Talk:Tidal force

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Why is the Weyl Tensor (or its simplification of differentiating Newton's force law equation) necessary?

Similar point: Some physicists, e.g., Richard Feynman in Lectures on Physics, vol 1, attribute the ocean bulge on the 'side' of the earth opposite the moon, to the centrifugal effect due to the rotation of the earth around the earth-moon center of mass. This seems to be a more cogent explanation than some mysterious 'cinching at the waist causes the the other parts to bulge' notion, especially where we are dealing with fluids rather than solids. Alternatively, a 'gravity warps the very fabric of space' argument may make some sense here.

1 Tidal forces causing the bulge on the other side
- 1.1 Centrifugal force not necessary
- 1.2 Formula
2 Additional effect of rotation?
3 Roche limit?

[edit] Tidal forces causing the bulge on the other side

I'm surprised no one before Feynman seems to have suggested the obvious centrifugal force from the earth's orbit (one rotation per month) around the center of mass of the moon and the earth. Common geography texts and even encyclopedia britannica are missing this.

This centrifugal force is obviously most significant as it is what balances the gravitation pull from the moon. Without it the earth and moon would fall in on each other. I would think Newton took this into account in his 1686 paper?

It is the DIFFERENCE between this centrifugal force and the the moon's gravitational pull on any material object on the earth's surface that make up the tidal forces "ellipsoid" (similarly for the earth-sun "ellipsoid". From the average values that balance each other, the centrifugal force increases linearly with the distance from, and in a direction away from the earth-moon center of mass. The gravitational pull decreases as one over the square of the distance from the moon.

[edit] Centrifugal force not necessary

While convienient to ease explanation, centrifugal force (or rotation) is not necessary for tidal forces to occur. If the moon was falling "directly towards" earth, the tidal forces would also be felt. (Of course, as the two bodies come closer, tidal forces varies).

possible, but would the tidal effect on earth as is, be as strong without the centrifugical force? or am i missing the point here

[edit] Formula

(William M. Connolley 09:35, 27 Aug 2004 (UTC)) Having pondered this a bit, I don't think I believe the formula. Certainly the one on the page has a typo: its literally meaningless to have the formula immeadiately followed by "<< r". Also, dr is undefined. Check back to before it was TeX'd: a comma is now missing.

But also, I'm not sure the intended formula is accurate either. I think that if the outside grav field is:

a = GM/R^2

(a = acceleration; M mass of central body; R distance from) then the "tidal force" between the center and edge of a body radius r at diatance R is:

delta-a = GM ( 1/R^2 - 1/(R-r)^2)

which is then *approximately* equal to:

delta-a = GM ( 2/R^3 ) . r

if r << R (modulo a sign convention or two...).

A clear derivation can be found at fr:Discuter:Sphère de Hill.

Urhixidur 14:47, 2004 Aug 27 (UTC)

A look at the history shows an interesting migration in the formula over time, but it was never actually right; an extra factor of 2 got inserted early on. I rewrote most of the article, and I hope it's right now. Fpahl 21:24, 6 Oct 2004 (UTC)

I think the article needs to explain better why there is a force in two directions instead of simply adding together in one direction. --ShaunMacPherson 03:19, 17 Oct 2004 (UTC)

[edit] Additional effect of rotation?

The article says "For two bodies rotating about their barycenter, the variation in centripetal force required for the rotation adds to the tidal force." I do not believe this. There is no "additional" effect of rotation. The high tides on both sides of the Earth are both perfectly well explained by the tidal effects of the Moon's gravity, and there is no additional effect caused by the fact the the Earth and the Moon are in orbit about their barycenter.

It is perhaps misphrased. What is means is, if you're trying to work out whether the body is going to fall apart or not, the rotation comes into it. William M. Connolley 14:30, 21 December 2005 (UTC).

Here's a thought experiment. Consider the positions of the Earth and the Moon at a particular moment in time. If the Earth and the Moon had somehow arrived at those positions and were stationary at the time (but of course free-falling towards each other), then the ocean tides on Earth would be exactly the same as normal. Someone else has made this point above under "Centrifugal force not necessary".

Next, suppose that the Earth and the Moon had somehow arrived at those positions but were passing each other at a high speed, way in excess of their actual orbital speeds about their barycenter. Again, the ocean tides on Earth would be exactly the same as normal.

The fact that in reality, the Earth and the Moon are, in a sense, passing each other at just the right speeds to keep them in roughly circular orbits about their barycenter makes no difference to the ocean tides on Earth.

Where do you get the formula

$F_t = \omega^2mr + \frac{GMmr} {R^3}$

from? It's not consistent with the earlier one:

$F_t = \frac{2GMmr} {R^3}$

for $ω = 0$ .

Also, you say that r is "the distance from the reference body's center along the axis". But surely, in the $ω 2 m r$ term, r has to be the distance from the center of rotation, ie. the barycenter. And where does the $ω 2 m r$ term come from? You can't say it comes from the fact that the object is moving in a circle. It's the force that makes the object move in a circle, but its origin must be something else.

So I don't see where you got this formula from, and I think this entire section on "Additional effect of rotation" is unnecessary and, moreover, incorrect.

I'm sorry if all this seems to break the rule on civility, but I think this section is wrong - I don't know how else to say it. I've presented my reasoning, let's discuss it.

Occultations 21:02, 20 December 2005 (UTC)

Gosh no that was very civil! But I think you've misunderstood - see top comment. William M. Connolley 14:30, 21 December 2005 (UTC).

I'm glad you weren't too offended. But a few questions ...

What have I misunderstood, and in what way?

Do you believe that there is an additional effect on tides that "adds to the tidal force" when two objects are rotating around their barycenter? (I don't.)

How do you arrive at the formula with the $ω 2 m r$ term?

How come it gives a different answer from the earlier formula when $ω = 0$ ? Which one is correct? (I think the earlier one is always correct.)

Occultations 13:06, 22 December 2005 (UTC)

OK, I think I know what the problem is but I may have misinterpreted you. So: this is nothing at all to do with tides, really. What it is, is about the forces that thend to break up the smaller body (for simplicity). Regardless of the relative motions of the bodies (in Newtonian physics...) there is the std tidal forces. *In addition* if the smaller body is rotating, there are additional forces (centrifugal). As it happens, the most common situation is for the two bodies to be tide-locked which affects omega.

having said that, I haven't checked the formula, and I don't see why it should be labelled as F_t. William M. Connolley 16:35, 22 December 2005 (UTC).

Ok, I think I understand what you're saying. The Earth and the Moon are rotating about their barycenter, and so there's a centrifugal effect - water anywhere on the Earth seems to weigh less than normal because of "centrifugal force", or, to avoid mentioning fictitious forces, because some of the net force on it is "used up" as the centripetal force making it move round in a circle once every sidereal month.

This is presumably where your

ω 2 r

term comes from. And in that term,

r

can only be the distance from the centre of rotation, ie. the barycenter of the Earth/Moon system.

So how big is this effect? One rotation per sidereal month is about 2.7×10^-6 radians/s. So

ω 2 r

at the surface of the Earth facing away from the Moon (where

r

is about 1.75 Earth radii) is about 8×10^-5 m/s², and at the sub-lunar point on the surface of the Earth (where

r

is about 0.25 Earth radii) it's about 1×10^-5 m/s². In comparison, the tidal effect of the Moon at these two points is $\frac{2GMr} {R^3}$ , which is about 1×10^-6 m/s². The centrifugal effect is much bigger than the tidal effect. So if there really is an "additional effect of rotation", then the high tide at the surface of the Earth facing away from the Moon would be about 8 times bigger than the high tide at the sub-lunar point. This is not what is observed.

For a theoretical approach, see Tides#Alternative explanation, which says "the centrifugal force is uniform and does not contribute to the tides".

Where you talk about the forces that tend to break up a body, here the rotation is obviously important, any rocky body will fly apart if it's rotating fast enough. But what rotation are we talking about? In the case of the Earth, it's its daily rotation relative to the fixed stars. Its monthly rotation around the barycenter of the Earth/Moon system does not contribute (look again at the picture at Tides#Alternative explanation - "the Earth has no "rotation" around this point. It just "displaces" around this point in a circular way"). This effect, where "centrifugal force" tries to pull a body apart, has nothing to do with tidal forces.

So I've presented theoretical and experimental arguments that there is no "additional effect of rotation" on tides or tidal forces. Do you agree?

--Occultations 21:24, 23 December 2005 (UTC)

Umm, I think we're getting closer, but not there yet. To repeat: the centrifugal effect has nothing whatever to do with tides (err). Tides and tidal forces are differenct. The only effect is on the break-up point of the smaller body, assuming that the smaller body has no internal cohesive forces other than gravity. Tidal forces tend to break the body up. So do centrifugal forces. William M. Connolley 21:55, 23 December 2005 (UTC).

Yes, I think we're getting there. You're saying that rotation produces a centrifugal effect, which has nothing to do with tides or tidal forces. But the centrifugal force does add to the tidal force to produce a net force that's larger than either of them individually, and this net force is what tries to break to body up.

But there's still one problem. When you add the

ω 2 m r

term to $\frac{2GMmr} {R^3}$ , how do you manage to end up with $\omega^2mr + \frac{GMmr} {R^3}$ ? A coefficient of 2 has disappeared. Is this a typo?

--Occultations 14:31, 24 December 2005 (UTC)

- - - - -

Let's define what the tidal force (say from the Sun) is. It is minus the force that the ground must exert in order to hold some small body on its surface, after we subtract the force that it would exert anyway due to effects unrelated to the Sun (Earth's gravity etc).

Since you're talking about the centrifugal force, you're probably working in a reference frame rotating around the Sun together with the Earth. OK, let's do so.

Let's define:

$r$ is the radius of the Earth

$R$ is the Sun-Earth distance

$F t$ is the tidal force which was calculated in the article without rotation

In the rotating reference frame, the body is moving in a circle around the center of the Earth with velocity $v = ω r$ (similarly to how the North pole points towards the Sun in the summer, and away from the Sun in the winter). Let's say the body is on the far side of the Earth. Its (centripetal) acceleration is $a = - ω 2 r$ , where we defined the direction away from the Sun as positive.

The forces acting on the body are:

(1) Gravity from the Sun = [Sun's gravity at the center of the Earth] $+ F t = - m ω 2 R + F t$

(2) Centrifugal force = $m ω 2 (R + r)$

(3) Coriolis force = $- 2 m ω 2 r$

(4) Force from the ground = [usual force] $-$ [tidal force]

(5) Forces unrelated to the Sun

Note that (5) and the first term of (4) cancel each other, by the definition of the tidal force.

So the net force is $Σ F = F t - m ω 2 r -$ [tidal force]

Using $Σ F = m a$ we get [tidal force] = $F t$

What do you think?

Happy New Year!

Yevgeny Kats 09:41, 1 January 2006 (UTC)

dances*

[edit] Roche limit?

Should the Roche limit be introduced on this page? It mentions break-up of celestial objects due to tidal forces in several places, but Roche limit is relegated to a "see also" at the bottom of the page. It ought to be fairly easy to give it a 1 sentence introduction somewhere in the text...? JulesH 11:40, 26 September 2006 (UTC)

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