David Levy

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?

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. 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? 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.

Thanks Let's start by pg. 27. Please advice your feedback.

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? 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" .

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.

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!

That is incorrect! Based on Newton – The mass outside the center is also important. 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; 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)…

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?

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,

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)!

Sorry, this is a severe mistake! Let me use your following answer: 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". Let's assume that there are no spiral arms and that it is a perfect system: So those stars set a perfect ring around the black hole. Actually, let assume that there are two perfect rings. Let's position S2 (for example) in a rotate motion between those two rings. If I understand you correctly, you believe that the inwards ring will attracts S2 while the outwards ring will not. Please explain why.

Thanks. Well, based on Newton: " "Every point mass attracts every single other point mass by a force pointing along the line intersecting both points" Therefore, every point mass (star for example) attracts every other point mass (another star) by a force which is called gravity force. So, if we try to calculate the gravity force of S2 for example, we need to take in account all the gravity force vectors which attract this star. (Inwards to the center plus outwards from the center). Hence, it should be the sum of the black hole gravity force on S2, plus The gravity force vectors of all the stars (or clusters) between S2 and the Black hole, minus The gravity force vectors of all the stars (or clusters) from S2 and outwards The outcome of all those gravity force vectors will be the equivalent gravity force which attracts that star. This equivalent gravity force should be in full balance with the proper motion of that star. I have explained this issue in the following thread: http://www.scienceforums.net/topic/87496-newton-gravity-for-spiral-galaxy/ The equivalent gravity force can be represented by a virtual hoster mass. So, the scientists found the proper motion of S2 (and also for 6,000 stars). However, as stated, this proper motion is balanced with the equivalent gravity force. Therefore, the scientists actually found the virtual hoster mass for S2 (or the virtual hoster of the 6000 stars). Never the less, this is not the real Black hole mass. So, the 4.1 Million Solar mass isn't the real mass of the Sagittarius A*, it is only the virtual hoster mass for S2 which the scientists found in their modeling. If we move from the core of the Sagittarius A* outwards we will find different gravity forces. Therefore, it is expected that for any radius vector there will be a specific virtual hoster mass. We could actually detect it if our calculations were accurate. Unfortunately, the error factor of the calculation is quite high. In the 6000 stars it is 16.5% (+0.2/- 0.4 out of 3.6), while for S2 it is 30% (+/- 0.6 out of 4.1). Therefore, it is overlap due to wide error factor.

What does it mean? "The point mass of Sagittarius A* is not sufficient to explain the velocity data." Please see the following: Orbital speed http://en.wikipedia.org/wiki/Orbital_speed "The orbital speed of a body, generally a planet, a natural satellite, an artificial satellite, or a multiple star, is the speed at which it orbits around the barycenter of a system, usually around a more massive body. Mean orbital speed." It should be quite easy to extract the mass of the black hole. In this formula there is no room for: "extended mass". What is the source for this requirement? Unless, there were different B.H. mass results for different stars. Is it possibale to get the real velocity data and further explanation about thier calculations? Why? They have analyzed 6000 stars. Is it not enough? Don't you think that it should be accurate enough? Really? Let's look again: "They imply a best-fit black hole mass of 3.6 (+0.2/-0.4) x 10^6 solar masses." So, the B.H. should be 3.2 min to 3.8 max. It isn't in the range of 4.1.(based on S2). However, (+0.2/-0.4, which is 0.6 out of 3.6 - 16.6%) by itself indicates that they have got different results from different stars. Please remember that those stars are located " within ~1.0 pc of the supermassive black hole Sgr A"

Thanks It is stated: "We measured the proper motions of more than 6000 stars within ~1.0 pc of the supermassive black hole Sgr A*". "They imply a best-fit black hole mass of 3.6 (+0.2/-0.4) x 10^6 solar masses." However, based on S2 it is: "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16]" So, there is a difference in the B.H. Mass (as I have expected). Can you please explain the following: "The point mass of Sagittarius A* is not sufficient to explain the velocity data. In addition to the black hole, the models require the presence of an extended mass of 0.5-1.5x10^6 solar masses in the central parsec." Why the science need extended mass in the center? Please be aware that this extended mass is 41% of the B.H. mass (up to 1.5 out of 3.6). Why this extended mass isn't part of the black hole itself? Where exactly this extended mass is located? Is it possible that the science get different black hole mass for different stars? Therefore, this extended mass is needed to balance between the different results If so, this is great news!

Based on Newton, the Galaxy rotation curves is quite simple! I will explain it later on.

Let's look again on the following diagram: http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg Based on S2 it is estimated that the total mass of the black hole is as follow: "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16] or about 8.2×1036 kg" In order for the science to prove this mass, they need to verify the black hole mass also based on the other stars. So, if based on S1 (for example) the mass of the black hole isn't exactly the same, than it proves that this isn't the real total mass of the black hole. Please be aware that as those stars are quite close together, it is expected that their equivalent host mass might be quite similar, but not the same. So we have to set an accurate calculation!

Newton's law of universal gravitation http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation "In Spiral galaxies the orbiting of stars around their centers seems to strongly disobey to Newton's law of universal gravitation." I assume that spiral galaxy fully obey to Newton low of universal gravitation. The answer had already been given by Newton: "Every point mass attracts every single other point mass by a force pointing along the line intersecting both points." In other words, every point mass (star for example) attracts every other point mass (another star) by a force which is called gravity force. Therefore, the equivalent gravity force vector which attracts any star should be the sum of the total gravity force vectors from all the stars in the system. Based on Newton, the gravity force is: "The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them" In the solar system for example, we can use Newton gravity formula to calculate the gravity force which any planet, moon or sun attract the Earth. We should find that the total gravity force vectors of all the other planets are neglected with regards to the gravity force vector from the sun. (As the sun includes over than 99% of the total mass in the whole solar system) Hence, the Gravity force on Earth is based on the gravity force with the sun. However, Spiral galaxy isn't Solar system! In Spiral galaxy the total mass of the whole stars in the galaxy is much higher than the mass of the supper massive black hole. Based on Wiki, the milky way black hole (For example) is 4.1 million M☉, while the total mass of the Milky way galaxy is: 0.81.5×1012 M☉. This is a ration of about 4 to 1,000,000. Therefore, the mass of the Milky way black hole is almost neglected with regards to the total mass in the galaxy (about 400 billion stars). This is the opposite scenario from a solar system. Therefore, we need to make the correct calculation in order to verify the real gravity results in spiral galaxy. Based on Newton, the total gravity force on any star in spiral galaxy is the sum of all the gravity force vectors which attracts this star in the galaxy. That's mean that we need to verify the gravity contribution of the black hole with the gravity contribution of the stars in the spiral arm (and even consider the contribution of all other stars in the galaxy). The outcome of this calculation will be called the equivalent gravity force vector. For each star in the galaxy we need to calculate its own equivalent gravity force vector. This gravity force has a direct impact on the rotation energy of that star. In order to understand how to calculate the equivalent gravity force, let's use a very simple system – Sun and Earth (without any other planets or moons) 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 1/8 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. 6. 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, let's go back to the Spiral galaxy. In the Milky Way there are 400 Billion stars in about seven spiral arms. Therefore, in each arm there are about 70 Billion stars (assuming that all stars are located in the arms and the arms are equally). Our sun is located in one of those arms. To make it simpler, let's assume that all the stars in that arm are located in a long straight line and ignore the gravity influence of the other arms. In order to calculate the equivalent gravity force that attracts the sun, we need to verify the following: -The gravity force vector that the black hole attracts the sun. -The gravity force vectors of all the stars in the arm which are located between the sun and the black -The gravity force vectors of all the stars in the arm which are located between the sun and the outwards side of the arm. Please be aware that all those vectors contribute a negative gravity force. The equivalent gravity vector is the sum of all the gravity vectors. However, it's quite difficult to calculate 70 billion gravity vectors. There is a simple way. The rotation energy force of the sun is excellent indication for the equivalent gravity force vector that attracts the sun in the galaxy. (There must be full balance between those two forces) The rotation energy is a direct outcome of the Sum mass and its velocity. The radius of the sun is about 26,000 light year, Therefore, it's quite easy to calculate its virtual hoster equivalent mass. The Sun revolves around this virtual hoster and complete one cycle in 240 million years. Actually, any star in the galaxy revolves around its own virtual hoster. Each virtual hoster has a unique Equivalent mass which fits to that specific star. Now, let's move the sun inwards and outwards in the spiral arm inorder to verify how the equivalent gravity force (and as an outcome – the virtual hoster equivalent mass) should be. So, if the sun is located at the most outwards side of the arm, all the gravity vectors forces will be in the same direction. Therefore, it should have the greatest gravity force. As we will move the sun inwards in the arm, the stars at the outwards side (with relevant to the sun location) will contribute a negative gravity force. Hence, the equivalent gravity force vector will be decreased as the sun moves inwards in the spiral arm. Therefore, the maximal gravity force (Maximal virtual hoster' equivalent mass) will be achieved by placing the sun at the most outwards side of the arm. The minimal gravity force (Minimal virtual hoster' equivalent mass) will be achieved by placing it at the most inwards side of the arm. That is the opposite scenario from a typical solar system. This proves Newton gravity low for spiral galaxy. It is also explains why the star velocity increases as its location is further from the center of the spiral galaxy.

Thanks Yes, sure. Newton might have an answer to this issue.

Please see the following orbit diagram: http://en.wikipedia.org/wiki/File:Oort_cloud_Sedna_orbit.svg It shows that most of the planets in the solar system have quite symmetrical cycle. (O.K. – it is not fully symmetrical – but it looks symmetrical.) However, Sedna has clearly an elliptical orbit which looks similar to S14. So the one of the further planets in the solar system, orbits in a similar way as the closest one in the spiral galaxy. We have to understand the source for that.

Regards to the elliptical center orbits http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg How come that all of those stars have an elliptical orbits? S14 for example, is located very close to the center and its elliptical shape is the sharpest one. How could it be? This is the opposite scenario from any solar system. In the solar system we can find elliptical orbits only with stars that are far away from the sun. Those stars have low gravity force due to their distance. Therefore, it is quite logical that Pluto (for example) has an elliptical orbit. However, in the spiral galaxy, those stars which are located quite close to the center have elliptical orbits, while a star which is located 26,000 lY (sun – for example) has a symmetrical cycle orbit. It shows that the gravity force in the spiral arm is much greater than the gravity force in the center. This is amazing phenomena.

Based on Newton's law of universal gravitation http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation "Every point mass attracts every single other point mass by a force pointing along the line intersecting both points." In other words, every point mass attracts every other point mass by a gravity force. Therefore, all the mass in the spiral arm should also have an effect on S2. Do you agree? O.K. If spiral galaxy was a perfect symmetrical system, and the Black hole will be located just in the center, than we could claim that the net power on the center is zero. However, a normal spiral galaxy is not a perfect system. Even the black hole is not located just at the center. Therefore, the net force isn't zero. Please also be aware that S2 rotates between the center and the inwards side of the spiral arms. Therefore, as it moves close to the arms, the net force should be much more than zero.

Thanks So, do you mean that S2 see (feel) more gravity than the black hole by itself (assuming that there are stars inside the rotation cycle of S2)? Why the mass outside can be ignored? If the inside mass effects the gravity on S2, why the mass outside has no effect? What about the mass ratio between the spiral arms to the Back hole? How can we justify a ratio of 4/one million?

http://en.wikipedia....sive_black_hole "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16] " However, the mass of the whole milky way galaxy is : 0.8–1.5×1012 M☉. This is astonishing ratio of four/one million... (On 4 Kg of mass in the black hole, there are 1,000,000 Kg (1,000 Tons) of mass in the spiral arms. Actually, the black hole is like an engine which rotates the whole spiral galaxy while spiral arms are similar to the biggest propeller in the universe. It is similar to an engine of 4 Kg which rotatse a propeller of 1,000,000 kg (or 1,000 tons)? In this kind of ratio, the spiral arms should be disconnected from the galaxy! Is it real? How could it be?

What do you realy know??? As you have already showed us your sever lack of knowledge, it's better for you to keep your "wisdom" and be silence. Never the less, if you insist…Would you kindly find even one example which contradicts this statement….

Spiral galaxy by definition is a disc system. The receding of the moon and the Earth is valuable information for our understanding of disc system activity. Why do you think so??? This is a pure imagination. Do you have any prove for this statement??? Why a galaxy which is spinning at high velocity and move in space at ultra speed could gain any mass from space??? There are many examples that spiral galaxies are losing stars & mass. Never the less, There isn't even one prove that a spiral galaxy could collect mass from space. Yes it is. I was expecting that in any disc system the stars should move outwards. So far, the science had found a data for only two disc objects in the universe. Noon & Earth. The science couldn't find even one measurement which might contradict my expectation. The main problem is that the science do not agree nor except the results of their measurements. They try to verify if there is any mistake. As you have stated: " This is an interesting open issue. The most likely explanation is that astronomers are doing something wrong in their measurements, their data analyses, or both. This is the explanation astronomers themselves invoke. The problem is that nobody can see what's being done wrong." Why the science can't agree or accept those results? Why they reject the Idea that stars in a disc system are moving outwards??? Based on logical point – Out of billion over billion stars in a disc system, the science could only find that two stars are moving outwards. As there is no even one evidence which contradicts this data, the science must assume that the stars in a disc system are moving outwards!!!