1

New Theories / Re: How gravity works in spiral galaxy?

« on: Yesterday at 16:11:42 »

You can answer that question yourself. Just ask yourself "does this create new energy?" If the answer is "yes", then you've contradicted the law of conservation of energy.

Well, the answer is clearly YES, while there is no contradiction with the law of conservation of energy .
In order to understand it, let's start with the following:
http://www.isustainableearth.com/energyefficiency/tidal-power-facts-understanding-how-tidal-energy-works
Quick Fact on How Tidal Energy Slows Earth Rotation
"During our research, we found an interesting detail surrounding tidal energy and the effect its generation could have on the Earth.  Naturally the movement of tides causes a loss of mechanical energy that is associated with the Earth’s rotation.  According to some studies, our planet has actually lost 17 percent of its rotational energy in the past 620 million years."
So, our scientists assume that the Tidal energy is taken from the rotational energy of the objects.
This is incorrect assumption.
I can prove that there could be a tidal energy without any need for any rotation.
Let's look at the Moon.
Its face is locked with the Earth.
Therefore, The tidal bulge doesn't cross its surface. It is a frozen Tidal Bulge. Hence, there is no new energy due to Tidal and therefore, the moon is a frozen object.
That is correct, as long as the object is only affected by one main orbital object.
However, let's assume that around the moon we will set four objects. Each object will be in the size of 1/1000 Moon mass and each one orbits at a different radius (R, 1.2R 1.4R, 1.6R) and different direction.
Let's assume that the Moon and all of those four objects have totally lost their rotational energy.
So, the question is:
Could it be that a tidal energy can be created at this moon although it has totally lost its rotational energy?
The answer should be -YES!!!
If you need an explanation - I will be happy to give.
So, we can see that without any need for rotation, tidal forces can set new energy.
That is exactly the case with our SMBH.
Let's assume that it doesn't rotate at all.
Let's even start with the assumption that it is a totally frozen object.
However, there are so many objects that are orbiting around that SMBH (Lets also assume that none has a rotation energy).
So, each object orbits at a different radius and even at a different orbital direction.
Therefore, Each one of them sets a small moving tidal bulge on the SMBH.
Each tidal bulge must set some small tidal energy as it cross the SMBH surface.
The Impact of millions tidal's bulge that are crossing that surface at different directions and amplitude must cause Ultra high tidal energy at the SMBH.
Therefore, even if the SMBH was frozen, after some time it Must be melted from the Ultra high energy of that tidal activity.
Therefore, Tidal energy is a NEW energy.
It is a direct outcome from gravity force.
So, as long as the gravity is there, and there are more than just one orbital object - we should get new tidal energy even if the rotation of all the objects is Zero!.

2

New Theories / Re: How gravity works in spiral galaxy?

« on: 10/11/2019 20:14:55 »

Your model posits the creation of energy, violating the "energy cannot be created" part of conservation of energy.

My model is based on Tidal renewable Energy:
https://energyfive.net/2017/10/07/tidal-energy/
"Tidal energy sources come from the gravitational movements of sun and moon as these creates high and low tides. When you think of it, it means that the ocean wave energy sources are renewable as compare to fossil fuels and nuclear reserves, moon and sum gravitational fields is not though of to be ceased in near future"
So, Tidal is based on Gravitational fields, it is renewable and it is not though of to be ceased in near future.
Therefore:
As long as there is gravitational field - there is Tidal.
As long as there is Tidal - there is renewable energy source.
https://ui.adsabs.harvard.edu/abs/2010Natur.468..952B/abstract
"Tidally driven flow in the Earth's liquid core develops internal shear layers, which distort the internal magnetic field and generate electric currents."
As long as there is Tidal renewable energy source - There is renewable Magnetic Field
As long as there is renewable Magnetic Field - There is renewable electromagnetic energy
So what is the problem with that?
Why can't we use that renewable energy source for the pair particle creation?
Do you still see any contradiction with conservation of energy?

3

New Theories / Re: How gravity works in spiral galaxy?

« on: 10/11/2019 17:06:10 »

I will wait until you have answered this:
What is the definition of the law of conservation of energy?.

The Law of Conservation of Energy Defined
https://www.thoughtco.com/law-of-conservation-of-energy-605849
"The law of conservation of energy is a physical law that states energy cannot be created or destroyed but may be changed from one form to another."
So, where is the problem with my model?

The SMBH's ultra high gravity force sets a tidal impact on the SMBH itself.
That tidal force generates extra heat/extra core rotation that increases the ultra high electromagnetic field.
Some of that electromagnetic energy is used to transformed the virtual pair particle into real pair particle.

4

New Theories / Re: How gravity works in spiral galaxy?

« on: 08/11/2019 10:00:49 »

No, the question is about conservation of energy. This is a question that you seem to be afraid of answering because you know it kills your model.

OK
Let me start with the conservation of energy in my model.
Please look again at the following image by Newton:
https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg
Let's assume that in the center we have a SMBH and there is a virtual canon up above.
That virtual cannon fire a virtual pair of particles at the exact magic velocity.
I hope that you agree that so far there is no violation in the energy conservation.
The Ultra gravity force of the SMBH keeps that virtual pair in the orbital path with about the speed of light.
We all know that around the SMBH there is a very strong electromagnetic field.
At some point of time that virtual pair is converted by the electromagnetic field/energy into real particle pair.
So, due to energy conservation, some energy from the electromagnetic had been transformed into this new born particle pair.
Now we need to find the source for that electromagnetic field:
The SMBH has a very strong gravity force that holds million (or even billions) of objects in orbital cycles around it.
Those objects set strong tidal force on the SMBH that is converted to extra heat and rotates its core/or even layers in the core.
This activity increases the electromagnetic field/energy of the SMBH.
Therefore, the conservation of energy works as follow:
The SMBH's ultra high gravity force sets a tidal impact on the SMBH itself.
That tidal force generates extra heat/extra core rotatation that increases the ultra high electromagnetic field.
Some of that electromagnetic energy is used to transformed the virtual pair particle into real pair particle.
Any objection?

5

New Theories / Re: How gravity works in spiral galaxy?

« on: 07/11/2019 16:58:59 »

It seems like you are now claiming that "energy cannot be created or destroyed" is not the definition of the law of conservation of energy. Is this what you are saying?

The question is: What Newton is claiming by its Cannon ball example?
Please look again at that image:

https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg

Newton shows that any orbital object must be boosted horizontally at the magic velocity in order to keep it in an orbital path.
Nothing is random in orbital system. You can't just hope to boost an orbital object at a random velocity and wish that that somehow those objects will set an orbital system
This is IMAGINATION
If that velocity is too low it will fall directly to the main object, if it is too high it will be ejected outwards.
Newton gave us the clear understanding that potential energy can't be transformed into Kinetic energy.
So, the outcome is quite simple.
1. There is no way to increase the Orbital kinetic energy by decreasing the potential energy.
2. Therefore, there is no way to increase the orbital velocity of the object by decreasing the potential energy. This is the most important issue in all our discussions so far!!!
3. Hence, if an object orbits at the magic velocity which keeps it in the orbital path, it can't increase that velocity (even if it falls in). In other words - Orbital objects couldn't increase their orbital velocity as they decrease their radius to the main object
4. Therefore, due to the conservation energy (kinetic orbital energy) if the orbital object is decreasing its radius while it can't increase its orbital velocity, it breaks the magic orbital velocity and therefore it must collide with the main object.
5. In the same token - if the orbital object is increasing its radius while it can't decrease its orbital velocity, it breaks the magic orbital velocity and therefore it must ejected outwards.
Do you agree with all of that?
If you don't agree than you don't accept Newton's cannon ball explanation for gravity

6

New Theories / Re: How gravity works in spiral galaxy?

« on: 07/11/2019 04:00:18 »

We are not focusing on that right now as per our prior agreement to focus purely on how your model deals with conservation of energy.

If you wish to focus on my model or any sort of model, don't you think that first you must know how the gravitational conservation of energy really works?
Newton have proved with his cannon example that there is no energy transformation between potential energy to orbital kinetic energy. Therefore, that kind of assumption is a fatal error.
He also proved that in order to set a stable orbital bonding you must first set the orbital object at that "magic velocity".
Actually, we have to ask how could it be that there is any sort of change in the orbital radius.
It is a severe contradiction to the orbital magic velocity.
We all know that the moon is drifting outwards.
Due to the conservation of energy, (and as there is no transformation of energy between the orbital kinetic energy to potential energy), it shouldn't change its current kinetic orbital energy.
Therefore, theoretically, it can't increase or decrease its orbital velocity.
So, once it's drifted inwards, the current orbital velocity will be too low to keep it in the orbital path and therefore, it MUST fall in.
In the same token, once it is drifted outwards, it should eventually be ejected from the Earth.
That shows that your understanding about conservation of energy is very poor.
So how can you speak in the name of the conservation of energy in my model while your current understanding/assumption in this key issue is totally incorrect?

7

New Theories / Re: How gravity works in spiral galaxy?

« on: 02/11/2019 15:35:53 »

This sounds like you are attacking existing orbital models instead of defending your own model. As I said before, evidence against an existing model is not evidence in favor of your model: https://en.wikipedia.org/wiki/False_dilemma

Right now, we are to focus our discussion on the energetic mechanism behind your model as per our mutual agreement.

Dear Kryptid
Newton have proved by it's "orbital cannon" that Potential energy can't be transformed into orbital kinetic energy.
Do you agree with that?
Please answer: Yes or No
This is the most important issue in our discussion on the energetic mechanism for orbital systems.
How could it be that there are so many orbital systems in our galaxy/universe while there is no way to increase any orbital kinetic energy by potential energy.
If the potential energy can't be used as a source of orbital kinetic energy, than what could be the source for that?
Newton have found that "if the cannon ball was fired with exactly the right velocity", and gaining the "magic orbital velocity" "it would be placed in orbit around the Earth".
So, we all understand that a canon can set the requested energy that is needed for a stable orbital system.
What kind of "canon" the Universe is using in order to set all of those orbital systems in the galaxy?
How that "canon" could set the magic velocity that perfectly fits any orbital system?
What is the real source for the orbital kinetic energy in our universe?

8

New Theories / Re: How gravity works in spiral galaxy?

« on: 02/11/2019 08:03:58 »

It doesn't matter how important you think it is, we both agreed that we will focus on conservation of energy until that particular issue is resolved. So answer my question: is orbital energy a form of energy?

Yes it is.
In order to get better understanding about the gravitational conservation of energy let me use the following example for Newton’s orbital cannon:
https://www.sciencelearn.org.nz/images/269-newton-s-orbital-cannon
"Newton reasoned that, if the cannon ball was fired with exactly the right velocity, the ball would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. It would be placed in orbit around the Earth."
What do we understand from that:
1. Potential energy: it is stated clearly that the ball is "always falling in the gravitational field" to the Earth (actually to the center of the Earth.
That proves that the potential energy of the ball is converted constantly to a falling kinetic energy.
2. Orbital kinetic energy - It is also stated that "if the cannon ball was fired with exactly the right velocity" "It would be placed in orbit around the Earth"
So, if its orbital kinetic energy can exactly compensate its free falling energy (due to its potential energy) "It would be placed in orbit around the Earth".
In other words:
If we shoot that ball at the correct velocity (with correct orbital kinetic energy) high enough from the earth surface than although the ball is always converts its potential energy to in falling kinetic energy "It would be placed in orbit around the Earth" - forever (assuming that there is no friction as air),.
So, the ball keeps its orbital velocity/Orbital kinetic energy.
The Potential energy has one mission. To force that ball to meet the earth, while the orbital kinetic energy keeps that ball in the same shooting velocity.
Therefore, when we discuss on conservation of energy in orbital cycle, we actually discuss on conservation energy of orbital kinetic energy.
That proves that the potential energy can't be converted into orbital kinetic energy.
On the contrary. the potential energy is constantly forcing the ball to fall directly to earth while the kinetic energy keeps it in a constant orbital cycle.
Therefore, I wonder what is the added value of Total Kinetic Energy?
Our scientists claim that
Total Orbital Kinetic energy = Orbital kinetic energy + Potential energy
However, based on Newton’s orbital cannon it seems to me that we should claim that:
Total Orbital Kinetic energy = Only the Orbital kinetic energy
Total energy = Orbital kinetic energy + Potential energy
Hence, due to the Momentum conservation in space (no friction) there is a conservation of energy only  in the Orbital Kinetic energy.
Now, let's assume that we gave that Newton's cannon ball the correct orbital velocity in order to keep it at constant orbital cycle around the earth.
As there is no friction, the orbital velocity should stay at the same orbital radius forever.
The value of the requested velocity is:
V^2 = MG/r
That velocity is called the "magic velocity"
https://www.scienceabc.com/nature/universe/what-is-orbital-velocity.html
"Newton realized that when the ball is propelled at a certain, magic velocity, it will never fall."
It is also stated:
"One can infer from the expression that first, the velocity decreases with r, the orbit’s distance from the center of Earth. This means that satellites orbiting closer to Earth’s surface must travel faster than satellites orbiting further away. The moon, which lies almost 385,000 km away, races around us at 1.002 km/s, while the International Space Station (ISS), merely 400 km away, completes a lap every 1.5 hours, racing at 7.67 km/s. Secondly, the velocity is independent of the mass m of the orbiter, which means that a multimillion-ton moon or a single-gram nail must travel at the same velocity to achieve orbit around Earth (at the same distance, r, that is)."
It is clear that as the ball will be placed higher from the Earth the requested orbital velocity should be lower and vice versa.
Hence, if the ball or Moon is already in a stable orbital cycle and we push it inwards (to the earth) its current orbital velocity would be too low to keep it in a stable orbital velocity (at a shorter radius) and therefore it should eventually fall in.
In the same token, if the ball or Moon is in a stable orbital cycle and we pull it outwards (from the earth) it's current orbital velocity would be too high too keep it in a stable orbital velocity (at a higher radius) and it should eventually escape from the gravitational Earth.
https://www.scienceabc.com/wp-content/uploads/2018/02/Newtons-cannonball-3.jpg

Conclusion -
I have proved that the potential energy can't increase the orbital kinetic energy.
At the maximum, due to the momentum conservation or energy conservation, that energy will stay as is forever.
Therefore, the assumption that a practical can increase its orbital velocity to almost the speed of light as it falls into the SMBH is a pure imagination.

 

9

New Theories / Re: How gravity works in spiral galaxy?

« on: 31/10/2019 07:21:38 »

No, it doesn't. Potential energy is energy. It isn't a velocity. It isn't a vector.

Potential energy is converted into falling kinetic energy that creats a velocity vector between the two objects.

If two objects are falling towards each other in a straight line, yes, some other force would have to be added in order to get an orbit out of it.

Thanks
So, you agree that Vp represents the direct falling velocity between the two objects.

Because stars and particles will have their own random trajectories and velocities before getting anywhere near the black hole

Actually, I agree with you.
Therefore, I have stated that there might be some initial sort of orbital kinetic energy.

Let's assume that the objects have already some low orbital kinetic energy (too low for a stable orbital cycle)
So, due to the Conservation of momentum they will keep that orbital velocity.

However, even if there is some sort of random orbital velocity - Vk, and that velocity can't hold a stable orbital momentum, than sooner or later the object should fall in (if it is too low) or just cross by and go away if it is too fast.

No they won't. The black hole's gravity will accelerate them. Momentum will be conserved because some of the momentum will be transferred to the black hole via gravitational force

There is no orbital acceleration.
The only acceleration is due to potential energy.
As we all agree that potential energy set the falling in kinetic energy and that energy represents a velocity Vp of a falling object.
So, there is acceleration in the Vp vector as
Ep = MGR
But there is no acceleration in the orbital velocity due to that potential energy.

It won't necessarily force them to get closer each cycle. It all depends on the specifics.

As long as there is a stable orbital momentum with regards to the falling vector (due to potential energy) than the orbital object will get back to the same distance every full orbital cycle.

Probability alone strongly suggests that any given star or particle will not be traveling directly towards the black hole on a straight path. Conservation of momentum won't let the particle or star suddenly stop its current trajectory and fall in a straight line towards the black hole. It will follow a curved path instead.

This is incorrect assumption.
We can see it from our location on Earth
There are many objects/asteroids that orbit around the sun.
From time to time the earth cross their orbital path.
In this case, some of the objects are colliding/falling directly to the earth. We see them as meteorites.
However, if I understand it correctly, none really is attracted by the earth and starts its stable orbital cycle around the earth instead of the sun.
Hence, the Conservation of momentum won't let any object/asteroid/particle suddenly to change its current trajectory/velocity and starts its orbital cycle around the Earth or any other main object including the BH/SMBH.

Enough of this. No more irrelevant talk. You have consistently avoided the key issue. So answer this question: Is orbital energy a form of energy?

This is the most important issue in all our discussion so far!!!
I have proved that random trajectories or velocities around the object can't hold a stable orbital cycle by itself.
This proves that our assumption that somehow objects start to orbit around each other is totally INCORRECT.
Our scientists must look for better explanation why there are so many objects that orbits around each other in a stable orbital cycle
Again – you can't convert any falling in kinetic energy into orbital kinetic energy!!!
However, If you are a satellite and you have full control on the power, than there is a possibility to approach the main object at the correct velocity and phase and find the correct window to get a stable orbital cycle, but this can't be done randomly (you can ask NASA about it)
Therefore, the assumption that somehow by random falling/crossing by activity, objects should orbit around each other - is the most fatal mistake of our scientists.
Hence, this issue is much more important than any other open issue including the conservation of energy.

10

New Theories / Re: How gravity works in spiral galaxy?

« on: 31/10/2019 01:40:09 »

If we can't convert Vp to Vk how could it be that Potential energy is part of the total orbital energy?
How can we discuss on conservation of orbital energy, if in the reality, a potential energy can't be transformed into orbital kinetic energy?

It can be converted. It's happening constantly in eccentric orbits. As a planet approaches perihelion (the closest point to the Sun in its orbit), it speeds up. This speed increase comes from a conversion of the planet's gravitational potential energy into kinetic energy. Once it passes perihelion, it starts moving further away from the Sun and begins to convert its kinetic energy back into potential energy, slowing it down.

Dear Kryptid
It seems to me that you didn't understand me correctly.
You discuss on a stable orbital cycle - with eccentric orbit. (Every full cycle, the distance between the objects is constant).
I don't discuss on this case.
Let's start with the meaning of Gravitational Field
https://en.wikipedia.org/wiki/Gravitational_energy#/media/File:Gravitational_field_Earth_lines_equipotentials.svg
"Image depicting Earth's gravitational field. Objects accelerate towards the Earth, thus losing their gravitational energy and transforming it into kinetic energy."
As you can see from any direction, the vector is pointing to the center of mass.
It is also stated:
https://en.wikipedia.org/wiki/Potential_energy
Common types of potential energy include the gravitational potential energy of an object that depends on its mass and its distance from the center of mass of another object,"
Therefore, Vp represents a Velocity vector between the center of mass of the two objects.
Hence,at any distance (even if we call it - perihelion), the Vp vector will be on the direct line between the two center of mass.
If they don't have any orbital velocity, is there any possibility for the Vp to be shifted by 90 degrees in order to be transformed into new orbital velocity Vk?
I assume that if we could add wings to the object than this might help.
So, if one of the objects is an airplane or shuttle, it could convert its Vp into Vk.
Another possibility is by adding external force as magnetic field or rocket.

t's unwise to assume that all stars and particles will be traveling in a straight line towards the black hole. Conservation of momentum and the initial trajectories and velocities are relevant.

Why not?
Let's assume that the objects have already some low orbital kinetic energy (too low for a stable orbital cycle)
So, due to the Conservation of momentum they will keep that orbital velocity.
However, as the Vp (Falling velocity due to potential energy) is forcing the objects to get closer every cycle, eventually - they must collide with each other.

If you have an idea how to convert the Vp by 90 degrees to - new Vk (orbital velocity), without external force or wings - than please show it mathematically.

11

New Theories / Re: How gravity works in spiral galaxy?

« on: 30/10/2019 04:57:03 »

You haven't proven anything until you've done that math. You're simply assuming that it's higher than the star's binding energy. Even if you did prove it to be the case, it wouldn't matter because it has nothing to do with the fact that your model violates conservation of energy.

Let's try to understand the real meaning of Orbital energy conservation especially with regards to potential energy.
Let's assume that in the whole Universe there are only two objects.
Due to the potential energy, the gravity force will convert the potential energy into falling kinetic energy.
Hence, they will move directly to each other until the final collision. The Vector Vp (velocity due potential energy) will be in the direct line between the two objects.
We all know that in orbital kinetic energy, the Velocity Vk (Velocity due to orbital kinetic energy) must be vertically to Vp.
Therefore, do you agree that the potential energy can ONLY be converted into falling kinetic energy (which is represented by Vp).
We know that in the open space objects maintain their momentum.
So, if Vp represents a velocity vector between the two objects, Do you see any possibility to convert that Vp to Vk?
If we can't convert Vp to Vk how could it be that Potential energy is part of the total orbital energy?
How can we discuss on conservation of orbital energy, if in the reality, a potential energy can't be transformed into orbital kinetic energy?

In the same token, let me ask how the accretion disc had been formed?
Let's assume that once upon a time there was a BH or a SMBH without any accretion disc.
Do you agree that any in falling atom or star should fall directly to that BH/SMBH?
So, how could it be that suddenly the Vp (in falling velocity vector due to potential energy) which was in a direct line between the Atom/star to the BH/SMBH is shifted by 90 degrees in order to be converted into a Vk (orbital velocity vector) at almost the speed of light?

12

New Theories / Re: How gravity works in spiral galaxy?

« on: 28/10/2019 05:19:06 »

I'm finished talking about this hypervelocity star issue. It doesn't have anything to do with conservation of energy.

The logic is very simple.
Does your model create energy? Yes.
Does the law of conservation of energy forbid the creation of energy? Yes.
Is the law of conservation of energy a law of physics? Yes.

Yes it is.
You discuss on the name of the gravitational conservation of energy in the galaxy.
However, in order to discuss about that key issue, do you really understand how gravity works in the galaxy?
So, we already know the conservation of energy between the potential energy to kinetic energy.
We all know that if we want to lift an object from the surface of the earth, we need to set a kinetic energy.
Therefore, if we shoot a bullet up to the air at very high velocity, that velocity represents a kinetic energy.
That kinetic energy is transformed into potential energy.
Therefore, as the bullet goes higher its velocity is decreasing.
At some point, (if we ignore the friction) when the potential energy is equal to the First moment kinetic energy, the bullet should stop completely and than fall back.
That is how gravity works on earth.
So, let's go back to our galaxy.
As this hyper star is climbing above the center of the galaxy, it increases its potential energy.
Therefore, its kinetic energy should go down as it goes higher and higher.
As it is moving away from the center of the galaxy, every moment it must gain higher potential energy and therefore, it also must decrease its kinetic energy due to the conservation of energy.
Therefore if we want to verify the first moment kinetic energy, we must add to that  4.97125 x 10^41 joules, all the kinetic energy that had been transformed into potential energy.
If you set this calculation you would surly find that the requested current potential energy represents 10^n times the current kinetic energy. 
Therefore, the Kinetic energy at the first moment due to the supernova, should be in the range of
4.97125 x 10^(41+n) joules
That first kick energy must be significantly higher than the 5.6741 x 10^41 joules

4.97125 x 1041 joules is smaller than 5.6741 x 1041 joules, so you are objectively wrong.

Therefore, by using your idea of the conservation of energy, I have proved that the requested first moment kinetic energy that is needed for this Hyper star, must break it to it's atoms.
We can ask also, why none of those Hyper stars don't fall back to the center of the galaxy?
Actually, the galaxy is crossing the space at a speed of 600Km/s.
It radius is more than 50,000 Ly.
We already know that there are more stars in the open space outside the galaxies than the total number in all the galaxies.
So, any star outside the galaxy had quite high potential energy (with regards to the galaxy center)
Therefore, it is expected that all the stars in the path of the milky way should fall in.
Similar to the falling meteorites on Earth.
How many fall in stars do we see?
One Million? One Thousand? or none?
How could it be that not even a single star can fall into the galaxy?
Don't they have any basic knowledge about the conservation of energy?
How could it be that they refuse to convert their high potential energy to Kinetic energy and fall in? 
Sorry, the outcome is very clear - Our scientists don't have a clue how gravity really works in the galaxy.
Once you understand how gravity works in the galaxy, then let's discuss about the gravitational conservation of energy in the galaxy.

13

New Theories / Re: How gravity works in spiral galaxy?

« on: 27/10/2019 06:38:23 »

Thanks Kryptid

The question then becomes: can enough kinetic energy be imparted to the star to speed it up to 1,200 kilometers per second from 200 kilometers per second without giving it enough energy to destroy it in the process? The mass of the star is estimated at 9.9425 x 1029 kilograms, which we put into the kinetic energy equation Ek = 0.5mv2. So we get Ek = (0.5)(9.9425 x 1029)(1,000,000)2 = 4.97125 x 1041 joules. It's close, but this demonstrates that the star can indeed acquire the needed energy from the supernova to speed up to 1,200 kilometers per second without being destroyed. Much of the star probably was blown off in the process, actually, since O-type subdwarfs are thought to be the result of red giants having their outer layers removed by some powerful process.

Sorry. I totally disagree
A star isn't an elastic ball that can observe the kick energy and restore it for its kinetic movement.
It is a solid object that keeps its ball shape due to gravity force.
Did you try to calculate the energy that holds any atoms in the star?
If an atom at the surface of the star will get that kind of energy, how it should react?
It is quite clear to me that if you kick a star with an energy of 4.97125 x 10^41 joules, that energy is much stronger than the total gravity energy that keeps the atoms together in order to set the ball shape.
However, Supernova is not just a pure energy. It comes with big broken objects.
We all know the outcome of one big object collision with the earth 65 million year ago (and it was not due to supernova, just a free falling object).
If some of the big broken objects from the supernova hit a nearby star at that ultra energy, they should cross though the star even before the star can open his eyes.
So, in a few moments the whole star should be broken to pieces by that ultra energy that comes with big broken objects.
It is similar to shooting a ballet into a water melon.
Try to do it and verify the outcome.
Therefore, supernova can't just kick a nearby star in order to gain the 1200Km/s without breaking it to pieces.


There is another issue - Star density.
We have a solid observation that in our aria the star density is 512 stars per 100 Ly.
So, if the supernova took place in the same radius as we are, it should affect several hundreds of stars.
Therefore, even if we accept the impossible mission and somehow it could deliver the requested energy, than why do we see only one lonely star at 1200Km/s?
Where are all the other nearby stars that were affected by that mighty supernova energy?

14

New Theories / Re: How gravity works in spiral galaxy?

« on: 26/10/2019 20:04:23 »

When you are in space twirling your test mass around on a string both your hand and the test mass will revolve around their mutual centre of gravity if you make an effort to maintain your hand away from the centre of gravity it will absorb energy from the system there will also be a small amount of energy radiated away in the form of gravitational waves.

At the moment that you change the location of your hand - you actually change the energy.
As you change the energy you get different orbital velocity.
Our scientists think that somehow a matter can fall in and set a new star or even a new galaxy.
This is a fiction.
If you have a floating matter in space, it will never ever merge and form a new star.
The only way to set a star is by a gas cloud that orbits near a SMBH.
Even so, the matter isn't falling inwards to the center of the gas cloud in order to form a single new born star.
In reality, the matter orbits around the center of the gas cloud.
As it orbits (under the impact of a nearby SMBH's gravity force), it crystallized and theoretically could form several stars per gas cloud (that orbit around the center of the cloud).
However, as they emerge from the gas cloud, they are actually orbiting around their common center of mass, (which is probably the center of the gas cloud).
This center represents the VHP (virtual host point) which I have described in the past.

15

New Theories / Re: How gravity works in spiral galaxy?

« on: 26/10/2019 17:56:16 »

The answer is in that very link:
"The fastest-known star in the Milky Way is on a path out of the galaxy, and new research suggests it was a supernova that gave it the boot".

This is fantasy.
If we set a ball on atomic bomb, is there a chance that the atomic bang will rock the ball to the moon?
Sorry, a supernova near a star should break it to pieces.
In any case, we discuss on orbital energy transformation.
Our scientists were positively sure that all of those hyper star had been ejected from the galaxy due to binary star.
They didn't mention any word about supernova.
Now that they find a star at 1200Km/s than suddenly they look for better alternative.
That proves that they really don't have a basic clue how gravity really works in the galaxy.
If they have stated that it is due to binary star and the observation contradicts this statement - It is expected that they should set this idea in the garbage.
You can't just invent new idea every time that new observation contradicts your current assumption.
If our scientists give an explanation for something and this explanation can't meet the observation - they should say: Sorry we have an error in our assumption. Clear and loud!
Once they find that a hyper star at that ultra velocity - they have to say, sorry our binary star idea can't meet this observation.
We see again and again that our scientists have a fatal error in understanding how gravity really works in the galaxy.
So far they couldn't offer any real theory that covers everything we see.
They could not even give a valid explanation for the star density and thickness of the spiral arms.
I have proved that the density wave is a pure fiction.
So, how can you claim that your current understanding in gravity is the ultimate one?
Sorry - if you wish to show that your understanding in relevant, than please give full explanation for whatever we see.
You can't just disqualify my understanding without offering real alternative.

Let me ask you again:
Please show how your current theory about gravity meets the density and thickness of the spiral arms

I did give you such an equation. The gravitational potential energy equation shows exactly that: how much energy is released by allowing an object to fall into a gravitational field or how much energy is consumed moving an object up against a gravitational field.

Yes and no.
As long as we convert the potential energy into falling kinetic energy - there is energy conservation.
However, once we deal with orbital kinetic energy - there is no energy conservation between the potential energy to the orbital kinetic energy.
I have already proved that:
If you decrease the distance/radius between the objects by twice you actually increase the Potential energy by twice and get only half of the orbital kinetic energy.
So, we see clearly that the sum of the total orbital energy changing as we change the radius.
You actually have agreed with that.
Therefore, theoretically, we can agree that as long as the radius is fixed, the gravitational energy is fixed.
However, as I have already claimed, this statement is not fully correct.
Based on my understanding, the orbital kinetic energy is decreasing over time.
Therefore, the radius/distance between the orbital objects is increasing over time.
So, the increased radius between the earth/moon systems is not due to tidal but it is due to that "over time" phenomenon.
I wonder how Einstein and Newton didn't discover that phenomenon.
I know that by now you don't agree with that, but sooner or later our scientists would find that this is fully correct.

Both Newtonian and relativistic physics allow energy to be converted from potential to kinetic by a gravitational field, so what you are saying is false: https://en.wikipedia.org/wiki/Gravitational_energy#Newtonian_mechanics

Again - this doesn't cover the orbital kinetic energy.

Nowhere do any of Newton's laws state that gravitational fields can generate new energy.

He also didn't say that the orbital kinetic energy needs a transformation of energy.
He had offered the formula for gravity force.
Based on that gravity force we can extract the orbital velocity and the orbital kinetic energy.
So, for any orbital radius, there is a specific orbital velocity.
That orbital velocity is a direct outcome of the gravity force.
Hence, the orbital kinetic energy is also a direct outcome of that orbital velocity.
If the orbital is new, than the orbital kinetic energy is new!!!
Newton didn't ask for any transformation in orbital kinetic energy.

16

New Theories / Re: How gravity works in spiral galaxy?

« on: 26/10/2019 06:48:13 »

Gravitational sling-shotting speeds one body up at the expense of slowing another body down. If you knew how conservation of energy worked (which you have abundantly demonstrated that you don't), you would have known that. Both stars can't make each other go faster as that would violate conservation of energy. If one speeds up, that is because it has extracted energy from the other star via gravitational interactions, which makes the other one slow down

So you claim that: "Both stars can't make each other go faster as that would violate conservation of energy."
We know that the average orbital velocity of the stars around the galaxy is about 200 Km/s
I have found a  runaway star at 1200Km/s:
https://www.space.com/28737-fastest-star-galaxy-strange-origin.html
The runaway star, US 708, is traveling at 745 miles per second (1200 km/s) — that's  26 million miles per hour (43 million km/h) —making it the fastest star in the Milky Way ever clocked by astronomers, according to the new research"
It is also stated:
"The monster black hole at the center of the Milky Way has the gravitational muscle to fling a star on a one-way-track out of the neighborhood, and many other hypervelocity stars are thought to originate from there. But US 708 didn't start its journey near the galactic center, the new research shows."
Based on your advice it must come from a binary star in the galactic disc (as it is far away from the SMBH).
However, you have stated that: : "Both stars can't make each other go faster as that would violate conservation of energy."
So, how can we convert the energy of 200Km/s to 1200Km/s?

Our scientists must find a real explanation for all the observations!!!
If we can't explain the observations by the theory that we believe - than this theory is not relevant.
Therefore, If we can't explain the G star's density and thickness - than we have to look for better theory.
Let's assume that 100,000 professors and scientists believe in the same assumption/theory.
Does it mean that this assumption is correct?
If you all believe in that assumption - than you have to explain the observation.
So, a theory can't be correct just because all of you believe in that theory.
There must be full correlation between any theory to all observations.
Theory D is the only valid theory that gives a perfect explanation for everything we see.
It meets all the observations by -100%.
Actually, there is a simple way to verify which theory is real.
You can tell me what is your expectation based on your theory/understanding, and I will tell you the expectation by theory D.
Let's start by the nearby G stars density.
I can tell you by 100% that at any 50LY radius in the nearby aria in the Orion arm - the density MUST be 64 G stars.
If you go out the arm - the density must drop to zero.
So, there are only two possibilities for star density - or 64 per 50Ly or Zero. Nothing in between (unless we are located exactly at the edge of the arm or the bridge).
Would you kindly advice what would be your expectation for the G star density around us (take radius of 2000 Ly?
If we will cross the Orion arm, what might be the star density?
Why don't we ask NASA to verify my expectation?
If we will discover that theory D is correct, do you agree to reconsider your objection against this theory?

Your model is automatically wrong because it violates conservation of energy. Is gravitational energy a form of energy? Yes. Therefore, it must obey the law of conservation of energy.

Gravity force is a FORCE!!!
Newton didn't call it Gravity energy
Gravity is not just an energy transformation.
A force can generate new energy.
Evey orbit of a star consume energy.
If I will give you a rope with a ball at the end, don't you need to generate new force/energy in order to keep it orbiting around you?
In the same token -  Every orbit of a star consumes Force and energy.
However, as the gravity force comes for free (Due to Newton , there is no need to fuel the gravity force or set any energy transformation), the energy due to this force also comes for free.
So, far you couldn't show any formula which links the gravity force into energy transformation.
In electromagnetic - energy transformation is must.
In gravity force - As Newton and Einstein didn't specify any request for gravity energy transformation - your assumption is totally incorrect.
Therefore, the statement that gravity force can't generate new energy is a severe violation of Newton law

17

New Theories / Re: How gravity works in spiral galaxy?

« on: 24/10/2019 06:04:48 »

Dear krptid

Our scientists have developed some ideas/assumptions about gravity and energy.
For example you have stated:

An object won't change its orbital radius unless it is gaining energy from an outside source or losing it to an outside source. A spacecraft won't spontaneously increase its orbital radius from the Earth, for example. If the astronauts want to increase the orbital radius, they will have to add energy by firing the engines.

Again, you have it backwards. Adding energy from an outside source will increase the orbital radius. You can't say, "I want the  orbital radius to increase, so the energy required to do that pops up out of nowhere". The radius won't increase unless the object in orbit is supplied with energy from another source that already existed. Look at my example with the spaceship.

However, spacecraft is not a star.
We need to verify the stars activity in the galaxy and see if those assumptions are correct.
The only way to verify that is by our observation on the galaxy.
The thickness of the spiral arms and the density of the stars at any section give us a perfect understanding.
So, if you look at our Milky Way spiral arms you would see immediately that stars can't migrate inwards.
Actually, the stars in the spiral arms MUST drift outwards.
We all see many stars that are ejected at ultra velocity from the galactic disc.
https://curiosity.com/topics/this-hypervelocity-star-was-ejected-from-the-milky-way-curiosity
"Every once in a while, the Milky Way ejects a star. The evicted star is typically ejected from the chaotic area at the center of the galaxy, where our Super Massive Black Hole (SMBH) lives. But at least one of them was ejected from the comparatively calm galactic disk, a discovery that has astronomers rethinking this whole star ejection phenomenon."
In order to explain this phenomenon our scientists claim that it is due to Binary star:
"When a star is kicked out of the galaxy, it's usually one star from a binary pair. Scientists think that as a binary pair get too close to the SMBH and its overwhelming gravity, the hole captures one of the stars. The other star is shot out into space in a "gravitational slingshot." The black hole has to be a supermassive one because only they have powerful enough gravity to accelerate these run-away stars to such high velocities."
The specifically claim: "the hole captures one of the stars"
So, please show me those stars that were captured by the SMBH?
If we see that stars are ejected outwards than we also must see stars that pushed inwards.
We see quite many of those Hypervelocity Star that were Ejected from the Milky Way.
Did we ever see one single Hypervelocity Star that was pushed inwards to the SMBH in the Milky Way?
The answer is very clear - No No and No.
Those Hypervelocity Star were not ejected due to the Binary affect.
They were ejected from the galaxy as they have moved to far away from the requested density in the arm/galaxy
Those 64 stars per 50 Ly is a living factor for our solar system in the galaxy.
Our sun is located near the edge of the Orion arm. If we will dare to move even 30 LY away from the edge of the arm, we will lose the gravity bonding of all our nearby stars.
At that moment we will be boosted outwards from the galaxy and will never come back again.
The galaxy doesn't take back any matter or stars from outside.
All the stars and matter that we see in the galaxy had been created by our mighty SMBH.
So, please - would you kindly answer the questions about the thickness and G stars density in the spiral arms?
How the "density wave theory" can still be valid under those observations?

18

New Theories / Re: How gravity works in spiral galaxy?

« on: 23/10/2019 06:35:19 »

Let's verify the observations:
At a radius of 100Ly around the sun there are 512 Stars.
That volume represents 8 times the volume of 50Ly.
So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:
512 / 8 = 64 stars per 50Ly.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)
This is not a random activity!!!
A density wave can't generate that kind of fixed density.

That density isn't fixed. The stars are moving relative to each other. It’s also a very bad assumption to calculate the density of stars within 100 light-years of the Sun and extrapolate that to the density of all of the other stars in the spiral arm. What does that have to do with anything, anyway?

Dear Kryptid

I would like to remind you that the title of this discussion is:
"How gravity works in spiral galaxy?"
We will continue our discussion about the Energy.
However, the density of G stars in spiral arm and the thickness of the galactic disc are very important for our understanding - "How gravity works in spiral galaxy?".

So, do you agree that there is clear observation for 64 Stars per radius of 50 Ly in the radius of 100 Ly around the sun?
If we will discover that this is correct for any location in the 1000 Ly segment around the sun (as long as you are in the Orion arm), If we also discover that outside the arm the density drop to Zero - What can we learn from that?

There is another issue which is a direct outcome of theory D.
As we go outwards from the galactic center the arm gets thinner and thinner.
If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.
That sets the end point of the spiral arms.
Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.

Actually our scientists have discovered that:
Close to the Ring (3KPC) - the thickness of the arms is 3000Ly
At our current location  (8KPC) - the thickness of the arms is 1000Ly
At the end of the spiral arms (12KPC?) - the thickness of the arms is less than 500Ly

So how the density wave theory can explain all of those observations?

19

New Theories / Re: How gravity works in spiral galaxy?

« on: 20/10/2019 15:22:12 »

Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?

There are 133 stars visible to the naked eye within that radius, which is less than 10% of the stars in total.  Map of the 133 here: http://www.icc.dur.ac.uk/~tt/Lectures/Galaxies/LocalGroup/Back/50lys.html
It says there are an estimated 1800 stars in 1300 systems (so much for binary being prevalent).

Your problem is that you get your data from hand-selected google pages instead of where scientists get it: From looking at the stars.

Sorry if I was not clear enough.
I'm speaking about G stars
http://solstation.com/stars3/100-gs.htm
As many as 512 or more stars of spectral type "G" (not including white dwarf stellar remnants) are currently believed to be located within 100 light-years or (or 30.7 parsecs) of Sol -- including Sol itself. Only around 64 are located within 50 light-years (ly),

20

New Theories / Re: How gravity works in spiral galaxy?

« on: 20/10/2019 12:54:04 »

Dear Kryptid
Thanks again for your great support!

1. Potential energy/gravity force for nearby stars

If you think that the Earth/moon potential energy has still an impact at long distance of 1000Ly, than why the potential energy of nearby stars (at a range of less than 100 Ly - there are 512 stars at that range) have no impact on the Sun orbital movement?

They do. The Sun is far from sitting still, you know.

Would you kindly advice what our scientists really understand from that nearby Potential energies/gravity forces?
Let's verify the observations:
At a radius of 100Ly around the sun there are 512 Stars.
That volume represents 8 times the volume of 50Ly.
So, if we try to calculate the average stars density per radius volume of 50 Ly the outcome will be:
512 / 8 = 64 stars per 50Ly.
Surprisingly, At a radius of 50Ly around the sun there are exactly 64 stars?
That shows that the density of stars in the Orion arm is exactly 64 Stars per 50Ly (or 512 Stars per 100Ly)
This is not a random activity!!!
A density wave can't generate that kind of fixed density.
The whole idea of density wave is that in some arias there is higher density of stars while in other arias there are less density of stars.
This is not the case in the Orion arm and not in any other spiral arm.
I would love to know the real 50 Ly radius volume densities of stars per distance from the galactic center - in the arm.
Based on my theory, we should find that as we go closer to the galactic center the density is higher.
As we go further away from the galactic center - the density is lower.
However - at any spiral arm - at any distance from the galactic center, the density of stars is FIXED.

Outside the arm - the density MUST be ZERO!
However, the arm might have any kind of shape and there are bridges between the arms.
I can promise you that as we cross the edge of the arm - there will be ZERO stars there!!!
So, at the nearby aria around the sun - the density could be: or 64 per 50Ly radius or ZERO.

There is another issue which is a direct outcome of theory D.
As we go outwards from the galactic center the arm gets thinner and thinner.
If we go further enough from the galactic center - the stars starts to be disconnected from the arms at the galactic disc.
That sets the end point of the spiral arms.
Our scientists don't have a basic clue how the spiral arm really works as they totally ignore the real impact of that density and why the arm is thinner and thinner as we move outwards.
If they would try to understand the real impact of the Potential energy/gravity force in short range - less than 500 or 1000Ly, they should get to the same conclusions which I have got.
If we could go back in time - let's say 100 M years ago - we would find that all the nearby stars are still there.
So, any star that is moving away from us today - should come back in the future.
While any star that is moving in our direction, should move away in the Future.
How is it possible?
The density wave doesn't give an answer for all of that.
However, the Potential energies and the gravity forces between nearby stars give a perfect solution.
They are all gravity bonded IN THE ARM itself!!!
So please - show me how our scientists translate that nearby potential energy/gravity force into real understanding about the structure of spiral arms.

2. Rotational energy conservation/transformation after the BBT
You claim that stars can't just decrease or increase their orbital radius without external energy.
If that is correct - than how the whole galaxies had been formed after the Big bang while there were no rotational energy at all?
At the early days - 380M years after the Big Bang we have only got matter.
No stars, No galaxies, No orbital objects no main body, no kinetic Energy No rotation energy - only matter everywhere.
So, if we take a cube of one billion LY - in that cube there was no Kinetic energy, no spin energy, no rotation energy - Nothing)
However, Today at the same volume cube there is so high rotational energy. It is everywhere. Try to take the same cube today and calculate the total rotation energy (of all the stars, galaxies) in that cube.
From Zero rotation energy in age of 380 M years we have got so high total rotation energy today.
So, how your idea of rotational energy conservation could be correct if in the early days there was no rotation energy at all???
Please - If you offer a random activity, than why the same idea for random gravity force/Energy that creates the whole stars, planets, moons, BH, SMBH, and even unlimited no of galaxies without any request for early rotational energy transformation/conservation in the Early days, can't work today even for just one pair of particle???
How can anyone believe that somehow in the early days of our universe the whole galaxies have been formed without any need for rotational energy transformation but today even for one pair of particle you insist for rotational energy transformation?
So please - would you kindly show how the whole orbital activities in the early galaxy had been formed out of nothing (no orbital activities after the matter creation - age of 380 MY after the Big bang), while we have to assume that the same rotational energy conservation should works also for those early days.

3. Moon/sun gravity force

Because the strength of Earth's gravity as felt by the Moon is higher than what it feels from the Sun. That's because distance matters as much as mass does. If the Sun was heavy enough, the Moon would feel it pulling on it more strongly than the Earth does. In that case, the Moon would indeed orbit the Sun.

So, you don't agree that the Sun/Moon gravity force is stronger than the earth/moon force?
https://www.universetoday.com/116158/why-doesnt-the-sun-steal-the-moon/
"If you’re up for some napkin calculations, you little mathlete, by using Newton’s law of gravity, you find that even with its greater distance, the Sun pulls on the Moon about twice as hard as the Earth does.
So why can’t the Moon escape the Earth?"
The answer is:
"So, yes, the Sun is totally trying to rip the Moon away from the Earth, but the Earth is super clingy.
The speed of the Moon around the Earth is about 1 km/s. At the Moon’s distance from the Earth, the escape velocity is about 1.2 km/s. The Moon simply isn’t moving fast enough to escape the Earth."
However, I have asked why the Moon had started to orbit around the Earth at its first day?
Our scientists claim:
https://www.space.com/29047-how-moon-formed-earth-collision-theory.html
"The formation of the moon has long remained a mystery, but new studies support the theory that the moon was formed from debris left from a collision between the newborn Earth and a Mars-size rock, with a veneer of meteorites coating both afterward."
I claim that this story is incorrect because:
a. Meteorites can't form new Moon. They only can set some sort of a ring, as we see around Saturn.
b. If one of the meteorite was as big as the moon, it had to follow the gravity force of the Sun at the moment of the impact (as at that moment the gravity force of the sun was stronger than the earth).


4. Could it be that the "Hill sphere" is the answer for why gravity works locally?

There is nothing "useless" about it. The problem is that 1,000 light-years is well outside of Earth's Hill sphere, so the Moon could not actually orbit at that distance because there are so many other stars and planets around that would be pulling on the Moon much more strongly than the Earth. If the Universe was empty of all matter except for the Earth and Moon, however, such an orbit could indeed exist because there would no longer be such a Hill sphere.

Wow
Thanks for that information about the Hill sphere:
https://space.stackexchange.com/questions/3015/how-large-is-the-earths-gravitational-sphere-of-influence-and-how-can-it-be-cal
"The Hill sphere (aka the Roche sphere) looks at things from the perspective of energy rather than force.
The Hill sphere asks a rather different question: Given a smaller body orbiting a larger body, can an even smaller body orbit the small body? "
Hence, as we monitor the rotational energy rather than gravity force we might find the answerfor why gravity works locally and why the moon orbits around the earth instead of orbiting directly around the Sun.

5. "Over time"

I wonder why our scientists do not add the "over time" impact to the gravity formula.
You can if you want to, but no amount of waiting will change the total orbital energy until energy is lost or gained by the object in orbit. Again, energy cannot be created or destroyed. You can't win an argument against nature.

Well, I think differently
Every orbit has a shape of spiral.
In the nature nothing comes back exactly to the same point (same orbital radius).
It might fall in or it might drift out.
Unfortunately, our scientists have missed that key element of "Over time".
I can promise you that by 100% all the stars in the spiral arms are drifting outwards over time.
After every cycle around the galactic center - the sun is drifting outwards.
If we could come back in one billion years from now we might find that our sun had been ejected from the arm and from the galactic disc.