[Dave Lev (OP)]

Thanks Halc

Quote
I 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.

Comets are irrelevant to our discussion as they are constantly orbiting the sun:
https://solarsystem.nasa.gov/asteroids-comets-and-meteors/comets/overview/?page=0&per_page=40&order=name+asc&search=&condition_1=102%3Aparent_id&condition_2=comet%3Abody_type%3Ailike
"Comets are cosmic snowballs of frozen gases, rock and dust that orbit the Sun. When frozen, they are the size of a small town... There are likely billions of comets orbiting our Sun in the Kuiper Belt and even more distant Oort Cloud."
Therefore, we can't use comets as example for falling in objects even if their orbital elliptic shape is very sharp.

Quote
So, 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.

I really don't understand that explanation.
Please look at the following image for:
1. Free Falling Object Dropped From a Known Height
https://owlcation.com/stem/Solving-Projectile-Motion-Problems-Applying-Newtons-Equations-of-Motion-to-Ballistics
We see that the final velocity (at the collision point) is v = √(2gh)
It is also very clear that this velocity vector is horizontal to the Earth.
So, how this Horizontal falling in velocity vector could be transformed to orthogonal velocity vector (orbital velocity)?
Would you kindly use Newton formula & mathematics to prove that unbelievable idea? 

[Dave Lev (OP)]

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.

Well, if we can show or prove that the comet had completely lost its orbital velocity due to collision, but now it regain it as it falls in, than, yes that could be a good example.
So, do we have any sort of evidence for current immediate outcome due to comet collision?
Why do you claim: "A reasonably circular orbit is suddenly altered, stopping the comet in place."
Why the comet had stopped in place?
Do you see any sort of collision or sudden orbital lost?
Don't you agree that we actually see a comet in a normal orbital cycle that just got to its maximal radius and then comes back?
That activity doesn't give any indication for sudden orbital velocity lost.
This is a very normal activity for any orbital cycle as the orbital object gets to its maximal radius.
So, if you don't have a clear evidence for a comet collision (and the direct outcome from that collision) than the comet is just irrelevant for our discussion.
In the article it is stated clearly:
"Comets are cosmic snowballs of frozen gases, rock and dust that orbit the Sun".
So, it is just in a constant and stable orbital cycle around the Sun, while its eccentric could even be close to one.
Therefore, the comet should come again and again to the same minimal radius and the same maximal radius after every full orbital cycle.
Hence, there is no change in the potential energy or orbital kinetic energy due to a direct collision.
Again - we try to understand the impact of collision that force the orbital object to lose suddenly its orbital velocity.
As the comet is already in a constant orbital status, it can't give any indication for the outcome due to our main focus on suddenly orbital velocity lost due to collision.
In any stable orbital system, with any sort of eccentric (zero to almost one) the orbital object gets to the same point (radius) after every full orbital cycle.
Therefore, do you agree by now that the eccentric can't give any indication for orbital lost energy due to collision?

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.jpg
The 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.

That could be a normal explanation for a any orbital cycle with eccentric close to one.
So, we only see a comet in a stable orbital cycle around the sun.
Therefore, that Vector rotation is a very normal outcome from any orbital system at any sort of eccentric
Again - this doesn't give any indication for a sudden orbital lost due to collision.

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.

So, do you agree that  the velocity of an object that falls in (to a planet) must be represented by a totally vertical velocity vector (with reference to the planet)?
If so, again - how that "totally vertical" velocity vector can be transformed to orthogonal velocity vector (orbital velocity vector)?

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.

Yes, that is correct.
However, the exact formula for the vertical falling in velocity is not so important.
It is very important that any falling in must be vertical to the planet.

Quote
Would 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.

This answer isn't clear.
How do you convert the "totally vertical" velocity vector due to falling in, to orthogonal velocity vector?

[Kryptid]

Hence, there is no change in the potential energy or orbital kinetic energy due to a direct collision.

But there is a change of potential and orbital kinetic energy due to the immense change in orbital radius of the comet over time.

[Dave Lev (OP)]

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This 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.

Dear Halc
So you claim that:
S = Orbital velocity
Vf (falling in velocity vector)= Scos(θ)
Vo (Ortogonal velocity vector) = Ssin(θ)
That could be correct as long as we monitor the velocity vectors.
However, in reality Vf is a direct outcome from gravity force, while Vo is the first moment orthogonal velocity of the object (as was explained by Newton cannon ball.
So, just if there is a full match between the Vf and Vo (orthogonal) we can get the magic velocity that we can call Orbital velocity.
Therefore Vf in reality has no impact on Vo (orthogonal) and vice versa.
However, both of then set the orbital velocity.
Therefore, the 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.

Let's go back to Krypptid example.
Please remember that we have stated the due to the collision the orbital object had totally lost its orbital velocity.
At that moment, as there is no orbital velocity, R is actually represents H.
So, just after the collision, the object is located at H, its orbital velocity (Vo) is zero, its falling in velocity vector Vf is zero and also its orthogonal velocity is zero.
So, We have the main object (let's assume that it is a planet), and there is also the object.
You have stated that if the orbital object is small enough, most of the time it should collide with the planet.

So 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.

I need to understand why do you claim "only most of the time" and how it could gain any orthogonal velocity as it falls in if its orbital velocity just after the collision is zero.
In order to understand that, let's assume that we have an object with mass of 1,000Km.
Let's drop it at a high of 10,000Km above the planet.
So at that moment, H = 10,000Km, Vo =0 , Vf =0.
I assume that we all agree that (if we ignore the air) it should fall horizontally to the planet.
Therefore, as it falls in it accelerates, its falling velocity - Vf (horizontally to the planet) is increasing over time, while its Vo should be zero.
So, even if Vf will get to it's maximal velocity, Vo must be zero.
Now, if we take much heavier object and set it at a higher distance from the planet,
Do you see any possibility that it will increase its Vo as it falls in?

But there is a change of potential and orbital kinetic energy due to the immense change in orbital radius of the comet over time.

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.
Every full orbital cycle it should get to the same minimal and maximal radius.
Therefore, the potential energy or the kinetic energy at the minimal or the maximal radius will always be the same after one full orbital cycle.
We discuss on totally different issue.
We want to understand what will be the real impact due to a collision with the orbital object.
First you have stated:

(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.

Than Halc has claimed that:

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.

Than you have accepted this explanation.
I don't agree with none of you.
I claim that
S (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)

Let's look at Newton cannon ball explanation:
http://www.hscstudyguides.com.au/hsc-physics-course-summary/hsc-physics-course-summary-space/rocket-launches-orbital-motion/
"Newton envisaged a cannon firing a projectile horizontally from the Earth’s surface. Ignoring air resistance, the projectile would prescribe a parabola, eventually falling back to Earth. However, as the speed of the projectile is increased, the projectile will take progressively longer to hit the ground, because although gravity is pulling towards the centre of the field, the Earth’s surface is falling away from the projectile at the same time due to its horizontal motion. Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth."
It is stated clearly:
"gravity is pulling towards the centre of the field"
The Vf is a direct outcome due to gravity. therefore, Vf is in a directly towards the centre of the field (or the center of the planet if you wish)
The "horizontal motion" is represented the Orthogonal velocity Vo
It is stated clearly that in order to "travelling in a circle around the Earth" there must be a full match between Vf and Vo.

With regards to ellipse orbit, It is stated clearly:
"As the velocity increases even more, the circle becomes an ellipse"
So, when you think about that ellipse orbital cycle (with eccentric greater than zero, less than one), the orbital object must get a boost in its orthogonal velocity.
It is also stated:
" if the speed is increased enough, the trajectory becomes hyperbolic. At this point, the projectile has enough velocity to leave the gravitational field."
At that moment the eccentric is greater than one.
So, your both assumption that the object can set even one full orbital cycle after loosing significantly it orthogonal velocity (or even zero) is totally wrong.
From now on you have to argue with Newton.

[Kryptid]

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.

Temporary or not, it is still a demonstration of an increase in orbital kinetic energy caused by gravity. Something which you claimed doesn't happen.

[Dave Lev (OP)]

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.

This is incorrect.
You are missing a key element in gravity.
Almost Every object in the galaxy orbits around something due to gravity force.
This something could be a planet, Star, a center mass of several objects, the center of the galaxy and so on.
The moon for example is orbiting around the Earth, while it is totally ignore the Sun Gravity.
Please be aware that the Moon/sun gravity force is more than twice with regards to the Moon/Earth gravity force.
The Earth orbits around the Sun, while it is totally ignore the gravity force of the center of the galaxy, while the Sun orbits around the center of the galaxy at a speed of over than 600 Km/s.
In the same token, none of the star in the galaxy care about the idea that the galaxy is also crossing the space at over than 600 km/s.
So, each object in the galaxy is actually "gravity bonded" in its current orbital system and almost totally ignore the impact of other gravity forces (with the exception of tidal forces).
So, with regards to the moon - it is "gravity bonded" to Earth. Hence, wherever the Earth goes, the moon goes with it.
The mass of the Moon is totally none relevant.
Therefore, it was a mistake to claim that:
"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."
Please be aware that in our theoretical discussion, we only have reduced the current orbital velocity of the object to Zero.
In that theoretical concept we also didn't break of the orbital object from the current gravity force.
So, we didn't break the gravity force between the Moon/Earth.
We have just reduced the moon orbital velocity to zero. (due to some imaginary collision or any idea which you like).
Therefore, It should be clear to all of us that as long as the moon is still "gravity bonded" with the Earth, it must fall directly to the center of the Earth. It can't miss the Earth as it falls in. Therefore, there is no way for the moon to gain any sort of orbital velocity as it falls in.

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.

The Earth can move at the speed of light. As long as the satellite is "gravity bonded" with the earth, it should fall in directly to the center of the earth without any ability to gain any sort of orbital velocity.

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.

So, please let's focus only on circular orbit and a frame of two-body system.

Quote
The 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.

Thanks
So let's agree
V (orbital velocity Vector) = Vf (falling velocity vector) + Vo (orthogonal vector)

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.

As we discuss on "circular orbit and a frame of two-body system" do you agree that the V, Vf and Vo are fixed?

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So, 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.

I hope that you also agree by now that if the moon had totally lost its orbital velocity, while it is still in a two body system with the earth, it must fall in directly to the center of the earth.
Therefore, Its Vf at the moment of the collision with Earth must be maximal, while its orbital velocity should be zero.

48: If the orbit is circular, Vf is zero, which hardly sounds like a 'full match' with Vo.

This is incorrect.
Every orbital object is actually falling in constantly.
https://www.quora.com/Why-do-things-orbit-the-earth-instead-of-falling-towards-its-gravity
"objects in orbit around the earth are in a perpetual state of “falling” toward the earth. If the object's velocity were to change (in this case, decrease), then the force of gravity would cause the “falling” to have a greater influence, resulting in the object's eventual plummet into the atmosphere and toward the surface."
I know that you really dislike those kind of answers from Quora, but I think that they are perfectly correct.
Let's look at the Moon.
This moon is actually falling constantly directly to the center of the earth, therefore its Vf is not zero.
However, as it also move forward at the orthogonal velocity - Vo, that movement compensate the falling in velocity.
Therefore, it keeps itself with the same radius (again, let's assume that the orbital is a cycle with eccentric =0).
So, even as Vf is greater than Zero, due to the full magic fit with Vo we get that magic orbital velocity that keeps it in the orbital cycle (while we might  think that Vf is zero).

47: V is orbital velocity whether or not there is this 'full match' between these two component vectors.

If there is no full match between the Vo and the Vf there will be no orbital system.
Please - we focus only on a pure orbital cycle with eccentric = 0.
That brings us to Newton cannon ball:
As was clearly stated by Newton, the magic orbital velocity is

"Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth."

Hence, If we can boost the (Vo -orthogonal velocity) of an object to the magic velocity that perfectly fit with its Vf, than this object will set a pure orbital cycles around the main mass.

[Dave Lev (OP)]

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.

I mean free objects in space. We are all part of Earth mass.

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.

It seems that you didn't understand my message:
Let's look at the following two options:
1. The moon is in the Earth frame
You have already agreed that:

54 Our moon is gravity bonded with Earth...

In this case, if you stop it, you stop it with reference to earth.
Therefore, it should be clear that it is similar to the idea that we just drop the Moon at its current location.

In that frame, the planet is always moving, so when you stop the moon, the planet keeps on going.

I disagree. We stop the moon with reference to Earth.
In this case, it should collide directly with the earth.
2. The moon is in another orbital frame.
Let's assume that the moon orbits around the Sun instead around the Earth ("Gravity bonded" with the Sun).
So, yes in this case it could miss the Earth.
However, as it is "gravity bonded" with the Sun, it will continue to orbit around the Sun.
Therefore, there is no way for the Moon to start orbiting around the earth unless it really collide with the Earth and break itself to several objects.

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.

This is totally incorrect.
The Earth is crossing again and again the orbital path of the asteroids that actually orbit the Sun.
So, far we can clearly assume that it had crossed the orbital path of Million over billions asteroids.
How many of them had been really captured by the earth gravity and set a circular orbital path around the earth?
You have told me that we have discovered one object and even this one is not orbiting in a circular orbit.
So, do you agree that the chance to set any sort of orbital path even with eccentric that close to 1 (Very sharp elliptical orbit) is less than one to million over billions.
However, the chance to set a circular orbital cycle is virtually ZERO!!!
I still claim that all the comets were orbiting the Sun from day one.
However, the shape of their orbital cycle (eccentric close to one) shows that they are in their way to totally disconnect from the Solar system.
So, they can't be used as a prove or evidence for setting an orbital cycle but on the contrary - they are a clear indication that the orbital system is losing "gravity bonded" over time.
Therefore, it is quite clear to me that those comets should eventually be ejected from the Solar frame.
I wonder if we can find a proof for that.
In other words, I claim that an eccentric orbital system can be converted to circular orbit!!!

That brings me again to Newton
It is quite clear that you refuse to accept Newton explanation.
http://www.hscstudyguides.com.au/hsc-physics-course-summary/hsc-physics-course-summary-space/rocket-launches-orbital-motion/
So, Please let me know which one from the following explanations you reject:
http://www.hscstudyguides.com.au/wp-content/uploads/2016/06/figure-5.jpg

Do you agree that for a Circular Orbit - A magic orbital velocity is needed?

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.

Somehow you insist to take an eccentric orbital system and try to understand how the forces works based on location in the orbital cycle.
This is not relevant to our discussion.
We want to understand how external forces can affect the orbital system.
Actually we wish to understand how orbital system can be created from the first stage.
Especially with regards to the Accretion disc around the SMBH.
Do you agree that all the atoms/particles in that disc/ring are actually orbiting at a perfect circular orbit at almost the speed of light?
If so, what is the chance that a falling star from outside could be break down to its atoms/particles and start orbiting in that velocity in a pure circular orbit?
I claim that the chance for that is less than a ZERO!!!
The accretion disc can't be created based on falling in matter.
This MUST be a direct outcome from Newton!

[evan_au]

The most powerful gamma-ray burst yet was seen this year.

GRB are thought to originate when:
- A very massive star collapses to a black hole or neutron star
- A burst of matter is ejected from its poles, at very close to the speed of light
- This matter runs into nearby gas, emitting a narrow beam of high-energy gamma rays
- We happen to be in line with this narrow beam of gamma rays

In this case, the gamma rays had energy up to 1 TeV (the LHC only operates up to 14 TeV)

Of course, the small amount of matter ejected along the poles is more than compensated by the large amount of matter falling into the black hole/neutron star.
See, for example: https://www.nasa.gov/feature/goddard/2019/hubble-studies-gamma-ray-burst-with-highest-energy-ever-seen

[Dave Lev (OP)]

Please look at the following article:
Interstellar Misfit ‘Oumuamua Could Be Dormant Comet
http://astrobob.areavoices.com/2017/12/19/interstellar-misfit-oumuamua-could-be-dormant-comet/
"This diagram shows the orbit of the interstellar asteroid / comet ‘Oumuamua as it passes through the solar system. Unlike all other asteroids and comets observed before, this body is not bound by gravity to the Sun. It has come from interstellar space and will return there after its brief encounter with our star system. Its hyperbolic orbit is highly inclined and it does not appear to have come close to any other solar system body on its way in"

We have excellent example for a "body that is not bound by gravity to the Sun" which comes very close to the sun but eventually that body is ejected from the solar system without setting even one full orbital cycle around the sun.

That is a key indication that object that is not bounded by gravity to the main mass might not be able to set even one full orbital cycle.
Somehow you wish to believe that this is feasible.
However even if I agree with this none realistic assumption, you and Kryptid have confirmed that  the orbital cycle must be very elliptical at the first falling/orbital stage.

In the following message you claim that the time that it takes to convert an elliptical orbit to circular orbit is almost infinite:

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.

You also add one more limitation: " in fact a perfect circular orbit can only be achieved with a 2-body system,"
When we look at ANY accretion disc around the SMBH at the center of spiral galaxy we clearly see that ALL of them have a perfect CIRCULAR orbit.
None of the accretion disc in the whole UNIVERSE shows any signs of eccentricity.

Even if we believe to our scientists that all the matter in the accretion disc is coming from outside, than somehow we must see around any accretion disc eccentric orbit for the matter that had fallen inwards in less than infinity.
I still don't understand how any falling in star or gas cloud can set a perfect circular orbit around the SMBH at a velocity of 0.3c, even after the infinity.

However, what is your intention for "infinity"?
Is it more than 13.8 Billion years or more than 10 billion years?
What is the requested Minimal/maximal age of the spiral galaxies in order to get that pure circular orbit of the plasma in the accretion disc?

If nothing was falling inwards during the last 10 billion years, while all of the accretion discs are clearly ejecting matter, how could it be that they are still full with matter/plasma (at orbital velocity of 0.3c)?

[Dave Lev (OP)]

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.

How can you compare Saturn's rings to accretion ring?
Can you prove that the Saturn's rings are coming/falling in from outside?
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:

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.

How could it be that something is falling in and set a circular/ or almost circular orbit?
3.How could it be that the orbital velocity of the plasma is 0.3c?

Let's look again on Oumuamua:

Oumuamua isn't in orbit since it has positive energy.  It is moving at greater than escape velocity at all times.

Oumuamua is a perfect example for an object that is coming/falling in from outside into the center of the main mass. We clearly see that it comes close to the sun, but then it is ejected outwards for never return.
You claim:

Oumuamua isn't in orbit since it has positive energy.  It is moving at greater than escape velocity at all times.

I agree with that.
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.
However, when we discuss about the matter in the accretion disc (that our scientists claim for coming from outside) you suddenly forget that in this case the matter must also set an orbital path with very high eccentric.

Conclusions

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.

[Halc]

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.

I can loop a strap around a hula-hoop (or a rubber ring for that matter) and drag it away by pulling on the strap.  Try that on the rings of Saturn and all you get is a mild disturbance of a couple bits of gravel.  The hula hoop, if spun, all rotates at the same angular rate.  The angular rate of the material in Saturn's rings moves at a rate that is a function of each bit's distance from Saturn.

These distinctions are true because the hula hoop and rubber ring are objects and the rings are not.  For this reason, the rings of Saturn are in a stable orbit and a ring object cannot be.  A ring object cannot be in a stable orbit about a mass at its center regardless of any spin (or lack of it).

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?

Fluid friction.  They hit each other if in eccentric orbits.  Energy is lost to friction until the motion is nearly circular.

Why it isn't elliptical as you have explained:

Anything infalling obviously was elliptical at some time.

3.How could it be that the orbital velocity of the plasma is 0.3c?

Apparently the temperature at that radius is enough to form plasma.

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.

It is departing, not falling in.  It already passed said center of the main mass a couple years ago, and obviously missed it, so your comment shows a lack of reading comprehension.

However, if Oumuamua isn't in orbit since it has positive energy than any falling in object should behave in a similar way.

64 Not the ones with negative energy, like Earth.

Therefore, it proves that there is no way to set any orbital path for an object that falls in.

65 Moon is currently falling in (getting closer), but it is in orbit, so obviously wrong.

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.

66, mostly for the 'Hence'. It may or may not have positive energy, and it may or may not keep it.

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.

Something can transition from positive to negative energy by losing some of that energy, and thus fall into orbit from a non-orbiting trajectory.

You have offered a comet as an example. We clearly see that the eccentricity of the comet is almost close to one.

0.6 might be more typical.

However, when we discuss about the matter in the accretion disc (that our scientists claim for coming from outside) you suddenly forget that in this case the matter must also set an orbital path with very high eccentric.[/quote]Not so.  Look at Saturn's rings.  That material never had a very eccentric orbit that I know of, but it also never had positive energy.  Accretion material rarely forms from positive energy material, but it can.

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.

But Oumuamua could have fallen into an orbit had a sufficient force acted on it to slow it down.  It just so happens that nothing ever did in this case.

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.

The comet was always in orbit, but a recent outside force altered it to a far more eccentric one, much like Kryptid's example.  The moon is currently falling in, yet its orbit isn't highly eccentric, so that part is wrong as well.  All these assertions, and every one of them has a counterexample.

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.

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.

[Dave Lev (OP)]

1. Collision

Quote
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.

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.

How do you know that: "Most material that gets near the accretion disk collides with it, and that slows the material down, circularizing its path"
We can clearly see billions of Spiral galaxies. In each galaxy there is an accretion disc.
Can you please give us one single example to this imagination???
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.
There is an explanation for that, but as expected, our scientists had claimed that there was an error in their instruments.
They didn't want to deal with the impact of their verification; therefore they have solved it by the "error factor"
In any case, let's assume that in the next time it would collide with the accretion disc.
What would be the impact of that collision?
Don't forget that the total mass in the accretion disc is about three sun mass. Therefore, the mass dissipation in the whole accretion disc is very low.
Did you try to set a simulation for that kind of collision?
Why the low dissipation mass in the accretion disc shouldn't be ejected in all directions due to a collision with the massive object as S2 that comes in with great energy?
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.
S2 has elliptical orbit. Its orbital path is quite similar to a comet path.
You claim that the eccentricity of the comet is 0.6. so what is the eccentricity of S2?
Based on Kepler, the same energy that brings S2 or comet near the main mass should take it outwards.
So, now you claim that this energy due to kepler law can be converted into heat?
Sorry, this isn't science this is science fiction.
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?
If you still believe in your imagination, then please prove it.
Let's assume that we accept this kind of imagination.
Let's assume that we even agree to ignore kepler law and somehow S2 is acting according to our wishful list.
Let's assume that it collides with the accretion disc and coverts its "kepler" energy to heat as you wish to believe..
Even in this case, there must be severe interruption in the accretion disc due to the 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?
What could be the eccentricity of the accretion disc due to that collision?
So, if for example the eccentricity of the objects/atoms in that accretion disc is less than 0.01, it is quite clear that due to that collision, the objects/atoms in the accretion disc should get high eccentricity.
But in reality - we don't see that.
So, out of the Billion accretion discs that we monitor, why we do not see any signs of falling in stars?
Do you agree that we have never ever seen any signs for this kind of interruption?
If we don't see and it and we will never see, what should be the outcome?
Actually, you hope that somehow a star can come closer and stay closer.
You have already rejected the explanation by Newton about the cannon ball.
Now you reject the impact of the orbital energy due to kepler law. You hope that somehow this outwards energy can be converted to heat.
In other words, you hope that orbital objects can reduce their average orbital radius.
Let's verify the chance for that:

2. Statistical chance to reduce the average orbital radius
We 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
If so, the chance for any orbital mass/matter to decrease its average orbital radius is less than 1/10^20
However, you WANT to believe that somehow all the matter in the accretion disc must come from outside.
Therefore, you hope that the matter from outside the accretion disc (It can be a star, gas cloud or even a rock) is reducing its average orbital radius around the SMBH and finely gets into the accretion disc.
You don't care about the simple evidence that the chance to decrease the average radius is less than 1/10^20.
You are using that very low chance as a leading concept for your unrealistic idea.
You hope that somehow the accretion disc eats its food from outside without any evidence for that and with a clear contradiction with Newton and Kepler law.

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.
All of that mass is coming from the accretion disc.
So, how could it be that a total mass of only three sun mass in the accretion disc can supply 10,000 sun mass to the molecular jet stream?
Our scientists hope that this mass is a direct outcome from eating process in the past.
That of course is imagination, as the accretion disc can't eat and it will never ever eat even one atom from outside.
All the matter in the accretion disc is coming from inside!!!
I have many other examples that shows that you have a severe misunderstanding about the accretion disc
However, I'm quite sure that you won't let them to change your idea.
It is clear to me that you don't let the evidences to confuse you.
Why you are not using your deep knowledge in science to see the reality?
How can you still believe it that science fiction?

[Halc]

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.

Link please, because I've heard no such thing.  Cannot comment on things made up.
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.

let's assume that in the next time it would collide with the accretion disc.
What would be the impact of that collision?

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.

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?

I suppose it would.

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.

There are 3 laws, and I don't see a violation of any of them, especially since none of the laws deal with colliding material.

You claim that the eccentricity of the comet is 0.6.

There is no 'the comet', so no, I don't claim that.

so what is the eccentricity of S2?

.885  Look it up.  Not hard to do.

So, now you claim that this energy due to kepler law can be converted into heat?

No, I didn't claim that.

Due 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.

It is similar to truck that collide with a bicycle.

Agree, but which of Kepler's laws describe a truck-bicycle collision?

How can anyone believe that a bicycle can drain the energy from a truck due to collision?

Oh, Newton's laws have something to say about that.  Try the 3rd law.  Once again you seem to be positing energy from nowhere when the bicycle gains energy but the truck doesn't lose any.

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?

No fireworks because S2 does not collide with anything.  It gets no closer than about 30 au.  I'm sure there would be a light show if S2 got that close.

2. Statistical chance to reduce the average orbital radius
We 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.

Statistics has nothing to do with that.  A great number of moons are drifting inward.  More than half of Jupiter's moons are moving inward. Depends how many of them you count.

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

No.  Unrealistic number you made up. Venus probably isn't drifting inward, and it masses well over that unreasonably small percentage of the total.  The moon and every other moon and planet will eventually drift inward (if not destroyed before then), so statistically over time, they spend more time going inward than outward.  I'd say 9/10, not 1/10²⁰.

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.

67.  This is you heavily twisting out of context a number you saw in a article.  The article did not say any such thing.

[Dave Lev (OP)]

Quote
Due 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.

https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion
Based on Second law of Kepler:
https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#/media/File:Kepler-second-law.gif
"The same (blue) area is swept out in a fixed time period. The green arrow is velocity. The purple arrow directed towards the Sun is the acceleration. The other two purple arrows are acceleration components parallel and perpendicular to the velocity."
S2 has an elliptical orbit with eccentricity of 0.885:

Quote
So what is the eccentricity of S2?

.885  Look it up.  Not hard to do.

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. Each Atom from S2 would have to obey to the Kepler's law and must continue the momentum outwards.
Therefore, due to the collision of S2 with the accretion disc, it would take significantly portion of the mass from the accretion disc outwards.
Try to set a simulation and verify that this is correct by 100%.

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

Ok.
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.
In any case, as the eccentricity of S2 is 0.885 and the eccentricity of the accretion disc is 0.01, than if S2 will be part of the accretion disc then the eccentricity of the accretion disc after the collision should be high above 0.01 (it might be lower than 0.885).
So, in order to verify a star collision with the accretion disc, we only need to find an accretion disc with high eccentricity.
Our scientists could find an Earth size gas cloud falling into the SMBH at a galaxy of 1 Billion LY away.
Based on that high verification, we should have the tools to see any severe interrupt in the accretion disc eccentricity of billions spiral galaxies.
Do we find any interruption in the accretion disc out of the billions spiral galaxies that we can monitor?
If there was a collision, we should also see fireworks.
Do we see any fireworks at any galaxy in the whole universe?
If we don't see any fireworks and any interruption in the eccentricity of the accretion disc than it proves that stars have never ever collide with any accretion disc in the whole Universe.

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.

Why can't we expect for collision at the eclipse?
Our scientists assume that S2 orbits around the SMBH.
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?
However, the total mass in the accretion disc is only 3 sun mass?
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"
https://phys.org/news/2012-05-ghostly-gamma-ray-blast-milky-center.html
However, they didn't give an answer if the accretion disc can hold 10,000 sun mass at once and why the jet stream is nicely and constant (without any interruption) as we can see in the following image:
https://phys.org/news/2012-05-ghostly-gamma-ray-blast-milky-center.html
If the accretion disc had Shoved in the past 10,000 suns at once, then at that moment the outwards stream from the accretion disc must be significantly higher than today, while there is only three sun mass in that accretion disc.
We had to see a severe difference in the thickness of the jet stream as the mass in the accretion disc is reducing.
However, the image of the jet proves that the molecular jet stream is quite constant.
Therefore, it isn't possible that the Milky way' accretion disc had Shoved 10,000 suns at once!!!
There is one more issue: If that was correct, than how could it be that out of the billions spiral galaxies (that we can monitor and which also must supply over 10,000 sun mass to their molecular jet stream), we can't see even one accretion disc with fireworks (due to star collision), with severe interruption in the eccentricity of the accretion disc or with a total of 10,000 sun mass in the accretion disc???

A 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.
Is it one more wrong assumption which is based on our scientists' wishful list (as usual)?

[Halc]

Due 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.

OK, you've chosen Kepler's 2nd law.  How does your statement above followed from that law?

Therefore, due to kepler, the same energy that brings S2 with eccentricity of 0.885 MUST also take it out.

Kepler's 2nd law makes no mention of energy.  It also doesn't mention collision with another object.  Again, you draw a conclusion from a random unrelated fact.  Not saying the conclusion is wrong, just that it doesn't follow from the law you've seemingly randomly selected.

Please look at the (blue) area!!!
Every atom in S2 must obey to the Second law of Kepler

68: Not true.  It certainly isn't true of every atom on Earth for instance.  The law isn't about atoms or other components of an object in orbit.  My mailbox does not obey Kepler's 2nd law.

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.

69: There you go with the 'Therefore' again.  Your statement doesn't follow from any of the above.  Not the 2nd law, and not from the other statements.  You have zero grasp on logic.  It's like saying 2+2=4 (no incorrect), therefore grass is green.  It just doesn't follow from 2+2=4 even if yes, grass is green.

Therefore, due to the collision of S2 with the accretion disc, it would take significantly portion of the mass from the accretion disc outwards.

70: Again, the 'therefore' is misplaced.  It doesn't follow from the preceding statements.

Kepler's 2nd law does not predict that a bicyclist would likely get severely injured by collision with a fast moving bus.

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.

71: Kepler's 2nd law (nor the 1st or 3rd) does not say this.  It doesn't follow from that law.  You have zero grasp of logic.

If there was a collision, we should also see fireworks.
Do we see any fireworks at any galaxy in the whole universe?

Indeed we do, almost everywhere except our own galaxy, which currently has a notoriously inactive rate of new matter falling in.

Why can't we expect for collision at the eclipse?

I've seen many eclipses, and never seen anyone worried that the sun and moon were colliding.  No, nobody has seen S2 line up with Sgr-A. You made that up.

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.

Word salad.  'located in a 2D' and 'orbital 2D' are meaningless phrases.  I have no idea what you think you are claiming.

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?

No.  'Pathetically weak' was the term I believe was quoted.  It's another thing you've made up.

If so, how could it be that accretion of only 3 sun mass can supply a constant stream of 10,000 sun mass.

That's right.  It can't, so the 10,000 figure you made up is obviously wrong.

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"

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.

If the accretion disc had Shoved in the past 10,000 suns at once

This is mass falling from the outside into the black hole.  The accretion disk is not particularly involved in this, although certainly there would be a disk of some proportion during this kind of activity.  But nobody is positing a disk of that mass.

A great number of moons are drifting inward.  More than half of Jupiter's moons are moving inward.

How do we know that?

Because there is no known force adding to their energy, while there is very much a known force subtracting from it.

Planets like Jupiter are moving outward mostly because the gravity of the sun is weakening as its mass is radiated away.

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.

[Kryptid]

Moon is heading outward on average.  There is a force adding to its total energy.

And, by contrast, we know that Phobos is drifting towards Mars. It's been measured. Dave seems to be ignoring that.

Both of the neutron stars in the Hulse-Taylor binary are drifting inward towards each other as well: https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary

These two white dwarfs are drifting towards each other: https://iopscience.iop.org/article/10.1088/2041-8205/757/2/L21/meta

As are these white dwarfs: https://arxiv.org/abs/1910.11389

The extrasolar planet WASP-12b has been measured to have a decaying orbit: https://arxiv.org/pdf/1812.02438.pdf

[Dave Lev (OP)]

Dear Halc
In that article it is stated:
"These jets probably flickered on and off as the supermassive black hole alternately gulped and sipped material," said Finkbeiner. It would take a tremendous influx of matter for the galactic core to fire up again. Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required."

Do you agree that the estimated mass of the molecular jet stream is 10,000 Sun mass?
However, can you please explain how the suppermassive black hole can gulped and sipped material of 10,000 Sun mass outwards from its core below its event of horizon?
Do you see any possibility for that activity?
You have already claimed that this is impossible mission.
How can you now agree for that Idea?
What kind of force is needed to sipped material outwards from the SMBH?
In any case, if matter is ejected outwards from the SMBH, than why do we even think that the accretion disc must get its mass from outside?
Why the same SMBH that can "gulped and sipped material" to the molecular jet stream, can't also gulped and sipped material directly to the accretion disc?
Actually, based on the diagram the molecular jet stream has a stable density.
Let's set a simple calculation:
the total mass in the jet stream is 10,000 sun mass.
The size of that jet stream is 27,000 LY
So, in average density of each one LY segment of that jet stream is:
10,000 / 27,000 = 0.27 sun mass per 1 LY
We have already discussed the speed of that jet stream.
You have stated that the estimated speed is 0.8 c
So, in only one year the total mass that is requested to be supplied to the jet stream is:
0.27 * 0.8 = 0.216 sun mass per year.
I wonder what is the estimated mass ejection per year from the accretion disc.
Do you agree that it should be 0.216 sun mass per year?

[Janus]

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?

Actually not all the moon's move away.  Phobos orbits close enough to Mars to be spiraling in,  There are outer moons of all the gas giants that orbit retrograde and thus spiral in.
You don't need oceans for there to be tidal drag.   The Moon also produces tidal bulges in the Earth itself, it just isn't as large as the Ocean bulges.   These bulges are capable of producing a tidal drag just like the ocean ones do.   
Having oceans increases the drag, but not having them would not eliminate it entirely. 

The degree of tidal drag is determined by a factor known as the Love number ( Named after Augustus Love),  which in turn depends on the density and rigidity of the bodies involved.   The only way you could avoid tidal drag entirely is if you only used perfectly rigid bodies.  Since no astronomical bodies are perfectly rigid, you are always going to have some degree of tidal drag.

[Dave Lev (OP)]

Quote from: Dave Lev on Today at 16:33:06
Do 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.

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.

So, would you kindly advice why they have used the size of 10,000 sun mass? Why not 1,000 or 100,000 sun mass?
As it is stated: "Finkbeiner estimates that a molecular cloud weighing about 10,000 times as much as the Sun would be required"
What is the meaning of "required"?
In the following article it is stated:
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"astronomers studying Sgr A* (the supermassive black hole at the centre of the Milky Way Galaxy) were surprised to notice that less than 1% of the gas and dust drawn into its gravitational field ever get consumed – almost everything else gets ejected."
So, if less than 1% of the falling gas is drawn into the SMBH, while 10,000 sun mass are required to fall in, than it is a clear indication that at least 9,900 sun mass is used for the molecular jet stream.
What is the estimated mass in the molecular jet stream?
Do we have any clue about the total mass ejection from the accretion disc per year?

[Dave Lev (OP)]

And, by contrast, we know that Phobos is drifting towards Mars. It's been measured. Dave seems to be ignoring that.

Would you kindly direct me to that article?
I would like to understand how our scientists have measured that Phobos is drifting towards Mars.