🚀 go-pugleaf

In article <scottj-211292091243@iamac-1.dml.georgetown.edu> scottj@magic.dml.georgetown.edu (John L. Scott) writes:
> The high tide facing the moon is, of course, explained by the gravitational
> pull of the moon.  However, there have been at least two explanations for
> the high tide on the opposite side of the earth:
> 
>    A.  The moon stretches the earth
>    B.  Centrifugal pseudo-force
> 
> So... which is it, A, B, both, or neither?
> 
Neither.  Well, sort of A, I guess -- it depends what you
mean by "strethes the earth".  Bad ascii graphics follow:

			       _______
                              /       \
          __                 /         \
         /  \                A    B    C
         \__/                \         /
         moon	              \_______/
			        earth

Points A, B, and C all experience some amount of "pull" from
the moon.  It's easy to see (as you point out above) that
point A is pulled more than B, the center of the earth.  It's
also easy to see from the above diagram that point C feels
less pull than B;  therefore C "rises" relative to B, just as
A does.

Is that what you mean by "stretches the earth"?

Shag


-- 
       Rob Unverzagt        |
  shag@aerospace.aero.org   |   The will to a system is a lack of integrity.
unverzagt@courier2.aero.org |                        - Friedrich Nietzsche 
   18348@rhodes.aero.org    |

In article <1h50c7INNees@news.aero.org>, shag@aero.org (Robert M.
Unverzagt) wrote:
> 
> In article <scottj-211292091243@iamac-1.dml.georgetown.edu> scottj@magic.dml.georgetown.edu (John L. Scott) writes:
> > The high tide facing the moon is, of course, explained by the gravitational
> > pull of the moon.  However, there have been at least two explanations for
> > the high tide on the opposite side of the earth:
> > 
> >    A.  The moon stretches the earth
> >    B.  Centrifugal pseudo-force
> > 
> > So... which is it, A, B, both, or neither?
> > 
> Neither.  Well, sort of A, I guess -- it depends what you
> mean by "strethes the earth".  Bad ascii graphics follow:
> 
> 			                            _______
>                               /       \
>           __                 /         \
>          /  \                A    B    C
>          \__/                \         /
>          moon	                \_______/
> 			                             earth
> 
> Points A, B, and C all experience some amount of "pull" from
> the moon.  It's easy to see (as you point out above) that
> point A is pulled more than B, the center of the earth.  It's
> also easy to see from the above diagram that point C feels
> less pull than B;  therefore C "rises" relative to B, just as
> A does.
> 
> Is that what you mean by "stretches the earth"?
> 
> Shag

Thanks for the reply.  I got "stretches the earth" from somebody else's
post.  I took it to mean that the earth's shape was actually deformed to
the point that it had a noticeable effect of the oceans.  I was skeptical
of that idea.

So, are you saying that the moon pulls the earth away from the water on the
far side?  That certainly sounds reasonable to me.

--John L. Scott

In article <scottj-221292163050@iamac-1.dml.georgetown.edu> scottj@magic.dml.georgetown.edu (John L. Scott) writes:
>In article <1h50c7INNees@news.aero.org>, shag@aero.org (Robert M.
>Unverzagt) wrote:
>> 
>> In article <scottj-211292091243@iamac-1.dml.georgetown.edu> scottj@magic.dml.georgetown.edu (John L. Scott) writes:
>>> The high tide facing the moon is, of course, explained by the gravitational
>>> pull of the moon.  However, there have been at least two explanations for
>>> the high tide on the opposite side of the earth:
>>> 
>>>    A.  The moon stretches the earth
>>>    B.  Centrifugal pseudo-force
>>> 
>>> So... which is it, A, B, both, or neither?
>>> 
>> [ explanation deleted ]
>>
>Thanks for the reply.  I got "stretches the earth" from somebody else's
>post.  I took it to mean that the earth's shape was actually deformed to
>the point that it had a noticeable effect of the oceans.  I was skeptical
>of that idea.
>
>So, are you saying that the moon pulls the earth away from the water on the
>far side?  That certainly sounds reasonable to me.
>
>--John L. Scott

Maybe I can add something to this discussion. This explanation, while 
technically correct, applies mostly to solid bodies. For example, a solid
satellite comig close to a massive planet will be disrupted by the differential
force between the center and the close and far sides of the satellite.

However, this effect is small and can't explain the tides by itself. 
The principal effect is, strangely, the force perpendicular to the axis
between the two bodies. If you compute the effective (vectorial) force
on the surface of the ocean (i.e. the gravitational pull from the moon
minus the force that keeps the Earth-Moon system together), you will notice 
that this force is oriented outward when you are along E-M axis, inward
when you are perpendicular to it, and parallel to the surface of the earth 
when at 45 degrees.

This last force is directed toward E-M axis and can easily move the water
since it is not opposed to the Earth's gravity. This effect is supposed
to be the main cause of the observed tidal bulge.

(for a simple mathematical derivation, see Zeilik and Smith, 
"an introduction to astronomy and astrophysics")

-----
Jean-Remi Duquet
Physics dept.
Universite Laval
Quebec

In article <scottj-211292091243@iamac-1.dml.georgetown.edu>, scottj@magic.dml.georgetown.edu (John L. Scott) writes:
>The high tide facing the moon is, of course, explained by the gravitational
>pull of the moon.  However, there have been at least two explanations for
>the high tide on the opposite side of the earth:
>
>   A.  The moon stretches the earth
>   B.  Centrifugal pseudo-force
>
>So... which is it, A, B, both, or neither?

Both.  By definition, a tide is caused by the moon stretching the Earth (it'd
be pretty hard to tell there was a tide if there were no stretching, no?). 
Now, the Earth is always accelerated toward the moon by the gravitational force
of the moon.  The average acceleration is equal to that felt by the center of
gravity of the Earth.  However, since gravitational force goes as 1/r^2, the
part of the Earth nearer the moon feels a larger acceleration (causing a bulge
toward the moon) and the part further from the moon feels a smaller
acceleration (causing a bulge away from the moon).  If you call the distance
from the center of the moon to the center of the Earth R the radius of the
Earth r, and the mass of the moon M, then the side near the moon feels an
acceleration of:

	   GM
	-------
	(R-r)^2

the center of the Earth feels an acceleration of

	GM
	---
	R^2

and the side furthest from the moon feels an acceleration of

	  GM
	-------
	(R-r)^2

Since the Earth is held together by its own gravity, the whole Earth is
accelerated at the same rate.

This means that the side of the Earth closest to the moon feels an acceleration
from the moon, RELATIVE TO THE REST OF THE EARTH, of

	  GM	  GM	 (2Rr - r^2)GM	  2rGM
	------  - ---  = ------------- ~= ----
	(R-r)^2	  R^2	  (R-r)^2*R^2	   R^3

while the side furthest from the moon feels a relative acceleration of

	  GM	  GM	 (-2Rr - r^2)GM	   -2rGM
	------  - ---  = -------------- ~= -----
	(R+r)^2	  R^2	  (R+r)^2*R^2	    R^3

These relative accelerations are tides, and, to a first order approximation,
you get equal bulges on both sides.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.

In article <scottj-221292163050@iamac-1.dml.georgetown.edu>, scottj@magic.dml.georgetown.edu (John L. Scott) writes:
>Thanks for the reply.  I got "stretches the earth" from somebody else's
>post.  I took it to mean that the earth's shape was actually deformed to
>the point that it had a noticeable effect of the oceans.  I was skeptical
>of that idea.

You shouldn't've been.  There are tides on land, too, on the order of a couple
of feet.  If the Earth's crust were perfectly rigid, the ocean tides, relative
to the land, would be even higher.  Remember, too, that the oceans are part of
the Earth.  The tides are noticeable simply because the oceans are easier to
stretch than the crust.

>So, are you saying that the moon pulls the earth away from the water on the
>far side?  That certainly sounds reasonable to me.

Yes.  Please see another article I recently posted for the mathematics of it.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.

In article <1992Dec23.163859.16008@cerberus.ulaval.ca>, duquet@astro.phy.ulaval.ca (Jean-Remi Duquet) writes:
>Maybe I can add something to this discussion. This explanation, while 
>technically correct, applies mostly to solid bodies. For example, a solid
>satellite comig close to a massive planet will be disrupted by the differential
>force between the center and the close and far sides of the satellite.
>
>However, this effect is small and can't explain the tides by itself. 

Er, yes, it does.

>The principal effect is, strangely, the force perpendicular to the axis
>between the two bodies. If you compute the effective (vectorial) force
>on the surface of the ocean (i.e. the gravitational pull from the moon
>minus the force that keeps the Earth-Moon system together), you will notice 
>that this force is oriented outward when you are along E-M axis, inward
>when you are perpendicular to it, and parallel to the surface of the earth 
>when at 45 degrees.

Water being a fluid, though, is incapable of passing compressional forces along
the surface, so the effect of the compressional force you're talking about is
only a second-order effect at the locus of a high tide.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.