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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.
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.
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?
Based on this scenario you can increase the velocity of some particles, but is it enough?
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.
The formula for potential energy is:EP = M G HEP1 (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 = 4,927,500In other words, we have increased the potential energy by 4,927,500.
QuoteQuote from: Dave Lev on Yesterday at 16:12:34The formula for potential energy is:EP = M G HEP1 (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 = 4,927,500In 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.
Quote from: Dave Lev on Yesterday at 16:12:34The formula for potential energy is:EP = M G HEP1 (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 = 4,927,500In other words, we have increased the potential energy by 4,927,500.
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?
if we lift the particle from 2-day light to 27,000- light years, what is the correct increased in the potential energy?
If we lift it from 2-day light
If so, don't you agree that this hypothetical idea by itself breaks the fundamental law of science that is called: "conservation energy"
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?
How wrong it could be?What is the correct calculation?
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?
I intend to calculate that.
However, we know that at the accretion disc, the particle temp is increasing to several billions or trillions of degrees.
QuoteQuote from: Dave Lev on Today at 06:32:04How 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.htmlTake 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).
Quote from: Dave Lev on Today at 06:32:04How wrong it could be?What is the correct calculation?
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).
QuoteQuote from: Dave Lev on Today at 06:32:04So 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.
Quote from: Dave Lev on Today at 06:32:04So 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 with this explanation?
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)/rU = (-(-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 joulesAt 27,000 light-years, the gravitational potential energy is:U = (-GMm)/rU = (-(-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.
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
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.
Don't you see that external real energy is needed?What is the source of this extra energy?
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.
But, if you answer thisQuote from: Bored chemist on Yesterday at 08:13:53If 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.
QuoteQuote from: Dave Lev on Today at 05:58:05So, 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.
Quote from: Dave Lev on Today at 05:58:05So, 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.
and the black hole's spin
I didn't include dark matter in that calculation because I don't know how much there is or how it is distributed.
I don't know how much there is or how it is distributed.
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..
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?
If I understand you correctly, some of the SMBH's spin & heat energies are transformed to those falling particles.
However, how the SMBH's spin energy could be transformed into those falling particles?
This magnetic fields could easily increase dramatically the velocity of the falling particles and also their heat.
However, the only question is - why the SMBH spins.
QuoteQuote from: Dave Lev on Yesterday at 16:12:41However, 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
Quote from: Dave Lev on Yesterday at 16:12:41However, how the SMBH's spin energy could be transformed into those falling particles?
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.
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)/rU = (-(-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 joulesWhat 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.
You are aware that this quasar isn't in the Milky Way galaxy, aren't you?