[Kryptid]

Sorry, just by increasing the velocity, you don't change the temp.
Technically, we could cross the space at the speed of light without increasing our temp.

You are comparing apples and oranges. Temperature is a measure of the average velocity of the particles that make up an object (the atoms in the spaceship, for example), not the average velocity of the object itself (the spaceship as a whole).

Only if we increase the friction with something, we can increase the temp.
For example, when the space shuttle returns back to Earth while crossing the atmosphere it increases dramatically its temp.

That's due to adiabatic compression, not friction.

Yes, fusion could increase the temp to a level of millions degrees. But is it good enough for Billions and trillions?
Don't you think that an extra heat power is needed?

My invoking of fusion wasn't meant to explain the trillions of degrees. It was specifically to show that there is indeed an energy source available to help explain your potential energy-kinetic energy argument.

Based on this scenario you can increase the velocity of some particles, but is it enough?

As long as they are moving faster than the escape velocity, yes.

Even if we assume an ejected velocity of four times faster (higher than the speed of light) they still must slow down dramatically and maybe get to a distance of 10 or 100 day lights.

Have you ever heard of the concept of an escape velocity? What you are saying is simply not true.

The formula for potential energy is:
EP = M G H
EP1 (at 2 day light) = M G (2 Day light)
EP2 (at 27000 Light years) = M G (27000 Light years) = MG ( 27000 * 365 Light days) = MG (9,855,000 day light)
Therefore, the potential energy had been increased by:
EP2 / EP1 = 9,855,000 / 2 = 4,927,500
In other words, we have increased the potential energy by 4,927,500.

You are misusing that equation. The equation is specifically for circumstances where the gravitational pull of the main body does not vary much (the value of "G" changes with altitude). So you can use it quite well for measuring the potential energy change of an object lifted, say, one meter above the Earth's surface. But when you are talking about astronomical distances, it's no longer sufficient. So your answer is very wrong.

[Dave Lev (OP)]

Quote from: Dave Lev on Yesterday at 16:12:34
The formula for potential energy is:
EP = M G H
EP1 (at 2 day light) = M G (2 Day light)
EP2 (at 27000 Light years) = M G (27000 Light years) = MG ( 27000 * 365 Light days) = MG (9,855,000 day light)
Therefore, the potential energy had been increased by:
EP2 / EP1 = 9,855,000 / 2 = 4,927,500
In other words, we have increased the potential energy by 4,927,500.

You are misusing that equation. The equation is specifically for circumstances where the gravitational pull of the main body does not vary much (the value of "G" changes with altitude). So you can use it quite well for measuring the potential energy change of an object lifted, say, one meter above the Earth's surface. But when you are talking about astronomical distances, it's no longer sufficient. So your answer is very wrong.

How wrong it could be?
What is the correct calculation?
Let's verify:
1. Do you agree that based on the current hypothetical theory/idea of the modern science, it is feasible for a particle to fall from 2-day light into the MW SMBH' accretion disc and then be ejected to 27,000 Ly without adding any real external energy?
2. If We lift an object/Atom/Particle from 2-day LY to 10-day light, do you confirm that its potential energy had been increased by 5?
3. If we lift it from 2-day light, to 2 Light year (2 * 365 days) do you confirm that its potential energy had been increased by 365?
4. If we lift it from 2-day light, to 200 Light year do you confirm that its potential energy had been increased by 36,500?

So please, based on your understanding, if we lift the particle from 2-day light to 27,000- light years, what is the correct increased in the potential energy?
Do you agree that it should be at least more than 10,000 times?
If so, don't you agree that this hypothetical idea by itself breaks the fundamental law of science that is called: "conservation energy"
Why the science community can beak this important science law in their hypothetical theory without getting any penalty?

[Bored chemist]

OK, let's see if we can explain why you are doing the wrong maths.

If I take a rock and throw it straight up from the surface of the earth at 20,000 meters per second, how high does it go?

[Dave Lev (OP)]

 

OK, let's see if we can explain why you are doing the wrong maths.

If I take a rock and throw it straight up from the surface of the earth at 20,000 meters per second, how high does it go?

Would you kindly answer directly my question:

if we lift the particle from 2-day light to 27,000- light years, what is the correct increased in the potential energy?

[Origin]

If we lift it from 2-day light

Light days are the distance light travels in a day, day lights are something that can get scared out of you.

If so, don't you agree that this hypothetical idea by itself breaks the fundamental law of science that is called: "conservation energy"

No.  There will be conversion of KE to PE.  In other words the particles slow down as the PE increases.

[Bored chemist]

if we lift the particle from 2-day light to 27,000- light years, what is the correct increased in the potential energy?

It's the definite integral of the product of the gravitational force and the distance moved between r= 2 light days and r= 27000 light years.
That force is a function of r which we can probably approximate as k/r^2

Where k is a constant which you have not told us, and without which we can not answer your question.

But, if you answer this

If I take a rock and throw it straight up from the surface of the earth at 20,000 meters per second, how high does it go?

you will, at least, learn what integral you are meant to be doing.

so, rather than insisting that I answer your impossible question, perhaps you could try answering my question- to which the answer is interesting and also calculable.

[Kryptid]

How wrong it could be?
What is the correct calculation?

I'll try to get to that later when I have more time. In the meantime, you can look at the proper equation here: http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

Take note how the webpage explains that U = mgh is the equation used in limited circumstances where "g" can be assumed as constant (such as when you are close to the Earth's surface). The more accurate equation is U = (-GMm)/r. It works over large distances where "g" is variable (which is definitely the case over light-years).

1. Do you agree that based on the current hypothetical theory/idea of the modern science, it is feasible for a particle to fall from 2-day light into the MW SMBH' accretion disc and then be ejected to 27,000 Ly without adding any real external energy?

Yes, because the energy from the quasar is more than sufficient.

2. If We lift an object/Atom/Particle from 2-day LY to 10-day light, do you confirm that its potential energy had been increased by 5?
3. If we lift it from 2-day light, to 2 Light year (2 * 365 days) do you confirm that its potential energy had been increased by 365?
4. If we lift it from 2-day light, to 200 Light year do you confirm that its potential energy had been increased by 36,500?

No, none of those things are correct. The relationship is not linear.

So please, based on your understanding, if we lift the particle from 2-day light to 27,000- light years, what is the correct increased in the potential energy?

I intend to calculate that.

Do you agree that it should be at least more than 10,000 times?

No.

[Origin]

I intend to calculate that.

That will be helpful for others reading the thread but of course Dave will ignore it just like he has ignored the hundreds of other corrections to his pseudoscience.

[Origin]

However, we know that at the accretion disc, the particle temp is increasing to several billions or trillions of degrees.

You do now know that is incorrect, right?  No one thinks accretion disks are trillions of degrees.  As far as quasars go the brightness temperature can be absurdly high but the actual temperature is not.
Do you fully accept this?

[Dave Lev (OP)]

Quote from: Dave Lev on Today at 06:32:04
How wrong it could be?
What is the correct calculation?

I'll try to get to that later when I have more time. In the meantime, you can look at the proper equation here: http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

Take note how the webpage explains that U = mgh is the equation used in limited circumstances where "g" can be assumed as constant (such as when you are close to the Earth's surface).

That is very clear

The more accurate equation is U = (-GMm)/r. It works over large distances where "g" is variable (which is definitely the case over light-years).

In the article it is stated:
"the force approaches zero for large distances, and it makes sense to choose the zero of gravitational potential energy at an infinite distance away. "
This could be correct as long as you ignore the existence of the dark matter.
From the sun gravity point of view, it is located near the surface of the big sphere which includes the SMBH + the dark matter.
Take out the idea of the dark matter, and you can't explain how the sun holds itself in the galaxy.
You can't just claim that the dark matter is there to hold the Sun in its orbit around the galaxy, but it isn't there when it comes to the jet stream.
Surprisingly, the sun is located exactly 27,000 Ly from the SMBH.
Therefore, as you claim for dark matter in our galaxy that is needed to hold the sun in its orbital motion - then the same dark matter must work identically on every object at the same distance.
Hence, the 27,000 LY dark matter spheres must work identically on the sun as it works at hydrogen atom that is located at the edge of the jet stream.
So, now it should be very simple to extract the potential energy at 27,000 LY.
Base on the orbital motion of the Sun, we can extract the effective dark matter mass at that 27,000 LY spheres.
Let's call it = M dark 27000 Ly = the effective dark matter mass at 27000 LY.
With this effective mass we can easily extract the potential energy for the hydrogen atom at 27000 LY.
Do you agree with this explanation?

Quote from: Dave Lev on Today at 06:32:04
So please, based on your understanding, if we lift the particle from 2-day light to 27,000- light years, what is the correct increased in the potential energy?

I intend to calculate that.

Thanks
Please don't forget to add the dark matter impact.

[Kryptid]

The mass of the black hole is 886 million times that of the Sun. The Sun's mass is 1.9885 x 10²⁶ kilograms. That makes the black hole's mass 1.722 x 10³⁵ kilograms. This means that the gravitational potential energy of a proton at two light-days from the black hole would be:

U = (-GMm)/r
U = (-(-6.674 x 10^-11)(1.722 x 10³⁵)(1.673 x 10^-27))/(5.18 x 10¹³)
U = (-(1.149 x 10²⁵)(1.673 x 10^-27))/(5.18 x 10¹³)
U = (-(1.9227 x 10^-2)/(5.18 x 10¹³)
U = -3.71 x 10^-16 joules

At 27,000 light-years, the gravitational potential energy is:

U = (-GMm)/r
U = (-(-6.674 x 10^-11)(1.722 x 10³⁵)(1.673 x 10^-27))/(2.554 x 10²⁰)
U = (-(1.149 x 10²⁵)(1.673 x 10^-27))/(2.554 x 10²⁰)
U = (-(1.9227 x 10^-2)/(2.554 x 10²⁰)
U = -7.52 x 10^-23 joules

What this means is that a proton falling from 27,000 light-years down to 2 light-days would gain (-3.71 x 10^-16) - (-7.52 x 10^-23) = -3.09999248 x 10^-16 joules of kinetic energy. In turn, this means it would take 3.09999248 x 10^-16 joules to lift that proton from 2 light-days to 27,000 light-years.

Now how much kinetic energy does a proton traveling at 99% the speed of light have? Using this calculator: https://www.omnicalculator.com/physics/relativistic-ke The answer is approximately 9.087 x 10^-10 joules. That is over 2.9 million times the energy needed to lift the proton from 2 light-days to 27,000 light-years. This would reduce the proton's kinetic energy to 9.0869969 x 10^-10 joules. The decreases the speed of the proton by such a tiny amount that the calculator can't even tell the difference between 99% the speed of light and that new speed. So your claim that the jet has to slow down significantly is just plain wrong.

Do you agree with this explanation?

You would need to know how much dark matter is present and in what distribution in order to account for it.

[Dave Lev (OP)]

The mass of the black hole is 886 million times that of the Sun. The Sun's mass is 1.9885 x 1026 kilograms. That makes the black hole's mass 1.722 x 1035 kilograms. This means that the gravitational potential energy of a proton at two light-days from the black hole would be:

U = (-GMm)/r
U = (-(-6.674 x 10-11)(1.722 x 1035)(1.673 x 10-27))/(5.18 x 1013)
U = (-(1.149 x 1025)(1.673 x 10-27))/(5.18 x 1013)
U = (-(1.9227 x 10-2)/(5.18 x 1013)
U = -3.71 x 10-16 joules

At 27,000 light-years, the gravitational potential energy is:
U = (-GMm)/r
U = (-(-6.674 x 10-11)(1.722 x 1035)(1.673 x 10-27))/(2.554 x 1020)
U = (-(1.149 x 1025)(1.673 x 10-27))/(2.554 x 1020)
U = (-(1.9227 x 10-2)/(2.554 x 1020)
U = -7.52 x 10-23 joules

Thanks for the calculations.
Do appreciate your efforts.
However, Why do you ignore the impact of the dark matter at 27,000 LY ?
If a proton is located near the Sun, do you think that the gravitational potential energy of  -7.52 x 10^-23 joules would be good enough to hold it in its orbital motion around the galaxy?
Why the dark matter works on the Sun motion, but it has no impact on the jet stream?

In any case, once we ignore the dark matter, then I fully agree with your calculation

What this means is that a proton falling from 27,000 light-years down to 2 light-days would gain (-3.71 x 10^-16) - (-7.52 x 10^-23) = -3.09999248 x 10^-16 joules of kinetic energy. In turn, this means it would take 3.09999248 x 10-16 joules to lift that proton from 2 light-days to 27,000 light-years.

I also agree with the following understanding:

Now how much kinetic energy does a proton traveling at 99% the speed of light have? Using this calculator: https://www.omnicalculator.com/physics/relativistic-ke The answer is approximately 9.087 x 10-10 joules.

However, there is a key problem.
The Gravitational potential energy of a proton at two light-days from the black hole is: U = -3.71 x 10^-16 joules
So, how this starting Gravitational potential energy of a proton at 2 day light  3.71 x 10^-16 joules, could be transformed into a kinetic energy of 9.087 x 10^-10 joules at the accretion disc.

Therefore, the starting Gravitational potential energy of a proton had been increased by 2.45 Million times at the accretion disc:
= 9.087 x 10^-10 / 3.71 x 10^-16 = 2.45 * 10^6

Is it realistic?
Don't you see that external real energy is needed?
What is the source of this extra energy?

[Kryptid]

Thanks for the calculations.
Do appreciate your efforts.
However, Why do you ignore the impact of the dark matter at 27,000 LY ?
If a proton is located near the Sun, do you think that the gravitational potential energy of  -7.52 x 10^-23 joules would be good enough to hold it in its orbital motion around the galaxy?
Why the dark matter works on the Sun motion, but it has no impact on the jet stream?
In any case, once we ignore the dark matter, then I fully agree with your calculation

I didn't include dark matter in that calculation because I don't know how much there is or how it is distributed.

So, how this starting Gravitational potential energy of a proton at 2 day light  3.71 x 10^-16 joules, could be transformed into a kinetic energy of 9.087 x 10^-10 joules at the accretion disc.

I've already addressed that: redistribution of energy. That particle isn't by itself. It's interacting with other particles (and the black hole's spin) while it's in the accretion disk. Some of those particles will gain a lot of energy from the black hole's spin as well as from the heat in the accretion disk. Other particles lose energy in the process and enter the black hole. So the total energy of the system is the same.

[Bored chemist]

Don't you see that external real energy is needed?
What is the source of this extra energy?

Yes, we see it.
And we already explained it. It comes from other particles
But you don't seem able to understand that.

I've already addressed that: redistribution of energy. That particle isn't by itself. It's interacting with other particles (and the black hole's spin) while it's in the accretion disk.

But if you drop a whole lot of particles in there, the energy gets shared among them as heat.
And that means there's a distribution of velocities.
A small fraction of those velocities will be much higher than average.
Friction can actually increase the speed of a particle.

Effectively, you are saying water in a puddle can not evaporate because it is not at the boiling point.

[Bored chemist]

But, if you answer this

Quote from: Bored chemist on Yesterday at 08:13:53
If I take a rock and throw it straight up from the surface of the earth at 20,000 meters per second, how high does it go?
you will, at least, learn what integral you are meant to be doing.

so, rather than insisting that I answer your impossible question, perhaps you could try answering my question- to which the answer is interesting and also calculable.

[Dave Lev (OP)]

Quote from: Dave Lev on Today at 05:58:05
So, how this starting Gravitational potential energy of a proton at 2 day light  3.71 x 10^-16 joules, could be transformed into a kinetic energy of 9.087 x 10^-10 joules at the accretion disc.

I've already addressed that: redistribution of energy. That particle isn't by itself. It's interacting with other particles (and the black hole's spin) while it's in the accretion disk.

How an interaction with other particles could increase the energy of a single particle by 2.45 Million times?
Technically, 2.45 million particles must lose completely their energy in order to get just one particle with 9.087 x 10^-10 joules.
Hence, in order to get just one particle at the correct energy at the accretion disc, 2.45 million particles should lose their entire energies.
It means that for any particle in the accretion disc, 2.45 million of falling particles lose their life.
Please also be aware that all particles in the accretion disc without exception orbits at almost the speed of light.
Therefore, it is expected that any falling particle would gain that velocity even before it enters the accretion disc.
Is it feasible that out of 2.45 million falling particles only one will get enough energy to enter the accretion disc?

However, your following answer is much more realistic scenario:

and the black hole's spin

If I understand you correctly, some of the SMBH's spin & heat energies are transformed to those falling particles.
The transformation of heat energy from the SMBH to the falling particles is acceptable.
However, how the SMBH's spin energy could be transformed into those falling particles?
Actually, in the nature, if the object spins then technically it could create dynamo.
If it has a dynamo then is can generate magnetic fields.
This magnetic fields could easily increase dramatically the velocity of the falling particles and also their heat.
So, this could be an excellent solution for the missing energy of the falling particles.
The SMBH could easily increase the particle energy by to 9.087 x 10^-10 joules.
However, the only question is - why the SMBH spins.

I didn't include dark matter in that calculation because I don't know how much there is or how it is distributed.

Well, I fully agree with you that we should totally ignore the dak matter idea.
You have proved in your calculation that the proton energy at the accretion disc is good enough for each particle to get at the 27000 Ly without any difficulties while it keeps its speed of light velocity.
However, if there is dark matter, then it could decrease dramatically the motion of this jet stream.
We don't observe any decrease in the stream motion.
Therefore, the jet stream is an ultimate observation that there is no dark matter.

Never the less, it isn't feasible that there is dark matter for the sun and there is no dark matter for the jet stream.
If we claim that there is a dark matter for the sun, then we have to accept the idea that there must be also dark matter for the jet stream.
We can't just hold the stick from both sides at the same time.
Please take a decision if there is a dark matter or not.

Assuming that there is a dark matter:

I don't know how much there is or how it is distributed.

I can help to overcome this issue.
As I have already explained, the sun is also located at 27,000 LY from the SMBH.
By simple calculation based on its orbital motion we can extract the effective mass of the dark matter.

Base on the orbital motion of the Sun, we can extract the effective dark matter mass at that 27,000 LY spheres.
Let's call it = M dark 27000 Ly = the effective dark matter mass at 27000 LY.
.

With this effective mass we can easily extract the potential energy for the proton at 27000 LY.
Then we can verify if the kinetic energy of 9.087 x 10^-10 joules at the accretion disc is good enough for that potential energy.

[Kryptid]

How an interaction with other particles could increase the energy of a single particle by 2.45 Million times?
Technically, 2.45 million particles must lose completely their energy in order to get just one particle with 9.087 x 10^-10 joules.
Hence, in order to get just one particle at the correct energy at the accretion disc, 2.45 million particles should lose their entire energies.
It means that for any particle in the accretion disc, 2.45 million of falling particles lose their life.
Please also be aware that all particles in the accretion disc without exception orbits at almost the speed of light.
Therefore, it is expected that any falling particle would gain that velocity even before it enters the accretion disc.
Is it feasible that out of 2.45 million falling particles only one will get enough energy to enter the accretion disc?

Not all of the energy need come directly from the particles interacting with each other. I don't know what proportion of it comes from that versus being accelerated by the black hole's spin.

If I understand you correctly, some of the SMBH's spin & heat energies are transformed to those falling particles.

It's transferred to the falling particles.

However, how the SMBH's spin energy could be transformed into those falling particles?

It's something called the ergosphere. Anything falling into it must move with the black hole's rotation: https://en.wikipedia.org/wiki/Ergosphere

This magnetic fields could easily increase dramatically the velocity of the falling particles and also their heat.

No. I've already pointed out that magnetic fields do not increase the speed of charged particles. They only change their direction.

However, the only question is - why the SMBH spins.

It's acquiring angular momentum from the accretion disk. The mass of much of the accretion disk is falling into the black hole, never to be seen again (except in the form of extremely weak Hawking radiation). However, the rotational kinetic energy of the consumed matter is still accessible through the ergosphere. That energy can then be used to boost some of the matter from the accretion disk that hasn't fallen in yet into the jets.

As far as dark matter goes, we can make some assumptions to get some numbers. Current modelling puts the Universe as being composed of 5% matter and 26.8% dark matter (by mass). That would make dark matter 5.36 times more abundant than matter. So we can add the black hole's mass to 5.36 times its mass (6.36) and get an approximation for how much gravity the proton is having to move against. Since gravitational potential energy increases linearly with mass, then we just multiple the numbers calculated before by 6.36. In that case, we get a potential energy at 2 light-days of -2.35956 x 10^-15 joules and a potential energy of -4.78272 x 10^-22 joules at 27,000 light-years. That's an energy change of 2.359559521728 x 10^-15 joules by moving out to 27,000 light-years.

That would decrease the kinetic energy of the proton to 9.0869764044 x 10^-10 joules. The resulting velocity is still above 98.9% the speed of light. So the speed barely changes at all and thus the jet would not be expected to slow down significantly even if with dark matter present.

[Dave Lev (OP)]

Quote from: Dave Lev on Yesterday at 16:12:41
However, how the SMBH's spin energy could be transformed into those falling particles?

It's something called the ergosphere. Anything falling into it must move with the black hole's rotation: https://en.wikipedia.org/wiki/Ergosphere

Thanks for this important article:
It is stated:
https://en.wikipedia.org/wiki/Ergosphere
"As a black hole rotates, it twists spacetime in the direction of the rotation at a speed that decreases with distance from the event horizon.[3] This process is known as the Lense?Thirring effect or frame-dragging."
What is the meaning of "Lense?Thirring effect"?
https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession
"In general relativity, Lense?Thirring precession is a relativistic correction to the precession of a gyroscope near a large rotating mass such as the Earth. It is a gravitomagnetic frame-dragging effect.
What is gravitomagnetic frame-dragging effect?
https://en.wikipedia.org/wiki/Gravitoelectromagnetism
Gravitoelectromagnetism, abbreviated GEM, refers to a set of formal analogies between the equations for electromagnetism and relativistic gravitation;
In the following explanation we get a direct answer to the formation and structure of the Jet stream:
"Indirect validations of gravitomagnetic effects have been derived from analyses of relativistic jets. Roger Penrose had proposed a mechanism that relies on frame-dragging-related effects for extracting energy and momentum from rotating black holes.[3] Reva Kay Williams, University of Florida, developed a rigorous proof that validated Penrose's mechanism.[4] Her model showed how the Lense?Thirring effect could account for the observed high energies and luminosities of quasars and active galactic nuclei; the collimated jets about their polar axis; and the asymmetrical jets (relative to the orbital plane).[5][6] All of those observed properties could be explained in terms of gravitomagnetic effects.[7] Williams' application of Penrose's mechanism can be applied to black holes of any size.[8] Relativistic jets can serve as the largest and brightest form of validations for gravitomagnetism.
So, the gravitomagnetism represents analogies between electromagnetism and relativistic gravitation.
Based on this understanding we have a mechanism that relies on frame-dragging-related that effects for extracting energy and momentum from rotating black holes.
In other words, based on this concept of the gravitomagnetism the rotating BH energy is used for the Jet stream energy.
Hence, the gravitomagnetism energy of the rotation SMBH is good enough for the jet stream creation including its collimated structure.
We don't need more than that.
The SMBH rotation energy can easily increase the energy of any falling particle by its gravitomagnetism energy. There is no need to use 2.45 Million falling particles for to gain the requested energy for just one particle.
With This SMBH' gravitomagnetism energy any falling particle would get the energy that is needed for it to move at almost the speed of light in the accretion disc.

Dark matter

As far as dark matter goes, we can make some assumptions to get some numbers. Current modelling puts the Universe as being composed of 5% matter and 26.8% dark matter (by mass). That would make dark matter 5.36 times more abundant than matter. So we can add the black hole's mass to 5.36 times its mass (6.36) and get an approximation for how much gravity the proton is having to move against. Since gravitational potential energy increases linearly with mass, then we just multiple the numbers calculated before by 6.36. In that case, we get a potential energy at 2 light-days of -2.35956 x 10-15 joules and a potential energy of -4.78272 x 10-22 joules at 27,000 light-years. That's an energy change of 2.359559521728 x 10-15 joules by moving out to 27,000 light-years.

Sorry, your calculation isn't fully correct.
The mass inside of the Sun's orbit is Mgalaxy = 10^11 MSun
https://sites.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect24/lecture24.html
Dark Matter
"We can estimate the mass of the Milky Way by using Kepler's laws of motion.
Kepler's laws are valid anytime two massive objects move around a common center of mass.
In a system like a galaxy, the Sun only feels the gravitational attraction from the parts of the galaxy which are closer to the center than the Sun is.
The gravitational attraction between the Sun and the rest of the galaxy is as though the inner part of the galaxy were compressed to a point at the center of the galaxy.
Kepler's law gives us the sum of the mass of the galaxy and the Sun once we know the orbital period and distance
Mgalaxy + MSun = (4 &pi2/G)R3/P2
But the mass of a star is tiny compared to the mass of the galaxy, so we can drop the mass of the Sun.
Kepler's law for motion in the galaxy:
Putting in the numbers for the Sun's motion, we find that the mass inside of the Sun's orbit is Mgalaxy = 10^11 MSun"
Hence:
Mgalaxy at 27,000 Ly = 10^11 MSun

Therefore, by reusing your calculation:

The mass of the black hole is 886 million times that of the Sun. The Sun's mass is 1.9885 x 1026 kilograms. That makes the black hole's mass 1.722 x 1035 kilograms. This means that the gravitational potential energy of a proton at two light-days from the black hole would be:

At 27,000 light-years, the gravitational potential energy is:

U = (-GMm)/r
U = (-(-6.674 x 10-11)(1.722 x 1035)(1.673 x 10-27))/(2.554 x 1020)
U = (-(1.149 x 1025)(1.673 x 10-27))/(2.554 x 1020)
U = (-(1.9227 x 10-2)/(2.554 x 1020)
U = -7.52 x 10-23 joules
What this means is that a proton falling from 27,000 light-years down to 2 light-days would gain (-3.71 x 10-16) - (-7.52 x 10-23) = -3.09999248 x 10-16 joules of kinetic energy. In turn, this means it would take 3.09999248 x 10-16 joules to lift that proton from 2 light-days to 27,000 light-years.

By using Mgalaxy at 27,000 Ly = 10^11 MSun instead of M sun we get:
U (with dark matter at 27000 LY) = -7.52 x 10^-23  * 10^11 joules = -7.52 x 10^-12 joules.

With regards to the kinetic energy of a proton:

Now how much kinetic energy does a proton traveling at 99% the speed of light have? Using this calculator: https://www.omnicalculator.com/physics/relativistic-ke The answer is approximately 9.087 x 10^-10 joules. That is over 2.9 million times the energy needed to lift the proton from 2 light-days to 27,000 light-years.

Please remember that in the Milky Way accretion disc, the particles move at 0.3c
KE = 0.5 * m * v^2
As v = 0.3c
KE (for v= 0.3c) = KE (for v=c) * 0.3^2

Therefore, the kinetic energy of a proton at the MW' accretion disc is:
9.087 x 10^-10 joules * 0.3^2 = 8.17 x 10^-11 joules.
The ratio between the proton energy at the accretion disc to the potential energy at 27,000 Ly is as follow:
With dark matter it is: 8.17 x 10^-11 joules / 7.52 x 10^-12 joules = 10.8
Without dark matter it was over than 2.9 million.
If we add to our calculation the energy lose as the magnetic fields bands the motion of the proton in the direction of the SMBH magnetic poles, as the proton tries to escape from the SMBH gravitational force and other energy dissipation as friction, compression, turbulence, we may find that this 10.8 is clearly not good enough.
Please remember, the estimated mass of the SMBH is just 4 million solar mass while the dark matter Mgalaxy at 27,000 Ly is 10^11 (100 Billions) solar mass.
Therefore, there is high possibility that based on the dark matter concept, the Jet stream woun't get to that 27,000 LY without decreasing significantly its starting velocity as it cross more and more dark mass.

[Kryptid]

You are aware that this quasar isn't in the Milky Way galaxy, aren't you?

[Origin]

You are aware that this quasar isn't in the Milky Way galaxy, aren't you?

Who knows what he thinks he knows.  He doesn't even seem to know that the jet stream is a weather phenomenon found on Earth.
I don't think he even actually knows what a quasar is!