# Thread: moon pulls the earth from both sides?

1. Yikes, I see that you are becoming frustrated with me. Let me be clearer on my notation in the g = -k*r hypothetical gravity. Here r is the distance to the center of the Moon. Then you look at differentials around the center of the Earth, and get delta g = -k*delta r, so now delta g is the "tidal force" (in acceleration units, of course) and delta r is the distance to the center of the Earth. How is this not aximsymmetric? It is in fact *isotropic* in its effects on Earth. No tides there.
And yes, the BA argument contrasts *two* points. This is my whole point, that's not enough! You need the 2D geometry, 1D is never enough. You need at least one more point, typically the sides of the Earth. But you can eliminate the need for geometry altogether by using the centrifugal force argument, i.e., argue that the true form of gravity does not yield a solid-body Kepler's law, whereas it does for a gravity that is linear in distance (which by the way sounds like it would yield tides by compressing the near and far points of the Earth, if you stick to the two-point analysis and forget the need to bring in at least one more point for comparison). I am sure you will see what I mean as soon as I find the right words to express myself. And thank you for welcoming me to the forum, this has been a very informative debate, I hope others might still be reading up to this point if they have similar confusions!

2. Originally Posted by Ken G
Yikes, I see that you are becoming frustrated with me.
Just a little. no problem though
Let me be clearer on my notation in the g = -k*r hypothetical gravity. Here r is the distance to the center of the Moon. Then you look at differentials around the center of the Earth, and get delta g = -k*delta r, so now delta g is the "tidal force" (in acceleration units, of course) and delta r is the distance to the center of the Earth. How is this not aximsymmetric? It is in fact *isotropic* in its effects on Earth. No tides there.
When you have an equation g = -k*r, and take a derivative, the differential that results is indeed delta g = -k*delta r, but the delta r is still relative to the center of the moon, not the Earth.

The radial distance from the moon r1 is obviously different from the radial distance from the Earth r2, and delta r1 does not equal delta r2. In other words g = -k*r1 becomes delta g = -k*delta r1, but not delta g= -k*delta r2.
And yes, the BA argument contrasts *two* points. This is my whole point, that's not enough! You need the 2D geometry, 1D is never enough.
The complete picture would be 3D. My argument is that the BA explanation conveys the essence, which is all that we should require. The complete explanation is not even found in textbooks on the subject.

3. Originally Posted by hhEb09'1
In other words g = -k*r1 becomes delta g = -k*delta r1, but not g= -k*delta r2.
I don't understand, as vectors, delta r1 is indeed just the same as delta r2.
The complete picture would be 3D. My argument is that the BA explanation conveys the essence, which is all that we should require.
Yes, the complete picture is 3D, but no, the 1D subset does *not* convey the essence. The minimum is 2D, not 1D. 1D does not tell us if the tidal effect is axisymmetric or not, and hence it is insufficient to test if it will generate tides. We never seek complete explanations, but we do seek the minimum required to actually convey the essence. 1D falls short-- as it would mislead us in the case g = -k*r.
Hey, we made it to page 3! But the fact that we are both still here obviously proves that we both want to get to the truth of the matter.

4. Originally Posted by Ken G
I don't understand, as vectors, delta r1 is indeed just the same as delta r2.
Sorry to be so late responding. That statement gave me pause, plus I got kicked out by the custodial staff.

r1 is a radius from the moon, and r2 is a radius from the Earth, so the delta r's are only equal in special circumstances, but that caused me to go back over all the posts and figure out what was going on. (so, that's a good thing )

After a few not-quite-so-straightforward derivations, I think I understand what is going on. So, I only have two or three questions left. When you have a central force, do you mean, directed towards a point? I ask because I think it affects the values, and it seems like you mean towards an axis, not towards a point, I'm not sure.
Yes, the complete picture is 3D, but no, the 1D subset does *not* convey the essence. The minimum is 2D, not 1D.
I've always said the minimum was 3D, or even 4D, but what I meant about the essence was that the explanation uses the difference between the gravity vector at one point and the gravity vector at another point.
1D does not tell us if the tidal effect is axisymmetric or not, and hence it is insufficient to test if it will generate tides.
But, neither does the 2D.
Hey, we made it to page 3! But the fact that we are both still here obviously proves that we both want to get to the truth of the matter.
True! and I want to repeat what I said earlier: we're glad to have you on the board, and contributing your expertise.

5. Originally Posted by hhEb09'1
r1 is a radius from the moon, and r2 is a radius from the Earth, so the delta r's are only equal in special circumstances
I mean r1 and r2 in the vector sense, not as magnitudes. It's too bad I don't know how to use vector notation here. So the delta rs you mention are *always* the same thing. I do mean a central force directed to a point, but since it suffices to analyze tides in the 2D equatorial plane, the distinction is not crucial. You are right that the full story is in 3D, but if the question is simply will you have tides, it's easiest to imagine an axis with no tilt and just look at the equator. That is sufficient to explain the basic cause. But you need points that define two independent directions along the equator, so it can't be done in 1D, not even the essence of it. And I appreciate the welcome, it is an excellent venue for exchanging ideas!

6. i heard from one member that moons planetary effect on earth is not that great to displace the earth's own gravity, and in this connection how one can be say that earth's part being lifted by the moon, you mean moon is aggregating its own gravity as better to the last?

7. Originally Posted by suntrack2
i heard from one member that moons planetary effect on earth is not that great to displace the earth's own gravity, and in this connection how one can be say that earth's part being lifted by the moon, you mean moon is aggregating its own gravity as better to the last?
I think what you mean to say is that the moon's gravity is not strong enough to overcome earth's at sea level, and therefore cannot lift things off the earth's surface. If so, then that is a common misconception.

Remember, both the moon and earth orbit around their mutual center of gravity, or their barycenter--the rightmost animation in that link is pretty similar to the earth-moon system. Anyway, the point is that the earth isn't stationary, it is free falling around the barycenter. And as far as it is solid, every part of the earth is forced to follow a similar, but displaced, path as the gravitational center of the earth. Set part of the earth's surface free and it will follow a slightly different path due to the gravity of the moon. It is that difference that gives rise (ho ho) to the tidal force.

8. Or another way to think of it, the shape of the Earth's surface is set by a balance between gravity squishing the Earth, including the centrifugal forces of rotation and orbit around the barycenter as ficticious gravities, and pressure trying to expand it. Pressure acts perpendicular to the surface only, so the shape must perturb until there is no sideways component of gravity plus centrifugal forces. If you interfere with that balance via the Moon's gravity, even the slightest bit, it could induce a sideways force and require adjustment of the surface. If so, it is the Earth's on pressure forces that cause the movement you observe, so it is not necessary for the Moon's gravity to overcome anything.
Note this explanation deals with the general issue of what controls the shape of the Earth's surface, and what allows the Moon to alter it. It is different from the analysis of the way in which the Moon's gravity causes tides, i.e., what type of gravity would not cause tides. All that is contained in the above posts, where it becomes clear that the centrifugal force of rotation is irrelevant to the question of ocean tides, and the centrifugal force of orbit is merely a convenience for explaining why tides appear, not a necessary cause of them.

9. Originally Posted by Ken G
I mean r1 and r2 in the vector sense, not as magnitudes. It's too bad I don't know how to use vector notation here. So the delta rs you mention are *always* the same thing.
I mean in the vector sense too, so I'm not sure about this. How do you mean that then?
I do mean a central force directed to a point, but since it suffices to analyze tides in the 2D equatorial plane, the distinction is not crucial.
If so (central force), then I am not yet convinced that there is no tide--the potential differs across the surface of the sphere I think.
You are right that the full story is in 3D, but if the question is simply will you have tides, it's easiest to imagine an axis with no tilt and just look at the equator. That is sufficient to explain the basic cause.
But couldn't you devise an ad hoc gravity field that would have the forces equal at all four points, yet still have tides? What happens if you just have four equal-sized bodies equally spaced in a plane falling around a rotating planet whose rotation axis is perpendicular to the plane?

10. This is the algebraic proof that a gravity proportional to distance, like a spring force, yields no tides. g = -k*r is the gravity acceleration. To look for tides, you subtract the g at the center of the Earth, call it g_o, which is at a vector distance r_o from the Moon. So:
delta g = g - g_o = -k*(r - r_o) = -k* delta r.
That's it. But now note that delta g would work just like the gravity from the Earth itself (except k is larger for the larger mass). The point being, it is isotropic at the surface of the Earth, just like the Earth's gravity would be. That yields no tides.
As for your question about whether an ad hoc gravity might require more than just 4 points (which are really any two that don't line up with the Earth center, the other two are just symmetrically across), I'll have to whip out some mathematics of linear functions. In general, *any* law of gravity that is at all smooth (i.e., differentiable, but I think we can take that as given) can be written to first order in the (small) quantity delta r (this is a vector Taylor expansion):
delta g = [L]*delta r, where [L] is a linear transformation, call it a matrix. If we restrict our attention to the equatorial plane, any linear transformation [L] that has the same action on two linearly independent directions must have that same action on the entire equator. This suffices to eliminate the potential for tides, because what is happening in the polar direction will not break the axisymmetry.
So there is no (smooth) ad hoc gravity that requires more than just the analysis along two independent directions around the planet equator, to determine if there will be tides. But you do need the two independent directions-- one by itself does not cut the mustard. This is the flaw in the usual "sound byte" explanation you see absolutely everywhere.

11. Originally Posted by Ken G
This is the algebraic proof that a gravity proportional to distance, like a spring force, yields no tides. g = -k*r is the gravity acceleration. To look for tides, you subtract the g at the center of the Earth, call it g_o, which is at a vector distance r_o from the Moon. So:
delta g = g - g_o = -k*(r - r_o) = -k* delta r.
That's it. But now note that delta g would work just like the gravity from the Earth itself (except k is larger for the larger mass). The point being, it is isotropic at the surface of the Earth, just like the Earth's gravity would be. That yields no tides.
I don't understand that. If your r is measured from the center of the moon then it is not the same on the near and far side of the earth, so -k*(r - r_o) will not be the same all over earth's surface. If r is from the center of the earth then it is the same everywhere on the surface of the earth, but that doesn't tell us anything about the tidal force of the moon.

What am I missing?

12. True, r-r_o is not the same, as a vector, over the surface of the Earth. But note what it is-- it's the displacement from the center of the Earth! So r-r_o is the transformation from a Moon-centered coordinate, r, to an Earth-centered coordinate, r-r_o. So over the surface of the Earth, -k*(r-r_o) is going to point toward the center of the Earth, and be a constant magnitude. That's what I meant by isotropic over the surface.

13. Originally Posted by Ken G
True, r-r_o is not the same, as a vector, over the surface of the Earth. But note what it is-- it's the displacement from the center of the Earth! So r-r_o is the transformation from a Moon-centered coordinate, r, to an Earth-centered coordinate, r-r_o. So over the surface of the Earth, -k*(r-r_o) is going to point toward the center of the Earth, and be a constant magnitude. That's what I meant by isotropic over the surface.
I see what you mean, I think. But aren't you forgetting that the direction of the earth's gravity will swap direction, and start to decrease as you move along the moon-earth axis from the moon and pass through earth?

EDIT:
I have a feeling I'm completely missing your point, Ken G , maybe this'll at least clarify how I'm going wrong. This is how I pictured it. Each number is the gravity vector pointing left and decreases linearly from each body. The first line is the moon's; second, the earth's; and third, the net g.
Code:
```M  10   9   8   7   6   5   4   3   2   1   0  -1?
-15 -16 -17 -18 -19   E  19  18  17  16  15
g  -5  -7  -9 -11 -13      23  21  19  17  15```
As you can see, three units from the earth's center we have vectors pointing towards the earth's center but one has a magnitude of 9 and the other 19. What we do get, though, is a constant difference in magnitude of 10 between the near and far side, is that what you meant?
Last edited by worzel; 2005-Oct-18 at 11:23 PM.

14. The problem worzel is you are imagining that gravity is decreasing linearly, because your intuition tells you gravity must decrease. But I meant the case where gravity *increases* linearly with distance. Pretty unphysical, I realize, but that's the case that would yield no tides, like a spring force.

15. Originally Posted by Ken G
The problem worzel is you are imagining that gravity is decreasing linearly, because your intuition tells you gravity must decrease. But I meant the case where gravity *increases* linearly with distance. Pretty unphysical, I realize, but that's the case that would yield no tides, like a spring force.
Ah, sorry, I should read more carefully.

I (tentatively) think I see what you mean now. The difference between the moon's g at earth's center and other positions along the moon-earth axis yeilds an equation that is identical (except for a constant) to earth's own g. At first I thought this would probably only work in that dimension, but a bit doodling leads me to believe that it does work in all three - giving the bizzare conclusion that with linearly increasing g, the presence of the moon just strengthens the earth's own g field (as you say), and by a constant factor, no matter how far apart they are! I think it still qualifies as a tidal force, but as it doesn't vary over time for a given place on earth it doesn't cause any tides

It is already in your equation anyway, but an easy way to see this is to note that the g due to a body M at position p is the vector from p to M, so the difference between the g due to M at two different positions is just the vector between those two positions.

16. Yes, you have it right on worzel. And I agree that some tidal forces don't actually produce tides, the nomenclature is a little slippery there. Note it would be a weird universe-- it would have to have finite mass, and the local squishing would be the same everywhere, regardless of the relationship to the actual masses present.

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