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

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.

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.

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