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I was watching a BBC documentary on space last night. It was talking about gravity, and it said that the reason we only ever see one side of the moon, is because the earths gravity is strong enough to actually stretch the moon into a longer shape, creating a 7m bulge on the moon.

This bulge travelled around the moon. It was described on the documentary as something along the lines of, if you were there, you would see the ground make a 7m high wave as it rotated. The bulge then acted a brake, and the moons spin slowly ground to a halt until where we are today when it's spin is very very slow so it looks like we only see one side.

So where is this giant death bulge tidal wave on earth? Shouldn't we been seeing the ground rising up a few meters in relation to the suns position, demolishing and killing everything in its path?

I assume the documentary grossly simplified what actually happened, or I misunderstood! I'm a physics noob, please go easy on me, chances are I'm not going to understand any formulas that use anything but addition or subtraction.

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    $\begingroup$ Ever heard of tide? $\endgroup$
    – Marek
    Commented Mar 23, 2011 at 12:56
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    $\begingroup$ @Marek yes, but if the earths gravity can stretch rock 7m then why doesn't this happen on land on earth? The doc made it seem like it would be observable travelling as well with just the eye $\endgroup$
    – Tom Gullen
    Commented Mar 23, 2011 at 13:03
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    $\begingroup$ Tom: the doc is undoubtedly misleading: I very much doubt even a 7m land tidal bulge would be observable travelling --- everything goes up 7 m over the period of 1/4 day (whatever length moon days were). It's like the entire landscape being on a very, very slow elevator. You wouldn't notice it. I expect it would put some strain on the moon's crust, but I don't know what observable effects that would have had $\endgroup$
    – Peter Shor
    Commented Mar 23, 2011 at 13:51
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    $\begingroup$ Anna notes that gravity falls by $1/r^2$, but that means that tidal forces go $F_T \propto d/dr F_G$ fall by $1/r^3$. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 23, 2011 at 15:53
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    $\begingroup$ @Tom: to emphasize what Peter said, if you were on the ancient moon, you wouldn't see the ground make a wave because everything on the ground would move up and down in unison with the ground itself, and similarly on Earth. This bulge you're talking about isn't capable of destroying anything. If the documentary implied otherwise, it was wrong. $\endgroup$
    – David Z
    Commented Mar 23, 2011 at 16:43

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In the ocean, it's called the tides, and it happens twice a day. You can see it if you're standing on the seashore, but on the deep ocean it happens so gradually you wouldn't even notice it. On land the tidal bulge only rises up a third of a meter or so (not seven meters, because the moon's gravity is much less than the earth's), and, again, it happens so gradually we don't even notice it. I suspect that even the seven meter bulge on the moon would happen so gradually you wouldn't notice it, although I suspect (this is pure speculation) it would put some extra strain on the crust, but I don't know what effects that would cause.

UPDATE

I didn't even notice that you said "sun" and not "moon". The sun causes tides, too, but because the sun is so much farther than the earth, they're weaker than the moon's. From wikipedia, the moon causes 70% of the tidal forces and the sun causes 30% of the tidal forces. When they're lined up, you have especially high tides, called "spring tides," and when they're 90° from each other, you have especially low tides, called "neap tides." Since "neap" comes from old English, and is a word that only applies to tides, it's clear that sailors and fishermen have been noticing this for a long time.

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    $\begingroup$ Just to elaborate on Prof. Shor's excellent answer: We notice the ocean tides because of the difference in the water and land bulge. This is due to both the different "stiffness" of water and rock, and also due to the different velocities of wave propagation. Because the tides drive the ocean above their resonant frequency, the tides lag the position of the sun/moon system by almost 90 degrees. Because sound propagates through the earth much faster, I think the earth tides are in phase with the position of the sun/moon. $\endgroup$
    – Anonymous Coward
    Commented Mar 24, 2011 at 23:01
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    $\begingroup$ @Anonymous Coward: I don't understand the dynamics, but isn't it true that if the earth tides are in phase with the position of the moon, they wouldn't slow down the moon, which means that energy wouldn't be conserved. I'd think this should only happen if rock is perfectly elastic. It doesn't seem to me that rock should be elastic enough for this to be a good approximation. --- Looking on the web, I found the statistics that the average lag for ocean tides is 4 hours and the average lag for earth tides is 2 hours. I have no idea whether this is accurate. $\endgroup$
    – Peter Shor
    Commented Mar 26, 2011 at 1:41
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    $\begingroup$ Shor: You are correct. I'll do some work to try to improve my comment. For the dynamics, I think it's a reasonable approximation to think of it as a damped harmonic oscillator. The equatorial surface velocity of the earth's rotation is ~500 m/s. The speed of sound in rock is something like 10 km/s, so the solid earth is driven far below resonance, hence there will only be a small (but nonzero) phase lag. The exact lag will depend on the damping, of course. But I think the damping is surprisingly low: Nature 381 p595 (1996) claims the Q of the tidally driven earth is around 370! $\endgroup$
    – Anonymous Coward
    Commented Mar 27, 2011 at 23:53
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    $\begingroup$ Continuing the above comment: The same Nature reference also quotes a tidal lag of the solid earth much smaller than the 2 hour lag you mentioned above. As for the water, long-wavelength water waves in an average-depth-of-4 km ocean would have a velocity of 200 m/s, so depending on how close you are to the equator, you could be drive above or below resonance. A perfect at-resonance drive would give a ~6 hour lag, so a 4 hour lag seems reasonable. The above two models would predict that most tidal dissipation on earth is coming from water (rather than rock). $\endgroup$
    – Anonymous Coward
    Commented Mar 27, 2011 at 23:55
  • $\begingroup$ You're right ... it looks like the source I got the 2 hour estimate from is incorrect, and that the lag of the Earth tides is only a minute or so. The lag of the ocean tides varies greatly from place to place, because of the shape of the ocean and the resulting fluid dynamics. $\endgroup$
    – Peter Shor
    Commented Dec 25, 2011 at 20:58
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It would be good if you read an article about tides, and the bulge the sun/moon combination makes on the earth, seas and crust. Earth tides are interesting, the whole crust moves reaching 55 cms at places.

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We ARE/Have experienced slow down in the rotation of the earth on its axis just as the moon has. We used to have ~18 hour days, now it is ~24 hours (give or take the remainder amount).

CITATION Colorado State Univ. (with other citations including notes on neap tides) http://asa.chm.colostate.edu/archive/evolution/199607/0052.html

As for the "death bulge" there is also about a 1 meter tidal bulge in the rock/solid part of the earth.

"The height of a tidal bulge on a planet is proportional to the inverse cube of the distance between the planet and the object causing the tidal bulge. The torque which slows down the planet is proportional to the inverse sixth power of the distance." from here: http://www.exo.net/~pauld/physics/tides/tidalevolution.htm

And here is the citation from Ohio State Univ. http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html

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Even if there was a 7 meter bulge that traveled around the earth. I don't think you could perceive it. 7 meters tapering off over 1000s of miles would not be visible, and each square mile of crust wont change in location much to the next square mile.

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