Hello, back again good to see this discussion carrying on without me.

To go back to my tower example, what would happen if you increased the Earths rotation rate such that the centrifugal force on the man in the middle matched the gravitional attraction. The man in the middle now feels equal and opposite forces and if the tower disapeared he wouldn't move, (he is in orbit around the Earth). The man at the top of the tower however feels a slightly smaller gravitational pull (and in this case a slightly larger centrifugal force) so he starts to rise.

This is what happens with the far side of the Earth, at the centre of the Earth gravity and the centrifugal force are balanced so the centre of the Earth moves in a circle around the barycentre. The far side of the Earth feels the same centrifugal force, but a slightly lower gravitational force and thus feels a net force away from both the centre of the Earth and the moon. Though this is of course tiny compared to the self gravity of the Earth.


Simon

As near as I can tell, johnwitts agrees with what I said there, but my objections were against other things. When he said "Every point of the Earth is following the same 'orbit' around the Earth/Moon centre of mass," I pointed out that every point of the Earth does not orbit around the Earth/moon center of mass. Positions on the surface of the Earth orbit points that are offset by 6000 kilometers from the Earth/moon center of mass.

Also, I do understand the explanations of the tides that include centrifugal force in their analysis--I just do not think that it is necessary to include centrifugal force. That is why I defend the BA's choice to not.

So what causes a spacecraft to increase it's orbital height when you increase it's orbital velocity, or indeed to lower it's height when 'retro firing' along the orbital path.

That is also an interesting subject, but when you think about it, it's clear: you have to fire a rocket to get into space and then into orbit, but when that rocket rises, it loses some of its energy, and velocity.

I also have no doubt that if you calculate all the forces relative to the Centre of the Earth, you get the right answer but I don't think it good way to understanding the forces involved.

To give another example it is realtively simple to reformulate the equations of motion of the planets to be relative to the centre of the Earth and this would be just as accurate as the equations relative to the Sun or the Barycentre. In this frame you would see something very similar to Ptolemy's epicycles.

But although this is accurate it is not a good way of describing the fundemental physics involved in the system.

Similarly with the Earth centred discription of the tides. The fundemental thing it seems to me in understanding the tides, is to understand why the tidal force exerted by the moon causes part of the Earth to move away from the moon. Though I have no doubt this will fall out of the maths in the Earth centred case, (I have been desperatly trying to avoid digging out my old textbooks and going through it). I think the best way of understanding what is going on is with the centrifugal force (or inertia).

On an other matter, you don't need rockets to change you distance from a object you are orbiting, it simple a matter of velocity. The simplest example is the eliptical orbit, the distance is constantly changing but not a rocket motor insight.


Simon

If you're concerned with the best (or even better) way of describing the tidal effect, then none of the ways described in this thread fit the bill. A far simpler and more complete analysis would calculate the form of the equipotential surface around the Earth. But that's not the point.

Clearly, with the "Earth-centered" version, you do not have epicycles falling out of it, or any other complications, so there really isn't any problem with using it.

There is best as is most complete, but there is also best as in allowing physical understanding.


It is by no means obvious in the Earth centred frame why part of the Earth moves away from the moon. It is not enough to say that the force is less, as the tower example shows. If you do the maths for the Earth centred case, the centrifugal force is implicit as the the Earth-Moon distance remains constant. I think for understanding of what happens the centrifugal force has to be made explicit.

It is not enough to understand the forces acting, there has to be some consideration of the reaction to the force and this is best done in an inertial frame.

I keep coming back to the fact I don't think the description of tides in the book gives good explanation of why part of the Earth moves away from the moon.

Simon

It's not necessarily an Earth-centered frame, for one thing. This is not a geocentrist argument. Not that there's anything wrong with that.

I don't see where the tower example shows that. You may have missed this post, since it came right after you left for vacation.

We certainly disagree about that. If the Earth were simply falling straight "down" (no centrifugal force), the tidal stretching would still be there. Of course, the Earth is simply falling.

In any frame the dstance from the far side of the Earth to the moon increases and this has to be explained.


I saw it, the point I was getting at is that in the stationary case, despite the difference in force both men remain where they are. But if you spin the system at just the right rate the man in the middle stays put but the man at the top moves up. If the tower is rigid he will find himself pushed against the ceiling.

In the falling case the tidal force would still be there, but all of the Earth would be moving towards the moon. I come back again it not just the force, but the reaction to the force. If you place two particles at 1AU from the sun, the force they feel is identical, but the way they behave depends on their initial velocity. They could crash into the Sun, orbit or fly off into deep space.

I am not sure I would agree that the Earth was simply falling. Although one way of thinking about an orbit is say that it is falling and missing. This however does have flaws, it does not explain, for example, why an object after passing pericentre moves away from the object it is orbiting. Another way of thinking about an orbit, is that the particle is trying to get away from the object and is always being pulled back, or to put it another way a balance between gravity and inertia.

Of course I wouldn't claim that the second way of thinking about an orbit is anymore 'right' than the first, but I do think it enables a better understanding of the effect of tidal forces.

Simon

But that's a personal standpoint--for me, it's not true at all. The point is, BA's presentation is not bad astronomy, from a physics standpoint. As you say, neither is better than the other.

Avoiding use of the term centrifugal force in the discussion because of its perception as a "forbidden term" would be bad, in my opinion, but I don't think that is the case here. Nor is it a matter of insisting upon a geocentrist position to the exclusion of all others.

I suspect that we are going to have to agree to disagree on this, but I come back again to the fact part of the Earth moves away from the moon, it gains potential energy. In order to understand tides this has to be explained and differences in force won't do it.


Simon

I thought we did agree that both explanations are OK? No disagreement there.

Sorry I meant disagree about the best way of explaining it, rather than which is true or not.

Now I was adding personal knowledge, but I read the description as saying, because the far side of the Earth feels a negative force relative to the centre of the Earth, it moves away from the centre of the Earth. Therefore as the distance between the centre of the Earth and the Moon is fixed the far side of the Earth moves away from the Moon.

Now thinking about it a bit more I realize that the fact the Earth-Moon distance is fixed, must imply a centrifugal force and that force must be proportional to the mass of the moon and hence the tidal force, so if I could be bothered to do the maths everything would come out right.

But, and in my mind at least it remains a big but, I don't think this is the best way of gaining physical understanding of what is going on. The discription implies that movement of the tides is simple due to the diferential gravity and this is not the whole answer. There has to be motion for the tidal force to have any effect, either falling or spinning. I know that the decription makes clear that the Earth is in freefall, but I don't think it makes clear enough how important the motion is.


On the otherhand the book and the discussion here have greatly expanded my understanding of tides. The last time I have looked at it was as a small part of a solar system course eight years ago that I knew I wasn't going to be examined on and to be honest I was more worried about Lagrange points and the disturbing function. So I have to thank you GrapesofWrath (can I call you Grapes) and all the others who have taken part in this discusion.

Simon

Well, I'm thinking of changing my nom anyway, to make it easier to type.

Different styles are not really bad astronomy, though.

Well, the distance is not fixed--but the important part, from the perspective of the Earth, is that two opposite sides move away from each other.

No.

No, by assuming a circular orbit has no real effect on the basic argument.

He is the root of my problem, I think the important bit is that it moves away from the moon. The effect of the tidal force is move part of the body away from the object exerting the force. This is so counter intuitive I think it needs explanation and I don't think the book does this.

Simon

<a name="20020409.5:49"> page 20020409.5:49 A.M.5:49 aka "_ALWAYS_"
On 2002-03-28 11:35, GrapesOfWrath wrote:
Gravity does not always act towards
HUb' 5:51 A.M. AGREE absolutly [NOT]!
maybe examples later I've been on way to long today
[/quote] Math Just takes time {no way around it}

OK, there is where I guess we disagree.

I think I gave an example earlier in this thread. Just look at the Earth/moon system. For us, standing on the Earth, gravity seems to point a couple tens of kilometers away from the center of mass of the Earth, but it misses the center of mass of the Earth/moon system by thousands of kilometers.

<a name="20030319.p24"> page 20030319.p24 aka push 2 look 4
March 19, 2003 8:02 P.M.
in an effort to find a path to actual
data for Pacific Northwest Tides
i've pushed this thead up..
:::50:::::::::::**********************:::::::::
:::30:::::**?***::::::::::::::::::::::******:::
:::10:::**:|::160+200*cos(t),:50*sin(t):::::**:
::::0::*-(Sun)----25yr/*--in AU units---------*
:::20:::**:|::HUb's ORBIT of planet X:::::::**:
:::40:::::*******::::::::::::::::::::*******:::
:::50::::::::::::********************::::::::::
::::::50:::0:::50:::100::150:200::250::300::350
to look for the eXpected delay then advance time's of
the actuall hi tides compared to Astronomical calculations

Here's the link to an extensive discussion of tides:
http://www.jal.cc.il.us/~mikolajsawi...and_tides.html
Enjoy
MS