Donal wrote:

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

The moon are a lot closer than the sun. Therefore the gravitational
force of the moon varies more over the earth's surface. It is the
variation in the gravitational force and not the force in itself that
creates the tides.

The moons pull on a water-molecule directly under the moon is larger
than
on a molecule on the far side of the earth, actually it is larger than
"the average pull on the whole earth", and here the moon pulls away
from the earth.
On the far side of the earth (seen from the moon) the gravitation from
the moon is less than average and at this point the moon pulls toward
the earth.


On the far side the tide is "high", ... just the same as at the near side.
If the moon's gravity was pulling the water, then you would expect LW to be
opposite the moon.


Quite so. The tidal gravity force is in the direction of the moon. This
is the potential energy in the system. So, there must be another force
present. The moon has kinetic energy in it's orbital velocity. From
Newton's first law:

F=m r omega^2

It is the difference in the two forces (and the resulting energy minima)
that causes two tides. Simple no? Why invoke something new like
"differential gravity"? Could it be to avoid saying inertial force?

Cheers

"Nav" wrote in message
...


Donal wrote:

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

The moon are a lot closer than the sun. Therefore the gravitational
force of the moon varies more over the earth's surface. It is the
variation in the gravitational force and not the force in itself that
creates the tides.

The moons pull on a water-molecule directly under the moon is larger
than
on a molecule on the far side of the earth, actually it is larger than
"the average pull on the whole earth", and here the moon pulls away
from the earth.
On the far side of the earth (seen from the moon) the gravitation from
the moon is less than average and at this point the moon pulls toward
the earth.


On the far side the tide is "high", ... just the same as at the near side.
If the moon's gravity was pulling the water, then you would expect LW to be
opposite the moon.


Quite so. The tidal gravity force is in the direction of the moon. This
is the potential energy in the system. So, there must be another force
present. The moon has kinetic energy in it's orbital velocity. From
Newton's first law:

F=m r omega^2

It is the difference in the two forces (and the resulting energy minima)
that causes two tides. Simple no? Why invoke something new like
"differential gravity"? Could it be to avoid saying inertial force?

Before I thought you were just arguing philosophically how much we should credit
centrifugal force, but now it appears you haven't really looked at the math at
all. The reason why "differential gravity" is invoked is because it represents
the differing pull of the Moon on differing parts of the Earth. Although this
force is all obviously towards the Moon, when you subtract off the centrifugal
force this is what is left. It is this differing pull that causes the two
tides.

This force can be viewed as having two components, one is on the Earth-Moon (or
Earth-Sun) axis, and the other is off axis. The on axis component varies
inversely with R cubed, outwards from the center of the Earth, and is usually
cited as cause of the two bulges. The other component is pulling towards the
Moon, but since its off-axis, from the an observer on Earth sees this as
somewhat downward. When the total gravitation force is subtracted from this,
the result is a downward pull. This is greatest at "low tide points," half way
around from the bulges, but also in a ring around Earth that includes the poles.
The net resulting idealized surface is a "prolate ellipsoid," squeezed around
the middle and out at the ends.

As a simple "reductio ad absurdum" consider that the tides at the poles are
always "Low" because the Moon is always pulling a bit downward on the poles.
If you only consider the centrifugal force, there is no component of that which
pulls the poles inward.

You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making it
simple for young children, it doesn't really explain what's going on.

Jeff Morris wrote:

"Nav" wrote in message
...


Donal wrote:


"Peter S/Y Anicula" wrote in message
.dk...


The moon are a lot closer than the sun. Therefore the gravitational
force of the moon varies more over the earth's surface. It is the
variation in the gravitational force and not the force in itself that
creates the tides.

The moons pull on a water-molecule directly under the moon is larger
than
on a molecule on the far side of the earth, actually it is larger than
"the average pull on the whole earth", and here the moon pulls away

from the earth.

On the far side of the earth (seen from the moon) the gravitation from
the moon is less than average and at this point the moon pulls toward
the earth.


On the far side the tide is "high", ... just the same as at the near side.
If the moon's gravity was pulling the water, then you would expect LW to be
opposite the moon.


Quite so. The tidal gravity force is in the direction of the moon. This
is the potential energy in the system. So, there must be another force
present. The moon has kinetic energy in it's orbital velocity. From
Newton's first law:

F=m r omega^2

It is the difference in the two forces (and the resulting energy minima)
that causes two tides. Simple no? Why invoke something new like
"differential gravity"? Could it be to avoid saying inertial force?


Before I thought you were just arguing philosophically how much we should credit
centrifugal force, but now it appears you haven't really looked at the math at
all. The reason why "differential gravity" is invoked is because it represents
the differing pull of the Moon on differing parts of the Earth. Although this
force is all obviously towards the Moon, when you subtract off the centrifugal
force this is what is left. It is this differing pull that causes the two
tides.


Again you say that gravity causes the two tides -but I say that is not
correct. It is the DIFFERENCE between the centripetal term and the
gravity that causes the tide (How many times do I have to say this?).
You may try to malign me by saying I've not looked at the math but I
have -much closer than you have I think. Here's something for you to try
during coffee:

1) How big is Differential gravity? (1e-6 m/s/s ? It's on the web)
2) How big is the difference in centrifigal acceleration on each side of
the barycenter?

Centrifugal acceleration = r omega^2
The moon orbits the earth every 28 days so omega = 2 pi/ (28 * 24 * 60 *
60).

The barycenter is about at 3/4 r (1/4 r under the earth surface) so the
difference in r from one side to the other makes the imabalance.

The difference in centrifugal acceleration is therefo

1.5 * r * (2 pi/ 28 * 24 * 60 * 60)^2

r is 6.4e^6 m

Get out your calculator and work it out for yourself. It's easy and I
think you'll be surprised at the answer.

Or is my equation for centrifugal acceleration wrong?



You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making it
simple for young children, it doesn't really explain what's going on.


But I never tried to explain it all by intertial forces, Jeff. I always
said it was the _difference_ between inertial forces and gravity. I'd
say it's you who is trying to explain it all by gravity instead!

What this shows is that without land masses to block tidal flow the
tides would be much bigger than they are. That differential gravity
produces the "correct" answer really just shows how it is not the
correct sole explanation (if it had predicted higher tides that would be
expected). Perhaps you would like to think about that approximation used
in the gravity term that "allows" it to cancel the centripetal term. Is
it correct to use this when we are dealing with very very small
fractions of the total acceleration??

Cheers

Peter S/Y Anicula wrote:

Nav 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 trouble is that the gravity difference does not "explain" the two
tides -it may seem to but that is not the case. The correct explanation
resides in the difference between the inertial force and gravity.
Interestingly, it predicts larger tides than are observed (and predicted
by the differerntial model). However, that is because the land masses
and friction reduce the tide height (to what is actaully observed). That
differential gravity appears to produce the "right" answer shows how
shallow (pardon the pun) that "explanation" really is!

Does this make sense?

Cheers

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

Before I thought you were just arguing philosophically how much we should
credit
centrifugal force, but now it appears you haven't really looked at the math
at
all. The reason why "differential gravity" is invoked is because it
represents
the differing pull of the Moon on differing parts of the Earth. Although
this
force is all obviously towards the Moon, when you subtract off the
centrifugal
force this is what is left. It is this differing pull that causes the two
tides.


Again you say that gravity causes the two tides -but I say that is not
correct. It is the DIFFERENCE between the centripetal term and the
gravity that causes the tide (How many times do I have to say this?).
You may try to malign me by saying I've not looked at the math but I
have -much closer than you have I think. Here's something for you to try
during coffee:

1) How big is Differential gravity? (1e-6 m/s/s ? It's on the web)

roughly correct

2) How big is the difference in centrifigal acceleration on each side of
the barycenter?

Centrifugal acceleration = r omega^2
The moon orbits the earth every 28 days so omega = 2 pi/ (28 * 24 * 60 *
60).

The barycenter is about at 3/4 r (1/4 r under the earth surface) so the
difference in r from one side to the other makes the imabalance.

The difference in centrifugal acceleration is therefo

1.5 * r * (2 pi/ 28 * 24 * 60 * 60)^2

r is 6.4e^6 m

Get out your calculator and work it out for yourself. It's easy and I
think you'll be surprised at the answer.


So, you you're claiming the lunar tidal forces are 65 times the accepted values.

Now, get out your calculator and run the same numbers for the Sun. The distance
to the Sun (and the E-S barycenter) is 1.5e^11 meters. The result is about 100
times less than your result for the Moon. So you're claiming that the Sun has
negligible effect on the night time tides?



Or is my equation for centrifugal acceleration wrong?


Actually, applying it in this context is your problem. Centrifugal acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.




You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making
it
simple for young children, it doesn't really explain what's going on.


But I never tried to explain it all by intertial forces, Jeff. I always
said it was the _difference_ between inertial forces and gravity. I'd
say it's you who is trying to explain it all by gravity instead!

Gravity is actually the only force at work. Any explanation must be consistent
with that. Centrifugal forces is "ficticious," it doesn't really exist. The
reality is that the Earth is in free fall towards the E-M barycenter.



What this shows is that without land masses to block tidal flow the
tides would be much bigger than they are.

Absoulute nonsense! The land masses build up the tides, they don't reduce
them. And there is no major landmass on the equator for almost half of the
Earth' circumfrence - there is plenty of room for the tides to fully devolope.
Your theory predicts island in the Pacific would be hit by 100 foot tides every
day.

That differential gravity
produces the "correct" answer really just shows how it is not the
correct sole explanation (if it had predicted higher tides that would be
expected).

Double talk - you made up a silly explanation for why we don't have 100 foot
tides and then fault the accepted explantion for not predicting the same thing.

Perhaps you would like to think about that approximation used
in the gravity term that "allows" it to cancel the centripetal term. Is
it correct to use this when we are dealing with very very small
fractions of the total acceleration??

Perhaps you'd like to explain why your approach shows that the Sun has
negligible contribution to the tides. Sorry Nav, this is looking like the
Constellation all over again.

-jeff

"Donal" wrote in message
...
"Martin Baxter" wrote in message
Donal wrote:
So why does the moon seem to have a greater impact on the

tides?

Well duh! Remember F=G*(m'*m")/(d^2),


Emmm... Huh?

What the hell does that mean in English?

Did you not take physics in school?

Sadly, I didn't!

With hindsight, I suspect that I had a poor teacher who managed to make the
subject appear much duller than it really is.

My high school physics teacher was possibly the worst teacher I ever had - a
true nut case who shouldn't have been left alone with children. Fortunately I
found much better teachers in college.



I've still got a suspicion that if we expand your equation, we will find
that the sun has a greater gravitational influence on the earth than the
moon does.

Yes, its does. The direct gravitational pull of the Sun is enormous, much
larger than the Moon's. However, the tides are caused by the difference in pull
between the near side and the far side. Since the Moon is a lot closer, that
difference is more significant. If you remember any calculus, you'd know that
differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of
the Sun's distance becomes a tiny number compared to the Moon's. The net
result is that the Moon's effect on the tides is 2.2 stronger than the Sun's.

Donal,

I hope you're satisfied! For a damned 1/4 of a point!? All this
differential and centrifugally, how does this Knowledge help a sailor to
ride the tides? That was the original question. Remember?
I hope you're satisfied (g)

Now, I wonder if Scot did any Tide Riding while he has been on Vacation
Cruise?

Ole Thom

"Jeff Morris"

Yes, its does. The direct gravitational pull of the Sun is enormous, much
larger than the Moon's. However, the tides are caused by the difference in pull
between the near side and the far side. Since the Moon is a lot closer, that
difference is more significant. If you remember any calculus, you'd know that
differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of
the Sun's distance becomes a tiny number compared to the Moon's. The net
result is that the Moon's effect on the tides is 2.2 stronger than the Sun's.

Yeah Donal!

Have ya ever seen any body on earth affected by a full sun?

Sheeze... even a 13 year old girl knows it the moon that shakes things up.


Joe

Jeff Morris wrote:




So, you you're claiming the lunar tidal forces are 65 times the accepted values.


No I'm not saying that. The tidal forces are what they are ("accepted
values?"). I can see you are very stubborn. The point is that the
outward component due to rotation is much larger than the apparent
outward force due to the change in distance and gravity. It's a fact
-you've calculated it for yourself!


Or is my equation for centrifugal acceleration wrong?



Actually, applying it in this context is your problem. Centrifugal acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.



It is a much larger force than differential gravity but you want to
ignore it? You are wrong Jeff, it does vary across the surface of the
earth Jeff -the Barycenter is at ~3/4 r! On the moon side it's ~1/7 as
large. Finally, (repeating yet again) it is the ____DIFFERENCE______
between inertial and gravity forces that make the tides. To say it's
only "differential gravity" (I shudder at that term) is clearly wrong -
this was a simple proof.

Cheers

OK, Nav, its clear you're not going to get this without some help. You keep
claiming the centrifugal force varies across the Earth. However, that is not
the case. Your assumption is that the Earth is rotating around the E-M
barycenter, and that because that is offset from the Earth center, the
centrifugal force is unbalanced. (Or more precisely, you claim the "r" in the
centrifugal force equation is different on the near and far sides of the Earth.)

However, if we remove the daily rotation, the Earth does not move around the
barycenter quite like you think. Only the center of the Earth describes a
circle around the barycenter. A point on the surface rotates around a point
offset by an Earth radius towards that point. Thus, the "r" in the centrifugal
force equation is the same for all points on the Earth.

I know this is hard concept to grasp at first, but its really quite simple once
you see it. To help visualize, rub your hand around your tummy, holding it
horizontal. The center of your hand rotates around the center of your stomach,
perhaps with a two inch radius. Your fingertips will also describe a two inch
circle, offset to the side. All points on your hand will describe the same
circle, and feel the same centrifugal acceleration.

Given that, your argument falls apart. The centrifugal force is exactly the
same on all points of the Earth, and (not by coincidence) is exactly opposite
the net gravitational force. What is left over is the differential gravity.


Now that's a simple proof.

So tell us Nav, why did you chose to ignore the Sun's contribution? You deleted
my comments that following your arguments, the Sun's contribution is 1% of the
Moon's; this is clearly at variance with reality.



"Nav" wrote in message
...


Jeff Morris wrote:




So, you you're claiming the lunar tidal forces are 65 times the accepted
values.


No I'm not saying that. The tidal forces are what they are ("accepted
values?"). I can see you are very stubborn. The point is that the
outward component due to rotation is much larger than the apparent
outward force due to the change in distance and gravity. It's a fact
-you've calculated it for yourself!


Or is my equation for centrifugal acceleration wrong?



Actually, applying it in this context is your problem. Centrifugal
acceleration
is constant, it doesn't vary across the surface of the Earth as you claim.
Remember, it doesn't even exist, its actually a reference frame shift.



It is a much larger force than differential gravity but you want to
ignore it? You are wrong Jeff, it does vary across the surface of the
earth Jeff -the Barycenter is at ~3/4 r! On the moon side it's ~1/7 as
large. Finally, (repeating yet again) it is the ____DIFFERENCE______
between inertial and gravity forces that make the tides. To say it's
only "differential gravity" (I shudder at that term) is clearly wrong -
this was a simple proof.

Cheers