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Gravitation constant or not


Timo Moilanen

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37 minutes ago, Timo Moilanen said:

By "low altitude" you mean on the Earth's surface, of course.

This is simply applying the geometry of the Earth to calculate the gravity on the Earth's surface. How is this relevant?

It does nothing to support your claim that gravity increases with altitude.

If anything, it is the basis of yet more evidence you are Wong. The values calculated this way have been confirmed by measurements on the ground and from space.

43 minutes ago, Timo Moilanen said:

Mark the add . my coarse model do this without empiric input ( other than earth densities  not a single one)

The only "empiric" input on that page is the measured size and shape of the Earth. If your model does not include that then it cannot possibly be correct.

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35 minutes ago, Strange said:

By "low altitude" you mean on the Earth's surface, of course.

Traditionally surface and hills ,mountains . 

1 minute ago, Strange said:

You need evidence this happens. (It doesn't.)

I presented one of the most credible sources there is to you . Read it thoroughly :angry:

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4 minutes ago, Timo Moilanen said:

Traditionally surface and hills ,mountains . 

You are not making any sense.

And you have still not provided any evidence that the existing laws of gravity are wrong. Or that gravity increases with altitude.

You have not acknowledged that your replacement for G has the wrong dimensions. This appears to be because you don't have sufficient mathematical understanding to know what this means or why it is a problem.

I can't see much point in this thread staying open 

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Just now, Strange said:

I can't see much point in this thread staying open 

I suppose you have the upper hand . Providing evidence is hard enough (my destiny) disclaiming them ( I would not even try (without any alternative) and additional evidence

For true participants of physic discussion https://sites.ualberta.ca/~unsworth/UA-classes/210/notes210/B/210B3-2008.pdf  

page 7 . Sorry I remembered page number wrong earlier

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20 minutes ago, Timo Moilanen said:

Providing evidence is hard enough

Again, if even you are not able to find any evidence at all to support your hypothesis, then it suggests that no such evidence exists. 

Therefore the rational conclusion must be that your hypothesis is wrong. However much it appeals to you, at some point you will have to admit there is no evidence for it.

One the other hand, we have two very good theories of gravity which are supported by large amounts of evidence. Do you see the difference?

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6 minutes ago, Timo Moilanen said:

I suppose you have the upper hand . Providing evidence is hard enough (my destiny) disclaiming them ( I would not even try (without any alternative) and additional evidence

 

You have posted your original paper abstract as 5 pictures.

This is good as I have just been able to print them out as 5 readable pages.

So I now propose to check through the maths properly, something I have only skimmed up till now.

Several things immediately come to notice.

Please confirm if you agree with them.

1) Your treatment appears purely classical. Relativity of gravity is not involved.

2) Because you have not numbered your equations as is standard practice and they are pictures, it is impossible to quote them directly here and difficult to refer to them.

3) I see that on page (picture) 3 of your paper you have introduced the mole and Avogadro's constant. I would would like to discuss the circularity of that part of your argument, but I cannot extract it or refer to it directly.

4) You have introduced 'energy' into your thesis, and also the special relativity relationship between energy and mass.

5) You have not discussed Gauss' theorem in your 5 pages, but did show a lack of understanding of it in the 6 pages of this thread. This needs proper discussion.

6) There has been considerable discussion about measurements that I have not participated in, but you have not made clear that bodies are subject to other influences (forces) which affect the measurement of the local attraction experienced between bodies. These are additional to the standard force due to the Universal Gravitational Constant (whatever you wish to call it). So if you are seeking to measure any variation of G in the real world, you have to acount for these.

7) One of you misunderstandings is that you have not shown Forces (however generated) as vector quantities and therefore you have ignored the directional properties of vectors. These are also affected by other acting forces. This is linked to your misunderstanding of Gauss.

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8 minutes ago, Timo Moilanen said:

I just sent you https://sites.ualberta.ca/~unsworth/UA-classes/210/notes210/B/210B3-2008.pdf page 7 ,. The Bouguer correction, I don't manage to copy the text and formulas here 

At least you have been a little more specific about what you mean.

That correction is for the fact that when you go up a hill, there is some extra mass underneath you. This is (an approximation) based on simple Newtonian gravity. You can't use the results from Newtonian gravity to prove that Newtonian gravity is wrong!

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7 minutes ago, studiot said:

1) Your treatment appears purely classical. Relativity of gravity is not involved.

Yes it is purely classical (geometrical) no relativistic elements added so far

10 minutes ago, studiot said:

4) You have introduced 'energy' into your thesis, and also the special relativity relationship between energy and mass.

This is only an other quality of energy/mass c^2/2 is as common that come

12 minutes ago, studiot said:

5) You have not discussed Gauss' theorem in your 5 pages, but did show a lack of understanding of it in the 6 pages of this thread. This needs proper discussion.

Last year we discussed this and agreed that I disagree  with Gauss on this point (for gravity not electromagnetism)

I wrote earlier that gravitation is not additive , the feature of the "particles" must be summed up (no opinion to what they are)

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26 minutes ago, Timo Moilanen said:

I wrote earlier that gravitation is not additive , the feature of the "particles" must be summed up

What is the difference between "adding" and "summing"?

The mass of the Earth is the sum of the masses of all the atoms that make it up. Is that what you mean?

The gravity of the Earth is the sum of the gravity of all the atoms making up the Earth. Is that what you mean?

How is this "not additive"?

(Note neither statement is exactly correct because it doesn't take account of the binding energy when the atoms make up molecules and the molecules make up larger structures. And the second statement is only correct for a Newtonian model; which yours is.)

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Tgrav6.jpg.cdf1e8e9a157196b388398ca3c8bf7c4.jpg

 

This is just plain bad mathematics.

 

Tgrav7.jpg.b2cc55d7e3c51f275520e7fd9469fabf.jpg

 

This definition and argument is circular

 

Tgrav8.jpg.36182bbeb12fbc92e169a8c8a2ff42b7.jpg

 

This is only true if G is constant, something you are trying to disprove !

You are  relying on the constancy of G or T in any summation or integral where you take G or T outside the summation or integral.
If they are not constant, then your calculation would be different.

It is very difficult to separate out calculations involving G and T from your pictures, since you have mixed them intimately.

 

 

Edited by studiot
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21 minutes ago, studiot said:

6) There has been considerable discussion about measurements that I have not participated in, but you have not made clear that bodies are subject to other influences (forces) which affect the measurement of the local attraction experienced between bodies. These are additional to the standard force due to the Universal Gravitational Constant (whatever you wish to call it). So if you are seeking to measure any variation of G in the real world, you have to acount for these.

Bodies are not influenced by anything else than other bodies , but their shape influence their gravitational fields shape and strength in different direction an distances (they can not be assumed as point masses ) Even spheres have shape that affect gravity field both inside and out. On longer distances negligible impact (Moon at distance 385*10^6 m expire 1/1.000043 less earth gravity than very "pure" classical view point , because earth is not a point.

All old measurements are valid , and will give a very stable constant calculated "my way"

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3 minutes ago, Timo Moilanen said:

Bodies are not influenced by anything else than other bodies

The fact that the Earth is spinning affects both the magnitude and the direction of the attraction experienced by any body in the Earth's gravitational field.

 

Edited by studiot
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3 minutes ago, Strange said:

What is the difference between "adding" and "summing"?

By summing  I mean adding every miniscule component separately , adding is like taking two spheres and saying this is now a point with double the mass

1 minute ago, studiot said:

The fact that the Earth is spinning affects both the magnitude and the direction of the attraction experienced by any body in the Earth's gravitational field.

I have not gone into these details yet since this is little compared to +17 % mass

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1 minute ago, Timo Moilanen said:

By summing  I mean adding every miniscule component separately , adding is like taking two spheres and saying this is now a point with double the mass

So you are adding all the masses of the atoms (or other components) together to get the total mass and the total gravitational effect.

So how is this "not additive"?

And how is this any different from Newtonian gravity?

This is why your arguments are so confusing: you say you can't add all the masses but instead you have to add all the masses. It makes no sense.

7 minutes ago, Timo Moilanen said:

their shape influence their gravitational fields shape and strength in different direction an distances  (they can not be assumed as point masses )

True. Only spherically symmetrical object can be treated as paint masses. (This may be what leads to your mistaken belief that G is defined for spheres.)

8 minutes ago, Timo Moilanen said:

Even spheres have shape that affect gravity field both inside and out.

True. But in this case, the effect is that the gravitational force is the same in all directions and they can be treated as point masses.

11 minutes ago, Timo Moilanen said:

On longer distances negligible impact (Moon at distance 385*10^6 m expire 1/1.000043 less earth gravity than very "pure" classical view point , because earth is not a point.

You don't seem to understand the concept of "approximation".

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I think this argument about point masses is a red herring.

Whether a body is a sphere or not, it has a centre of mass.

Mechanics requires we use this centre in our formulae.

The (solid) sphere is just one of the easiest for calculation.

Edited by studiot
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6 minutes ago, studiot said:

I think this argument about point masses is a red herring.

Agreed. But it seems to be important to Timo (and may be a key part of his misunderstandings).

4 minutes ago, Timo Moilanen said:

Point of the figure is the difference in adding the pieces 

You have not explained how adding things is "not additive".

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5 minutes ago, studiot said:

The (solid) sphere is just one of the easiest for calculation.

My first solution was to calculate via distance between "gravitational mass points " Mas pieces farther away and beside the common axis do less work and the more nearby parts compensate this only to a degree . On long distances these factors go to "1.000"

The method gave same constant and that was long before I found the as someones call relativity coupling (mass is energy and the half is as v^2/2)

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15 minutes ago, Timo Moilanen said:

My first solution was to calculate via distance between "gravitational mass points " Mas pieces farther away and beside the common axis do less work and the more nearby parts compensate this only to a degree .

You seem to be trying to integrate the gravitational mass of all the points in an object. As you get the wrong result you have obviously made an error somewhere. Hopefully, someone like studiot will work through it to see where you have gone wrong (I don't have the patience - or probably even the math skills any more).

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13 minutes ago, Strange said:

You have not explained how adding things is "not additive".

Ti =my equality to G  Traditionally dM*G*cos a /s^2 is said to be gravity of one piece and is added  up to MG/r^2 

I do the same addition but divide by Ftot/Fm  to" compensate" for both cosa and the s^2 average at same time . Ftot= (sigma)dM/s^2

And this give the force F/m as I write  that is 2/3*Ti*M/R^2 at surface and TiM/r^2 for very far away . so summing up is from 2/3 to1 times  adding

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!

Moderator Note

Six pages in and the OP is trotting out the same answers that don’t address the questions, and has provided no evidence of this alleged difference in values between reality and Newtonian prediction. We’re done.

Do not open another thread on this topic

 
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