Newton Gravity Force In Spiral Galaxy

[davdan]

Gravitation

Let's start by understanding the power of gravity

By Wiki: http://en.wikipedia.org/wiki/Law_of_gravity

"Gravitation or gravity is a natural phenomenon by which all physical bodies attract each other. Gravity gives weight to physical objects and causes them to fall toward one another.

Gravity is the weakest of the four fundamental forces of nature.

On the other hand, gravity is the dominant force at the macroscopic scale, that is the cause of the formation, shape, and trajectory (orbit) of astronomical bodies, including those of asteroids, comets, planets, stars, and galaxies. It is responsible for causing the Earth and the other planets to orbit the Sun; for causing the Moon to orbit the Earth; for the formation of tides; for natural convection, by which fluid flow occurs under the influence of a density gradient and gravity; for heating the interiors of forming stars and planets to very high temperatures; for solar system, galaxy, stellar formation and evolution; and for various other phenomena observed on Earth and throughout the universe."

Therefore, we shouldn't give up on this critical power!

 

Newton solution for spiral galaxy

The answer for the problem had already been given by Newton:

Newton's law of universal gravitation http://en.wikipedia.org/wiki/Newton%27s_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 (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: 0.8–1.5×10¹² 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 lets use a very simple orbiting system

 

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) + 1/4 x F (From the second star) = 3/4 x F.

Therefore, the equivalent Virtual hoster mass is 3/4 x M.  

Hence, by using a virtual hoster at a mass of 3/4 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) - 1/4 x F (From the second star) = 1/4 x F.

Therefore, the Equivalent virtual hoster mass is  1/4 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.

 

 

Gravity force in spiral galaxy

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 we move the sun 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.

 

Evidence

There is an evidence for this Theory. It is called S2

Super massive black hole

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

"The star S2 follows an elliptical orbit with a period of 15.2 years and a pericenter (closest distance) of 17 light-hours (1.8×10¹³ m or 120 AU) from the center of the central object"

S2 is located between the Milky way black hole center to the inwards side of the spiral arms.

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

Therefore, in one side S2 is attracted by the gravity force of the Black hole. In the other side it is attracted by the mass in the spiral arms. (With regards to the spiral arms - the nearby section of the inwards spiral arms to S2 has higher gravity effect).  However, those gravity force vectors have opposite polarities. The sum of those vectors should give us the equivalent gravity force vector on S2.

Hence, S2 rotation energy should be based on that equivalent gravity force. However, the scientists have neglected completely the gravity force vectors of the spiral arms. They have estimated that rotation energy should base only on the black hole gravity force. Therefore, they have concluded that the total mass of the black hole is 4.1 million M_☉.

"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"

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.  So, how could it be that an engine of 4 Kg will rotate a propeller of 1,000,000 kg (or 1,000 tons)? This is absolutely illogical.

In this kind of ratio, the spiral arms should be disconnected from the galaxy!

Therefore, this 4.1 million M_☉ represents the equivalent hoster mass.

In order to extract the real mass of the rotatable supper massive black hole, we must add the gravity force vector which each point of mass in the spiral arms attracts S2.

It's quite clear that the total mass of this black hole should be significantly higher than this relatively low mass. Actually all the "missing mass" is located at the SMBH.

 

 

Ejected star from spiral arm:

Let's assume that one star had been drifted out from the spiral arm and try to figure what should be the outcome:

We all know that the gravity force is : " inversely proportional to the square of the distance between them"

Therefore, the nearby stars contribute significant portion of the equivalent gravity force which attracts this star in the galaxy. If a miserable star is drifted out from the spiral arm, (not inwards or outwards in the arm – but just out of the arm) its equivalent gravity force should be decreased.

Hence, there will be no balance between its rotation energy to its new decreased equivalent gravity force. Therefore, it will be kicked out from the arm and eventually from the galaxy.

This also proves that there are no stars in between the spiral arms!

With regards to our solar syatem -It's better for us to keep our position in the spiral arm.

 

 

Spiral Galaxy – Orbital speed of star

 

By wiki:

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

"If disc galaxies have mass distributions similar to the observed distributions of stars and gas then, the orbital speed would always decline at increasing distances in the same way as do other systems with most of their mass in the centre, such as the Solar System or the moons of Jupiter".

This is incorrect!

As I have already proved spiral galaxy isn't solar system. In a solar system, most of the mass is in the center, while in spiral galaxy there are significant portion of mass outside the center. In order to evaluate the orbital speed vs. distance, let's assume that all the stars in spiral arms keep their position and do not drift them inwards or outwards. Therefore, it is like a rigid spiral disc. In this case, it is expected that all points of mass should complete one cycle at the same time.

The circumference of a circle for is

 

Where: C is circumference, r is Radius and π is a dimensionless constant approximately equal to 3.14159.

Let's use the following example:

For first star  - R1 = 10,000 ly

C = 2 x 3.14 x 10,000 = 62,800 ly

For second star  – R2 = 20,000 ly

C = 2 x 3.14 x 20,000 = 125,600 ly 

Both stars complete one cycle in the same time (rigid spiral disc). The circumference of second star is double than the first one, therefore, its orbital speed must be doubled with regards to the first one.

Hence, the orbital speed must increase at increasing distance, assuming that the stars keep their position in the spiral arms.

 

 

Rotation Curve problem in spiral galaxy

By wiki:

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

"galaxy rotation problem is the discrepancy between observed galaxy rotation curves and the theoretical prediction, assuming a centrally dominated mass associated with the observed luminous material. When mass profiles of galaxies are calculated from the luminosity profiles and mass-to-light ratios in the stellar disks, then they do not match with the masses derived from the observed rotation curves and the law of gravity."

"The rotation curve of a disc galaxy (also called a velocity curve) is a plot of the magnitude of the orbital velocities (i.e., the speeds) of visible stars or gas in that galaxy versus their radial distance from that galaxy's centre, typically rendered graphically as a plot."

Let's see the following diagram:

http://en.wikipedia.org/wiki/File:M33_rotation_curve_HI.gif

R1 = 10,000 ly the observation velocity is 90 km/s.

R2 = 20,000 ly the observation velocity is 105 km/s.

Based on our calculations, in a rigid spiral disc the orbital velocity must increase at increasing distance. At a double distance, the orbital speed must be doubled.

So, if R1 = 10,000 Ly and observation orbital velocity is 90 km/s, than for R2 = 20,000 Ly the expected speed must be 90 x 2 = 180 km/s.

However, this isn't the case. We need to explain why the speed is only 105 km/s while based on our expectation from rigid spiral disc it should be 180 km/s.

The answer is quite simple - Spiral galaxy disc isn't rigid disc.

We must give some freedom to the stars to drift further away in the arm.

So we need to decrease the speed by 75 km/s. (180 – 105)

This is achievable by decreasing the circumference of a circle which the star will have to go in that "same period of time –T.

The ratio between 75 to 180 is 0.41666.

So if we will decrease the distance in the circumference of a circle from 125,600 ly by:

125,600 x 0.41666 = 52,332 ly.

We will achieve the 105 km/s as observed.

Hence, in the same period of time – T,  the star should move in total:

125,600 – 52,332 = 73,268 ly (instead of 125,600 ly).

So, the star will start at R=10,000 Ly, set a complete movement of 73,268 Ly and get to the point where R = 20,000 at the same time -T. By doing this movement the star speed at R=20,000 will be exactly 105 km/s as observed.

In this case, the star will drift inwards from 20,000 Ly to 10,000 Ly in the same time - T.

However, the mathematical calculation proves that spiral galaxy isn't a rigid disc. The stars must drift outwards in the arm while the whole arm rotates around the center of the galaxy.

This activity sets the unique shape of spiral arms.

This proves that all stars in spiral arms must drift outwards. Hence, in the future, our solar system should move further away from the center of the Milky Way galaxy. 

Edited May 27, 2017 by davdan

[mrg]

TL:DR.  Cheers, MrG

[Turtle]

...

However, the mathematical calculation proves that spiral galaxy isn't a rigid disc. The stars must drift outwards in the arm while the whole arm rotates around the center of the galaxy.

This activity sets the unique shape of spiral arms.

This proves that all stars in spiral arms must drift outwards. Hence, in the future, our solar system should move further away from the center of the Milky Way galaxy. 

Our Sun drifts up & down across the galactic plane and I see no mention of this motion in your discourse. How do you account for this?

[davdan]

Our Sun drifts up & down across the galactic plane and I see no mention of this motion in your discourse. How do you account for this?

 

Thanks for the question

 

Our scientists have no clue about this up & down movement.

The sun is not just moving up and down. It orbits around its virtual center point of mass.

Actually, all the stars in the spiral galaxy must orbit around some center point of mass

.

In order to understand the sun movement, let's look at the solar system.

The moon and the Earth orbits around their center of mass, and this center orbits the sun.

However, we see it as a moon orbits the Earth and the Earth orbits the Sun.

Let's assume that the Earth is invisible.

In this case we should see a moon which orbits the Sun while it moves up and down.

 

In the same token the sun orbits its unseen center of mass, while its center of mass follows the equivalent local gravity of force in the spiral arms.

Therefore, it looks as the sun moves up and down but in reality it has a perfect orbit around this unseen center of mass.

Edited May 27, 2017 by davdan

[Turtle]

Thanks for the question

Our scientists have no clue about this up & down movement.

Au contraire. We explored the question in depth some number of years ago here: >> Our Galactic Plane

 

 

The sun is not just moving up and down. It orbits around its virtual center point of mass.

Actually, all the stars in the spiral galaxy must orbit around some center point of mass.

In order to understand the sun movement, let's look at the solar system.

The moon and the Earth orbits around their center of mass, and this center orbits the sun.

However, we see it as a moon orbits the Earth and the Earth orbits the Sun.

Let's assume that the Earth is invisible.

In this case we should see a moon which orbits the Sun while it moves up and down.

 

In the same token the sun orbits its unseen center of mass, while its center of mass follows the equivalent local gravity of force in the spiral arms.

Therefore, it looks as the sun moves up and down but in reality it has a perfect orbit around this unseen center of mass.

:lol: But you said "However, Spiral galaxy isn't Solar system!" which implies solar system dynamics are not an analog for galaxy dynamics. You can't have it both ways.

[davdan]

Gravity is gravity

It works the same for the moon as it works for the Sun.

Please see the following diagram:

 

 

http://www.biocab.org/Motions_of_the_Solar_System.jpg

 

 

the blue line show the apparent motion of the solar system.

 

The orange solar system represents its center of mass.

 

 

So, if you change this center of mass with the Earth you would get identical apparent motion for the Moon.

 

 

Hence, the Moon center of mass is the Earth, while the Sum has its own virtual Center of mass. So simple and clear!

 

[Turtle]

Gravity is gravity

It works the same for the moon as it works for the Sun.

Please see the following diagram:

 

http://www.biocab.org/Motions_of_the_Solar_System.jpg

 

the blue line show the apparent motion of the solar system.

 

The orange solar system represents its center of mass.

 

So, if you change this center of mass with the Earth you would get identical apparent motion for the Moon.

 

Hence, the Moon center of mass is the Earth, while the Sum has its own virtual Center of mass. So simple and clear!

Please stop formatting the font & size; it makes it difficult to read your reply when responding.

 

I think there is also a language barrier here which makes your [apparent] arguments unintelligible.

 

Anyway, where did you get that diagram? I can only determine its origin as your personal blog. ??

Moreover, you said "This proves that all stars in spiral arms must drift outwards. Hence, in the future, our solar system should move further away from the center of the Milky Way galaxy.", and in the diagram the line to the galactic center has arrows on both ends and a label 20KmS which seems to indicate the solar system is moving in and out with respect to galactic center. ??

 

I also point out that you claimed "Our scientists have no clue about this up & down movement." when I asked about it, and now you present a diagram showing the up & down motion. I saw you spent some time reading the thread I linked to, so what is your claim now in light of those facts? (Not sure if your diagram label for this movement of 5-7KmS is in agreement with our findings or not.)

 

Anyway, I don't see any particular merit in what you have presented. It's as clear as mud. Good luck.

[davdan]

http://www.biocab.org/Coplanarity_Solar_System_and_Galaxy.html

[Turtle]

http://www.biocab.org/Coplanarity_Solar_System_and_Galaxy.html

Thanks for a proper attribution. Alas, this does nothing to abrogate your contradictions. When and if you have something of merit for us to consider you can point us to the journal in which it is published.