0 Members and 1 Guest are viewing this topic.
The radius increases due to the normal activity by gravity. Any orbital object increases it's orbital radius over time. This radius increasing is not due to any sort of Tidal. It is just normal gravity activity.
In any case, at that first day, the Earth/Moon gravity was much stronger than Sun/Moon gravity.Therefore, the moon had chosen to orbit around the Earth instead of around the Sun. It keeps orbit around the Earth due to element No.4 (Hysteresis phenomenon in gravity).
That VHP1 is valid for the Sun and for any new born star in the galaxy.So, each star in the galaxy orbits by definition around its unique VHP1 (that was the center of the gas cloud). It will continue to orbit around this VHP1 (due to element No.4 - Hysteresis phenomenon in gravity) and goes where ever the VHP1 goes.
Therefore, with related to gravity force, the nearby aria do not "see" the star itself, but they "see" or set the gravity impact due to its VHP1.
Therefore, when we set the gravity simulation in the spiral arm, we need to focus on the locations of the VHP1 of each star and its effective mass, (which is estimated to be much higher than the real mass value).This is a key element why the gravity force of the nearby G stars in the Orion arm can hold/bond them together in the arm, (although by monitoring the real location/mass/density we might think that it's not good/high enough).I will stop at this point in order to verify that this message (so far) is clear to you.
There is no hysteresis in the function you gave. It references a current recession rate, nothing from the past. It reduces force from Earth without reducing the force from the sun, so if the magnitude of the force determines which thing it orbits (it doesn't), then Newton's formula should have Earth hanging onto the moon harder than your formula suggests.
If the moon is receding from Earth, is it taking its VHP with it? If so, why does it still orbit Earth and not this moved VHP?
QuotePlease see the following excellent image (It was not set by our scientists - but it is correct by 100%):http://www.biocab.org/Motions_of_the_Solar_System.jpg The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).If that picture were in any way representative of the actual motion, the orange ball moves at 217 km/sec and circles the galaxy in 200 million years. The blue dot path appears to be about 8 times longer, so if it is supposed to keep pace with this orange VHP, it would need to move about 8 times as fast, or 1700 km/sec. Instead it labels the speed at 5-20 km/sec, or about 1% of the speed it needs. The sun's speed is neither, so that picture is not an "excellent image".I am not commenting on your idea here, just commenting that the solar system cannot follow a path like that around the galaxy. Get your evidence from real data, not from drawings made by somebody with no credentials.
Please see the following excellent image (It was not set by our scientists - but it is correct by 100%):http://www.biocab.org/Motions_of_the_Solar_System.jpg The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).
Quote from: Halc on 17/03/2019 14:06:57There is no hysteresis in the function you gave. It references a current recession rate, nothing from the past. It reduces force from Earth without reducing the force from the sun, so if the magnitude of the force determines which thing it orbits (it doesn't), then Newton's formula should have Earth hanging onto the moon harder than your formula suggests.Sorry, I don't understand your question.
In any case, I have introduced five elements as follow:1. Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2... no. 1. (Newton formula for gravity force for any time interval is: F=GMm/(R+tΔ)^2)
In any case, by the time that the star emerge from the gas cloud, each moon orbits directly around its plane, each planet orbits directly around the star while the star orbits around the center of the gas cloud which represents the VHP1 point of that star.Therefore, those moons and the planets (that orbits around the star) don't have a VHP as they orbit directly around the star.That should answer your following question:Quote from: Halc on 17/03/2019 14:06:57If the moon is receding from Earth, is it taking its VHP with it? If so, why does it still orbit Earth and not this moved VHP?
So, only the star with its very high mass (With all its planets, moons and other objects (as Oort cloud) that it had collected from the gas cloud) will be able to keep on orbiting that VHP1. Is it clear?
QuotePlease see the following excellent image (It was not set by our scientists - but it is correct by 100%):http://www.biocab.org/Motions_of_the_Solar_System.jpg The blue dots (Apparent motion of the solar system) shows the orbital motion of the sun around its virtual host point (Orange ball).This image just show the orbital movement of the sun around Its VHP1. You shouldn't monitor the blue dots in order to get an idea about the orbital velocity. We actually know this data. It sets one full cycle in 60 million years..
However, can you please advice what is the amplitude/radius of that orbit which our scientists call "wobbling".
Please - I don't think that we should de- estimate the ability of "somebody with no credentials".
For instance, the moon currently has an average recession rate of 3.8 cm/year, but that rate was larger in the past at times, and smaller at other times
The formula references nothing from these past rates, only the current rate, which isn't hysteresis.
In fact nothing in our solar system has a VHP since they all directly orbit some mass.
Perhaps the sun doesn't,
but you've not described how to compute the location of its VHP or how the motion of an object relates to that VHP.
There isn't a circle, so there's no radius. The sine-wave picture you showed was more representative, even if it also showed many more waves per lap than the 3.5 waves. Work it out. If the 5-7km/sec speed is anything near accurate, the amplitude of the sine wave in and out of the disk is about 200 LY. I think I've read that it is more than that, but that's what I compute from 6km/s and a 60 M Year period sin wave. If there is a VHP associated with that motion, the sun crosses it every 30 million years, and has maximum acceleration when it is furthest from it, not when closest to it as you would expect if there was an actual object with mass responsible for that motion. Orbital motion has a fixed axis, and your biocab picture does not depict motion about any axis.
what about the mass of some object that is far closer: Does that divert the sun? What if the mass is strong enough to pull the sun away from its VHP?
These are really trivial questions with Newtonian physics, and I only need that one formula, plus F=ma to compute the motion of anything. Force on the moon from the sun is greater than force from Earth, therefore the moon always accelerates towards the sun, even when between the two. It only accelerates straight towards Earth when Earth is between the two and the sun and Earth are pulling in the same direction. That's what it means for one force to be greater than the other. The moon always accelerates primarily towards the greater force. It does not mean that it directly orbits the thing with the greater force.
Anyway, I cannot make any of these observations with your physics since you don't tell me how to compute the location of a VHP or how to compute the motion in relation to that VHP. It is why I cannot predict where my 3 unit-masses will go. Your theory is absent and I can make no predictions.
Yes, I get the 6 points
You mentioned it having an elliptic orbit around it sort of like S2 does, except S2 orbits Sgr-A, not a VHP.
The picture is wrong in just about every way, but it shows what you want, so you take it for fact instead of actual measurements
Quote from: Halc on 18/03/2019 00:07:58For instance, the moon currently has an average recession rate of 3.8 cm/year, but that rate was larger in the past at times, and smaller at other timesHow do you know that?Do you have any valid measurements to prove this idea, or is it based on current concept?
Based on my understanding this recession rate of 3.8 cm/year was never ever larger in the past
This formula gives the data about the radius in the past or into the future.F=GMm/(R+tΔ)^2
But that information is not hysteresis.hysteresis means that if the moons orbit around the Earth than it will continue to do so even if there will be a nearby object with much stronger gravity force (sun).
Quote from: Halc In fact nothing in our solar system has a VHP since they all directly orbit some mass.That is correct.
Quote from: Halcbut you've not described how to compute the location of its VHP or how the motion of an object relates to that VHP.Thanks for the question.
The location of VHP1 (VHP1 is the first orbital level of any star in the galaxy) is based on the amplitude of the wobbling activity.
Quote from: HalcThere isn't a circle, so there's no radius. The sine-wave picture you showed was more representative, even if it also showed many more waves per lap than the 3.5 waves. Work it out. If the 5-7km/sec speed is anything near accurate, the amplitude of the sine wave in and out of the disk is about 200 LY. I think I've read that it is more than that, but that's what I compute from 6km/s and a 60 M Year period sin wave. If there is a VHP associated with that motion, the sun crosses it every 30 million years, and has maximum acceleration when it is furthest from it, not when closest to it as you would expect if there was an actual object with mass responsible for that motion. Orbital motion has a fixed axis, and your biocab picture does not depict motion about any axis.ThanksIf I understand you correctly, the sun is located about 200LY above the disc
Please look at the following diagram:https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.htmlSo the sun is located 200 LY above the disc and it is still moving up.
Based on your explanation in order for the sun to set full cycle it needs to get all the way down, crossing the disc to the same downwards amplitude (200LY) and then get back again to the same distance from the galactic plane.
Therefore, 60 million years is needed for just one cycle.
Sorry - this is a severe mistake.
The Sun has no need to cross the disc plane.
If it is currently high above the disc(200 LY) it will stay above the disc as long as needed.
It will just need to orbit around its VHP1 as all the other nearby stars do.
Based on hysteresis it is clear that the sun will continue to orbit around its VHP1 even if there will be a nearby mass with much stronger gravity force. So, the sun is going to orbit its VHP1 as long as it takes.
Quote from: HalcThese are really trivial questions with Newtonian physics, and I only need that one formula, plus F=ma to compute the motion of anything. Force on the moon from the sun is greater than force from Earth, therefore the moon always accelerates towards the sun, even when between the two. It only accelerates straight towards Earth when Earth is between the two and the sun and Earth are pulling in the same direction. That's what it means for one force to be greater than the other. The moon always accelerates primarily towards the greater force. It does not mean that it directly orbits the thing with the greater force.That is clear.
Quote from: HalcIt is why I cannot predict where my 3 unit-masses will go. Your theory is absent and I can make no predictions.My theory is based on the data which is available to all of us.
It is why I cannot predict where my 3 unit-masses will go. Your theory is absent and I can make no predictions.
I would recommend to look at the nearby stars and verify their relatively orbital velocity.If I understand it correctly, we see that their relatively maximal velocity is about 15 Km/sec in all direction. So, by average they are moving at about 7.5 Km/s around their VHP. That should also be correct for our sun. If that is correct, and assuming that the radius of VHP1 is 2LY, we can extract the expected cycle time for one full VHP1 cycle. Do you agree?
Quote from: Halc The picture is wrong in just about every way, but it shows what you want, so you take it for fact instead of using actual measurementsThanks
The picture is wrong in just about every way, but it shows what you want, so you take it for fact instead of using actual measurements
QuoteThis formula gives the data about the radius in the past or into the future.F=GMm/(R+tΔ)^2It's a formula for force, not radius. Kindly show an example of this formula for a real object (say the moon) since, as I said, it is totally unclear how to add two numbers of different units. You can do it for the current force between Earth and moon, since you supposedly know its Δ.
Does the sun bring its VHP up there with it? The VHP doesn't come down because it has no mass, and the sun is not attracted to the mass below it, but rather to the VHP which 'needs' to be up there for some purpose.
It seems to me that you still don't understand how VHP works.So, let me do it for one more time:The Sun does not bring it's VHP anywhere. It works the opposite way.The Sun' VHP1 brings the Sun everywhere it goes..
Each one of those stars orbits around its VHP1. If we are located 200Ly above the galactic disc, than all of us should stay where we are.
Based on 65 Stars per 50 LY sphere, I would assume that the orbital cycle of each star around its personal VHP1 should be in the range of 2LY (or less).
So, the Sun orbits around its personal VHP1.This VHP1 must obey to the gravity law in the galaxy and in the spiral arms.
We can compare [star density in spiral arms] to the flash in our arm. The density of the flash is almost the same. If we try to measure the density of flash outside our arm, we should find that as there is no arm, there is also no flash over there. Therefore, outside our arm the density of the flash is Zero.In the same token, outside the spiral arm, the star density should drop to ZERO!!!
Each one of those stars orbits around its VHP1. If we are located 200Ly above the galactic disc, than all of us should stay where we are..
If we could find the exact location of VHP1 for all the nearby stars, we might find that the relative velocity between the VHP1 is almost zero.
It is called flesh. Anyway, perhaps you could consult actual survey data to verify that the star density between arms is zero, because that would indeed sink the density wave theory, and finding a non-zero density would sink your idea. Pick a spot that doesn't have one of these inter-arm pipes doing a high-speed transfer
You suggest here and in a prior post that the VHP is only 2 light years from here (closer than any star) and that we trace a sort of orbital path around that virtual point, which is a complete violation of Newtonian forces. There is negligible mass inside that small circle (most of it being our own Oort cloud), and thus gravity cannot be the force that accounts for that motion.
If you can find a single star hanging between the arms by itself, than my theory is none relevant by definition.
But please, don't show me an image of far end galaxies. Just a clear image from the nearby aria.
Quote from: HalcYou suggest here and in a prior post that the VHP is only 2 light years from here (closer than any star) and that we trace a sort of orbital path around that virtual point, which is a complete violation of Newtonian forces. There is negligible mass inside that small circle (most of it being our own Oort cloud), and thus gravity cannot be the force that accounts for that motion.Why do you insist to ignore my explanation again and again.Why do you insist to see real object at VHP while I have stated clearly that it is Virtual point.No real object is there!!!
I wonder what I should say to open your eyes.So, if I will say that there is a dark matter, would it be more logical for you?
If so, think about virtual dark matter.You don't see it; you don't smell it and you don't feel it.It is there becouse I said that it is there.If our scientists can set a dark matter at the center of the galaxy, why can't I set a dark matter at the VHP?
If Newton can work for the dark matter while it is at the center of the galaxy, why it can't work for the dark matter at the VHP1?
I assume that anyone who accepts the idea of dark matter at the center of the galaxy should also accept the idea of dark matter at that VHP1.
https://www.astronomynotes.com/ismnotes/s8.htm is a nice note about the problems with the model of arms being objects, and why that theory doesn't produce observed results. It proposes several alternatives, density wave theory being only one of them.The reason I linked it is because it also says this near the front:"There are many stars that are also in-between the spiral arms, but they tend to be the dimmer stars (G, K, M-type stars).". My simple google search turned up countless sites that verify this, but none that state than the space between arms is devoid of stars.
3.Virtual Host Point (VPH) - We had also long discussion on VPH. You don't agree with that idea. it is clear to me. However, this is very important point in my theory.
Sorry.There is no real data in that article.
In this article they don't say even one word about the possibility that there are wide aria without even a single star.
Therefore, this article is none relevant for our discussion.
So, during the process of star forming, the star orbits around the center of the gas cloud, while in this gas cloud there is no matter in the center. Therefore, that center of the gas cloud represents the location of VHP1 for any new born star.However, this gas cloud also set a gravity impact/"connection" with its nearby aria, as the center of the gas cloud represents its center of gravity.So, the gas cloud is "connected" to its nearby aria by the gravity which is represented by the location of its center.However, that gravity center also represents the Sun VHP1 as it had been emerged from the gas cloud.
Due to hysteresis from now on the nearby aria "see" that VHP1 as the host center of the Sun.Is it clear to you?
If not, than please try to accept it as is.
VHP = Virtual host Point.There in nothing there. However, the sun directly orbits around that Virtual point. It also represents an effective mass due to the orbital cycle of the sun around that point.
You have agreed to accept this idea.
So please - take it as is and at the end of my introduction you are more than welcome to disagree.
We can discuss it later on after introducing the full theory.
If the VHP doesn't have mass, then it doesn't have effective mass either. It cannot attract anything on its own. Hence it has no effect at all.
So, let's start from the idea that there is one pure gas cloud that orbits around the SMBH (while the matter in the gas cloud rotates at high velocity around the center of the gas cloud).1. Do you agree that in this case we can set the gravity calculation between the centers of the gas cloud to the center of the SMBH?
2. Let's also assume the following: As the matter in the gas cloud rotate around the center of cloud, and due to the Huge gravity impact of the nearby SMBH, at the first phase it crystallized into small objects that continue to orbit around the center of the gas cloud (all over the gas cloud).
So, how the gravity between the SMBH and that gas cloud with all of those small objects work?Do you agree that it still works based on the center of the gas cloud to the center of the SMBH?
3. Let's assume that those small objects crystallized with each and set bigger and bigger objects in the cloud.Eventually we might find few significant objects that orbits around the center of that gas cloud.
Do you agree that the gravity still works based between the center of the gas cloud to the center of the SMBH?
Let's assume that eventually we might find that those big objects have merged with each other and set one massive object which still orbits around the center of that gas cloud. This object consume significant portion of the matter in the gas cloud, so even the matter that was in the center had drifted outwards and become part of this object.In this case, how the gravity works?
Do you agree that the gravity still works between the center of the SMBH to the center of the gas cloud even if there is a significant object which orbits around the center of that gas cloud (while the center of the gas cloud has no significant matter or even without any matter?
Anything that passes very near that point will not be diverted in the least by it. Virtual points do not have mass, and gravity acts only on mass.
The non-uniform gravity would pull the cloud apart and also make the inner parts orbit faster than the outer parts.
There is a reason rockets always immediately tilt to the East to get away from Earth. It takes much more energy to go straight up, and when you turn the engines off, the straight up rocket just falls right back down again. Motion to the side (especially the East) allows it to get to and stay at a desired altitude. Rockets bring their own reaction mass with them No idea how your stars do it, or the moon for that matter since you deny classic sources of that lateral force.
S2 is theorized to have formed from a cloud that came from a considerable distance.
You don't agree to accept the idea that VHP1 is actually the center of the gas cloud.
However, how do we know that the current hypothetical ideas about gas cloud, star forming activity, gravity forces near the SMBH and many others ideas are correct?
How do we know that the following message is correct?Quote from: HalcAnything that passes very near that point will not be diverted in the least by it. Virtual points do not have mass, and gravity acts only on mass.
Actually, few years ago, our scientists have discovered a star which is moving very close to the SMBH.They were expecting for fireworks as the SMBH gravity is going to smash that star.But surprisingly - that star had not been effected by the mighty SMBH gravity force.
So, time after time our scientists set an expectation and surprisingly - it doesn't work according to their expectations.
Quote from: HalcThe non-uniform gravity would pull the cloud apart and also make the inner parts orbit faster than the outer parts. Can you prove it?There are several gas cloud orbiting around the SMBH. Do you see there any cloud that can confirms this theory?
You discuss about rockets:Quote from: Halc There is a reason rockets always immediately tilt to the East to get away from Earth. It takes much more energy to go straight up, and when you turn the engines off, the straight up rocket just falls right back down again. Motion to the side (especially the East) allows it to get to and stay at a desired altitude. Rockets bring their own reaction mass with them No idea how your stars do it, or the moon for that matter since you deny classic sources of that lateral force.Based on this answer, due to the ultra high gravity force of the galaxy, there is no way for a star to be ejected out without an engine/rocket.
Surprisingly, somehow stars can escape with any rocket
"But researchers from the University of Michigan have identified one hypervelocity star that appears to have been ejected from the stellar disk rather than the galactic bulge."If there was a dark matter, why that dark matter couldn't hold that star in the galaxy/disc?
How could it be ejected without any engin?
I have a solid evidence why the current theory about the the density wave and dark matter are a fatal errors:Please look at the following image of the galactic disc:https://www.quora.com/How-big-is-the-Milky-WayWe see that as we move further away from the center of the galaxy, the disc became thinner and thinner.I have explained why based on my theory we get exactly that disc shape.However, is it possible to get that shape based on the current theory?
It is clear that as we move further away from the center, the impact of that Inclination is more visible
For pluto we see it very clear as it is located far away from all the other planets in the solar system.
So, it is clear that any orbital object has some sort of Inclination if it orbits around a central main host.
Therefore, if the stars in the galaxy were orbiting due to the gravity force of the dark matter we would expect to see that as the radius is longer, the disc should be thicker and thicker.
As we move further away the Inclination has more significant impact and therefore it is expected to see a thicker disc at the far end.
Quote from: HalcS2 is theorized to have formed from a cloud that came from a considerable distance.So, from where S2' came from?What does it mean: "considerable distance?"
Where S2' gas cloud was when S2 had been formed?
How far it was from the SMBH at that time?
What is the age of S2?
Have we ever monitored the process of new star formation in the gas cloud?
Have our scientists really monitored the galactic disc?
Did they really try to see if there are arias without stars?
You have stated that there are many unclear problems with the current theories/concepts.My theory fully meets all the evidences.
You have accepted the unrealistic idea of dark matter. But you are not willing to accept the idea of VHP that is vital for my theory.
QuoteDid they really try to see if there are arias without stars?The survey database is always being updated. Many stars have been charted for where they appear and luminosity, but lack distance, which requires multiple measurements. Plus, they only see the brighter objects. Between the arms is quite far away and low luminosity stars take a lot of time to see, so they can't survey the entire sky for them in a short period. It's a lot of work.
QuoteYou have stated that there are many unclear problems with the current theories/concepts.My theory fully meets all the evidences.You don't have a theory.Yes, any hindsight theory always meets all evidence, since that's all it needs to do is copy the evidence and say it predicts it.
QuoteSo, time after time our scientists set an expectation and surprisingly - it doesn't work according to their expectations.That happens all the time. Science would make no progress without it.