Milky way galaxy mass vs. black hole mass

[Strange]

Sorry, this is a severe mistake!

 

Well, an approximation certainly.

 

So, the extended mass is surrounding stars.

Those stars do not just hang there in one spot. They must be in full rotate motion around the Black hole. (Please remember that the black hole attracts any mass, even if it is called - "extended mass".

 

Correct. But if the stars are just moving around inside the orbit (of a given star) then it doesn't matter (they are just part of the "extended mass").

 

If they just move about outside the orbit of that star, then it doesn't matter (they have no effect on the orbit).

 

If stars move from inside to outside the orbital distance, or vice versa, then they change the mass of the "extended mass". But, given the number of stars involved, that is unlikely to be significant either.

 

Let's assume that there are no spiral arms and that it is a perfect system:

 

The spiral arms are irrelevant here: we are talking about a cluster of stars at the center of the galaxy.

 

 

If I understand you correctly, you believe that the inwards ring will attracts S2 while the outwards ring will not.

 

Looking at rings of stars is more complex (unless you ca simplify it to a 2D problem.

 

But the stars in the paper can be treated as a reasonably symmetrical spherical distribution.

[David Levy]

Correct. But if the stars are just moving around inside the orbit (of a given star) then it doesn't matter (they are just part of the "extended mass").

 

If they just move about outside the orbit of that star, then it doesn't matter (they have no effect on the orbit).

Sorry, but this is incorrect.

Let me explain it to you in the following way.

Any ring is a collection of stars.

Let's start by focusing in just one star in each ring and try to verify how it attracts S2. This star should be located in the same radius vector as S2. So as S2 rotates around the B.H., this star rotates at the same radius vector (with diffrent distance from the B.H).

Please remember the following statement from Newton:

"Every point mass attracts every single other point mass by a force pointing along the line intersecting both points"

Newton didn't specify an inwards or outwards. He just claimed – every point of mass attracts every other single other point of mass.

Therefore, any star at any ring should have a gravity force on S2.

Actually I have already explained it the other tread.

Let me just copy it for you as follow (I have used Sun and Earth system as an example for gravity force calculation):

 

"Gravity force in a Simple orbiting system

Let's use Sun and Earth (without any other planets or moons) as an example of gravity force. We know very well how to calculate the gravity force which attracts the Earth to the Sun. Let's call it as follow:

Gravity force – F, Radius – R, Sun mass – M.

In the following examples we will set the following:

- Split the sun into two equally stars. The mass of each star will be 1/2 x M.

- One star will be set at the same place as the Sun. This star will be called Hoster.

- The Earth will revolve around the hoster. The radius is – R.

-The second star will be moved on the vector of the radius to different distances from the Earth.

- So, while the Earth revolve around the hoster, the second star will keep its position on the on the same vector line of the radius R (but at different distances), so those two stars should be in one line with the Earth.

- We will calculate the gravity force vector that each star attracts the Earth. The sum of those forces should is the equivalent gravity force on Earth.

- This equivalent gravity force can be represented by Virtual hoster. Based on the radius R, we will calculate the Equivalent virtual hoster mass. The Idea is that instead of using two stars system, we will find the Equivalent mass which can replace those two stars.

1. Let's keep the second star at the same place as the Hoster. In this case, each star attracts the Earth at a gravity force of 1/2 x F. The equivalent gravity force vector on Earth is as follow:

Equivalent gravity force vector = 1/2 x F (From the Hoster) + 1/2 x F (From the second star) = F.

As stated, this equivalent gravity force vector could be represented by a virtual hoster. It's equivalent mass will be M (as the sum). This is quite clear, as those two stars together, have the same mass as the Sun.

2 . Let's move the second star further away from the hoster, so its distance will be doubled than the current distance between the Earth to Hoster. In this case, the gravity contribution of second star will be decreased by four. (Please remember that - The gravity force is inversely proportional to the square of the distance between any star and Earth). Therefore, its contribution to the gravity on earth will be 1/4 x F.

Hence, the new equivalent gravity force vector on earth is:

Equivalent gravity force vector = 1/2 x F (From the first star) + 1/8 x F (From the second star) = 5/8 x F.

Therefore, the equivalent Virtual hoster mass is 5/8 x M.

Hence, by using a virtual hoster at a mass of 5/8 x M we can get the same gravity force on earth as the two stars system.

3. Let's position the second star at the opposite direction from the hoster with regards to Earth. Hence, the Earth will be placed in between the hoster and the second star. The distance to each direction will be the same. In this case, the gravity force from each side of the earth will be half of F, but at the opposite polarity. Therefore, the sum of those gravity force vectors will be zero. Hence, the equivalent virtual hoster mass is zero.

4. Let's position the second star at the opposite direction from the hoster with radius of 2 x R (doubled distance of R). The new equivalent gravity force on Earth is:

Equivalent gravity force = 1/2 x F (From the first star) - 1/8 x F (From the second star) = 3/8 x F.

Therefore, the Equivalent virtual hoster mass is 3/8 x M.

5. Let's position the second star at the opposite direction from the hoster with radius of 1/2 x R (Half distance of R). The new equivalent gravity force on Earth is:

Equivalent gravity force = 1/2 x F (From the first star) - 2 x F (From the second star) = -1.5 x F.

This time, we have got a negative gravity force. That's mean that the Earth should disconnect from the hoster.

5. Let's split the second star to several stars with different mass at each one. Let's position those divided stars in the same vector line as R but at different random locations of both sides of earth.

The Equivalent force vector should be the sum of all gravity force vectors that those stars and the Hoster attract the Earth.

However, the stars in one side should contribute gravity force in one direction while the stars in the side of the Earth should contribute a gravity force in the opposite direction. Therefore, in order to get the equivalent force vector, we need to sum all vectors in one direction and subtract it from the sum of all vectors in the other direction.

Based on the equivalent gravity force, we can calculate the equivalent virtual hoster mass.

This equivalent virtual hoster mass can replace all of this system."

 

 

Now, it should be clear that there is a gravity force vector on S2 from a star which is located in the inwards ring or the outwards ring.

In the same token, there are gravity force vectors on S2 from any number of stars (it could be an infinit stars in each ring) in the inwards ring or the outwards ring.

There is no cancelation of the gravity force from a specific ring.

We just need to add the total gravity force vectors on S2 from both rings (inwards and outwards rings)!

Edited January 28, 2015 by David Levy

[Strange]

Sorry, but this is incorrect.

Let me explain it to you in the following way.

Any ring is a collection of stars.

 

We are not talking about a ring of stars. We are talking about a roughly spherical cluster.

[David Levy]

No, it must be a ring.

However, let's assume that it is: a "roughly spherical cluster".

Let's also assume that we only have the B.H., this roughly spherical cluster and S2.

The B.H is located in the center, than this roughly spherical cluster and at the outwards side we will position S2.

So, you agree that S2 will be attracted by the B.H. gravity + by the roughly spherical cluster gravity.

Now, let's call back our good friend – Newton.

I assume that you have heard about his third low:

http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law

"For every action, there is an equal and opposite reaction"

"The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object"

So, you agree that this roughly spherical cluster attracts S2 by a gravity force – F (as it is located in the inwards side).

However, based on our dear Newton third low, S2 must also attract this roughly spherical cluster with the same force – F (but at the opposite direction).

Therefore, by definition - this roughly spherical cluster had been attracted by the B.H. gravity force plus by S2 gravity force (Although S2 is located in the outwards side of the roughly spherical cluster).

In other words, it doesn't matter if you are in or out – you should fill the gravity force from any direction.

If you still do not agree – than you have to argue with Newton,

Edited January 29, 2015 by David Levy

[Mordred]

It took me a while to relocate this particular article. It is probably one of the more extensive and explanative articles I've read on galaxy rotation curves including the mass to luminosity relations. Of particular interest for your calculations is the region mass breakdown on the chart of page 25.

 

The other key importance is the application of the virial theorem.

 

http://arxiv.org/abs/1307.8215

 

 

You probably will find this article a great aid.

 

Particularly since it covers the problems with using strictly the Kepler laws in rotation curves

In particular the influence and distribution of the dark matter halo

Edited January 29, 2015 by Mordred

[Strange]

No, it must be a ring.

However, let's assume that it is: a "roughly spherical cluster"

 

Which it is.

 

Let's also assume that we only have the B.H., this roughly spherical cluster and S2.

The B.H is located in the center, than this roughly spherical cluster and at the outwards side we will position S2.

So, you agree that S2 will be attracted by the B.H. gravity + by the roughly spherical cluster gravity.

 

S2 is right at the centre of the cluster, orbiting very close to the black hole.

 

I'm not quite sure what point you are trying to make. All I am pointing out is that the orbit of S2 (near the centre of the cluster) will be largely unaffected by the stars in the cluster - as demonstrated by Newton.

 

S2 (near the centre of the cluster) will only be affected by the mass of the black hole and (to a tiny degree) other stars nearby.

 

Stars further from the black hole (more than about 1/3rd of a parsec) are noticeably affected by the inner stars as well as the black hole. (As shown in the paper.)

 

A star at the edge of the cluster will be affected by all the stars in the cluster plus the black hole.

Edited January 29, 2015 by Strange

[David Levy]

 

S2 is right at the centre of the cluster, orbiting very close to the black hole

It seems that you have missed the point.

I have use S2 as an example for Newton low.

So please, try to focus on my explanation on Newton lows and let me know if it is clear for you.

 

With regards to roughly spherical cluster;

Would you kindly explain what is it? How far is it from the B.H?

If it is made by stars (or mass), why do you claim that it is not a ring of stars (or mass)?.

If it isn't rotating around the B.H, what kind of force holds it in its position?

[Strange]

It seems that you have missed the point.

I have use S2 as an example for Newton low.

 

And I have tried to use it to explain Newton's Shell Theorem.

 

To understand the orbit of S2, all that matters is the mass of the central black hole. The mass of S2 is much smaller and can be ignored. The gravitational forces of the surrounding stars cancel out.

 

With regards to roughly spherical cluster;

Would you kindly explain what is it? How far is it from the B.H?

 

This is all described in the paper you quoted from. It is a dense cluster of stars at the centre of the galaxy (it seems that most spiral galaxies have a central nucleus like this). It is about 3 parsecs in radius (I think, I can't be bothered to go and read the paper again right now) surrounding the balck hole.

 

If it is made by stars (or mass), why do you claim that it is not a ring of stars (or mass)?.

 

Because it is a globular cluster.

 

If it isn't rotating around the B.H, what kind of force holds it in its position?

 

The stars are all orbiting around the black hole (or, more accurately, the centre of mass of the balck hole and the cluster). They have different speeds and directions but the overall rotation is parallel to the plane of the galaxy.

[pzkpfw]

David, imagine you are floating in space with a mass in front, and another five equal masses the same distances behind, left, right, above and below you. That is, you are exactly in the middle of these six equal masses.

 

Clearly all their effects will cancel out. e.g. The mass in front "pulls you" forwards, but the mass behind "pulls you" equally backwards.

 

Now, using rockets move yourself towards the front mass, and then reverse to stop. You might think that now you are closer to the front mass (and further from the rear mass), it's "force" would be stronger and you'd be pulled towards it - that you'd begin to drift forwards.

 

But that doesn't happen; because the masses that were left, right, above and below you are now also a bit behind you. Their "force" in the rearwards direction adds to that of the "force" from the rear mass.

 

Newton, who could do calculus, figured out (proved mathematically) that inside a shell of equal density, these "forces" all cancel each other out. e.g. if Earth was a thin shell of ultra dense material, so that it's surface gravity was exactly the same as it is now, but it was hollow - you could be anywhere inside that shell and you'd feel no "force" of gravity in any direction. That's the Shell Theorem.

 

And it's why, as we can treat the Galaxy as homogenous (all the stars more or less evenly spread), S2 can pretty much ignore the stars further out than it. It's the masses (such as the black hole) inside its orbit that matter most.

[David Levy]

 

 

To understand the orbit of S2, all that matters is the mass of the central black hole.

That is incorrect!

Based on Newton – The mass outside the center is also important.

The gravitational forces of the surrounding stars cancel out.

No! Those gravity force vectors have a significant effect on S2.

Please try to read again my explanations about Newton lows.

With regards to roughly spherical cluster;

This is all described in the paper you quoted from. It is a dense cluster of stars at the centre of the galaxy (it seems that most spiral galaxies have a central nucleus like this). It is about 3 parsecs in radius (I think, I can't be bothered to go and read the paper again right now) surrounding the balck hole.

Sorry, I couldn't find answers to my questions about this cluster.

What do you mean by – surrounding the black hole?

Does it mean that this cluster rotate around the B.H?

As the cluster is made by stars, it is expected that after several rotations around the B.H. we should get a ring of stars (as in any disc system). Why not?

If you insist that the cluster keeps its structure of "dense stars", what kind of force keep those stars together? Is it Newton or some dark power?

If it keeps its structure, and still rotate around the B.H, than it should have some effect on the B.H.

Please remember that its mass could be up to 1.5 Million solar mass. It is almost 40% of the B.H. itself. Therefore, they should have significant effect on each other. Actually it sounds like a binary star:

"A binary star is a star system consisting of two stars orbiting around their common center of mass"

In this case, those two objects should orbit around their common center.

But this is not the case. So, it means that there is no cluster (or did I miss something)…

[Strange]

That is incorrect!

Based on Newton – The mass outside the center is also important.

 

No. I am telling you what Newton actually said (proved): http://en.wikipedia.org/wiki/Shell_theorem

 

What do you mean by – surrounding the black hole?

Does it mean that this cluster rotate around the B.H?

 

Yes. As stated in the paper that you quoted from.

 

As the cluster is made by stars, it is expected that after several rotations around the B.H. we should get a ring of stars (as in any disc system). Why not?

 

Because your expectation is wrong?

 

If you insist that the cluster keeps its structure of "dense stars", what kind of force keep those stars together? Is it Newton or some dark power?

 

Just gravity and orbital velocity.

 

In this case, those two objects should orbit around their common center.

But this is not the case.

 

This is the case. (Except there are not two objects, there are thousands) The paper says that the stars in the cluster and the black hole all orbit the barycenter (i.e. their common center of mass).

[David Levy]

 

 

No. I am telling you what Newton actually said (proved): http://en.wikipedia.org/wiki/Shell_theorem

O.K. I see your point.

However, this theory isn't applicable in our case.

It is sated:

"In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body."

"A corollary is that inside a solid sphere of constant density the gravitational force varies linearly with distance from the centre, becoming zero by symmetry at the centre of mass:

Hence, if the stars could set a solid constant density in a perfect ring shape around the center, tail to noise, than we could potentially use this theory.

The reality is not even close.

Please look again on the following rotation rings of S1 to S14 (Just as an example for the reality).

http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg

Each one of those stars has its own ring. There are no two stars in the same ring. It is not even a symmetrical rings but elliptical. At any given time, each star is located at different distance from the center and has different velocity.

Is this a spherically symmetrical body?

This is far away from a " a spherically symmetrical body" or " a solid sphere of constant density".

How could you even compare those two systems?

The shell theorem is ideal theory. The science take this ideal theory and try to imply in the reality on a totally different system.

This is a sever mistake.

The science actually destroys Newton lows, by Newton ideal theory…

Now I can understand why the science estimates that:

"In Spiral galaxies the orbiting of stars around their centers seems to strongly disobey to Newton's law of universal gravitation".

We have to follow Newton lows! There is no short cut. We must calculate the gravity effect on each star by all the mass in the system.

As Newton had already declared:

"Every point mass attracts every single other point mass by a force pointing along the line intersecting both points."

Once we do it, we could see how simple it is, and why Newton lows perfectly fits spiral galaxy activity!

[Strange]

Hence, if the stars could set a solid constant density in a perfect ring shape around the center, tail to noise, than we could potentially use this theory.

 

If it was a ring then we could not.

 

If it was a reasonably even, spherical distribution of stars (which it is) then it is a useful approximation. Once could try and calculate the effect of ever star in the galaxy but this would be (a) impossible and (b) would not make a significant difference.

 

Science deals with approximations like this all the time.

http://en.wikipedia.org/wiki/Spherical_cow

 

Please look again on the following rotation rings of S1 to S14 (Just as an example for the reality).

http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg

Each one of those stars has its own ring. There are no two stars in the same ring. It is not even a symmetrical rings but elliptical. At any given time, each star is located at different distance from the center and has different velocity.

Is this a spherically symmetrical body?

 

No. They are indivudal stars. The cluster can be approximated as a spherically symmetrical body.

 

Once we do it, we could see how simple it is, and why Newton lows perfectly fits spiral galaxy activity

 

Perhaps you can show your calculations to support this, as it appears to disagree with everyone else's.

[David Levy]

 

The cluster can be approximated as a spherically symmetrical body.

Well, we all know that stars in a cluster could be in a different mass, different radius from the center, different rotation direction, different cycle shape (elliptical for example) and it might contain some sporadically clouds of dust.

One star could be in the size of a moon while the other could be in the size of 100 solar mass.

However, it is believed that it is good approximation to assume that it is symmetrical body.

Technically, we could also claim that if we ignore the issue of mass, than by approximation, mouse is quite simmilar to an elephant.

I have no intention to argue about approximation.

I just wonder why the science prefers to use this shell theorem instead of focusing on the real Newton lows.

I can prove that by using Newton lows we can explain the activity of spiral galaxy. However, we need to abandon this idea of shell theorem.

[Strange]

I just wonder why the science prefers to use this shell theorem

 

It is used when it is a reasonable approximation (as the results in the paper you quote from shows it to be). Accounting for all the stars in the galaxy (and all the toehr galaxies in the universe?) would be infeasibly hard. And, we know, from Newton's work, that it wouldn't make a significant difference.

 

 

instead of focusing on the real Newton lows.

 

The shell theorem, as you can see, is derived from Newton's laws.

 

I can prove that by using Newton lows we can explain the activity of spiral galaxy.

 

Go on then.

[David Levy]

 

David, imagine you are floating in space with a mass in front, and another five equal masses the same distances behind, left, right, above and below you. That is, you are exactly in the middle of these six equal masses.

Clearly all their effects will cancel out..

Thanks Pzkpfw.

With your permition, please let me answer you as follow:

Imagine you are floating in space with a mass in front, and another five different masses, different distances behind, left, right, above and below you. That is, you are approximation in the middle of these six different masses

Clearly by approximation, all their effects will cancel out.

Do you agree?

 

Go on then.

Thanks

Would you kindly read my reply in pg. 27.

S2 is using just as an example for star. You can call it under any name.

So please just see if you agree with this explanation of Newton low.

Please, no "roughly spherical cluster" and no "shell theorem" .

Edited January 30, 2015 by David Levy

[Strange]

Thanks Pzkpfw.

With your permition, please let me answer you as follow:

Imagine you are floating in space with a mass in front, and another five different masses, different distances behind, left, right, above and below you. That is, you are approximation in the middle of these six different masses

Clearly by approximation, all their effects will cancel out.

Do you agree?

 

You don't give enough information to agree or disagree.

 

IF you carefully arrange the distances so that the forces due to each mass is equal then: agree.

 

If you just have random different masses and distances, then: disagree.

Would you kindly read my reply in pg. 27.

S2 is using just as an example for star. You can call it under any name.

So please just see if you agree with this explanation of Newton low.

Please, no "roughly spherical cluster" and no "shell theorem" .

 

I don't have time to go through this now (but I agree that the shell theorem is irrelevant in the example you give because you just have three objects). I will give a more detailked reply later.

[pzkpfw]

Thanks Pzkpfw.

With your permition, please let me answer you as follow:

Imagine you are floating in space with a mass in front, and another five different masses, different distances behind, left, right, above and below you. That is, you are approximation in the middle of these six different masses

Clearly by approximation, all their effects will cancel out.

Do you agree?

Strange is right, that there isn't enough information there; but since I can see what you're trying to do with that, I'll just go ahead and (simplify, and) say "no".

 

The thing is, if you're going to reject the shell theorem because it's an approximation, you can't just substitute that with another approximation. If you're going to insist on ignoring the shell theorem, you need to do something like actually calculate the effects of the individual stars outside the orbit of S2, not just talk your way into a claim that supports your idea.

 

Go ahead.

[David Levy]

 

Strange is right, that there isn't enough information there; but since I can see what you're trying to do with that, I'll just go ahead and (simplify, and) say "no".

 

The thing is, if you're going to reject the shell theorem because it's an approximation, you can't just substitute that with another approximation. If you're going to insist on ignoring the shell theorem, you need to do something like actually calculate the effects of the individual stars outside the orbit of S2, not just talk your way into a claim that supports your idea.

 

Go ahead.

Thanks

Let's start by pg. 27.

Please advice your feedback.

Edited January 31, 2015 by David Levy

[ACG52]

 

 

Let's start by pg. 27.

Please advice your feedback.

Your premise that stars are in a ring is simply incorrect. We know that by direct observation. This central bulge is certainly not a ring.

 

[Strange]

I said I would come back to your post 27. But there is so much here, I will just pick up on a couple of points.

 

- So, while the Earth revolve around the hoster, the second star will keep its position on the on the same vector line of the radius R (but at different distances), so those two stars should be in one line with the Earth.

 

This is physically impossible, in terms of orbits: the second star cannot remain between Earth and the first star because their orbital velocities will not allow it. However, you can calculate the sum of forces at the instant they aligned as you describe.

 

 

2 . Let's move the second star further away from the hoster, so its distance will be doubled than the current distance between the Earth to Hoster. In this case, the gravity contribution of second star will be decreased by four. (Please remember that - The gravity force is inversely proportional to the square of the distance between any star and Earth). Therefore, its contribution to the gravity on earth will be 1/4 x F.

 

If the second star is now twice as far from "Hoster" as Earth, then the Earth is midway between them. It will feel equal force from both. I assume this is not what you intend to describe, but I don't know what you intend.

 

But, yes, you can calculate the force contributed by each object based on mass and distance and perform a vector sum (taking into account the angles) to get the total force. So what?

 

There is no cancelation of the gravity force from a specific ring.
We just need to add the total gravity force vectors on S2 from both rings (inwards and outwards rings)!

 

I don't know what "rings" you are talking about, as you example just had three bodies: Earth and two stars.

 

Apart from stating the (obvious) fact that you can do a vector sum of the forces of multiple objects, what is the point of this long exposition?

[David Levy]

This is physically impossible, in terms of orbits: the second star cannot remain between Earth and the first star because their orbital velocities will not allow it. However, you can calculate the sum of forces at the instant they aligned as you describe.

It is only a theory. The main idea is to keep the forces in the same vector line. Please try to look at that just as an example for Newton gravity force.

If the second star is now twice as far from "Hoster" as Earth, then the Earth is midway between them. It will feel equal force from both. I assume this is not what you intend to describe, but I don't know what you intend.

No. Assuming that the distance between the Hoster and earth is R, than the distance between the second star and the earth is 2R. Is it clear?

But, yes, you can calculate the force contributed by each object based on mass and distance and perform a vector sum (taking into account the angles) to get the total force. So what?

Apart from stating the (obvious) fact that you can do a vector sum of the forces of multiple objects, what is the point of this long exposition?

This is Newton. I'm just giving few examples for the effect of Newton gravity. This is only the first step in getting the full picture.

Assuming that you ask the following question:

What is the velocity - if it takes you 5 hours to fly with Delta air lines from N.Y to L.A.

And the answer is – I don't fly with delta…

So, with this kind of negative approach it seems that we are just wasting our time.

Edited February 1, 2015 by David Levy

[Strange]

No. Assuming that the distance between the Hoster and earth is R, than the distance between the second star and the earth is 2R. Is it clear?

 

So you were moving the second star away from Earth? So it is further away than the first start? That wasn't very clear. A diagram would have been useful.

So, with this kind of negative approach it seems that we are just wasting our time.

 

You said you were going to explain, mathematically, the rotation of stars in a galaxy. Instead of that long rambling and confusing post 27, you could have just written [math]f = G\frac{m_1 m_2}{r^2}[/math] and [math]\vec{f} = \vec{f_1} + \vec{f_2}[/math]. And then gone on to explain the orbital velocities of stars.

[David Levy]

 

You said you were going to explain, mathematically, the rotation of stars in a galaxy. Instead of that long rambling and confusing post 27, you could have just written [math]f = G\frac{m_1 m_2}{r^2}[/math] and [math]\vec{f} = \vec{f_1} + \vec{f_2}[/math]. And then gone on to explain the orbital velocities of stars.

O.K.

It seems that you have got the idea.

However,

If we focus on a star in spiral arm (Sun for example) –

Its equivalent gravity force vector should be represented by the following:

[math]\vec{f} =\vec{f_c}+\vec{f_n}+\vec{f_m}[/math].

While

(center)- Representing the gravity force vectors by the center – mainly by the black hole.

(n) - Representing gravity force vectors of all the stars in its spiral arm.

(m) - Representing gravity force vectors of all the other stars in the spiral galaxy (Please be aware that this vector is quite neglected).

Therefore, the equivalent gravity force vector on a star in the outwards side of the arm is stronger than the one in the inwards side of the arm.

Is it clear?

Edited February 7, 2015 by David Levy

[Strange]

Therefore, the equivalent gravity force vector on a star in the outwards side of the arm is stronger than the one in the inwards side of the arm.

 

You need to sum all the relevant forces to show that.