[Kryptid]

Mp means the mass of Pluto
Mm - mass of Mercury

What unit are you measuring those in?

I made a calculation using the example of Mercury and Pluto - this is the proof, since centrifugal forces are always equal to gravitational and it is possible to determine the accuracy of attractive forces from them.

So when your calculations say that the force on Pluto is 0.00347 Mn, what unit is that?

Everything is distorted here, you have to look at the published article

I found a downloadable PDF file called "foundation of vortex gravitation, cosmology and cosmogony". That's what I've been looking at.

Nothing is broken. The mass of ether turns into the mass of the nucleon.

Oh, so the total mass of the ether decreases as the mass of the planet increases? I guess that would work out.

[sorlov (OP)]

Итак, когда ваши расчеты говорят, что сила на Плутоне равна 0,00347 МН, что это за единица?

Absolute physical units are not required here. That is, the value of 0.00347 Mp means that the force of solar gravity acting on Pluto is equal to 0.00347 of the mass of Pluto. In the same arbitrary units, I determined centrifugal force and compare them with gravitational ones. Centrifugal forces are a standard of actual force and they are always equal to gravitational forces. In other words, 0.00347 is the acceleration of gravity in Pluto's orbit caused by solar gravity. What matters here is not the absolute values ​​of the forces of gravity, but their differences with centrifugal forces. As you can see, the equation of vortex gravity is an order of magnitude more accurate than the classical ones.

[sorlov (OP)]

О, значит, общая масса эфира уменьшается по мере увеличения массы планеты? Я думаю, что это сработает.

Of course, the mass of ether "decreases", but it is infinite. If we subtract any finite quantity from infinity, then in the end we will still get the same infinity.

[Kryptid]

That is, the value of 0.00347 Mp means that the force of solar gravity acting on Pluto is equal to 0.00347 of the mass of Pluto.

Okay, if the mass of Pluto is 1.303 x 10²² kilograms, then 0.00347 multiplied by that would be 4.521 x 10¹⁹ kilograms of force (4.434 x 10²⁰ newtons). However, I calculated that the gravitational force on Pluto is 4.95755 x 10¹⁶ newtons using the Newtonian equation. So the two values still do not match up.

[sorlov (OP)]

Хорошо, если масса Плутона составляет 1.303 х 10 22 килограмма, то 0.00347 умноженное на это будет 4.521 х 10 19 килограммов силы (4.434 х 10 20 ньютонов). Однако я рассчитал, что гравитационная сила на Плутоне составляет 4,95755 х 10 16 ньютонов, используя уравнение Ньютона. Таким образом, эти два значения по-прежнему не совпадают.

You're right. I did not convert kilometers to meters, because I did not need it, since I compared the force of gravity with centrifugal force. But it does not change anything. I did not need to find absolute values. If you translate my results into meters, then you need to take 3 zeros from them and this will be your value. Take a closer look at the units in the calculation in my article.

[Kryptid]

If you translate my results into meters, then you need to take 3 zeros from them and this will be your value.

4.434 x 10²⁰ newtons divided by 10³ equals 4.434 x 10¹⁷ newtons, which is still 8.9 times larger than the Newtonian value.

EDIT: Actually, I made an error with my first calculation. I assumed the semi-major axis was being used, not perihelion:

F_Pluto = G(Mm/r²)
F_Pluto = (6.674 x 10^-11)(((1.9885 x 10³⁰)(1.303 x 10²²))/(4.43682 x 10¹²)²)
F_Pluto = (6.674 x 10^-11)((2.591 x 10⁵²)/(1.9685 x 10²⁵))
F_Pluto = (6.674 x 10^-11)(1.31623 x 10²⁷)
F_Pluto = 8.7845 x 10¹⁶ newtons

Your calculation is still too large by a factor of 5.

[sorlov (OP)]

Ваш расчет все еще слишком велик в 5 раз.

Your mistake is that in the calculation you took the distance to Pluto 4.44 x 10 to the 12th degree.
I define gravity at the top of the minor axis. The distance to this point is 5.9 x 10 to the 12th degree. The force of gravity (at this distance) is - 5 x 10 to the 16th degree.
Read carefully the calculations in my article again.

[Kryptid]

The distance to this point is 5.9 x 10 to the 12th degree.

Oh, so that means that my original calculation was correct all along:

FPluto = (6.674 x 10^-11)(((1.9885 x 10³⁰)(1.303 x 10²²))/(5.906 x 10¹²)²)

So your calculations are still off by about 8.9-fold.

The force of gravity (at this distance) is - 5 x 10 to the 16th degree.

Actually, your calculated force is 4.434 x 10¹⁷ newtons:
 

4.434 x 10²⁰ newtons divided by 10³ equals 4.434 x 10¹⁷ newtons, which is still 8.9 times larger than the Newtonian value.

[sorlov (OP)]

Фактически, ваша расчетная сила составляет 4.434 x 10 17 ньютонов:
 

What are you confusing. My equation in a flat system produces the same result as Newton.
In this case = 5 x 10 to the 16th degree of Newtons.
But in the spatial system, this value in my equation is multiplied by my gravitational coefficient, equal to the cosine of the deflection angle in the cube. Pluto at this point it is 0.908.
 That is, 5x10 to the 16th degree is multiplied by 0.908. We get 4.54x10 in the 16th degree.
The centrifugal forces at this point are 4.47 x 10 to the 16th degree - and this is the actual force.
I hope that you can determine by which equation of gravity its value is closer to the value of centrifugal forces, and therefore to the truth.

[Kryptid]

What are you confusing. My equation in a flat system produces the same result as Newton.
In this case = 5 x 10 to the 16th degree of Newtons.
But in the spatial system, this value in my equation is multiplied by my gravitational coefficient, equal to the cosine of the deflection angle in the cube. Pluto at this point it is 0.908.
 That is, 5x10 to the 16th degree is multiplied by 0.908. We get 4.54x10 in the 16th degree.
The centrifugal forces at this point are 4.47 x 10 to the 16th degree - and this is the actual force.
I hope that you can determine by which equation of gravity its value is closer to the value of centrifugal forces, and therefore to the truth.

Okay, I see now that I misinterpreted what you posted before. I see what you are trying to do with your math now. I may need to do some more math of my own, so I'll get back to you later.

EDIT: I did some calculations for Venus and Pluto using their semi-major axes. The standard Newtonian equation seems to be highly accurate (measurement or rounding errors alone could potentially account for the differences).

Venus:

F_centrifugal = (mv²)/r
F_centrifugal = ((4.8675 x 10²⁴)(3.502 x 10⁴)²)/(1.08208 x 10¹¹)
F_centrifugal = ((4.8675 x 10²⁴)(1.2264004 x 10⁹)/(1.08208 x 10¹¹)
F_centrifugal = (5.969503947 x 10³³)/(1.08208 x 10¹¹)
F_centrifugal = 5.51669 x 10²² newtons

F_{gravitational} = G((Mm)/r²)
F_{gravitational} = (6.674 x 10^-11)((1.9885 x 10³⁰)(4.8675 x 10²⁴))/(1.08208 x 10¹¹)²
F_{gravitational} = (6.674 x 10^-11)((1.9885 x 10³⁰)(4.8675 x 10²⁴))/(1.1708971264 x 10²²)
F_{gravitational} = (6.674 x 10^-11)(9.67902375 x 10⁵⁴)/(1.1708971264 x 10²²)
F_{gravitational} = (6.45978045075 x 10⁴⁴)/(1.1708971264 x 10²²)
F_{gravitational} = 5.51695 x 10²² newtons

Those two values are very, very close (the calculated centrifugal force is 99.995% that of the gravitational force).

Pluto:

F_centrifugal = (mv²)/r
F_centrifugal = ((1.303 x 10²²)(4,743)²)/(5.90638 x 10¹²)
F_centrifugal = ((1.303 x 10²²)(2.2496049 x 10⁷))/(5.90638 x 10¹²)
F_centrifugal = (2.9312351847 x 10²⁹)/(5.90638 x 10¹²)
F_centrifugal = 4.962828644 x 10¹⁶ newtons

F_{gravitational} = G((Mm)/r²)
F_{gravitational} = (6.674 x 10^-11)((1.9885 x 10³⁰)(1.303 x 10²²))/(5.90638 x 10¹²)²
F_{gravitational} = (6.674 x 10^-11)((1.9885 x 10³⁰)(1.303 x 10²²))/(3.48853247044 x 10²⁵)
F_{gravitational} = (6.674 x 10^-11)((2.5910155 x 10⁵²))/(3.48853247044 x 10²⁵)
F_{gravitational} = (1.7292437447 x 10⁴²)/(3.48853247044 x 10²⁵)
F_{gravitational} = 4.9569375 x 10¹⁶

Those two values are also very close (99.88%).

[sorlov (OP)]

F centrifugal = (mv 2 ) / r
f centrifugal = ((1.303 x 10 22)(4,743)2)/(5.90638 х 1012)
Фцентробежного = ((1.303 х 1022)(2.2496049 х 107))/(5.90638 х 1012)
Фцентробежного = (2.9312351847 х 1029)/(5.90638 х 1012)
Фцентробежного = 4.962828644 х 1016 ньютонов

Incorrect raw data
The speed at the top of a small strip at Pluto is 4581 m/s.
The radius of curvature at the top of the minor axis should be determined taking into account that the orbit of Pluto is not a circle, but an ellipse and it is determined by the formula Rb = a2 / b = 6098.48 x 10 in 9 m. Where a is the semimajor axis, in is the semimajor axis. Then the centrifugal forces are 4.47 x 10 in 22. Therefore, the discrepancy according to your data is even greater - 23%.
The discrepancies along Venus are small, since its orbit hardly deviates from the gravitational plane. Therefore, the classical force of gravity is almost no different from vortex gravity and it makes no sense to look for discrepancies here. In my article I chose Pluto and Mercury, since their orbits have maximum deviations from the gravitational etheric plane.

[Kryptid]

Incorrect raw data
The speed at the top of a small strip at Pluto is 4581 m/s.

I'm using the average orbital speed because I'm using the average distance from the Sun (the semi-major axis).

The radius of curvature at the top of the minor axis should be determined taking into account that the orbit of Pluto is not a circle, but an ellipse and it is determined by the formula Rb = a2 / b = 6098.48 x 10 in 9 m. Where a is the semimajor axis, in is the semimajor axis.

Where do you get that equation for centrifugal force from?

[sorlov (OP)]

Откуда вы взяли это уравнение для центробежной силы?

My centrifugal force equation is classic. In the orbit of each planet, the speed and radius of curvature change. Therefore, the magnitude of the classical force changes.
Read Kepler's laws and ellipse geometry.

[Kryptid]

My centrifugal force equation is classic. In the orbit of each planet, the speed and radius of curvature change. Therefore, the magnitude of the classical force changes.
Read Kepler's laws and ellipse geometry.

Can you provide a reference? The reason I ask is because I've looked for such equations and have had trouble finding them.

Also, you seem to be under the impression that the centrifugal force and gravitational force will always be equal in all parts of a planet's orbit. That would be true if the orbit was circular. This would not be true in the case of an elliptical orbit. The fact that the planet changes its distance from the Sun over time shows that one of the forces acting on it is stronger than the other at certain points and weaker at different points.

When the planet passes the semi-major axis on its way out to aphelion, the gravitational force becomes weaker but the centrifugal force becomes weaker at a faster rate. So the gravitational force exceeds the centrifugal force and starts to decelerate the planet (deceleration in the sense of its rate of moving away from the Sun, I mean), until the ratio of gravitational force to centrifugal force reaches its maximum at aphelion. So we should expect a calculation of centrifugal force at aphelion to be lower than gravitational force there.

The planet stops receding from the Sun and the gravitational force takes over, accelerating it back towards the Sun. Once it passes the semi-major axis again, the centrifugal force exceeds the gravitational force and the acceleration turns back into a deceleration, with the planet no longer approaching the Sun when it reaches perihelion. Perihelion should therefore be the point where the centrifugal force to gravitational force reaches its maximum and begins to accelerate the planet away from the Sun once more. The only places in the orbit where centrifugal force and gravitational force should be equal would be at the semi-major axis (a point crossed four time in the orbit).

[sorlov (OP)]

Можете ли вы предоставить ссылку? Причина, по которой я спрашиваю, заключается в том, что я искал такие уравнения и у меня были проблемы с их поиском.

Кроме того, вы, по-видимому, находитесь под впечатлением, что центробежная сила и гравитационная сила всегда будут равны во всех частях орбиты планеты. Это было бы верно, если бы орбита была круговой. Это было бы неверно в случае эллиптической орбиты. Тот факт, что планета со временем меняет свое расстояние от Солнца, показывает, что одна из сил, действующих на нее, сильнее другой в определенных точках и слабее в разных точках.

Когда планета проходит полу-большую ось на своем пути к афелию, гравитационная сила становится слабее, но центробежная сила становится слабее с более быстрой скоростью. Таким образом, гравитационная сила превышает центробежную силу и начинает замедлять планету (замедление в смысле скорости ее удаления от солнца, я имею в виду), пока соотношение гравитационной силы к центробежной силе не достигнет своего максимума на афелии. Поэтому мы должны ожидать, что расчет центробежной силы в афелии будет ниже, чем сила гравитации там.

Планета перестает удаляться от Солнца, и гравитационная сила берет верх, ускоряя ее обратно к Солнцу. Как только он снова проходит через полу-большую ось, центробежная сила превышает гравитационную силу, и ускорение превращается обратно в замедление, причем планета больше не приближается к Солнцу, когда она достигает перигелия. Таким образом, перигелий должен быть точкой, где центробежная сила достигает своего максимума и гравитационная сила начинает ускорять планету от солнца еще раз. Единственные места на орбите, где центробежная сила и гравитационная сила должны быть равны, были бы на полу-главной оси (точка, пересекаемая четыре раза на орбите).

You have a gross error. Centrifugal force is a reactive force and it is always equal to the active force, in our case, the force of gravity. Centrifugal force is like a shadow, according to Newton's third law. It arises only under the condition that a moving point is rejected by an external force (in our case, gravitational). If gravitational forces were directed equally in all directions, according to Newton, then this point would move in a circle. But if it moves along an ellipse, then the forces of gravity change their values. This is the first proof of my theory. Gravity is created by a flat vortex disk. The orbital planes of the planets have small deviations from this disk. Moving by inertia, they cross it in two places (perihelion and aphelion). The force of gravity reaches its maximum value here, and the orbits - the maximum curvature. When moving away from the aphelion or perihelion, the planets move away from the gravitational disk, which means the forces of gravity decrease and the orbits straighten. All these explanations are detailed in my article.

[Kryptid]

Centrifugal force is a reactive force and it is always equal to the active force, in our case, the force of gravity.

If I have a ball on a string and I start to hurl it around in a circle with an unfaltering grip, then we can say that the centrifugal force and centripetal forces are equal. However, if I have a loose grip on the string, then the centrifugal force will slowly pull the string out of my hand. In that case, this must mean that the centripetal force was weaker than the centrifugal force.

[sorlov (OP)]

Centrifugal force is a reactive force and it is always equal to the active force, in our case, the force of gravity.

If I have a ball on a string and I start to hurl it around in a circle with an unfaltering grip, then we can say that the centrifugal force and centripetal forces are equal. However, if I have a loose grip on the string, then the centrifugal force will slowly pull the string out of my hand. In that case, this must mean that the centripetal force was weaker than the centrifugal force.

There was an inaccurate translation.
If your string is pulled, it means that your centripetal force has weakened due to outside interference. In your case, due to a communication failure. In this case, the centrifugal force decreases to the same extent. You understand, centrifugal force is a conditional, reactive force that does not exist on its own. Only gravity can create centrifugal force. The centrifugal force itself cannot change unless the centripetal force changes. These explanations can be found on the Internet.

[Halc]

When the planet passes the semi-major axis on its way out to aphelion, the gravitational force becomes weaker but the centrifugal force becomes weaker at a faster rate. So the gravitational force exceeds the centrifugal force and starts to decelerate the planet (deceleration in the sense of its rate of moving away from the Sun, I mean), until the ratio of gravitational force to centrifugal force reaches its maximum at aphelion. So we should expect a calculation of centrifugal force at aphelion to be lower than gravitational force there.

Sorry Kryptid, but I'm actually with sorlov on this one.  The two forces are equal and opposite, one acting on the planet (thus slowing it down on its way out to aphelion), and the other acting on the primary (star presumably), slowing it down as well. The planet slows down not because the forces are not equal, but because the force is not acting tangentially to its motion.

[Kryptid]

Sorry Kryptid, but I'm actually with sorlov on this one.  The two forces are equal and opposite, one acting on the planet (thus slowing it down on its way out to aphelion), and the other acting on the primary (star presumably), slowing it down as well. The planet slows down not because the forces are not equal, but because the force is not acting tangentially to its motion.

I realize (due to Newton's third law), that the gravitational force on the planet must be equal and opposite to the gravitational force on the Sun. Likewise, the centrifugal force on the planet and the Sun must also be equal and opposite. But must this mean that the gravitational force between the Sun and planet are always equal to the centrifugal force between the Sun and planet? How can the planet and Sun accelerate away from each other if there is no net force present between them? F=ma seems to demand a net force in order for an acceleration to be present.

[Halc]

I realize (due to Newton's third law), that the gravitational force on the planet must be equal and opposite to the gravitational force on the Sun. Likewise, the centrifugal force on the planet and the Sun must also be equal and opposite. But must this mean that the gravitational force between the Sun and planet are always equal to the centrifugal force between the Sun and planet?

There is an equal and opposite force, per Newtons 3rd law. You can call it a centrifugal force if you like, but I never thought that made sense in any but a rotating frame. It's just a force of gravity, one mass acting on another, each accelerating the other.

How can the planet and Sun accelerate away from each other if there is no net force present between them?

They never accelerate away from each other.  Acceleration vector is always straight towards the other in a 2 body system.  They move away from each other (by inertia), but since they're slowing down while doing that, they're not accelerating that way.

F=ma seems to demand a net force in order for an acceleration to be present.

There is a net force on each mass, hence each of them accelerates.  If there is a string, tension on that string represents the one force accelerating each object.  The system as a whole has balanced forces and the center of gravity of that system is thus entirely inertial (non accelerating).