75 lines
4.9 KiB
Plaintext
75 lines
4.9 KiB
Plaintext
|
The Geocentric Testimony of our Tides
|
|||
|
|
A brief study of the near and far side effects of gravity
|
|||
|
|
robert.bennett@rcn.com
|
|||
|
|
There are abundant attempts that try to resolve why the Earth’s high tides occur when the Sun or Moon are at both the near side and far side. Also an issue is why the Moon is more than twice as important as the Sun in determining local tidal ranges.
|
|||
|
|
Geostatics: Let us try to find the acceleration at points A and B with respect to the center of the Earth O, due to only the gravitational influence of Moon and Earth, as shown in the figure.
|
|||
|
|
|
|||
|
|
O and X are the center of the Earth and Moon respectively. Let the radius of Earth be RE, distance between Earth and Moon be d, and the mass of Earth and Moon be mE and mM respectively. Acceleration to the right is positive. Note that RE << d = distance OX.
|
|||
|
|
|
|||
|
|
Acceleration of point B aB is: aB = −GmE/RE2+ GmM/(d−RE)2 ~= −GmE/RE2 + GmM(1+2RE/d)/d2
|
|||
|
|
|
|||
|
|
Acceleration at point A is: aA = GmE/RE2 + GmM/(d+RE)2 ~= GmE/RE2 + GmM (1-2RE/d)/d2
|
|||
|
|
|
|||
|
|
And aO is:
|
|||
|
|
|
|||
|
|
aO = GmM/d2
|
|||
|
|
|
|||
|
|
Thus, the accelerations of point A and B with respect to O are:
|
|||
|
|
|
|||
|
|
aAO = aA−aO = GmE/RE2 + GmM (1-2RE/d)/d2 - GmM/d2 ~= GmE/RE2 + GmM (1 - 2RE/d)/d2 - GmM/d2 ~= GmE/RE2 - 2GmM RE/d3
|
|||
|
|
|
|||
|
|
aBO = aB−aO = - GmE/RE2 + GmM (1+2RE/d)/d2 - GmM/d2 ~= - GmE/RE2 + GmM (1+2RE/d)/d2 - GmM/d2 ~= - GmE/RE2 + 2GmM RE/d3
|
|||
|
|
|
|||
|
|
Replacing the Moon with the Sun will repeat the same calculations...with mM => mS and d => D The combined effect of Moon and Sun – acting in the same straight line – is
|
|||
|
|
aAO = aA−aO ~= GmE/RE2 - 2GmMRE/d3 - 2GmSRE/D3 aBO = aB−aO ~= - GmE/RE2 + 2GmMRE/d3 + 2GmS RE/D3
|
|||
|
|
So aAO = -aBO and aAB = aAO – aOB = 2AO = 2GmE/RE2 - 4GmM RE/d3- 4GmS RE/D3 This is the acceleration difference from the near to the far side, caused by the gravitational attraction of the Earth, Moon and Sun...to order(RE/d and RE/D)
|
|||
|
|
Since aBO = −aAO , on both sides of the Earth, water will be trying to accelerate equally from the center of the Earth, causing the same tides on both sides of the Earth.
|
|||
|
|
|
|||
|
|
What is the ratio of the lunar acceleration to the solar acceleration of the tides....aM/aS?
|
|||
|
|
|
|||
|
|
Based only on the static gravitational forces above, the ratio is
|
|||
|
|
|
|||
|
|
aM/aS = [2GmMRE/d3]/[ 2GmSRE /D3] = (mM/mS)*(D3/d3)
|
|||
|
|
|
|||
|
|
In MKS units:
|
|||
|
|
|
|||
|
|
mM = 7.35*1022 kg mS = 2*1030 kg d = 3.84*108 m D = 1.5*1011 m
|
|||
|
|
|
|||
|
|
aM/aS ~= 2.35 theoretically... due to geostatics.
|
|||
|
|
|
|||
|
|
Here are the values of the Moon/Sun tidal force ratio, aM/aS, as cited by online sources:
|
|||
|
|
|
|||
|
|
Physics forum 2.16
|
|||
|
|
|
|||
|
|
NOAA
|
|||
|
|
|
|||
|
|
2.5
|
|||
|
|
|
|||
|
|
Hyperphysics 2.27
|
|||
|
|
|
|||
|
|
Wiki
|
|||
|
|
|
|||
|
|
2.22
|
|||
|
|
|
|||
|
|
NJIT
|
|||
|
|
|
|||
|
|
2.2
|
|||
|
|
|
|||
|
|
<Average> 2.27
|
|||
|
|
|
|||
|
|
Gravity only – no dynamic central forces = 2.35 ..a difference of 3%
|
|||
|
|
|
|||
|
|
The Moon’s acceleration of the tidal water is more than twice that of the Sun at the New Moon alignments. The geostatic prediction is consistent with a cosmic model where the Earth is stationary. Only gravitational forces produce the tides.
|
|||
|
|
Geokinematics: The Moon causes no Centrifugal Acceleration on the Earth’s tides because the Earth doesn’t orbit the Moon....the Moon orbits the earth. But the Sun causes a Centrifugal Acceleration on the Earth in the consensus heliocentric model: On the far side: CAS = V2/D = (30+.47)2/1.5*108 = 6.2*10-3 m/s2
|
|||
|
|
V is the earth’s orbital speed D is the AU. On the near side CAS = V2/D = (30-.47)2/1.5*108 = 5.8*10-3 m/s2 At the center O: CAS ~= 6.0*10-3 m/s2
|
|||
|
|
How does CAs compare with aS, CAS/aS...the Sun’s centrifugal acceleration of the earth compared to its gravitational acceleration ? aS = 2GmS RE/D3 ~= 5.1 * 10-7 m/s2 where G = 6.67*10-11 CAs/as ~= 6.0*10-3/ 5.1 * 10-7 = 1,170 Result – the central force caused by the Sun compared to the gravitational acceleration toward the Sun is almost 1200 times larger! Were this true, the tides would be a thousand times higher than reality... This is wildly beyond the size of aS and aM, whose values do give the correct tidal range!
|
|||
|
|
Summary 1. Since gravity alone explains the tides - the first conclusion is that CAs does not exist – CAs is zero. 2. The second conclusion is that the Earth does not orbit...Geocentrism.
|
|||
|
|
|
|||
|
|
This result adds to the burgeoning experiments that test and confirm the geocentric hypothesis.
|
|||
|
|
Static gravity alone accounts for the observation of the double tides on opposite sides of the Earth and the different range of the lunar and solar tides.
|
|||
|
|
When the centrifugal acceleration of the Earth’s [alleged] orbital speed is added to the gravitational accelerations, the HelioC theory is exposed as fictitious. There is no evidence of a centrifugal acceleration ...or force... in the tidal behavior.
|
|||
|
|
This agrees with other tests, like Newton’s Bucket, Sagnac’s rotor and R. Wang’s linear Sagnac version. They also support an immobile Earth.
|
|||
|
|
Conversely, there are no proofs by testing by scientific method or realistic interpretation of the tests that the Earth orbits the Sun.
|
|||
|
|
The rise and fall of the tides around the world is a semi-diurnal repetitive demonstration of the Earth’s central position in the universe.
|
|||
|
|
|