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QuoteIn 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.What do you mean by "take it to infinity"? How can you take anything to infinity?
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.
When did Newton show anything that was counter to what is said on this page? The method mentioned on the page used Newtonian math to arrive at the result:
That's just it, though. The bulges aren't symmetrical. Using the equations on that page, you can find that the difference between nadir and zenith (the far bulge and the near bulge, respectively) is around 5% for the Earth, about 1.4% for the Moon and 1.3% for Io.
Let me offer the following:You agree that it works perfectly in a spherically symmetric cycle.
1. If we divide the spherically symmetric cycle into two halves -One in the front and one in the back - Do you agree that the center of mass should stay at the same point?
3. If so, let's use this spited spherically symmetric cycle to represent the symmetric bulge (one in front and one in the back)So, one spherically symmetric cycle will represent the moon minus the mass of the bulge, while the bulge will be represented by splited spherically symmetric cycle - one in the front and one in the back.Why the two spherically symmetric cycles can't fully represents the moon + the bulge with the center of mass at the same point as before?
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?
Quote from: HalcA pair of bulges like that indeed has no effect on the center of mass of the moon. Nobody claimed otherwise.So, if you agree that the Bulge do not change the total mass of the moon and it also do not change the center of mass of the moon, than how can you claim that there is a change in the total gravity force between the planet and the moon?
A pair of bulges like that indeed has no effect on the center of mass of the moon. Nobody claimed otherwise.
OkI 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/r².
an = -(2RGm)/(r3)(1-(3R/2r)
F = (GMm)/r²
The difference in size of the bulges actually does change over time because of the eccentricity of the orbit. The difference will be larger at perigee and smaller at apogee.
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?
Quote from: Dave Lev 3but it has to be taken from the rotation momentum and not from the total orbital energy.It can be taken from rotation as well, but it can also definitely be taken from orbital energy. I already explained in a prior post how a braking effect from Jupiter lowers the total orbital energy of Io.
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.
QuoteAs I have already explained this formula is INCORRECT.Your "explanation" is what was incorrect.
But even if you don't like the formula I provided,
I posted such an example calculation at the end of post #862. That increase in force
As a matter of fact, I'll do exactly that. For the case of a 100 kilogram sphere orbiting 10,000,000 meters above Earth's surface:F = (GMm)/r²F = ((6.674 x 10−11(5.97237 x 1024)(100))(10,000,000)2F = (3.985959738 x 1016)/1014F = 398.6 newtonsNow I'll consider a scenario where there are two spheres, each with a mass of 50 kilograms. One is orbiting at 11,000,000 meters and the other is orbiting at 9,000,000 meters:F = (GMm)/r²F = ((6.674 x 10−11)(5.97237 x 1024)(50))(11,000,000)2F = (1.992979869 x 1016)/(1.21 x 1014)F = 164.7 newtonsF = (GMm)/r²F = ((6.674 x 10−11)(5.97237 x 1024)(50))(9,000,000)2F = (1.992979869 x 1016)/(8.1 x 1013)F = 246 newtonsAdd those two forces together and you get a total force of 410.7 newtons, which is larger than the 398;6 newtons of the single sphere scenario. So asymmetrical tidal bulges aren't even necessary to get an increase in total force.
Gravity is all about mass and center of mass.
quote]but it has to be taken from the rotation momentum and not from the total orbital energy./quote]It can be taken from rotation as well, but it can also definitely be taken from orbital energy.
QuoteSo, I agree with your idea that the tidal energy should be taken from somethingThen we can end this thread here and now, since that was the point all along. You can't make an infinite energy factory using tidal force
the rotational momentum of Jupiter is transferred to the orbital momentum of Io
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.
Then let's start off by considering each sphere in isolation. Now we have two separate objects that are not connected to each other. The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 246 newtons.The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 164.7 newtons. The sum total of the forces acting on both spheres is therefore 410.7 newtons.Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters). If your reasoning is correct, this now-singular object is suddenly and magically feeling 12.1 newtons of force less than it did when the objects were separate (because you claim that these two tethered spheres should feel the same force as a single sphere of identical mass and center of mass). How can the mere addition of a string cause the Earth to pull on the spheres less than it did before?The answer is that it can't: strings don't have magical antigravity properties. We can therefore safely conclude that your reasoning is incorrect.
QuoteWe take something - gravity force, convert it to infinite something by Taylor series and then use finite something just in order to prove other something - Tidal negative impact on total rotation energy.Then you must think calculus itself is wrong, since it also involves infinite sums.
We take something - gravity force, convert it to infinite something by Taylor series and then use finite something just in order to prove other something - Tidal negative impact on total rotation energy.
QuoteHowever, some of the energy for that creation is taking also from the gravity force.No, it isn't. Gravity does not create net energy.
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.
QuoteIn other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.No it doesn't. Conservation of energy won't let you. Give it up already. What part of "energy cannot be created or destroyed" do you not understand?
You realize that would make me right, not wrong, don't you?Attraction between the Earth and Sun:F = (GMm)/r2F = ((6.674 x 10−11(5.97237 x 1024)(1.9885 x 1030))(149,598,023,000)2)F = (7.926080939013 x 1044)/(2.2379568485508529 x 1022)F = 3.5418308070902173002872574484922 x 1022 newtonsAttraction between the Moon and Sun:F = (GMm)/r2F = ((6.674 x 10−11(7.342 x 1022)(1.9885 x 1030))(149,982,422,000)2)F = (9.7437510158 x 1042)/(2.2494726908986084 x 1022F = 4.3315711523075275820924842308266 x 1020Sum of the two forces: 3.5851465186132925761081822908005 x 1022 newtonsAttraction between the Sun and a hypothetical Earth-Moon combination object:F = (GMm)/r2F = ((6.674 x 10−11(6.04579 x 1024)(1.9885 x 1030))(149,602,691,137)2)F = (8.023518449171 x 1044)/(2.2380965195362677778850495607477 x 1022)F = 3.5849742757445814209639045480079 x 1022 newtons.
The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms.
The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms.
Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters).
QuoteTaylor series is an excellent tool to get close to the solution.Then what was that nonsense you were saying when you called it "mathematical fiction"?
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.
Gravity doesn't contribute any energy. All it can do is transfer or transform energy that already exists.Every single bit of that kinetic energy will come from that black hole's rotation, reducing the black hole's rotation (and therefore its total energy) by the same amount.
Do you mean that if the black hole's rotation is zero, than its gravity force should also be zero?
Do you really mean the gravity force is only some sort of energy transformation?
If so, why Newton didn't call it Gravity Field instead of Gravity Force?
You have to agree that once an object have lost its rotation energy, than it also should lose its gravity force.How could it be?
It is a severe contradiction to the Newton Gravity Law.Newton didn't set any connection between the gravity force to the rotation energy.If there is a formula that connects the rotation energy to gravity force - Would you kindly offer it?
We all know that a function of energy is force times distance.Therefore, Gravity force by itself can generate energy.
That energy is a direct outcome of the force.
Therefore, the force is not a transformation.
It generates energy based on Newton gravity force/Law.
Can we claim that the moon has no ability to converts its gravity force into energy just because it does not rotate any more?
If an object will come closer to the moon, its movement will not be affected by the Moon's gravity force?
Tidal forces and gravity are not the source of energy, rotation is.
No, what I'm saying is that energy transformation is all that gravity can do. It can't create energy.
QuoteYou have to agree that once an object has lost its rotation energy, than it also should lose its gravity force.How could it be?It will not lose all of its gravity. The rotation of an object contains kinetic energy. Due to mass-energy equivalence, that energy also has a mass. That resulting mass increases the strength of the gravitational field. This additional gravity will be extremely tiny compared to the gravity produced by the object's rest mass in most cases (like for planets and stars). In the case of a black hole with the maximum possible rotation rate, however, 29% of its mass-energy will be contained in that rotation (which is fairly significant). So if some of the rotation (and therefore energy) of the hole is extracted, the overall energy content (and therefore mass) of the hole is reduced.
You have to agree that once an object has lost its rotation energy, than it also should lose its gravity force.How could it be?
I never said that an object's entire gravitational field is generated by rotation,
If that was true, then we would have recognized long ago that conservation of energy was wrong. But that's not true. Force doesn't generate energy. It only changes its form.
It will be affected, but the object's total energy will be unchanged. All that is happening is that some of the object's gravitational potential energy is being converted into kinetic energy.
SorryI see it differently.
Gravity force and tidal creates energy.
Rotation is a direct outcome of gravity force.
The Kinetic energy is directly affected by the orbital velocity.
So, Newton gravity force which had been set ONLY by mass and radius (and the constant G), has a direct impact on the orbital velocity and that orbital velocity sets the Kinetic orbital energy.
Therefore, The kinetic energy is a direct outcome of the gravity force and not the other way.
So, even the potential energy is a direct outcome of mass and distance (or radius). As the gravity force is a direct outcome of mass and radius - the potential energy is also an outcome of gravity force.
I disagreeGravity force can create new energy
How did you get the 29% for the BH?
What should be the gravity lost in the Moon which have already lost its rotate energy?
Please show the formula which links the rotate energy to gravity force.
Again - Please show the formula which links between the two.
It seems to me that you are using the conservation of energy at the wrong place.
I have already discussed about Newton formula that shows that Gravity generates energy.
If I understand you correctly, you claim that the total rotation energy (Kinetic + potential) is fixed.
So, if we take an object to the infinity - we can also claim that its potential energy is infinite.
Just based on this idea we can claim that the energy in our Universe is infinity.
In reality - if it is far enough, it will not be affected by the gravity force any more.
So, although the potential energy should be infinite
the gravity force between those two objects at a distance of infinite is virtually zero
and therefore there is no meaning for that Potential energy.
On the other way, I could claim that the virtual pair are coming from the infinity.
So, at the moment of creation, their infinite potential energy had been transformed into Kinetic energy.
I assume that if you give the possibility to transform energy between kinetic and potential than you should also agree with that sort of idea.
Not according to the law of conservation of energy, it doesn't. Do you think you are some kind of higher authority than the laws of physics themselves?You are right in the sense that gravity is converting gravitational potential energy into kinetic energy.
Therefore, as Ek is not equal to Ep any change in the radius also set a change in Et.
If the total rotation energy is changing due to radius change - where do you get the idea of law of conservation of energy in gravity force???
Do you see any error in this simple example?
In any case, let me remind you that we have started this discussion as I have stated that there is no energy transformation in Gravity.
I have used the moon example -It had totally lost its rotate ability as it is fully locked with the Earth.
Even so, any object that has a wish to orbit around the moon must fully obey to the same Ep, Ek and Et as I have used.So, with the ability to rotate/spin or without it, the outcome due to gravity force is the same.
Therefore, there is no need for transformation of energy from the rotate/spin energy into the orbital rotation energy and vice versa.
This idea is a fiction and I have just proved it!
How those scientists that set laws of physics can't get that simple outcome?
Right, but the radius doesn't change spontaneously. Energy either has to be added to the orbit or lost from it in order for such a radius change to occur.
So it still obeys conservation of energy.
An increase in the orbital radius requires a net input of energy,
The total energy remains the same both before and after the orbit change.
and that required amount of energy comes from the black hole's spin
The Moon is still rotating, it just takes the same amount of time to complete one rotation as it takes to complete one orbit around the Earth.