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.