To claim that an explanation is wrong you should either demonstrate
that the result is wrong or that the logic is wrong. You have done
neither.

You mix to different models and claim that because they are different
the "other" model is wrong, or rather that it doesn't explain
the reality (the far bulge) - which it clearly does (or at least
claims to do) within its own framework.

I think you have become hypnotised (or is it paralysed?) by the
centrifugal force, that is part of the model you prefer.

So which part of the explanation do you disagree with?

Peter S/Y Anicula

P.S.
The problem is not that I don't understand the math or "your" model, I
think I do, and in some contexts I even prefer it, but the problem
aparrently is that you don't understand the differential gravity
explanation.




"Nav" skrev i en meddelelse
...
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

In a way that's correct but a bit colorful. It is the excess of
centripetal force over gravity that causes the high tide on the far side

Cheers

Scout wrote:
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.

Jeff Morris wrote:

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.


YES!

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.

This is where I take issue with such sloppy thinking. Gravity acts ONLY
toward the center of the system. It cannot be relabelled in any way to
make an "outward" force. As a physicist, do you think subtracting an
inertial force from gravity and call it "new gravity" is an acceptable
idea? I say the two are quite different forces and need to be identified
and kept separate. If the force due to the radial acceleration played no
part in the final equation why would it be needed in the derivation at
all? What happens if omega=0 and the body is in free fall? Are there
still two tides?

Anyway, there's no point to my constantly repeating myself so we'll just
have to agree to disagree. In case anyone else is interested, here's a
quote from a teaching site that explains it all quite simply.

http://ijolite.geology.uiuc.edu/04Sp...tes/Lect29.pdf

" TIDES
Periodic rise and fall of water level along coastlines related to the
phases of the Moon.
One cause: The moon’s gravity.
BUT...If this were the only cause, we would have only one high tide
per day (i.e., we would have
high tide when we are position closest to the moon). However, there
are often two high tides and
two low tides per day- why?
The other issue: Centrifugal/centripetal forces.
1. The moon orbits around the earth, and this makes the earth
“wobble”. More precisely, the earth
and moon both orbit around their center of mass.
Analogy: Imagine a large adult and small child holding hands and
spinning around. The
child spins in a wide circle, but the adult also moves in a circle,
albeit a smaller one
because the adult is so much heavier.
2. Because the earth is orbiting (moving a circle slightly) there is a
“centrifugal force”. Or, more
accurately, there is a centripetal force required to move the earth in
that orbit.
Cause for Tides: imbalance of two forces:
1. G = Gravitational attraction between Earth and the Moon (and Sun).
2. C = Centripetal force = Force needed to make the Earth revolve
around the center of mass
(center of gravity) of the Earth-Moon system.
Here’s the imbalance part:
1. Gravity from Moon stronger on side facing moon, weaker on the side
away from it
2. Centripetal force is the same everywhere on earth
3. G and C are exactly balanced at center of the earth
4. They are not balanced in other places that are closer to moon or
farther away from it
• Example: On the side away from the moon, the centripetal force
needed to keep the
ocean water in its orbit is greater than the gravitational force from
the moon
• Net result is like a force away from the moon (a centrifugal “force”)
5. So wherever C and G do not balance, there is force that is large
enough to make water
flow
6. Water flows and forms tidal bulges "


Cheers

Sigh. Look you can say what you like but it is the imbalance of force
between the centrifigal and gravitational forces that causes the two
tides. Now you can a call that "differential gravity" but why? Doesn't
that lead to the confusion that is so apparent? For the last time,
gravity, as a perectly defined force, only acts toward the center of the
system, it cannot _by itself_ produce two tides. If gravity alone could
do it why is the rotation of the system important (what happens in free
fall?). Have a look at the teaching site I posted for a clear simple
explanation.

That's all I'm going to say on this subject.

Cheers


Peter S/Y Anicula wrote:


P.S.
The problem is not that I don't understand the math or "your" model, I
think I do, and in some contexts I even prefer it, but the problem
aparrently is that you don't understand the differential gravity
explanation.

I'm a colorful kinda guy ~ but in a manly sorta way of course!
Scout

"Nav" wrote in message
...
In a way that's correct but a bit colorful. It is the excess of
centripetal force over gravity that causes the high tide on the far side

Cheers

Scout wrote:
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.

"Nav" wrote:

What happens if omega=0 and the body is in free fall? Are there
still two tides?


If the rotation of the moon was stopped, and the moon and the earth
was "falling" towards each other, there would still be two bulges, as
far as I can figure out.

(You wouldn't have any centrifugal powers, but you would have an
acceleration.)

Peter S/Y Anicula


"Nav" wrote:

What happens if omega=0 and the body is in free fall? Are there
still two tides?

Anyway, there's no point to my constantly repeating myself so we'll
just
have to agree to disagree. In case anyone else is interested, here's
a
quote from a teaching site that explains it all quite simply.

http://ijolite.geology.uiuc.edu/04Sp...tes/Lect29.pdf

" TIDES
Periodic rise and fall of water level along coastlines related to
the
phases of the Moon.
One cause: The moon's gravity.
BUT...If this were the only cause, we would have only one high
tide
per day (i.e., we would have
high tide when we are position closest to the moon). However,
there
are often two high tides and
two low tides per day- why?
The other issue: Centrifugal/centripetal forces.
1. The moon orbits around the earth, and this makes the earth
"wobble". More precisely, the earth
and moon both orbit around their center of mass.
Analogy: Imagine a large adult and small child holding hands and
spinning around. The
child spins in a wide circle, but the adult also moves in a
circle,
albeit a smaller one
because the adult is so much heavier.
2. Because the earth is orbiting (moving a circle slightly) there
is a
"centrifugal force". Or, more
accurately, there is a centripetal force required to move the
earth in
that orbit.
Cause for Tides: imbalance of two forces:
1. G = Gravitational attraction between Earth and the Moon (and
Sun).
2. C = Centripetal force = Force needed to make the Earth revolve
around the center of mass
(center of gravity) of the Earth-Moon system.
Here's the imbalance part:
1. Gravity from Moon stronger on side facing moon, weaker on the
side
away from it
2. Centripetal force is the same everywhere on earth
3. G and C are exactly balanced at center of the earth
4. They are not balanced in other places that are closer to moon
or
farther away from it
. Example: On the side away from the moon, the centripetal force
needed to keep the
ocean water in its orbit is greater than the gravitational force
from
the moon
. Net result is like a force away from the moon (a centrifugal
"force")
5. So wherever C and G do not balance, there is force that is
large
enough to make water
flow
6. Water flows and forms tidal bulges "


Cheers

One standard way to approach this problem is to consider that the Earth is in
freefall towards the Moon (or Sun) and then to compute the gravitational forces
caused by the pull on different points on the Earth, from both distance and
angle differences. This yields exactly the same results as other approaches.
While obviously all of the force is towards the Moon, when using the "freefall
method" you subtract the net gravitational attraction, and what's left over is
the differential gravity.

This is the model I mentioned early on, where the "near" high tide is falling
faster towards the Moon, and the "far" high tide is falling slower.




"Peter S/Y Anicula" wrote in message
...
"Nav" wrote:

What happens if omega=0 and the body is in free fall? Are there
still two tides?


If the rotation of the moon was stopped, and the moon and the earth
was "falling" towards each other, there would still be two bulges, as
far as I can figure out.

(You wouldn't have any centrifugal powers, but you would have an
acceleration.)

Peter S/Y Anicula


"Nav" wrote:

What happens if omega=0 and the body is in free fall? Are there
still two tides?

Anyway, there's no point to my constantly repeating myself so we'll
just
have to agree to disagree. In case anyone else is interested, here's
a
quote from a teaching site that explains it all quite simply.

http://ijolite.geology.uiuc.edu/04Sp...tes/Lect29.pdf

" TIDES
Periodic rise and fall of water level along coastlines related to
the
phases of the Moon.
One cause: The moon's gravity.
BUT...If this were the only cause, we would have only one high
tide
per day (i.e., we would have
high tide when we are position closest to the moon). However,
there
are often two high tides and
two low tides per day- why?
The other issue: Centrifugal/centripetal forces.
1. The moon orbits around the earth, and this makes the earth
"wobble". More precisely, the earth
and moon both orbit around their center of mass.
Analogy: Imagine a large adult and small child holding hands and
spinning around. The
child spins in a wide circle, but the adult also moves in a
circle,
albeit a smaller one
because the adult is so much heavier.
2. Because the earth is orbiting (moving a circle slightly) there
is a
"centrifugal force". Or, more
accurately, there is a centripetal force required to move the
earth in
that orbit.
Cause for Tides: imbalance of two forces:
1. G = Gravitational attraction between Earth and the Moon (and
Sun).
2. C = Centripetal force = Force needed to make the Earth revolve
around the center of mass
(center of gravity) of the Earth-Moon system.
Here's the imbalance part:
1. Gravity from Moon stronger on side facing moon, weaker on the
side
away from it
2. Centripetal force is the same everywhere on earth
3. G and C are exactly balanced at center of the earth
4. They are not balanced in other places that are closer to moon
or
farther away from it
. Example: On the side away from the moon, the centripetal force
needed to keep the
ocean water in its orbit is greater than the gravitational force
from
the moon
. Net result is like a force away from the moon (a centrifugal
"force")
5. So wherever C and G do not balance, there is force that is
large
enough to make water
flow
6. Water flows and forms tidal bulges "


Cheers

"Nav" wrote

Sigh. Look you can say what you like but it is the imbalance of
force between the centrifigal and gravitational forces that causes
the two tides.

That's the explanation that you advocate. That is one out of several
explanation models.

Now you can a call that "differential gravity" but why? Doesn't
that lead to the confusion that is so apparent?

You are certainly confused. You don't seem to understand the
"Differential model". You claim that it doesn't explain the two bulges
when it clearly does. I like the model.

For the last time, gravity, as a perectly defined force, only acts
toward the center of the system, it cannot _by itself_ produce two
tides. If gravity alone could do it why is the rotation of the
system important (what happens in free fall?). Have a look at the
teaching site I posted for a clear simple explanation.

If the moon stopped its rotation around the earth and was "falling"
toward the earth, there would still be two bulges.
Anyone who claims otherwise clearly haven't understood what is
happening.

Peter S/Y Anicula

"Nav" wrote

Sigh. Look you can say what you like but it is the imbalance of
force between the centrifigal and gravitational forces that causes
the two tides.

That's the explanation that you advocate. That is one out of several
explanation models.

Now you can a call that "differential gravity" but why? Doesn't
that lead to the confusion that is so apparent?

You are certainly confused. You don't seem to understand the
"Differential model". You claim that it doesn't explain the two bulges
when it clearly does. I like the model.

For the last time, gravity, as a perectly defined force, only acts
toward the center of the system, it cannot _by itself_ produce two
tides. If gravity alone could do it why is the rotation of the
system important (what happens in free fall?). Have a look at the
teaching site I posted for a clear simple explanation.

If the moon stopped its rotation around the earth and the earth and
the moon was "falling" toward each other, there would still be two
bulges.
Anyone who claims otherwise clearly haven't understood what is
happening.

Peter S/Y Anicula

Peter S/Y Anicula wrote:

"Nav" wrote


Sigh. Look you can say what you like but it is the imbalance of
force between the centrifigal and gravitational forces that causes
the two tides.


That's the explanation that you advocate. That is one out of several
explanation models.


Now you can a call that "differential gravity" but why? Doesn't
that lead to the confusion that is so apparent?


You are certainly confused. You don't seem to understand the
"Differential model". You claim that it doesn't explain the two bulges
when it clearly does. I like the model.


For the last time, gravity, as a perectly defined force, only acts
toward the center of the system, it cannot _by itself_ produce two
tides. If gravity alone could do it why is the rotation of the
system important (what happens in free fall?). Have a look at the
teaching site I posted for a clear simple explanation.


If the moon stopped its rotation around the earth and the earth and
the moon was "falling" toward each other, there would still be two
bulges.

YES but how big would they be (Hint: Smaller than the tides?)

Cheers