Dicussion of sandokhan's claims about the Moon in Q&A

Since discussion about posts in Q&A is sometimes discouraged, this thread addresses some of the "interesting" statements recently made in the Q's about the Moon thread in the Q&A section.

Here's the post in question, quoted in its entirety here. Another was posted as this was being composed.

Official science theory

Libration occurs because the Moon's rotation has a constant angular speed, while the Moon's elliptical orbit has a varying angular speed.

But that explanation cannot be correct since it is very easy to demonstrate the Moon does not orbit the Earth in shape of an elliptical curve.

The theoretical foundation of modern astronomy's understanding of the Moon's orbit is tidal locking.

However, the oceanic tides of the Earth ARE NOT caused by the Moon's supposed gravitational influence.

The most devastating demonstration that the tides have nothing to do with the Moon:

http://immanuelvelikovsky.com/NewtonEinstein&Veli.pdf

Chapter 1, section Tidal Theory, Gravity and Mathematics, pages 9-24

See also: http://milesmathis.com/tide.html

Let us also remember that modern science cannot explain at all atmospheric tides:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1707294#msg1707294


The true shape of the Moon is a disk, as can be seen in the T. Legault/F. Bruenjes photographs posted starting here:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1786946#msg1786946

Let's discuss these.

Libration occurs because the Moon's rotation has a constant angular speed, while the Moon's elliptical orbit has a varying angular speed.

That's the largest contributor to it. There are also additional small amounts due to the tilt of the Moon's axis of rotation is not perfectly normal to its orbital plane (akin to what causes the seasons on the Earth), and a small amount of diurnal parallax since the Earth's diameter is roughly 1/30 the distance to the Moon.

But that explanation cannot be correct since it is very easy to demonstrate the Moon does not orbit the Earth in shape of an elliptical curve.

Let's see this demonstration. Since it's supposedly easy, it should be clear and concise, and not take walls of arcane bafflegab.

The theoretical foundation of modern astronomy's understanding of the Moon's orbit is tidal locking.

Um, no. The theoretical foundation for understanding the Moon's orbit is Keplerian orbital mechanics, same as everything else in the solar system (and beyond). Tidal locking has caused the Moon's period of rotation to be the same is its orbital period; that's all.

However, the oceanic tides of the Earth ARE NOT caused by the Moon's supposed gravitational influence.

The most devastating demonstration that the tides have nothing to do with the Moon:

http://immanuelvelikovsky.com/NewtonEinstein&Veli.pdf

Chapter 1, section Tidal Theory, Gravity and Mathematics, pages 9-24

Tl;dr. I don't really see the relationship with lunar libration or, for that matter, the fundamentals of the Moon's orbit. Besides... Velikovsky; a guy with "interesting" ideas with little, if any, basis in fact.

See also: http://milesmathis.com/tide.html

Miles Mathis is the fellow who, among other seriously bizarre things, claims that pi = 4. Seriously.

Let us also remember that modern science cannot explain at all atmospheric tides:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1707294#msg1707294

So many words, so little useful information, but, again, how does this affect lunar libration?

The true shape of the Moon is a disk, as can be seen in the T. Legault/F. Bruenjes photographs posted starting here:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1786946#msg1786946

How can you tell the difference between the silhouette of an opaque sphere and the silhouette of a perfectly face-on opaque disk in a photograph? The answer? You can't. We can easily tell from other observations (librations, for instance), that the Moon is spherical, not a disk, however.

Another post:

You need to update your knowledge of the libration motion of the moon:

http://www.thetruthdenied.com/news/2015/01/06/what-is-lunar-libration-what-is-the-far-side-of-the-moon/

The polar motion of the libration phenomenon cannot be explained by modern astrophysics.

Sure it can. It's because the Moon's axis of rotation is not exactly perpendicular to its orbital plane. Straightforward. Next!

And when you come here with your videos you better make sure that they were not faked in the first place:

http://www.conspiracyoutpost.com/topic/2303-nasa-says-earth-is-round-wheres-the-proof/?do=findComment&comment=74910

In fact the real time photographs posted here show a Moon in the shape of a disk just like this image:

That image looks like the surface of a sphere to me, not a disk, because of the progressively foreshortened presumably-circular features toward the edge.

The outline of a sphere in a photographic image on a 2D surface (like, say, a photograph) is a circle[nb]Unless it's terribly distorted.[/nb]. Do you really not understand this, or are you intentionally trying to deceive anyone you can?

And you certainly haven't done your homework on the impossibility of a spherically shaped sun in the first place:

Impossibility of a round Sun shape:

By "round", I presume you mean "spherical". Circles are round. The photos you yourself provide clearly show a "round" sun. Sorry, but the spherical sun is not an impossibility. Even if you want to think so.
 

"The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun.

You seem be conflating mass with weight, a common error among people unfamiliar with the basics of physics.

Grams are units of mass, not weight. A gram of mass on the surface of the Earth has the same mass as a gram of mass on (what's defined as) the surface of the Sun or anywhere else, although their weight can differ vastly, from nothing (in free-fall) to enormous. What you apparently mean (although it's kind of hard to tell) is that the weight of a gram of mass at the surface of the Sun would be 27.47 times the weight of a gram of mass on the surface of the Earth. That is, a mass of one gram has a weight at the surface of the Earth of about 0.0098 newton; at the surface of the Sun, a gram would weigh about 0.274 newton. If you don't understand the distinction, please ask for an explanation; if you do understand the distinction, please try to use the correct terminology.

At any rate, in case you  haven't noticed, the Sun is not physically the same as the Earth. The Sun is a very hot (compared to the Earth, anyway) very massive (compared to the Earth) ball (as in sphere) of plasma. The Earth is mostly solid or semi-solid ball (sphere) of rock with a cool (compared to the Sun) atmosphere. Their properties are quite different. You are basically carrying on about the conditions at the Sun being different than what we find on earth. Well, no shit, Sherlock!

Thus the attraction by the solar mass is 10,000 times greater than the repulsion of the solar light. Recourse is taken to the supposition that if the pull and the pressure are calculated for very small masses, the pressure exceeds the pull, one acting in proportion to the surface, the other in proportion to the volume.

Let's see... you're comparing a proxy for pressure (which is properly force per unit area, but "2.75 milligrams per square centimeter" is mass per unit area) with an apparently made-up mass per unit volume (which is density). Please explain whatever it is you think you're trying to say in a coherent manner if you want anyone to understand it. If you don't want anyone to understand it, then carry on; you've been exposed.

But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Possibly because there's more going on in sunspots than just light emission, like intense magnetic fields ejecting plasma away from the Sun?

Because of its swift rotation, the gaseous sun should have the latitudinal axis greater than the longitudinal, but it does not have it. The sun is one million times larger than the earth

The Sun's diameter is about 100 (10²) times the Earth's diameter, so its volume is about (10²)³, or 10⁶ (one million) time the Earth's volume.

and its day is but twenty-six times longer than the terrestrial day; the swiftness of its rotation at its equator is over 125 km. per minute;

The Sun has about 100 times (not a million... nice try!) the diameter of the Earth, and therefore 100 times the circumference, but spins 26 times slower, so its tangential velocity at the equator is a little less than 4 times greater. OK. So?

at the poles, the velocity approaches zero. Yet the solar disk is not oval but round: the majority of observers even find a small excess in the longitudinal axis of the sun. The planets act in the same manner as the rotation of the sun, imposing a latitudinal pull on the luminary.

Indicating that there's a lot going on inside the Sun that's not going on inside the planets. You point this out yourself in the next paragraph.

Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

Near the polar regions of the sun, streamers of the corona are observed, which prolong still more the axial length of the sun.

Well, there you go. Energy emission from the Sun is not uniform. No competent solar physicist would claim otherwise, since it's clear that it isn't.

You recognize it's not simple. Why the oversimplification then?

If planets and satellites were once molten masses, as cosmological theories assume, they would not have been able to obtain a spherical form

Au contraire. If it has low shear strength (liquids have very low shear strength), material naturally flows to the state where it has the lowest energy, which, in the case of an inverse-square field like gravity, is a sphere. In other words, they wouldn't do otherwise under the influence of gravity alone. Why do you think otherwise?

especially those which do not rotate, as Mercury or the moon (with respect to its primary)."

Both Mercury and the Moon rotate with respect to their primary, but fairly slowly. If the Moon didn't rotate, we would not always see the same side.

Solar Atmosph. Pressure as a Function of Depth (official science information)^{[citation, please]}

Depth (km) % Light from this Depth Temperature (K) Pressure (bars)

0 99.5 4465 6.8 x 10-3
100 97 4780 1.7 x 10-2
200 89 5180 3.9 x 10-2
250 80 5455 5.8 x 10-2
300 64 5840 8.3 x 10-2
350 37 6420 1.2 x 10-1
375 18 6910 1.4 x 10-1
400 4 7610 1.6 x 10-1

This table indicates that the solar atmosphere changes from being almost completely transparent to being almost opaque over a distance of about 400 km. Notice also that in this region the temperature drops rapidly as we near the surface, and that the pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.

"Official science information." LOL. If you have a source, name it.

And? The Earth's atmosphere changes from being almost completely transparent to being almost opaque quite abruptly under some conditions (like, say, clouds) without extensive changes in temperature or pressure.

Did you notice that those temperatures are considerably higher than the surface of the earth? They are considerably higher than the average atmospheric temperature at the Surface of the Earth, which is less than 300K. The Sun isn't the Earth. Trying to compare them the way you're doing is way too simplistic to be meaningful.

Subscribed

Only a troll would not know that the moment of inertia is based on the concept of mass.

Wrong.

An object's angular momentum is determined by its moment of inertia, not mass, and rate of spin.

Here is the definition of the moment of inertia:

The sum of the products of the mass and the square of the perpendicular distance to the axis of rotation of each particle in a body rotating about an axis.


Here is what he wrote earlier:

An object's angular momentum is determined by its moment of inertia, not mass, and rate of spin.


The troll had no idea that the moment of inertia is defined in terms of mass.

No idea whatsoever.


When this fact was brought to his attention, during the course of the same day, he wrote:

Mass is a factor, yes.

NOBODY is ever going to take you seriously on this forum anymore.


One month later, here he is at it again.

 The theoretical foundation for understanding the Moon's orbit is Keplerian orbital mechanics, same as everything else in the solar system (and beyond).

But Kepler faked HIS ENTIRE SET OF DATA relating to any kind of an elliptical orbit.

“After detailed computational arguments Donahue concluded the results
reported by Kepler . . . were not at all based on Brahe’s observational data; rather
they were fabricated on the basis of Kepler’s determination that Mars’s orbit was
elliptical. Donahue reasons that Kepler must have gone back to revise his earlier
calculations that were made prior to his understanding that the orbit of Mars was
actually elliptical. Thus, anyone who cared to check Kepler’s tables would find
numbers that are consistent with the elliptical orbit [he] postulated for Mars and
would be inclined to believe that the numbers represented observational data. In
fact, they were computed from the hypothesis of an elliptical orbit and then
modified for measurement error; such data, if they were truly observations, would
be prime facie evidence of the theories’ correctness.

“So Donahue . . . realized that the theory was not obviously derivable from the
observations, . . . ‘Not only would the numbers be confused, but Kepler saw clearly
that no satisfactory theory could come from such a procedure. . . [Instead], he chose
a short cut.’ He became so convinced of what drove these physical processes that he subjectively projected his personal nonobservational-based belief onto the reporting scene to convince others in the scientific community of the validity of his theories.”

Thus, the very first law of planetary motion was built not on observation but on theory
and the mathematics was then employed to prove the theory not test it.


http://adsabs.harvard.edu/full/1988JHA....19..217D

Kepler's fabricated figures, by W.H. Donohue

The scholar, William H. Donahue, said the evidence of Kepler's scientific fakery is contained in an elaborate chart he presented to support his theory.

The discovery was made by Dr. Donahue, a science historian, while translating Kepler's master work, ''Astronomia Nova,'' or ''The New Astronomy,'' into English. Dr. Donahue, who lives in Sante Fe, N.M., described his discovery in a recent issue of The Journal of the History of Astronomy.

The fabricated data appear in calculated positions for the planet Mars, which Kepler used as a case study for all planetary motion. Kepler claimed the calculations gave his elliptical theory an independent check. But in fact they did nothing of the kind.

''He fudged things,'' Dr. Donahue said, adding that Kepler was never challenged by a contemporary. A pivotal presentation of data to support the elliptical theory was ''a fraud, a complete fabrication,'' Dr. Donahue wrote in his paper. ''It has nothing in common with the computations from which it was supposedly generated.''


Thus, the notion that a planet orbits the Sun in an elliptical orbit was a simple fabrication, based on fudged data.


No planet could have attained an elliptical orbit in the first place, the easiest demonstration:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1779395#msg1779395


No wonder that someone who thinks that the moment of inertia has nothing to do with mass, accepts KEPLER'S FAKED DATA as "theoretical foundation".


If you have no idea that the definition of the moment of inertia does include mass, HOW IN THE WORLD are you going to understand the definition of METRIC MASS (WEIGHT)?

"The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun.



http://csep10.phys.utk.edu/astr162/lect/sun/photosphere.html

Solar Atmosph. Pressure as a Function of Depth (official science information)

Depth (km) % Light from this Depth Temperature (K) Pressure (bars)

0 99.5 4465 6.8 x 10-3
100 97 4780 1.7 x 10-2
200 89 5180 3.9 x 10-2
250 80 5455 5.8 x 10-2
300 64 5840 8.3 x 10-2
350 37 6420 1.2 x 10-1
375 18 6910 1.4 x 10-1
400 4 7610 1.6 x 10-1

This table indicates that the solar atmosphere changes from being almost completely transparent to being almost opaque over a distance of about 400 km. Notice also that in this region the temperature drops rapidly as we near the surface, and that the pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.


Since you cannot explain this very straightforward fact, that THE GASES IN THE SOLAR ATMOSPHERE ARE UNDER A VERY LOW PRESSURE, THUS THEIR VERY PRESENCE THERE BEING A CLEAR DEFIANCE OF THE LAWS OF PHYSICS (CENTRIFUGAL FORCE WHICH WOULD DISPERSE THE GASES IMMEDIATELY), you are forced to resort to massive shitposting, where you (of all individuals) bring into question mass vs. weight.

The context of the discussion makes it very clear what is meant, the very same definition accepted by modern science:

the weight of a mass is the force of gravity on the mass


Thus the attraction by the solar mass is 10,000 times greater than the repulsion of the solar light. Recourse is taken to the supposition that if the pull and the pressure are calculated for very small masses, the pressure exceeds the pull, one acting in proportion to the surface, the other in proportion to the volume.

There is no need for any kind of an explanation UNLESS you are one of those persons who thinks that the moment of inertia is not related to mass.


But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Possibly because there's more going on in sunspots than just light emission, like intense magnetic fields ejecting plasma away from the Sun?

You CANNOT bring magnetism into any kind of discussion involving gravity.

Do you understand what your statement means?

Modern gravitational theories EXCLUDE magnetism from any kind of calculations.


Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

Near the polar regions of the sun, streamers of the corona are observed, which prolong still more the axial length of the sun.


DO YOU UNDERSTAND ENGLISH?

DO I HAVE TO SPELL IT FOR YOU?


Here we go again.

Solar Atmosph. Pressure as a Function of Depth (official science information)[citation, please]

Depth (km) % Light from this Depth Temperature (K) Pressure (bars)

0 99.5 4465 6.8 x 10-3
100 97 4780 1.7 x 10-2
200 89 5180 3.9 x 10-2
250 80 5455 5.8 x 10-2
300 64 5840 8.3 x 10-2
350 37 6420 1.2 x 10-1
375 18 6910 1.4 x 10-1
400 4 7610 1.6 x 10-1

This table indicates that the solar atmosphere changes from being almost completely transparent to being almost opaque over a distance of about 400 km. Notice also that in this region the temperature drops rapidly as we near the surface, and that the pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.

"The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun.


As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.


THE FACT THAT THE GASES OF THE SOLAR ATMOSPHERE ARE UNDER A VERY WEAK PRESSURE IS A CLEAR DEFIANCE OF NEWTONIAN MECHANICS, SINCE THE CENTRIFUGAL FORCE OF ROTATION WOULD DISPERSE IMMEDIATELY THE SAME GASES.

No one in the scientific community can explain HOW these gases stay there in clear defiance of the laws of physics.



If the Moon didn't rotate, we would not always see the same side.

The so-called synchronous rotation of the Moon cannot be explained at all by modern science, especially if we bring into play the recession rate of the Moon paradox.

The very fact that you do see the same side, means it is not rotating at all.


Au contraire. If it has low shear strength (liquids have very low shear strength), material naturally flows to the state where it has the lowest energy, which, in the case of an inverse-square field like gravity, is a sphere. In other words, they wouldn't do otherwise under the influence of gravity alone. Why do you think otherwise?

But it does not.

Hydrogen gas in outer space does not clump together. *Harwit’s research disproves the possibility that hydrogen gas in outer space can clump together. This is a major breakthrough in disproving the Big Bang and related origin of matter and stars theories. The problem is twofold: (1) The density of matter in interstellar space is too low. (2) There is nothing to attract the particles of matter in outer space to stick to one another. Think about it a minute; don’t those facts make sense?

This point is so important (for it devastates the origin of stars theory) that *Harwit’s research should be mentioned in more detail:

*Harwit’s research dealt with the mathematical likelihood that hydrogen atoms could stick together and form tiny grains of several atoms, by the random sticking of interstellar atoms and molecules to a single nucleus as they passed by at a variable speed. Using the most favorable conditions and the maximum possible sticking ability for grains, Harwit determined that the amount of time needed for gas or other particles to clump together into a size of just a hundred-thousandth of a centimeter in radius—would take about 3 billion years! Using more likely rates, 20 billion years would be required—to produce one tiny grain of matter stuck together out in space. As with nearly all scientists quoted in our 1,326-page Evolution Disproved Series (which this book is condensed from), *Harwit is not a Creationist (*M. Harwit, Astrophysical Concepts, 1973, p. 394).

To make the entire discussion crystal clear, here is the official set of data used by modern day astrophysicists:



Fraknoi, Morrison Wolff, Voyages throught the Universe (award-winning astronomy educator (Fraknoi) and two distinguished research scientists (Morrison at NASA and Wolff at NOAO).


The pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.

"The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth."


Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.


No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.



For those who do not understand the physics involved, let us go directly to the textbook on the subject.

The Moon generally has one hemisphere facing the Earth, due to tidal locking.

Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit Earth.

But tides, oceanic and atmospheric have nothing to do with the supposed gravitational force exerted by the Moon.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1707294#msg1707294


"It has been known now for two and a half centuries, that there are more or less daily variations in the height of the barometer, culminating in two maxima and two minima during the course of 24 hours. The same observation has been made and puzzled over at every station at which pressure records were kept and studied, but without success in finding for it the complete physical explanation."

NATIONAL WEATHER SERVICE DATA:


The most basic change in pressure is the twice daily rise and fall in due to the heating from the sun. Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m. The magnitude of the daily cycle is greatest near the equator decreasing toward the poles.

http://oceanservice.noaa.gov/education/yos/resource/JetStream/atmos/pressure.htm

Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m.


BAROMETER PRESSURE PARADOX

One maximum is at 10 a.m., the other at 10 p.m.; the two minima are at 4 a.m. and 4 p.m.

The heating effect of the sun can explain neither the time when the maxima appear nor the time of the minima of these semidiurnal variations.

If the pressure becomes lower without the air becoming lighter through a lateral expansion due to heat, this must mean that the same mass of air gravitates with changing force at different hours.


Lord Rayleigh: ‘The relative magnitude of the latter [semidiurnal variations], as observed at most parts of the earth’s surface, is still a mystery, all the attempted explanations being illusory.’



Currently, the barometer pressure paradox CANNOT BE EXPLAINED AT ALL.

SUNSPOTS DEFY NEWTONIAN MECHANICS

The best works which describe the relation between solar activity/solar tides and the orbits of planets (heliocentrical theory):

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20070025111.pdf

http://ozwx.plasmaresources.com/wilson/Syzygy.pdf



The Sun exhibits a variety of phenomena that defy contemporary theoretical understanding.

Eugene N. Parker


It is not coincidence that the photosphere has the appearance, the temperature and spectrum of an electric arc; it has arc characteristics because it an electric arc, or a large number of arcs in parallel.

British physicist C. E. R. Bruce


It is likely that the problem of the dynamics of the explosions affecting the prominences will only be solved when the electrical conditions obtaining in the chromosphere and inner corona are better understood.

Italian solar astronomer Giorgio Abetti


Observations give a wealth of detail about the photosphere, chromosphere and the corona. Yet we have difficulty in matching the observations with a theory.

Solar Interior & Atmosphere, J.-C. Pecker


The modern astrophysical concept that ascribes the sun’s energy to thermonuclear reactions deep in the solar interior is contradicted by nearly every observable aspect of the sun.

Ralph E. Juergens


The above-referenced papers do explain the precise relation between solar activity/tides and orbits of the planets but fail to show HOW these planets could possibly influence sunspots dynamics.


Graphics and article by Rolf Witzsche:



The resulting interaction has noticeable effects that become visible in the sunspots on the Sun!

These effects that are observed, however, present a paradox, because what is observed shouldn't really happen.

In order to solve the paradox, let me begin by saying that the story, which the planets tell us, begins with the planets' alignment with each other, as their alignment has an apparent effect on the Sun. 



Many people are puzzled by the planets having an effect on the sunspots. When the major planets are in line with each other their combined effect weakens the sunspots. 

For example, it has been recognized by researchers that the intensity of the sunspots decreases when the sunspots come into the view of the Earth. This shouldn't be possible. 

Our tiny earth, shouldn't have such a strong effect, to affect the sunspots over such a great distance. The planets are shown in scale here, by size, but not by distance. The Earth is about 100 times as distant from the Sun as the Sun is wide. And Neptune is over 3,000 times as distant. None of the planets should affect the Sun. Definitely not the Earth, with the Sun being 332,000 times more massive. Still, in spite of what one would expect from this interrelationship, an effect is being observed.

What has been observed doesn't make any sense then, right? 

It is not possible that one of the smallest planets in the solar system affect our gigantic Sun so strongly that the results become visible in the dynamics of the sunspots, some of which are thousands of kilometers wide? 

The difference between even the combined mass of planets, and the mass of the Sun, is so immensely great that the planets altogether contain only 14 one hundredth of a percent of the mass of the solar system, with the Sun containing all by itself 99.86% of it.

How then is it possible that this minuscule portion of mass that the planets represent, affect the gigantic mass of the Sun in any way at all?

As one observer has asked recently: 'We have been noticing that active sunspots deteriorate as they are facing earth, and then power back up as they turn away. What would cause this?'

The gravitational effect of the Earth is simply too minuscule to affect something as powerful as the sunspots on the Sun. 

Even if the alignment of Jupiter with the Earth was considered as a potential cause, with Jupiter having a 12 times greater gravitational affect on the Sun, than the Earth has, the combined result will still not be sufficient for any effect whatsoever to be noticeable. This applies to all the major planets in the same manner. The combined 'tidal' action of the Earth and Jupiter together, on the Sun, would still be only six one-thousands as strong of the tidal effect of the Moon on the Earth. This, too, is far too little for an effect to be observable. Thus the question remains, what has the power to produce the change of the sunspots on the Sun, and other similar phenomena?

The continuing paradox implies that a different principle, other than mass via gravity interaction, is in operation that produces the observed effects.

One researcher has hinted in such a direction when he found it puzzling, for example, that the observed effect of Saturn on the solar sunspots is nearly as extensive as that of Jupiter, with Jupiter being 3 times more massive and only half as distant from the Sun!

He recognized that the observed effect in this case does not really agree with the assumption, and thereby puts into doubt the entire base of assumptions. 

In puzzling over how to resolve the enigma the researcher noted, that like the Sun, Saturn and Jupiter are both strong sources for cosmic radio emissions. Based on this knowledge, he reasoned that perhaps other forces of energy, other than gravitational attraction, might be involved in affecting the sunspots in relationship to the solar system.


An extraordinary work which takes an in-depth look at how MAGNETISM was left out of the gravitational theories which were put forward as basic hypotheses in the 17th and 18th centuries.

No one can explain, even within the forged/faked history of these centuries, why the force of magnetism was not immediately put in a central place as the main force of universal gravitation.

http://www.gsjournal.net/old/science/ricker9.pdf


Professor John Cox, McGill University:

That's just horrible. No sandokhan, I can't have any more of this stupid, way too long copy pasta. Enough. Just stop, or stay on topic, condense your posts and say things in your own words. Until you do that, you'll just be ignored.

<long copy-paste of an earlier tantrum over being corrected about moment of inertia and angular momentum>

The mass of a body and the rate at which it spins, in classical physics, determines an object's "angular momentum."

This is still wrong. It seems like you would have wanted people to forget that you didn't know that mass and moment of inertia are not the same thing. Oh, well. Nothing new here, just a copy-paste of part of his meltdown later in that thread.

The theoretical foundation for understanding the Moon's orbit is Keplerian orbital mechanics, same as everything else in the solar system (and beyond).

But Kepler faked HIS ENTIRE SET OF DATA relating to any kind of an elliptical orbit.
...

Thus, the very first law of planetary motion was built not on observation but on theory
and the mathematics was then employed to prove the theory not test it.

http://adsabs.harvard.edu/full/1988JHA....19..217D

Kepler's fabricated figures, by W.H. Donohue
...

That's one man's opinion.

Donohue's paper doesn't seem to be widely cited; do you know if anyone else has reached at a similar conclusion?

Even if Kepler's 1609 publication is messier and less complete than what would currently be accepted, the fact remains that his elliptical model for orbits works quite well making accurate predictions about real objects, and that is what makes a model useful (and, likely, reasonably close to correct). Slag Kepler all you want, you haven't cast doubt on the elliptical-orbit model.

Thus, the notion that a planet orbits the Sun in an elliptical orbit was a simple fabrication, based on fudged data.

No, the model that a planet orbits the Sun in an elliptical orbit is used because it gives accurate answers. Details about how it came into being four centuries ago might be an interesting side topic, but that doesn't affect how well Keplerian orbits have been demonstrated to work, time and time again, for centuries.

No planet could have attained an elliptical orbit in the first place, the easiest demonstration:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1779395#msg1779395

That's not a demonstration. That's a declaration that elliptical orbits are impossible. Nothing more.

Nothing in that post justifies the claim that elliptical orbits are impossible.

No wonder that someone who thinks that the moment of inertia has nothing to do with mass, accepts KEPLER'S FAKED DATA as "theoretical foundation".

I said moment of inertia is not the same as mass; you keep misunderstanding or intentionally trying to misrepresent this. If you think you can show otherwise, how about continuing in the thread where you originally had the confusion that mass is the same as moment of inertia, and at least try to stay on topic here for a change. Thanks!

Anyway, we were talking about libration and the elliptical orbit of the Moon.

Libration occurs because the Moon's rotation has a constant angular speed, while the Moon's elliptical orbit has a varying angular speed.

But that explanation cannot be correct since it is very easy to demonstrate the Moon does not orbit the Earth in shape of an elliptical curve.

Does this mean that you can't demonstrate (easily or otherwise) that "the Moon does not orbit the Earth in shape of an elliptical curve"? Were we supposed to forget that claim?

If you ignore this, we can only presume your claim was bogus.

<blah, blah, blah> HOW IN THE WORLD are you going to understand the definition of METRIC MASS (WEIGHT)?

I understand the difference between mass and weight, and how the terms are sometimes misused. Do you? What you write is not clear.

"The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun.

Here's an example of your sloppy mixing of mass and weight. On earth a gram is a gram, but on the Sun a gram is 27 grams? When you say "one gram at the surface of the Sun" are you talking about a gram (of mass), or some lesser mass that would weigh the same as one gram of mass weighs on the surface of the earth. You are trying to get by with "a gram is not a gram except when it is a gram." It's possible to parse out what you're getting at here, but why not just say it clearly by avoiding sloppy terminology in the first place.

What's the average density of the atmosphere of the Sun? Your example mentions 1 gm/cm³ (in red, yet), which is the density of liquid water at 4°C, but is this meaningful here? The average density of the Sun is about 1.4 gm/cm³[nb]http://hyperphysics.phy-astr.gsu.edu/hbase/solar/sun.html[/nb], and that includes the very dense core. Earth's atmosphere has a density of 0.001225 gm/cm³ (at STP - 15°C and 1 atm)[nb]https://en.wikipedia.org/wiki/Density_of_air[/nb]. The Sun's atmosphere is exceedingly tenuous because it's (wait for it...) exceedingly hot.

DO YOU UNDERSTAND ENGLISH?

No need to shout, but, yes, I understand English, especially when it's clearly written. I do have trouble understanding gibberish.

<table>

Since you cannot explain this very straightforward fact, that THE GASES IN THE SOLAR ATMOSPHERE ARE UNDER A VERY LOW PRESSURE, THUS THEIR VERY PRESENCE THERE BEING A CLEAR DEFIANCE OF THE LAWS OF PHYSICS (CENTRIFUGAL FORCE WHICH WOULD DISPERSE THE GASES IMMEDIATELY),

Handwaving. Let's see the numbers that show a clear defiance of the laws of physics.

you are forced to resort to massive shitposting, where you (of all individuals) bring into question mass vs. weight.

Now, now... let's keep it nice here. You're the one who keeps using mass and force incorrectly.

The context of the discussion makes it very clear what is meant, the very same definition accepted by modern science:

Why not just use correct terminology in the first place so what you're trying to say is obvious rather than implied from context? Do you not want readers to easily understand what you mean?

the weight of a mass is the force of gravity on the mass

That's one definition. Another is:

There is also a rival tradition within Newtonian physics and engineering which sees weight as that which is measured when one uses scales. There the weight is a measure of the magnitude of the reaction force exerted on a body. Typically, in measuring an object's weight, the object is placed on scales at rest with respect to the earth, but the definition can be extended to other states of motion. Thus, in a state of free fall, the weight would be zero. In this second sense of weight, terrestrial objects can be weightless. Ignoring air resistance, the famous apple falling from the tree, on its way to meet the ground near Isaac Newton, is weightless.

Thus the attraction by the solar mass is 10,000 times greater than the repulsion of the solar light.

That's a meaningless statement. The ratio of pressure (force per unit area) to force (mass times acceleration of gravity here) tells you the area the force is distributed over, which you have never mentioned.

Recourse is taken to the supposition that if the pull and the pressure are calculated for very small masses, the pressure exceeds the pull, one acting in proportion to the surface, the other in proportion to the volume.

Meaningless unless you know the area.

There is no need for any kind of an explanation UNLESS you are one of those persons who thinks that the moment of inertia is not related to mass.

Are you still on that? You were the one who thought mass and moment of inertia were the same until told otherwise, remember? I'm surprised you don't want to put that embarrassment behind you. Being shown wrong still rankles, I guess, and you can't resist picking at the scab.

But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Possibly because there's more going on in sunspots than just light emission, like intense magnetic fields ejecting plasma away from the Sun?

You CANNOT bring magnetism into any kind of discussion involving gravity.

Why not?[nb]Note: I am not saying that gravity and magnetism are the same thing. They are independent, but can both act on matter.[/nb] Magnetic fields exist and can exert force. The Sun has very strong magnetic fields.

Do you understand what your statement means?

Modern gravitational theories EXCLUDE magnetism from any kind of calculations.

Any kind of calculations? How about the calculations needed to determine magnetic field strength needed to overcome gravity and levitate, say, a train?

Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

That might be the case if gravity and rotation were the only two factors at play. They're obviously not (to most people, at least, maybe not you). You say so yourself:

Near the polar regions of the sun, streamers of the corona are observed, which prolong still more the axial length of the sun.

See. You just said it again. What do you think causes those streamers and elongation along the axis?

DO YOU UNDERSTAND ENGLISH?

DO I HAVE TO SPELL IT FOR YOU?

Just try to explain what you're trying to say coherently. If you could do that you wouldn't need to shout.

Here we go again.

Solar Atmosph. Pressure as a Function of Depth (official science information)[citation, please]

<table>

Was it really necessary to repeat that whole thing again? Thanks, BTW, for the citation, though. Was that so hard?

THE FACT THAT THE GASES OF THE SOLAR ATMOSPHERE ARE UNDER A VERY WEAK PRESSURE IS A CLEAR DEFIANCE OF NEWTONIAN MECHANICS, SINCE THE CENTRIFUGAL FORCE OF ROTATION WOULD DISPERSE IMMEDIATELY THE SAME GASES.

You must have calculated the centrifugal force since you state this as a fact. Can you show us your calculation, please?

No one in the scientific community can explain HOW these gases stay there in clear defiance of the laws of physics.

I told you that you need to go back to high school to learn physics.

You say a lot of things that are, quite frankly, nonsense. Just a reminder from the earlier copy of this same claim: can we see the math showing "clear defiance of the laws of physics"?

If the Moon didn't rotate, we would not always see the same side.

The so-called synchronous rotation of the Moon cannot be explained at all by modern science, especially if we bring into play the recession rate of the Moon paradox.

Tidal locking. That was easy!

Another "paradox". So far, it seems like your "paradoxes" are simply lack of understanding on your part. See the one about elliptical orbits earlier in this post for an example. I doubt I've seen them all, though, so there might be one or more that is an actual paradox.

The very fact that you do see the same side, means it is not rotating at all.

Nope. It means the period of rotation and revolution are the same. This is (or at least used to be) grade-school stuff.

Au contraire. If it has low shear strength (liquids have very low shear strength), material naturally flows to the state where it has the lowest energy, which, in the case of an inverse-square field like gravity, is a sphere. In other words, they wouldn't do otherwise under the influence of gravity alone. Why do you think otherwise?

But it does not.

<ramble about an interesting looking but off-topic paper on theoretical astrophysics and the origin of the universe>

You're getting a little ahead of yourself. The statement my reply addressed presumed planets and satellites are already molten masses. Here's what you asserted:

If planets and satellites were once molten masses, as cosmological theories assume, they would not have been able to obtain a spherical form

Anything about the form gravity would impose on molten bodies? Hint: you already asserted it would be a sphere when you were talking about the Sun.

[Edit] Changed quote to footnote, and clarified comment about pressure, force, and area.

To make the entire discussion crystal clear, here is the official set of data used by modern day astrophysicists:

<Screenshot of web page.>

Fraknoi, Morrison Wolff, Voyages throught the Universe (award-winning astronomy educator (Fraknoi) and two distinguished research scientists (Morrison at NASA and Wolff at NOAO).

<repeat of earlier commentary>

No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.

[nb]Why not just include a link to the page? Here it is: http://csep10.phys.utk.edu/astr162/lect/sun/photosphere.html[/nb]

There is plenty that's not known about the Sun, but what defiance of Newtonian mechanics, specifically, are you asserting? Do you have references to actual scientific papers that clearly state what you claim here?

For those who do not understand the physics involved, let us go directly to the textbook on the subject.

Did you omit something, or are you changing the subject from the Sun to tides on the Earth? If it's the latter, where's the textbook?

The Moon generally has one hemisphere facing the Earth, due to tidal locking.

Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit Earth.

But tides, oceanic and atmospheric have nothing to do with the supposed gravitational force exerted by the Moon.

Atmospheric tides appear to have only a small component associated with the Moon. Ocean tides are clearly influenced by the Moon, and the best current explanation, which is consistent with conventional physics, is that they are driven primarily by the Moon's gravity.

<atmospheric tides stuff>

There are a lot of things about atmospheric science that aren't well understood yet, and may never be; it's a very complex topic. Does that mean all that is known about planetary motion is wrong? No.

Science can't explain everything. No person who is even moderately well educated denies that.

The quote in which you stated clearly that the moment of inertia definition is not related to mass is a clear indication of your level of understanding of basic physics.

It seems like you would have wanted people to forget that you didn't know that mass and moment of inertia are not the same thing.

No such thing ever happened.

https://www.theflatearthsociety.org/forum/index.php?topic=66455.msg1779816#msg1779816

You mean the one that considers mass but not the distribution of that mass? That one is wrong.

Really?

DEFINITION #1.

Angular momentum = cross product of the position vector and the momentum (mv).

This is the definition I chose.


DEFINITION #2.

Angular momentum is proportional to the moment of inertia and the angular speed.


For you to state that one of them is wrong, is ludicrous.


You are a troll.


At each point of our discussion, I must bring up the point that you do not understand plain English.

Donohue's paper doesn't seem to be widely cited.

https://www.google.ro/search?q=W.+Donohue+kepler+fabricated+figures&oq=W.+Donohue+kepler+fabricated+figures&aqs=chrome..69i57.13699j0j8&sourceid=chrome&ie=UTF-8

https://www.google.ro/search?q=W.+Donohue+kepler+fabricated+figures&oq=W.+Donohue+kepler+fabricated+figures&aqs=chrome..69i57.13699j0j8&sourceid=chrome&ie=UTF-8#q=W.+Donohue+kepler+fudged+data


Even if Kepler's 1609 publication is messier and less complete than what would currently be accepted, the fact remains that his elliptical model for orbits works quite well making accurate predictions about real objects.

You still don't get it.

Kepler faked the ENTIRE data, each and every calculation: nothing was related to any kind of an elliptical orbit, no measurement, no astronomical observation; he simply conjured up an elliptical orbit that did not exist.

He faked the entire work Nova Astronomia.

“After detailed computational arguments Donahue concluded the results
reported by Kepler . . . were not at all based on Brahe’s observational data; rather
they were fabricated on the basis of Kepler’s determination that Mars’s orbit was
elliptical.

Thus, anyone who cared to check Kepler’s tables would find
numbers that are consistent with the elliptical orbit [he] postulated for Mars and
would be inclined to believe that the numbers represented observational data. In
fact, they were computed from the hypothesis of an elliptical orbit and then
modified for measurement error; such data, if they were truly observations, would
be prime facie evidence of the theories’ correctness.


No such thing as an elliptical orbit: Kepler was forced to fake, fudge, invent a totally new set of data, to push forward his heliocentrical agenda.


The fabricated data appear in calculated positions for the planet Mars, which Kepler used as a case study for all planetary motion. Kepler claimed the calculations gave his elliptical theory an independent check. But in fact they did nothing of the kind.

''He fudged things,'' Dr. Donahue said, adding that Kepler was never challenged by a contemporary. A pivotal presentation of data to support the elliptical theory was ''a fraud, a complete fabrication,'' Dr. Donahue wrote in his paper. ''It has nothing in common with the computations from which it was supposedly generated.''


Thus, the notion that a planet orbits the Sun in an elliptical orbit was a simple fabrication, based on fudged data.



Here is the most basic demonstration that indeed an elliptical orbit for a planet would be impossible.


No one can explain how a planet could be captured at either the aphelion or the perihelion points on an ellipse.

The ellipse, as a logical-looking orbit, falls apart from the very start.

"But, let's build that ellipse again, starting from aphelion. Let us draw the whole thing, just accepting that an ellipse must somehow be created, since we have evidence of them in the solar system. Finally, let us look for the "equivalent" circular orbit.


Meaning that if we have the same planet with the same initial velocity and we want to put it into a circular orbit, where do we put it? Turns out that the circle is completely outside the ellipse, and that it has a lot greater area. Remember that the only way we can explain the planet in ellipse beginning to dive toward the sun as we move it past aphelion is that its velocity is not great enough to keep it in circular orbit."


Remember, we find ourselves AT THE APHELION POINT WHERE THE GRAVITATIONAL FIELD CANNOT INFLUENCE THE TANGENTIAL VELOCITY OF THE PLANET.

That is, the Sun could not capture another planet at either the perihelion or the aphelion points, the simplest and most obvious way to describe the capture theory.

"Therefore, to put it into a stable circular orbit, we must move it further away from the sun at aphelion. If we do that then this distance becomes the radius of the circle, and we have our circular orbit. As you can see from this simple illustration, the path of the ellipse never crosses the path of the "equivalent" circle. If that is true, then the planet in ellipse can never reach a point where its perpendicular velocity overcomes the centripetal acceleration produced by the gravitational field. It never achieves a temporary escape velocity. No, it simply spirals into the sun. Its orbital velocity increases, yes. The "orbital velocity" continues to increase until the planet burns up in the sun's corona."



This is the simplest demonstration which at the present time no physicist can dispute.

There are other more involved demonstrations, starting here:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1776706#msg1776706


Your drivel on the mass/weight paragraph amounts to nothing at all.



Professor John Cox, McGill University


How about the calculations needed to determine magnetic field strength needed to overcome gravity and levitate, say, a train?

You seem not to understand the history of newtonian mechanics.

From the very first publication of the Principia, electromagnetism was never allowed to enter any kind of a calculation purporting to planetary orbit computation.

No matter how much astrophysicists would like to connect plasma physics/electromagnetism with gravity they simply cannot: the definition attributed to Newton is the only one allowed to stand in the textbooks.

IF magnetism was allowed to enter calculations then the entire paradigm of science based on attractive gravity would be obliterated at once.


This is the third time you have FAILED to provide an explanation for the barometric pressure paradox, a total defiance of tidal theory.

"It has been known now for two and a half centuries, that there are more or less daily variations in the height of the barometer, culminating in two maxima and two minima during the course of 24 hours. The same observation has been made and puzzled over at every station at which pressure records were kept and studied, but without success in finding for it the complete physical explanation."

NATIONAL WEATHER SERVICE DATA:


The most basic change in pressure is the twice daily rise and fall in due to the heating from the sun. Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m. The magnitude of the daily cycle is greatest near the equator decreasing toward the poles.

http://oceanservice.noaa.gov/education/yos/resource/JetStream/atmos/pressure.htm

Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m.


BAROMETER PRESSURE PARADOX

One maximum is at 10 a.m., the other at 10 p.m.; the two minima are at 4 a.m. and 4 p.m.

The heating effect of the sun can explain neither the time when the maxima appear nor the time of the minima of these semidiurnal variations.

If the pressure becomes lower without the air becoming lighter through a lateral expansion due to heat, this must mean that the same mass of air gravitates with changing force at different hours.


Lord Rayleigh: ‘The relative magnitude of the latter [semidiurnal variations], as observed at most parts of the earth’s surface, is still a mystery, all the attempted explanations being illusory.’



Currently, the barometer pressure paradox CANNOT BE EXPLAINED AT ALL.

It would be very easy to calculate the total centrifugal force exerted on the photosphere and the cromosphere, once we had at our disposal the total mass of each such layer of the sun.

Then, we would need the angular velocity of the sun and the corresponding radius.

The TOTAL force would be huge.

I tried to find the total mass for the photosphere and cromosphere, but none were available, even on the Nasa websites.


It would be a straightforward calculation: the total force is enormous.

In fact, here is an online centrifugal force calculator:

http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html




The pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.

"The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth."


Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.


No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.

Let us now get an estimate of this centrifugal force.

Photosphere/Chromosphere density:

http://www.swpc.noaa.gov/sites/default/files/images/u33/Chapter_2_.pdf

Volume of a spherical shell:

http://www.emathzone.com/tutorials/geometry/spherical-shell.html

Radius photosphere: 700,000 km = 700,000,000 m (outer)

Radius chromosphere: 705,400 km = 705,400,000 m (outer)

Thickness of photosphere shell = 400 km

Thickness of cromosphere shell = 5000 km


Estimated mass of the chromosphere:

33,507,780,986,235,477.5 kg

Estimated mass of the photosphere:

2,463,007,232,981,157.17 kg


Sun rotation speed, equator:

https://books.google.ro/books?id=BrbSAgAAQBAJ&pg=PA14&lpg=PA14&dq=sun+surface+rotation+speed+1.9+km/s&source=bl&ots=J33J1iUt43&sig=UMTba3SpbIKTh0owe3_wHKT68D0&hl=ro&sa=X&ved=0ahUKEwjF8vDY8bjNAhWJhywKHX12AKUQ6AEIVjAG#v=onepage&q=sun%20surface%20rotation%20speed%201.9%20km%2Fs&f=false


http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html

Total centrifugal force exerted on the photosphere:

12,702,080,158,659.96 N


Total centrifugal force exerted on the chromosphere:

171,481,555,656,804.75 N


Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.


No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.

Sandokhan hasn't even read the paper by Donohue, which clearly states that Kepler DIDN'T fake the ENTIRE data.

Kepler DIDN'T fake the ENTIRE data.

This is hilarious!

You are ADMITTING that Kepler faked his data?

Here is Dr. W. Donahue's celebrated paper on Kepler's humongous, total faking of his data:

http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1988JHA....19..217D&classic=YES

Nova Astronomia is the very basis of modern astrophysics.

The data on Mars is the foundation of the entire work, Nova Astronomia.


THE DATA ON MARS WAS TOTALLY FAKED/FUDGED BY KEPLER, the entire chapter 53, the very basis of the whole treatise.


The fudging is terrifying, unbelievable, sickening.


Kepler faked and fudged each and every number used in his graphics, tables to INVENT an elliptical orbit that did not exist in reality.


http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1988JHA....19..217D&db_key=AST&page_ind=17&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES

Donahue asks:

How could Kepler perpetrate such a patent fraud?


The fabricated data appear in calculated positions for the planet Mars, which Kepler used as a case study for all planetary motion. Kepler claimed the calculations gave his elliptical theory an independent check. But in fact they did nothing of the kind.

''He fudged things,'' Dr. Donahue said, adding that Kepler was never challenged by a contemporary. A pivotal presentation of data to support the elliptical theory was ''a fraud, a complete fabrication,'' Dr. Donahue wrote in his paper. ''It has nothing in common with the computations from which it was supposedly generated.''

But when Dr. Donahue started working through the method to make sure he understood the basis for Kepler's chart, he found his numbers disagreeing with those of the great astronomer. After repeatedly getting the wrong answers for the numbers displayed on Kepler's chart, Dr. Donahue started trying other methods. Finally, he realized that the numbers in the chart had been generated not by independent calculations based on triangulated planetary positions, but by calculations using the area law itself.

''He was claiming that those positions came from the earlier theory,'' Dr. Donahue said. ''But actually all of them were generated from the ellipse.''

“After detailed computational arguments Donahue concluded the results
reported by Kepler . . . were not at all based on Brahe’s observational data; rather
they were fabricated on the basis of Kepler’s determination that Mars’s orbit was
elliptical. Donahue reasons that Kepler must have gone back to revise his earlier
calculations that were made prior to his understanding that the orbit of Mars was
actually elliptical. Thus, anyone who cared to check Kepler’s tables would find
numbers that are consistent with the elliptical orbit [he] postulated for Mars and
would be inclined to believe that the numbers represented observational data. In
fact, they were computed from the hypothesis of an elliptical orbit and then
modified for measurement error; such data, if they were truly observations, would
be prime facie evidence of the theories’ correctness.


The entire Nova Astronomia is a SCIENTIFIC FRAUD, no elliptical orbit was measured, observed at all, the data is all fake.

This is hilarious!

You are ADMITTING that Kepler faked his data?

Here is Dr. W. Donahue's celebrated paper on Kepler humongous total faking of his data:

http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1988JHA....19..217D&classic=YES

Nova Astronomia is the very basis of modern astrophysics.

The data on Mars is the foundation of the entire work, Nova Astronomia.


THE DATA ON MARS WAS TOTALLY FAKED/FUDGED BY KEPLER, the entire chapter 53, the very basis of the whole treatise.


The fudging is terrifying, unbelievable, sickening.


Kepler faked and fudged each and every number used in his graphics, tables to INVENT an elliptical orbit that did not exist in reality.


http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1988JHA....19..217D&db_key=AST&page_ind=17&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES

Donahue asks:

How could Kepler perpetrate such a patent fraud?


The fabricated data appear in calculated positions for the planet Mars, which Kepler used as a case study for all planetary motion. Kepler claimed the calculations gave his elliptical theory an independent check. But in fact they did nothing of the kind.

''He fudged things,'' Dr. Donahue said, adding that Kepler was never challenged by a contemporary. A pivotal presentation of data to support the elliptical theory was ''a fraud, a complete fabrication,'' Dr. Donahue wrote in his paper. ''It has nothing in common with the computations from which it was supposedly generated.''

But when Dr. Donahue started working through the method to make sure he understood the basis for Kepler's chart, he found his numbers disagreeing with those of the great astronomer. After repeatedly getting the wrong answers for the numbers displayed on Kepler's chart, Dr. Donahue started trying other methods. Finally, he realized that the numbers in the chart had been generated not by independent calculations based on triangulated planetary positions, but by calculations using the area law itself.

''He was claiming that those positions came from the earlier theory,'' Dr. Donahue said. ''But actually all of them were generated from the ellipse.''

“After detailed computational arguments Donahue concluded the results
reported by Kepler . . . were not at all based on Brahe’s observational data; rather
they were fabricated on the basis of Kepler’s determination that Mars’s orbit was
elliptical. Donahue reasons that Kepler must have gone back to revise his earlier
calculations that were made prior to his understanding that the orbit of Mars was
actually elliptical. Thus, anyone who cared to check Kepler’s tables would find
numbers that are consistent with the elliptical orbit [he] postulated for Mars and
would be inclined to believe that the numbers represented observational data. In
fact, they were computed from the hypothesis of an elliptical orbit and then
modified for measurement error; such data, if they were truly observations, would
be prime facie evidence of the theories’ correctness.


The entire Nova Astronomia is a SCIENTIFIC FRAUD, no elliptical orbit was measured, observed at all, the data is all fake.

I thought I had ALREADY said that I think it's possible that he did. Uh... yes, I have. And it still means very little.

"The fudging is terrifying, unbelievable, sickening."


HAHAHAHAHAHAHA!!! This is hilarious! No, your logic is sickening. Kepler realized that elliptical orbits made sense, and he formed a hypothesis. He tried to INDEPENDENTLY VERIFY it, but it wasn't very easy, so he probably faked the verification, not the "basis" of it, just like many other scientists back then did. Eventually, it didn't matter that much, because his hypothesis was later verified anyway. Even Donahue agrees that it's not that serious. That's only IF Donahue is right.

"The entire Nova Astronomia is a SCIENTIFIC FRAUD, no elliptical orbit was measured, observed at all, the data is all fake."


It's not, it has been explained to you countless times. The only thing that was probably faked was the independent verification of the hypothesis, and that doesn't even matter any more, because it's been verified now. You're 400 years late.

No wonder you are a RE.

Your pretzel logic is noted.

Kepler realized that elliptical orbits made sense.

HOW in the world could he have realized such a fact, when he had NOTHING going for him to form such a wild hypothesis?

Where is the astronomical data which could have supported such a theory?

There is none whatsoever.


There is a very basic reason why Kepler chose ellipses and it has nothing to do with observational data, but with sun worship.

Who alone appears, by virtue of his dignity and power, suited…and worthy to become the home of God himself, not to say the first mover” (On the Motion of Mars, Prague, 1609, Chapter 4 - Kepler describing the Sun).

Tycho Brahe's observations/data were based on the concept of epicycles.

Such a model excluded from the very start the concept of attractive gravitation which the conspirators had in mind to try to fool the entire world.

That is why they chose the mathematical equivalent of an epicycle: the ellipse.

But there was no real data on which to base such a hypothesis, so Kepler had to fake the ENTIRE data on Mars, the basis of Nova Astronomia.


Don't worry, Newton chose to do the same thing himself.

There is no such as thing as Newtonian mechanics.

Better yet, it should be called INDIAN MECHANICS.


Because Newton copied his laws of motion from the Naya Vaisesika Sutra.

https://archive.org/stream/thevaiasesikasut00kanauoft#page/n7/mode/2up

The force on a body is the resultant of gravity and the work done against it. V.S 5.1.13

In the absence of all other forces gravity exists. V.S 5.1.7

Action is opposed by an equivalent opposite reaction - V.S 5.1.16-18

Newton's laws of motion copied from the Naya Vaiseshika Sutra.

Suppose that the mass of an object is 'm' and in time interval 't', the velocity of the object changes from 'u' to 'v' due to the force acting on it. Then,

Initial momentum = mu
Final momentum = mv
Change in momentum = m(v-u)

Therefore, the rate of change of momentum = m(v-u)/t = ma (from Kanada's first law)

From Kandas second law,
force is proportional to the rate of change of momentum.
Or, p k ma
Or, p = kma (where k is a constant)

If m=1 and a=1, then
1 = k*1*1 or k = 1
Or, p = ma

Therefore, unit force is the one that produces unit acceleration in an object of unit mass.

Prashastpada

http://www.abovetopsecret.com/forum/thread120045/pg1 (translation of Naya Vaiseshika Sutra verses on mechanics)


And Newton and Leibniz did even more: THEY COPIED ALL OF THEIR RESULTS IN CALCULUS FROM THE INDIAN TEXTS, without understanding what they were doing.


http://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1574605#msg1574605


Origin of Calculus: How Mathematical Analysis Was Imported to India, Italy, France and England

http://www.hinduwisdom.info/Yuktibhasa.pdf

A relevant epistemological question is this: did Newton at all understand the result he is alleged to have invented? Did Newton have the wherewithal, the necessary mathematical resources, to understand infinite series? As is well known, Cavalieri in 1635 stated the above formula (the infinite series expansion for the sine function) as what was later termed a conjecture. Wallis, too, simply stated the above result, without any proof. Fermat tried to derive the key result above from a result on figurate numbers, while Pascal used the famous “Pascal’s” triangle long known in India and China. Though Newton followed Wallis, he had no proof either, and neither did Leibniz who followed Pascal. Neither Newton nor any other mathematician in Europe had the mathematical wherewithal to understand the calculus for another two centuries, until the development of the real number system by Dedekind.

The next question naturally is this: if Newton and Leibniz did not quite understand the calculus, how did they invent it? In the amplified version of the usual narrative, how did Galileo, Cavalieri, Fermat, Pascal, and Roberval etc. all contribute to the invention of a mathematical procedure they couldn’t quite have understood? The frontiers of a discipline are usually foggy, but here we are talking of a gap which is typically 250 years.


Here is another step by step demonstration:

Other pieces of circumstantial evidence include:

James Gregory, who first stated the infinite series expansion of the arctangent (the Madhava-Gregory series) in Europe, never gave any derivation of his result, or any indication as to how he derived it, suggesting that this series was imported into Europe.

http://www.muslimheritage.com/article/kerala-mathematics-and-its-possible-transmission-europe


This is RE science at its best: Kepler fudged his entire work, Nova Astronomia, and Newton copied his "laws" of motion from Indian sutras.


Because it's been verified now.

But it hasn't been verified, even at least once, in the past 400 years.

We have been given a fantasy world to play in, with no connection to the real world.

The Biefeld-Brown effect precisely defies newtonian mechanics.

Maxwell's original set of equations prove the existence of ether.

The Miller and Galaev experiments prove the existence of ether.

The missing orbital Sagnac effect proves the Earth is stationary.

The Allais effect defies newtonian mechanics.


Newton simply copied his "laws" from indian sutras.

Kepler simply faked his entire set of data to match an ellipse.

RE science at its "best".

sandokhan's post is, as often, pretty long and rambles quite a bit. Here's the first part - an irrelevancy he'd be better off avoiding with an ad-hom thrown in for good measure.
 

It seems like you would have wanted people to forget that you didn't know that mass and moment of inertia are not the same thing.

No such thing ever happened.

Really?

The mass of a body and the rate at which it spins, in classical physics, determines an object's "angular momentum."

You clearly used mass when the proper parameter is moment of inertia. It would really be better for everyone - especially you - if you just admitted "oops..." and moved on. It happens. No biggie. Next best would be to just let it drift into oblivion. Keep bringing it up, and I'll keep reminding you (and everyone else reading) what you actually said. It's up to you.

https://www.theflatearthsociety.org/forum/index.php?topic=66455.msg1779816#msg1779816

Note all that bluster that was after your obvious goof was shown to you.

At each point of our discussion, I must bring up the point that you do not understand plain English.

"Must." LOL! You're confusing want with need. You keep bringing it up because your arguments are weak and you want to throw in a gratuitous insult because you can't do much else.

Next: the continuing saga of "Kepler is a Poopyhead", or "What sandokhan Thinks He Knows About Elliptical Orbits is... um... [impolite word]"

No mistake on my part.

There are two definitions in use for the angular momentum.

I used this one:

DEFINITION #1.

Angular momentum = cross product of the position vector and the momentum (mv).

This is the definition I chose.


DEFINITION #2.

Angular momentum is proportional to the moment of inertia and the angular speed.


For you to state that one of them is wrong, is ludicrous.


On the other hand, you had no idea that the definition of the moment of inertia is based on the concept of mass.


This is what you wrote:

An object's angular momentum is determined by its moment of inertia, not mass, and rate of spin.


The troll had no idea that the moment of inertia is defined in terms of mass.

No idea whatsoever.


When this fact was brought to his attention, during the course of the same day, he wrote:

Mass is a factor, yes.


Why am I bringing this up each and every time?

So that everyone can see at a glance the total fakery of your messages: no matter which thread you choose to shitpost, you start to point the way forward, to tell others how wrong they are... and yet, here you are not knowing the most basic facts of physics.

Again, this is what you wrote:

An object's angular momentum is determined by its moment of inertia, not mass, and rate of spin.

... continued from previous.

Donohue's paper doesn't seem to be widely cited.

https://www.google.ro/search?q=W.+Donohue+kepler+fabricated+figures&oq=W.+Donohue+kepler+fabricated+figures&aqs=chrome..69i57.13699j0j8&sourceid=chrome&ie=UTF-8

https://www.google.ro/search?q=W.+Donohue+kepler+fabricated+figures&oq=W.+Donohue+kepler+fabricated+figures&aqs=chrome..69i57.13699j0j8&sourceid=chrome&ie=UTF-8#q=W.+Donohue+kepler+fudged+data

You did get several pages of search-engine hits, and I didn't explicit say "cited in scholarly publications", so I'll concede that point.

Let's look at one of the first Google hits. It lists that paper by Donahue in the bibliography, so it showed up in the results, but I don't see any actual citation from that paper in the text; I could have missed it, but the tone of the page doesn't seem to concur with your conclusions from the Donohue paper.

It did say this at the end of the section titled The 'War with Mars':

The actual process of calculation for Mars was immensely laborious - there are nearly a thousand surviving folio sheets of arithmetic - and Kepler himself refers to this work as 'my war with Mars', but the result was an orbit which agrees with modern results so exactly that the comparison has to make allowance for secular changes in the orbit since Kepler's time.

This section is immediately followed by another aspect of the topic you seem to have difficulty grasping and may be the root source of your outrage:
 

Observational error

It was crucial to Kepler's method of checking possible orbits against observations that he have an idea of what should be accepted as adequate agreement. From this arises the first explicit use of the concept of observational error. Kepler may have owed this notion at least partly to Tycho, who made detailed checks on the performance of his instruments (see the biography of Brahe).

Even if Kepler's 1609 publication is messier and less complete than what would currently be accepted, the fact remains that his elliptical model for orbits works quite well making accurate predictions about real objects.

You still don't get it.

Kepler faked the ENTIRE data, each and every calculation: nothing was related to any kind of an elliptical orbit, no measurement, no astronomical observation; he simply conjured up an elliptical orbit that did not exist.

<more ranting about Kepler>

He faked the entire work Nova Astronomia.

<more ranting>

Thus, the notion that a planet orbits the Sun in an elliptical orbit was a simple fabrication, based on fudged data.

While I disagree with your assessment of Kepler's work, it really doesn't matter. Your whole premise boils down to faulty logic.

Stripped of the excessive verbiage, you're saying "Kepler cheated when he calculated that Mars follows an elliptical orbit with the Sun at one focus. Therefore Mars does not follow an elliptical orbit with the Sun at one focus."

That's an obvious non-sequitur. Even if the first part is true, it doesn't mean the second part is true.

[I see that you're going on about this while this post is being composed. I'll look at your new post at some point and see if there's anything new and meaningful.]

Unfortunately (for you), using Kepler's elliptical orbits as the basis for predicting the positions of orbiting bodies (including Mars) works exceedingly well in reality, and is used successfully every day.

Here is the most basic demonstration that indeed an elliptical orbit for a planet would be impossible.

No one can explain how a planet could be captured at either the aphelion or the perihelion points on an ellipse.

How can you know this? Have you heard from everyone who is an expert in orbital mechanics, without exception, that they cannot explain this? Even if that were true, did you know there are an infinite number of points between aphelion and perihelion along an elliptical orbit?

What follows is the copy-paste from sandokhan's post in Flat-Earth Believers that he included in his post. The link to the original has been added to the quote block here for the convenience of the reader.

The ellipse, as a logical-looking orbit, falls apart from the very start.

"But, let's build that ellipse again, starting from aphelion. Let us draw the whole thing, just accepting that an ellipse must somehow be created, since we have evidence of them in the solar system. Finally, let us look for the "equivalent" circular orbit.

Meaning that if we have the same planet with the same initial velocity and we want to put it into a circular orbit, where do we put it? Turns out that the circle is completely outside the ellipse, and that it has a lot greater area. Remember that the only way we can explain the planet in ellipse beginning to dive toward the sun as we move it past aphelion is that its velocity is not great enough to keep it in circular orbit."

Remember, we find ourselves AT THE APHELION POINT WHERE THE GRAVITATIONAL FIELD CANNOT INFLUENCE THE TANGENTIAL VELOCITY OF THE PLANET.

Gravity is normal to the tangential velocity at that point, so the magnitude of the tangential velocity doesn't change at that instant. It does continuously influence the tangential velocity by constantly changing its direction. Once the planet has passed aphelion, gravity is no longer normal to the tangential velocity, so both its magnitude and direction change.

That is, the Sun could not capture another planet at either the perihelion or the aphelion points, the simplest and most obvious way to describe the capture theory.

As already noted, those are simply two points out of an infinite number. There is no reason to assume that those are the only places where capture could or could not occur.

Your argument is based on an oversimplification, however. In order for two bodies to be captured into a closed (elliptical or circular) orbit from an open (hyperbolic or parabolic) orbit, it takes the presence of at least one other body to perturb the open orbit. There is no requirement that the capture occur at any particular point on the ellipse; the size, shape (eccentricity), and orientation of the ellipse will depend on the initial conditions (the state vector - location and velocity of one body relative to the other) at the time of capture.

Two-body orbits are elegant, orderly, and relatively simple. Three-body, and, more generally, N-body, orbital mechanics are messy, sometimes chaotic, and can be very complex. If you get away from your computer long enough to go outside and actually look at what's in the night sky, you can see that there are far more than just two bodies in the universe.

"Therefore, to put it into a stable circular orbit, we must move it further away from the sun at aphelion. If we do that then this distance becomes the radius of the circle, and we have our circular orbit. As you can see from this simple illustration, the path of the ellipse never crosses the path of the "equivalent" circle."

The size, shape, and orientation of the ellipse will depend on the state vector at the time of capture. Note that a circle is simply the special case of an ellipse with zero eccentricity.

[Quotation in original text continues]

"If that is true, then the planet in ellipse can never reach a point where its perpendicular velocity overcomes the centripetal acceleration produced by the gravitational field. It never achieves a temporary escape velocity. No, it simply spirals into the sun. Its orbital velocity increases, yes. The "orbital velocity" continues to increase until the planet burns up in the sun's corona."

That statement is incorrect. As the planet moves from aphelion to perihelion its speed increases and reaches maximum at perihelion. At perihelion its tangential velocity is too high to be in circular orbit (the "equivalent" circular orbit would be smaller, and contained completely within the ellipse) so the distance to the Sun starts to increase and the speed drops until it returns to aphelion. No "death spiral".

http://i113.photobucket.com/albums/n206/dharanis1/sothic3_zps7k5nssjg.jpg

This is the simplest demonstration which at the present time no physicist can dispute.

This is a clear demonstration that whoever wrote that unattributed quote knows little about the subject. If you believe what it says, that demonstrates that you know little about the subject.

There are other more involved demonstrations, starting here:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1776706#msg1776706

From that post:



"v₁ = v₂"

Not in an elliptical orbit. No need to go on.

Next: "Mass, Weight, Magnetism, and Gravity... Can't They Just Get Along?"

v1 does equal v2 at the aphelion and at the perihelion in an elliptical orbit.

This is why nobody can explain how the innate motion of a planet can change by magic to suit the faked/fudged elliptical orbit falsified by Kepler.

"The orbiter must retain its innate motion throughout the orbit, no matter the shape of the orbit. If it did not, then its innate motion would dissipate. If it dissipated, the orbit would not be stable. Therefore, the orbiter always retains its innate motion over each and every differential. If we take the two most important differentials, those at perihelion and aphelion, and compare them, we find something astonishing. The tangential velocities due to innate motion are equal, meaning that the velocity tangent to the ellipse is the same in both places. But the accelerations are vastly different, due to the gravitational field. And yet the ellipse shows the same curvature at both places."

To make the ellipse work, you have to vary not only the orbital velocity, but also the tangential velocity. To get the correct shape and curvature to the orbit, you have to vary the object's innate motion. But the object's innate motion cannot vary. The object is not self-propelled. It cannot cause forces upon itself, for the convenience of theorists or diagrams. Celestial bodies have one innate motion, and only one, and it cannot vary.


You have failed to address the fact that the Sun could not have captured any planet, either at the aphelion or the perihelion points.


Again, do you understand English?


AT BOTH THE APHELION AND AT THE PERIHELION POINTS, THERE IS NO GRAVITATIONAL INTERACTION BETWEEN THE SUN AND THE TANGENTIAL VELOCITY OF THE PLANET.


Can you understand this much?


HOW then would the sun capture any planet at these points? It cannot.

That is why the elliptical orbit model falls apart from the very start.

There is no reason to assume that those are the only places where capture could or could not occur.

These are the MOST SIMPLE points one can investigate.

If these two points fail, the whole scheme falls apart.

Stop wasting everybody's time with your massive shitposts.

YOU HAVE FAILED TO ADDRESS THE POINTS RAISED IN MY MESSAGE.

v1 does equal v2 at the aphelion and at the perihelion in an elliptical orbit.

Nope.
https://www.quora.com/Why-does-an-orbiting-body-move-faster-around-the-perihelion-and-slow-down-near-the-aphelion
http://curious.astro.cornell.edu/about-us/41-our-solar-system/the-earth/orbit/85-how-fast-does-the-earth-go-at-perihelion-and-aphelion-intermediate
http://cseligman.com/text/history/kepler2.htm

... continued from previous.

Your drivel on the mass/weight paragraph amounts to nothing at all.

http://i113.photobucket.com/albums/n206/dharanis1/auro_zpsmpjrmuvs.jpg

Given the snarky response it looks like you still haven't figured out why a force can't be compared to a pressure without knowing something about area, or why the distinction between force and mass is even important. Oh well... all I can do is try.

Let's see... in sandokhan physics^TM:

mass = moment of inertia
weight = mass
ergo
weight = moment of inertia

Right? It's what you were claiming until sometime in May, or did mass change from being the same as moment of inertia to being the same as weight at the same time?

You CANNOT bring magnetism into any kind of discussion involving gravity.

How about the calculations needed to determine magnetic field strength needed to overcome gravity and levitate, say, a train?

You seem not to understand the history of newtonian mechanics.

You seem to be tap dancing around the question.

Before you go on and on about what you think I don't know or don't understand, would you mind explaining how to solve that problem since it appears to involve both gravity (which along with its mass determines the weight of the train) and magnetism (to determine the repulsive force necessary to offset the weight). I don't see why this cannot be done. After all, you're just comparing forces.

From the very first publication of the Principia, electromagnetism was never allowed to enter any kind of a calculation purporting to planetary orbit computation.

No matter how much astrophysicists would like to connect plasma physics/electromagnetism with gravity they simply cannot: the definition attributed to Newton is the only one allowed to stand in the textbooks.

As I understand it, research into a unified theory that includes gravity and the other fundamental forces is still being worked on but has not been successfully accomplished yet. Not being able to unify electromagnetism with gravity is a technical issue (as in, no one has found a model that works), not doctrine.

IF magnetism was allowed to enter calculations then the entire paradigm of science based on attractive gravity would be obliterated at once.

That's a stretch.

"Allowed". LOL. If an effect is so small that it makes no practical difference, it can be (and usually is) ignored. When it makes a significant difference it has to be considered.

If you think magnetism is not used in models of the Sun, you need to be finding better material to read.

This is the third time you have FAILED to provide an explanation for the barometric pressure paradox

I don't know enough about atmospheric science to explain how the atmosphere behaves in detail. I've never claimed otherwise.

a total defiance of tidal theory.

Before you can say something defies theory, you must be familiar with what the theory predicts.

So, given the accepted values for the distance to and mass of the Moon (that's probably a good enough starting point; add in the Sun and other bodies if you think they're significant), size and mass of the Earth, and properties of the atmosphere (density as a function of altitude comes to mind as probably the most significant). Please calculate what you think what tidal theory predicts, and show your calculations and supporting data. Please state any assumptions you are making and the values you're using, and show your work. If using a value that is not widely known, please show the source.

How does this effect compare with measured data and other known effects like diurnal heating and cooling of the atmosphere and confounding factors like uneven distribution of heat which causes circulation?

Until you can clearly show there is a problem, why should we believe you just because you say there's one? Your "paradoxes" are a dime a dozen[nb]Since this is an international forum and regionalisms can sometimes be perplexing, "a dime a dozen" means "so common they're not worth much".[/nb].
 

You still don't get it.

Lord Rayleigh has performed each and every calculation you required a long time ago, in order to get to the bottom of the barometric pressure paradox.

Here are his conclusions:

‘The relative magnitude of the latter [semidiurnal variations], as observed at most parts of the earth’s surface, is still a mystery, all the attempted explanations being illusory.’

The barometric pressure paradox cannot be explained at all by modern science: a total defiance not only of attractive gravitation but also of tidal theory (atmospheric tides).

BAROMETER PRESSURE PARADOX

One maximum is at 10 a.m., the other at 10 p.m.; the two minima are at 4 a.m. and 4 p.m.

The heating effect of the sun can explain neither the time when the maxima appear nor the time of the minima of these semidiurnal variations.

If the pressure becomes lower without the air becoming lighter through a lateral expansion due to heat, this must mean that the same mass of air gravitates with changing force at different hours.


I have calculated the required centrifugal force for both the photosphere and the cromosphere:

https://www.theflatearthsociety.org/forum/index.php?topic=67119.msg1792328#msg1792328

Total centrifugal force exerted on the photosphere:

12,702,080,158,659.96 N


Total centrifugal force exerted on the chromosphere:

171,481,555,656,804.75 N


Do you understand now why magnetism cannot be even mentioned by modern science in relation to gravity?

If magnetism is brought into play to explain how the photosphere and the cromosphere do not disperse at once in space, then given these huge factors, ALL NEWTONIAN LAWS PERTAINING TO THE SCIENCE OF GRAVITY WILL HAVE TO BE MODIFIED DRASTICALLY.

Then the "nuclear furnace" sun becomes an ELECTRIC sun.

Then the big bang scenario flies out the window.

Then we can explain the faint young sun paradox.

And certainly the shape of the Earth will be brought into the limelight, since magnetism cannot be invoked to explain how one trillion billion gallons of water stay glued next to the surface of a sphere.

frenat... you have forgotten about the missing orbital Sagnac effect.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1782182#msg1782182

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1784780#msg1784780

That is why your links amount to nothing.

Ask the people there to explain the fact that GPS satellites DO NOT REGISTER the orbital Sagnac effect: you won't get any answer back.

frenat... you have forgotten about the missing orbital Sagnac effect.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1782182#msg1782182

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1784780#msg1784780

That is why your links amount to nothing.

Ask the people there to explain the fact that GPS satellites DO NOT REGISTER the orbital Sagnac effect: you won't get any answer back.

I'm not responding to anything about an orbital Sagnac effect nor do I care.  I simply responded to the fact that velocity at aphelion is NOT equal to velocity at perihelion.

Was it really necessary to separately link two post that immediately followed each other?

Do not expect a serious answer. All sandokhan can do is copy and paste things he does not even read himself.
When I pointed out that he even contradicts himself in all his copy and pasting, he chose to ignore the matter (see https://www.theflatearthsociety.org/forum/index.php?topic=66738.30 )

It would be very easy to calculate the total centrifugal force exerted on the photosphere and the cromosphere, once we had at our disposal the total mass of each such layer of the sun.

It's probably not as easy as you think, but, fortunately, it isn't necessary.

Then, we would need the angular velocity of the sun and the corresponding radius.

The TOTAL force would be huge.

I tried to find the total mass for the photosphere and cromosphere, but none were available, even on the Nasa websites.

It's not needed for analysis, anyway.

It would be a straightforward calculation: the total force is enormous.

Enormous? So is the Sun. Straightforward? Maybe not.

In fact, here is an online centrifugal force calculator:

http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html

I have a calculator and know the formula, but it's nice to have a handy tool to check the answer, so thanks!

Since we're really interested in the comparison of centripetal acceleration to the acceleration of gravity, instead of trying to determine the centripetal force on the entire photosphere all at once, which isn't trivial but not really meaningful, anyway, calculate the centripetal acceleration at, say, the equator, since it's greatest there. From that you can calculate the centripetal force on a given mass (1 kg might be convenient) if you want.

So...
Acceleration of gravity at photosphere
g_sol = (27.47)(9.8 m/sec²) [nb]Using your number 27.47 for the ratio of the weight of one gram mass on the Sun and on the Earth.[/nb]
g_sol = 269 m/s²

Radius at top of photosphere
R_sol = 700,000 km = 700,000,000 m [given below]

Rotation period (equator, sidereal)
Oops... the figure originally used was the synodic period of rotation, not the intended sidereal period of rotation, both given in sandokhan's reference. This error produced an equatorial speed about 8% too small and a centripetal acceleration about 13% too small (both percentages lessened in the calculations because of rounding).

T_sol = 26.24 day[nb]https://books.google.ro/books?id=BrbSAgAAQBAJ&pg=PA14&lpg=PA14&dq=sun+surface+rotation+speed+1.9+km/s&source=bl&ots=J33J1iUt43&sig=UMTba3SpbIKTh0owe3_wHKT68D0&hl=ro&sa=X&ved=0ahUKEwjF8vDY8bjNAhWJhywKHX12AKUQ6AEIVjAG#v=onepage&q=sun%20surface%20rotation%20speed%201.9%20km%2Fs&f=false (your reference)[/nb]
= (26.24 day) (86400 sec/day)
T_sol = 2,267,000 sec

T_sol = 24.47 day[nb]ibid.[/nb]
= (24.47 day) (86400 sec/day)
T_sol = 2,114,000 sec

V_sol = 2 pi R_sol / T_sol
= 2 pi 700,000,000 m / 2,267,000 s
= 4,398,000,000 m / 2,267,000 s
V_sol = 1940 m / s (call it 2000 m/s)

= 2 pi 700,000,000 m / 2,114,000 s
= 4,398,000,000 m / 2,114,000 s
V_sol = 2080 m / s (call it 2100 m/s)

The original calculation here produced an answer too large by two orders of magnitude; this was possibly from anticipating the conversion to percent, but I'm really not sure. Expanding the steps this time.
Centripetal acceleration
a_c = v² / r
a_c = 0.57 m/s²

a_c = (2100 m/s)² / 700,000,000 m
= 4,410,000 m²/s² / 700,000,000 m
a_c = 0.0063 m/s²

a_c / g_sol = 0.57 m/s² / 269 m/s²
a_c / g_sol = 0.00212

a_c / g_sol = 0.0063 m/s² / 269 m/s²[/s]
a_c / g_sol = 0.00002342

Centripetal acceleration is only 1/42,700 the acceleration of gravity.

So centripetal acceleration at the top of the photosphere at the solar equator is 0.212% 0.002342% of the acceleration of gravity at the equator. It would be less elsewhere.

Force of gravity on 1 kg at the photosphere on the equator = (1 kg) (269 m/s²)
 = 269 newtons.
Centripetal force on 1 kg at the photosphere on the equator = (1 kg) (0.57 m/s²)
 = 0.57 newtons.

Centripetal force on 1 kg at the photosphere on the equator = (1 kg) (0.0063 m/s²)
 = 0.0063 newtons.

Meh.

I don't see that exactly ripping the Sun apart or causing it to form a disk. That's orders of magnitude less than the effect on the Earth (0.3% if I recall correctly), but the Sun is probably "squishier".

Let's see your approach:

Let us now get an estimate of this centrifugal force.

Photosphere/Chromosphere density:

http://www.swpc.noaa.gov/sites/default/files/images/u33/Chapter_2_.pdf

Why didn't you give the values here? Giving the reference is always good, but that's a 10-page document. If you won't give the value, at least please provide the page number, figure, or table they came from. The idea is to make it as easy as you can for your readers to see what you're doing; if you make it difficult, it looks like you're trying to deceive.

Volume of a spherical shell:

http://www.emathzone.com/tutorials/geometry/spherical-shell.html

Showing the formula here would be good form. You copy and paste a lot of other stuff; why not this?

Here's how it's done:
The volume, v, of a sphere is v = (4/3) pi r³ where r is the radius of the sphere and pi is the ratio of a circle's diameter to radius, the well-known constant 3.14159265...

The volume of a spherical shell is the volume of a larger sphere with radius r_L minus the volume of a smaller sphere r_S.

v_shell = (4/3) pi r_L³ - (4/3) pi r_S³
v_shell = (4/3) pi (r_L³ - r_S³)

Of course, it's not necessary to explain how a formula is derived in detail if can be expected to be reasonably well known to the intended audience, but giving details can sometimes help with understanding of the material and make it less mysterious.

Radius photosphere: 700,000 km = 700,000,000 m (outer)

Radius chromosphere: 705,400 km = 705,400,000 m (outer)

Thickness of photosphere shell = 400 km

Thickness of cromosphere shell = 5000 km

Estimated mass of the chromosphere:

33,507,780,986,235,477.5 kg

Estimated mass of the photosphere:

2,463,007,232,981,157.17 kg

Sun rotation speed, equator:

https://books.google.ro/books?id=BrbSAgAAQBAJ&pg=PA14&lpg=PA14&dq=sun+surface+rotation+speed+1.9+km/s&source=bl&ots=J33J1iUt43&sig=UMTba3SpbIKTh0owe3_wHKT68D0&hl=ro&sa=X&ved=0ahUKEwjF8vDY8bjNAhWJhywKHX12AKUQ6AEIVjAG#v=onepage&q=sun%20surface%20rotation%20speed%201.9%20km%2Fs&f=false

That page has a bunch of numbers. How about stating which one you selected (and why, since there are several possible choices) in addition to showing the source?

http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html

Formula? Don't make this as mysterious as you can.

It's a = v² / r where a is the centripetal acceleration, and v is the speed of a point at radius r because of the rotation.

Total centrifugal force exerted on the photosphere:

12,702,080,158,659.96 N

Can you show how you arrived at this number?

Total centrifugal force exerted on the chromosphere:

171,481,555,656,804.75 N

Can you show how you arrived at this number? If the previous explanation was well enough organized and detailed, this could be at least somewhat abbreviated since it's a similar calculation.

Until you can show your input and calculations, those numbers are just meaningless (impressively large on a human scale, but still meaningless) numbers.

Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

Nope. Centripetal (or centrifugal... they're often interchanged) force is a very small fraction (small as in about 1/500) of the acceleration of gravity. See the analysis earlier in this post. This is the power of math... it can cut through the BS in a hurry! If someone isn't clear and won't explain how they arrived at a conclusion, ask for details.

No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.

I don't believe you. Please show the details how you arrived at this conclusion.

[Edit to correct the calculations. I apologize for any confusion.]

v1 does equal v2 at the aphelion and at the perihelion in an elliptical orbit.

If this were true it would be a huge problem for Keplerian elliptical orbits. Unfortunately (for you) it's not true.

Believe it or not, sandokhan, just because you declare something to be a fact doesn't make it a fact. Please review the links in frenat's post for a start.

This is why nobody can explain how the innate motion of a planet can change by magic to suit the faked/fudged elliptical orbit falsified by Kepler.

"The orbiter must retain its innate motion throughout the orbit, no matter the shape of the orbit. If it did not, then its innate motion would dissipate. If it dissipated, the orbit would not be stable. Therefore, the orbiter always retains its innate motion over each and every differential. If we take the two most important differentials, those at perihelion and aphelion, and compare them, we find something astonishing. The tangential velocities due to innate motion are equal, meaning that the velocity tangent to the ellipse is the same in both places. But the accelerations are vastly different, due to the gravitational field. And yet the ellipse shows the same curvature at both places."

Thanks for your opinion and the unattributed quote. Anybody can say anything at any time, especially on the Internet. Can you show that any of this is true?

To make the ellipse work, you have to vary not only the orbital velocity, but also the tangential velocity. To get the correct shape and curvature to the orbit, you have to vary the object's innate motion.

Yes. That's exactly what happens naturally (without any need for second-person involvement). And?

But the object's innate motion cannot vary.

Says who?

The object is not self-propelled. It cannot cause forces upon itself, for the convenience of theorists or diagrams.

Yep. Who has said otherwise?

Celestial bodies have one innate motion, and only one, and it cannot vary.

Says who?

You have failed to address the fact that the Sun could not have captured any planet, either at the aphelion or the perihelion points.

If by "you", do you mean me? If so, then, no, that's not correct. Please read this post again, if necessary. It's late in the post; a search for second occurrence of the term "oversimplification" (without the quote marks) in the post should get you close to the explanation.

Again, do you understand English?

I understand English fine - it's my native language and both my wife and daughter are editors for technical publications, so it's kind of a "thing" in our family. Do you care to debate the fine points of the Oxford Comma?

AT BOTH THE APHELION AND AT THE PERIHELION POINTS, THERE IS NO GRAVITATIONAL INTERACTION BETWEEN THE SUN AND THE TANGENTIAL VELOCITY OF THE PLANET.

Without actually diagramming it and on first read, other than being all capitalized and having an unnecessary comma, that appears to be a properly-formed English sentence. It's factually incorrect even though it's more or less properly formed.

In other words, it's written in English but it's still wrong.

Part of your problem is that you assume that "I disagree with what you wrote because it is wrong, and here's why" means "I don't understand English". If people constantly misunderstand what you write, maybe you should write more clearly. If they get the gist of what you write and explain why they think it's wrong, maybe you should try to understand what they are saying and answer it succinctly, clearly, and logically if you still disagree.

Can you understand that?

Can you understand this much?

You betcha! I understand you're wrong. Again. Gravity affects the velocity in an elliptical orbit at all points.

HOW then would the sun capture any planet at these points? It cannot.

In the presence of a third (or more) perturbing body (or bodies), it can happen. Unless you profess that there are no more than two bodies in the universe, then, yes, it's possible. Have you gone outside and looked at the night sky yet?

That is why the elliptical orbit model falls apart from the very start.

There is no reason to assume that those are the only places where capture could or could not occur.

These are the MOST SIMPLE points one can investigate.

Perhaps those are the simplest points to investigate. Do you think the universe operates for your convenience?

If these two points fail, the whole scheme falls apart.

Who said they fail? They're just two points among an infinite number of points, so their likelihood is pretty small, but they are no more or less likely than any other two points.

Stop wasting everybody's time with your massive shitposts.

<ostentatious throat clearing>

YOU HAVE FAILED TO ADDRESS THE POINTS RAISED IN MY MESSAGE.

All of them? Perhaps. In your verbose and obfuscated ramblings it's likely something was overlooked or simply dropped in the sheer volume of balderdash. Which point(s) in particular are you concerned about? Specifically and succinctly, please. If you think there are many, can you limit them to no more than to no more than two or three per post? If necessary, please pace the posts so that there are no more than a few pending at any time before bringing up new issues. This will help ensure something isn't missed again.

[Edit] Added the part about the volume of balderdash in sandokhan's posts.

You have totally failed to address the barometer pressure paradox.

You have totally failed to address the impossibility of a spherically shaped sun facts.

You have totally failed to address the aphelion/perihelion solar capture paradox.


You were given a very simple task: to explain how it is possible for the sun to capture a planet at the simplest points on an ellipse, the aphelion and the perihelion.

You have surreptitiously tried to circumvent the matter by unloading a massive shitpost.


Let us get back to the issue discussed here.

Gravity affects the velocity in an elliptical orbit at all points.

But it does not.


At both the aphelion and the perihelion points, there is no gravitational interaction between the sun and the tangential velocity of a planet.


Says who?

Einstein, that's who.

The unit measuring rod thus appears a little shortened in relation to the system of co-ordinates by the presence of the gravitational field, if the rod is laid along a radius. With the tangential position, therefore, the gravitational field of the point of mass has no influence on the length of a rod.

A. Einstein (The Foundation of the Generalised Theory of Relativity, 1916)


The sun could not capture any planet at both the aphelion and the perihelion points.

It is as simple as that.

That is why the elliptical orbit model falls apart from the very start.



I have clearly indicated how the centrifugal forces computations were done.

I have provided each and every figure needed from the rotational speed of the Sun (a single page reference, very easy to spot the data) to the density of both the photosphere and the cromospheres (page 11 of the pdf reference).

Then, it is a very simple task to get the final results (I have even provided an online centrifugal force calculator):

Total centrifugal force exerted on the photosphere:

12,702,080,158,659.96 N


Total centrifugal force exerted on the chromosphere:

171,481,555,656,804.75 N


Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.


No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.


Centripetal (or centrifugal... they're often interchanged) force is a very small fraction (small as in about 1/500) of the acceleration of gravity.

You still don't get it.

Since the gases in the both the photosphere and the cromosphere are under a very low pressure, you cannot bring gravity into the discussion anymore.

Those gases then will be subject to the full centrifugal force computed above.

That is why no scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.

Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.