What if for one split second the moon disappeared?

[DrPizza]

Originally posted by: Smilin
Allow me to modify your swimming pool analogy for a moment.

Imagine a swimming pool held a foot off the ground by chains. Over the course of 24 hours you slowly lower and raise each chain so that each end of the swimming pool touches the ground once ever 24 hours. If viewed from within the pool, yes you will see the water level rise and fall on each end but it would take a 24 time lapse view to actually see it happening.

Now cut one of the chains while the opposite end of the pool is resting on the ground.

Tidal waves. See?

The same sort of effect would happen to the solid earth in addition to the oceans. Not only would water rise on the dropped end it would likely slosh out of the pool.


I don't believe your brick wall analogy applies in this situation. Unlike a brick wall, the Earth is elastic and can change shape. If you dug your finger into a big sphere of jello and pulled until the whole sphere became egg shaped you would get an idea of what the moons force of gravity does to the earth. If you
suddenly let go, shock waves will travel through the jello as it snaps back to a spherical shape.

I like your analogy. So, lets scale it to the Earth. We'll use a swimming pool that's 4 feet deep. Now, tell me again, how far does one side have to be lifted by the chain? Just a rough approximation by me put it at less than 1/100th of an inch.

As far as the brick wall analogy goes, The brick wall is also elastic. Just very slightly so. Again, I think it's a matter of scale. Continental drift is currently being measured with a precision of 1 millimeter (maybe even more precise now). I'd have to believe that the elastic stretching of the earth due to the moon's gravitational effects are less than that precision. So, you're saying that snapping back a millimeter or less is going to cause earthquakes?? (or, when an earthquake is felt along the San Andreas fault, they're measuring the ground movement as less than a millimeter?)

edit: I realized after replying that I'd also have to scale up the newly created waves in the swimming pool, and arguably they might be high enough to be considered tsunamis. I wanted to attempt to test this. If you take a container of water, and strike the bottom of it, you are going to create waves. - As your pool strikes the ground with a force, as Newton stated, and we all know, the ground strikes the pool with an equal force. This impulse is going to create waves. I experimented with a container of water, tilted it so the water was just at the rim on one side. Dropping the raised side resulted in water splashing over the sides. Then, to eliminate this impulse as much as possible, I repeated this experiment. However, the second time, I put the container of water on a platform suspended by 3 strings. I tilted the platform until the water was just about to spill out of one side. Upon releasing the platform, a couple of drops still spilled out, but far far less than in the first experiment. Nonetheless, I realized I wasn't able to completely remove the impulse either.

Then, as I typed this, the following idea came to mind. Since water is diamagnetic (repelled by a magnet), it would be possible to simulate this situation by using magnets to repel water to one side of the container, simulating high tide on the side opposite the magnet. Removal of the magnetic force on the water would be completely analagous to pushing the bob of a pendulum and suddenly releasing the force pushing the pendulum. The pendulum will NOT swing higher than it's release point. Likewise, the resulting wave travelling back across the water will not be higher than the high tide.

Anyone care to find flaws in the preceding argument? I love learning, and would enjoy being proven wrong. (and learning something new)

 

[jagec]

Originally posted by: L00PY

For the earth to instantly "snap" back into the shape of a moon-less earth would require the addition of a force in the opposite direction to the moon's gravitational pull. Put your hand in that pool and create a bulge in the water by moving your hand. When you take your hand out of the water it doesn't suddenly snap to a flat surface. Nor will that bulge ever get larger. It just continues onwards until it runs out of momentum.

The bulge is caused by the gravitational force of the moon being balanced out by the gravitational attraction of that bit of earth to the rest of the earth. Remove the force of the moon, and suddenly you have a net attractive force.

 

[Jeff7]

Instead of the moon shielding the earth, the earth makes a great shield for the moon since it's not only larger, it's got a stronger gravity pull.

True - the near side of the moon has far fewer craters than does the far side. The far side always faces away, so it's exposed more directly to small objects flying throughout the solar system. The near side has far fewer craters, plus it has evidence of past melting. This may have been from one thing I read, that part of the moon may be remnants of where the Pacific Ocean now is. Perhaps the near-side's melted areas are from the heat released in that devastating impact?

What is it do you suppose that keeps the earth in such a stable rotation? Granted the earth "swings" to and fro on it's 21.5 degree axis but if there were no moon to keep her stable the earth instead of spinning like a well balanced top it would imediately begin to spin just a toy top that looses speed and starts to wobble uncontrolably. Ifthe moon just flat disappeared the seasons would be no more, there would be climatic changes that would make Global Warming look like bambi and in a quite a short amount of time humanity would be extinct.

I don't think it's the moon that does that. Examples:
Mercury's tilt is very nearly 0 degrees. No moons.
Venus' tilt is about 2.7 degrees - strange, because something screwed up during its development. Its direction of rotation is opposite that of the other planets.
Mars - 25.19 degrees. It has two moons, which are mere chunks of rock. Phobos, 17x12 miles, and Deimos, around 6 miles wide. They'd hardly have any effect on its axis tilt, so that tilt of Mars' likely came from large asteroid impacts.
Jupiter - 3.1 degrees. It's got a bunch of moons - some of them are close to Mercury in size, but puny compared to the planet.
And...I don't feel like checking out the others right now. I will mention Uranus though, since its axis is something like 80 degrees off. That must have been from some massive collision back during the solar system's formation. Moons won't do that.

Aaaaanyway, the moon. If the moon disappeared, like just for a fraction of a second, there would probably be "ripples" in the ocean, due to the sudden disappearance of the gravitational attraction that causes the tides. They'd grow to large waves at the coastlines of the world. No, I don't have sophisticated computers to figure out how high the waves would be, but I doubt they'd go unnoticed. If the moon vanished, and stayed that way, you'd get the effects above (making tidal power plants useless:Q), but other than that, I can't think of anything else. Well, no more eclipses anyway...anything else?

 

[DrPizza]

Originally posted by: Description

That'd be like suggesting that if I push against a brick wall, I can't knock it over, but if I remove the force I'm exerting rapidly (or even instantaneously), the wall will collapse.

It could. Resonant frequency.

And the Earth is not like a brick wall, it is very flexible on a large scale. Closer to a stack of bricks with no mortar.

As far as being very flexible on a large scale, the scale of the stretching is quite small. The equatorial bulge is only 21km (IIRC) Thus, while the earth is an oblate spheroid, (as proposed by Newton), it barely deviates from spherical. (Jupiter on the other hand has a bulge of about 7%)

 

[Smilin]

I like your analogy. So, lets scale it to the Earth. We'll use a swimming pool that's 4 feet deep. Now, tell me again, how far does one side have to be lifted by the chain? Just a rough approximation by me put it at less than 1/100th of an inch.

Although very rough that's probably a closer estimate to the scales we are talking about. A little more accurate would be a very large pool (say an olympic swimming pool) that was a few inches deep. The rise and fall of the chains would be very small, say 1/1000 of an inch. With this in mind how tall do you think wave would be if instead of rocking the pool 1/1000 of an each every 24 hours you just dropped the end of the pool? Would it be 3/1000 of an inch? Or would it be 1/1000 of an inch, but a very very long wave. To put the scale back in perspective tides are going to rise and fall about as much as a person is tall. Whether you end up with a 20 foot tall wave or a 5 foot wave that's several hundred miles long the net result is the same: You'll have a tidal wave that will reach 1/4 of the way across the US from the East coast.

The big thing to remember is the rise and fall of tides happens over the course of 24 hours. This is going to happen instantly.

As far as the brick wall analogy goes, The brick wall is also elastic. Just very slightly so. Again, I think it's a matter of scale. Continental drift is currently being measured with a precision of 1 millimeter (maybe even more precise now). I'd have to believe that the elastic stretching of the earth due to the moon's gravitational effects are less than that precision. So, you're saying that snapping back a millimeter or less is going to cause earthquakes?? (or, when an earthquake is felt along the San Andreas fault, they're measuring the ground movement as less than a millimeter?)

Continental drift and earth tides are completely separate. Using continental drift for this is a bad assumption. They may measure continental drift in millimeters per year, but earth tides cause the ground to rise and fall over six inches every 24 hours. The ball of jello analogy can give you a rough estimate of what happens when several hundred thousand square miles of earth suddenly drop six inches.

You have a planetary body exerting enough gravetational force to change the shape of the earth itself and form tides. I assure you that if it suddenly disappears you are going to see more than a ripple.

 

[L00PY]

Originally posted by: jagec
The bulge is caused by the gravitational force of the moon being balanced out by the gravitational attraction of that bit of earth to the rest of the earth. Remove the force of the moon, and suddenly you have a net attractive force.

Nope, the bulge is caused by the difference in distance from the moon. The side closer to the moon feels a pull approximately equal to g*1.13x10^-7 (using an estimate of 6.37x10^6m for the earth's radius and an average distance of the moon of 3.84x10^8m). This is just physics and can be read up in Schad's Physical Science 1996 or your favourite physics text book.

And the scale for these pool analogies is off by three or four orders of magnitude.

Also, the effect of the moon's gravitational pull on the earth is large enough to raise it by as much as half a meter depending on location.

 

[jagec]

Originally posted by: L00PY

Originally posted by: jagec
The bulge is caused by the gravitational force of the moon being balanced out by the gravitational attraction of that bit of earth to the rest of the earth. Remove the force of the moon, and suddenly you have a net attractive force.

Nope, the bulge is caused by the difference in distance from the moon. The side closer to the moon feels a pull approximately equal to g*1.13x10^-7 (using an estimate of 6.37x10^6m for the earth's radius and an average distance of the moon of 3.84x10^8m). This is just physics and can be read up in Schad's Physical Science 1996 or your favourite physics text book.

that's what I said, you just said it clearer.

 

[DrPizza]

I don't doubt at all what you've said about the amount of the bulge.
I'm curious now though... how do they account for that buldge when measuring the amount of continental drift?

incidentally, how is my scale off by 3 or 4 orders of magnitude? I gave a pool depth of 4 feet, average depth of the ocean is about 4 miles. Therefore, the scale is 1 foot per mile, or 1 ft per 5280 feet. I used a tide table from Maine to get an average tide of around 10 feet. 10 feet would therefore scale to about 10/5280 feet (around .0019 feet). Multiply by 12, and it's just a little over 2 hundredths of an inch.

Of course, to be to scale, my 4 foot deep pool needs to be a couple thousand feet across. However, I was trying to refute the lift one side 1 foot off the ground analogy. It's also why the couple of inches deep pool was a little better. But then, the pool would be lifted an even smaller fraction above the ground.

 

[L00PY]

Originally posted by: jagec

Originally posted by: L00PY

Originally posted by: jagec
The bulge is caused by the gravitational force of the moon being balanced out by the gravitational attraction of that bit of earth to the rest of the earth. Remove the force of the moon, and suddenly you have a net attractive force.

Nope, the bulge is caused by the difference in distance from the moon.

that's what I said, you just said it clearer.

The gravitational force of the earth doesn't play an important role in the bulge created by the moon. It is not balancing out anything. The real reason behind the bulge is gravity and the size of the earth in relation to the distance to the moon.

"Remove the force of the moon" and there is no difference in distance from the moon which means there is no force -- not a "net attractive force".

And as for scale issues, let me start off by saying that I don't like the whole pool analogy. If anything, the pool effects are more due to momentum than to the forces we're talking about. The force of gravity upon the pool is the same before and after you've dropped it -- you haven't removed anything. In addition to not removing anything, you're "adding" the ground beneath the pool to get it to come to a sudden halt. A more accurate example would be to just stop moving the chains when the pool is level. The water will continue to go back and forth for a while but it won't gain any height.

But back to scale. I thought you said you were scaling the pool to the earth -- not the depth of the oceans. If scaled to the earth, the size is definately off. If you're scaling the pool to the ocean depths, your scale might not be off by several orders of magnitudes. You can't really use tide tables from a single location to estimate the effect of the moon though -- local currents can play as large role in tides as the moon or the sun.

Finally, the bulge is probably irrelevant when it comes to continental drift because the bulge is in constant motion across the lines of longitude, making it around the earth in around a day. By taking the measurements around the same hour of the day, you could remove most of the effect of the bulge when measuring continental drift. The additional error from the bulge is probably negligible compared to other sources. Observed drift is already off by 20-50% of the predicted drift due to plate tectonics.

 

[Smilin]

The pool analogy is getting stretched for sure.

It would be like dropping one end however, not stoping the rocking when it was level. As the moon pulls tides up from the surface the water would, in the absense of gravity roll back level again. It actually does continuously. As the moon orbits, water behind the orbit (a small percentage of the entire bulge) settles back down some as the moon gets more distant. New water is pulled up in front to maintain the bulge. A slow 'wave' rolls around the earth as it rotates. If the gravity suddenly disappears altogether then all the water in the bulge will suddenly return to a spherical shape. The returning water will be lead by a wave that although not tall will be very long. If you know how tsunamis work you will realize just how devastating this would be. As the wave hits shallow shores and finally goes up onto land the wave will grow tall enough to level buildings with little loss in energy.

 

[jagec]

Originally posted by: L00PY

The gravitational force of the earth doesn't play an important role in the bulge created by the moon. It is not balancing out anything. The real reason behind the bulge is gravity and the size of the earth in relation to the distance to the moon.

OK. What I did was draw a control volume around the bulge. Basically, the bulge feels two attractive forces: One between itself and the moon, and one between itself and the rest of the earth. If the gravitational force of the earth is ignored, that would imply the bulge would tear itself off from the rest of the earth, float across space, and land on the moon, which is obviously not the case.

In order for there to be equilibrium, forces must balance. You can't have the moon's gravity be the only force acting on that section of the earth.

 

[DrPizza]

Originally posted by: Smilin
The pool analogy is getting stretched for sure.

It would be like dropping one end however, not stoping the rocking when it was level. As the moon pulls tides up from the surface the water would, in the absense of gravity roll back level again. It actually does continuously. As the moon orbits, water behind the orbit (a small percentage of the entire bulge) settles back down some as the moon gets more distant. New water is pulled up in front to maintain the bulge. A slow 'wave' rolls around the earth as it rotates. If the gravity suddenly disappears altogether then all the water in the bulge will suddenly return to a spherical shape. The returning water will be lead by a wave that although not tall will be very long. If you know how tsunamis work you will realize just how devastating this would be. As the wave hits shallow shores and finally goes up onto land the wave will grow tall enough to level buildings with little loss in energy.

Smilin, you just changed my mind! (and I'm going to go back to the swimming pool analogy in a moment! heh heh)

I'm still not convinced of the earthquakes, as I don't think the solid ground would "instantly" spring back to a spherical shape. (there has to be some change in time, albeit .0001 seconds or 20 seconds.) As the center of the bulge on the earth is not directly at the point nearest to the moon, but rather lags behind (I'll provide links in a moment), I believe that's plenty of evidence to assume that the amount of time to "spring" back to a spherical shape will be more on the magnitude of an hour. (or, is this evidence that the speed of gravity is far less than c?? GASP!)

Water on the other hand... (here's the swimming pool again)
Lets assume for simplicity that the bulge is in the middle of the ocean. (approximately 1 meter - the land bulges about 30 cm)

So, lets put a waterballoon in the middle of the swimming pool, and push it up a bit to represent the bulge from the moon's gravity. Now, lets remove the force holding up the balloon (and pop it... sorry, I couldn't get the water to stay up there without putting it in something) There's a lotta water in that bulge, attempting to have a circular surface rather than a bulge. When it reaches the shallow shores, as you pointed out Smilin, a lotta stuff is gonna get wet! (of course, the coastline is far less regular than the perimeter of the pool and would have a huge effect on the size of the waves at various locations)

 

[jagec]

Originally posted by: DrPizza

I'm still not convinced of the earthquakes, as I don't think the solid ground would "instantly" spring back to a spherical shape. (there has to be some change in time, albeit .0001 seconds or 20 seconds.) As the center of the bulge on the earth is not directly at the point nearest to the moon, but rather lags behind (I'll provide links in a moment), I believe that's plenty of evidence to assume that the amount of time to "spring" back to a spherical shape will be more on the magnitude of an hour. (or, is this evidence that the speed of gravity is far less than c?? GASP!)

well it doesn't have to "spring back" instantly to cause earthquakes. The way it would cause earthquakes is that, when the force was removed and the earth started to return to spherical form, rocks would catch and slip as it went, causing the earthquakes. I don't believe there would be many (or any?) large earthquakes, though.

 

[Smilin]

Neither the water from tides nor the earth itself would snap back instantly. It would snap back at something aproximating the speed of a shockwave through that particular medium (something around the speed of sound through rock/iron/water). Someone with some knowledge on earthquakes could tell us.

I'm not sure of the exact speed but it would be measured in the low thousands of miles per hour. That tidal wave is gonna suck.

Normal earthquakes are just the grinding of plates on the surface. These normal quakes are similar to two patches of film grinding against each other on the surface of some cooling warm pancake syrup.

Now when the moon disappears you're going to be shifting the entire earth down to the core. Expect some eruptions. The Earth will probably resemble Io for a while. As a matter of fact, Io is probably a good example of this whole thing. It has constant volcanic activity from being reshaped by the gravity of Jupiter and it's other satellites.

I think we lose track of just how small we are. We're just some organic film on the surface of this big rock we call Earth. Viewed from space, the moon disappearing would appear a trivial event. On the ground the event might just be enough to obliterate life.

 

[DrPizza]

Originally posted by: Smilin
Neither the water from tides nor the earth itself would snap back instantly. It would snap back at something aproximating the speed of a shockwave through that particular medium (something around the speed of sound through rock/iron/water). Someone with some knowledge on earthquakes could tell us.

I'm not sure of the exact speed but it would be measured in the low thousands of miles per hour. That tidal wave is gonna suck.

Normal earthquakes are just the grinding of plates on the surface. These normal quakes are similar to two patches of film grinding against each other on the surface of some cooling warm pancake syrup.

Now when the moon disappears you're going to be shifting the entire earth down to the core. Expect some eruptions. The Earth will probably resemble Io for a while. As a matter of fact, Io is probably a good example of this whole thing. It has constant volcanic activity from being reshaped by the gravity of Jupiter and it's other satellites.

I think we lose track of just how small we are. We're just some organic film on the surface of this big rock we call Earth. Viewed from space, the moon disappearing would appear a trivial event. On the ground the event might just be enough to obliterate life.

If the earth was able change its shape at the speed of a shockwave through that particular medium, then why is it that the bulge is NOT directly below the moon? Plus, while the moon may play a role in earthquakes and volcanos, remember that this bulge occurs twice per day (every 12 hours 24 minutes, if I recall correctly) So, the earth is already oscillating a foot or two.

If I'm not mistaken, the volcanoes on Io are caused by the difference in gravitational force (between it and Jupiter) from one side of that moon to the other.

From a site I was referred tothis is hard to do since I can't space or tab things out)

Moon <------ <---- <--
The <---'s represent the forces on the earth exerted by the moon,
for the side closest, the center, and the side away from the moon.
If you were to look at the relative forces, using the center of the earth as a reference point, (represented as an x) then you'll notice that relative to the center the forces look like this:
<--- x -->
That's why there are two bulges on the earth. (but, as I said above, the bulge isn't actually directly in line with the moon, it's playing catch-up i.e. high tide isn't when the moon is highest overhead)

 

[Pixle]

We would have no night light.

 

[rimshaker]

You'd weigh more for a split second.

Gravitons don't interact instantaneously, but for the sake of this example, it can be safely assumed.

 

[DrPizza]

I'm not quite sure what you mean by we'd weigh more for a split second. But, actually, you can do a simple calculation of F = Gm1m2/r^2 for the gravitational force of attraction on your human body. Actually, you'd have to do that calculation twice, for 2 distances - when you're on the side of the earth nearest the moon, and when you're on the side of the earth away from the moon.

I'm not going to repeat the calculations - but I was checking those done by someone else last night. IIRC, the moon's effect on your weight varies by (the weight of either 25 milligrams or 25 micrograms on the surface of the earth, I don't recall exactly) for someone with a body mass of 70kg. Or, as he put it another way, the moon's gravitational force of attraction on you is about the same as that of a pea held above your head.
(although, now that I think about it, I believe I may repeat the calculations when I have some more spare time to verify them. Even 25 micrograms sounds like it might be high (for the pea)

 

[rimshaker]

Well, I said you'd weight more, but I didn't mean by a significant amount.

The earth exerts a gravity on the moon and vice versa. Take away the moon, and the earth's gravitational effects on itself increases. Does this even make any sense? maybe i've been drinking too much (root) beer lately.

 

[L00PY]

Ok, here's an online text by Robert Reid, Dynamical Oceanography. Part I: Fundamental Principles. The part we're interested in is under "Geostatics and Dynamics: The Earth as a Reference System". Many of the real world approximations we've been using are written out and calculated there. A revealing bit supporting my position can be read at the end of "Tidal forces associated with the moon". The "tidal waves" caused by the moon disappearing would be approximately equal to four times the size of the "tidal waves" created if the forces causing the Gulf Stream currents stopped moving water around the Atlantic.

jagec: The "control volume you drew" doesn't make sense. Here's another way to see the bulges. The earth is currently in free fall towards the moon. The side closest* to the moon is falling faster than the center of the earth (because it is closer to the moon) and bulges towards the moon. The side furthest* from the moon is falling slower than the center of the earth (because it is further from the moon) and bulges away from the moon. These bulges are caused by the distance to the moon and because the force of gravity falls off with distance. Or if that doesn't show why your argument is wrong, how do you "draw a control volume" around the bulge on the side furthest from the moon? Is the force of the moon's gravity somehow pushing that side of the earth away?

*The earth's rotation causes the bulges to be offset slightly.

 

[jagec]

<sigh>

My control volume makes PERFECT sense. I don't know how you keep misinterpreting it. You say I'm wrong, then reiterate the EXACT same thing I'm saying. Please, tell me what you're having trouble understanding about my picture.

1) forces balance, or else we get collisions
2) all individual particles exert gravitational effects on all other individual particles, falling off with distance.

 

[L00PY]

Originally posted by: jagec
The bulge is caused by the gravitational force of the moon being balanced out by the gravitational attraction of that bit of earth to the rest of the earth. Remove the force of the moon, and suddenly you have a net attractive force.

.
.
.

OK. What I did was draw a control volume around the bulge. Basically, the bulge feels two attractive forces: One between itself and the moon, and one between itself and the rest of the earth. If the gravitational force of the earth is ignored, that would imply the bulge would tear itself off from the rest of the earth, float across space, and land on the moon, which is obviously not the case.

In order for there to be equilibrium, forces must balance. You can't have the moon's gravity be the only force acting on that section of the earth.

There are two outward bulges on earth -- one on the side closest to the moon and one on the side furthest from the moon. On the side furthest from the moon, the earth bulges outwards too. That bulge is not caused by the gravitational force of the moon being balanced out by the gravitation attraction of that bit of earth to the rest of the earth. The force of gravity from the earth and force of gravity from the moon are both pulling that bulge in the same direction and your control volume makes no sense.

 

[jagec]

Originally posted by: L00PY
There are two outward bulges on earth -- one on the side closest to the moon and one on the side furthest from the moon. On the side furthest from the moon, the earth bulges outwards too. That bulge is not caused by the gravitational force of the moon being balanced out by the gravitation attraction of that bit of earth to the rest of the earth. The force of gravity from the earth and force of gravity from the moon are both pulling that bulge in the same direction and your control volume makes no sense.

look, the control volume was never meant to be a dynamic model of why the earth is shaped the way it is. it simply shows that right now, forces are balanced (the earth isn't tearing in half), but when the moon was removed, there would be a force (attractions between the atoms that make up the earth) which would start pulling it back into a sphere. It's not supposed to explain the shape, it's not supposed to be fully accurate, it's supposed to be a simple way of showing that the earth would try to pull itself back together if the moon was removed.

One of the basic tenants of engineering is that you can draw a control volume around any system in equilibrium, and the sum of everything entering and exiting is zero.

 

[L00PY]

Look jagec, your misapplication of a control volume is clouding the issue. Let's return to discussion of the moon disappearing.

Right now there are already forces attempting to move the bulges back into place. Both the earth's gravity and rotation currently move each bulge back into its "natural" form in roughly 6 six hours. This work is shown by the fact that the earth's rotation is decreasing. There is work being done by the earth right now in moving the bulges across the earth (to keep their position static relative to the moon) and this work is causing neither tremendous tsunamis nor enormous earthquakes. This work being done by the earth is at times opposed by the moon's gravitation pull and at times in conjunction with the moon's gravitation pull -- depending on which side of the bulge of which bulge you happen to be on. That is, the moon's gravity both helps form a bulge and return the bulge to "normal".

If you remove the force provided by the moon, the force of the earth's gravity and the earth's rotation will move the bulges back into its natural form. For there to be tidal waves or earthquakes when the moon bulges recess, there needs to be some additional work done beyond the current recession. No one has suggested a force that is added to the system by the removal of the moon. As I've said before, there is no additional force will instantly push the bulge down into normal shape. The work already being done by the earth right now in moving the bulges will unbulge the earth.

Please feel free to point out the flaw in my logic. I'd love to hear an explanation of where addiitonal energy is introduced into the system.

 

[jagec]

Originally posted by: L00PY
If you remove the force provided by the moon, the force of the earth's gravity and the earth's rotation will move the bulges back into its natural form. For there to be tidal waves or earthquakes when the moon bulges recess, there needs to be some additional work done beyond the current recession. No one has suggested a force that is added to the system by the removal of the moon. As I've said before, there is no additional force will instantly push the bulge down into normal shape. The work already being done by the earth right now in moving the bulges will unbulge the earth.

AND THAT'S ALL I WAS TRYING TO SAY...that if you removed the force provided by the moon, the other forces would move the earth back to its natural form. That's it. Punto.

Hey, if you want to call me wrong again and feel smug about it, by all means do it, but know that it's not my picture that's wrong so much as my explanation of the picture.