Remember the previous essays, where the tidal forces are basically the gravitational pull, that is?

Here I get - when I turn to reading up on tides - quite another formula for

*tidal*forces. My source here is a book, collection being Bibliothèque Scientifique, author being J. Rouch, his qualities being Honorary Director of Musée Océanographique and Member of L'Académie de la Marine. The title is Les Marées, and the publisher is Payot, in Paris. I turn to pages 49 and 50. French in left and my English translation in right column of below table:

Comme la Lune et le Soleil ont très sensiblement le même diamètre apparent, ils pouvaient penser que les deux astres étaient de même grosseur. | As the Moon and the Sun very visibly have the same apparent diameter, they could think that the heavenly bodies were of same size. |

Mais lorsque les astronomes établirent que le Soleil était en réalité 26 millions de fois plus gros que la Lune, | But when astronomers established that the Sun is really 26 million times as huge as the Moon, |

cette prééminence de la Lune a pu paraître surprenante. | this preeminence of the Moon could have appeared astonishing. |

Comme nous le verrons, les forces génératrices de la marée sont directement proportionnelles à la masse de l'astre qui les provoque | As we shall see, the generating forces for tides are directly proportional to the mass of the heavenly body provoking them |

et inversement proportionnelles au cube de sa distance. | and inversely proportional to the cube of its distance. |

Wait a minute!

Are the forces provoking tides gravitation or something else? Supposedly gravitation right!

And we agree between this formula and Newton's above that the gravitational forces are directly proportional to the masses - except that here J. Rouch is leaving out compounding them by the mass of earth, presumably including that in some gravitational constant.

But the inverse proportionality, is it with the square or the cube of the distance?

Certes, si le Soleil et la Lune étaient tous deux à la même distance de la Terre, | Certainly, if Sun and Moon were at the same distance from Earth, |

la marée produite par le Soleil serait bien 26 millions de fois plus forte que celle produite par la Lune. | then the tide produced by the Sun would be 26 million times stronger than that produced by the Moon. |

Mais la Lune est 389 fois plus près de la Terre que le Soleil ... | But the Moon is 389 times closer to Earth than the Sun ... |

OK, I calculated a mean distance from taking mean distance of Aphelium and Perihelium of Earth and taking the mean of Apogee and ... for Moon and got a factor of 497.

^{8}. Here the factor is only 389 times. Did I count wrong? Did science change? Or are we talking of mean distance of syzygies being a different thing than I calculated, does he start from largest distance of Moon? I do not know.

Note also that J. Rouch is giving the facts in order of explanation. As order of discovery goes (same relation as detective novel to criminal's autobiography, for instance) it would be atrocious to set hugeness or mass of Sun (did he even distinguish volume from mass=volume times density?)

*before*the distance, since it is calculated

*by*the distance, just as those of the Moon are.

... et le cube (!) de 389 est égal, en chiffres ronds, à 59 millions. | ...and the cube (!) of 389 is equal, in round numbers, to 59 millions. |

La Lune, par suite de sa proximité de la Terre, | The Moon, by reason of its closeness to Earth, |

a donc un avantage qui peut être grossièrement évalué au rapport 59/26, soit environ 2,25. | has then an advantage which grossly can be evaluated to the ratio 59/26, or 2.^{25} to round it off. |

Ce rapport est appelé | This ratio is called |

rapport des actions moyennes de la Lune et du Soleil. | ratio of mean actions of Moon and Sun. |

149 597 870 km = half long axis of the Sun (of earth according to heliocentrics)

384 399 km = half long axis of the Moon

389.

^{1734109610067}ratio of distances

151,455.

^{9437990246}square of the ratio

58,942,626.

^{25858494}cube of the ratio

If you wonder why decimals are fewer and fewer, it is because the calculator on the computer has a given space to all of the numerals.

"The Sun has a mass of 1.9891×10

^{30}kg

"The Earth has a mass of 5.9736 x 10

^{24}kg

"So the Sun is about 332,900 heavier than the Earth ..." source

"Finally, let’s take a look at mass. The mass of the Moon is 7.347 x 10

^{22}kg. But the Earth is much more massive. The mass of the Earth is 5.97x 10

^{24}kg. This means that the mass of the Moon is only 1.2% of the mass of the Earth. You would need 81 objects with the mass of the Moon to match the mass of the Earth."

Read more: source

Which also gives us the absolute (as believed) and therefore relative masses of Sun and Moon.

Now, it seems tides are not directly caused by gravitation only but indirectly only - by some formula derived from the F=GMm/R

^{2}but not identical to it. So, cube of distance

*is*correct for solar and lunar tidal influences, it would seem. And it is 58.

^{9}million times.

Well, here we get to a sticky part: how do you derive that? Page 59:

Pour évaluer la grandeur des forces qui représentent l'action d'un astre M sur une particule P placée sur la surface de la Terre, | To evaluate the greatness of forces that a heavenly body M act on a particle P on the surface of Earth |

on applique à la particule | you apply to the particle |

en plus de l'accélération que lui imprime l'astre M | above the acceleration which heavenly body M exercises on it |

une accélération égale et de signe contraire à celle que l'astre M imprime à la Terre elle-même. | An acceleration equal and of opposite sign to what the heavenly body M impresses on Earth itself. |

Wow!

If particle P is between hb M and Earth ... why?

If particle P has Earth between itself and hb M ... again why?

Because, "greatness" here obviously mean "greatness and direction", as usually with Newtonic vectors.

Somehow I missed something. I actually missed it when I was a Heliocentric in grades nine and ten discussiing this with a retired Physics teacher who sometimes replaced the younger science teacher.

I was already then suspicious of the case why spring tides at newmoons occur on BOTH sides of the earth according to the diagrams of spring tides and neap tides I had been looking at.

When given explanation that gravity displaces water between Earth and Moon+Sun more than Earth's solids but Earth's total solids more than waters behind it, I though "ok - but what about the neap tide case?" At the halfmoon crescant or the halfmoon decrescant?

Let us see if I can make sense of the following explanation as I slowly translate it from French to English:

Considerons le cas très simple des points C et D de la Terre, | Let's consider the very simple case of points C and D on Earth, |

sur la ligne qui joint le centre de l'astre perturbateur | on the line which joins the centre of the perturbing heavenly body |

(la Lune par exemple) | (the Moon for instance) |

au centre de la Terre. | to the centre of the Earth. |

The referral is now to figure 9, on which I see A and B as points on the Poles and C and D as opposite points of the equator. C has the heavenly body in zenith, D in nadir.

Leaving the text aside, since I do not have the book now, I cite content from memory.

Let r be the distance from centre of Earth to centre of Moon. Let M be the mass of the Moon. Let a be distance between centre and surface of Earth. This mass imposes on any particle in place C between itself and centre of Earth a disruption equalling the gravitational pull on point C minus the gravitational pull on Earth (centre of mass), and on any point D beyond the centre of Earth a disruption equalling gravitational pull on Earth (at centre of mass) minus the gravitational pull on point D.

How so? And how come the author instead of gravitational pull used the words "accélération"?

First question: since centre of Earth is further from Moon than point C and closer to Moon than point D, the same centre and all solids directly attached to it (or even semisolidly more attached to it than waters, if we count the theory of magmas floating inside earth) is more affected by gravitation than detachables or floatables on point D but less affacted than floatables on point C or detachables there either. Using the Newtonian formula F=GMm/r

^{2}we get first to acknowledge that:

F(point D) < F < F(point C)

Then we replace with other side of formula, making a the addition or subtraction for radius:

GMm/(r+a)

^{2}< GMm/r

^{2}< GMm/(r-a)

^{2}

Since G - gravitational constant and m - mass of Earth - are equal in all, we can bracked them out:

Gm { M/(r+a)

^{2}< M/r

^{2}< M/(r-a)

^{2}}

Or:

M/(r+a)

^{2}< M/r

^{2}< M/(r-a)

^{2}

Which implies two possible subtractions:

M/r

^{2}- M/(r+a)

^{2}= disruption for point D.

M/(r-a)

^{2}- M/r

^{2}= disruption for point C.

Now, why does the author use the word "acceleration" rather than "gravitational pull"? In Newtonian physics ... ouch!

I just forgot about the counterforces! It is as great as (Gm)(M/r

^{2}) so maybe, just maybe the equations will be equal anyway.

But the question for a philosopher is not just if the mathematics fit, but also if the interpretation on which one bases the mathematics fits the reality. So, I will not just go on with the mathematics, I will try to analyse the meaning of it. The meaning of what I just said, as well as the meaning of the other thing that J. Rouch said.

*If I can*, that is.

What I mean is simply taking accelerations produced by Moon without taking counterforces into account and making subtractions to account for differences.

What he means is that there is at C a counterforce equal to the pull of the Moon ... somewhere? ... which counterforce is equal to Moon's pull on centre of mass of Earth. Why?

Does this difference of my intuitive following of his concepts and the way he expresses them mean I misunderstand his so to speak physical and ontological physics? Or is he merely saying it a different way in order to get an opposite sign on other side in order to get forces of opposite directions because he is calculating vectors rather than just sizes of forces? Is it just a difference of mathematical expression of physics in formulas?

Why is this an important question for the philosopher?

I am not writing this essay to get a physicist carreer instead of a writer's and composer's carreer. I am writing it to understand if possible and if possible make my readers understand whether the physics involved - the ontological part, what you can describe in words that can be intuitively understood - make sense, and if they make sense as well from a Geocentric perspective.

OK. What does Newton say when he claims that every action of a force provokes an equal and opposite counterforce?

I tried to figure it out, here is what I came up with: if Moon pulls Earth, and if Earth does not move towards Moon, this means Earth exercises a counterforce to the Moon, equal in size (M/r

^{2}) to that exercised by Moon and opposite in direction. Does this mean Earth is all the time vibrating back and forth? And where would such a force come from? Is Newton not attributing to matter a power of obstinacy which of itself it has not? Is this even a good time to ask the question? And is this the kind of question that physicists deal with when asked by common people, or do they leave such questions out even then?

Whichever formula we chose, and whether they are equivalent or not (I have not yet done the maths, will do) I would like to know:

1°) Is it a counterforce manifested in small tugging back and forth as vibration or is it a difference of displacement in acceleration as seen in absolute space;

2°) can both explanations be used for a Geocentric cosmography (as in serious cosmography, not just mathematical model) and for a Heliocentric / Acentric one;

3°) and is the maths involved the same in both explanations?

Getting back to writing this essay after a few tries at recuperating the algebraic steps that J. Rouch goes through to derive the inverse cube rule for tidal forces from the inverse square rule for gravitational pull in total. Remember, the difference between on the one hand the gravitational pull on solid earth masses, between sea and sea on opposite sides of Earth, which are everywhere equal to a gravitational pull where the masses of Earth and Moon are divided by the square of distance from centre of Moon to centre of Earth, because solids are connected to centre and, on the other hand, the gravitational pull on loose things like water on one or other side, where the masses of Earth and Moon are divided by the square of distance between centre of Moon and that one side of the Earth, that difference rather than the gravitational pull as such, is the tidal force. And similar as to Moon and its tidal forces, so also Sun and its tidal forces.

So, let us take either M/r

^{2}- M/(r+a)

^{2}or M/(r-a)

^{2}- M/r

^{2}. I have been stuck with the first of these some time till today continuing, and here is what I came up with.

First of all, we need to get a common denominator. For 1/3 - 1/4, we need to get a denominator which can be translated to .../3 as well as to .../4 and we get it by multiplying these, which have no factor in common: .../12. But such a multiplication must be made on the enumerator side as well. To translate 1/3 into .../12 we multiply both enumerator and denominator by 4, like this: 4/12. And to translate 1/4 into .../12 we multiply by 3, like this: 3/12. So 1/3 - 1/4 = 4/12 - 3/12 = 1/12. Now, apply this to the formula M/r

^{2}- M/(r+a)

^{2}!

I get M((r+a)

^{2}- r

^{2})/r

^{2}(r+a)

^{2}.

Now for any (a+b)

^{2}- a

^{2}we remember the conjugate rule, which states that a

^{2}- b

^{2}= (a+b)(a-b). Obviously this is no way near giving the same geometry of the one side and the other side of the equation, but it is giving the same sum in arithmetic by diverging paths. When we apply algebra to natural phenomena one or other way of describing the mathematics may be a good way to describe the things, but it might be the other way that makes calculation simpler.

In the conjugate rule, insert (r+a)

^{2}for a

^{2}and r

^{2}for b

^{2}. When we simplify the two sides we get a very funny-looking and even seemingly paradoxical formula:

(r+a)

^{2}- r

^{2}

*=(r+a+r)(r+a-r)*

= 2ra + a

^{2}.

I doubted the result very much, as I found it paradoxical that r gets squared on one side but not other, but I made a table of rs and as with whole number values. For one on either variable, both formulas give three. For 2 and 1 they give 5. For two and two they give 12. Try 3 and 1, 3 and 2, 3 and 3, 4 and 1, 4 and 2, 4 and 3 and 4 and 4 yourself. Whatever values you put into the formulas, wherever r is equal to or greater than a, it remains true that:

(r+a)

^{2}- r

^{2}= = 2ra + a

^{2}

So, I can now simplify

M((r+a)

^{2}- r

^{2})/r

^{2}(r+a)

^{2}

into

M(2ra + a

^{2})/r

^{2}(r+a)

^{2}.

But I still cannot get to the inverse cube rule we are looking for.

Now, since I am not seriously doubting that J. Rouch got the algebra right insofar as algebra is ever right (i e in equivalent numbers or sizes or other values), I will simply give what J. Rouch offered as an explanation and hope people better in algebra than I can make some sense of it:

M/(r-a)

^{2}- M/r

^{2}

= M/r

^{2}[1/(1 - a/r)

^{2}- 1]

=2Ma/r

^{3}(with a/r being very small)

And remember how we excluded mass of earth and gravitational constant from the calculation, here he brings them back again in one quantity he calls f.

On either side of Earth, on equator close to and far from Moon, we have now a perturbing force, a difference of gravitational pull, equal to:

(2Ma/r

^{3})f

But to get it on other side, instead of starting from

M/(r-a)

^{2}- M/r

^{2}

we start from

M/r

^{2}- M/(r+a)

^{2}

which means we get the perturbing force in opposite direction, it would seem!

Now, what does this mean physically? I mean in descriptions with verbs like pull and push and so, rather than in physicists' algebra?

One

*could*say: the Moon makes the water on Moon side rush up from the Earth a little faster than it makes the Earth rush up from the waters on the side opposite the Moon. A Heliocentric would add: for as long as Earth is in that rotational position, until it is another angle of Earth and water that is aligned straight line with Moon. A Geocentric believing this explanation (such as Sungenis unless he takes my other interpretation of the formula) would say until Moon rotates around Earth and makes this happen to another angle of Earth and water aligned with itself. Meanwhile, in either explanation, the parts where this happened an hour earlier or so are dealing with a similar formula but a smaller a and therefore a smaller tidal force. When they reach or Moon reaches a position where they are not aligned with the Moon but stand side by side with centre of Earth with same distance to Moon as it, then we reach the lowest point of the tidal forces.

That was one interpretation. Now let us get to the actual words of J. Rouch.

"Let r be the distance of centre of Earth from centre of Moon. The attraction of Moon on centre of Earth is, in virtue of the law of universal gravitation [F=GMm/r

^{2}], proportional to M/r

^{2}, M being the mass of the Moon. The distance r is in the mean constant. [He would have been wiser to say that it is in any given moment equal to itself, since that is very much more relevant]. The attraction of Moon on Earth, since the distance r remains constant, must be counterbalanced by a centrifugal force directed in opposite direction of attraction and equal to M/r

^{2}."

The obvious problem is: where does J. Rouch get this centrifugal force from? It may be a postulate to have it, in order to make distance of Earth and Moon remain constant, but in astronomy that is taken care of by the speed of the Moon rotating around Earth. Not by any force on Earth directed opposite way. In Sungenis' system, there would be a pull, and the gravitational forces of the Universe (all stars taken together rotating daily around the Earth) would pull the Earth immediately back into position in absolute centre. Such a pull and pull back would make Earth vibrate with a vibration strongest on a line through the Earth, aligning centre of Earth with centre of Moon and second strongest on a line aligning centre of Earth with centre of Sun, except at newmoon and fullmoon when these two lines coincide.

So, for water on point C, it is pulled upwards by the Moon and shaken back down by Earth pulling back with a slightly less force. For water on point D, it is pushed upwards by the Earth's backpull from the Moon and pulled down by Moon's slightly less great pull. And for either, the pushing or pulling up is slightly greater than the pulling or shaking down.

Note here, that though the Mathematics is quite equivalent on either side of the Earth on points aligned with centres of Earth and Moon, the physical causality is not the same, but in fact inversed.

Hans-Georg Lundahl

Georges Pompidou Library, Paris

St Geoffroy, bishop and abbot

8-XI-2012

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