Tide error on you web pages.
 

Hi Phil

Keep up the good work! But there is an error on your pages where you
describe how the moon forms two tides. You cannot explain this correctly
without noting that the earth-moon pair rotate about a common point. It
is this rotation that causes the two tides. If you like, you can explain
it as a centrifugal force acting to throw water out while gravity pulls
the water to water the moon. Since the gravity term is weaker on the
outside of the earth the centrifugal term dominates and the result is
two tides.

Many regards

Mark Cannell
Professor,
University of Auckland.

You make it sound as if the gravitational forces explains the bulge
under the moon and the centrifugal forces explains the bulge on the
side of the earth that turns away from the moon.
That is not right.

The gravitational difference alone can explain that there are bulges
on both side of the earth. That's why it is sometimes the only factor
mentioned when trying to keep the explanation simple.
The centrifugal element can only explain that there is a bulge on the
part of the earth that turns away from the moon.
That is why it is one of the elements (and there are others), that is
sometimes left out of the explanation.

While I think that in some cases it is a good idea to include the
centrifugal element in the explanation, I don't know exactly how many
elements one should include to make it a good explanation - but I
haven't yet seen a complete explanation in a popular publication.

Peter S/Y Anicula
Sailor
The seven seas


"Nav" skrev i en meddelelse
...
Hi Phil

Keep up the good work! But there is an error on your pages where you
describe how the moon forms two tides. You cannot explain this
correctly
without noting that the earth-moon pair rotate about a common point.
It
is this rotation that causes the two tides. If you like, you can
explain
it as a centrifugal force acting to throw water out while gravity
pulls
the water to water the moon. Since the gravity term is weaker on the
outside of the earth the centrifugal term dominates and the result
is
two tides.

Many regards

Mark Cannell
Professor,
University of Auckland.

Well Peter, I have to disagree there. The gravitational force acts only
toward the center of mass of the system. This cannot by itself produce
two bulges. To clarify this, try imagining the forces of gravity in 2D
on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the far
side and more water as you move toward the moon... -two bulges would not
be present.

Cheers

Peter S/Y Anicula wrote:
You make it sound as if the gravitational forces explains the bulge
under the moon and the centrifugal forces explains the bulge on the
side of the earth that turns away from the moon.
That is not right.

The gravitational difference alone can explain that there are bulges
on both side of the earth. That's why it is sometimes the only factor
mentioned when trying to keep the explanation simple.
The centrifugal element can only explain that there is a bulge on the
part of the earth that turns away from the moon.
That is why it is one of the elements (and there are others), that is
sometimes left out of the explanation.

While I think that in some cases it is a good idea to include the
centrifugal element in the explanation, I don't know exactly how many
elements one should include to make it a good explanation - but I
haven't yet seen a complete explanation in a popular publication.

Peter S/Y Anicula
Sailor
The seven seas

We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...
Well Peter, I have to disagree there. The gravitational force acts
only
toward the center of mass of the system. This cannot by itself
produce
two bulges. To clarify this, try imagining the forces of gravity in
2D
on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the
far
side and more water as you move toward the moon... -two bulges would
not
be present.

Cheers

You might make a case that the centrifugal explanation is easier for some people
to understand, but claiming that gravity doesn't cause the tides is just plain
bogus! What force are you really proposing? The Tide Fairy? The only force
at work here is gravity. It must be possible to explain the tides completely
(not counting local effects) simply by looking at gravity.

The Moon's gravity affect each portion of the Earth differently. You can divide
the total force into two components, one that affects the Earth equally, and the
other represents the differences. The sum of all of the differences nets out to
zero, so the first force can be thought of as the force that pulls the Earth
around the Moon-Earth center. When you subtract that out, what you're left with
are the differential forces that push out the bulges, and pull down the poles.

One conceptual problem with the differential view is that it appears that the
far side bulge is being pushed away from the Moon. That its not really the
case: it is being pulled toward the Moon, but with less force than the rest of
Earth. Its only when you subtract off the large primary force, which maintains
the orbit, that it appears that the far side bulge is being pushed away from the
Moon.



"Nav" wrote in message
...
Well Peter, I have to disagree there. The gravitational force acts only
toward the center of mass of the system. This cannot by itself produce
two bulges. To clarify this, try imagining the forces of gravity in 2D
on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the far
side and more water as you move toward the moon... -two bulges would not
be present.

Cheers

Peter S/Y Anicula wrote:
You make it sound as if the gravitational forces explains the bulge
under the moon and the centrifugal forces explains the bulge on the
side of the earth that turns away from the moon.
That is not right.

The gravitational difference alone can explain that there are bulges
on both side of the earth. That's why it is sometimes the only factor
mentioned when trying to keep the explanation simple.
The centrifugal element can only explain that there is a bulge on the
part of the earth that turns away from the moon.
That is why it is one of the elements (and there are others), that is
sometimes left out of the explanation.

While I think that in some cases it is a good idea to include the
centrifugal element in the explanation, I don't know exactly how many
elements one should include to make it a good explanation - but I
haven't yet seen a complete explanation in a popular publication.

Peter S/Y Anicula
Sailor
The seven seas

The gravitational force acts only toward the center of mass of the
system. This cannot by itself produce two bulges.

When you say that, you are mixing two explanations. That doesn't work.

We can certainly look at the gravitational force from the moon and the
gravitational force of the earth separately, and then ad the two, to
have a look at the combined forces. If you do not include part of the
rotation element, it works just fine.

If you only look at the gravitational forces, you can explain the two
bulges!

It is an abstraction. Not the "truth". Even if you include the
rotation it is still an incomplete abstraction. We are discussing
different incomplete models. We haven't yet reached anything near the
"truth".

When discussing different models it is important not to mix elements
casually.
I'm surprised that a mere sailor have to teach this to a professor.

Peter S/Y Anicula

"Nav" skrev i en meddelelse
...
Well Peter, I have to disagree there. The gravitational force acts
only
toward the center of mass of the system. This cannot by itself
produce
two bulges. To clarify this, try imagining the forces of gravity in
2D
on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the
far
side and more water as you move toward the moon... -two bulges would
not
be present.

Cheers

Peter S/Y Anicula wrote:
You make it sound as if the gravitational forces explains the
bulge
under the moon and the centrifugal forces explains the bulge on
the
side of the earth that turns away from the moon.
That is not right.

The gravitational difference alone can explain that there are
bulges
on both side of the earth. That's why it is sometimes the only
factor
mentioned when trying to keep the explanation simple.
The centrifugal element can only explain that there is a bulge on
the
part of the earth that turns away from the moon.
That is why it is one of the elements (and there are others), that
is
sometimes left out of the explanation.

While I think that in some cases it is a good idea to include the
centrifugal element in the explanation, I don't know exactly how
many
elements one should include to make it a good explanation - but I
haven't yet seen a complete explanation in a popular publication.

Peter S/Y Anicula
Sailor
The seven seas

Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:

We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...

Well Peter, I have to disagree there. The gravitational force acts

only

toward the center of mass of the system. This cannot by itself

produce

two bulges. To clarify this, try imagining the forces of gravity in

2D

on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far

side and more water as you move toward the moon... -two bulges would

not

be present.

Cheers

Jeff Morris wrote:

You might make a case that the centrifugal explanation is easier for some people
to understand, but claiming that gravity doesn't cause the tides is just plain
bogus!

What are you talking about? I never said that gravity was not a part of
the equation.


Let me repost:
"The gravitational force acts only toward the center of mass of the
system. This cannot by itself produce two bulges."

Note the "by itself".

I'll repeat myself, the key is to understanding the _two tides_ problem
is that the system is rotating and "centrifugal" forces are balanced
only at the centers of the masses by gravity.

Cheers

Peter S/Y Anicula wrote:

The gravitational force acts only toward the center of mass of the

system. This cannot by itself produce two bulges.

When you say that, you are mixing two explanations. That doesn't work.

We can certainly look at the gravitational force from the moon and the
gravitational force of the earth separately, and then ad the two, to
have a look at the combined forces. If you do not include part of the
rotation element, it works just fine.

If you only look at the gravitational forces, you can explain the two
bulges!


Well you keep saying that but it is not so. Unless you unclude the fact
that the system is rotating you cannot make two bulges on opposite
sides. Jeff posted a URL, have a read and then you will see the problem
-I hope.

It is an abstraction. Not the "truth". Even if you include the
rotation it is still an incomplete abstraction. We are discussing
different incomplete models. We haven't yet reached anything near the
"truth".

I was under the impression that gravitational models are very accurate
indeed. How else could we shoot a probe through the Cassini divsion?
Is'nt that near some sort of "truth"?

Cheers

"Nav" wrote in message
...


Jeff Morris wrote:

You might make a case that the centrifugal explanation is easier for some
people
to understand, but claiming that gravity doesn't cause the tides is just
plain
bogus!

What are you talking about? I never said that gravity was not a part of
the equation.


Let me repost:
"The gravitational force acts only toward the center of mass of the
system. This cannot by itself produce two bulges."

Note the "by itself".

That's exactly the point - gravity is the only force at work here. Gravity does
cause the bulges. The centrifugal forces are a "fiction" caused by the
accelerating reference frame. Why is it accelerating? Because of gravity!



I'll repeat myself, the key is to understanding the _two tides_ problem
is that the system is rotating and "centrifugal" forces are balanced
only at the centers of the masses by gravity.


That is a simplified way to look at it. If it helps your understanding, fine.
Your problem, however, is that you're insisting that this is the *only* way to
understand the problem. The are numerous correct ways to look at this. You
don't have to use centrifugal force to explain the far bulge.

Frankly, for me, it doesn't help at all, because the centrifugal force is
constant throughout the Earth. If it produces the bulge on the far side, how
can it also produce a bulge in the opposite direction on the near side? The
answer, of course, is that you have to add the centrifugal force to
gravitational force. which is different throughout the Earth. The resulting
force is exactly the same as the differential gravity from the other model. Why
is this? Because the centrifugal force is a "fiction" - it is simply the
opposite of the net gravitational force that causes the Earth to rotate around
the Earth-Moon system. In the differential model you subtract this out, in the
centrifugal model you add it.

So I have trouble thinking of centrifugal force as pushing out the far bulge;
for me the bulge is caused because the far side receives less pull from the Moon
than the rest of the Earth. However, arguing that one model is more correct
than the other is like arguing whether A+B=C or A=C-B.

"Nav" wrote in message
...


Peter S/Y Anicula wrote:

The gravitational force acts only toward the center of mass of the

system. This cannot by itself produce two bulges.

When you say that, you are mixing two explanations. That doesn't work.

We can certainly look at the gravitational force from the moon and the
gravitational force of the earth separately, and then ad the two, to
have a look at the combined forces. If you do not include part of the
rotation element, it works just fine.

If you only look at the gravitational forces, you can explain the two
bulges!


Well you keep saying that but it is not so. Unless you unclude the fact
that the system is rotating you cannot make two bulges on opposite
sides. Jeff posted a URL, have a read and then you will see the problem
-I hope.

There is no problem. The "differential" explanation starts by subtracting out
the total, net gravitational force and looking at just the differences at
various places on the Earth. What is the effect of this net component? It
accelerates the Earth towards the center of the Earth-Moon system. Thus, when
looking at the left over differences, you're already accounting for the rotation
of the Earth in this way.

As I mentioned in my other post, the net gravitational force subtracted out is
simply the opposite of the centrifugal force you've mentioned. To my mind,
neither of these causes the bulges, its when you subtract (or add its negative)
and looking at the differences around the Earth that you get the answer.

Jeff Morris wrote:
"Nav" wrote in message
...


Jeff Morris wrote:


You might make a case that the centrifugal explanation is easier for some

people

to understand, but claiming that gravity doesn't cause the tides is just

plain

bogus!

What are you talking about? I never said that gravity was not a part of
the equation.


Let me repost:
"The gravitational force acts only toward the center of mass of the
system. This cannot by itself produce two bulges."

Note the "by itself".


That's exactly the point - gravity is the only force at work here. Gravity does
cause the bulges. The centrifugal forces are a "fiction" caused by the
accelerating reference frame. Why is it accelerating? Because of gravity!



I'll repeat myself, the key is to understanding the _two tides_ problem
is that the system is rotating and "centrifugal" forces are balanced
only at the centers of the masses by gravity.



That is a simplified way to look at it. If it helps your understanding, fine.
Your problem, however, is that you're insisting that this is the *only* way to
understand the problem. The are numerous correct ways to look at this. You
don't have to use centrifugal force to explain the far bulge.

Frankly, for me, it doesn't help at all, because the centrifugal force is
constant throughout the Earth. If it produces the bulge on the far side, how
can it also produce a bulge in the opposite direction on the near side? The
answer, of course, is that you have to add the centrifugal force to
gravitational force. which is different throughout the Earth. The resulting
force is exactly the same as the differential gravity from the other model. Why
is this? Because the centrifugal force is a "fiction" - it is simply the
opposite of the net gravitational force that causes the Earth to rotate around
the Earth-Moon system. In the differential model you subtract this out, in the
centrifugal model you add it.

So I have trouble thinking of centrifugal force as pushing out the far bulge;
for me the bulge is caused because the far side receives less pull from the Moon
than the rest of the Earth. However, arguing that one model is more correct
than the other is like arguing whether A+B=C or A=C-B.


Good points. Well A=C-B ;-)

But, lets open this can of worms a bit further. I take and largely agree
with most of your view, but it is the kinetic energy in the system that
is powering the tides. If you locked the moon to the earth with a big
pole you would not have two tides would you? The mass is the same and so
is it's center... Gravity still works... but, just one tidal bulge.


Cheers

If the center of mass was the only factor involved, wouldn't the bulge be on
one side of the earth only?
Scout

"Nav" wrote in message
...
Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:

We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...

Well Peter, I have to disagree there. The gravitational force acts

only

toward the center of mass of the system. This cannot by itself

produce

two bulges. To clarify this, try imagining the forces of gravity in

2D

on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far

side and more water as you move toward the moon... -two bulges would

not

be present.

Cheers

"Nav" wrote in message
...

[deleted stuff where we unfortunately seem to agree]


Good points. Well A=C-B ;-)

But, lets open this can of worms a bit further. I take and largely agree
with most of your view, but it is the kinetic energy in the system that
is powering the tides. If you locked the moon to the earth with a big
pole you would not have two tides would you? The mass is the same and so
is it's center... Gravity still works... but, just one tidal bulge.


I don't think its fair to do this - you can mathematically eliminate effects by
shifting the reference frame, but "locking" objects together is changing the
problem at a more fundamental level. In this case, how to you "lock" the Earth?
In fact, the crux of this problem is that different parts of the Earth are
actually acting somewhat independently.

However, this brings up an interesting point. At some point in the distant
future the tides will be eliminated. How will this happen? Because the tides
lag the Moon the high tide is not directly under the Moon, but offset. This
creates soon torque that is transferring energy from the Earth to the Moon. The
result is that the Earth is slowing down, and the Moon's orbit is increasing.
This will continue (some say) until the Earth's rotation slows down to match the
Moon, and the bulge stays under the Moon. The Earth and Moon will at that point
be locked together. Because the Moon is smaller, it has already assumed this
orientation WRT the Earth.

If we work this backwards we find the in the distant past the Moon's orbit was
much closer to the Earth, and the Earth's day much shorter. Exactly how much
depends on what other theory you're trying to support or disprove. However, we
do know the effect is real - the measurement using equipment left behind by the
astronauts shows the distance increasing about 4 cm a year, and the Earth's day
lengthening by 1.5 milliseconds a century.

well then, you'd have to use a telescoping pole! ; )
Scout

"Jeff Morris" wrote in message
...
"Nav" wrote in message
...

[deleted stuff where we unfortunately seem to agree]


Good points. Well A=C-B ;-)

But, lets open this can of worms a bit further. I take and largely agree
with most of your view, but it is the kinetic energy in the system that
is powering the tides. If you locked the moon to the earth with a big
pole you would not have two tides would you? The mass is the same and so
is it's center... Gravity still works... but, just one tidal bulge.


I don't think its fair to do this - you can mathematically eliminate
effects by
shifting the reference frame, but "locking" objects together is changing
the
problem at a more fundamental level. In this case, how to you "lock" the
Earth?
In fact, the crux of this problem is that different parts of the Earth are
actually acting somewhat independently.

However, this brings up an interesting point. At some point in the
distant
future the tides will be eliminated. How will this happen? Because the
tides
lag the Moon the high tide is not directly under the Moon, but offset.
This
creates soon torque that is transferring energy from the Earth to the
Moon. The
result is that the Earth is slowing down, and the Moon's orbit is
increasing.
This will continue (some say) until the Earth's rotation slows down to
match the
Moon, and the bulge stays under the Moon. The Earth and Moon will at that
point
be locked together. Because the Moon is smaller, it has already assumed
this
orientation WRT the Earth.

If we work this backwards we find the in the distant past the Moon's orbit
was
much closer to the Earth, and the Earth's day much shorter. Exactly how
much
depends on what other theory you're trying to support or disprove.
However, we
do know the effect is real - the measurement using equipment left behind
by the
astronauts shows the distance increasing about 4 cm a year, and the
Earth's day
lengthening by 1.5 milliseconds a century.

Yes, so...

Cheers


Scout wrote:

If the center of mass was the only factor involved, wouldn't the bulge be on
one side of the earth only?
Scout

"Nav" wrote in message
...

Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:


We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...


Well Peter, I have to disagree there. The gravitational force acts

only


toward the center of mass of the system. This cannot by itself

produce


two bulges. To clarify this, try imagining the forces of gravity in

2D


on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far


side and more water as you move toward the moon... -two bulges would

not


be present.

Cheers

Jeff Morris wrote:

"Nav" wrote in message
...

[deleted stuff where we unfortunately seem to agree]


Good points. Well A=C-B ;-)

But, lets open this can of worms a bit further. I take and largely agree
with most of your view, but it is the kinetic energy in the system that
is powering the tides. If you locked the moon to the earth with a big
pole you would not have two tides would you? The mass is the same and so
is it's center... Gravity still works... but, just one tidal bulge.



I don't think its fair to do this - you can mathematically eliminate effects by
shifting the reference frame, but "locking" objects together is changing the
problem at a more fundamental level.

I don't see it that way, the explanation for the two tides based on
differential gravity alone does not care whether the earth is "moon
locked" at (say) an L point -and that why it is not the correct
explanation in my opinion. Of course it all comes down to gravity and
the energy of the system but the simplest close answer should consider
the rotation as well.

In this case, how to you "lock" the Earth?
In fact, the crux of this problem is that different parts of the Earth are
actually acting somewhat independently.

However, this brings up an interesting point. At some point in the distant
future the tides will be eliminated.

(Well not really, unless you ignore the Sun). But I think this point
reinforces what I've been trying to get across, without considering the
rotation(s) about the center of mass you don't get a two tide situation.
Any description that does not explicitly consider the relative motion
will not generate two tides -do you agree?

How will this happen? Because the tides
lag the Moon the high tide is not directly under the Moon, but offset. This
creates soon torque that is transferring energy from the Earth to the Moon. The
result is that the Earth is slowing down, and the Moon's orbit is increasing.
This will continue (some say) until the Earth's rotation slows down to match the
Moon, and the bulge stays under the Moon. The Earth and Moon will at that point
be locked together. Because the Moon is smaller, it has already assumed this
orientation WRT the Earth.

If we work this backwards we find the in the distant past the Moon's orbit was
much closer to the Earth, and the Earth's day much shorter. Exactly how much
depends on what other theory you're trying to support or disprove. However, we
do know the effect is real - the measurement using equipment left behind by the
astronauts shows the distance increasing about 4 cm a year, and the Earth's day
lengthening by 1.5 milliseconds a century.


I never looked it up but would have guessed the rate of slow down would
be larger than that. From that number you can calculate the energy cost
of the tidal forces... Here's a thought, at current rate of energy
consumption growth how long before even this energy source would be
insufficient for our needs? :)

Cheers

I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...
Yes, so...

Cheers


Scout wrote:

If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...

Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:


We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...


Well Peter, I have to disagree there. The gravitational force acts

only


toward the center of mass of the system. This cannot by itself

produce


two bulges. To clarify this, try imagining the forces of gravity in

2D


on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far


side and more water as you move toward the moon... -two bulges would

not


be present.

Cheers

I knew I saw this somewhere
Scout

The following diagram shows how the moon causes tides on Earth:




In this diagram, you can see that the moon's gravitational force pulls on water in the oceans so that there are "bulges" in the ocean on both sides of the planet. The moon pulls water toward it, and this causes the bulge toward the moon. The bulge on the side of the Earth opposite the moon is caused by the moon "pulling the Earth away" from the water on that side.

If you are on the coast and the moon is directly overhead, you should experience a high tide. If the moon is directly overhead on the opposite side of the planet, you should also experience a high tide.

During the day, the Earth rotates 180 degrees in 12 hours. The moon, meanwhile, rotates 6 degrees around the earth in 12 hours. The twin bulges and the moon's rotation mean that any given coastal city experiences a high tide every 12 hours and 25 minutes or so.





"Scout" wrote in message ...
I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...
Yes, so...

Cheers


Scout wrote:

If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...

Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:


We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...


Well Peter, I have to disagree there. The gravitational force acts

only


toward the center of mass of the system. This cannot by itself

produce


two bulges. To clarify this, try imagining the forces of gravity in

2D


on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far


side and more water as you move toward the moon... -two bulges would

not


be present.

Cheers

Yup. That's about it. As I said a while back, the Earth is "falling" towards
the Moon as the two rotate around their common center. The near part of the
Earth falls a bit faster, the far part falls a bit slower. The result is the
two bulges.

Nav has been asking what happens if we prevent the Earth from "falling" but
somehow still had the Moon's gravity. Then we would have higher tide on the
near side, and low (but not as low as normal) tide on the far side.



"Scout" wrote in message
...
I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...
Yes, so...

Cheers


Scout wrote:

If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...

Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:


We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...


Well Peter, I have to disagree there. The gravitational force acts

only


toward the center of mass of the system. This cannot by itself

produce


two bulges. To clarify this, try imagining the forces of gravity in

2D


on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far


side and more water as you move toward the moon... -two bulges would

not


be present.

Cheers

I think if the earth moon-earth system were not rotating there would be
one tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:

I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...

Yes, so...

Cheers


Scout wrote:


If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...


Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:



We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...



Well Peter, I have to disagree there. The gravitational force acts

only



toward the center of mass of the system. This cannot by itself

produce



two bulges. To clarify this, try imagining the forces of gravity in

2D



on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far



side and more water as you move toward the moon... -two bulges would

not



be present.

Cheers

but the only way the system wouldn't rotate is if there was no gravity, so I'm
not sure what your point is.



"Nav" wrote in message
...
I think if the earth moon-earth system were not rotating there would be
one tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:

I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...

Yes, so...

Cheers


Scout wrote:


If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...


Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:



We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...



Well Peter, I have to disagree there. The gravitational force acts

only



toward the center of mass of the system. This cannot by itself

produce



two bulges. To clarify this, try imagining the forces of gravity in

2D



on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far



side and more water as you move toward the moon... -two bulges would

not



be present.

Cheers

"Nav" wrote in message
...

I don't think its fair to do this - you can mathematically eliminate effects
by
shifting the reference frame, but "locking" objects together is changing the
problem at a more fundamental level.

I don't see it that way, the explanation for the two tides based on
differential gravity alone does not care whether the earth is "moon
locked" at (say) an L point -and that why it is not the correct
explanation in my opinion. Of course it all comes down to gravity and
the energy of the system but the simplest close answer should consider
the rotation as well.

The rotation is certainly considered in the differential explanation. The "net
force" which is subtracted out in the differential model is the force which
casues the Earth to rotate around the moon. It is exactly oppostite the
"fictional" centrifugal force.




In this case, how to you "lock" the Earth?
In fact, the crux of this problem is that different parts of the Earth are
actually acting somewhat independently.

However, this brings up an interesting point. At some point in the distant
future the tides will be eliminated.

(Well not really, unless you ignore the Sun). But I think this point
reinforces what I've been trying to get across, without considering the
rotation(s) about the center of mass you don't get a two tide situation.
Any description that does not explicitly consider the relative motion
will not generate two tides -do you agree?

How will this happen? Because the tides
lag the Moon the high tide is not directly under the Moon, but offset. This
creates soon torque that is transferring energy from the Earth to the Moon.
The
result is that the Earth is slowing down, and the Moon's orbit is
increasing.
This will continue (some say) until the Earth's rotation slows down to match
the
Moon, and the bulge stays under the Moon. The Earth and Moon will at that
point
be locked together. Because the Moon is smaller, it has already assumed
this
orientation WRT the Earth.

If we work this backwards we find the in the distant past the Moon's orbit
was
much closer to the Earth, and the Earth's day much shorter. Exactly how
much
depends on what other theory you're trying to support or disprove. However,
we
do know the effect is real - the measurement using equipment left behind by
the
astronauts shows the distance increasing about 4 cm a year, and the Earth's
day
lengthening by 1.5 milliseconds a century.


I never looked it up but would have guessed the rate of slow down would
be larger than that.

I also thought that at first glance. But consider: 1.5 millisecs/century is
1500 seconds/100 mil years, or a bit under a half hour per 100 mil years. So
when was the distance zero? About 5 billion years ago! Of course, the math is
a lot more complex than that, but it shows how long a time a billion years is.


From that number you can calculate the energy cost
of the tidal forces... Here's a thought, at current rate of energy
consumption growth how long before even this energy source would be
insufficient for our needs? :)

Well, there are a few places tapping tidal energy. But I think the big score
would be to tap the thermal energy in the Earth's core. Why should Iceland be
the only place that gets a free ride?

Let me try to be clear. A gravity only argument is usually based on the
equation F=m1m2/r^2. This is a monotonic field function so that water
would only ever flow toward the point closest to the center of mass of
the system (note that the water's potential energy is proportional to
F(r).r). It cannot create two tidal bulges for that requires two
potential energy minima. Now couple that equation with F=mr omega^2 and
you get the local force on a mass of water (mw) as mw.ms/r^2 - mw r
omega^2 where ms is the system mass and r the distance to the center of
mass. Now the potential energy of water has a new local minima away from
the center of mass (plot 1/r + r^2 ; r 0). This gives the tidal bulge
on the far side of the planet. This is only a local minimum, so that if
the earth did not constantly move water to it by it's own rotation that
bulge would gradually disappear. If you like, water gets stuck at the
far side local minumum as the earth rotates.

So, as I see it, without explictly including the rotation(s) you don't
get two tides -OK?

Cheers

Jeff Morris wrote:

but the only way the system wouldn't rotate is if there was no gravity, so I'm
not sure what your point is.



"Nav" wrote in message
...

I think if the earth moon-earth system were not rotating there would be
one tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:


I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...


Yes, so...

Cheers


Scout wrote:



If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...



Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:




We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...




Well Peter, I have to disagree there. The gravitational force acts

only




toward the center of mass of the system. This cannot by itself

produce




two bulges. To clarify this, try imagining the forces of gravity in

2D




on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far




side and more water as you move toward the moon... -two bulges would

not




be present.

Cheers

If I understand correctly, I think you're saying that if the e-m system stop
rotating, the water on the far side of the planet would eventually "fall" or
stabilize at fixed levels, with low tide being farthest from the moon ;
i.e., net forces would eventually equal zero? I guess I can see that.
Does inertia provide the rest of the answer (i.e., the water on the far side
tends to stay at rest (temporarily) while the earth accelerates toward the
moon?)
Scout


"Nav" wrote in message
...
I think if the earth moon-earth system were not rotating there would be one
tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:

I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of
earth, and also pulls the earth away from the water on the far side,
resulting in a low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...

Yes, so...

Cheers


Scout wrote:


If the center of mass was the only factor involved, wouldn't the bulge
be on one side of the earth only?
Scout

"Nav" wrote in message
...


Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:



We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...



Well Peter, I have to disagree there. The gravitational force acts

only



toward the center of mass of the system. This cannot by itself

produce



two bulges. To clarify this, try imagining the forces of gravity in

2D



on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far



side and more water as you move toward the moon... -two bulges would

not



be present.

Cheers

Yes, the extra force not considered in the "gravity only" explanation is
from Newton's second law. This extra term is required to make 2
energy minima where water will want to reside as the earth rotates.

Cheers

Scout wrote:

If I understand correctly, I think you're saying that if the e-m system stop
rotating, the water on the far side of the planet would eventually "fall" or
stabilize at fixed levels, with low tide being farthest from the moon ;
i.e., net forces would eventually equal zero? I guess I can see that.
Does inertia provide the rest of the answer (i.e., the water on the far side
tends to stay at rest (temporarily) while the earth accelerates toward the
moon?)
Scout


"Nav" wrote in message
...

I think if the earth moon-earth system were not rotating there would be one
tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:


I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of
earth, and also pulls the earth away from the water on the far side,
resulting in a low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...


Yes, so...

Cheers


Scout wrote:



If the center of mass was the only factor involved, wouldn't the bulge
be on one side of the earth only?
Scout

"Nav" wrote in message
...



Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:




We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...




Well Peter, I have to disagree there. The gravitational force acts

only




toward the center of mass of the system. This cannot by itself

produce




two bulges. To clarify this, try imagining the forces of gravity in

2D




on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far




side and more water as you move toward the moon... -two bulges would

not




be present.

Cheers

Nav -
The two forces you are talking about are exactly the same. The centrifugal
force IS =m1m2/r^2. They are NOT two different forces. The centrifugal
force is the fictional force that is exactly opposite the real force (as viewed
from a non-accelerating system). Even if you calculate from a centrifugal
force point of view, you end up deriving the angular velocity in terms of
F=m1m2/r^2.

For anyone who wants to follow this through, here's a pretty complete version.
http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual parts of
the Earth, then to add in the centrifugal force and the Earth's pull and the
Earth's daily rotation component. However, the centrifugal force is derived
from the total gravitation pull, so that component is certainly not ignored.
As is often the case, it gets messy before terms start to cancel, but the end
result is that the "differential gravity" is exactly symmetrical on the near and
far side from the Moon. It would not be fair to say the the near side component
is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the Moon.
It is when you subtract out the net pull (or add the centrifugal) that this
becomes symmetrical on both sides.


"Nav" wrote in message
...
Let me try to be clear. A gravity only argument is usually based on the
equation F=m1m2/r^2. This is a monotonic field function so that water
would only ever flow toward the point closest to the center of mass of
the system (note that the water's potential energy is proportional to
F(r).r). It cannot create two tidal bulges for that requires two
potential energy minima. Now couple that equation with F=mr omega^2 and
you get the local force on a mass of water (mw) as mw.ms/r^2 - mw r
omega^2 where ms is the system mass and r the distance to the center of
mass. Now the potential energy of water has a new local minima away from
the center of mass (plot 1/r + r^2 ; r 0). This gives the tidal bulge
on the far side of the planet. This is only a local minimum, so that if
the earth did not constantly move water to it by it's own rotation that
bulge would gradually disappear. If you like, water gets stuck at the
far side local minumum as the earth rotates.

So, as I see it, without explictly including the rotation(s) you don't
get two tides -OK?

Cheers

Jeff Morris wrote:

but the only way the system wouldn't rotate is if there was no gravity, so
I'm
not sure what your point is.



"Nav" wrote in message
...

I think if the earth moon-earth system were not rotating there would be
one tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:


I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in
a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...


Yes, so...

Cheers


Scout wrote:



If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...



Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:




We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...




Well Peter, I have to disagree there. The gravitational force acts

only




toward the center of mass of the system. This cannot by itself

produce




two bulges. To clarify this, try imagining the forces of gravity in

2D




on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far




side and more water as you move toward the moon... -two bulges would

not




be present.

Cheers

I didn't realize until I made an HTML post (my very small graphic of the
moon-earth and tide) that Google would just ignore it and post nothing. I
thought Google would at least post it as text.
Something new everyday I guess.
Scout

"Jeff Morris" wrote in message
...
Yup. That's about it. As I said a while back, the Earth is "falling"
towards
the Moon as the two rotate around their common center. The near part of
the
Earth falls a bit faster, the far part falls a bit slower. The result is
the
two bulges.

Nav has been asking what happens if we prevent the Earth from "falling"
but
somehow still had the Moon's gravity. Then we would have higher tide on
the
near side, and low (but not as low as normal) tide on the far side.



"Scout" wrote in message
...
I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of
earth,
and also pulls the earth away from the water on the far side, resulting
in a
low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...
Yes, so...

Cheers


Scout wrote:

If the center of mass was the only factor involved, wouldn't the bulge
be
on one side of the earth only?
Scout

"Nav" wrote in message
...

Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:


We can certainly look at the gravitational force from the moon and
the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...


Well Peter, I have to disagree there. The gravitational force acts

only


toward the center of mass of the system. This cannot by itself

produce


two bulges. To clarify this, try imagining the forces of gravity in

2D


on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far


side and more water as you move toward the moon... -two bulges would

not


be present.

Cheers

Now I see why you don't follow my point.

The centrifigal force is m1m2/r^2 _only) along the orbital path (which
passes through the center of mass of course). But the diameter of the
earth puts the water well away from the center of mass so that the
gravitational force at the planet surface is _not_ balanced by the
centrifugal force (except at two points on the surface). Near the moon,
the centrifigal force is less than gravity to water gets pulled there.
On the outside of the system, centrifugal force is greater than the
gravitational force so water tries to move outwards.

I hope you can see my point now.

Cheers

Jeff Morris wrote:

Nav -
The two forces you are talking about are exactly the same. The centrifugal
force IS =m1m2/r^2. They are NOT two different forces. The centrifugal
force is the fictional force that is exactly opposite the real force (as viewed
from a non-accelerating system). Even if you calculate from a centrifugal
force point of view, you end up deriving the angular velocity in terms of
F=m1m2/r^2.

For anyone who wants to follow this through, here's a pretty complete version.
http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual parts of
the Earth, then to add in the centrifugal force and the Earth's pull and the
Earth's daily rotation component. However, the centrifugal force is derived
from the total gravitation pull, so that component is certainly not ignored.
As is often the case, it gets messy before terms start to cancel, but the end
result is that the "differential gravity" is exactly symmetrical on the near and
far side from the Moon. It would not be fair to say the the near side component
is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the Moon.
It is when you subtract out the net pull (or add the centrifugal) that this
becomes symmetrical on both sides.


"Nav" wrote in message
...

Let me try to be clear. A gravity only argument is usually based on the
equation F=m1m2/r^2. This is a monotonic field function so that water
would only ever flow toward the point closest to the center of mass of
the system (note that the water's potential energy is proportional to
F(r).r). It cannot create two tidal bulges for that requires two
potential energy minima. Now couple that equation with F=mr omega^2 and
you get the local force on a mass of water (mw) as mw.ms/r^2 - mw r
omega^2 where ms is the system mass and r the distance to the center of
mass. Now the potential energy of water has a new local minima away from
the center of mass (plot 1/r + r^2 ; r 0). This gives the tidal bulge
on the far side of the planet. This is only a local minimum, so that if
the earth did not constantly move water to it by it's own rotation that
bulge would gradually disappear. If you like, water gets stuck at the
far side local minumum as the earth rotates.

So, as I see it, without explictly including the rotation(s) you don't
get two tides -OK?

Cheers

Jeff Morris wrote:


but the only way the system wouldn't rotate is if there was no gravity, so

I'm

not sure what your point is.



"Nav" wrote in message
...


I think if the earth moon-earth system were not rotating there would be
one tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:



I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of earth,
and also pulls the earth away from the water on the far side, resulting in

a

low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...



Yes, so...

Cheers


Scout wrote:




If the center of mass was the only factor involved, wouldn't the bulge be
on one side of the earth only?
Scout

"Nav" wrote in message
...




Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:





We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...





Well Peter, I have to disagree there. The gravitational force acts

only





toward the center of mass of the system. This cannot by itself

produce





two bulges. To clarify this, try imagining the forces of gravity in

2D





on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far





side and more water as you move toward the moon... -two bulges would

not





be present.

Cheers

Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy minima due
to the system rotation. Note the first two terms of the force balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include the
centripetal force term (mr omega^2) across the diameter of the earth is
simply not a correct analysis. Furtherore, if that "explanation" is used
to show two tides it's bogus (it is clear the earth's radius does not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:



For anyone who wants to follow this through, here's a pretty complete version.
http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual parts of
the Earth, then to add in the centrifugal force and the Earth's pull and the
Earth's daily rotation component. However, the centrifugal force is derived
from the total gravitation pull, so that component is certainly not ignored.
As is often the case, it gets messy before terms start to cancel, but the end
result is that the "differential gravity" is exactly symmetrical on the near and
far side from the Moon. It would not be fair to say the the near side component
is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the Moon.
It is when you subtract out the net pull (or add the centrifugal) that this
becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered except
that for simplicity I did not consider the earth's rotation. It also
supports my quick and dirty proof or there being two energy minima due
to the system rotation.

I see your point, but I keep looking at the final answer. When all the terms
are balanced, and the minor effects ignored, what is the left is 2GmMr
cos^2/R^3, which comes from the radial component of the moon's gravity on a
piece if the Earth m. All of the other forces, including all of the
centrifugal forces have been balanced out. The cos^2 term is the only thing
left that varies with latitude, which means that explains why the bulges are at
the equator, and the pole's tides are depressed.


"Nav" wrote in message
...
Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy minima due
to the system rotation. Note the first two terms of the force balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include the
centripetal force term (mr omega^2) across the diameter of the earth is
simply not a correct analysis. Furtherore, if that "explanation" is used
to show two tides it's bogus (it is clear the earth's radius does not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:



For anyone who wants to follow this through, here's a pretty complete
version.
http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual parts
of
the Earth, then to add in the centrifugal force and the Earth's pull and the
Earth's daily rotation component. However, the centrifugal force is derived
from the total gravitation pull, so that component is certainly not ignored.
As is often the case, it gets messy before terms start to cancel, but the
end
result is that the "differential gravity" is exactly symmetrical on the near
and
far side from the Moon. It would not be fair to say the the near side
component
is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the
Moon.
It is when you subtract out the net pull (or add the centrifugal) that this
becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered except
that for simplicity I did not consider the earth's rotation. It also
supports my quick and dirty proof or there being two energy minima due
to the system rotation.

Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of gravity
and the centripetal term. You seem to have missed the importance (and
cause) of the term that raises the radius of the earth (r) to the fourth
power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:

I see your point, but I keep looking at the final answer. When all the terms
are balanced, and the minor effects ignored, what is the left is 2GmMr
cos^2/R^3, which comes from the radial component of the moon's gravity on a
piece if the Earth m. All of the other forces, including all of the
centrifugal forces have been balanced out. The cos^2 term is the only thing
left that varies with latitude, which means that explains why the bulges are at
the equator, and the pole's tides are depressed.


"Nav" wrote in message
...

Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy minima due
to the system rotation. Note the first two terms of the force balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include the
centripetal force term (mr omega^2) across the diameter of the earth is
simply not a correct analysis. Furtherore, if that "explanation" is used
to show two tides it's bogus (it is clear the earth's radius does not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:



For anyone who wants to follow this through, here's a pretty complete

version.

http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual parts

of

the Earth, then to add in the centrifugal force and the Earth's pull and the
Earth's daily rotation component. However, the centrifugal force is derived
from the total gravitation pull, so that component is certainly not ignored.
As is often the case, it gets messy before terms start to cancel, but the

end

result is that the "differential gravity" is exactly symmetrical on the near

and

far side from the Moon. It would not be fair to say the the near side

component

is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the

Moon.

It is when you subtract out the net pull (or add the centrifugal) that this
becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered except
that for simplicity I did not consider the earth's rotation. It also
supports my quick and dirty proof or there being two energy minima due
to the system rotation.

"Scout" wrote in message ...
If I understand correctly, I think you're saying that if the e-m system stop
rotating, the water on the far side of the planet would eventually "fall" or
stabilize at fixed levels, with low tide being farthest from the moon ;
i.e., net forces would eventually equal zero? I guess I can see that.
Does inertia provide the rest of the answer (i.e., the water on the far side
tends to stay at rest (temporarily) while the earth accelerates toward the
moon?)
Scout


"Nav" wrote in message
...
I think if the earth moon-earth system were not rotating there would be one
tide. Gavity pulls all objects toward the center of mass. So without
rotation the water would be deepest near the moon.

Cheers

Scout wrote:

I was hoping you could solve this riddle.
But I'll toss in my oversimplified guess: the moon's gravity attracts the
water closest to it resulting in high high tide on the moon side of
earth, and also pulls the earth away from the water on the far side,
resulting in a low high tide on the side farthest from the moon.
Scout


"Nav" wrote in message
...

Yes, so...

Cheers


Scout wrote:


If the center of mass was the only factor involved, wouldn't the bulge
be on one side of the earth only?
Scout

"Nav" wrote in message
...


Yes, you can. Where is the center of mass of the earth moon system?

Cheers

Peter S/Y Anicula wrote:



We can certainly look at the gravitational force from the moon and the
gravitational force of the earth seperatly, and then ad the two, to
have a look at the combined forces.

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...



Well Peter, I have to disagree there. The gravitational force acts

only



toward the center of mass of the system. This cannot by itself

produce



two bulges. To clarify this, try imagining the forces of gravity in

2D



on a piece of paper. In all cases, water would be pulled toward the
center of the Earth-Moon pair. This would lead to less water on the

far



side and more water as you move toward the moon... -two bulges would

not



be present.

Cheers









This is fascinating. The more I think about this, the more I see the
planet as a huge impellor in a centrifugal pump, piling water up on
the far side of the planet, while the planet itself acts as a dam of
sorts, preventing the water from spilling back to the low tide areas.
Scout

If you're not prepared to derive the tides from first principles, you're not a
real sailor!

In the old days, this is what we had to pack in our ditch bag:
http://ftp.arl.army.mil/ftp/historic...gif/eniac1.gif

but it was a step up from this puppy:
http://co-ops.nos.noaa.gov/predmach.html

OzOne wrote in message ...

Jeez, I'm glad I can just look at a tide chart :-)
Or use J tides
http://vps.arachnoid.com/JTides/index.html
unless of course I lived in the UK where GovCo has decided that they
OWN tide data ....


On Tue, 05 Oct 2004 18:09:58 +1300, Nav
scribbled thusly:

Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of gravity
and the centripetal term. You seem to have missed the importance (and
cause) of the term that raises the radius of the earth (r) to the fourth
power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:

I see your point, but I keep looking at the final answer. When all the
terms
are balanced, and the minor effects ignored, what is the left is 2GmMr
cos^2/R^3, which comes from the radial component of the moon's gravity on a
piece if the Earth m. All of the other forces, including all of the
centrifugal forces have been balanced out. The cos^2 term is the only
thing
left that varies with latitude, which means that explains why the bulges
are at
the equator, and the pole's tides are depressed.


"Nav" wrote in message
...

Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy minima due
to the system rotation. Note the first two terms of the force balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include the
centripetal force term (mr omega^2) across the diameter of the earth is
simply not a correct analysis. Furtherore, if that "explanation" is used
to show two tides it's bogus (it is clear the earth's radius does not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:



For anyone who wants to follow this through, here's a pretty complete

version.

http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual
parts

of

the Earth, then to add in the centrifugal force and the Earth's pull and
the
Earth's daily rotation component. However, the centrifugal force is
derived
from the total gravitation pull, so that component is certainly not
ignored.
As is often the case, it gets messy before terms start to cancel, but the

end

result is that the "differential gravity" is exactly symmetrical on the
near

and

far side from the Moon. It would not be fair to say the the near side

component

is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the

Moon.

It is when you subtract out the net pull (or add the centrifugal) that
this
becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered except
that for simplicity I did not consider the earth's rotation. It also
supports my quick and dirty proof or there being two energy minima due
to the system rotation.







Oz1...of the 3 twins.

I welcome you to crackerbox palace,We've been expecting you.

You mean you don't have one of those on your boat? Shame on you, mind
you rough water plays havoc when the weights start swinging.

Cheers

Jeff Morris wrote:
If you're not prepared to derive the tides from first principles, you're not a
real sailor!

In the old days, this is what we had to pack in our ditch bag:
http://ftp.arl.army.mil/ftp/historic...gif/eniac1.gif

but it was a step up from this puppy:
http://co-ops.nos.noaa.gov/predmach.html

OzOne wrote in message ...

Jeez, I'm glad I can just look at a tide chart :-)
Or use J tides
http://vps.arachnoid.com/JTides/index.html
unless of course I lived in the UK where GovCo has decided that they
OWN tide data ....


On Tue, 05 Oct 2004 18:09:58 +1300, Nav
scribbled thusly:


Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of gravity
and the centripetal term. You seem to have missed the importance (and
cause) of the term that raises the radius of the earth (r) to the fourth
power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:


I see your point, but I keep looking at the final answer. When all the

terms

are balanced, and the minor effects ignored, what is the left is 2GmMr
cos^2/R^3, which comes from the radial component of the moon's gravity on a
piece if the Earth m. All of the other forces, including all of the
centrifugal forces have been balanced out. The cos^2 term is the only

thing

left that varies with latitude, which means that explains why the bulges

are at

the equator, and the pole's tides are depressed.


"Nav" wrote in message
...


Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy minima due
to the system rotation. Note the first two terms of the force balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include the
centripetal force term (mr omega^2) across the diameter of the earth is
simply not a correct analysis. Furtherore, if that "explanation" is used
to show two tides it's bogus (it is clear the earth's radius does not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:




For anyone who wants to follow this through, here's a pretty complete

version.


http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on individual

parts

of


the Earth, then to add in the centrifugal force and the Earth's pull and

the

Earth's daily rotation component. However, the centrifugal force is

derived

from the total gravitation pull, so that component is certainly not

ignored.

As is often the case, it gets messy before terms start to cancel, but the

end


result is that the "differential gravity" is exactly symmetrical on the

near

and


far side from the Moon. It would not be fair to say the the near side

component


is caused by one force, and the far side by another. In fact, all of the
"latitude dependent" forces are caused by the differential pull from the

Moon.


It is when you subtract out the net pull (or add the centrifugal) that

this

becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered except
that for simplicity I did not consider the earth's rotation. It also
supports my quick and dirty proof or there being two energy minima due
to the system rotation.






Oz1...of the 3 twins.

I welcome you to crackerbox palace,We've been expecting you.

But now you really know why there are usually two tides, which put you
in a more knowledgeable position that 99.9% of the population and most
sailors. Doesn't that make you feel ... empowered?

Cheers M ;-)

OzOne wrote:

Jeez, I'm glad I can just look at a tide chart :-)
Or use J tides
http://vps.arachnoid.com/JTides/index.html
unless of course I lived in the UK where GovCo has decided that they
OWN tide data ....

Ok, are you now ready to admit that you were mistaken, and that the
"differential gravitational explanation" does in fact explain that
there is a bulge on the side of the earth that turns away from the
moon ?

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...
Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of
gravity
and the centripetal term. You seem to have missed the importance
(and
cause) of the term that raises the radius of the earth (r) to the
fourth
power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:

I see your point, but I keep looking at the final answer. When
all the terms
are balanced, and the minor effects ignored, what is the left is
2GmMr
cos^2/R^3, which comes from the radial component of the moon's
gravity on a
piece if the Earth m. All of the other forces, including all of
the
centrifugal forces have been balanced out. The cos^2 term is the
only thing
left that varies with latitude, which means that explains why the
bulges are at
the equator, and the pole's tides are depressed.


"Nav" wrote in message
...

Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not
interested
in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy
minima due
to the system rotation. Note the first two terms of the force
balance
equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include
the
centripetal force term (mr omega^2) across the diameter of the
earth is
simply not a correct analysis. Furtherore, if that "explanation"
is used
to show two tides it's bogus (it is clear the earth's radius does
not
cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:



For anyone who wants to follow this through, here's a pretty
complete

version.

http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on
individual parts

of

the Earth, then to add in the centrifugal force and the Earth's
pull and the
Earth's daily rotation component. However, the centrifugal force
is derived
from the total gravitation pull, so that component is certainly
not ignored.
As is often the case, it gets messy before terms start to cancel,
but the

end

result is that the "differential gravity" is exactly symmetrical
on the near

and

far side from the Moon. It would not be fair to say the the near
side

component

is caused by one force, and the far side by another. In fact,
all of the
"latitude dependent" forces are caused by the differential pull
from the

Moon.

It is when you subtract out the net pull (or add the centrifugal)
that this
becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered
except
that for simplicity I did not consider the earth's rotation. It
also
supports my quick and dirty proof or there being two energy minima
due
to the system rotation.

I'm going to give you the benefit of that doubt and hope you are not
just trolling. I'm sorry if you can't understand the maths. It is not
"differential gravity" -the maths are clear and unambiguous on this
point: Differentiate the gravity field equation and you just get a
monotonic function of distance from the center of system mass so that
water would only ever move in one direction, namely toward the center of
the system. It is the centripetal term that introduces the extra force
required to make a second tidal bulge. So, you need to include rotation
about the center of mass in any explanation of two tides.

If you still don't follow my argument (and accept the veracity of the
maths) then I can't help you.

Cheers

Peter S/Y Anicula wrote:
Ok, are you now ready to admit that you were mistaken, and that the
"differential gravitational explanation" does in fact explain that
there is a bulge on the side of the earth that turns away from the
moon ?

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...

Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of

gravity

and the centripetal term. You seem to have missed the importance

(and

cause) of the term that raises the radius of the earth (r) to the

fourth

power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:


I see your point, but I keep looking at the final answer. When

all the terms

are balanced, and the minor effects ignored, what is the left is

2GmMr

cos^2/R^3, which comes from the radial component of the moon's

gravity on a

piece if the Earth m. All of the other forces, including all of

the

centrifugal forces have been balanced out. The cos^2 term is the

only thing

left that varies with latitude, which means that explains why the

bulges are at

the equator, and the pole's tides are depressed.


"Nav" wrote in message
...


Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not

interested

in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy

minima due

to the system rotation. Note the first two terms of the force

balance

equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include

the

centripetal force term (mr omega^2) across the diameter of the

earth is

simply not a correct analysis. Furtherore, if that "explanation"

is used

to show two tides it's bogus (it is clear the earth's radius does

not

cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:




For anyone who wants to follow this through, here's a pretty

complete

version.


http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on

individual parts

of


the Earth, then to add in the centrifugal force and the Earth's

pull and the

Earth's daily rotation component. However, the centrifugal force

is derived

from the total gravitation pull, so that component is certainly

not ignored.

As is often the case, it gets messy before terms start to cancel,

but the

end


result is that the "differential gravity" is exactly symmetrical

on the near

and


far side from the Moon. It would not be fair to say the the near

side

component


is caused by one force, and the far side by another. In fact,

all of the

"latitude dependent" forces are caused by the differential pull

from the

Moon.


It is when you subtract out the net pull (or add the centrifugal)

that this

becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered

except

that for simplicity I did not consider the earth's rotation. It

also

supports my quick and dirty proof or there being two energy minima

due

to the system rotation.

I just want to make sure I'm understanding this correctly. Sorry if I'm
repeating myself here, but when you say "centripetal" aren't you implying
that the planet acts as a huge impellor in a centrifugal pump, "throwing" or
piling water up on the far side of the planet, while the planet itself acts
as a dam of sorts, preventing the water from spilling back to the low tide
areas.
Scout


"Nav" wrote in message
...
I'm going to give you the benefit of that doubt and hope you are not just
trolling. I'm sorry if you can't understand the maths. It is not
"differential gravity" -the maths are clear and unambiguous on this
point: Differentiate the gravity field equation and you just get a
monotonic function of distance from the center of system mass so that
water would only ever move in one direction, namely toward the center of
the system. It is the centripetal term that introduces the extra force
required to make a second tidal bulge. So, you need to include rotation
about the center of mass in any explanation of two tides.

If you still don't follow my argument (and accept the veracity of the
maths) then I can't help you.

Cheers

Peter S/Y Anicula wrote:
Ok, are you now ready to admit that you were mistaken, and that the
"differential gravitational explanation" does in fact explain that
there is a bulge on the side of the earth that turns away from the
moon ?

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...

Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of

gravity

and the centripetal term. You seem to have missed the importance

(and

cause) of the term that raises the radius of the earth (r) to the

fourth

power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:


I see your point, but I keep looking at the final answer. When

all the terms

are balanced, and the minor effects ignored, what is the left is

2GmMr

cos^2/R^3, which comes from the radial component of the moon's

gravity on a

piece if the Earth m. All of the other forces, including all of

the

centrifugal forces have been balanced out. The cos^2 term is the

only thing

left that varies with latitude, which means that explains why the

bulges are at

the equator, and the pole's tides are depressed.


"Nav" wrote in message
...


Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not

interested

in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy

minima due

to the system rotation. Note the first two terms of the force

balance

equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include

the

centripetal force term (mr omega^2) across the diameter of the

earth is

simply not a correct analysis. Furtherore, if that "explanation"

is used

to show two tides it's bogus (it is clear the earth's radius does

not

cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:




For anyone who wants to follow this through, here's a pretty

complete

version.


http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on

individual parts

of


the Earth, then to add in the centrifugal force and the Earth's

pull and the

Earth's daily rotation component. However, the centrifugal force

is derived

from the total gravitation pull, so that component is certainly

not ignored.

As is often the case, it gets messy before terms start to cancel,

but the

end


result is that the "differential gravity" is exactly symmetrical

on the near

and


far side from the Moon. It would not be fair to say the the near

side

component


is caused by one force, and the far side by another. In fact,

all of the

"latitude dependent" forces are caused by the differential pull

from the

Moon.


It is when you subtract out the net pull (or add the centrifugal)

that this

becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered

except

that for simplicity I did not consider the earth's rotation. It

also

supports my quick and dirty proof or there being two energy minima

due

to the system rotation.

Sorry, nav, I still can't really buy your argument. It is obvious that you
can't ignore the centrifugal force component if that's the mathematical approach
you're taking. Clearly, if you ignore a major component, the final answer will
be wrong.

However, when all the forces are summed up and canceled out, there is one major
force "left standing," and that's the "differential gravity" from the moon (and
also sun, of course). "Differential Gravity" is often defined as the pull from
the moon, after the centrifugal force has been removed.

As I pointed out before, the final answer is

2GmMr cos^2/R^3

which comes from the radial component of the moon's gravity on a piece if the
Earth m. This is the force that is used to define the "equipotential shape" of
the Earth that has the two bulges. If you trace through the math, this
component is not part of the centrifugal force, so it isn't fair to say that
centrifugal force is the cause of the two bulges.

You're quite correct that we can't ignore centrifugal force, although we could
of course use a more difficult approach that doesn't use it. It is, after all,
a fictional force. But it makes sense to use it because we are seeking an
Earth-centric answer. And the centrifugal force is a dominant force, so it
certainly can't be simply left out. However, the centrifugal force from the
Earth-Sun system is much greater than that of the Earth-Moon, yet the resulting
tidal forces are smaller. Why? because the differential forces are smaller
from the distant Sun.





"Nav" wrote in message
...
I'm going to give you the benefit of that doubt and hope you are not
just trolling. I'm sorry if you can't understand the maths. It is not
"differential gravity" -the maths are clear and unambiguous on this
point: Differentiate the gravity field equation and you just get a
monotonic function of distance from the center of system mass so that
water would only ever move in one direction, namely toward the center of
the system. It is the centripetal term that introduces the extra force
required to make a second tidal bulge. So, you need to include rotation
about the center of mass in any explanation of two tides.

If you still don't follow my argument (and accept the veracity of the
maths) then I can't help you.

Cheers

Peter S/Y Anicula wrote:
Ok, are you now ready to admit that you were mistaken, and that the
"differential gravitational explanation" does in fact explain that
there is a bulge on the side of the earth that turns away from the
moon ?

Peter S/Y Anicula


"Nav" skrev i en meddelelse
...

Ok, then we are almost there :)! The final answer on that good site
shows that it is the _imbalance_ between the radial component of

gravity

and the centripetal term. You seem to have missed the importance

(and

cause) of the term that raises the radius of the earth (r) to the

fourth

power.

h~Mr^4 cos^2 theta/ER^3.

Cheers



Jeff Morris wrote:


I see your point, but I keep looking at the final answer. When

all the terms

are balanced, and the minor effects ignored, what is the left is

2GmMr

cos^2/R^3, which comes from the radial component of the moon's

gravity on a

piece if the Earth m. All of the other forces, including all of

the

centrifugal forces have been balanced out. The cos^2 term is the

only thing

left that varies with latitude, which means that explains why the

bulges are at

the equator, and the pole's tides are depressed.


"Nav" wrote in message
...


Jeff,

That approach is the same as the one I offered except that for
simplicity I did not consider the earth's rotation (I was not

interested

in the height of the tide, just the number of energy minima). It
supports my "quick and dirty" proof of there being two energy

minima due

to the system rotation. Note the first two terms of the force

balance

equation are :

-Gm1m2/r^2 + mr omega^2 -exactly as I said.

The conclusion for the question remains as I said -it's the system
rotation _plus_ gravity that makes two tides. You can't say the
centripetal term is equal to the gravity term because that is not
generally so.

Thus, any gravity based tide "explanation" that does not include

the

centripetal force term (mr omega^2) across the diameter of the

earth is

simply not a correct analysis. Furtherore, if that "explanation"

is used

to show two tides it's bogus (it is clear the earth's radius does

not

cancel out).

I hope we can agree now?

Cheers

Jeff Morris wrote:




For anyone who wants to follow this through, here's a pretty

complete

version.


http://www.clupeid.demon.co.uk/tides/maths.html

The approach is to take the force the Moon's pull exerts on

individual parts

of


the Earth, then to add in the centrifugal force and the Earth's

pull and the

Earth's daily rotation component. However, the centrifugal force

is derived

from the total gravitation pull, so that component is certainly

not ignored.

As is often the case, it gets messy before terms start to cancel,

but the

end


result is that the "differential gravity" is exactly symmetrical

on the near

and


far side from the Moon. It would not be fair to say the the near

side

component


is caused by one force, and the far side by another. In fact,

all of the

"latitude dependent" forces are caused by the differential pull

from the

Moon.


It is when you subtract out the net pull (or add the centrifugal)

that this

becomes symmetrical on both sides.



Jeff, That apparoach is exactly the same as the one I offered

except

that for simplicity I did not consider the earth's rotation. It

also

supports my quick and dirty proof or there being two energy minima

due

to the system rotation.