Airy’s Failure And The Sagnac’s Experiment
Airy’s “Failure” was that to observe a particular star he did NOT have to increase the tilt of his telescope when he filled it with water. This animation demonstrates what happens, proves that the Earth is stationary and therefore at the centre of the universe – as the Bible maintains.
Sagnac’s experiment proved the existence of the aether, thus demolishing Einstein’s Theory of Relativity. Complaints by two experienced physicists that when they were at university, they were never informed about this important experiment. When other experiments – Airy’s Failure, Michelson-Gale’s experiment (also not taught at universities!) and of course the famous Michelson-Morley experiment- a completely different picture of the cosmos emerges – with the Earth at the centre of the universe.
See Also:
How Did The Kings of Astronomy Get it So Wrong? Part II: Einstein and the Still Earth
Gerrard Hickson PORTRAIT OF THE AUTHOR, BY HIMSELF *** It will be remembered how Hipparchus failed to get an angle to the stars 2,000 years ago, and arrived at the conclusion that they must be infinitely distant; and we have seen how that hypothesis has been handed down to us through all the centuries without …
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Perihelion shift of Mercury
There is a model of Mercury. A long rod penetrates a true sphere and at the both ends of the rod, weights are set. This model is rotating horizontally and is moving on the orbit of Mercury (two planes fall on). Main forces acting on the weights are gravity of the sun and inertial force (centrifugal force). And each force acting on the outside weight and inside weight is different.
Inertial force pulls the orbit to the outside. But actual orbit of Mercury is pulled to the inside. Gravity of the sun acting on the two weights is inversely proportional to the square of the distance (not come out even. not plus minus zero). In Mercury, the action of gravity will be superior.
Now, there are plural modelsof Mercury. Length of rods and mass of weight each is different. These are revoleved around separately on the real orbit of Mercury. Maybe, all will be explained by Newton’s theory (including 575 arcsec).
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Perihelion shift of Mercury
Let’s reconsider the main cause of perihelion shift again. On Mercury or Venus, main cause will be the size of sphere. On Earth or Mars, effect of satellite is added. On asteroids each, effects of size is negligible. On Jupiter or Saturn each, the powerful and unstable effect of the other will act. On Uranus or Neptune each, slight and unstable effect of the other all planets will act. Anyway, common view on Mercury is wrong.
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Perihelion shift of Mercury
The perihelion shift of the earth is 11.45 arcsec / year. Main cause will be its size (size of sphere). It is the same to Mercury. In addition, the earth has a moon as a satellite that Mercury does not have. The inertial force of the moon and gravity of the sun acting on the moon are also powerful. And like Mercury, effect of other planets must be slight and unstable.
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Perihelion shift of Mercury
The perihelion shift of the Earth is 11.45 arcsec / year. Main cause will be its size (size of sphere). It is the same to Mercury. In addition, the Earth has a moon as a satellite that Mercury does not have. The inertial force of the moon and gravity of the sun acting on the moon are also powerful. And as Mercury, effect of other planets must be slight and unstable.
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Perihelion shift of Mercury
Let’s reconsider the main cause of perihelion shift again. On Mercury or Venus, main cause will be the size of sphere. On Earth or Mars, effect of satellite is added. On asteroids each, effects of size is negligible. On Jupiter or Saturn each, the powerful and unstable effect of the other will act. On Uranus or Neptune each, slight and unstable effect of the other all planets will act. Anyway, common view on Mercury is wrong.
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Perihelion shift of Mercury
In Mercury, the non-uniformity of the Sun’s gravity (in the size of Mercury) will be the main cause of perihelion shift. Even in artificial satellites, the effects of non-uniformity of the Earth’s gravity (the position of the center of mass and the center of gravity are different) are also mentioned.
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Moon and earth (an essay)
As moon passes overhead, high tide (one of two high tides a day) will come after a short delay. But why ? Why is seawater with a low specific gravity bulged ? Newton imagined that moon will continue falling. Earth will continue falling also. And seawater will cotinue falling too. So, it doesn’t matter how specific gravity is.
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Moon and earth (rewritten again)
Allow me to rewrite again.
Moon and earth are supposed to be two-body problem. Also it is supposed that the common center of gravity of moon and earth is situated outside the earth. And the orbit of earth is a perfect circle, and earth is a perfect sphere. In the illustration, the surface of earth closest to moon is A and the surface farthest from moon is B. At the two points A and B, the strength of resultant forces of moon’s gravity and centrifugal force (of earth) will be the same (act to opposite directions). Otherwise, earth cannot stay on orbit of perfect circle. This will explain that the level of high tide twice a day are the same.
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Pelihelion shift of Mercury (an essay)
Some wide binaries are separated by one light years. And many wide binaries are at most (as much as) by 1000 au. These motion will be treated as mass points (a point). On the other hand, many close binaries are detected. What are physicists who repeat nonsense on perihelion shift of Mercury ?
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Moon and Earth (rewritten)
Suppose the moon and the earth is two-body problem. And imagine, the earth is revolving around the common center of gravity with the moon. The orbit is a perfect circle. If lunar attractive force acting on the center of gravity of the earth is action, the centrifugal force of the earth is a reaction. And the strength of the two will be equal. This will be also true for the earth as a whole.
In an illustration, the earth is drawn next to moon. Imagine two points on the surface of the earth closest to the moon and farthest from the moon. The difference between lunar attractive force and the centrifugal force of the earth at above two points will be almost equal and therefore the resultant force will also be almost equal. This will explain that the level of high tides twice a day are almost equal.
Note: Is the law of action and reaction valid for celestial bodies on elliptical orbits?
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Moon and earth
Moon and earth are supposed to be two-body problem. In the illustration, the surface of earth closest to moon is A and the surface farthest from moon is B. At the two points A and B, the strength of resultant forces of moon’s gravity and centrifugal force (of earth) will be the same (act to opposite directions). It will be proved by the sameness of high tides.
Also it will be proved by stability of orbits of each body. 89
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Moon and earth (rewritten again)
Allow me to rewrite again.
Moon and earth are supposed to be two-body problem. Also it is supposed that the common center of gravity of moon and earth is situated outside the earth. And the orbit of earth is a perfect circle, and earth is a perfect sphere. In the illustration, the surface of earth closest to moon is A and the surface farthest from moon is B. At the two points A and B, the strength of resultant forces of moon’s gravity and centrifugal force (of earth) will be the same (act to opposite directions). Otherwise Hu is propagated, earth cannot stay on orbit of perfect circle. This will explain that the level of high tide twice a day are the same.
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Perihelion shift of Mercury (rewritten)
Mercury is moving on the revolution orbit. The sun’s gravity is equal to the centrifugal force. Because the two are action and reaction. Following are some explanations. Centrifugal force follows Mercury’s mass. But in addition, gravity is affected by the size of Mercury (and acting position of gravity is different). These are caused by the non-uniformity of gravity (in the space occupied by Mercury). And after perihelion passage, orbit will be pulled inward (from its original orbit).
High tide level twice a day is the gsame. Gravity and centrifugal force caused by the moon will be action and reaction in total. Centrifugal force is not fictitious.
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Perihelion shift of Mercury
The perihelion shift of the earth is 11.45 arcsec / year. Main cause will be its size (size of sphere). It is the same to Mercury. In addition, the earth has a moon as a satellite that Mercury does not have. The inertial force of the moon and gravity of the sun acting on the moon are also considerable. And like Mercury, effect of other planets must be slight and unstable.
Value 5.75 arcsec/year seems to be an observed value. In a website, contribution of other planets to this value is shown. These are added simply !! And value 5.75 arcsec (and contributions) seems to be constant every year !! On these problems, further explanation seems not to be done.
I say again, other planets will not be main cause of this value 5.75 arcsec.
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Perihelion shift of Mercury
Perihelion shift moves forward constantly. It cannot be explained by gravity of other planets.
On asteroids, no perihelion shift will be observed. Some size is needed.
Cause of perigee movement of the moon is written to be the sun. But it will be the same phenomenon to perihelion shift of planets. Not acceptable.z
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Perihelion shif of Mercury
Mercury revolving is divided in two (hemisphere A facing the sun and the other B). Inertial force is A＜B and gravity is A＞B (center of gravity is not on the orbit).
Above must be the most natural explanation of perihelion shift of Mercury. Because the value of pergee movement of the moon is remarkable (around 8.85 years). On the other hand, value of asteroids will not be found. Common explanation is not acceptable.
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Perihelion shift of Mercury (again)
The main forces acting on Mercury are attractive force of sun and centrifugal force only. Hemisphere of Mercury facing sun is supposed to be A and the other hemisphere is B. The attractive force of the sun acting on the two will be A>B, and the centrifugal force will be B>A. From the look of the perihelion shift, the attractive force acting on Mercury as a whole will be slightly stronger.
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Perihelion shift of Mercury (an essay)
On planets, it is said that centrifugal force caused by orbital motion is balanced with attractive force of sun (and the action and reaction are equal). On Mercury, it will stand up also. But exactly ? Solar wind or some other solar-derived substances seem to go down slightly speed of orbital motion of Mercury (one of perturbation ?). If so, centrifugal force will be reduced accordingly. On the other hand, attractive force of sun will not be affected at all. This may be the main reason for the perihelion shift of Mercury. You
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Perihelion shift of Mercury (an essay)
On planets, it is said that centrifugal force caused by orbital motion is balanced with attractive force of sun (and the action and reaction are equal). On Mercury, it will stand up also. But exactly ? Solar wind or some other solar-derived substances seem to go down slightly speed of orbital motion of Mercury (one of perturbation ?). If so, centrifugal force will be reduced accordingly. On the other hand, attractive force of sun will not be affected at all. This may be the main reason for the perihelion shift of Mercury.
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Propagation of gravity
The propagation of gravity will be done in an instant. For this, here are two reasons. One reason is that two-body problem, many-body problem are true for celestial bodies. The other reason is that the whole solar system is in an uniform linear motion and planets are in elliptical revolution on their revolution planes.
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Centrifugal force : reconsidering
Two mass points a and b (with large difference of mass) are in motion of two-body problem (a large, b small). One picture is in outer space, and the knot of a,b is gravity (b is in a circular motion). The other picture is on a plane (no friction), and the knot is tension of a string (the mass of the string is zero). In the two pictures, the centrifugal force of b is a vector that is on the extension line of a,b. The reaction is gravity in one picture and tension of the string in the other picture. Where and how is the centripetal force?
Tension of the string between a and b is constant. Gravity is not constant.
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