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Cracky_Jack
2010-Jun-12, 01:28 PM
This old thread seems comes closest to answering a question I worry about. (http://www.bautforum.com/showthread.php/74661-Tides-and-centripetal-force-I-m-stuck?p=1249403#post12494039)

Consider a disc free-falling towards a central atrracting mass, with the disc's plane normal to the line from mass to disc. In its plane the disc will experience a compressive tidal force directed towards its centre. (There will also be a stretching tidal tension normal to its plane, but let's ignore this for the case of a thin disc.) If the disc were set spinning about its axis a centrifugal force would act outwards towards the circumference of the disc. At an appropriate angular speed these two forces might neatly cancel.

Would an intelligent ant on a disc therefore observe a rapidly rotating celestial sphere without experiencing any concomitant motion sickness? (Supposing the mass to be large enough and the ant on the spinning disc disc to be close enough to it.) A great mystery for that ant?.

Ken G
2010-Jun-13, 07:07 AM
Consider a disc free-falling towards a central atrracting mass, with the disc's plane normal to the line from mass to disc. In its plane the disc will experience a compressive tidal force directed towards its centre. (There will also be a stretching tidal tension normal to its plane, but let's ignore this for the case of a thin disc.) If the disc were set spinning about its axis a centrifugal force would act outwards towards the circumference of the disc. At an appropriate angular speed these two forces might neatly cancel. An interesting point that is completely true.


Would an intelligent ant on a disc therefore observe a rapidly rotating celestial sphere without experiencing any concomitant motion sickness? I think they might still get motion sickness (if ants can!), because I thought it was a disconnect between the fictitious forces and what the eye sees-- so you get sick if you feel fictitious forces but don't see anything strange (like below decks in a ship), but you can also feel sick if your eyes see something moving but you feel no fictitious forces (like watching a movie of a roller coaster ride). The ants might still get that latter effect.

Also interesting is that you can get rid of the centrifugal fictitious forces, but not the coriolis fictitious forces. So the ants would not feel any tendency to slide to the outside edge of the disk, but they could build Foucalt pendulums to tell them that the disk is spinning, and the universe is not spinning around them (notwithstanding currently debated issues surrounding Mach's principle and general relativity). Note you could also put the disk in orbit, rather than free-fall toward the gravity source, to maintain an equilibrium situation, and still play the game you are playing. It's a cute scenario, to be sure, and a nice problem to ask what the spin rate needs to be (the spin period must equal the orbital period divided by the square root of 2).

astromark
2010-Jun-13, 08:01 AM
Taking into account that a gyroscopic effect is sensed by the canals of the inner ear and 'vertigo' can effect the vision less and hidden.

and ants do not seem to be effected by such... Can you imagine a sea sick whale :)?

:" the spin period must equal the orbital period divided by the square root of 2.":... There go the ants..;...`';.

Ken G
2010-Jun-13, 08:16 AM
Taking into account that a gyroscopic effect is sensed by the canals of the inner ear and 'vertigo' can effect the vision less and hidden. Yes, that's another good point-- the inner ear uses the coriolis force as much (or more?) than centrifugal forces. So just what causes seasickness is rather complicated, and cancelling out centrifugal effecs might not be enough-- especially if the ants' eyes (which they don't have) see stars going around them for no apparent reason.

tusenfem
2010-Jun-13, 10:50 AM
Moved to its own thread, and the inner ear ant discussion is not to be pursued here.

DrRocket
2010-Jun-13, 02:00 PM
This old thread seems comes closest to answering a question I worry about. (http://www.bautforum.com/showthread.php/74661-Tides-and-centripetal-force-I-m-stuck?p=1249403#post12494039)

Consider a disc free-falling towards a central atrracting mass, with the disc's plane normal to the line from mass to disc. In its plane the disc will experience a compressive tidal force directed towards its centre. (There will also be a stretching tidal tension normal to its plane, but let's ignore this for the case of a thin disc.) If the disc were set spinning about its axis a centrifugal force would act outwards towards the circumference of the disc. At an appropriate angular speed these two forces might neatly cancel.

Would an intelligent ant on a disc therefore observe a rapidly rotating celestial sphere without experiencing any concomitant motion sickness? (Supposing the mass to be large enough and the ant on the spinning disc disc to be close enough to it.) A great mystery for that ant?.

I assume that we are talking about the ant as a point -- vanishingly small.

Tidal forces are due to the variation in gravity, (1/R^2) as function of distance from the central attracting mass. The centrifugal forces dut to spinning are due to rotational speed and distance r from the center of the disc. The centrigugal force will be radially symmetric about the axis of the disc, but the tidal forces will not. Think about a line joining the center of the disc to the center of the attracting mass -- the gravitational force plus the centrigugal force add on the end toward the attracting mass and subtract on the end away from it, while the gravitational force is slightly greater on the end towards the mass and slightly less on the end away from it. . So the centrifugal forces cannot precisely compensate for the tidal forces at all points of the disc simultaneously.

Any compensation that you set up will be the result of internal stresses in the disc, which do not affect the ant. So, any cancellation that you might set up does not affect the ant.

tusenfem
2010-Jun-13, 02:17 PM
Sorry my fault about the discussion of motion sickness of the intelligent ants.

DrRocket
2010-Jun-13, 02:23 PM
Sorry my fault about the discussion of motion sickness of the intelligent ants.



You are completely forgiven -- I didn't understand that part of it anyway.

grant hutchison
2010-Jun-13, 02:45 PM
]The centrigugal force will be radially symmetric about the axis of the disc, but the tidal forces will not.I think the OP sets up a situation in which the tidal forces are radially symmetrical about the axis of the disc: it stipulates that the plane of the disc is perpendicular to the line of fall of the disc.

Grant Hutchison

DrRocket
2010-Jun-13, 02:50 PM
I think the OP sets up a situation in which the tidal forces are radially symmetrical about the axis of the disc: it stipulates that the plane of the disc is perpendicular to the line of fall of the disc.

Grant Hutchison

That just means that the disc is flying towards the "sun" like a frisbee. You would only have radially symmetric tidal forces if the center of the disc were to coincide with the center of the "sun", which would not be the case in a free fall.

grant hutchison
2010-Jun-13, 03:40 PM
That just means that the disc is flying towards the "sun" like a frisbee. You would only have radially symmetric tidal forces if the center of the disc were to coincide with the center of the "sun", which would not be the case in a free fall.We seem to have different pictures in your heads. To me, the OP describes the situation in which the disc falls towards the "sun" with its plane orientated horizontally, like a falling leaf or a hovering frisbee.

Grant Hutchison

Jeff Root
2010-Jun-13, 03:42 PM
A Frisbee flies in its own plane. The disk of the OP is horizontal in
the gravity field and falling vertically.

Also, a spinning disk in orbit would have to experience forces different
from those experienced by a spinning disk falling straight down. For
one, the orientation of the orbiting disk's plane has to change as it
orbits.

-- Jeff, in Minneapolis

Ken G
2010-Jun-13, 04:51 PM
Also, a spinning disk in orbit would have to experience forces different
from those experienced by a spinning disk falling straight down. For
one, the orientation of the orbiting disk's plane has to change as it
orbits.
True enough, but that's a minor complication (you could, for example, supply some lateral torque on the disk to get it to precess at the necessary rate). The interesting feature of the OP is that the spin of the disk is maintained such that the centrifugal force on objects (like ants) attached to the disk cancels the lateral tidal stresses, which is indeed possible everywhere on a (small) disk. For example, this means that buildings could be built on the disk without having to tilt them sideways to avoid being toppled by the tidal stresses (the buildings would also have a negative weight due to the radially stretching tidal stresses, so you'd have to nail them down!). The coriolis forces would still be there, however.

In terms of space-station kinds of applications, you could imagine a disk that was spinning to cancel the lateral tidal forces, but which was orbiting at something less than the Keplerian orbital rate, say by catching stellar radiation to produce an outward force on it. Then you could achieve a kind of "normal" gravity on the outer face of the disk, and the spin of the disk could avoid lateral tidal forces so you could avoid having water flow to the center of the disk (shaping the disk to follow an equipotential would work too, this is all if we insist on the disk being flat). It's just kind of a cute fact you can use spin to cancel tides in that geometry, I doubt there's too much practical importance, but the OPer wanted to use the scenario to gain some kind of insight.

DrRocket
2010-Jun-13, 04:52 PM
We seem to have different pictures in your heads. To me, the OP describes the situation in which the disc falls towards the "sun" with its plane orientated horizontally, like a falling leaf or a hovering frisbee.

Grant Hutchison

In that case, and to a first approximation (say a thin disc of small radius) the distance to the "sun" and the gravitational field would be uniform across the surface and hence there would be no tidal effect.

Edit. I re-read the OP, more carefully, and your picture seems to be correct.

To a higher degree of accuracy there would be some sort of bending moment imparted by the small tidal effect. I will have to think about that some more.

DrRocket
2010-Jun-13, 05:22 PM
Ok lets take the picture with the frisbee approaching the sun like a pancake falling towards a griddle. Also lets' idealize the disc and assume it is rigid, so we don't have to worry about bending moments or deformation that would cause it to form a slight "dome".

Consider two extremes.

Case 1 A disc of very large radius.

We will consider things near the outer edge. Conditions hear the center are included in the considerations of case 2.

With a very large radius, the edge of the disc is a long long way from the sun. So the gravitational field and hence any second order effects like a tidal effect are negligible at the edge. So an ant near that outer edge would be in free fall relative to that local gravitational field, unless there is some adhesion to the disc. The disc on the other hand experiences a gravitational field that is the integral of the attraction of it infitesimal parts under the local gravitational field, which does vary. So the net acceleration on the disc is greater than the local acceleration due to the gravitational field at the edge. The disc then falls away from the ant. In the meantime the centrifugal force also forces the ant radially outward. So long as the ant is in contact witt the disc, while the gravitational force has a component towards the center (simply due to trgonometry and not particularly to tidal effects) . Now one could cause those radial components to balance one another with the proper selection of geometry and rotational speed. So the ant would, at least for an instant, experience no acceleration radially outward from the center of the disc --- in essence gravity would be the "string" tie him to the center of the disc. This situation is highly dependent on the disc radius, the rotational speed of the disc and the distance from the "sun" and would not be maintained without a continuous adjustment of the rotatinal speed of the disc. Tidal forces are not an issue, as the ant is still viewed as a point, and the disc is rigid.

So momentarily the ant would be in equilibrium, but that would change as the disc falls towards the sun without active management of the rotational speed of the disc.

Case 2. A disc of very small radius

In this case the gravitational force can be considered normal to the disc at all points and uniform. The disc and ant are in free fall together and any rotation of the disc simply causes the ant to fly radially outward. Tidal forces are not an issue because the spatial extent of the disc allows the gravitational field to be considered as uniform across the disc.

Ken G
2010-Jun-13, 05:55 PM
Case 2. A disc of very small radius

In this case the gravitational force can be considered normal to the disc at all points and uniform. The disc and ant are in free fall together and any rotation of the disc simply causes the ant to fly radially outward.No. As I said, a rotation with a period that is the square root of 2 shorter than the orbital period at the distance in question will suffice to cancel the lateral tidal stress, so no ant will ever experience any tendency to slide to the edge of the disk if the rotation is set to be that rate. Yes, that will require "active management" of the rotation rate in the falling scenario (though not in the orbiting scenario).

DrRocket
2010-Jun-13, 05:59 PM
No. As I said, a rotation with a period that is the square root of 2 shorter than the orbital period at the distance in question will suffice to cancel the lateral tidal stress, so no ant will ever experience any tendency to slide to the edge of the disk if the rotation is set to be that rate. Yes, that will require "active management" of the rotation rate in the falling scenario (though not in the orbiting scenario).

Are you considering the disc to be in some orbit other than a simple free fall towards the center of the "sun" ?

That was not my understanding, but does sound like an interesting case.

Ken G
2010-Jun-13, 06:17 PM
Are you considering the disc to be in some orbit other than a simple free fall towards the center of the "sun" ?
I'm imagining one free fall where it is moving radially toward the Sun, and another where it is in a circular orbit around the Sun (just to make it last longer!). My answer applies either way.

DrRocket
2010-Jun-13, 06:42 PM
I'm imagining one free fall where it is moving radially toward the Sun, and another where it is in a circular orbit around the Sun (just to make it last longer!). My answer applies either way.

I am only looking at the former case. I don't think we are in conflict,

I do not see any significant tidal effects in the situation as I have idealized it, for the reasons stated. But to be clear what I am considering as tidal effects are phenomena due to the gradient of the gravitational field. There are certaily effects due to the field and the trigononetry relating lines on the plane of the disc to radii centered at the "sun", and the gravitational field does certainly obey the inverse-square law.

Ken G
2010-Jun-13, 07:09 PM
I am only looking at the former case. I don't think we are in conflict,

I do not see any significant tidal effects in the situation as I have idealized it, for the reasons stated. But the point is, if the disk rotates at the rate I mentioned, there will not be any tendency for any ants to slide off its edge, no matter how small the disk is. So that is a bit of a correction that is required here-- there really is a "cute" solution implied by the OP, and it really does have centrifugal forces balancing tidal forces.


But to be clear what I am considering as tidal effects are phenomena due to the gradient of the gravitational field. There are certaily effects due to the field and the trigononetry relating lines on the plane of the disc to radii centered at the "sun", and the gravitational field does certainly obey the inverse-square law.I'm not sure what you mean here-- we are all talking about gradients in the gravitational field, but that is a vector field that includes both the inverse-square business, and the trigonometric effects too. They're all in the gradients in the vector-field components, and all at the same order as the centrifugal forces of the rate of spinning I mentioned.

DrRocket
2010-Jun-13, 07:26 PM
I'm not sure what you mean here-- we are all talking about gradients in the gravitational field, but that is a vector field that includes both the inverse-square business, and the trigonometric effects too. They're all in the gradients in the vector-field components, and all at the same order as the centrifugal forces of the rate of spinning I mentioned.

There are two effects here. One is just the resolution of the gravitational force into components normal and tangential to the plane of the disc. That is what I mean by trigonometric effects.

Then there is the spatial gradient of that force, the difference in the gravitational force at the top of your head and at your feet. That is what I understand by "tidal force" the net effect of which in this case is tensile stress in your body (if you were in free fall).

And since I only condidered the case of free fall along a radius of the "sun" I don't know what to do with a spin rate tied to the orbital period -- that would be zero spin.

astromark
2010-Jun-13, 07:36 PM
.,';'.; ----0---- Oops !

So lets forget the ants... I am facing some confusion understanding this subject and want an answer.. ?
Does the gyro effect play any part of this ?
I can visualize a disk approaching a gravity well and think that some force would be detected as the forward movement would accelerate while the receding edge might slow. A twisting distortion. If we were to up scale this to converging galactic scales then I can see a massive distortion and twisting as the core area would twist as the outer rim might not. Is any of this what the OP is asking... ?

And whats going on here ? I only mentioned ants and motion sickness because it was part of the OP... I thought I saw a question.

publius
2010-Jun-13, 08:54 PM
This is indeed a rather cute scenario. In thinking about it, I tried to think what was "really happening" defined in terms of a coordinates centered on the gravitational source, and then of course had to chide myself, because the picture in free-falling "tidal coordinates" is just a good as any other, and may be simpler.

Consider the tidal tensor, which can be roughly written GM/r^3 *Diag(-2, 1, 1) in spherical coordinates, and where the minus sign indicates "stretching" (I forget which way it naturally works out, if the minus sign should be on the 2 or the others :sigh:) If we imagine a point at some 'r' radius, we can construct a local cylindrical coordinate system where the z-axis is radial, and the, "rho" and "theta", call them axes are in the plane tanget to the sphere at that point. "Rho" is the cylindrical radial coordinate.

We thus have a central tidal acceleration in the plane of GM/r^3 *rho (for small rho, of course). That seems to nicely let a test particle rotate around the observer at that point. Of course as the observer falls, r decreases and so the apparent central force will increase with time.

Also, in the case of doing this with an observer in orbit, the disc would have to *also rotate about an axis normal to the orbital plane* in order to keep the plane of the disc aligned tangent to the sphere around the gravitational source. The period of that rotation would be the same as the orbital period, which is of the same order as the other rotation period, so the net rotation would be something more complicated. I don't think it would possible to cancel out the tidal forces nicely in that situation, although I could be wrong. The second component of rotation necessary to maintain alignment would expose the edge of the disc to the stretching radial tidal component. EDIT: No it wouldn't, as would keep the disk at the same 'r' -- but it would be an additional rotation with the centrifugal force not in the plane of the disc.

And of course that ignore the GR geodetic precession effects of orbits, :lol: . That would be insignificant for most situations, but you can imagine what strange additional effects that would produce for this problem in a strong field in a tight orbit around a black hole. Then add strong frame dragging from a rotating hole, and you can only imagine what the possibilities might be there.


-Richard

Ken G
2010-Jun-14, 12:39 AM
And since I only condidered the case of free fall along a radius of the "sun" I don't know what to do with a spin rate tied to the orbital period -- that would be zero spin.
The period I mentioned was the Keplerian period at that radius, whether or not there was really an orbit at all. That sets the scale of the necessary spin, even if we have radial infall. But publius has pointed out that the precession is also at that scale, so I've oversimplified-- it isn't possible to completely cancel the lateral tidal effects in all directions at once, my solution only canceled the lateral tidal forces in the "north-south" direction (the direction perpendicular to the orbital plane in the orbiting case), though it works for the radial infall with one correction: forget the "square root of 2" bit, I was using the radial gradient not the lateral gradient. The required spin period is the same as the Keplerian period (for an orbit), or the Keplerian period at that radius (for radial infall). In the free-fall case, there's no problem with the precession, and you can indeed cancel the lateral tidal forces-- you just have to keep spinning faster, and of course you can only do it for so long before you hit!