# 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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You can still shove the "balderdash" up your ass, though.

3. ### Janus58Valued Senior Member

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Which part do you consider impossible? That the mass only increases by the cube if it increases equally in all three dimensions, or that the tidal bulge only increases its height?

5. ### invert_nexusZe do caixaoValued Senior Member

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A thought that occurred to me while driving to work this morning.

Everyone is going on about the effect of the tidal bulge and whatnot on the Earth... But what about its effect on the passing body? The models so far provided have all been one-sided.

What effect upon the math would figuring the tidal effects on both the Earth and the passing body? (Don't forget Luna as well....) With three interacting bodies, the math likely becomes rather nasty?

7. ### superluminalI am MalcomRValued Senior Member

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For reference, here is a previous post of mine:

The relative acceleration created by the moon on the earth (tidal effect) is:

(G * M<sub>moon</sub> * 2 * R) / D<sup>3</sup>.

The tidal effect is therefore directly proportional to the mass of the "body".

This means that the analysis in the above quoted post is a reasonable approximation. The following is therefore true:

The gravitational gradient caused by a body with enough mass to "pull" on a protion of the earth hard enough to stop it within an hour would tidally destroy the earth. Please examine the following wiki article.

http://en.wikipedia.org/wiki/Roche_limit

and you can estimate what a body that is 1.85e12 times more gravitationally powerful will do, given that it must "pull" on the tidal bulge that much more than the mood currently does.

8. ### invert_nexusZe do caixaoValued Senior Member

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Yes.
The Earth.

That's why I brought in the other body. It would be experiencing tidal effects and whatnot as well.

How would this affect the energy being mentioned in this system? Would it be shared between all participating bodies? Or would each body experience the aforementioned energy?

9. ### superluminalI am MalcomRValued Senior Member

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The immensely larger body would hardly notice the presence of the earth as it tore us apart.

10. ### invert_nexusZe do caixaoValued Senior Member

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And, by the way, what led me to thinking along these lines (from a non-earth vantage point) was mention of an article in the newest issue of Science on the podcast I was listening to on the way to work.

http://www.sciencemag.org/cgi/content/summary/313/5787/588b

The paper provides a new model explaining the discrepancies in the shape of the moon taking into consideration its spin and tidal effects from the Earth.

An interesting notion is that early in the moon's history, it most likely had an eccentric orbit and would thus grow and shrink in the skies. Also, it probably orbited the Earth several times every Earth day. Would have made for a striking picture.
This aspect is mostly off-topic of course.

11. ### superluminalI am MalcomRValued Senior Member

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Still, would have been extremely cool to see.

12. ### invert_nexusZe do caixaoValued Senior Member

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Oh. Yeah.
Doh!

I was thinking in terms of Venus and had failed to account for the fact that a Venus-sized planet would be unable to do the job.

So.
Let's forget about the extreme of mandating that the Earth had to be stopped and started for a moment and just consider the effects of a Venus-size mass passing within a reasonably close range.

Venus would experience just as much tidal effect as the Earth.

A point to consider, of course, is that in this scenario, Venus would be pretty much completely molten with practically no atmsophere. This would mean there wouldn't be any crust to flex...

Oh well. Just a passing thought.

13. ### invert_nexusZe do caixaoValued Senior Member

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Although... even though the large planet wouldn't be perturbed by the effect of the near-earth... would it lessen the effects on the Earth?

What I'm asking is if the energy you postulated being spread out on the Earrth would actually be shared between all objects in the system. Even though the large object wouldn't experience damage from this, would it soak up some of the energy you had been postulating wreaking havok on the Earth?

Even if this were the case, I'm sure the leftover energy would be enough to pretty much wipe out the planet. That's practically a given given the scales concerned.

But, I'm just thinking about being as precise as possible here.

14. ### superluminalI am MalcomRValued Senior Member

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The rotational kinetic energy of the earth must be transferred to another body (via tidal slowing) or directly dissipated as heat (giant sized space clamps<sup>TM</sup>). Transferring the energy to another body by "grabbing" it tidally is not an option since the amount of "pull" requires such a gravitational gradient that the earth is completely destroyed. The energy is shared by both members in this system, but far from equally.

You need to somehow decelerate the rotation of the earth. You can only do this by applying a sufficient torque somehow. This is what MK seems to be missing. The torque required does not change depending on the method. You either have to grab the planet with a giant space wrench, or grab some part of it with a gravitational differential. Either way, you destroy the planet.

15. ### MetaKronRegistered Senior Member

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Draw the ellipse that represents the Earth at maximum tidal effect. What is the degree of the sharpest bend? How does that compare to the normal curvature? The biggest problem with the heating theory is that we are talking about flexing a fraction of a degree for each kilometer. You can take an iron rod and bend it double, representing a very much more extreme degree of flexing, and it doesn't suddenly get red hot and vaporize, even though this heating takes place over seconds instead of hours.

I'm not at all sure how short the radius of curvature has to be to exceed the tolerances of the materials involved. We're talking about curvatures that can only be detected on the ground with sensitive instruments even if the radius of curvature is down to 100 kilometers.

16. ### invert_nexusZe do caixaoValued Senior Member

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Now carry your analogy a bit further, Metakron.

That iron rod is now several kilometers thick.

A sudden flexing of the rod bends it to, let's say, a 90 degree angle.

How much heat is produced in the section that has bent? One fact is immediately apparent, there is a lot more mass involved in this kilometers thick rod.

Question: Is this heat from flexing caused by friction? What, exactly, is the mechanism?

17. ### MetaKronRegistered Senior Member

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Show me where the amount of heat generated per mass is increased.

18. ### superluminalI am MalcomRValued Senior Member

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Stop, stop, stop. Forget the heat. You're deflecting the bigger issue now. Torque. The torque necessary to stop the earth is well defined. How will you apply this torque MK?

19. ### invert_nexusZe do caixaoValued Senior Member

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That's why I asked where the heat is derived. My thought is that it is caused by friction. And in this scenario, the more mass you have, the more mass you have rubbing against itself when it flexes. Thus, more mass flexing means more heat generated.

This is all speculation.

Sounds like we need a torque converter...

20. ### superluminalI am MalcomRValued Senior Member

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Yes. A magic torque converter...

21. ### MetaKronRegistered Senior Member

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All of friction is collisions between molecules. One of the reasons that I have so much trouble with the flexing theory is that it involves collisions at very low differences in velocity. These don't translate into much heating, that's why the Earth's moon can flex us day after day and we don't wind up with a molten ball of lava. In fact I find the idea quite questionable that there can be much of a buildup of heat in the interior of a planet when a small fraction of a degree is added each day. That's a discussion that's guaranteed to divert us.

How do we get the torque? So many bazillion tons of mass projecting on both ends of the ovoid, plus the narrowing around the waist, decelerating the whole assembly at a small fraction of a gee for an hour or so. We aren't just talking about one mass pulling the Earth into line, we are talking about the combined gravities of the two planets, doubling the forces dragging the rotation of the Earth to a temporary stop or the Earth to an orientation that makes it look like its rotation has stopped.

One other thing we haven't factored in is the ballerina effect. If a spinning ballerina extends her arms her rotation slows. Just the act of stretching the globe slows down its rotation.

22. ### superluminalI am MalcomRValued Senior Member

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You need to show the math. I guarantee you that the forces you are talking about are well beyond the destruction stress of the earth. Also, you keep talking as if the gravitational forces causing the stopping are distributed throughout the two bodies. They aren't. In order for a torque to arise you must have an imbalance of forces about the axis of the rotating object.

I thought of another way to illustrate the problem. You can scale the problem down. The energy and forces invloved will be the same, only the time scales will be different. Imagine the earth is a 2m diameter sphere made of lead.

The density of lead at room temp is 11.34 g cm<sup>−3</sup> or

The mass of this sphere is:

m<sub>s</sub> = 4/3 &pi; 100<sup>3</sup> x 11.34 g cm<sup>−3</sup> = 4.75x10<sup>7</sup>g or 4.75x10<sup>4</sup>kg

Conservation of angular momentum gives:

L<sub>e</sub> = L<sub>s </sub>or I<sub>e</sub> &omega;<sub>e</sub> = I<sub>s</sub> &omega;<sub>s</sub>

so that &omega;<sub>s</sub>= I<sub>e</sub> &omega;<sub>e</sub> / I<sub>s</sub>

where:

I<sub>e</sub> = 2/5 m<sub>e</sub> r<sub>e</sub><sup>2</sup> = 2/5 x 5.974x10<sup>24</sup> x (6.378x10<sup>6</sup>)<sup>2</sup> = 9.72x10<sup>37</sup>kg m<sup>2</sup>
&omega;<sub>e</sub> = 7.29 × 10<sup>-5</sup> rad/s

I<sub>s</sub> = 2/5 m<sub>s</sub> r<sub>s</sub><sup>2</sup> = 2/5 x 4.75x10<sup>4</sup> x 1 = 1.9x10<sup>4</sup>kg m<sup>2</sup>

Therefore,

&omega;<sub>s</sub> = 9.27x10<sup>37</sup> x 7.29x10<sup>-5</sup> / 1.9x10<sup>4</sup> = 3.55x10<sup>29</sup> rad/sec

This is 3.55x10<sup>29</sup> / 2 &pi; rev per second which is 5.66 x10<sup>28</sup> rpm

Obviously this is impossibly fast, but let's pretend.

So you are suggesting that by deforming the shape of this sphere a little you can grab it gravitationally and stop it's rotation within 1/24th of a rev? MK, you are seriously underestimating the energies and forces involved here. I would think that after all of these examples you would at least acknowledge these facts.

As for the ballerina, I estimate that you would have to stretch the earth out into a barbell shape at least 10 times it's current diameter to get any significant effect.

23. ### MetaKronRegistered Senior Member

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If we can "scale the problem down" as you say, we can work with the mathematics at the scale of the Earth. That lead ball analogy doesn't make a lot of sense. With one the size of the Earth we have a lot more options.