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.