A bank of rocket motors would have to go on the outside of the body pushing in, and would be the same as the track - no farside bulge. A tether only exerts its centripetel force on a single point on the body, which means it's not comparable to a gravitational orbit - the effects on the body would be different.

Orbit is freefall, and the tides are the same either way.

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SeanF

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I will maybe get to the low tides in the fullness of time – but first things first. I’m not proposing a theory, I’m critiquing Mik Sawicki ’s.

I’ve already pointed out some of the Bad Mathematics in the Sawicki paper so it’s unlikely I’d get the same answers - and any claims to validity are just a tad premature.

How truly bizarre.

Anyway, if anyone else out there has genuine concerns about asymmetric tides, push gravity and suchlike and you’d appreciate a nudge in the right direction I’d be glad to help – for a while anyway.

Oh, please. Let's not be nitpicking spelling errors now, shall we?

From the NASA website:

"Orbit" and "freefall" are the same thing.

In other words, you only want to talk to people who agree with you. Bizarre, indeed.

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SeanF

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You mean like milli360??? We’ve been going at it for nearly a month and he doesn’t agree with a thing I say. Don’t talk such nonsense.
Certainly I’d prefer to talk to people with a genuine concern about the Sawicki paper, but you don’t have to agree with me. If you don’t agree let’s have a well-tempered debate, but one thing I’m not up for is surly argument – as you discovered.

If you’re interested in my viewpoint by all means question me – you are assured of a polite response. Meantime let me ask you a question.

The implication of what you say here is that NASA’s elevator passengers, falling towards the ground, are actually in orbit - they’re in ‘freefall’, orbit and freefall are the same thing, ipso facto they’re in ‘orbit’.
Is this the meaning that you intended to convey?

Milli360 doesn't agree with anything anybody says. Watch, he'll probably disagree with this!

And you feel that "(sic)"ing a spelling error and making statements like "How truly bizarre" are a good way to do this?

Okay, my statement was too general when taken out of context - but are you deliberately being obtuse? No, of course not all freefalls are orbits*, but still . . . orbit is freefall. You're arguing that a square is not a rectangle - don't take a disagreement to imply that all rectangles are squares.

*One of the definitions of orbit is "The path of a body in a field of force surrounding another body." If I wanted to nitpick, I could point to this definition and say that, yes, the falling elevator does qualify - but that would be just as irrelevant to the subject, anyway.

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I went back through all your posts to this thread, and I could only find two where you do actual calculations, this one and this too. Were there others that I missed? In the first one, you mistakenly try to apply the tidal calculations to a dripping tennis ball (the drip is not a tidal bulge)--and even then, you make a correction later. The second one was also a mistake on your part, you misunderstood what angle was being used in the sine calculation.
Hah, I once agreed with Beskeptical, even. And there were others.
No, you're on solid ground here. Excepting atmospheric and lithospheric friction, it will become an orbit. Even falling straight down, if you could drill a path through the Earth, you'd end up in an oscillating motion not unlike an orbit--in fact, the period of the orbit would match that of the ellipse of the same semimajor axis.

Told ya!
"On solid ground." Good one.

I never thought about it that way (although we have discussed the "oscillating through the hole" thing on this board before), but you're right - falling really can be considered in orbit. It's just an orbit that happens to intersect the surface at a couple points.

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Well, not quite. After all, once you're under the surface of the Earth, the gravitational pull/curvature approaches zero as you approach the center. That's going to complicate things, so it's not going to be as simple as a regular ellipse. Another thing I've wanted to calculate but not gotten around to (as long as you let me assume uniform density!)....
But the period will almost certainly be different, assuming there isn't some cancelled term that stretches the perigee as much as it slows down the period.

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We can't let you do that John, because it's the fact the density of the Earth is not uniform that means gravity actually increases as you approach the core.

As far as I understand it though, unform density or not, as you approach the centre you will be undergoing acceleration, and leaving the centre you will be decellerating. Calculating the actual rate of these is tricky though.

You may be familiar with the formula for period of an orbit about a mass, which is sqrt( (2pi)^2 r^3 /(GM) ). Using r=6378km, the radius of the Earth, M=5.976x10^24kg, G=6.672x10^-11 N m^2 /kg^2, and pi=3.1415926535, we get T=84.47 minutes for an orbit right at the surface of the Earth.

For the hole drilled through the center of the Earth, using that model of uniform density, the motion becomes simple harmonic motion, since the acceleration is just gx/r, g is just GM/r^2. k/m is g/r, and omega zero is sqrt( g/r ). The period is 2 pi divided by omega zero: sqrt( (2pi)^2 r^3 /(GM) ), which is the same formula used above.
Which is another problem with that "orbit". But it's close enough, and the point has been made that freefall has many characteristics of an orbit.

I'd just like to quote a snippet from that link, which was a post of tracer's:
Notice, we were both disagreeing with JS Princeton--but with good reason!

Yes. You appear to have missed this one (same link):

By claiming that they are the same Sawicki
(a) makes Da(s) general which is wrong, it applies only at point F and
(b) incorrectly locates the differential gravity mean at the centre of the earth, which is again wrong.

These errors combine to mask a tidal force asymmetry in Sawicki’s model which reaches 6% in the lunar tide calculation (Da(m) equ.4).

If you’re happy with these asymmetries that’s fine, but if anyone wants to know how to get rid of them give me a shout.

I answered it before at the same time I answered the rest of the post at that link.
No, that's not true, as I said before. The second sentence after his equation three says "Note that in the second step in Eq.(3) we neglected corrections from higher order R/ds terms."

If that is your last objection to Sawicki's math, then I have answered all of your objections. You can no longer say you've "already pointed out some of the Bad Mathematics in the Sawicki paper."
Go ahead and post it. Let's have a look at it.

Certainly.

The net acceleration at the far (F), centre (O) and close (C) points on earth in the Earth – Moon system is:

a(n) = w^2D(b) – GM/D(m)^2

where D(b) is distance to the barycentre, M is mass of the moon, and D(m) is distance to the moon.

Data: mean distance earth-moon = 3.844e08 m
lunar sidereal period = 2360591.5 sec
mass of moon = 7.353e22 kg
radius of earth = 6.378e06 m
earth centre to barycentre = 4.6867e06 m

Using the above data and equation for net acceleration gives:

at F: 4.626035e-05 ms^-2
at O: 0 ms^-2
at C: -4.63163e-05 ms^-2

As can be seen, the net acceleration at the centre of the earth is zero, as required, and at the far and close points the net acceleration is equal to 0.1%, well within the uncertainty in the data.

Using Sawicki’s equ.4 I get 1.0745e-06 and 1.1299e-06 – a difference of 5% (ouch!).

What this shows is that simple mechanics predicts a symmetrical tide to with 0.1% or less.
Sawicki’s ‘differential’ scheme, which ignores centripetal acceleration, predicts a tidal asymmetry of 5%.

You choose.

If anybody would like clarification on any of this I’ll be very glad to help.
If anybody wants to argue with it please do me the courtesy of first checking the maths yourself and telling me your results.

That first term on the right hand side is the centrifictional force term.

That is similar to how Stacey starts his analysis, except he uses potential instead of force. I was checking out the Wolfram pages on tides, and actually found Stacey's diagram there.
Your first clue should have been the fact that your answers are fifty times bigger than those (which, BTW, don't seem to come from Sawicki's equation four at that link). You still have the once-per-month rotational component in your answers, which gives an extremely large bulge all the way around the Earth equator. That swamps the tidal component--which is why your percentage is fifty times less.

It also misrepresents the actual size of the tide--that centrifictional bulge appears at all points along the equator, and does not act as a tide.
I'll choose Stacey over Sawicki over yours. But, as I've said before, if you're careful, any method can arrive at the same values. For your case you have to subtract the rotation of the Earth, w^2D, where w=(2pi)/month and D is the radius of the Earth. That value is ((2*3.14159)/(2360591.5 sec )^2 x 6.378e06 m, or 4.5186e-5 m/s/s. If you subtract that from your two answers, at F: 4.626035e-05 ms^-2, and at C: -4.63163e-05 ms^-2, you get at F: 1.07435e-06 and at C: -1.13030e-06, essentially the same as what you say Sawicki would get--without as much work.

In other words, the difference between your two answers is the same as the difference between the other two values, but you increased yours by a large factor that is not relevant to the calculation of the tide.

You seem to be saying (and this is not contentious) that the (monthly rotation) centripetal acceleration is uniform at all point around the equator so it should be factored out, whereas I say that this centripetal acceleration, because it is centred on the barycentre is lopsided so it should remain in. That would be fair wouldn’t it?
We're still not in agreement about centripetal acceleration are we?
O.k. then, some time ago you sent this diagram http://mensware.home.mindspring.com/dan/centrif.jpg with this explanation:
Just so I understand what the diagram means, can you perhaps elaborate a bit, for instance how does it show that centripetal acceleration is the same at all points and maybe how this is affected by an eccentric orbit – such as the earth around the barycentre?