Why doesn't the moon leave earth's orbit?

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22 years 1 week ago #3760 by Jim
Replied by Jim on topic Reply from
Never mind the minute details and look at the basic statement that the tidal force is proportional to the 3rd power of the distance to the moon and sun. The moon is nearer by 14% than it is at its most distant point in its orbit. Do the math; 1.14x1.14x1.14= a much higher tide than is observed. About the 3.8cm data-if you get a reply you are better at this than I am.

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22 years 1 week ago #3331 by Atko
Replied by Atko on topic Reply from Paul Atkinson
…and how do you make the leap from 1.14^3 to a tidal amplitude? We need to combine the gravitational force and the centrifugal force when figuring out <b>actual heights</b> of tides due to the Moon's effect. The formula can be derived as -

Height ~ (M*r^4*cos^2a)/(E*R^3)

Where

M = Moon Mass
E = Earth Mass
a = Angle of the radial component of the gravitational force of the Moon (assume =0, i.e. high tide)
r = Earth's radius
R = Distance

Plugging in the numbers -

At Apogee -

~ (7.353E+22*6357E+3^4*1)/(6E+24*4.1E+08^3)
~ 0.3m

At Perigee -

~ (7.353E+22*6357E+3^4*1)/(6E+24*3.56E+08^3)
~ 0.44m

This gives a variation of 0.44/0.3 = 1.485, which is naturally from the above, 1.14^3.

This is the actual variation in height of the tide in the Earth's oceans due to the Moon's influence. Tides vary more in coastal waters, basins, creeks and rivers due to compression effects. Even the ocean's tides can rise and fall enormously due to air pressure, currents and other variables. And of course there's also the Sun's slightly smaller force to add to all this.

The bottom line though, is there is no anomaly in the observed differences between spring and neap tides, in fact the variation fits the formula. The force is basically acting over a very short amplitude in terms of sea height.


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22 years 1 week ago #3375 by Jim
Replied by Jim on topic Reply from
The 1.14 3rd power tide effect is not the spring/neap tide. This effect is not even included in the model as far as I know(I could be wrong). The 3rd power law is a I stated before proportional to distance cubed and so the effect would be to generate tides 40% more energentic than observed. And so I conclude the law is false(unless or until proven otherwise). You can assume as do all other experts on tidal theory that I'm wrong. Then you learn nothing-prove it one way or the other.

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22 years 1 week ago #3761 by Atko
Replied by Atko on topic Reply from Paul Atkinson
Whatever, the formula gives the values for the variation in tide heights excluding Sun, tectonics, air streams, pressure etc ad nauseum - i.e. the Moon's effect - and meets your 48% more energetic requirement. So I don't see where your issue lies? Tide variation in the open sea due to the Moon's influence is a fraction of a metre, matches predictions, and that's all there is to it.

I don't assume you're wrong per se because there's no disagreement between what you're saying and the third power law predicts. What is your expectation of how this "more energetic" state will manifest itself?

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22 years 1 week ago #3624 by Jim
Replied by Jim on topic Reply from
When the moon is near the tide will be 48% higher than when the moon is far. This has nothing to do with any of the details you are talking about. This effect is not observed.

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22 years 1 week ago #3337 by Jim
Replied by Jim on topic Reply from
I reread the last few posts on this subject and it seems that some confusion about tidal flow entered a while back. The spring tide is a high tide that occurs at the new and full moon. The neap tide is a high tide that occurs at the 1st and 3rd quarter moon. These are all high tides. (The low tides are not honored like high tides and I don't think any low tide is noted in models). So, the fact that the spring tides are 40% higher than the neap tides is not a result of the distance between the Earth and moon as is claimed in the earlier post.

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