"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.

"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 in message
...
"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.

Sorry! This doesn't make any sense at all. How does the water on the far
side(of the earth) know that there is a different pull on the other side?
Even if you were correct, then there would be a high tide facing the moon, a
low tide at right angles to the moon, and a much lower *high* tide opposite
the moon. The reality is that the HW opposite the moon is only fractionally
smaller.

Centrigugal force explains why there is a high tide on the oppisite side of
the Earth from the moon - if you consider that the two bodies are rotating
around a common centre.

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.

Tsk...tsk.

The moon only has a stronger effect on tides because the Earth and Moon are
an orbiting pair.

Where's Gilligan when you need him?



Regards


Donal
--

"Donal" wrote in message
...

"Jeff Morris" wrote in message

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.

Sorry! This doesn't make any sense at all. How does the water on the far
side(of the earth) know that there is a different pull on the other side?

It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you
say, they are falling towards each other. Because the gravitational field varies, the
near side falls faster than the middle; and the far side falls slower. Hence, they
bulge out from the middle. That's actually all that is needed to explain the tides;
its so simple a lot of people have trouble getting it.



Even if you were correct, then there would be a high tide facing the moon, a
low tide at right angles to the moon, and a much lower *high* tide opposite
the moon. The reality is that the HW opposite the moon is only fractionally
smaller.

Well, you're right that there are low tides at right angles, but the way the math
works out the far side high tides are virtually the same. The magnitude of the
differing pull 4000 miles closer to the Moon is about the same as 4000 miles further
out. (Though I'm curious now just how much they do differ from each other ...)


Centrigugal force explains why there is a high tide on the oppisite side of
the Earth from the moon - if you consider that the two bodies are rotating
around a common centre.

OK. Centrifugal force is the explanation for children. Its kind of like the Tooth
Fairy of physics. The problem is that while CF can explain the "outward force" needed
for the far bulge, its still the differing gravitational force that defines the size
and shape of the tidal force. And CF isn't needed at all if you can accept the "free
fall) argument.



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.

Tsk...tsk.

The moon only has a stronger effect on tides because the Earth and Moon are
an orbiting pair.

The Earth and Sun are also an orbiting pair. There is no qualitative difference, only
quantitative.



Where's Gilligan when you need him?


Gilly is an educated man. I'm sure he agrees with me.

jeffie, run this by your wife before you respond again.

"Jeff Morris"
Date: 10/15/2004 9:46 PM Eastern Daylight Time
Message-id:


"Donal" wrote in message
...

"Jeff Morris" wrote in message

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.

Sorry! This doesn't make any sense at all. How does the water on the far
side(of the earth) know that there is a different pull on the other side?

It doesn't "know" anything. Because the Earth and Moon are an "orbiting
pair," as you
say, they are falling towards each other. Because the gravitational field
varies, the
near side falls faster than the middle; and the far side falls slower.
Hence, they
bulge out from the middle. That's actually all that is needed to explain the
tides;
its so simple a lot of people have trouble getting it.



Even if you were correct, then there would be a high tide facing the moon,
a
low tide at right angles to the moon, and a much lower *high* tide opposite
the moon. The reality is that the HW opposite the moon is only
fractionally
smaller.

Well, you're right that there are low tides at right angles, but the way the
math
works out the far side high tides are virtually the same. The magnitude of
the
differing pull 4000 miles closer to the Moon is about the same as 4000 miles
further
out. (Though I'm curious now just how much they do differ from each other
...)


Centrigugal force explains why there is a high tide on the oppisite side of
the Earth from the moon - if you consider that the two bodies are rotating
around a common centre.

OK. Centrifugal force is the explanation for children. Its kind of like the
Tooth
Fairy of physics. The problem is that while CF can explain the "outward
force" needed
for the far bulge, its still the differing gravitational force that defines
the size
and shape of the tidal force. And CF isn't needed at all if you can accept
the "free
fall) argument.



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.

Tsk...tsk.

The moon only has a stronger effect on tides because the Earth and Moon are
an orbiting pair.

The Earth and Sun are also an orbiting pair. There is no qualitative
difference, only
quantitative.



Where's Gilligan when you need him?


Gilly is an educated man. I'm sure he agrees with me.

It looks like tides are another topic where jaxie is in over his head.


"JAXAshby" wrote in message
...
jeffie, run this by your wife before you respond again.

"Jeff Morris"
Date: 10/15/2004 9:46 PM Eastern Daylight Time
Message-id:


"Donal" wrote in message
...

"Jeff Morris" wrote in message

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.

Sorry! This doesn't make any sense at all. How does the water on the far
side(of the earth) know that there is a different pull on the other side?

It doesn't "know" anything. Because the Earth and Moon are an "orbiting
pair," as you
say, they are falling towards each other. Because the gravitational field
varies, the
near side falls faster than the middle; and the far side falls slower.
Hence, they
bulge out from the middle. That's actually all that is needed to explain the
tides;
its so simple a lot of people have trouble getting it.



Even if you were correct, then there would be a high tide facing the moon,
a
low tide at right angles to the moon, and a much lower *high* tide opposite
the moon. The reality is that the HW opposite the moon is only
fractionally
smaller.

Well, you're right that there are low tides at right angles, but the way the
math
works out the far side high tides are virtually the same. The magnitude of
the
differing pull 4000 miles closer to the Moon is about the same as 4000 miles
further
out. (Though I'm curious now just how much they do differ from each other
...)


Centrigugal force explains why there is a high tide on the oppisite side of
the Earth from the moon - if you consider that the two bodies are rotating
around a common centre.

OK. Centrifugal force is the explanation for children. Its kind of like the
Tooth
Fairy of physics. The problem is that while CF can explain the "outward
force" needed
for the far bulge, its still the differing gravitational force that defines
the size
and shape of the tidal force. And CF isn't needed at all if you can accept
the "free
fall) argument.



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.

Tsk...tsk.

The moon only has a stronger effect on tides because the Earth and Moon are
an orbiting pair.

The Earth and Sun are also an orbiting pair. There is no qualitative
difference, only
quantitative.



Where's Gilligan when you need him?


Gilly is an educated man. I'm sure he agrees with me.

Jeff,
Remember that I first posted that very same sentiment, and even provided a
graphic. I still believe that to be true, but have modified my internal
model, giving allowance for the centrifugal force. I'm not a physicist, but
the way I'm seeing it, there is a middle ground in this discussion. I'm
curious to know if you're discounting centrifugal force as a contributor to
the far bulge.
Scout


"Jeff Morris" wrote in message
...

"Donal" wrote in message
...

"Jeff Morris" wrote in message

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.

Sorry! This doesn't make any sense at all. How does the water on the
far
side(of the earth) know that there is a different pull on the other side?

It doesn't "know" anything. Because the Earth and Moon are an "orbiting
pair," as you
say, they are falling towards each other. Because the gravitational field
varies, the
near side falls faster than the middle; and the far side falls slower.
Hence, they
bulge out from the middle. That's actually all that is needed to explain
the tides;
its so simple a lot of people have trouble getting it.



Even if you were correct, then there would be a high tide facing the
moon, a
low tide at right angles to the moon, and a much lower *high* tide
opposite
the moon. The reality is that the HW opposite the moon is only
fractionally
smaller.

Well, you're right that there are low tides at right angles, but the way
the math
works out the far side high tides are virtually the same. The magnitude
of the
differing pull 4000 miles closer to the Moon is about the same as 4000
miles further
out. (Though I'm curious now just how much they do differ from each other
...)


Centrigugal force explains why there is a high tide on the oppisite side
of
the Earth from the moon - if you consider that the two bodies are
rotating
around a common centre.

OK. Centrifugal force is the explanation for children. Its kind of like
the Tooth
Fairy of physics. The problem is that while CF can explain the "outward
force" needed
for the far bulge, its still the differing gravitational force that
defines the size
and shape of the tidal force. And CF isn't needed at all if you can
accept the "free
fall) argument.



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.

Tsk...tsk.

The moon only has a stronger effect on tides because the Earth and Moon
are
an orbiting pair.

The Earth and Sun are also an orbiting pair. There is no qualitative
difference, only
quantitative.



Where's Gilligan when you need him?


Gilly is an educated man. I'm sure he agrees with me.

"Scout" wrote in message
...
Jeff,
Remember that I first posted that very same sentiment, and even provided a
graphic. I still believe that to be true, but have modified my internal
model, giving allowance for the centrifugal force. I'm not a physicist, but
the way I'm seeing it, there is a middle ground in this discussion. I'm
curious to know if you're discounting centrifugal force as a contributor to
the far bulge.
Scout

I've always said that Centrifugal Force can be used as part of the explanation, as
long as you end up with the same answer. There are several different ways of looking
at this, all valid. (I hope I can get through this without mangling the terms too
badly ...)

The problem with Centrifugal Force is that it is a "fictional force." It is only
needed if you work in a non-inertial, or accelerating reference frame. If you are in
a car going around a curve, your reference frame is accelerating towards the center of
the curve, and thus you feel a Centrifugal Force in the opposite direction. To an
outside observer, the CF doesn't exist, the only force is the car pulling the
passenger around the turn. The outside observer can analyze the situation completely
without invoking CF. (The passenger feels CF push him outward, the observer sees the
car pull the passenger inward.)

In the Earth-Moon system there is gravity pulling both the Earth and Moon around
curves. Because the gravity acts on all objects, we don't notice ourselves being
pulled around. The magnitude of the Centrifugal force is to small to notice, but in
that reference frame it exists. To the outside observer, we're just in freefall,
being pulled inward by gravity.

The problem with CF arises when you look carefully at the math. One pitfall Nav fell
into was trying to calculate CF as a function that varies with the distance to the
barycenter. However, all points on the Earth do not rotate around the barycenter,
only the center does. Other points describe the same circle around nearby points, so
that all points on Earth feel the same Centrifugal Force. (This is a tough concept to
explain in words; its easier to do it graphically. Consider a plate wobbling around a
point but with no rotation - each point on the plate describes the same circle.)

BTW, Nav provided two commonly used formulas, one for gravity and the other for CF.
Although they look quite different, you should appreciate that they are the same,
since the angular velocity is determined by the gravitational force. The CF will be
the same (with the opposite sign) as the gravitational pull at the Earth's center.

Since the CF is a constant force, it can't describe the two bulges in opposite
directions. It is gravity itself that varies with distance. The differential force
can be derived either by subtracting the average gravitational force which causes the
freefall at the center of the Earth, or it can be derived by adding the centrifugal
force. Since the two are the same, except for the sign, the math is identical.

So take your pick, either explanation works, and I'm sure there are others. However,
I hope you can appreciate that explanations like "gravity creates the inner bulge,
centrifugal force creates the outer bulge" makes physicists wince!

"Jeff Morris" wrote
[snip]
So take your pick, either explanation works, and I'm sure there are
others. However,
I hope you can appreciate that explanations like "gravity creates the
inner bulge,
centrifugal force creates the outer bulge" makes physicists wince!

Yes, I can, as I've watched my physics teaching partner wince quite a bit
this week as we discussed this thread. He was quick to cover our whiteboard
with formulas and drawings. It's an interesting thread though, and
notwithstanding my oversimplified analogies, I've learned a lot from it. By
the way, I saw that same wince from a black history professor when I
suggested that the Civil War was fought to free the slaves, and then again
when I suggested to an ancient lit professor that The Odyssey has all the
earmarks of an Arnold Schwarzenegger movie. Probably explains why I like a
good fart joke.
Scout

Well just to confuse things a bit mo
Even if we only focus on the tide generating potential, there is a
cupple of things that we haven't discussed yet, and one of them has to
do with rotation: "The Coriolis freqency". The other thing one could
include is the "parallax".
I mention this just to make clear that the two models discussed above
both are incomplete.

Peter S/Y Anicula

"Jeff Morris" skrev i en meddelelse
...
"Scout" wrote in message
...
Jeff,
Remember that I first posted that very same sentiment, and even
provided a
graphic. I still believe that to be true, but have modified my
internal
model, giving allowance for the centrifugal force. I'm not a
physicist, but
the way I'm seeing it, there is a middle ground in this
discussion. I'm
curious to know if you're discounting centrifugal force as a
contributor to
the far bulge.
Scout

I've always said that Centrifugal Force can be used as part of the
explanation, as
long as you end up with the same answer. There are several
different ways of looking
at this, all valid. (I hope I can get through this without mangling
the terms too
badly ...)

The problem with Centrifugal Force is that it is a "fictional
force." It is only
needed if you work in a non-inertial, or accelerating reference
frame. If you are in
a car going around a curve, your reference frame is accelerating
towards the center of
the curve, and thus you feel a Centrifugal Force in the opposite
direction. To an
outside observer, the CF doesn't exist, the only force is the car
pulling the
passenger around the turn. The outside observer can analyze the
situation completely
without invoking CF. (The passenger feels CF push him outward, the
observer sees the
car pull the passenger inward.)

In the Earth-Moon system there is gravity pulling both the Earth and
Moon around
curves. Because the gravity acts on all objects, we don't notice
ourselves being
pulled around. The magnitude of the Centrifugal force is to small to
notice, but in
that reference frame it exists. To the outside observer, we're
just in freefall,
being pulled inward by gravity.

The problem with CF arises when you look carefully at the math. One
pitfall Nav fell
into was trying to calculate CF as a function that varies with the
distance to the
barycenter. However, all points on the Earth do not rotate around
the barycenter,
only the center does. Other points describe the same circle around
nearby points, so
that all points on Earth feel the same Centrifugal Force. (This is
a tough concept to
explain in words; its easier to do it graphically. Consider a plate
wobbling around a
point but with no rotation - each point on the plate describes the
same circle.)

BTW, Nav provided two commonly used formulas, one for gravity and
the other for CF.
Although they look quite different, you should appreciate that they
are the same,
since the angular velocity is determined by the gravitational force.
The CF will be
the same (with the opposite sign) as the gravitational pull at the
Earth's center.

Since the CF is a constant force, it can't describe the two bulges
in opposite
directions. It is gravity itself that varies with distance. The
differential force
can be derived either by subtracting the average gravitational force
which causes the
freefall at the center of the Earth, or it can be derived by adding
the centrifugal
force. Since the two are the same, except for the sign, the math is
identical.

So take your pick, either explanation works, and I'm sure there are
others. However,
I hope you can appreciate that explanations like "gravity creates
the inner bulge,
centrifugal force creates the outer bulge" makes physicists wince!