# Could the earth stop spinning one day??

Discussion in 'Astronomy, Exobiology, & Cosmology' started by Vega, Jul 14, 2006.

1. ### MetaKronRegistered Senior Member

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Superluminal, one very important thing that you are missing is that the tidal bulge is much larger. You have a fourth power sort of effect going. The mass of the bulge is roughly inversely proportional to the square of the distance, and the gravitational pull is roughly inversely proportional to the square of the distance. That means that when you have one tenth the distance you have ten thousand times the drag.

3. ### superluminalI am MalcomRValued Senior Member

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Can you please demonstrate this for me? Possibly with some pictures and some math? Otherwise your statements sound like evasive gibberish.

5. ### superluminalI am MalcomRValued Senior Member

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And the kinetic energy dissipated during this "drag" goes where? And by what mechanism?

7. ### MetaKronRegistered Senior Member

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I'm going for the mechanism that transfers the kinetic energy to the intruding object.

When the object is closer, the tidal bulge becomes larger. If its height is linearly proportional to the gravitational pull from the object, then it is inversely proportional to the square of the distance from the object. The mass involved in the induced bulge is probably proportional to the cube of the height of the bulge. If these things hold true then the force working against the rotation of the Earth is inversely proportional to the fifth power of the distance of the intruder from the Earth. Thus one tenth the distance yields 100,000 times the drag.

You haven't shown me that the kinetic energy must be turned into heat. You have no real idea how to estimate how much, so you don't have even a back-of-the-envelope calculation. You don't have to do that and you don't have to try to make it look like you've proven anything, either. I can prove that some of it goes to accelerating the intruder, and you can prove that some of it is turned to heat, but neither one of us can prove how much goes to which. I am not going to yield to your winning personality or any credentials that you may have. I am also not going to yield to your threat to find my IP and try to stalk me.

You have yet to impress me.

8. ### superluminalI am MalcomRValued Senior Member

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There seem to be some mistakes here. I'll deal with them later.

Oh brother.

If that was my purpose I'd send you a private picture of my nether regions. Then you would have to be impressed.

9. ### MetaKronRegistered Senior Member

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I'm telling you what it looks like. I don't know if what I said was "mistakes" so much as an incomplete picture of the problem. I think that it will turn out that the tidal drag against the rotation of a planet is roughly inversely proportional to the fifth power of the distance. It's because the mass of the bulge is roughly proportional to the cube of the height of the bulge, a very simple fact of solid geometry. The height of the bulge is proportional to the pull against it, which is proportional to the inverse of the square of the distance. Half the distance, the gravitational pull is four times as much, the mass of the bulge is eight times as much, the result is thirty-two times the drag against the rotation of the Earth. A tenth of the distance gives you a hundred thousand times the drag.

One thing that is true about Venus that should be impossible is the fact that Venus always keeps the same face to the Earth. It is tidally locked with the Earth. We might assume that this is impossible at thirty million miles and neither being the satellite of the other and this fact may imply a much closer relationship at one time. A fluid Venus would have been much more affected than the Earth, too.

10. ### superluminalI am MalcomRValued Senior Member

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There is a definite way to calculate the size of tidal bulges and the amount of torque applied by one body on another. I'm workin on it. As for Venus being tidally locked to Earth - I don't think that's been shown conclusively. Do you have a link?

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12. ### OphioliteValued Senior Member

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Humbug. Utter balderdash. Where do you get your "facts"?

13. ### leopoldValued Senior Member

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In the 1960's, radar pulses were bounced off of Venus while at its closest distance to the Earth, and it was discovered that its rotation period, its day, was 243.09 +/- 0.18 earth days long, but it rotated on its axis in a backwards or retrograde sense from the other planets.

14. ### TheoryofrelativityBannedBanned

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PJ I have replied to your pm.

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16. ### eburacum45Valued Senior Member

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Tidal flexing is what keeps Io erupting, and keeps the subsurface water ice of Europa fluid. Both those moons would be cold , inactive worlds without the internal heating they get friom tides. You are proposing an effect many, many magnitudes greater. The biosphere of Earth would be sterilised by such an effect. Good luck with that.

17. ### eburacum45Valued Senior Member

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Venus and Earth might (or might not) have a resonance, which is a little different to a tidal lock. From here
http://history.nasa.gov/SP-345/ch8.htm

18. ### MetaKronRegistered Senior Member

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Then you should have no trouble finding a model that proves this.

19. ### Janus58Valued Senior Member

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No, the mass would only increase by the cube if the bulge increase equally by all three directions. Since the bulge increases only by its height and not in the West and East or North and South directions, it's volume and therefore mass, only increases linearly with height. Another case of your trying to bend the facts to meet your needs
And what you so conviniently ignore, is that even if these things held, this would also increase the tidal force working to pull the tidal bulges free of the planet by the same factor ( you can add into this the fact that as the tidal bulges grow, they get further from the center of the Earth, which decreases the Earth's hold on them, making it even easier for the tidal forces to pull them free). Since it has already been shown that this force is magntudes larger than the force acting to slow the Earth, your argument does not change the fact that the Earth's oceans would be ripped away by tidal forces before the Earth could be stopped in 24 hrs.

20. ### OphioliteValued Senior Member

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Here is a sequence of observations -

Metakron: One thing that is true about Venus that should be impossible is the fact that Venus always keeps the same face to the Earth.

Ophiolite: Humbug. Utter balderdash. Where do you get your "facts"?

Metakron: I put up the links. Put the balderdash up your ass.

Outcome: Here is the link posted by metakron - http://en.wikipedia.org/wiki/Tidal_locking

Here is the relevant quote (the emphasis is mine) - A curious aspect of Venus' orbit and rotation periods is that the 583.92-day interval between successive close approaches to the Earth is almost exactly equal to 5 Venusian solar days (precisely, 5.001444 of these), making approximately the same face visible from Earth at each close approach.

Keeping the same face towards the Earth as the Moon (almost) does, is not the same as making approximately the same face visible at each close approach.

Your own source demonstrates that you were talking, as I pointed out, humbug and balderdash. Please be so kind as to acknowledge this.

21. ### LaikaSpace BitchRegistered Senior Member

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So how big do you think the tidal bulge was MetaKron? And how was such strain accommodated over such a short time-scale without leaving a scar? And all the other questions I've asked which remain unanswered, and which I cannot be bothered to write for the umpteenth time.

22. ### Walter L. WagnerCosmic Truth SeekerValued Senior Member

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(tune playing in background)

You've got to know when to hold 'em, and when to fold 'em!

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