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.