star S2. The star is one of the brightest in the sample, and we observed a full orbit (see Figure 13). In 2002 S2 passed its pericenter, thus changing quickly in velocity throughout a period of a few months. These data are particularly useful for constraining the potential of the MBH. "
"Figure 13. Top: S2 orbital data plotted in the combined coordinate system and fitted with a Keplerian model in which the velocity of the central point mass and its position were free-fit parameters. "
So, In Figure 13 we see clearly the elliptical path of S2 as we see it from our location.
However, they don't say even one word about the real orbit plane of S2.
Therefore, technically, it could have a perfect circular orbit, while we see it as elliptical - as we might not have a direct top view location.
Please look at the following diagram rotation of S2:
https://en.wikipedia.org/wiki/Sagittarius_A*#/media/File:Galactic_centre_orbits.svg
The MBH is located just at the bottom side of this elliptical path.
Based on this diagram the distance from the MBH to the bottom side of the S2 elliptical orbit is about one square, while the distance from the MBH to the top side is more than 8 squares.
Therefore, I would expect to see significant change in R0. (Distance to MBH)
However, in "Table 4. Results for the Central Potential from Orbital Fitting":
There are six measurements of S2. The value of R0 is in between 8.80 Kpc (Measurement nu. 5) to 6.63Kpc (measurement nu. 6).
So, we really don't have any information about the top/bottom sides of S2 elliptical rotation path.
Can someone explain it?
How could it be that there is a relatively minor R0 change in the table (6.63 / 8.80), while there is a significant change in the diagram (1 / 8)?
Why they don't try to verify the real orbital plane of S2?

Shaula
2016-May-12, 04:17 PM
So, let me ask the following:
1. What does the diagram represent? Is it what we see from our location, or does it represnt the real orbit plane of each star?
2. Do you mean that for each star they have verified the correct plane and based on that they have set the diagram? So they actually set a 3D rotation on 2D while each orbit represents a central top view.
3. Why it is not specify in the article what is the shift between each orbit plane to our location view?
1) Read the article. It says exactly what the diagram represents in the caption for it. "Stellar orbits of the stars in the central arcsecond for which we were able to determine orbits. In this illustrative figure, the coordinate system was chosen such that Sgr A* is at rest."
2) Read the article. It says exactly what the state vector representation of each orbit was.
3) Why would they? They have defined a co-ordinate system and used it. Why should they then provide transformations into arbitrary systems?


I have tried to verify the orbital plane in that article. The word "plane" is located in only 5 places in the whole article:
So? If you read a paper on domestic cats you probably won't find many examples of the word "Kitty". Use their terminology and read the paper - it explains the motion models used and how they turned them into orbits. You seem to have an image in your head about how they 'should' do it - I'd suggest you try to understand how they did do it rather than search fruitlessly for your own method in their paper.


I was quite surprised to discover that most of those stars orbits around some virtual center – which is not the MBH?
Why? People have been saying throughout this thread that the orbits are affected by more than just one object.


How could it be? Why they do not orbit around the MSB (which is located at 0)
Because there are other masses in the area. Can you rephrase your question? I must not be understanding it. You seem to be surprised that a body orbiting a cluster of objects doesn't orbit one of them.


How do we know for sure that this diagram represents the real orbit plane for each star?
Because ... I already told you. I pointed out how they had done this. I pointed out the models they had used. If you are confused then I'd suggest you point to sections of the paper you are struggling with - because the answers to your questions are all in it so far.


Therefore, technically, it could have a perfect circular orbit, while we see it as elliptical - as we do not have a direct top view location.
No. Because it would not be consistent with the data. Look at the 6DOF stuff they did. A circular orbit would simply not reproduce the results.


Why they don't try to verify the real orbital plane of S2?
They implicitly did via the orbit modelling. They just didn't use your non-standard approach.

Hornblower
2016-May-12, 05:05 PM
1) Read the article. It says exactly what the diagram represents in the caption for it. "Stellar orbits of the stars in the central arcsecond for which we were able to determine orbits. In this illustrative figure, the coordinate system was chosen such that Sgr A* is at rest."
2) Read the article. It says exactly what the state vector representation of each orbit was.
3) Why would they? They have defined a co-ordinate system and used it. Why should they then provide transformations into arbitrary systems?


So? If you read a paper on domestic cats you probably won't find many examples of the word "Kitty". Use their terminology and read the paper - it explains the motion models used and how they turned them into orbits. You seem to have an image in your head about how they 'should' do it - I'd suggest you try to understand how they did do it rather than search fruitlessly for your own method in their paper.


Why? People have been saying throughout this thread that the orbits are affected by more than just one object.


Because there are other masses in the area. Can you rephrase your question? I must not be understanding it. You seem to be surprised that a body orbiting a cluster of objects doesn't orbit one of them.


Because ... I already told you. I pointed out how they had done this. I pointed out the models they had used. If you are confused then I'd suggest you point to sections of the paper you are struggling with - because the answers to your questions are all in it so far.


No. Because it would not be consistent with the data. Look at the 6DOF stuff they did. A circular orbit would simply not reproduce the results.


They implicitly did via the orbit modelling. They just didn't use your non-standard approach.
Let me add that in the region shown here, the total mass of the stars is a tiny fraction of that of the central black hole. Thus the orbits will be close approaches to Kepler orbits around the central mass, except in the rare cases of extremely close encounters with one another.

A circular orbit nearly edge on will look like an ellipse with the black hole at its geometric center, that is, the midpoint of its major axis. None of these orbits come even close. S13's true ellipse would have a minor axis about 90% of the length of the major axis, so a moderate inclination with an orientation that foreshortens the major axis but not the minor could make it look circular from this viewpoint.

Assuming that the calculated shape of S2's orbit is accurate, the star will spend most of its time between 6 and 8 squares from the origin, and it conceivably could have gone unobserved during the short interval during which it whipped around periapsis. Nevertheless, with sufficiently good data from the points at which it was observed, the experts working on this project could fit the ellipse as shown with reasonable confidence. I have no motive to second-guess them.

Dave Lee
2016-May-12, 05:33 PM
1) Look at the 6DOF stuff they did. A circular orbit would simply not reproduce the results.

Can you please direct me to that info?

Dave Lee
2016-May-12, 06:19 PM
With regards to my question:
"I was quite surprised to discover that most of those stars orbits around some virtual center – which is not the MBH?"
The answer was:



Why? People have been saying throughout this thread that the orbits are affected by more than just one object.
However, the total mass of the stars is a tiny fraction of that of the central black hole. Please see the following message:

Let me add that in the region shown here, the total mass of the stars is a tiny fraction of that of the central black hole.
So, the MBH must have full control on all the stars in the aria.
We can use the solar system as an example - as all planets and moons in the solar system are a tiny fraction the sun mass.
So, all the Planets are orbiting the Sun, while the moons are orbiting the planets.
Therefore, let's look again on the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f19
How can we justify the local rotations of stars in this diagram (for example: S29, S97, S33 and many others?)
There is no visible central mass in the center of their orbit. (Not even anywhere inside the orbit cycle)
So, how can we justify those local orbits?
It is similar like a moon which orbits an invisible point in the solar system. Is it feasible?
So, the questions are as follow:
If the total mass of the stars is a tiny fraction of the central black hole, how could it be that those stars rotate around some invisible point in space?
Could it be that they rotates around some Black Hole?
If there are black holes, (and quite many of them), then could it be that the total mass of the stars + all the black holes aren't so tiny fraction with related to the MBH?

However, if there were BH then we had to see some strong gravity lensing in their spot (and maybe some sort of gravity interaction with the MBH.
Do we see it? If we don't see it, and there are no B.H. then how can we justify those local orbits???

Hornblower
2016-May-12, 08:43 PM
With regards to my question:
"I was quite surprised to discover that most of those stars orbits around some virtual center – which is not the MBH?"
The answer was:


However, the total mass of the stars is a tiny fraction of that of the central black hole. Please see the following message:

So, the MBH must have full control on all the stars in the aria.
We can use the solar system as an example - as all planets and moons in the solar system are a tiny fraction the sun mass.
So, all the Planets are orbiting the Sun, while the moons are orbiting the planets.
Therefore, let's look again on the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f19
How can we justify the local rotations of stars in this diagram (for example: S29, S97, S33 and many others?)
There is no visible central mass in the center of their orbit. (Not even anywhere inside the orbit cycle)
So, how can we justify those local orbits?
It is similar like a moon which orbits an invisible point in the solar system. Is it feasible?
So, the questions are as follow:
If the total mass of the stars is a tiny fraction of the central black hole, how could it be that those stars rotate around some invisible point in space?
Could it be that they rotates around some Black Hole?
If there are black holes, (and quite many of them), then could it be that the total mass of the stars + all the black holes aren't so tiny fraction with related to the MBH?

However, if there were BH then we had to see some strong gravity lensing in their spot (and maybe some sort of gravity interaction with the MBH.
Do we see it? If we don't see it, and there are no B.H. then how can we justify those local orbits???
You appear to be interpreting those distorted ellipses on the display linked in post 70 as orbital motion of stars around some points other than the central black hole, when I cannot tell what that display is supposed to mean. It looks like a flat surface projection of the surface of an entire sphere. With no text and no links back to the source, I don't have the foggiest idea what to make of it. At least the display linked in post 60 is clearly a view of the small region around the black hole. Comparison with other sources linked in this thread corroborate this. The ellipses shown are consistent with Keplerian orbits around the central mass, with various amounts of inclination of the orbital planes and orientation of the lines of apsides. I stand by what I wrote in my previous post.

cjameshuff
2016-May-12, 11:06 PM
Your questions appear to be rooted in the same misconceptions that have been corrected over and over. There is no confusion between stars in elliptical orbits around a black hole and stars in circular orbits around "virtual centers", because the motions of stars in circular orbits are different, no perspective can make them appear the same, and stars don't choose to orbit random points in space. There is no assumption that we are viewing the orbits from above, and there is no requirement for it, the math they use does not assume an orbital plane, they fit for that as well as for the other orbital parameters.

You seem to again be trying to force everything into some conception of objects following circular orbits around "virtual centers". You may as well be trying to explain things in terms of a flat Earth. Your framework is broken, and you need to throw it away instead of trying to wedge everything into it.

Reality Check
2016-May-13, 12:23 AM
So, the MBH must have full control on all the stars in the aria.
Only if we ignore the existence of the rest of the galaxy, Dave Lee, as has been pointed out. For that matter why are you ignoring the effect of the other stars that orbit the MBH?
The MBH has a large mass compared to the stars. The MBH has a tiny mass compared t the rest of the galaxy. Looking at mass alone the rest of the galaxy "must have full control on all the stars in the aria". But gravitation does not depend on mass along! Gravitation also depends on distance (an inverse square law). What has the most influence over the stars is a complex issue. We have observed the stars for some decades, see that stars are in orbits around the MBH and thus are currently mostly influenced by the MBH. In the past the stars may have not orbited the MBH. In the future the stars may even stop orbiting the MBH.

Reality Check
2016-May-13, 12:32 AM
Therefore, let's look again on the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f19
How can we justify the local rotations of stars in this diagram (for example: S29, S97, S33 and many others?)
There are no "local rotations" in that diagram to justify because you give no context for the diagram. Not even a caption!

This is Figure 19 in MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)

Figure 19. Orientation of the orbital planes of those S-stars for which we were able to determine orbits. The orientation of the orbits in space is described by the orbital angular momentum vector, corresponding to a position in this all sky plot, in which the vertical dimension corresponds to the inclination i of the orbit and the horizontal dimension to the longitude of the ascending node Ω. A star in a face-on, clockwise orbit relative to the line of sight, for instance, would be located at the top of the graph, while a star with an edge-on seen orbit would be located on the equator of the plot. The error ellipses correspond to the statistical 1σ fit errors only, thus the area covered by each is 39% of the probability density function.
Those ellipses are error ellipses not "local rotations".

You may want to look at "Table 7. Orbital Parameters of Those S-stars for Which We Were Able to Determine Orbits" for measured eccentricities, etc.

Hornblower
2016-May-13, 02:30 AM
Reality Check, thank you so much for posting a link to the complete paper, rather than the uncaptioned figures which we had before.

Assuming that the calculated shape of S2's orbit is accurate, the star will spend most of its time between 6 and 8 squares from the origin, and it conceivably could have gone unobserved during the short interval during which it whipped around periapsis. Nevertheless, with sufficiently good data from the points at which it was observed, the experts working on this project could fit the ellipse as shown with reasonable confidence. I have no motive to second-guess them. I stand corrected. The paper shows clearly that they observed S2 through its pericenter passage, and that Dave Lee's material to which I had responded must have been incomplete and/or out of context.

Reality Check
2016-May-13, 03:03 AM
Reality Check, thank you so much for posting a link to the complete paper, rather than the uncaptioned figures which we had before.
Actually the complete paper was linked to in the OP by Dave Lee - I just put the title in my link. As for S2's orbit - the paper is from 2009 so we have another 7 years of the orbit mapped and probably published somewhere.

Hornblower
2016-May-13, 03:11 AM
Actually the complete paper was linked to in the OP by Dave Lee - I just put the title in my link.

I stand corrected again. This thread has gone through so many twists and turns that I had forgotten what was in the OP. Dumb me!

Shaula
2016-May-13, 04:43 AM
Can you please direct me to that info?
Section 5 - Orbital Fitting. Section 6.4.


So, the MBH must have full control on all the stars in the aria.
The mass of the Sun is much larger than the mass of the Earth. So by your logic the Sun should have full control of the Moon. That is an extreme example but it covers the point I was making that you have ignored over and over. It is not a simple two body problem - there are other objects around.

The rest of your points have already been covered by other posters. You are hanging on to your own conceptual framework about how gravity works. And it is not a good enough one.

Dave Lee
2016-May-13, 05:27 AM
Only if we ignore the existence of the rest of the galaxy, Dave Lee, as has been pointed out. For that matter why are you ignoring the effect of the other stars that orbit the MBH?
The MBH has a large mass compared to the stars. The MBH has a tiny mass compared t the rest of the galaxy. Looking at mass alone the rest of the galaxy "must have full control on all the stars in the aria". But gravitation does not depend on mass along! Gravitation also depends on distance (an inverse square law). What has the most influence over the stars is a complex issue. We have observed the stars for some decades, see that stars are in orbits around the MBH and thus are currently mostly influenced by the MBH. In the past the stars may have not orbited the MBH. In the future the stars may even stop orbiting the MBH.

Thanks
Yes, this is an excellent explanation.
It is also correlated with the message from Shaula

It is not a simple two body problem - there are other objects
So,
1.It is not a simple two body problem - there are other objects which we shouldn't ignore.
2.We shouldn't ignore with the existence of the rest of the galaxy.
3.Gravitation also depends on distance (an inverse square law).
Therefore, let's look again on the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f19
If I understand it correctly there is no need to find any BH or some invisible mass in the center of those rotations. Each star orbits around the center of gravity which is a direct outcome from the mass in the galaxy including the MBH (but especially – from the close distance due to an inverse square law).
Is it correct?

Shaula
2016-May-13, 06:21 AM
Each star orbits around the center of gravity which is a direct outcome from the mass in the galaxy including the MBH (but especially – from the close distance due to an inverse square law).
Is it correct?
... Only to first order, as was first said back in post 14 and in fact by the preceding 3 points you listed in the post. You are still trying to apply this simplistic two body solution to a problem that is much more complex.

They roughly orbit the centre of mass of the objects inside their orbits - but said orbit is perturbed by the non-spherical nature of said mass distribution, by nearby stars and other influences. The size and nature of the orbits depend on what is around them, what interactions they have had in the relatively recent past and so on.

You are over-simplifying the situation for the level of detail you seem to want to get in to.

Dave Lee
2016-May-13, 12:41 PM
... Only to first order, as was first said back in post 14 and in fact by the preceding 3 points you listed in the post. You are still trying to apply this simplistic two body solution to a problem that is much more complex.

They roughly orbit the centre of mass of the objects inside their orbits - but said orbit is perturbed by the non-spherical nature of said mass distribution, by nearby stars and other influences. The size and nature of the orbits depend on what is around them, what interactions they have had in the relatively recent past and so on.

You are over-simplifying the situation for the level of detail you seem to want to get in to.

Thanks
However, would you kindly give a direct answer:

Is it correct that each star orbits around the center of gravity which is a direct outcome from the mass in the galaxy including the MBH (but especially – from the close distance due to an inverse square law).
Yes or no.

What do you mean by:

1. "... Only to first order" - What is the first order? order of what?
2. "They roughly orbit the centre of mass of the objects inside their orbits - but said orbit is perturbed by the non-spherical nature of said mass distribution, by nearby stars and other influences. The size and nature of the orbits depend on what is around them, what interactions they have had in the relatively recent past and so on." - I have no clue what does it mean. Please elaborate. Do you mean that only the mass inside the orbit has an effect on the center of Gravity?

Please - try to avoid from complex/indirect answer. Somehow, I need to understand your message. Please - simple answer to simple question.

Hornblower
2016-May-13, 01:33 PM
Thanks
However, would you kindly give a direct answer:

Is it correct that each star orbits around the center of gravity which is a direct outcome from the mass in the galaxy including the MBH (but especially – from the close distance due to an inverse square law).
Yes or no.

What do you mean by:

1. "... Only to first order" - What is the first order? order of what?
2. "They roughly orbit the centre of mass of the objects inside their orbits - but said orbit is perturbed by the non-spherical nature of said mass distribution, by nearby stars and other influences. The size and nature of the orbits depend on what is around them, what interactions they have had in the relatively recent past and so on." - I have no clue what does it mean. Please elaborate. Do you mean that only the mass inside the orbit has an effect on the center of Gravity?

Please - try to avoid from complex/indirect answer. Somehow, I need to understand your message. Please - simple answer to simple question.
We will do well to break a response down into small chunks, so that if you encounter a point you do not understand for whatever reason, we can resolve it without first having to tease it out of a profusion of confusion we can get by trying to cover too much in one post.

We are studying a region with a radius of about 1/8 of a lightyear. Here we have a central black hole of several million solar masses, and a few dozen stars totaling at most a few hundred solar masses. This looks like a good analogy to our own solar system, with the black hole analogous to the Sun and the stars analogous to the planets. We know from observation that our planets move in close approximations to Kepler ellipses around the Sun, and that the orbits slowly drift because of perturbations by the relatively slight gravitational effects of the planets on each other. The Sun overwhelmingly dominates the action here. I feel safe in saying that the black hole does likewise in its close-in territory, which appears to be corroborated by the findings in the paper linked in this thread. Let me pause for comments and/or questions before going any farther.

Dave Lee
2016-May-13, 02:17 PM
We will do well to break a response down into small chunks, so that if you encounter a point you do not understand for whatever reason, we can resolve it without first having to tease it out of a profusion of confusion we can get by trying to cover too much in one post.

We are studying a region with a radius of about 1/8 of a lightyear. Here we have a central black hole of several million solar masses, and a few dozen stars totaling at most a few hundred solar masses. This looks like a good analogy to our own solar system, with the black hole analogous to the Sun and the stars analogous to the planets. We know from observation that our planets move in close approximations to Kepler ellipses around the Sun, and that the orbits slowly drift because of perturbations by the relatively slight gravitational effects of the planets on each other. The Sun overwhelmingly dominates the action here. I feel safe in saying that the black hole does likewise in its close-in territory, which appears to be corroborated by the findings in the paper linked in this thread. Let me pause for comments and/or questions before going any farther.

I read your answer, as well as Shaula's, and I have no clue what do you really want to say.

I must tell you a real story about an examination which I had in the University. It was very difficult question - and I didn't know the correct answer.
As it was very important examination, I have started to give a very complex and indirect answer. They were quite confused from my long and indirect answer that they didn't ask me any further question and moved on to the next student.

However, this isn't examination in the University.

So, please - Yes or no.
Why is it so difficult to give a simple answer? For example:
Yes - that is correct.
Or
No - it isn't correct due to...

In any case, let me try to guess what do you really mean by these indirect answers:
I assume that your answer is no. (Otherwise you would probably say - Yes.)

So, If it is no, and based on your understanding, what is the source for that "local" orbit rotation?
Please - simple and direct answer.

grapes
2016-May-13, 02:17 PM
Thanks
However, would you kindly give a direct answer:

Is it correct that each star orbits around the center of gravity which is a direct outcome from the mass in the galaxy including the MBH (but especially – from the close distance due to an inverse square law).
Yes or no.

What do you mean by:

1. "... Only to first order" - What is the first order? order of what?
2. "They roughly orbit the centre of mass of the objects inside their orbits - but said orbit is perturbed by the non-spherical nature of said mass distribution, by nearby stars and other influences. The size and nature of the orbits depend on what is around them, what interactions they have had in the relatively recent past and so on." - I have no clue what does it mean. Please elaborate. Do you mean that only the mass inside the orbit has an effect on the center of Gravity?

Please - try to avoid from complex/indirect answer. Somehow, I need to understand your message. Please - simple answer to simple question.
I've read some of the responses, got more to read, but some of the confusion may come from this.

Objects do not typically orbit the center of mass, when the mass is extensive. As an extreme example, consider the satellites/modules in orbit around our moon, or earth. Their orbits are very far from centered around the center of mass of the earth/moon system. They'd have to be quite a ways away from both moon and earth in order for that to be the case.

01101001
2016-May-13, 02:29 PM
So, If it is no, and based on your understanding, what is the source for that "local" orbit rotation?

What is a "local" orbit rotation?

Dave Lee
2016-May-13, 02:44 PM
What is a "local" orbit rotation?

Please look again at the following diagram:
http://iopscience.iop.org/0004-637X/...e/apj297214f19
Please try to find S21 and S29 (as an example).
Do you see their orbit cycle?
Those stars do not rotate around the MBH. they rotate around some invisible point in space. (I call it - "local" orbit rotation).
Is it clear?

grapes
2016-May-13, 02:46 PM
Please look again at the following diagram:
http://iopscience.iop.org/0004-637X/...e/apj297214f19
Please try to find S21 and S29 (as an example).
Do you see their orbit cycle?
Those stars do not rotate around the MBH. they rotate around some invisible point in space. (I call "local" orbit rotation).
Is it clear?
Does that link work for everyone else?

Dave Lee
2016-May-13, 02:47 PM
I've read some of the responses, got more to read, but some of the confusion may come from this.

Objects do not typically orbit the center of mass, when the mass is extensive. As an extreme example, consider the satellites/modules in orbit around our moon, or earth. Their orbits are very far from centered around the center of mass of the earth/moon system. They'd have to be quite a ways away from both moon and earth in order for that to be the case.

Thanks
However, I don't understand how this example can offer a solution for the "local" orbit rotation. Would you kindly elaborate?

01101001
2016-May-13, 02:55 PM
Does that link work for everyone else?

It's a failed copy. Dave Lee means http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f19


Do you see their orbit cycle?

What "orbit cycles"? The circles/ellipses/curves around the stars are not orbit cycles.


Those ellipses are error ellipses not "local rotations".

What RC said.

grapes
2016-May-13, 03:02 PM
Thanks for the revised link


What "orbit cycles"? The circles/ellipses/curves around the stars are not orbit cycles.



What RC said.

Does that resolve the issue, Dave Lee?

Dave Lee
2016-May-13, 03:12 PM
Thanks for the revised link


Does that resolve the issue, Dave Lee?


Yes - Thanks

Shaula
2016-May-13, 03:13 PM
I must tell you a real story about an examination which I had in the University. It was very difficult question - and I didn't know the correct answer. As it was very important examination, I have started to give a very complex and indirect answer. They were quite confused from my long and indirect answer that they didn't ask me any further question and moved on to the next student.
Let me tell you a true story. Once upon a time I was posting on a forum dedicated to astronomy and science in general. On this forum was a poster who kept demanding simple and direct answers to quite complex questions. However every time they were given these answers they proceeded to misunderstand them. These misunderstandings were not possible when they were given complicated answers - but they seemed determined to repeatedly make these misunderstanding, to quote small sections of larger answers and papers out of context, to ignore large sections of papers and responses and to try to ignore any complexity of the systems in favour of an overly simplistic model of what was going on. After a while I decided that nothing I could say or do would change the simplistic model this person insisted on using so I stopped answering their questions. I was sad but used the now free time productively.

True story.

Hornblower
2016-May-13, 07:53 PM
However, would you kindly give a direct answer:

Is it correct that each star orbits around the center of gravity which is a direct outcome from the mass in the galaxy including the MBH (but especially – from the close distance due to an inverse square law).
Yes or no.In my opinion that question is too general to be appropriate for this special case in which we are so close to the black hole that its gravity overwhelms everything else as far as S2 and its sister stars are concerned. A more appropriate question might be, "Is it correct that each star in this sample orbits around the black hole in a good approximation of a Keplerian ellipse?" My answer to that would be yes. Next question, please?

AFJ
2016-May-13, 09:23 PM
Ah, just 1 more on hollow shells :)

This was partly answered earlier, but still eludes me.

Imagine a Dyson sphere around a star that is; a hollow non-spinning shell of uniform density. Would this construction be gravitationally stable?

Dave Lee
2016-May-13, 09:57 PM
Sorry about the misunderstanding regards Fig. 19.
However, I have some more questions.

Please look at the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
With regards to the location of Sgr A*:
It seems that they are not sure if it is S17 or Sgr A*, as it is stated: "At the position of Sgr A* some light is seen, which could be either due to Sgr A* itself or due to a faint, so far unrecognized star being confused with Sgr A*."
So, how could it be that the science doesn't know for sure the real location of the MBH?

In that diagram we see clearly quite high gravity lensing around S95, S96 and S97.
However, there are absolutely no special gravity lensing around Sgr A*.
We can easily thing that the gravity force of those stars is significantly higher than the MSB.
So;
1. If the mass of all stars in this area is a fraction of MSB, than how could it be that it is so difficult even to verify the real location of MSB?
2. Why we can't see the MSB gravity lensing effect?
3. If they are not sure about the real location of the MSB, than how could they draw the orbit diagram without this critical information?

cjameshuff
2016-May-13, 10:56 PM
Sorry about the misunderstanding regards Fig. 19.
However, I have some more questions.

Please look at the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
With regards to the location of Sgr A*:
It seems that they are not sure if it is S17 or Sgr A*, as it is stated: "At the position of Sgr A* some light is seen, which could be either due to Sgr A* itself or due to a faint, so far unrecognized star being confused with Sgr A*."
So, how could it be that the science doesn't know for sure the real location of the MBH?

From the bit you quoted, the location is known. How else could they speak of detecting light from that location?



In that diagram we see clearly quite high gravity lensing around S95, S96 and S97.

There is no gravity lensing visible in that image. The paper specifically states that:

Gravitational lensing might affect the measured positions. A quantitative analysis shows that the effects are very small except in unusual, exceptional geometric configurations.
...
In our data set, none of the stars gets close to the regime that gravitational lensing actually becomes important.

DaveC426913
2016-May-13, 11:01 PM
Ah, just 1 more on hollow shells :)

This was partly answered earlier, but still eludes me.

Imagine a Dyson sphere around a star that is; a hollow non-spinning shell of uniform density. Would this construction be gravitationally stable?

The shell would not be stable with respect to the central star, no. Newton's shell theorem works both ways. The star feels no net pull from the shell, and the shell feels no net pull from the star.

If they were not initially positioned with zero relative motion, the two would eventually drift together and the shell would be destroyed.

(This was a bit of a debacle in Larry Niven's Hugo and Nebula-winning Ringworld. The Ringworlds is 2 dimensional but same thing applied. Fans wrote to him, pointing out that the system was not gravitationally stable, forcning him to write a sequel - The Ringworld Engineers - that detailed the attitide jets for keeping the Ringworld from brushing its star.)

AFJ
2016-May-13, 11:04 PM
What do you mean by gravitationally stable?

The shell would not be stable with respect to the central star, no. Newton's sheel theorem works both ways. The star feels no net pull from the shell, and the shell feels no net pull from the star. If they were not set together with zero relative motion, the two would evntually drift together and hte shele would be destroyed.

Aaah, thank you. I thought this was so, but every time these things turn up as if there no gravitational problem, bugged me.

01101001
2016-May-13, 11:34 PM
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
[...]
In that diagram we see clearly quite high gravity lensing around S95, S96 and S97.

Me also, a problem. I see rings around those objects in Figure 1. What is your evidence that they portray effects of gravity lensing rather than, image-processing ringing artifacts, such as those of deconvolution?

grant hutchison
2016-May-14, 12:32 AM
(This was a bit of a debacle in Larry Niven's Hugo and Nebula-winning Ringworld. The Ringworlds is 2 dimensional but same thing applied. Fans wrote to him, pointing out that the system was not gravitationally stable, forcning him to write a sequel - The Ringworld Engineers - that detailed the attitide jets for keeping the Ringworld from brushing its star.)Actually, it's worse for the Ringworld. Once the ring and star are off-centre they attract each other. For a spherical shell, you can keep the relative velocity arbitrarily low and it stays that way; for a ringworld, any initial relative velocity, or any deviation from absolute centrality, is amplified by mutual gravitation.

Grant Hutchison

Hornblower
2016-May-14, 01:05 AM
Sorry about the misunderstanding regards Fig. 19.
However, I have some more questions.

Please look at the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
With regards to the location of Sgr A*:
It seems that they are not sure if it is S17 or Sgr A*, as it is stated: "At the position of Sgr A* some light is seen, which could be either due to Sgr A* itself or due to a faint, so far unrecognized star being confused with Sgr A*."
So, how could it be that the science doesn't know for sure the real location of the MBH?

In that diagram we see clearly quite high gravity lensing around S95, S96 and S97.
However, there are absolutely no special gravity lensing around Sgr A*.
We can easily thing that the gravity force of those stars is significantly higher than the MSB.
So;
1. If the mass of all stars in this area is a fraction of MSB, than how could it be that it is so difficult even to verify the real location of MSB?
2. Why we can't see the MSB gravity lensing effect?
3. If they are not sure about the real location of the MSB, than how could they draw the orbit diagram without this critical information?

That link gave me a blank screen. Please refer us directly to the paper. Chapter, figure and table numbers take us right there.

Swift
2016-May-14, 01:06 AM
Ah, just 1 more on hollow shells :)

This was partly answered earlier, but still eludes me.

Imagine a Dyson sphere around a star that is; a hollow non-spinning shell of uniform density. Would this construction be gravitationally stable?
Please no. There are plenty of Dyson sphere threads already, do not derail this thread. Please go find one of those threads or start your own.

01101001
2016-May-14, 01:21 AM
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
[...]
It seems that they are not sure if it is S17 or Sgr A*, as it is stated: "At the position of Sgr A* some light is seen, which could be either due to Sgr A* itself or due to a faint, so far unrecognized star being confused with Sgr A*."
So, how could it be that the science doesn't know for sure the real location of the MBH?

Why isn't it right there where labeled? S17 is there. Right beside/on it, they say, is the location of Sgr A*. The slight blob protrusion beside/on S17 is something. It could be light from Sgr A* or it could be light from an unrecognized star (in which case the black hole Sgr A* must not be contributing enough light to image). They say it is there because of their data. It may be imaged right beside/on S17, or some star may be imaged instead, but regardless Sgr A* is there.

Why do you think it is elsewhere? Where?

01101001
2016-May-14, 02:00 AM
That link gave me a blank screen. Please refer us directly to the paper. Chapter, figure and table numbers take us right there.

It's the paper from post 1, same source of all the questions.

MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)

The inage in question now is Figure 1.

Dave Lee
2016-May-14, 03:27 AM
Thanks

That is clear.

Dave Lee
2016-May-14, 04:01 AM
With regards to figure 10:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f10
It is stated:
"Figure 10. 2002 data of S2. The gray symbols show the measured positions, the errors are as obtained from the standard analysis and are not yet enlarged by the procedure described in Section 3.5. The black dots are the positions predicted for the observation dates using an orbit fit obtained from all data other than 2002."

So, the gray symbols show the measured position of S2.
While The black dots are the positions predicted for the observation dates using an orbit fit obtained from all data other than 2002
In other words, in order to fit S2 diagram to what is consider as an orbit fit, the science have actually moved the gray symbols to the black dots.
By that activity, they have actually shift the bottom side of the S2 orbit to a lower location.

Now, please look again at the following orbit diagram:
https://en.wikipedia.org/wiki/Sagittarius_A*#/media/File:Galactic_centre_orbits.svg

In this diagram the MBH is located very close to the bottom side of S2 orbit. However, it is inside the S2 elliptical cycle.
However, what should be the outcome without moving the dots?
In other words, could it be that based on the real verification (by the real gray symbols) the MSB was actually outside the S2 elliptical cycle?
So, in order to set the MSB inside the S2 Cycle, they have shift the Gray symbols to the black dots.
As it is stated: " The black dots are the positions predicted for the observation dates using an orbit fit obtained"
So, if I understand it correctly, instead of dealing with the question – How could it be that the MBH is outside the S2 Cycle?, they have moved the dots further down and got the requested orbit fit.
Bravo!
Is it real?

cjameshuff
2016-May-14, 11:46 AM
So, the gray symbols show the measured position of S2.
While The black dots are the positions predicted for the observation dates using an orbit fit obtained from all data other than 2002
In other words, in order to fit S2 diagram to what is consider as an orbit fit, the science have actually moved the gray symbols to the black dots.

No. They are comparing predictions not using the 2002 data to actual measurements during 2002 in order to find errors in either the model or the instruments that made the measurements. They proceeded with one data set that included the suspect data and gave larger error bars, and another that excluded it.

Are you just ignoring the parts of the paper that don't say what you want?

Hornblower
2016-May-14, 12:24 PM
So, the gray symbols show the measured position of S2.
While The black dots are the positions predicted for the observation dates using an orbit fit obtained from all data other than 2002
In other words, in order to fit S2 diagram to what is consider as an orbit fit, the science have actually moved the gray symbols to the black dots.Welcome to the wild and woolly world of astrometry, which abounds with error bars that often are as large as the magnitudes of what they are measuring. This is inevitable when we are pushing our observation techniques to their limits. Large portions of the paper are devoted to detailed explanations of the statistical methods for dealing with these uncertainties mathematically. If you wish to get simple answers, detailed discussion of this error-handling work is not for you at this point.

chornedsnorkack
2016-May-14, 02:16 PM
A circular orbit nearly edge on will look like an ellipse with the black hole at its geometric center, that is, the midpoint of its major axis. None of these orbits come even close. S13's true ellipse would have a minor axis about 90% of the length of the major axis, so a moderate inclination with an orientation that foreshortens the major axis but not the minor could make it look circular from this viewpoint.

Also about the speed along the orbit - a truly circular orbit would have uniform speed along the whole circle. A circular orbit looking like ellipse because inclined would have equal speed on near side and far side.
Whereas an elliptical orbit has higher speed near periapse and slower near apoapse. Therefore an elliptical orbit foreshortened into a circle by an appropriate inclination along major axis would have faster speed near periapse and slower near apoapse, and measuring these speeds would give away the true eccentricity and inclination of the orbit as well as the location of the black hole, even if it is otherwise unseen.

Dave Lee
2016-May-14, 03:21 PM
Please look at paragraph 3.5
It is stated clearly: "In 2002, S2 was positionally nearly coincident with Sgr A* and thus confused with the NIR counterpart of the MBH."
So, what is the meaning of "coincident"
By Google - "occurring together in space or time".
Hence, based on the measurements in 2002, S2 and MBH were almost at the same spot in space.
Therefore, we can claim that in 2002 the radius between MBH and S2 was almost zero!
How could it be? That contradicts the whole idea that S2 orbits the MBH.
I assume that the science could not accept this real measurement, so they have decided that there must be an error due to the increase brightness of S2:
"Given the importance of the 2002 data, we decided not to discard it completely but to estimate the astrometric error assuming a confusion event, given the measured increase in brightness."
So, if I understand it correctly, based on this error, the science have shift the measurement symbols of S2 in order to set the MBH inwards of S2 orbit cycle.
However, error can work at any direction.
So technically, if we will shift the measurement symbols to the other direction and at the same magnitude, then for sure the MBH would be completely outside the orbit cycle of S2.
Hence:
1.Is there any possibility to get the exact location of MBH with regards to the S2 orbit – without any adaptation or shifting. Just based on real measurements?
2. Do we have new data about S2 at pericenter?
3. What is the velocity of S2 at apocenter?
It is stated: " Given the maximum velocity of S2 at pericenter (v ≈ 8000 km s−1), the radius of the star (r = 11R☉; Martins et al. 2008)…"
So, the velocity of S2 at pericenter is v ≈ 8000 km s−1. But what is the velocity of S2 at apocenter?

Hornblower
2016-May-14, 03:35 PM
Please look at paragraph 3.5
It is stated clearly: "In 2002, S2 was positionally nearly coincident with Sgr A* and thus confused with the NIR counterpart of the MBH."
So, what is the meaning of "coincident"
By Google - "occurring together in space or time".
Hence, based on the measurements in 2002, S2 and MBH were almost at the same spot in space.
Therefore, we can claim that in 2002 the radius between MBH and S2 was almost zero!
How could it be? That contradicts the whole idea that S2 orbits the MBH.
I assume that the science could not accept this real measurement, so they have decided that there must be an error due to the increase brightness of S2:
"Given the importance of the 2002 data, we decided not to discard it completely but to estimate the astrometric error assuming a confusion event, given the measured increase in brightness."
So, if I understand it correctly, based on this error, the science have shift the measurement symbols of S2 in order to set the MBH inwards of S2 orbit cycle.
However, error can work at any direction.
So technically, if we will shift the measurement symbols to the other direction and at the same magnitude, then for sure the MBH would be completely outside the orbit cycle of S2.
Hence:
1.Is there any possibility to get the exact location of MBH with regards to the S2 orbit – without any adaptation or shifting. Just based on real measurements?
2. Do we have new data about S2 at pericenter?
3. What is the velocity of S2 at apocenter?
It is stated: " Given the maximum velocity of S2 at pericenter (v ≈ 8000 km s−1), the radius of the star (r = 11R☉; Martins et al. 2008)…"
So, the velocity of S2 at pericenter is v ≈ 8000 km s−1. But what is the velocity of S2 at apocenter?

You appear to be cherry-picking rather than evaluating the entire body of data points and the authors' explanations of what they considered unlikely possibilities for astrophysical reasons. I can find no reason to second-guess the statistical methods these experts used to deal with the uncertainties of individual data points and come up with a most probable set of orbital elements for S2.

cjameshuff
2016-May-14, 05:11 PM
Please look at paragraph 3.5
It is stated clearly: "In 2002, S2 was positionally nearly coincident with Sgr A* and thus confused with the NIR counterpart of the MBH."
So, what is the meaning of "coincident"
By Google - "occurring together in space or time".
Hence, based on the measurements in 2002, S2 and MBH were almost at the same spot in space.

No, they were simply in the same line of sight as seen from Earth. Another instance of you not caring what something actually means, only what you can twist it around to mean.



Therefore, we can claim that in 2002 the radius between MBH and S2 was almost zero!

No, you can't.



I assume that the science could not accept this real measurement

You assume a lot of nonsensical things, and repeat those assumptions despite your source information completely contradicting them and despite people here repeatedly correcting you. Yet again, you're picking out one little item that you can twist around and reinterpret to support your own preconceptions. And now, simply refusing to take things as they were written isn't enough, so you're slandering scientists and engineers everywhere with accusations of scientific misconduct?

Either Newtonian gravitation is fundamentally flawed and everything built on it...every satellite, every spacecraft, and most of astronomy...is a lie, or you do not understand gravity as well as you think you do. If you want to claim the former, there's specific forums for that. The latter can be easily remedied by admitting you were wrong and doing real research rather than hunting for bits that can be reinterpreted to support your preconceptions.

Dave Lee
2016-May-14, 06:44 PM
No, they were simply in the same line of sight as seen from Earth.

Sorry - That is just an impossible mission.
O.K.
Try to draw an elliptical cycle.
Set a gravity center at the plane of this cycle (Anywhere at the inwards side of the cycle)
Now, is there any possibility to see them in the same line of sight?
The answer is NO.
Unless -
1. We look at the elliptical cycle directly from the side - But then we should see just one line.
2. The center of gravity is not at the same plane as the Cycle – But then it can't be used as a center of gravity.
3. Any other idea?

cjameshuff
2016-May-14, 07:43 PM
Sorry - That is just an impossible mission.
O.K.
Try to draw an elliptical cycle.
Set a gravity center at the plane of this cycle (Anywhere at the inwards side of the cycle)
Now, is there any possibility to see them in the same line of sight?
The answer is NO.
Unless -
1. We look at the elliptical cycle directly from the side - But then we should see just one line.
2. The center of gravity is not at the same plane as the Cycle – But then it can't be used as a center of gravity.
3. Any other idea?

Wrong. We're talking about whether the two objects can appear "nearly coincident" to an instrument with finite angular resolution, and the specific case being considered is the pericenter passage of a star in a highly elliptical orbit. The answer is yes.

Hornblower
2016-May-14, 11:05 PM
Sorry - That is just an impossible mission.
O.K.
Try to draw an elliptical cycle.
Set a gravity center at the plane of this cycle (Anywhere at the inwards side of the cycle)
Now, is there any possibility to see them in the same line of sight?
The answer is NO.
Unless -
1. We look at the elliptical cycle directly from the side - But then we should see just one line.
2. The center of gravity is not at the same plane as the Cycle – But then it can't be used as a center of gravity.
3. Any other idea?

You appear to be fixated on one data point, out of many, which happens to appear to coincide with the black hole's position most likely as a result of imaging artifacts which are an occupational hazard in a project like this. This point would not necessarily be rejected, as there are others that individually fall outside the best-fit ellipse that obeys Kepler's laws with respect to the black hole. As I see it, that is curve-fitting 101. For astrophysical reasons the authors can have a high level of confidence that S2 is indeed orbiting SgrA* in a Keplerian ellipse or something very close to it, rather than some other previously undetected and unsuspected supermassive body.

Hornblower
2016-May-15, 12:22 AM
Here is an appropriate illustration for comparison with the astrometric work being discussed in this thread. I am posting it to show another example of the occupational hazards of this type of work, which are not grounds for questioning the competence or honesty of the observers.
21561
This diagram shows visual measurements of a double star that was near the resolution limits of the telescopes of the time, by various observers with various telescopes. The latest ones, which have the least scatter, were almost surely taken with the 36-inch Lick Observatory refractor, which greatly improved the precision of this type of work. I don't think anyone seriously assumed that the companion was actually zigzagging under the gravitational action of multiple unseen massive bodies. Much of the time atmospheric turbulence would have been turning the pair of spots into a blob an arcsecond or more across, and skilled observers were teasing out the true binary nature during fleeting moments of good seeing and getting rough measurements of it with a filar micrometer. Mr. Hussey was using the same sort of statistical criteria as the authors of the SgrA* project to determine how much weight to give the various data points and to fit an ellipse to them.

Note that the primary star is displaced from the line of apsides of the ellipse as drawn. There is nothing wrong with this. The true orbit is a much fatter ellipse that is steeply inclined to the plane of the sky, and the true line of apsides is rotated out of this plane. If I am not mistaken, the line through the primary star is the foreshortened projection of the true line of apsides.

The image is a copy of one in Sky Catalogue 2000.0 Volume 2, copyright 1985 Sky Publishing Co. I was taught when writing research papers in high school and college that including such excerpts for reference, with proper credit, is fair use.

Dave Lee
2016-May-15, 02:31 AM
Note that the primary star is displaced from the line of apsides of the ellipse as drawn. There is nothing wrong with this. The true orbit is a much fatter ellipse that is steeply inclined to the plane of the sky, and the true line of apsides is rotated out of this plane. If I am not mistaken, the line through the primary star is the foreshortened projection of the true line of apsides..

Sorry, I don't think that this is the case for S2 as it isn't a binary system. So please, if S2 and MSB appear "nearly coincident" then there is good chance that they are - "nearly coincident". Why always we need to fight with the evidence if we don't like it? Sometimes it's better to consider what is the impact of that evidence.


Wrong. We're talking about whether the two objects can appear "nearly coincident" to an instrument with finite angular resolution, and the specific case being considered is the pericenter passage of a star in a highly elliptical orbit. The answer is yes.

Sorry again - but I disagree.
There are two main possibilities to see MBH and S2 in the same line of sight, or "nearly coincident".
1. The distance is Zero or almost Zero - In this case S2 should collide with MSB, or the huge gravity force of the MBH could break it apart. - I assume that this isn't the case.
2. The MBH isn't located at the same plane as S2 cycle. Therefore we see them "nearly coincident". However, if they are not at the same plane, then the MSB isn't the real center of gravity for S2. - I assume that this could be the real reason for: "nearly coincident".

Reality Check
2016-May-15, 09:00 PM
Therefore, let's look again on the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f19
...Is it correct?
No because you are repeating your misinterpretation of Figure 19 again, Dave Lee: Those ellipses are error ellipses not "local rotations". (http://cosmoquest.org/forum/showthread.php?161160-The-galactic-center&p=2354337#post2354337)
You acknowledge your mistake later.

Reality Check
2016-May-15, 09:08 PM
Please look at the following diagram:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER, (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)

Figure 1. Finding chart of the S-star cluster. This figure is based on a natural guide star adaptive optics image obtained as part of this study, using NACO at UT4 (Yepun) of the VLT on 2007 July 20 in the H band. The original image with a FWHM of ≈75 mas was deconvolved with the Lucy–Richardson algorithm and beam restored with a Gaussian beam with FWHM=2 pixel = 26.5 mas. Stars as faint as mH = 19.2 (corresponding roughly to mK = 17.7) are detected at the 5σ level. Only stars that are unambiguously identified in several images have designated names, ranging from S1 to S112. Blue labels indicate early-type stars, red labels late-type stars. Stars with unknown spectral type are labelled in black. At the position of Sgr A* some light is seen, which could be either due to Sgr A* itself or due to a faint, so far unrecognized star being confused with Sgr A*.
This is stating they the know the "exact" position of Sgr A*. Also: At the position of Sgr A* some light is seen which could be an unknown star or Sgr A* itself.

Reality Check
2016-May-15, 09:17 PM
In that diagram we see clearly quite high gravity lensing around S95, S96 and S97.
No we do not, Dave Lee.

3.4.8. Gravitational Lensing
...
In our data set, none of the stars gets close to the regime that gravitational lensing actually becomes important. Therefore, we neglected the effect.

What we probably see in that figure are optical halos around the stars from the telescope.
1. There is no difficulty in determining the real location of the MSB.
2. Because we do not have the instruments (yet) with the resolution to observe gravitational lensing of background stars by Sgr A*. Observing gravitational lensing effects by Sgr A* with GRAVITY (http://arxiv.org/abs/1204.2103)
3. We know the real location of the MSB.

Reality Check
2016-May-15, 09:27 PM
With regards to figure 10:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f10
..
https://en.wikipedia.org/wiki/Sagittarius_A*#/media/File:Galactic_centre_orbits.svg

There is no such thing as a "S2 elliptical cycle", Dave Lee. S2 has an elliptical orbit. S2 and the other stars are orbiting Sgr A* . Thus Sgr A* is inside the orbits of the stars including S2.
There is no "movement" of "Gray symbols to the black dots". As already noted they are comparing 2002 predictions to their measurements. The Wikipedia diagram comes from a 2003 paper: Eisenhauer, F.; et al. (23 October 2003). "A Geometric Determination of the Distance to the Galactic Center". The Astrophysical Journal 597 (2): L121–L124. arXiv:astro-ph/0306220. Bibcode:2003ApJ...597L.121E. doi:10.1086/380188 (https://dx.doi.org/10.1086%2F380188).

Reality Check
2016-May-15, 09:38 PM
Hence, based on the measurements in 2002, S2 and MBH were almost at the same spot in space.
That is what happens when a star in a highly elliptical orbit gets close to its primary (Sgr A*), Dave Lee. We cannot determine the position of stars to an accuracy of millimeters :eek:. There will be uncertainty in their positions. Sometimes these uncertainties will overlap so much that the positions are described as coincident. The small fact that we have observed S2 during two entire orbits around Sgr A* (https://en.wikipedia.org/wiki/S2_(star)) tells us that it is in orbit around Sgr A*!
1. We have "the exact location of MBH with regards to the S2 orbit".
2. S2 (https://en.wikipedia.org/wiki/S2_(star)) was at periastron (nearest approach) in 2002 and has a 15.56 year period. So it will be next at periastron in 2017.
3. S2 (https://en.wikipedia.org/wiki/S2_(star)) has the information that you can use to work out orbital velocities.

Reality Check
2016-May-15, 09:58 PM
Sorry, I don't think that this is the case for S2 as it isn't a binary system.
Sorry Dave Lee but you are forgetting about several things.

The existence of Sgr A*!
S2 and Sgr A* are a binary system to a first approximation.
The phrase "nearly coincident" does not mean that S2 is not in orbit around Sgr A*.
The observed orbit of S2 shows that it is in orbit around Sgr A*. The existence of S2 shows that it has not been close enough to Sgr A* to be ripped apart.
We do not know distances exactly - all measurements in science have error limits. Thus the phrase "nearly coincident". The distance to S2 is 25900 ± 1400 light years. The distance Sagittarius A* to is 25,900 ± 1,400 light years.
FYI: it is the orbital parameters of S2 that shows how close it gets to Sgr A* and these are derived by observing its orbit.

MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)

3.5. S2 in 2002
There are several reasons why a star could change its apparent brightness.

In 2002, S2 was positionally nearly coincident with Sgr A* and thus confused with the NIR counterpart of the MBH. Typically, Sgr A* is fainter than mK = 17 and thus the extra-light from Sgr A* in quiescence is not sufficient to explain the observed increase in brightness of S2. However, Sgr A* is known to exhibit flares that can reach a brightness level that could account for the observed increase in brightness (Genzel et al. 2003a; Trippe et al. 2007). In that case, we would expect to see intra-night variability of S2 in the 2002 data. Assuming conservatively that we can determine the relative flux of S2 to ΔmK = 0.1 in each frame and given the brightness of S2 (mK ≈ 14), we estimate that we would have noticed any variations in Sgr A* that exceed mK ≈ 16.5. Since we did not observe any intranight variability, we exclude that flares from Sgr A* significantly contributed to the increased brightness of S2 in 2002.

Hornblower
2016-May-15, 10:10 PM
Now I am posting for the benefit of my fellow presumably well-informed responders in this thread, as well as for lurkers who may be reading. I stand by everything I have posted so far, on the basis of what I have seen so far in reading the paper, along with some consultation with Wiki and other sources when the authors used unfamiliar terms. My fellow responders and I can disregard Dave Lee's remarks just as astronomers can disregard spurious astrometric data points which are inevitable when they are pushing their equipment close to its limits. The astronomers can call on their theoretical understanding and the greater number of reliable data points to identify the spurious ones with reasonable confidence. I don't think Dave Lee's objections, as presented here, will seriously mislead anyone who has at least a rudimentary understanding of this sort of astrometric work, and readers who have no prior knowledge about this topic (we all need to start somewhere) are more than welcome to ask any of us to explain or clarify what we are posting.

Dave Lee
2016-May-16, 05:47 AM
Welcome Reality Check

Thanks for all your answers, although some of them had already been covered.
In any case,


S2 and Sgr A* are a binary system to a first approximation.
Based on Wiki: https://en.wikipedia.org/wiki/Binary_system
"binary system is a system of two objects in space (usually stars, but also brown dwarfs, planets, neutron stars, black holes, galaxies, or asteroids) which are close enough that their gravitational movement causes them to circle around each other (orbit) around a shared mass"
It is stated - Circle. So do you claim that S2 and MBH "circle around each other (orbit) around a shared mass"?
Is it real? The MSB mass is million times higher than S2. Do they circle around each other? Sorry – I would consider it as a mistake.

In any case, let's see if I understand it correctly:

S2 has an elliptical orbit. S2 and the other stars are orbiting Sgr A* . Thus Sgr A* is inside the orbits of the stars including S2.

So, based on the current theory - S2 and the other stars are orbiting Sgr A*
Therefore, I assume the science community was quite is shock when it was discovered that S2 and MBH are "nearly coincident", or in same line of sight:

No, they were simply in the same line of sight as seen from Earth.
.
We had already covered the impact of that issue - Please read it.
So, the science had to find a solution for this key evidence.
They could fix the theory accordingly to this new evidence. However, it seems to me that they have decided to fix the evidence:

We cannot determine the position of stars to an accuracy of millimeters
As it was just an issue of few millimeters, then what is the big deal if they will shift it by a few more millimeters?
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f10
This activity had been justified by well known scientists under the word - ERROR.
We also had discussed this issue. Please read it.

So, if we don't like what we see, then suddenly we claim that we might have some problems with our eyes, the focus is not good enough, the distance to too far away, the instruments are not fully accurate, and so on..
Please, why is it so difficult to except the simple evidence as is? Is there any possibility for us to understand that what we see represents the real universe (Even if we don't like it)?
Hence, if I understand it correctly – it was called "Fit orbit"
Therefore, by fitting the S2 orbit to the science expectation – they have got the requested "Fit orbit"


As it is stated:
http://www.eso.org/public/usa/news/eso0226/
Star Orbiting Massive Milky Way Centre Approaches to within 17 Light-Hours
" It turns out that earlier this year the star approached the central Black Hole to within 17 light-hours - only three times the distance between the Sun and planet Pluto - while travelling at no less than 5000 km/sec."

Please advice the following:
1. What is the distance between S2 and MBH after the error fix (Fit orbit)? Is it 17 light-hours?
2. What is the distance based on the measured 2002 data - without the impact of the error (based on gray symbols) - Is it Zero Light hours?.
3. What is the distance by adding the opposite direction at the same magnitude of the error. In other words - is it correct that if we shift the symbols to the other direction, the MBH should be outside the orbit cycle? Could it be 17 Light-hours outwards?

In any case. S2 is not there by itself.
Please look again at the following picture:
http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f1
S17 and MBH are located at the same spot or "nearly coincident".
However, the science claim that they don't know where is the MSB in that picture. In one hand they claim that the MBH light is stronger than any other star in the aria, but if the MSB is "nearly coincident" with another star, then suddenly they are not sure where it is.

This is stating they the know the "exact" position of Sgr A*. Also: At the position of Sgr A* some light is seen which could be an unknown star or Sgr A* itself.
It is similar to an elephant which is hidden by a mosquito.
And even if we can't see the MBH directly, we could verify its location by its gravity lensing.
So, the science could easily find the MSB location - if they really want.
The question is - Why they didn't want to know where it is?
So, sorry – We have to adjust the theory, not the evidence!!!

You have already offered the solution for that problem:

The MBH has a large mass compared to the stars. The MBH has a tiny mass compared t the rest of the galaxy. Looking at mass alone the rest of the galaxy "must have full control on all the stars in the aria". But gravitation does not depend on mass along! Gravitation also depends on distance (an inverse square law). What has the most influence over the stars is a complex issue. We have observed the stars for some decades, see that stars are in orbits around the MBH and thus are currently mostly influenced by the MBH. In the past the stars may have not orbited the MBH. In the future the stars may even stop orbiting the MBH.
Think about it.

Dave Lee
2016-May-16, 02:42 PM
The small fact that we have observed S2 during two entire orbits around Sgr A* (https://en.wikipedia.org/wiki/S2_(star)) tells us that it is in orbit around Sgr A*!
1. We have "the exact location of MBH with regards to the S2 orbit".
2. S2 (https://en.wikipedia.org/wiki/S2_(star)) was at periastron (nearest approach) in 2002 and has a 15.56 year period. So it will be next at periastron in 2017.
3. S2 (https://en.wikipedia.org/wiki/S2_(star)) has the information that you can use to work out orbital velocities.

Thanks
I couldn't find any real information about the orbital velocities at different locations of S2 at the elliptical cycle.
Please advice where can I find the requested info.

Reality Check
2016-May-16, 10:11 PM
In any case,
https://en.wikipedia.org/wiki/Binary_system shows that S2 and Sgr A* are a binary system like two stars orbiting each other are a binary system. A better explanation is Binary Star (https://en.wikipedia.org/wiki/Binary_star)

A binary star is a star system consisting of two stars orbiting around their common barycenter.


In any case, let's see if I understand it correctly:
Lots of misunderstanding follows, Dave Lee :eek:!

Based on actual measurements of the orbits of S2 and the other stars, S2 and the other stars are orbiting Sgr A*.
There was and will never be any shock to anyone who knows astronomy like the science community that S2 and Sgr A* were "nearly coincident" at one point in time in 2002.
This is almost high school level science. Measurements of orbits have error limits. Those error limits mean that when 2 objects are close together in the orbit they can be seen as "nearly coincident".
It is a implied insult to astronomers that any evidence was "fixed".
It is a fantasy that a MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta) figure is a shifting or fixing of any evidence.
You still cannot understand the bad scholarship of linking to figures.
"This activity had been justified by well known scientists under the word - ERROR." is gibberish.
Star Orbiting Massive Milky Way Centre Approaches to within 17 Light-Hours (http://www.eso.org/public/news/eso0226/) is a press release from 2002.
You still cannot understand the bad scholarship of linking to figures.
You cannot understand that there are no instruments that can detect gravitational lensing by Sgr A* yet.
You cannot understand that a close approach by S2 in 2002 is not shown in a different figure from a different year! The stars around Sgr A* move!.
You cannot understand what I wrote.
We have observed the stars for some decades, see that stars are in orbits around the MBH and thus are currently mostly influenced by the MBH. (http://cosmoquest.org/forum/showthread.php?p=2354336#post2354336) is not a solution for an imaginary problem. This is a statement of the fact that measurements of obits have shown that the orbits exist!

Reality Check
2016-May-16, 10:15 PM
I couldn't find any real information about the orbital velocities at different locations of S2 at the elliptical cycle.
Please advice where can I find the requested info.
No. I gave you a citation to the orbital parameters of S2. If you want these irrelevant velocities then you plug the numbers into the relevant equation.

Reality Check
2016-May-16, 10:25 PM
MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)

Figure 1. Finding chart of the S-star cluster. This figure is based on a natural guide star adaptive optics image obtained as part of this study, using NACO at UT4 (Yepun) of the VLT on 2007 July 20 in the H band. The original image with a FWHM of ≈75 mas was deconvolved with the Lucy–Richardson algorithm and beam restored with a Gaussian beam with FWHM=2 pixel = 26.5 mas. Stars as faint as mH = 19.2 (corresponding roughly to mK = 17.7) are detected at the 5σ level. Only stars that are unambiguously identified in several images have designated names, ranging from S1 to S112. Blue labels indicate early-type stars, red labels late-type stars. Stars with unknown spectral type are labelled in black. At the position of Sgr A* some light is seen, which could be either due to Sgr A* itself or due to a faint, so far unrecognized star being confused with Sgr A*.
There is a double blob. One of the blobs is S17. The other blob may be Sgr A* or an unrecognized star. Thus they label the two blobs with "S17/Sgr A* ?"

Reality Check
2016-May-16, 10:39 PM
MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)


We checked whether the residuals of the 2002 data, relative to an orbit fit to the data other than 2002, show some systematic trend (Figure 10) and found that indeed all points appear to be shifted systematically by 10 mas≈1 pixel toward the northeast. ...
Figure 10. 2002 data of S2. The gray symbols show the measured positions, the errors are as obtained from the standard analysis and are not yet enlarged by the procedure described in Section 3.5. The black dots are the positions predicted for the observation dates using an orbit fit obtained from all data other than 2002. The blue shaded areas indicate the uncertainties in the predicted positions resulting from the uncertainties of the orbital elements and of the potential, taking into account parameter correlations. The little ellipse close to the origin denotes the position of the fitted mass and the uncertainty in it. This plot shows that the S2 positions are dragged for most of the data by ≈10 mas to the northeast; they are not biased toward Sgr A*.
The authors are analyzing whether S2 could have been confused with another star. They conclude

From this analysis, it is clear that the weight of the 2002 data will influence the resulting orbit fits, since these points will systematically change the orbit figure at its pericenter. At the same time, we have no plausible explanation for the increase in brightness and the systematic residuals in the 2002 data; in particular a confusion event seems unlikely. Thus, it is clear that using the 2002 data will affect the results, but we cannot decide whether it biases toward the correct solution, or away from it. Therefore, we use in the following two options: (1) we include the 2002 data with the increased error bars; and (2) we completely disregard the 2002 data of S2.

Dave Lee
2016-May-17, 06:17 AM
Thanks again for all your efforts in answering my questions.

In order to be productive - let's focus on the key questions which I currently have, one by one.
Let's start with the following:

1. R0: What is the meaning of R0?
Let's start with the following image:
http://www.eso.org/public/usa/images/eso0226c/
in the image it is stated that the diameter of S2 cycle (at the narrow middle side of the elliptical cycle) is only 2 day light.
However, in the following article it is stated the R0 ≈ 8.1 kp.

http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075
It is stated:
"Figure 14. Fitted value of R0 for various scaling factors of the S2 2002 data, using a fit with the coordinate system priors. The factor by which the 2002 astrometric errors of the S2 data is scaled up strongly influences the distance. The mean factor determined in Figure 9 is ≈7, corresponding to R0 ≈ 8.1 kpc."
So what is the meaning of R0?

2. Velocities - It is critical for me to get the information about velocities. So far (thanks to you) I have only found the following:
https://en.wikipedia.org/wiki/S2_(star)
"this makes S2 the fastest known ballistic orbit, reaching speeds exceeding 5000 km/s (11,000,000 mph) or 1.67~% of the speed of light"
Please try to advice where can I find info about the orbital velocities at different locations of S2 at the elliptical cycle.

01101001
2016-May-17, 07:59 AM
In any case, let's assume that this is our error level – (as the scientist at that day drinks too much, so the focus was very poor).

Dave Lee, for someone who has been so wrong so many times about your interpretations of simple sentences regarding this topic's simple ideas, you would be wise to avoid being sarcastic about the work habits of the scientists involved, lest someone ascribe your many proven blunders to your own moral lapses.

Word to the wise: if you have more questions, you should ask them without adding your snarky, disrespectful comments.

I hope the kind people here refuse to assist you further until you apologize and exhibit reform. Earn their help.

Edit for clarfication: The phrase I quoted was removed as I wrote my reply.

tusenfem
2016-May-17, 11:29 AM
Thanks again for all your efforts in answering my questions.

In order to be productive - let's focus on the key questions which I currently have, one by one.
Let's start with the following:

1. R0: What is the meaning of R0?
Let's start with the following image:
http://www.eso.org/public/usa/images/eso0226c/
in the image it is stated that the diameter of S2 cycle (at the narrow middle side of the elliptical cycle) is only 2 day light.
However, in the following article it is stated the R0 ≈ 8.1 kp.

http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075
It is stated:
"Figure 14. Fitted value of R0 for various scaling factors of the S2 2002 data, using a fit with the coordinate system priors. The factor by which the 2002 astrometric errors of the S2 data is scaled up strongly influences the distance. The mean factor determined in Figure 9 is ≈7, corresponding to R0 ≈ 8.1 kpc."
So what is the meaning of R0?


Well, reading helps, in the paper (2nd link) it is stated that:
Our current best estimate for the distance to the Galactic center is R 0 = 8.33 ± 0.35 kpc.
In the figure (1st link) the minor axes of the ellipse is 2 light days.

Dave Lee
2016-May-17, 11:56 AM
Thanks

What is the size of the long axes of S2 ellipse?

With regards to Table 4:
http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075
http://iopscience.iop.org/0004-637X/692/2/1075/suppdata/apj297214t4_lr.gif
Table 4. Results for the Central Potential from Orbital Fitting


I'm not sure what is the meaning of R0 at this table.
The Min. value of R0 (for S2 only) is 6.63 Kpc and the max value is 8.8 Kpc.
So, what is the real meaning of this R0?
How could it be that the distance to the Galactic center is changing so dramatically?
Could it be that it represents the distance to S2?
If so, the difference between the minimal the maximal is:
8.8 – 6.63 = 2.17 Kpc.
One KPC is 3,261.63344 light years
In one year there are 365 days.
Therefore, the distance is:
2.17 x 3,261.63344 x 365 = 3,213,715 days light.
This distance represents the long axes of S2 ellipse.
It seems to me as unrealistic value, as the minor axes is only 2 days light.
If it is incorrect, what is the real value of the long axes of S2?

antoniseb
2016-May-17, 01:00 PM
... I'm not sure what is the meaning of R0 at this table.
The Min. value of R0 (for S2 only) is 6.63 Kpc and the max value is 8.8 Kpc.
So, what is the real meaning of this R0?...
It may be obvious to you if you think it through, R0 is the distance from Earth to S2. It has nothing to do with the size of the ellipse, only the apparent size from here.

Dave Lee
2016-May-17, 01:40 PM
It may be obvious to you if you think it through, R0 is the distance from Earth to S2. It has nothing to do with the size of the ellipse, only the apparent size from here.

Thanks

So in this case:
"R0 is the distance from Earth to S2."
The difference between the minimal to the maximal is:
8.8 – 6.63 = 2.17 Kpc.
One KPC is 3,261.63344 light years
In one year there are 365 days.
Therefore, the distance is:
2.17 x 3,261.63344 x 365 = 3,213,715 days light.
This value represents the difference between the minimal location with regards to the maximal location of S2 from Earth.
Therefore, it represents the long axes of S2 ellipse.

So, why do you claim : "It has nothing to do with the size of the ellipse"?

However, the long axes of S2 ellipse is 3,213,715 days light, while the minor axes is only 2 days light.

That is absolutely not logical.
Even if S2 will fly at a speed of light, it will take 3,213,715 days to cross the long axes.
So, please advice what could be the error in this calculation.
In any case, what is the value of the long axes of S2?

antoniseb
2016-May-17, 02:01 PM
... So, why do you claim : "It has nothing to do with the size of the ellipse"?...
Because the min and max values represent uncertainty, not observed variation in distance. Also, that uncertainty is much less now, as Tusenfem pointed out in post #133.
S2's observed orbital period is about 15 years, so it couldn't be orbiting a 6000 light year long ellipse in that time.

Hornblower
2016-May-17, 02:11 PM
Thanks

What is the size of the long axes of S2 ellipse?

With regards to Table 4:
http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075
http://iopscience.iop.org/0004-637X/692/2/1075/suppdata/apj297214t4_lr.gif
Table 4. Results for the Central Potential from Orbital Fitting


I'm not sure what is the meaning of R0 at this table.
The Min. value of R0 (for S2 only) is 6.63 Kpc and the max value is 8.8 Kpc.
So, what is the real meaning of this R0?
How could it be that the distance to the Galactic center is changing so dramatically?
Could it be that it represents the distance to S2?
If so, the difference between the minimal the maximal is:
8.8 – 6.63 = 2.17 Kpc.
One KPC is 3,261.63344 light years
In one year there are 365 days.
Therefore, the distance is:
2.17 x 3,261.63344 x 365 = 3,213,715 days light.
This distance represents the long axes of S2 ellipse.
It seems to me as unrealistic value, as the minor axes is only 2 days light.
If it is incorrect, what is the real value of the long axes of S2?

The estimates for R0 differ from one line item in Table 4 to the next because the authors have used different sets of data points to calculate them, in the process of looking for the possibility that there may be significant amounts of extended material that could perturb the orbits of S2 and the other stars. They differ from one another and from the overall best estimate which they gave in the abstract at the beginning of the paper.

In case there is any doubt, R0 is the distance from Earth to the center of the galaxy, and all evidence indicates that this black hole is at or very near the center. It has been estimated by various means over the past century, with the orbital studies of these close-in stars being the most recent. The best estimate in the abstract is the result of averaging all of these methods, with appropriate weight given to the various types of data according to the authors' confidence in their reliability. Once again, welcome to the rough and dirty world of astrometry, in which observational uncertainties are an unavoidable occupational hazard.

Dave Lee
2016-May-17, 02:12 PM
Because the min and max values represent uncertainty, not observed variation in distance. Also, that uncertainty is much less now, as Tusenfem pointed out in post #133.
S2's observed orbital period is about 15 years, so it couldn't be orbiting a 6000 light year long ellipse in that time.

Yes, sure. I fully agree.
So, as "the min and max values represent uncertainty", do we have now more updated/accurate data?
Why do we have to discuss about 2002 data?
What about 2016 data?

Hornblower
2016-May-17, 02:21 PM
Yes, sure. I fully agree.
So, as "the min and max values represent uncertainty", do we have now more updated/accurate data?
Why do we have to discuss about 2002 data?
What about 2016 data?

I would be interested in the data over the next two years, which will cover the next pericenter passage which is where the authors were finding anomalies in the year 2002 data which were difficult to account for. After that will be the next 15 years or so, giving us a chance to see if there is any measurable precession of the orbit. The authors noted that relativistic effects would cause a prograde precession in the absence of anything else, while a significant amount of extended unseen mass would cause a retrograde precession. We can reasonably expect present day observations to be more precise than the best available in 2002, with further improvement taking place after that. S2 is our best bet for continuing discoveries because of its short period and extremely close approaches to the black hole.

Dave Lee
2016-May-17, 02:47 PM
The estimates for R0 differ from one line item in Table 4 to the next because the authors have used different sets of data points to calculate them, in the process of looking for the possibility that there may be significant amounts of extended material that could perturb the orbits of S2 and the other stars. They differ from one another and from the overall best estimate which they gave in the abstract at the beginning of the paper.

In case there is any doubt, R0 is the distance from Earth to the center of the galaxy, and all evidence indicates that this black hole is at or very near the center. It has been estimated by various means over the past century, with the orbital studies of these close-in stars being the most recent. The best estimate in the abstract is the result of averaging all of these methods, with appropriate weight given to the various types of data according to the authors' confidence in their reliability. Once again, welcome to the rough and dirty world of astrometry, in which observational uncertainties are an unavoidable occupational hazard.

Thanks for the excellent explanation.
So, R0 indicates the distance from Earth to the center of the galaxy.
That is clear.
However, how can we know the distance from Earth to S2.

I assume that somewhere/somehow we keep this information.
So, I would mostly appreciate if there is any possibility to verify the distance/velocities at different locations in the S2 ellipse.

Dave Lee
2016-May-17, 03:05 PM
With regards to S2 plane

It's quite clear that if we look directly from a top view on the S2 plane, then the distance S2/Earth must be fix to all the ellipse.
However, if we are not located at the top view of S2 plane, then there must be a change in S2/Erath distance (even if it is a minor change).
I assume that from a distance of 8.1Kpc (1,190,265 light days) it is quite difficult to verify a change of a few light days - but this is very important.
So, the distance is a key element for our understanding about the real S2 plane.

Hornblower
2016-May-17, 03:24 PM
With regards to S2 plane

It's quite clear that if we look directly from a top view on the S2 plane, then the distance S2/Earth must be fix to all the ellipse.
However, if we are not located at the top view of S2 plane, then there must be a change in S2/Erath distance (even if it is a minor change).
I assume that from a distance of 8.1Kpc (1,190,265 light days) it is quite difficult to verify a change of a few light days - but this is very important.
So, the distance is a key element for our understanding about the real S2 plane.

Now we are getting somewhere. First things first. We have no way of independently determining the distance from us to S2 with enough certainty to tell whether it is exactly the same distance as the black hole, a few light-days closer, or a few light days farther. We make the realistic assumption that it is orbiting the Sgr A* black hole itself rather than something else that happens to be aligned with it. We fit an ellipse to the measured positions. If the orbit is not face on, the star's angular velocity at various parts of the apparent ellipse and the displacement, if any, of the black hole from the focus of the apparent ellipse enable experts to determine the inclination of the orbital plane to that of the sky and the orientation of the true line of apsides within that orbital plane. As always, there are uncertainties which the experts strive to reduce with more and better measurements. The orbital determination is always a work in progress. Double star observers have been doing just that for a couple of centuries, without the aid of computers for most of that period.

Dave Lee
2016-May-17, 03:33 PM
We fit an ellipse to the measured positions. If the orbit is not face on, the star's angular velocity at various parts of the apparent ellipse and the displacement, if any, of the black hole from the focus of the apparent ellipse enable experts to determine the inclination of the orbital plane to that of the sky and the orientation of the true line of apsides within that orbital plane.

Thanks
So, based on our knowledge/info, do we face on the S2 orbital plane?

antoniseb
2016-May-17, 03:47 PM
Thanks
So, based on our knowledge/info, do we face on the S2 orbital plane?
IIRC, it is inclined 48 degrees. This is something that can be determined fairly precisely.

Dave Lee
2016-May-17, 07:34 PM
IIRC, it is inclined 48 degrees. This is something that can be determined fairly precisely.

There are two axes: Minor and Long.
What is the incline in each one of those axes?
So far I didn't find any info about the size of the long axes.
Is it available anywhere? If yes, can you please share it with me? If no, then how did we get the result of the incline without all the info about the ellipse itself?

We make the realistic assumption that it is orbiting the Sgr A* black hole itself rather than something else that happens to be aligned with it. We fit an ellipse to the measured positions.
Is it possible to get info on the measured positions including the fit calculation?
What is the chance that it is a different shape of ellipse or even circular?

Dave Lee
2016-May-17, 08:27 PM
Let me set a simple calculation for a real circular orbit of S2:
It is stated:
https://en.wikipedia.org/wiki/S2_(star)
"this makes S2 the fastest known ballistic orbit, reaching speeds exceeding 5000 km/s (11,000,000 mph) or 1.67~% of the speed of light"
Now, let's assume that this is a constant speed in a circular orbit.
In this case, after 16 Years the circumference is:
16 x 0.0167x 365 = 97 days light.
The radius is:
97 / 3.14 / 2 = 15 days light.
So, based on the measured velocity we have got a circular orbit with a calculated radius of 15 days.
Could it be that based on this circular orbit and its orbital plane inclined, we see the current shape of S2 ellipse?

With regards to S2 velocity:
Please be aware that due to the incline, even if the real velocity of S2 is constant, we should see significant change in its velocity.
Therefore, I have asked more info about the velocities at different locations.

Please be aware that I don't claim the S2 has a circular orbit. I have set this simple calculation as a reference for maximal orbital cycle/Radius.

antoniseb
2016-May-17, 09:23 PM
... Please be aware that I don't claim the S2 has a circular orbit. I have set this simple calculation as a reference for maximal orbital cycle/Radius.
I'm not sure where you're going with all this... maybe just a bit of tourism, looking at extreme cases?
Anyway, just looking at various resources, S2 looks like its orbit is not very eccentric, perhaps 0.2 or 0.3.
If you are looking for something that moves very quickly for a small part of its orbit, check out S14.

Reality Check
2016-May-17, 10:50 PM
1. R0: What is the meaning of R0?
Unfortunately it means that you cannot understand your citations, Dave Lee.
No "R0" in your link The motion of a star around the central black hole in the milky way (http://www.eso.org/public/images/eso0226c/)
R0 is defined in MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta)

R0, the distance to the GC
So it is simply the distance to the Galactic center.



2. Velocities - It is critical for me to get the information about velocities. .
It is not critical to anyone else, Dave Lee, so do your own work.

Reality Check
2016-May-17, 10:53 PM
I'm not sure what is the meaning of R0 at this table.
The same as everywhere else in the paper: R0 = the distance to the Galactic center.

Reality Check
2016-May-17, 10:57 PM
What about 2016 data?
It is kind of hard for the paper published on February 23, 2009 you cited (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075)to discuss 2016 data :D!

Reality Check
2016-May-17, 11:19 PM
...Could it be that based on this circular orbit and its orbital plane inclined, we see the current shape of S2 ellipse?
The question answers itself - a elliptical (eccentricity > 0 and < 1) orbit can never be an circular orbit which has an eccentricity = 0. S2 has a measured eccentricity (e) = 0.881 ± 0.007 (https://en.wikipedia.org/wiki/S2_(star)).
Are you thinking that the orbit of S2 (and other S-stars?) is actually circular for some unknown reason but tilted to look elliptical, Dave Lee?
Astronomers know about orbital mechanics and how to reliably measure orbits. We can trust them not to confuse circular and elliptical orbits.

Dave Lee
2016-May-18, 06:53 PM
I have found some new data on S2:
http://www.solstation.com/x-objects/s2.htm
"S2 is apparently a "normal" star (ESO press release). It has about 15 times Sol's mass and seven times its diameter, and so, if it is a main sequence star, then S2 may be of spectral and luminosity type (early) B V (or even type O V). Given the frequency of supernovae at the galactic core, it is likely to be relatively more enriched than Sol with elements heavier than hydrogen ("metallicity"), based on its abundance of iron."
The star orbits the galaxy's central black hole at an average distance of 5.5 light-days that takes about 15.2 years to complete, at "an inclination of 46 degrees with respect to the plane of the sky" (MPE research introduction). Its highly elliptical orbit is fairly stable, as S2 would have to come 70 times closer to the hole (16 light-minutes) to be at risk of being destroyed by tidal forces from the hole's gravity. Early in 2002, S2 came very close to the black hole, coming within 17 light-hours or around three times the orbital distance of Pluto from the sun (or 39 AUs). At its nearest approach, it zooms around the black hole at speeds exceeding 11 million mph (5,000 km per second). Because of its extremely high eccentricity (e= 0.87) of orbit, however, S2 also moves as far away from the hole as 10 light-days. (See an infrared, time-lapse movie of the very fast motions of stars near the central black hole at Astronomy Picture of the Day.)

Based on this data the long axes is 10 light days.

However, that data isn't enough.
I would like to get more info about S2.


It is not critical to anyone else, Dave Lee, so do your own work.
Unfortunately, it seems that I'm the only one which is asking this data.
Without it, I can't even convince myself, so how can I convince anyone else.
So, let's assume that S2 Orbital motion is perfectly O.K.

However, let's look at other stars:

http://iopscience.iop.org/0004-637X/692/2/1075/downloadFigure/figure/apj297214f16

Somehow, it looks to me as fatal chaos.
Each star has its own orbit plane and at different orbital shape.
However, all of them must rotate around the MBH.

So let me ask the following:
1. How could it be that with all the gravity power of the MBH, those stars do not have a nice circular orbit at one plane as the planets in the solar system.
2. With all of this chaos, how could it be that they do not collide with each other?
3. Somehow it looks that the MSB is located at the inwards cycle of all the stars. However, do we know about any star in the galactic center that the MBH is out of his orbit cycle?
2. S13 - it looks like a circular orbit, while the MBH is not located at the center. Actually it is located at 1/4 of the diameter. In circular orbit the expectation is to see the MSB at the center (or close to the center). This is not the case for this star. So, how can we get an orbit fit?
3. S1 - It has a wide elliptical orbit, while the MSB is located almost at the center of the long axes and 1/3 of the short axes. In elliptical orbit the expectation is to see the MSB close to the edge of the long axes. This is not the case for this star. So, how can we set an orbit fit?
4. S6 - It has a narrow elliptical orbit, while the MSB is located directly on the line of the orbit cycle. It seems that there is high possibility for collision between S6 and MSB. It is also not located at the edge of the long axes. So how can we set an orbit fit?

Hornblower
2016-May-18, 07:03 PM
The question answers itself - a elliptical (eccentricity > 0 and < 1) orbit can never be an circular orbit which has an eccentricity = 0. S2 has a measured eccentricity (e) = 0.881 ± 0.007 (https://en.wikipedia.org/wiki/S2_(star)).
Are you thinking that the orbit of S2 (and other S-stars?) is actually circular for some unknown reason but tilted to look elliptical, Dave Lee?
Astronomers know about orbital mechanics and how to reliably measure orbits. We can trust them not to confuse circular and elliptical orbits.
Let me add to that. An inclined circle that projects as an apparent ellipse would have the black hole at the midpoint of the apparent major axis. In the case of S2 the black hole is clearly near one end. Although they did not bother to show the dates of the data points in the diagrams and tables, I can say with virtually 100% certainty that S2 was moving much faster around that end than around the opposite end.

To repeat, the mathematical techniques for fitting an inclined Keplerian ellipse to a set of observed positions have been known for a couple of centuries, and the great observers of the past did such fitting without the aid of computers as we know them until well into the 20th century. There is nothing new here.

Reality Check
2016-May-18, 09:31 PM
Without it, I can't even convince myself, so how can I convince anyone else.
You cannot convince yourself of what, Dave Lee? That astronomers do not know astronomy :p?
The rest of the post is mostly more assertions from ignorance.

The mass of the primary object does not magically make orbits circular.
This is especially true for SMBH in the center of galaxies. Stars will form in orbits with different eccentricity because the clouds they condense from will have different angular momentum. A circular orbit is the extremely rare case of angular momentum = 0.
The stars are light years apart and have different inclinations so they do not collide.
You may as well ask why the planets in the solar system so not collide :D!
A primary such as Sgr A* is always inside the orbit of objects orbiting it :eek!
There are millions of stars in the Milky Way (including the Sun) that are not orbiting Sgr A*. That includes many in the galactic center.
From this thread, what you imagine you see in a diagram is usually not what is happening, Dave Lee, especially with the ignorance of just linking to the diagram, ignoring its caption and description in the paper and ignoring the actual content of the paper!
Table 7. Orbital Parameters of Those S-stars for Which We Were Able to Determine Orbits (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075)

Dave Lee
2016-May-21, 03:59 AM
There are millions of stars in the Milky Way (including the Sun) that are not orbiting Sgr A*. That includes many in the galactic center.

So there are stars in the galactic center that are not orbiting Sgr A*.
Thanks for this important statement.
I assume that the answer for this phenomenon is as follow:


The MBH has a large mass compared to the stars. The MBH has a tiny mass compared t the rest of the galaxy. Looking at mass alone the rest of the galaxy "must have full control on all the stars in the aria". But gravitation does not depend on mass along! Gravitation also depends on distance (an inverse square law). What has the most influence over the stars is a complex issue. We have observed the stars for some decades, see that stars are in orbits around the MBH and thus are currently mostly influenced by the MBH. In the past the stars may have not orbited the MBH. In the future the stars may even stop orbiting the MBH.
Hence, even in the galactic center, "Gravitation does not depend on mass along! Gravitation also depends on distance (an inverse square law).
This gravitation of mass is:

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.)
Therefore:
1. 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.
2. It's only true for objects in circular orbits around the center of mass of a spherically symmetric system.
3. If the outer shell is not spherically symmetric, the gravitation does not cancel out.

Hence, if I understand it correctly:
A spherically symmetric system is considered only for stars in circular orbits around the center of mass. For any star in spherically symmetric system - the gravitation of masses is within that sphere.
However, as most of S.. stars do not have a circular orbits, then by definition they are not located at a spherically symmetric system.
Therefore, for those stars, we can claim that their gravitation of masses depends on:
As stated:

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.

cjameshuff
2016-May-21, 11:44 AM
So there are stars in the galactic center that are not orbiting Sgr A*.

You have been told repeatedly that you are assigning too much importance to that single object. It is the largest single object in the core, but it is just one object of many influencing the trajectories of objects passing through the core, and only contains a tiny fraction of the galaxy's mass.

Many of the stars in the bulge of the galactic center are on highly eccentric orbits, and of these, those that pass close to Sgr A* are above escape velocity for it. Others simply don't approach close to it. Only a tiny fraction of the core's stars are both close enough to Sgr A* and moving slowly enough to be caught in orbits around it. And the galaxy is not a uniform distribution of mass, most of the mass is condensed into stars, and many of those stars are orbiting other stars as they orbit within the bulge, or having their orbits within the bulge being deflected by close passes with other stars.




1. 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.
2. It's only true for objects in circular orbits around the center of mass of a spherically symmetric system.

No, it is true for all objects. Their orbit is a result of gravitation, not a cause.



A spherically symmetric system is considered only for stars in circular orbits around the center of mass. For any star in spherically symmetric system - the gravitation of masses is within that sphere.
However, as most of S.. stars do not have a circular orbits, then by definition they are not located at a spherically symmetric system.

No. In general, closed orbits around a point mass, such as the case of stars in orbit around Sgr A*, are elliptical. Circular orbits are just a special case, one that happens to be possible in all systems with radial symmetry.

Dave Lee
2016-May-21, 12:32 PM
With regards to "spherically symmetric systems";
In your first message you have stated that:

it's only true for objects in circular orbits around the center of mass of a spherically symmetric system.
Now you claim that:


No. In general, closed orbits around a point mass, such as the case of stars in orbit around Sgr A*, are elliptical. Circular orbits are just a special case, one that happens to be possible in all systems with radial symmetry.
So, based on the first answer I had the impression that a spherically symmetric system is considered only for stars with circular orbit.
Now you claim that Circular orbits are just a special case for elliptical orbit.
So, do you mean that spherically symmetric system is considered also for all stars with elliptical orbit? Hence, with regards to S. stars, it is considered that they rotate around a spherically symmetric system even if they have an elliptical orbit at any shape. Is it correct?

cjameshuff
2016-May-21, 01:37 PM
*particularly blatant cherrypicking snipped*

The "it's only true" bit referred to a specific statement in the post I was replying to, not to whatever you decide to apply it to after removing the quote from context.

In case anyone reading hasn't figured this out, please assume anything I said that is being quoted by Dave Lee is being distorted to mean something other than what I intended. Nothing I've written intentionally supports any of his ideas. He will invariably bring things around to some variation of "asking" if stars could actually be in circular orbits around some point in empty space, as opposed to elliptical or more complex orbits. The answer will always be "no".

If anyone's got genuine questions, ask away.

Dave Lee
2016-May-21, 02:43 PM
In case anyone reading hasn't figured this out, please assume anything I said that is being quoted by Dave Lee is being distorted to mean something other than what I intended.

I can promise you that I have no intention to distort your message. I have quoted as I really understood your valuable answers.
Sorry that I have quoted you without your permission.
However, if you don't like to be quoted, then please try to avoid answering my questions.

01101001
2016-May-21, 02:51 PM
However, if you don't like to be quoted, then please try to avoid answering my questions.

Everyone, please follow this advice.

Swift
2016-May-21, 07:06 PM
Closed pending moderator discussion