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.