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QuoteQuoteI really don't understand why do you see any possibility to convert the in falling velocity (or falling kinetic energy) to orthogonal velocity (or orbital velocity) at any sort of orbital objects.Look at comets, which have almost the exact sort of orbit you're describing here. They start out coming almost straight in (nearly pure 'falling' velocity as you put it), and suddenly as they pass the sun, the high velocity vector doesn't change much, but the sun is suddenly off to the side and that high velocity vector is now completely orthogonal to the force vector. The force vector (representing 'down') rotates around quickly but the velocity vector doesn't so much.
QuoteI really don't understand why do you see any possibility to convert the in falling velocity (or falling kinetic energy) to orthogonal velocity (or orbital velocity) at any sort of orbital objects.
QuoteQuoteSo, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?By rotation of the vectors. It is falling fast, and suddenly 'down' is a different direction as it passes by Earth. The direction of 'down' is changing all the time. Takes month to go all the way around, but less time in this new orbit you've given it.
QuoteSo, how could it be that this falling in vector can set or be transformed to any orthogonal velocity?
They are completely relevant since the orbits of the objects you describe (the ones that don't involve impacts) will orbit forever in their new highly eccentric orbits.The ones we see happen just like you describe: A reasonably circular orbit is suddenly altered, stopping the comet in place. It thus begins to fall into the inner solar system, achieving far higher velocities than it ever had before. This is exactly what I've been describing.
The picture does not depict an orbit. Find a picture of a comet orbit to see a real example of this so called unbelievable idea. Here's one:http://www.khadley.com/Courses/Astronomy/ph_205/topics/pluto/images/cometorbit.jpgThe comet moves clockwise. Draw a velocity vector at say the 1985 mark and notice it is pointed almost the same direction as the force vector, not perpendicular at all. So the comet is gaining speed as it fall nearly directly towards the sun. It reaches its highest speed as it swings behind the sun near where the word 'Mars' is written. At that perihelion point the velocity is entirely perpendicular to the force vector. It is no longer 'falling in' at all, but is moving faster than any other point in the orbit. That is because the velocity vector has rotated only about 80° from the 1985 point to the perihelion point, but the force vector has rotated around 170°, pointing nearly the opposite direction as it was in 1985. That's the vector rotation I'm talking about.
What velocity vector? The picture you asked me to look at shows a guy dropping a rock from a small height. The (implied, not depicted) velocity is totally vertical in the picture.
We see that the final velocity (at the collision point) is v = √(2gh)Has nothing to do with collision point. That formula assumes a uniform gravitational field g, which means it only works for short distances. Gravity force on our moon varies by its distance from the planet, so that formula doesn't work. Use the PE formula, and convert that figure into KE. That gives an accurate final velocity for dropping something from a large height.
QuoteQuoteWould you kindly use Newton formula & mathematics to prove that unbelievable idea?Use the formulas I indicate above, which give accurate velocity values at any point in the orbit. Those formulas do not give eccentricity for a given scenario, but they give what you asked: velocity resulting from dropping an object from altitude X to altitude Y.
QuoteWould you kindly use Newton formula & mathematics to prove that unbelievable idea?
Hence, there is no change in the potential energy or orbital kinetic energy due to a direct collision.
QuoteQuoteThis answer isn't clear. How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?The answer I gave yields orbital speed S, not component speeds. Orbital velocity V is the velocity with speed S tangential to the orbital path. If you want to break that velocity into your two components at any point in the orbit, find the angle between the acceleration (or force) vector and the velocity (or momentum) vector. What you call the 'falling in vector' Vf is Scos(θ) in the direction of the primary. The vector orthogonal to Vf (Vo) is Ssin(θ) in the direction perpendicular to Vf. It's that easy.
QuoteThis answer isn't clear. How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?
QuoteSo you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.Most of the time, yes.
So you agree that the small orbital object that had completely lost its orbital velocity should collide with the main object.
But there is a change of potential and orbital kinetic energy due to the immense change in orbital radius of the comet over time.
(3) The new orbit is closer to the planet than before, with the satellite's velocity having increased because gravity is stronger at this radius than it was at the original orbital radius.
The new orbit would still come back to the collision point every time unless a 2nd application of force (another collision or millions of years of tidal correction) circularizes the orbit like that. The new orbit would remain like the 2nd picture, not the third.
In a stable orbital cycle as the eccentric is greater than zero and less than one, that change in the energy (potential/kinetic) is actually a temporary change.
But the moon is massive, so Earth will be moving relatively quickly in its own orbit around the moon, so if the moon is halted, there's a chance the Earth gets out of the way in time. Probably not. As I said, only Charon has a chance of doing this. I haven't worked out the numbers.
If the planet is moving fast enough, the moon will miss it as it falls and go into that eccentric new orbit.
Most of the time the planet is barely moving. So you stop any typical satellite and it's going to drop pretty much straight down.
In my discussion, I assume it was halted in place in the frame in which it had a circular orbit (the frame of the two-body system). In any other frame, the moon's path was not a circle, but more like a helix.
QuoteQuoteThe formula should be as follow:S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)So, the orbital velocity is the outcome between the Vf and Vo. We called the orbital velocity V, not S. S is a speed, not a vector..With that terminology exception, your statement is actually correct, and I never said otherwise.
QuoteThe formula should be as follow:S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)So, the orbital velocity is the outcome between the Vf and Vo.
But keep in mind that these vectors are constantly changing over time for an object in orbit. They're not fixed values, and your argument seems to hinge on them being fixed. There is force on the orbiting thing, so there is always acceleration, and acceleration means all vectors (V, Vf, and Vo) are constantly changing.
QuoteQuoteSo, even if Vf will get to it's maximal velocity, Vo must be zero.If it falls straight down, they yes. It isn't in orbit in any scenario with an impact like that.
QuoteSo, even if Vf will get to it's maximal velocity, Vo must be zero.
48: If the orbit is circular, Vf is zero, which hardly sounds like a 'full match' with Vo.
47: V is orbital velocity whether or not there is this 'full match' between these two component vectors.
"Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth."
I don't. My mailbox doesn't. Newton's slow cannon ball doesn't. These objects are subject to gravity force, and yet they don't orbit. Yes, all objects orbit if you restrict 'objects' to things that are in orbit, which is like saying every red object is red.
51 Our own moon is currently stopped in its own frame (by definition), and in that frame, Earth is moving fast enough that the moon misses it and instead goes into an eccentric orbit. Thus my statement is exactly correct.
54 Our moon is gravity bonded with Earth...
In that frame, the planet is always moving, so when you stop the moon, the planet keeps on going.
There is no magic 'match' required for Vo and Vf. If you take a random asteroid from the asteroid belt and give it a random Vo and Vf with no regard to 'matching' them, so long as the sum of them (V) yields a kinetic energy that when added to its potential energy is a negative figure, the object will probably orbit the sun.
For example, Mercury's velocity is currently decreasing, and yet the force of gravity is decreasing because of this, or "causing the 'falling' to have a lesser influence" as that quote words it.
Most orbits tend to circularize over time via tidal effects, but it takes infinite time to finish the job in an ideal case, but in fact a perfect circular orbit can only be achieved with a 2-body system, and there is no 2-body system isolated from all other bodies.
Disks are not objects, and ring-objects cannot have a stable orbit. If they were objects, they would fall in. The rings are never perfectly circular due to the presence of other objects perturbing the motion. Saturn's rings are the best nearby example.
Oumuamua isn't in orbit since it has positive energy. It is moving at greater than escape velocity at all times.
Why do you claim that "Disks are not objects".In each disc there are many objects. Every stone, rock and even Atom is the disc is an object by itself.
So what is the source for all the Atoms in the accretion disc?1.Do you agree that the matter/plasma/Atoms in the accretion disc is coming from outside?2. If so, than why the orbital cycle of that matter is so circular orbit?
Why it isn't elliptical as you have explained:
3.How could it be that the orbital velocity of the plasma is 0.3c?
Let's look again on Oumuamua:Oumuamua is a perfect example for an object that is coming/falling in from outside into the center of the main mass.
However, if Oumuamua isn't in orbit since it has positive energy than any falling in object should behave in a similar way.
Therefore, it proves that there is no way to set any orbital path for an object that falls in.
Hence, it proves that even a gas cloud or a star that is falling into the direction of a SMBH should have a positive energy and should be ejected immediately outwards.
Even if you wish to belive that a falling object can set an orbital path, it is clear that the eccentricity of that orbit should be very high.
You have offered a comet as an example. We clearly see that the eccentricity of the comet is almost close to one.
1. If Oumuamua can't set an orbital path around the Sun, than any falling in matter to the direction of the SMBH can't also set an orbital path around the SMBH.
2. If you still hope that somehow a falling matter can set an orbital path, than based on your explanation about the comet the falling matter must set high eccentricity orbital path.
3. Hence, if the matter in the accretion disc was coming from outside than their orbital eccentricity had to be very high. However, as the orbital eccentricity of the matter in the accretion disc is close to zero (almost a perfect circular orbit), than it proves that this matter can't technically comes from a matter that is falling from outside the SMBH.
QuoteQuote3. Hence, if the matter in the accretion disc was coming from outside than their orbital eccentricity had to be very high. However, as the orbital eccentricity of the matter in the accretion disc is close to zero (almost a perfect circular orbit), than it proves that this matter can't technically comes from a matter that is falling from outside the SMBH.All wrong. Pretty much saying the words 'had to be', 'must', 'cannot', 'proves', etc. tends to generate false statements since there are usually exceptions. Not always, but as a rule of thumb.Most material that gets near the accretion disk collides with it, and that slows the material down, circularizing its path. All those eccentric orbit bits of sand and such that hit Earth? Notice that the collision immediately nearly circularizes their subsequent motion. The excess energy due to the orbit change is discarded as heat, which is why the accretion disk glows.
Quote3. Hence, if the matter in the accretion disc was coming from outside than their orbital eccentricity had to be very high. However, as the orbital eccentricity of the matter in the accretion disc is close to zero (almost a perfect circular orbit), than it proves that this matter can't technically comes from a matter that is falling from outside the SMBH.
Few years ago, our scientists were positively sure that they are going to see fireworks as S2 is going to collide with the SMBH.They even verify that S2 and the SMBH were in the same direct view line from us.Unfortunately for them and for you, there were no fireworks and no collision.S2 had passed very close to the SMBH without setting any sort of effect on the Accretion disc or on itself.
let's assume that in the next time it would collide with the accretion disc.What would be the impact of that collision?
Actually, don't you think that even if S2 comes too close to the accretion disc (without touching it), it should generate severe interruption in that disc?
How can you believe that due to that collision: " Notice that the collision immediately nearly circularizes their subsequent motion. The excess energy due to the orbit change is discarded as heat, which is why the accretion disk glows."Actually, this is a clear contradiction to Kepler law.
You claim that the eccentricity of the comet is 0.6.
so what is the eccentricity of S2?
So, now you claim that this energy due to kepler law can be converted into heat?
Due to kepler law, that energy should blow out the whole accretion disc.
It is similar to truck that collide with a bicycle.
How can anyone believe that a bicycle can drain the energy from a truck due to collision?
Don't forget that the orbital velocity of S2 might not be fully aligned with the orbital velocity of the accretion disc.So, we must have a severe interruption in the accretion disc.Where are the fireworks?
2. Statistical chance to reduce the average orbital radiusWe clearly know that all the planets in the solar system and almost all moons (with one or two exceptions) are increasing their average orbital radius per cycle.
The Moon is drifting outwards from the Earth over time. The Earth is drifting outwards from the Sun.So, they are both increasing their average orbital radius over time.The total orbital objects (Moons + Planets) mass in the solar system that are increasing their average orbital radius, is significantly higher than the total moons mass that are reducing their average orbital radius.I hope that you agree that the ratio between the total orbital mass in the solar system that are decreasing their average orbital radius to the total mass that are increasing their average orbital radius is less than 1 to 1/10^20
3. Mass ratio between the accretion disc to the molecular jet stream.We all know that the total estimated mass in the molecular jet stream above and below the accretion disc is about 10,000 sun mass.
QuoteQuoteDue to kepler law, that energy should blow out the whole accretion disc.It seems you're the one claiming Kepler's involvement here. Kindly show which of his laws says this.
QuoteDue to kepler law, that energy should blow out the whole accretion disc.
QuoteQuoteSo what is the eccentricity of S2?.885 Look it up. Not hard to do.
QuoteSo what is the eccentricity of S2?
S2 would be torn apart by tidal forces and some of it would form a new disk oriented with S2's orbit. The old disk would probably be absorbed by the new one
For S2 and Sgr-A to line up in our view, we'd have to cross the plane of S2's orbit, and that only happens every 100M years or so. It's called an eclipse when it happens and nobody expects collisions from an eclipse.
A great number of moons are drifting inward. More than half of Jupiter's moons are moving inward.
Quote from: HalcQuote from: DaveDue to kepler law, that energy should blow out the whole accretion disc.It seems you're the one claiming Kepler's involvement here. Kindly show which of his laws says this.Based on Second law of Kepler:"The same (blue) area is swept out in a fixed time period.
Quote from: DaveDue to kepler law, that energy should blow out the whole accretion disc.It seems you're the one claiming Kepler's involvement here. Kindly show which of his laws says this.
Therefore, due to kepler, the same energy that brings S2 with eccentricity of 0.885 MUST also take it out.
Please look at the (blue) area!!!Every atom in S2 must obey to the Second law of Kepler
Therefore, if S2 would be collided with the accretion disc, it is not going to lose its outwards energy due to Kepler's Second law even if it would it be torn apart.
Therefore, due to the collision of S2 with the accretion disc, it would take significantly portion of the mass from the accretion disc outwards.
Let's agree that S2 would be torn apart.However, due to kepler law the matter in the accretion disc should be blow away with the outwards momentum of all the atoms in S2.
If there was a collision, we should also see fireworks.Do we see any fireworks at any galaxy in the whole universe?
Why can't we expect for collision at the eclipse?
In this case, they are both located in a 2D.If S2 set an eclipse with the SMBH is MUST collide with it.Unless, the SMBH is not located at the orbital 2D of S2.Only in this case there is a possibility for eclipse without a collision.
In any case, with regards to the Molecular jet stream:Do you agree that the total mass in that jet stream is 10,000 solar mass which had been formed from a constant supply stream from the accretion disc?
If so, how could it be that accretion of only 3 sun mass can supply a constant stream of 10,000 sun mass.
In one of the article that I have found it was stated that "Shoving 10,000 suns into the black hole at once would do the trick"
If the accretion disc had Shoved in the past 10,000 suns at once
Quote from: HalcA great number of moons are drifting inward. More than half of Jupiter's moons are moving inward. How do we know that?
Did we really measure the drifting direction?Would you kindly show one (only one moon in the whole universe, except our moon) that we really measured its inwards drifting direction.
Moon is heading outward on average. There is a force adding to its total energy.
I assume that by now we all understand how the tidal friction set the thrust which is needed to push away the Moon from the Earth.However, the water bulges in oceans are very unique for the Earth/Moon system.We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?They don't have water. They don't have relatively big moon around them. They don't form those water bulges.So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?
QuoteQuote from: Dave Lev on Today at 16:33:06Do you agree that the estimated mass of the molecular jet stream is 10,000 Sun mass?...I agree to none of this. It's all nonsense you're making up sans evidence.
Quote from: Dave Lev on Today at 16:33:06Do you agree that the estimated mass of the molecular jet stream is 10,000 Sun mass?...
It can't, so the 10,000 figure you made up is obviously wrong.
Yes. They're positing 10,000 suns falling in, not jetting out. That much falling mass might be enough to power the more significant jets posited in the past.
And, by contrast, we know that Phobos is drifting towards Mars. It's been measured. Dave seems to be ignoring that.