Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:
On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.

Fresh Breezes- Doug King

"DSK" wrote in message
...
Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:
On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.

Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple way:
http://burro.astr.cwru.edu/Academics...ity/tides.html

Jeff Morris wrote:



Here's a site that describes the Differential Gravity in a fairly simple way:
http://burro.astr.cwru.edu/Academics...ity/tides.html


That page shows a (trivial) spatial differentiation of a gravity field.
Not a good explanation IMHO -the terms used are not described at all...
What is the "center of mass force" and why is that different from gravity?

The force vector diagram makes no sense at all. How to you add the vectors:

- -- ---

to the center of mass force

-- X --

to get

-- X --

(note the reversal in direction at the left!) Must be some new maths!
(If it were so why don't people fly of into to space at the equator?).

Even worse, the site then goes on to use the _differential_ expression
to calculate the ratio of forces between the moon and sun!

Cheers

Jeff, you really cannot explain two tides a day unless you also include
the centripetal forces of the earth moon pair -this is the key that is
seems repeatedly lost.

Cheers


Jeff Morris wrote:

"DSK" wrote in message
...

Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:

On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.


Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple way:
http://burro.astr.cwru.edu/Academics...ity/tides.html

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces. Here's a site that uses that approach:

http://co-ops.nos.noaa.gov/restles3.html


"Nav" wrote in message
...
Jeff, you really cannot explain two tides a day unless you also include
the centripetal forces of the earth moon pair -this is the key that is
seems repeatedly lost.

Cheers


Jeff Morris wrote:

"DSK" wrote in message
...

Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:

On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.


Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple
way:
http://burro.astr.cwru.edu/Academics...ity/tides.html

Yep it's spot on. I like the pointed quote:

"1. The Effect of Centrifugal Force. It is this little known aspect of
the moon's orbital motion which is responsible for one of the two force
components creating the tides."

The question is how many sites that try to explain the two tide problem
ignore this? Answer -almost all!!!! Even the one from your astronomy
Professor! That the tidal problem can be repeated incorrectly so many
times really annoys to me. I'd say it is not beyond the ability of most
children to understand the correct answer is it?

Cheers

Jeff Morris wrote:

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces. Here's a site that uses that approach:

http://co-ops.nos.noaa.gov/restles3.html


"Nav" wrote in message
...

Jeff, you really cannot explain two tides a day unless you also include
the centripetal forces of the earth moon pair -this is the key that is
seems repeatedly lost.

Cheers


Jeff Morris wrote:


"DSK" wrote in message
. net...


Peter, thanks for your educational posts.

Peter S/Y Anicula wrote:


On both sides the change in gravitational pull from the moon reduces
or counteracts the gravitational force of the earth on the
water-molecule(making it lighter, so to speak).
This should explain why there is to tides a day, one when the moon is
culminating and one when it is on the other side.

I would think that when the moon is on the opposite side, it's
gravitation effect would be cumulative, acting to depress the water
level. But it would be far less than when it's overhead, and the water
has been put in motion. My (relatively vague) understanding of the
science behind tides is that it's partly gravity and partly harmonics.


Gravity is the force that drives it, harmonics determines the timing.

Here's a site that describes the Differential Gravity in a fairly simple

way:

http://burro.astr.cwru.edu/Academics...ity/tides.html

Jeff,

I think the term centrifuigal is appropriate in this context. It is a
term in the system that can be appreciated without needing to consider
Newtonian forces. To understand centripetal forces is a lot harder than
just demonstrating the effect. similarly gravity can be demonstrated
without maths. Thus the explanation becomes really simple e.g.:

The earth-moon body rotates around a common point and water tries to
move away from the center (water in a bucket swung on a rope analogy).
The moon exerts gravity which is stronger on the side of the moon. Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


Cheers


Jeff Morris wrote:

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces.

You may be right, certainly proponents of this explanation use "centrifugal
force."

However, differential gravity can be explained a number of ways. For example,
the Moon's pull causes the Earth to accelerate towards the Moon. That portion
of the Earth closer feels more force, and thus falls faster; that portion on the
far side feels less force, and thus falls slower. These differences cause the
bulges on the near and far sides.

Remember that Centrifugal Force may be a handy explanation, but it is a
"fictional force" that only appears real to an observer in an accelerating frame
of reference. Therefore, whenever it is used to explain something, there must
be another explanation that works in a non-accelerating frame.



"Nav" wrote in message
...
Jeff,

I think the term centrifuigal is appropriate in this context. It is a
term in the system that can be appreciated without needing to consider
Newtonian forces. To understand centripetal forces is a lot harder than
just demonstrating the effect. similarly gravity can be demonstrated
without maths. Thus the explanation becomes really simple e.g.:

The earth-moon body rotates around a common point and water tries to
move away from the center (water in a bucket swung on a rope analogy).
The moon exerts gravity which is stronger on the side of the moon. Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


Cheers


Jeff Morris wrote:

As I said, I was leaving this as an exercise for the reader.

I have heard tides on the far side of the Earth described in terms of the
centrifugal force caused by the Earth's rotation around Earth-Moon system.
Although this is a consistent way of describing it, I've never liked using
"fictional" forces.

Jeff Morris wrote:

You may be right, certainly proponents of this explanation use "centrifugal
force."

However, differential gravity can be explained a number of ways. For example,
the Moon's pull causes the Earth to accelerate towards the Moon. That portion
of the Earth closer feels more force, and thus falls faster; that portion on the
far side feels less force, and thus falls slower. These differences cause the
bulges on the near and far sides.


I like it! That's a way of putting it I've not heard before and it is
quite elegant (provided the listener can accept that the Earth is
falling toward the Moon!)


Remember that Centrifugal Force may be a handy explanation, but it is a
"fictional force" that only appears real to an observer in an accelerating frame
of reference. Therefore, whenever it is used to explain something, there must
be another explanation that works in a non-accelerating frame.


I could be devious and say we are all in an accelerating frame! But you
are quite right about the artifice of a virtual force. Nevertheless,
children want to know about tides and for them centripetal force can be
experienced more easily than the idea they are on an earth that is
falling...

Cheers

Nav wrote:
The earth-moon body rotates around a common point and water tries to
move away from the centre (water in a bucket swung on a rope
analogy).
The moon exerts gravity which is stronger on the side of the moon.
Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


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 centrifugal element can only explain that there is a bulge on the
part of the earth that turns away from the moon.
That is why it is one of the elements (and there are others), that is
sometimes left out of the explanation.

While I think that in some cases it is a good idea to include the
centrifugal element in the explanation, I don't know exactly how many
elements one should include to make it a good explanation - but I
haven't yet seen a complete explanation in a popular publication.

Peter S/Y Anicula





"Nav" skrev i en meddelelse
...
Jeff,

I think the term centrifuigal is appropriate in this context. It is
a
term in the system that can be appreciated without needing to
consider
Newtonian forces. To understand centripetal forces is a lot harder
than
just demonstrating the effect. similarly gravity can be demonstrated
without maths. Thus the explanation becomes really simple e.g.:

The earth-moon body rotates around a common point and water tries to
move away from the center (water in a bucket swung on a rope
analogy).
The moon exerts gravity which is stronger on the side of the moon.
Thus
water forms two bulges on opposite sides and makes two tides as the
earth rotates...


Cheers