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

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #30 on: June 22, 2016, 02:36:16 AM »
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.
I do not think this is possible. Capturing a planet always involves at least two other celestial objects. As for our sun and our planet, I think we originated out of the same cloud of matter.
You have surreptitiously tried to circumvent the matter by unloading a massive shitpost.
A famous discussion tactic invented by sandokhan.
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)
Yes. A length of a rod is not affected in a tangential direction. Do you want to tell us that length equals velocity?
The sun could not capture any planet at both the aphelion and the perihelion points.
You might be true here. The sun alone can not do it. Tell me again how this is relevant.
That is why the elliptical orbit model falls apart from the very start.
Nope.

Have not the time to read through the rest right now, might adress it later.

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #31 on: June 22, 2016, 03:33:44 AM »
kami, you are a total nullity when it comes to physics.

Remember this?

Also, applying it to miles davis proves that it is quite justified, as every paper I have read of him demonstrates a complete lack of understanding.

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)


My messages are read by hundreds of thousands of people, they are that good.

Imagine what would happen if YOU were to open a separate thread bearing your name.


Is your comment supposed to be a joke?


What Einstein is telling you is that the gravitational field has no mechanism to influence that tangential vector.


By the way, did you pass your physics course in high school?


Einstein admitted that the gravitational field had no influence at the tangent.

It is as simple as that.


By the way, nice try in attempting to start a new page.

Tricks like this don't work with me.


Capturing a planet always involves at least two other celestial objects.

No wonder you think that Miles Davis writes papers on physics.



Diagram drawn by the founder of the capture theory: Dr. Michael Woolfson

http://astrogeo.oxfordjournals.org/content/41/1/1.12.full

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #32 on: June 22, 2016, 03:37:36 AM »
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.

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #33 on: June 22, 2016, 04:56:14 AM »
kami, you are a total nullity when it comes to physics.
Says the one who thinks that you can tell the distance to the sun by looking at a picture of it. (see https://www.theflatearthsociety.org/forum/index.php?topic=66738.30 ) Care to answer my questions in this thread, by the way?

Remember this?

Also, applying it to miles davis proves that it is quite justified, as every paper I have read of him demonstrates a complete lack of understanding.
[/quote]
Yes, although I already corrected this error. I obviously meant miles mathis. You know, the source of most of your copy+paste, who by the way claims that pi equals 4.
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)
Yes, Einstein talks about the contraction of length in a gravitational field, which is an effect of general relativity. Care to tell me how length equals velocity and acceleration? He never claims that gravity does not affect the rod, he just states that it does not contract it.

My messages are read by hundreds of thousands of people, they are that good.

Imagine what would happen if YOU were to open a separate thread bearing your name.
Did not bother to read through all of your copy+paste, honestly. I like this forum, but I also have some other things to do, you know?

Is your comment supposed to be a joke?
No.
What Einstein is telling you is that the gravitational field has no mechanism to influence that tangential vector.
Seriously? I must have missed that one. Your quote indicates that the gravitational field does not affect the length of a rod. I must have missed the part where he talked about the tangential vector, care to quote it for me?

By the way, did you pass your physics course in high school?
I am right now studying mathematics and physics and will finish my master's thesis this year. How about you?

Einstein admitted that the gravitational field had no influence at the tangent.

It is as simple as that.
Wrong. See above.

Capturing a planet always involves at least two other celestial objects.

No wonder you think that Miles Davis writes papers on physics.
Again, sorry for this, that was an error of mine. I meant miles mathis. And no, he does not write papers on physics, he produces hot, steaming amounts of bullshit, where he claims absurd things like pi=4, velocity=acceleration and so on.



Diagram drawn by the founder of the capture theory: Dr. Michael Woolfson

http://astrogeo.oxfordjournals.org/content/41/1/1.12.full
Yeah. Do you notice that this ball is not considered as a solid object but consists of several objects? Therefore my claim is still correct, capturing an object always involves other celestial objects.

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #34 on: June 22, 2016, 05:14:58 AM »
kami, you have no experience whatsoever when it comes to physics.

That much is clear.


Now, it is a well-accepted fact of physics that a gravitational field does not affect an object TANGENTIALLY.


I want to make sure you understand this.

When we place a rod RADIALLY in the field, of course there were be an effect, summarized by Einstein here:

The unit measuring rod appears, when referred to the co-ordinate-system, shortened by the calculated magnitude through the presence of the gravitational field, when we place it radially in the field.

Now, when we place the rod TANGENTIALLY TO THE FIELD, there will be NO EFFECT in the tangential direction which is normal to the direction of the source.

Again, Einstein:

The gravitational field has no influence upon the length of the rod, when we put it tangentially in the field.

https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity


It is as simple as this.


Obviously, you are not well versed in higher level physics, otherwise you would have not raised this point at all.


A capture of a planet involves ONLY two objects: a star (sun) and a proto-planet/star (exactly what is meant in the graphic, for your information, capture of a proto-star).

It is again obvious that you are trying to troll around this thread, by having invented ad hoc your own laughable "theory" on the subject.

Be sure to familiarize yourself with the work of Dr. M. Woolfson before embarrassing yourself beyond redemption here.



« Last Edit: June 22, 2016, 05:18:44 AM by sandokhan »

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #35 on: June 22, 2016, 05:24:15 AM »
kami, you have no experience whatsoever when it comes to physics.

That much is clear.
Some people might disagree with yo there.

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Now, it is a well-accepted fact of physics that a gravitational field does not affect an object TANGENTIALLY.
That one is true (considering a weak gravitational field at least), but it influences it radially, even on perihelion and aphelion, thus changing its velocity vector.

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I want to make sure you understand this.

When we place a rod RADIALLY in the field, of course there were be an effect, summarized by Einstein here:

The unit measuring rod appears, when referred to the co-ordinate-system, shortened by the calculated magnitude through the presence of the gravitational field, when we place it radially in the field.

Now, when we place the rod TANGENTIALLY TO THE FIELD, there will be NO EFFECT in the tangential direction which is normal to the direction of the source.

Again, Einstein:

The gravitational field has no influence upon the length of the rod, when we put it tangentially in the field.

https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity


It is as simple as this.
Keep repeating this. That does not change my statements.

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Obviously, you are not well versed in higher level physics, otherwise you would have not raised this point at all.
I can give you proof of my university courses in physics, if you want.
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A capture of a planet involves ONLY two objects: a star (sun) and a proto-planet/star (exactly what is meant in the graphic, for your information, capture of a proto-star).
Strange. When I look at the picture you posted, especially in the last step I see more than two objects. Really, really strange...

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #36 on: June 22, 2016, 05:35:02 AM »
kami, you are wasting everybody's time here.

I invited you to familiarize yourself with the work of Dr. Woolfson.

Did you do that?

No, here you are two minutes later unloading another shitpost.


Here is the capture of a proto-star, a single object, drawn by Woolfson:

http://www.droxley.freeserve.co.uk/img79.gif

AFTER this proto-star is captured, in his vision, and makes several orbits around the sun, then a filament of material from the star is tidally drawn to form several particles.

Then he drew the other diagram.

This should have been very obvious to you, but it was not.


When we place a rod RADIALLY in the field, of course there were be an effect, summarized by Einstein here:

The unit measuring rod appears, when referred to the co-ordinate-system, shortened by the calculated magnitude through the presence of the gravitational field, when we place it radially in the field.

Now, when we place the rod TANGENTIALLY TO THE FIELD, there will be NO EFFECT in the tangential direction which is normal to the direction of the source.

Again, Einstein:

The gravitational field has no influence upon the length of the rod, when we put it tangentially in the field.

https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity


It is as simple as this.


At both the aphelion and the perihelion points, there is no gravitational interaction between the sun and the tangential velocity of a planet.
« Last Edit: June 22, 2016, 05:36:35 AM by sandokhan »

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #37 on: June 22, 2016, 05:42:16 AM »
I do not need to read an entire article to find a mistake in your understanding (as I do understand orbital mechanics, contrary to your claims).
The protostar does not do several orbits around the sun and then a filament of material is drawn to the sun.

As the protostar moves past the sun it seperates into "material captured" and "depleted protostar", which are now two different objects. The "material captured" now orbits the sun while the "depleted protostar" continues on an escape trajectory.

As for your tangential BS, just keep posting, it will not change a thing.

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #38 on: June 22, 2016, 05:54:48 AM »
As for your tangential BS, just keep posting, it will not change a thing.

It is an universally accepted fact of physics that a gravitational field will not affect an object tangentially.

Einstein said so:

When we place a rod RADIALLY in the field, of course there were be an effect, summarized by Einstein here:

The unit measuring rod appears, when referred to the co-ordinate-system, shortened by the calculated magnitude through the presence of the gravitational field, when we place it radially in the field.

Now, when we place the rod TANGENTIALLY TO THE FIELD, there will be NO EFFECT in the tangential direction which is normal to the direction of the source.

Again, Einstein:

The gravitational field has no influence upon the length of the rod, when we put it tangentially in the field.

https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity


It is as simple as this.


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


I told you to familizarize yourself with the capture theory.

After Dr. Woolfson's initial publication of this theory, a very deep flaw in this hypothesis was found: the absolutely necessary collision theory.

"Even greater problems arise when we try to imagine how the earth was captured by the sun. How is an orbit like this created? How is any planetary orbit created? The textbooks never go there. By giving us the ball-on-a-string illustration, the book leaves the impression that the analogy is complete; that is, that the tangential velocity and the acceleration are conceptually connected in both instances. We are left with a fait accompli: since the two motions are tied to one another with the ball on a string, the two motions must be tied in the earth/sun example, and there is nothing to explain. But there is an awful lot to explain. To start with, in reality an orbit like this creates a hairline balance of two independent motions. The tangential motion and the centripetal motion must be perfectly balanced or the orbit will deteriorate immediately in one direction or another (inward or outward). Any satellite engineer knows this. There is one perfect distance that creates a stable orbit for a given velocity. Any other orbit requires the satellite to speed up or slow down—to make corrections. Obviously, the earth cannot make any corrections. It is not self-propelled. It cannot speed up or slow down. Therefore it must be taken to its optimum distance and kept there.

Now, think of the earth's orbit for a moment. Let's work backwards and see if we can imagine how the earth might get to that optimum distance, with just the optimum tangential velocity. If you reverse time, and conceptually back the earth out of orbit, you see that the only way you can do so is if you accelerate it out of there. If you keep the same velocity, it stays in orbit. If you decelerate, then it crashes into the sun. So you must accelerate the earth out of the orbit. But that means that unless the earth was ejected by the sun, it had to decelerate to reach its present position. If it is coming from outer space into the field of the sun, it must somehow decelerate in order to fall into its current position. But how can an object entering a gravitational field decelerate? It is getting closer to the sun: it should be accelerating. The only possibility appears to be a fortunate collision that accidentally throws it into the perfect spot. Even a planet ejected by the sun cannot reach any possible orbit, without a collision, since an ejected planet will not have any velocity tangential to the sun. There is no way to eject an object from the center of its future orbit with a velocity tangential to that orbit.

So, the unavoidable implication of historical theory is that all orbits must have been created by fortuitous collisions, either by planets arriving from outer space or being ejected by the sun. The problem is that planets arriving in orbits immediately after collisions are going to be damaged planets. Most likely they are going to be out of round. They are going to be missing chunks. This is a problem since imperfect planets create perturbations in orbits. Spins and wobbles are created, which cause uneven velocities and uneven forces. This should be fatal since the sort of orbit described by current theory is not correctable. There is no margin of error. Either the forces balance or they do not. If they do not, then the orbit should not be stable.

Some will interrupt here to point out that current theory provides that the earth was formed from a solar disc. It was not captured or ejected; it was simply always there, in some form. It congealed out of the nebula. But this answers nothing, for current theory fails to explain how this primordial disc of pre-planets or planetoids achieved its tangential motion in the first place (see below). Gravitational theory provides absolutely no mechanism, not even one as magical as gravity, to explain rotational motion in a gravitational field. It is the same question as to why galaxies rotate like wheels: they just do. We have a partial answer for why the stars don’t fly out into space: gravity. But we have no answer at all for why the stars move sideways to the gravitational field of the galaxy. If they weren’t captured, what set them in motion? The pat answer is “a spinning gravitational field”, but if you ask how a gravitational field imparts tangential velocity you get no answer. It is implied that the spin of the sun about its own axis somehow set the whole solar system to spinning, but this is mystical in the extreme. Almost no one thinks that the moon’s orbit is caused by the rotation of the earth about its own axis. No one thinks this because there is no mechanism to link the rotation of the earth to the orbit of the moon. There is no mechanism to link the orbit of the solar disc to the spin of the sun either, and yet it is accepted at face value."

« Last Edit: June 22, 2016, 05:57:00 AM by sandokhan »

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #39 on: June 22, 2016, 05:58:39 AM »
I see.. changing the topic and posting a wall of text.. nice work!

Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #40 on: June 22, 2016, 12:04:05 PM »
You have totally failed to address the barometer pressure paradox.

Can you explain succinctly how changes in barometric pressure relate to the tidal locking of the Moon, which is where this topic was brought in? From what you've said, the "paradox" seems to be that small daily variations in air pressure aren't fully understood, which is probably true, and that this somehow invalidates physics, which seems unlikely.

Until you can show something concrete like "we would expect the atmosphere to behave this way because of the the tidal effects of the Moon's gravity (and show why), but we don't see that" there's really nothing to address. In other words, I'm still waiting for you to show first that there is a problem and then exactly what the problem is.

Put up or shut up. Bloviating endlessly about it accomplishes nothing.

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

No, you have just totally ignored the answers.

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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 must have missed it.

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.

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

LOL!

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Let us get back to the issue discussed here.

Yes, please!

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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)

You're messing around with context here. I was asking where the unattributed quote about "innate motion" came from. Here's the question in context:

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But the object's innate motion cannot vary.

Says who?

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

Yep. Who has said otherwise?

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

Says who?

Would you mind answering the question? Where is that quote about innate motion from?

Anyway...

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

This is not correct. The tangential speed of a planet is not changed by the gravity of the Sun at the aphelion and perihelion points. Its velocity vector is changed, however, because it changes direction due to the force applied perpendicular to velocity. The fact that it changes direction is evidence that it's being accelerated by the Sun's gravity. Velocity is a vector, and vectors have both a magnitude (speed, in the case of velocity) and direction. An applied force on a mass will accelerate it; acceleration is a change in velocity, which can be a change in speed, direction, or both. That's pretty basic.

Before worrying about relativity you would be better served to learn about vectors first.

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

You've said that many times already. You've been shown why it's wrong. You haven't shown why it is true; repeating it does not make it true.

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It is as simple as that.

That is why the elliptical orbit model falls apart from the very start.
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This simply shows that you don't understand elliptical orbits.

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).

There are several rotational speeds given on that page. Which one did you choose? Why not just state what the data is rather than playing guessing games? What are you trying to hide?

Show you work, please. All of it, not just part of the data, a few intermediate answers, and final results. It looks like you're covering something up.

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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

We can only guess at how you arrived at those answers since you only hint at some of the input data and refuse to show the calculations used.

Is the following what you did?

1) Calculate the volumes of the chromosphere and photosphere using the spherical-shell formula and the stated radii and thicknesses.
2) Multiply these volumes by the respective densities from somewhere in the reference you cited to get the respective masses.
3) Convert solar rotation rate from rotations per day to rotations per minute or surface speed in meters per second.
4) Plug mass, radius, and rotation rate into one of the on-line calculators button.
5) Copy answer.

Is that it? If so, it's wrong. Think about it. Once you recognize why it's wrong, it should be facepalm obvious.

If that's not what you did, please explain your procedures along with the actual values you used so we can check and try to understand your work.

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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.

I don't get whatever you think your point is. It doesn't seem to have any basis in physics.

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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.

Why not?

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

How is all that mass not affected by the Sun's gravity?

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

Centrifugal force but not gravity? Again, why not? Because you say so?

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

Scientists don't have a problem with it. You, on the other hand, seem to have thoroughly confused yourself.

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

"Low gravitational pressure"? What does that mean? The acceleration of gravity (which you can't simply ignore even if you want to) is about 500 times the centrifugal acceleration according to my earlier calculations. Did you find an error in them? Actually, there are a couple of errors. Can you find them? I'll edit the post to add an erratum in a while.

I really don't see the problem with a nearly perfectly spherical sun when considering only gravity and rotation. The Sun rotates slowly and is quite massive so its gravity swamps centrifugal acceleration by orders of magnitude.

"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #41 on: June 22, 2016, 01:20:23 PM »
You have no experience dealing with stellar astrophysics.

That is why you are picking at the straws here.


The online centrifugal force website is not equipped to deal with such large numbers, everything had to be done by hand.

It is just a ballpark figure, not a precise, latitude by latitude calculation; the kind of estimate that suffices for this level of discussion.


Centrifugal force but not gravity? Again, why not? Because you say so?

Do you have a basic understanding of stellar astrophysics?

Could that be why you still don't get it?


Gravity has ALREADY been taken into full consideration.

NO further recourse can be made for gravity.

Gravity has already balanced out as much as was possible of the gaseous pressure, and still we are left with A VERY LOW PRESSURE.

That is, the solar gases in the photosphere and cromosphere are just standing there, with no explanation by modern science whatsoever.

As if this wasn't enough, we have the huge centrifugal force factor that is exerted each and every second on the photosphere and the cromosphere.


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.


The acceleration of gravity (which you can't simply ignore even if you want to) is about 500 times the centrifugal acceleration according to my earlier calculations. Did you find an error in them? Actually, there are a couple of errors. Can you find them? I'll edit the post to add an erratum in a while.

I really don't see the problem with a nearly perfectly spherical sun when considering only gravity and rotation. The Sun rotates slowly and is quite massive so its gravity swamps centrifugal acceleration by orders of magnitude.


You don't have the most basic understanding of stellar astrophysics.

It is a subject way beyond your level.


Gravity has already been taken into consideration.

Yet, those gases are under a very low pressure.

That is, no one can explain how the photosphere and the cromosphere can stay glued next to the surface of the sun, while they are also subjected to the full centrifugal force.


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 barometric pressure paradox is a clear defiance of attractive gravity: it defies newtonian mechanics.

It takes a single counterexample to destroy a hypothesis.

Moreover, if atmospheric tides have nothing to do with the influence of the gravity of the sun or that of the moon, it also means oceanic tides also cannot be explained using newtonian mechanics.


You have to explain the barometric pressure paradox, which no scientist can accomplish at the present time.


No capture can be accomplished by the sun in the first place: we have to deal with the collision theory paradox.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1777336#msg1777336

The quote about the innate motion, in Newton's own formulation can be found here:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1776803#msg1776803

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

Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #42 on: June 22, 2016, 01:33:04 PM »
Almost missed this little gem in the flurry.

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.

Then why did you say this?

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

There's nothing in your original assertion about position at all, just mass and rotation rate. How can you determine v (velocity) given only rate of rotation? The short answer: it can't be done.

This is someone desperately trying to find a definition, any definition, for angular momentum that doesn't mention moment of inertia explicitly.

DEFINITION #2.

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

Which is what I said. Do you see the word mass in that definition? Neither do I, so thanks for the validation!

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

Where did I say one of them was wrong?

I'm saying this is wrong:

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

because it is. You incorrectly used mass when you should have used moment of inertia. The definition you provide agrees with me. Why is that hard to grasp?

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

No. You made that up.

Quote
This is what you wrote:

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


That's right. The definition you provided agrees with me:

DEFINITION #2.

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

Note that mass is not present in that definition!

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

No idea whatsoever.

No. You must think repeating things makes them true, though.

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

Mass is a factor, yes.

That's because it is.

Are you saying mass is not a factor in moment of inertia?

Are you saying that mass is the same as moment of inertia?

If not either of these, what's your point?

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Why am I bringing this up each and every time?

Because you're a masochist?

Because you like to have everyone reminded that you blundered but can't admit making a mistake?

Quote
So that everyone can see at a glance the total fakery of your messages:

It's backfiring.

Quote
no matter which thread you choose to shitpost, you start to point the way forward, to tell others how wrong they are...

Feel better after your rant?

Quote
and yet, here you are not knowing the most basic facts of physics.

Hey, you were the one that didn't know when to use mass and when to use moment of inertia. I'm the one who had to point that out to you, remember?

Quote
Again, this is what you wrote:

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

And, again, it agrees with this:

DEFINITION #2.

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

I still don't see mass in the definition you provided. Do you?

"If you find yourself in a hole, stop digging."
 -- Will Rogers
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

*

Pezevenk

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #43 on: June 22, 2016, 02:52:59 PM »
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".

AAAH STOP! I can't stand those massive ass posts!!!

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


He DID have loads of observations, elliptical orbits could easily explain the motions of the planets. Kepler had already used data from Tycho Brahe to come up with an epicycle hypothesis, with which he wasn't very satisfied. He used the same data to form his hypothesis of ellipses. What he allegedly faked was the direct "check" of his hypothesis, measuring the distance of Mars from sun, to PROVE it was an ellipse, using triangulation:

"In chapter 53 of ''The New Astronomy,'' Kepler outlined a method he said he used to calculate the distances of the Earth and Mars from the Sun. The complex method was based on triangulation, which takes the distance between two points and then calculates the distance to a third by analyzing the angles in the triangle formed by the three points. The resulting distances, calculated for a variety of different dates, traced the geometry of planetary motions and gave hints that the orbits were far from circular.

Kepler cited these calculations as independent proof of the correctness of his claim for elliptical orbits. His findings were displayed in a large chart. [...] ''He was claiming that those positions came from the earlier theory,'' Dr. Donahue said. ''But actually all of them were generated from the ellipse.'' [...] Dr. Donahue, like many other experts, feels the episode does little to diminish Kepler's reputation.

''He had a difficult job trying to convince people that the ellipse was correct,'' he said. ''So he fudged a little. This doesn't take him down a notch. It was a small point in the argument.''"


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


Back then or now? Back then, the positions of planets in the night sky were common knowledge for astronomers. Everyone was trying to figure out how to explain them, and it was clear that they couldn't be explained by circular orbits, so silly theories like epicycles sprung. Well, ok, epicycles aren't that silly, but they were wrong. Eventually, Kepler realized that the orbits don't HAVE to be circular, and that ellipses could easily explain the motions observed. What was NOT available at the time was direct measurements of planets' distances from the sun, to directly show that they're ellipses. That's what Kepler allegedly faked, but subsequent observations have since confirmed that.

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


Uh, no. If you think this description you cited is proof, you should know that descriptions like that were very common back then.

"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."


So Kepler conspired for attractive gravity before attractive gravity theory was even a thing. Seems legit.

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


That's not the "mathematical equivalent" of an epicycle.

"There is no such as thing as Newtonian mechanics.

Better yet, it should be called INDIAN MECHANICS."


This is the second time you say that. It's still stupid. I don't even think Newton had access to this information. Still, why do you think 2 people can't discover the same thing independently? Besides, the ancient Indian version lacked the mathematical formalism of Newton's version, which made it clearer. That's not to say the Indian version is shit or something, it's actually a bit amazing for its time.

"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."


Oh, both Newton AND Leibniz? Lol. Anyway, I can't check that claim right now, I'll do it later when I have time.
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It is not a scientific fact, it is a scientific fuck!
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Read a bit psicology and stick your imo to where it comes from
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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #44 on: June 22, 2016, 08:00:12 PM »
You have no experience dealing with stellar astrophysics.

That is why you are picking at the straws here.

No. I'm questioning your basic premises and poor analysis.

Quote
The online centrifugal force website is not equipped to deal with such large numbers,

Really? Then why did you describe the process this way?

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):

[Emphasis added]

First you say you provided an online calculator. Later you say it doesn't work with these numbers. You really need to sanity check the things you say to see if they are at least close to right before saying them. It helps your "cred".

At any rate, just dumping your numbers for mass and radius (after removing the commas) and a rotation rate for the Sun in meters/second into the calculator worked for me. It returned numbers similar to yours.

The commas had to be removed from the numbers before submitting the data or it returns NaN (Not a Number, a.k.a. invalid result). Perhaps that was your problem? You should check this stuff before posting. Going through the bother of basic diligence helps you look like you know what you're doing, even, in some cases, if you don't.

Quote
everything had to be done by hand.

Oh, the humanity!

Quote
It is just a ballpark figure, not a precise, latitude by latitude calculation;

Didn't you say it is simple?

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

Yeah... you did say it is simple. Apparently this was before you realized it didn't work with your invalid input.
.
"Latitude by latitude calculation." You never mentioned this before. It sounds like the story is changing; you must have realized where you went wrong, once prodded!

Can we see calculations for, say, at least two or three latitudes? Please include the actual data you used (numbers and units) rather than make us guess what the actual values and your methodology are.

What are the final answers now?

Quote
the kind of estimate that suffices for this level of discussion.

That's a good approach and certainly enough for this sort of thing.

Quote
Centrifugal force but not gravity? Again, why not? Because you say so?

Do you have a basic understanding of stellar astrophysics?

Yes. You started out by stating as fact "You have no experience dealing with stellar astrophysics." Why are you asking now?

Quote
Could that be why you still don't get it?

Misunderstanding is always possible. This seems unlikely in this case.

Quote
Gravity has ALREADY been taken into full consideration.

Where?

Quote
NO further recourse can be made for gravity.

Thanks for the edict. Can you justify it?

Quote
Gravity has already balanced out as much as was possible of the gaseous pressure, and still we are left with A VERY LOW PRESSURE.

The pressure on the surface is caused by gravity. What you said is nonsense.

Quote
That is, the solar gases in the photosphere and cromosphere are just standing there, with no explanation by modern science whatsoever.

You're obviously mystified. Scientists, not so much.

It's unlikely that the gases are "just standing there", either. Evidence suggests that there is very violent motion going on.

Quote
As if this wasn't enough, we have the huge centrifugal force factor that is exerted each and every second on the photosphere and the cromosphere.

Huge compared to what? The weight of the gases in question? Centrifugal force is tiny compared to that. Can you show otherwise? Show, not handwave.

Quote
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.

Well, you obviously can't explain what's going on. That much is very clear. Assuming your inability applies to everyone else is a bit presumptuous, don't you think?

Quote
The acceleration of gravity (which you can't simply ignore even if you want to) is about 500 times the centrifugal acceleration according to my earlier calculations. Did you find an error in them? Actually, there are a couple of errors. Can you find them? I'll edit the post to add an erratum in a while.

I really don't see the problem with a nearly perfectly spherical sun when considering only gravity and rotation. The Sun rotates slowly and is quite massive so its gravity swamps centrifugal acceleration by orders of magnitude.


You don't have the most basic understanding of stellar astrophysics.

Why did you ask earlier?

Quote
It is a subject way beyond your level.

How can you tell? You're the one struggling with the basics.

Quote
Gravity has already been taken into consideration.

Where?

Quote
Yet, those gases are under a very low pressure.

OK. They have very low density. So?

Quote
That is, no one can explain how the photosphere and the cromosphere can stay glued next to the surface of the sun, while they are also subjected to the full centrifugal force.

"Full centrifugal force" oh, my! Do we have only partial force of gravity here? Why? Where did the rest go?

It's really simple if you don't confuse yourself with your own obfuscation. The acceleration of gravity is much greater than the centrifugal acceleration. Period. The photosphere and the coronosphere "stick" to the Sun because they weigh more than the centrifugal force from rotation. By a very large margin.

Quote
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.

You keep mentioning "gravitational pressure". I presume that's essentially just the weight of the gases since the centrifugal force is negligible (even at the equator, where it's greatest) divided by the area they cover (i.e. the surface area of the Sun) since pressure is force per unit area. Do you mean something different? If so, what?

Quote
The barometric pressure paradox is a clear defiance of attractive gravity: it defies newtonian mechanics.

It takes a single counterexample to destroy a hypothesis.

Not exactly. First of all, the counterexample has to be both valid and relevant before it can disprove a hypothesis.

Quote
Moreover, if atmospheric tides have nothing to do with the influence of the gravity of the sun or that of the moon, it also means oceanic tides also cannot be explained using newtonian mechanics.

Who said they have nothing to do with the influence of the gravity of the Moon or the Sun? I was asking how much of a contribution should be expected from these sources. So far all I've heard on this is crickets.

Atmospheric tides sound like a very complex problem in fluid dynamics; expecting a model that's accurate, yet simple enough for a non-expert (like me and, most likely, you) to understand is probably unrealistic.

Quote
You have to explain the barometric pressure paradox, which no scientist can accomplish at the present time.

No I don't. First you must demonstrate there is something that needs to be explained. You haven't done so yet.

If you can show there is no evidence for a lunar component to atmospheric tides that should realistically be expected, and explain why they should be expected (a link to a competent source would be fine), I'd be interested in hearing about that. Until then, you're just posturing.

Quote
No capture can be accomplished by the sun in the first place: we have to deal with the collision theory paradox.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1777336#msg1777336

Yet another "paradox"? Can you distill that tome into a succinct and coherent synopsis? Think of the abstract of an actual technical paper; explain what it is about, the general approach taken in analysis, and highlights of the data and findings. If it seems sufficiently useful or even mildly interesting then your full wall of text can be perused.

Quote
The quote about the innate motion, in Newton's own formulation can be found here:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1776803#msg1776803

The unattributed quote in question is not in that post.

I did get a laugh out of this from it, though: "The Principia is notorious for its lack of numbers and variables." You continually complain and laud others who complain about things like this.

That's a reasonable complaint, so do better. Meanwhile, you steadfastly refuse to provide the data you purport use in computations and reply with a flippant "look it up" when asked what the number was. See the top of this post for an excellent example: "a single page reference, very easy to spot the data" containing several possible values, but no hint which one you actually used.

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

Aha!! There's the source for the quote I was looking for.

"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."

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

Turns out, it's none other than our own sandokhan! Unless he's plagiarizing someone else. Note the quote marks in the first but not second blocks.

"Innate motion" is an archaic term that was replaced by inertia in more recent writings.

[Edit] Fix embedded quote and complete last sentence.
« Last Edit: June 22, 2016, 09:28:58 PM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #45 on: June 22, 2016, 09:55:53 PM »
That's not the "mathematical equivalent" of an epicycle.

You are not a mathematician.

An epicycle and an ellipse ARE mathematically equivalent, here is the demonstration:

http://www00.unibg.it/dati/bacheca/63/21692.pdf

The usual Newtonian model of the Solar system assumes that, for most purposes, it is sufficiently accurate to describe the orbits of each planet as an ellipse with its focus at the Sun. If each ellipse is replaced by the equivalent epicycle , the model is just as accurate.

It therefore follows that if Ptolemy had used an origin at the Sun then a simple model that represented the orbit of each planet by a simple offset epicycle would have represented the orbits of the planets just as accurately as the Newtonian model based on ellipses.

He (Kepler) gave the numerical values to the parameters of the ellipses that matched Brahe’s observations , and that settled the question ellipses were in, epicycles were out. He never stated and may not have noticed that every ellipse is an epicycle, so if he had calculated the epicycle parameters that corresponded to his ellipses the epicycles would have fitted the data just as well.



At each and every point of your message you have no idea what you are talking about.

He used the same data to form his hypothesis of ellipses.

But he COULD NOT! Don't you understand?

There were no ellipses in Tycho Brahe's formulation, NOR did the data match any kind of an ellipse.


Dr. Donahue, like many other experts, feels the episode does little to diminish Kepler's reputation.

Really?

“Almost 400 years later, William H. Donohue undertook the task of translating
Kepler’s 1609 Astronomia Nova into the English New Astronomy (Donohue 1992)
when in the course of his work he redid many of Kepler’s calculations, he was
startled to find some fundamental inconsistencies with Kepler’s reporting of these
same calculations (Donohue 1988). Writing of Donohue’s pathbreaking work in
The New York Times, William Broad (1990) summarized Donahue’s findings
saying that although Kepler claimed to have confirmed the elliptical orbit by
independent observations and calculations of the position of Mars, in fact Kepler
derived the data from the theory instead of the other way around . . .

“But a close study of Kepler’s New Astronomy . . . shows that the plotted points
[he used] do not fall exactly on the ellipse (of course, measurements rarely fall
exactly on a theoretical curve because they usually have random error sources
incorporated into them.) Curtis Wilson (1968), however, carries error argument
further. The lack of precision inherent in the method . . . would have forced Kepler
to use the plotted points only as a guide to his theorizing . . .
“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.
« Last Edit: June 23, 2016, 02:36:01 AM by sandokhan »

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #46 on: June 22, 2016, 10:10:48 PM »
The reason I did not mention the position vector was because it was easily discerned from the discussion at hand: the solar system.

The definition I used is absolutely correct.

Which is what I said. Do you see the word mass in that definition?

But you did not.

You explicitly stated that THE MOMENT OF INERTIA IS NOT RELATED TO MASS AT ALL.

This is what you said.

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

But the definition of the moment of inertia is directly based on the mass of an object.


At that point in time you had no idea about the fact.

Yet here you are demanding all sorts of calculations when you are unable to even state the definition of the moment of inertia correctly.


No. You made that up.

I did not.

This is what you wrote at the time:

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

A terrible and grievious mistake on your part: the moment of inertia is defined in terms of mass.


I was asking how much of a contribution should be expected from these sources.

Lord Rayleigh has already performed this calculations a long time ago, they are readily available on the net, please do your homework.

Here is what he stated:

‘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.




Here are Newton's thoughts on the innate motion.

In the Principia, it is stated explicitly that the earth had this velocity before it entered the orbit. Newton calls it the body's "innate motion."

The unbelievable deception used in the making of Kepler's works continued unabated with Newton's Principia.


I will be sent to the Principia, where Newton derives the equation a = v2/r. There we find the velocity assigned to the arc. True, but a page earlier, he assigned the straight line AB to the tangential velocity: "let the body by its innate force describe the right line AB". A right line is a straight line, and if Newton's motion is circular, it is at a tangent to the circle. So Newton has assigned two different velocities: a tangential velocity and an orbital velocity. According to Newton's own equations, we are given a tangential velocity, and then we seek an orbital velocity. So the two cannot be the same. We are GIVEN the tangential velocity. If the tangential velocity is already the orbital velocity, then we don’t need a derivation: we have nothing to seek! If you study Newton's derivation, you will see that the orbital velocity is always smaller than the tangential velocity. One number is smaller than the other. So they can't be the same.

The problem is that those who came after Newton notated them the same. He himself understood the difference between tangential velocity and orbital velocity, but he did not express this clearly with his variables. The Principia is notorious for its lack of numbers and variables. He did not create subscripts to differentiate the two, so history has conflated them. Physicists now think that v in the equation v = 2πr/t is the tangential velocity. And they think that they are going from a linear expression to an angular expression when they go from v to ω. But they aren't.

This problem has been buried ever since Newton used his new calculus to find the orbital or curved velocity given the tangential velocity. In The Principia, he is given the tangential velocity or straight-line motion, and he derives the orbital or curved motion from it, using his ultimate interval (like a limit). The velocity variable in a=v2/r must then be this new orbital velocity. So the old tangential velocity is lost. It has been buried from sight ever since. But in the ellipse, or any real orbit, we must continue to monitor the old tangential velocity, since we cannot allow it to vary without giving a mechanical explanation of that variation. If we see it varying in the ellipse, as I have shown, then we must ask how a planet can vary its innate motion to suit an orbit. How can either the planet itself, or the gravitational field, cause that velocity to vary? The planet cannot, because it is not self-propelled or self-correcting. The gravitational field cannot, because the gravitational field has no mechanism to influence that vector. Even Einstein admitted that the gravitational field had no influence at the tangent.

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #47 on: June 22, 2016, 10:24:32 PM »
Where?

Right here.

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 (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.


OK. They have very low density. So?

Do we have only partial force of gravity here? Why? Where did the rest go?

It's really simple if you don't confuse yourself with your own obfuscation. The acceleration of gravity is much greater than the centrifugal acceleration. Period. The photosphere and the coronosphere "stick" to the Sun because they weigh more than the centrifugal force from rotation. By a very large margin.


Do you want everybody to be laughing at you each time they read your catastrophic understanding of basic solar astrophysics?


The very low pressure means that gravity has ALREADY been taken into consideration: that is, gravity has already balanced out the enormous pressure of the gases of the sun.

There is nothing else that can be done in this respect.

Now, the fact that both the photosphere and the cromosphere stay glued next to the surface of a sphere while at the same time they are subjected to an additional enormous centrifugal force, running in the quadrillions of newtons, is what demonstrates clearly the impossibility of a spherically shaped sun.


The pressure on the surface is caused by gravity. What you said is nonsense.

The gases in both the photosphere and the cromosphere are under a very low pressure.

The force of gravity was taken into consideration, by having balanced out the pressure of the gases.

There is nothing else that can be done with solar gravity in this respect.


Can you understand these  basic facts of astrophysics?


In a star, the huge pressure of the gases is balanced out by gravity, this is the official explanation.


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.


Since there is NO FURTHER GRAVITATIONAL PRESSURE, the presence of  both the photosphere and the cromosphere next to the surface of the sun is totally unexplained, given the centrifugal force of rotation which also acts on the solar atmosphere, but is not included in the gravity/gaseous pressure equation at all.



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.


Gravity has already been balanced out by the gaseous pressure.

Now, we have the ADDITIONAL centrifugal force to deal with, which is not explained by modern astrophysicists.

No scientist at the present time can explain the defiance of newtonian mechanics by the gases in the solar photosphere.
« Last Edit: June 23, 2016, 12:40:26 AM by sandokhan »

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #48 on: June 23, 2016, 12:09:11 AM »
Now things are going to get much worse for you.

We have already seen how low the solar atmospheric pressure is for the photosphere.

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.

But the atmospheric pressure in the chromosphere is MUCH LOWER THAN THAT:



Moreover, the thickness of the chromosphere is 10,000 km (I only used a figure of 5,000 km in my calculations).


PRESSURE: 10-13 BAR = 0.0000000000001 BAR


The entire chromosphere will then be subjected to the full centrifugal force of rotation, as will the photosphere itself of course.


Completely unexplained by modern science.


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/chromosphere.
« Last Edit: June 23, 2016, 12:36:47 AM by sandokhan »

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #49 on: June 23, 2016, 09:49:10 AM »
Okay, to be honest, I have not found the motivation to read through your three copy+pastes. However, I do not need to.

That's not the "mathematical equivalent" of an epicycle.

You are not a mathematician.

An epicycle and an ellipse ARE mathematically equivalent, here is the demonstration:

http://www00.unibg.it/dati/bacheca/63/21692.pdf
You are obviously not a mathematician. Get down from your high horse, will you?
Do you know the definition of "equivalent"?
You claim that every ellipse had to be an epicycle and vice versa. This is not the case.

As for the rest, yes, ellipses can be modeled as epicycles. But the epicycles introduced by your worshipped Tycho Brahe were not ellipses.

So your first statement in your first post was wrong. Didn't bother to read the rest, to be honest.

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #50 on: June 23, 2016, 09:54:25 AM »
Okay, I did it again, read some more.
Moreover, the thickness of the chromosphere is 10,000 km (I only used a figure of 5,000 km in my calculations).

PRESSURE: 10-13 BAR = 0.0000000000001 BAR

The entire chromosphere will then be subjected to the full centrifugal force of rotation, as will the photosphere itself of course.

Completely unexplained by modern science.
Of course the pressure of the sun's "atmosphere" lowers at higher altitudes. As does earth's.
That does simply mean that there is not so much gas above those layers which could create pressure. Pressure does not equal force or acceleration, you know?

How about a little gedankenexperiment:
Picure earth had no atmosphere. Then a single molecule of air gets introduced. Pressure=0,0000000000000....BAR!
Does that mean that this molecule gets flung off into space? No! Why not? Gravity!

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #51 on: June 23, 2016, 10:09:00 AM »
You never passed any of your physics courses, did you?

Here is solar astrophysics 101.




This is what the official scientific establishment is telling us: gravity compression is balanced out by thermal/gaseous pressure.


But there is something very funny going on at the surface of the sun, which has puzzled astrophysicists for the past eighty years.

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.


Then you really have a big problem.






The entire chromosphere will then be subjected to the full centrifugal force of rotation, as will the photosphere itself of course.


Completely unexplained by modern science.


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/chromosphere.

Gravity has already been balanced out by the gaseous pressure.

Now, we have the ADDITIONAL centrifugal force to deal with, which is not explained by modern astrophysicists.

The centrifugal force would cause the sun to collapse into a disk in no time at all.



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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #52 on: June 23, 2016, 10:34:30 AM »
You never passed any of your physics courses, did you?

Here is solar astrophysics 101.
I would have a problem then.
Quote
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.
Why would you expect that? I would expect the atmospheric pressure of the sun to depend on altitude, as on earth. You talk about pressure at sea-level. What about pressure at 100km up? 200km?
Quote
The entire chromosphere will then be subjected to the full centrifugal force of rotation, as will the photosphere itself of course.
Yes. With the astonishingly fast rotation of 1 rotation every 26 days! Now that is a huge centrifugal force! It is 0,0054m/s^2 at the outer rim of the sun (700.000 km)! That is even less than the centrifugal force on earth! Now take into account that the sun has a much higher gravitational pull, tadaaa, sphere explained.
Quote
Completely unexplained by modern science.
I should publish my complicated calculations as soon as possible, then.
Quote
Gravity has already been balanced out by the gaseous pressure.
Why do you think that? Pressure does not equal force. Might have said that before.
Quote
Now, we have the ADDITIONAL centrifugal force to deal with, which is not explained by modern astrophysicists.

The centrifugal force would cause the sun to collapse into a disk in no time at all.
See above. Simply wrong.

Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #53 on: June 23, 2016, 11:14:12 AM »
In an effort to keep this short, repeats and silly fussing within the post are omitted. If there are still any readers, I'm sure most of them are tired of your continuous evasions and ridiculous interpretations of what was said.

The reason I did not mention the position vector was because it was easily discerned from the discussion at hand: the solar system.

Oh? What is the "position vector" for the solar system, including the rotating sun, orbiting (and rotating) planets, and some of the larger moons?

Anyone can see the context here.

I ain't buying "it was implied". That makes no sense at all given 1) the vast variation in the distances and sizes among the objects in the discussion at hand, from the radius of the Earth to the radius of Pluto's orbit, and 2) the different way the mass of a rotating sphere and the mass of an orbiting body are treated.

Nice try. At least that was something new.

Quote
The definition I used is absolutely correct.

Which one? The definitions for moment of inertia that you copied from elsewhere while back-pedaling? Those look right, and I don't think anyone has said otherwise. The one for angular momentum where mass was substituted for moment of inertia? That one's wrong.

Quote
Which is what I said. Do you see the word mass in that definition?

But you did not.

You explicitly stated that THE MOMENT OF INERTIA IS NOT RELATED TO MASS AT ALL.

No, you just misinterpreted what I said. I presume it was intentional so you would have something to whine about, but maybe not.

Quote
This is what you said.

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

But the definition of the moment of inertia is directly based on the mass of an object.

Exactly. Mass is already included in moment of inertia - you just said so (again) yourself - so saying "moment of inertia and mass" would be wrong. If you use mass and moment of inertia when calculating angular momentum you'll get the wrong answer. Understand yet? I'm running out of ways to explain this.

Quote
At that point in time you had no idea about the fact.

No. You made that up. If you can show otherwise, please do so. Simply repeating it over and over (and over...) doesn't make it true, it just gets tedious.

Quote
Yet here you are demanding all sorts of calculations

Trying to weasel out of doing any calculations? Noted.

Quote
when you are unable to even state the definition of the moment of inertia correctly.

That's just another of your unsubstantiated assertions until you can show some evidence[nb]Wishful thinking is not evidence.[/nb] that it is true. Do you have anything other than that one line you keep repeating because you don't understand it - the one that anyone reasonably astute would have understood. You, on the other hand, apparently didn't know that moment of inertia even existed. Here's the evidence for that conclusion:

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

If you knew that when calculating angular momentum the distribution of mass can be more important than the mass itself, why didn't you bother mention it?

I see two realistic possibilities: 1) You didn't know that mass distribution was  even a factor, or perhaps you knew it was in there somewhere but not important enough to be worth mentioning. 2) You simply made an error and used mass when you meant moment of inertia or something that refers to distribution of mass. The latter is just an "oops..." Mistakes happen. You should have opted for the graceful exit instead of clutching at straws in a failed effort to save face that just makes you look worse. Oh, well. If you keep bringing it up without adding anything new, I'll keep showing you why you're still wrong.

Are there any other possibilities?

Quote
A terrible and grievious mistake on your part:

::) Your histrionics are only mildly amusing.

That was still pretty long, but there was a lot of BS to address. Sorry about that.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #54 on: June 23, 2016, 12:07:51 PM »
No matter how much you try to whine about it, at that point in time, you had no idea that the definition of the moment of inertia involves mass.

You explicitly stated:

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

But the definition of the moment of inertia is directly based on the mass of an object.


You seem to be very confused.



Now take into account that the sun has a much higher gravitational pull, tadaaa, sphere explained.

There is no higher gravitational pull at the surface of the sun at all.

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.


In the chromosphere the pressure is 10-13 bar.


These are the facts accepted by each and every astrophysicist.


Then, both the chromosphere and the photosphere will be subjected to an enormous centrifugal force, for which I offered a ballpark estimate.

The masses of both the photosphere and the chromosphere run in the quadrilllions of kilograms.


Gravity has already been balanced out by the pressure of the gases.

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.

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #55 on: June 23, 2016, 12:32:25 PM »
Now take into account that the sun has a much higher gravitational pull, tadaaa, sphere explained.

There is no higher gravitational pull at the surface of the sun at all.
Really? Those 2*10^30(!!!)kg do not count? Why? Does gravity suddenly disappear close to the sun?

Quote
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.


In the chromosphere the pressure is 10-13 bar.


These are the facts accepted by each and every astrophysicist.
I do not debate those facts. I debate your conclusions. I see no reason why those pressures should contradict anything. The atmospheric pressure at 200km at earth is also about 10-13 bar. Does that mean anything? No!
Because pressure does not equal force.

Quote
Then, both the chromosphere and the photosphere will be subjected to an enormous centrifugal force, for which I offered a ballpark estimate.
I calculated it. It is diminishingly small. See above.
Quote
The masses of both the photosphere and the chromosphere run in the quadrilllions of kilograms.
You have no basic understanding of physics, do you? Since when is mass important when you calculate an acceleration?
Quote
Gravity has already been balanced out by the pressure of the gases.
Just because you repeat that does not make it true.
"Centrifugal force has already been balanced out by the pressure of the gases". Your turn.
Quote
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.
gravitational pressure is not even a correct physical term. gravity is an acceleration, how much do I need to tell you that pressure does not equal acceleration.
Did you take this from miles mathis again? Or which pothead wrote this?

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #56 on: June 23, 2016, 01:02:11 PM »
You have no idea what you are talking about.

Have you EVER taken any physics courses at all?

Quotes from RE textbooks:

Pressure (symbol: p or P) is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

For instance, the pressure-gradient force prevents gravity from collapsing Earth's atmosphere into a thin, dense shell, whereas gravity prevents the pressure gradient force from diffusing the atmosphere into space.

This occurs when external forces such as gravity are balanced by a pressure gradient force.


Does gravity suddenly disappear close to the sun?

This fact is what has puzzled astrophysicists for the past eighty years, as I have stated before.


The facts are very 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.


In the chromosphere the pressure is 10-13 bar.


"Centrifugal force has already been balanced out by the pressure of the gases". Your turn.

I told you, that you have no knowledge of astrophysics.

This is the very crux of our discussion here: the pressure of the gases is balanced out by gravity.

At no point in time has the centrifugal force ever been taken into consideration.

The atmospheric pressure at 200km at earth is also about 10-13 bar. Does that mean anything? No!

Update your knowledge of atmospheric physics.

https://www.st-andrews.ac.uk/~dib2/climate/winds.htm (geostrophic winds, gradient winds)

See also: http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/grad.rxml


I calculated it. It is diminishingly small. See above.

The centrifugal force is calculated as follows:

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


The centrifugal force for both the photosphere and the chromosphere runs in the quadrillions of newtons.

It is completely unaccounted for in modern astrophysics.


« Last Edit: June 23, 2016, 01:04:52 PM by sandokhan »

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sandokhan

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #57 on: June 23, 2016, 01:12:25 PM »
Because pressure does not equal force.

REFERENCE #1

https://books.google.ro/books?id=msZMEvEpxG8C&pg=PA173&lpg=PA173&dq=stellar+astrophysics+gravity+pressure+forces+balance&source=bl&ots=SU6xA__Fr6&sig=02GBf32AcLj8DP9CxHFeDoaSHxA&hl=ro&sa=X&ved=0ahUKEwi13MrI-b7NAhWKuRQKHbUsCGEQ6AEILjAB#v=onepage&q=stellar%20astrophysics%20gravity%20pressure%20forces%20balance&f=false

Temperature, pressure and gravitational forces in the envelope are then reduced until a balance is reestablished between pressure and gravity forces.


REFERENCE #2

http://www.astronomynotes.com/starsun/s7.htm

Pressure---the amount of force/area.


REFERENCE #3

http://science.jrank.org/pages/6502/Stellar-Structure.html

However, the gravity is opposed by the internal pressure of the stellar gas which normally results from heat produced by nuclear reactions. This balance between the forces of gravity and the pressure forces is called hydrostatic equilibrium, and the balance must be exact or the star will quickly respond by expanding or contracting in size. So powerful are the separate forces of gravity and pressure that should such an imbalance occur in the sun, it would be resolved within half an hour.


If we do add the centrifugal force of rotation, completely unaccounted for, the matter would be resolved in minutes.

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #58 on: June 23, 2016, 03:01:58 PM »
You have no idea what you are talking about.

Have you EVER taken any physics courses at all?

Quotes from RE textbooks:

Pressure (symbol: p or P) is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

For instance, the pressure-gradient force prevents gravity from collapsing Earth's atmosphere into a thin, dense shell, whereas gravity prevents the pressure gradient force from diffusing the atmosphere into space.

This occurs when external forces such as gravity are balanced by a pressure gradient force.
I do understand pressure. That's why I understand that the pressure decreases with increasing altitude while the gravity stays (roughly) constant.

Quote
Does gravity suddenly disappear close to the sun?

This fact is what has puzzled astrophysicists for the past eighty years, as I have stated before.
Citation, please. Some reknown astrophysicist who states that this fact does puzzle him.
Quote
"Centrifugal force has already been balanced out by the pressure of the gases". Your turn.

I told you, that you have no knowledge of astrophysics.
The thing is: I don't believe you.
Quote
This is the very crux of our discussion here: the pressure of the gases is balanced out by gravity.

At no point in time has the centrifugal force ever been taken into consideration.
Why would one consider one force acting on the gases and totally forgot another one? That does not make sense to me.
Quote
The atmospheric pressure at 200km at earth is also about 10-13 bar. Does that mean anything? No!

Update your knowledge of atmospheric physics.

https://www.st-andrews.ac.uk/~dib2/climate/winds.htm (geostrophic winds, gradient winds)

See also: http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/grad.rxml
What do those winds have to do with anything?
I admit, this value might change, but you can not compare atmospheric pressure of different celestial bodies at different altitudes, because pressure changes with altitude!
Quote
I calculated it. It is diminishingly small. See above.

The centrifugal force is calculated as follows:

http://www.engineeringtoolbox.com/centripetal-acceleration-d_1285.html
Thank you for this link. I do know how to calculate centripetal acceleration. Although you do not believe me, I indeed took some physics courses (although they would not be needed for his, seriously, this is school-level).
Take w=2*pi/(26days) and r=700.000km and you get the value a=0,0054m/s^2. That is a very small value.
Quote
The centrifugal force for both the photosphere and the chromosphere runs in the quadrillions of newtons.

It is completely unaccounted for in modern astrophysics.
The force exerted on what, exactly? Important is the acceleration, force is dependend on mass.

And regarding theother post: Yes, pressure is force per area, not force, not acceleration. I know that. It's not that hard.

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Kami

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Re: Dicussion of sandokhan's claims about the Moon in Q&A
« Reply #59 on: June 23, 2016, 03:07:28 PM »
Also, sandokhan, I think you misunderstand something. The pressure in the atmosphere (be it sun or earth) rather tries to push the particles away from the sun, not towards it. So this pressure (or rather the pressure gradient) has to be countered by gravity, not theother way around.