A "standard" tide prediction uses (at least) 32 factors. The most
important being the difference in graviton from the moon over the
earth surface. That leaves out at least 31 factors of practical
importance. One of these is the difference in centrifugal forces due
to the rotation around the mass-centre of the moon-earth system. There
are still 30 left. I was not trying to give the full explanation of
the tide generating forces.

I was answering a question as to why the influence of the sun on the
tides was smaller than the influence of the moon. I think that the
fact that while the gravitation of the sun is larger than that of the
moon, the difference of the gravitation over the earth surface is
smaller, answers that question fairly well.

Peter S/Y Anicula

"Nav" skrev i en meddelelse
...


DSK wrote:
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.


The key to understanding resides in where the center of mass of the
earth-moon system resides.

Cheers