Muzzle velocity is still finite, so leading the target would still be required.
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My job is to prove that the SMBH's gravity force contributes more energy than the energy that is needed to create those new particle pairs.
However, they also get kinetic energy from the SMBH's gravity force.
Therefore, we need to prove that:
Er is greater than Eout
For this simple calculation lets ignore the impact of the Ep
Er = Ek = 1/2 M c^2
Eout = 2M
Therefore, I have proved that only the Kinetic energy of the in falling antiparticle contributes much more energy that was needed to set the whole creation activity.
In this case, you have set all the objects at the same radius of 149,602,691,137.
I can promise you that if in your example you have used the same radius for all objects - you would probably get fully balanced forces.
Therefore, please use the minimal and maximal distance/radius from the Earth Moon to the Sun (maximal tidal) and then set your recalculation.
Mathematical fiction is a direct outcome if we use the error/delta in the Taylor series to prove something which is totally based on that error/delta.
But still this equation gives the same number whether the substance is a 150,000g/mole or 150g/mole which is not sensible.
As seen, visual perceiving does not need an exponential function of parameters.
The subject of Radar: We never see any think simultaneously because of the finite/limited value of light’s velocity theoretically. However we have not any problem for nearby objects. So, negligible effects are mentioned. You are right radar system never gives exact results; but it is functional for our local distances.
Your additional proofs are not my interest area.
I study astronomy, cosmology, light kinematics.
Therefore the root postulate of SR is not consistent.
You already tried that argument and it didn't work. The equation you offered predicted a linear relationship between velocity and red shift, whereas relativity predicts an exponential one. Relativity's prediction is the one with experimental support. You tried to save the equation by claiming that you have to use "universal velocity" instead of relative velocity, but that makes the problem worse. If we did live in a world with an absolute reference frame and red shift was dependent upon absolute velocity, then the radar guns used by police officers would have to be continually calibrated throughout the day and year to reflect that change in velocity over time due to the Earth's rotation and orbit.
If what you are claiming instead is that we have to measure the difference in the universal velocity of the car and the universal velocity of the radar gun, then that simplifies right back to relative velocity again. It doesn't matter if the Milky Way galaxy was sitting still or moving at 50% the speed of light in some absolute frame: the red shift detected by the radar gun would be the exact same because it's measuring relative velocity. That was known long before special relativity was even conceived of.
Time dilation is also far from the only observable prediction of relativity.
An illusion caused by the finite speed of light cannot explain the observation of quadrupolar gravitational waves.
An illusion caused by the finite speed of light cannot explain the observation of mass-energy conversion.
An illusion caused by the finite speed of light cannot explain the precession of mercury's orbit.
An illusion caused by the finite speed of light cannot explain the decay rate of neutron star's orbits.
Make the gravity forces calculations for the two scenarios. I promise you that you would be surprised to find that the results are almost identical to the systems that you have offered!!!
Taylor series is an excellent tool to get close to the solution.
Let's assume that this Taylor series represents the reality by 99%.
That by itself is an excellent estimation. However, there is still a delta or error of 1%.
In that formula, our scientists actually are focusing on that delta.
So, they take the delta and set a new formula for gravity which shows the connection between the tidal to the delta.
This was their severe mistake.
The SMBH's mighty gravity force accelerates those new pair creation into the speed of light.
So, that acceleration comes for free due to the SMBH's mighty gravity force.
That ultra high orbital velocity of the particles/plasma in the accretion disc is transformed back to the SMBH's and increases its internal rotation.
The ultra high velocity of the new created particles in the accretion disc - is transformed back into the SMBH and sets the compensation in its rotation velocity.
In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.
By inventing methods such as the scientific method.
It is some sort of mathematical fiction.
We take something - gravity force, convert it to infinite something by Taylor series and than use finit something just in order to prove other something - Tidal negative impact on total rotation energy.
It's not an issue that I don't like it.
It is just a totally incorrect formula.
We shouldn't use it - Never and ever.
So, it was a mistake to set two diffrent calculations - One for the object at 11,000,000 meters and other one for the object at 9,000,000 meters.
You had to find the center of mass of those two objects and than find the gravity force to the main body.
That shows you that the true is more important for me than to prove a key section in my theory.
So, how can we overcome that issue?
So, we can claim that some of the decreasing orbital energy is transformed back to the SMBH and maintain its orbital momentum.
My answer for that will be - "Gravity force".
However, some of the energy for that creation is taking also from the gravity force.
So, the SMBH's gravity force contributes some new energy to our Universe.
In any case, new energy must be created by the SMBH – For sure
That SMBH's extra new energy is used to create the mass in the accretion disc and then the SMBH is using some of this new mass to form stars planets and moons.
Philosophy could help prevent the requirement of brute force attempts to get results.
Rotation momentum, if transferred to the satellite, increases its total orbital angular momentum, and thus energy.
This is the case with Io, and our moon: the orbital radius is increasing as the rotational momentum of Jupiter is transferred to the orbital momentum of Io, an angular acceleration effect, not a braking effect.
1- In universe, everything has a motion. The world has also its motion according to Sun, galaxy, local cluster, ........etc.
We use the earth as a co-reference frame for physical and other events. However , If we want to analyze the motion relation of Earth and a celestial object (e.g. Fornax cluster) we must use an outer/common reference frame how includes both of them. And -as a first step- we have to consider the centre of local cluster as the equivalent partner of Fornax.
2- If claimed deformations are visual; that space-time is also an illusion
SR does not impose/effect the life.
Even, it cannot transform the astronomical parameters to respected/ useful / simultaneous values.
As I have already explained this formula is INCORRECT.
Gravity force should only be represented by the following formula:
but it has to be taken from the rotation momentum and not from the total orbital energy.
So, Tidal energy works on the Rotation momentum/energy and not on Total gravity energy!!!
So, I agree with your idea that the tidal energy should be taken from something
Do you agree that as there are changes in the gravity force between perigee to apogee, with or without tidal bulge the gravity should be go down over time?
a search for truth by thinking
but, a similar deformation never be happened in our life because of universal motion of the world.
As I say; if the theory SR is consist of this, it will become fairy tale for adults.
I agree, there are visual deformations. The reason of them is the finite/limited value of light's velocity.
Probably it may be also similar to muon experiment.
They had neglected to consider the speed of reference muons.
I'd be interested in some numbers from any relativity guru on this forum: what duration of travel might produce a reasonably measurable time difference between two clocks?
I was not aware about that difference in size between the front bulge to the rear bulge.
However - Do you agree that if they were exactly at the same size - the center of mass had to be located exactly at the center of the cycle as explained by Newton's shell theorem?
If so, let's look at the Earth.
It's front bulge is bigger by 5% from the rear one.
That for sure will set the center of mass closer to the moon (with regards to spherically symmetric bulge)
That will increase the gravity force between the Moon -Earth comparing to spherically symmetric bulge.
So, now we have stronger gravity force.
However, as there is no change in this ratio (assuming that it is constantly at 5%), there is also no change in the location of center of mass and therefore - there is no change in gravity force.
Hence, as long as the 5% stay, there is no change in the center of mass and no change is the gravity force.
Without a change in those two segments, there will be no change in the total rotation energy.
Therefore, the bulge by itself doesn't change the center of mass point, the gravity force or the total rotation energy. Only a change in the bulge ratio can change those values.
However, they can go up or down.
If the ratio be get to 4% the center of mass will be shifted backwards and therefore les gravity force and less total rotation energy.
If the ratio will get to 6%, the center of mass will be shifted inwards and therefore more gravity force and more total rotation energy.
Therefore - the bulge does not consume energy from the total rotation energy - it just change the location of center of mass and therefore, it sets the amplitude of that gravity force or the total energy based on the bulge rear/front ratio.
Because it does. Do the math. It is trivial. Compute the force on a 2m mass all at one point, and then a pair of 1m masses that are connected but separated by some distance, but have the same center of mass. The combined force on the two 1m masses will be larger than that of the 2m mass. All you need is F=GMm/r˛.
In order to get closer to the reality we must take this Taylor's series to the infinity.
If we don't take it to infinity, than by definition there is a room for error.
Any none perfectly assumption or set up could also cause some minor error.
Newton has used different approach.
I prefer to use Newton solution.
The conventional treatment of the tidal field of the Moon is an approximation that uses calculus differentials, As the distance of a moon from its parent decreases progressively, the assumptions of this first-order approximation become ever more unacceptable. It is then necessary to apply Newton's universal law of gravity to virtually all components of the body.
If the splited spherically symmetric cycle can represent the symmetric bulge