Jeff Morris wrote:
No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said:

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

Yes that's right for that site. The differential of the gavity equation
does not give a force it gives the rate of chnage of force with
distance. That's basic calculus.

You even produced a bogus formula to support your claim that centrifugal force varies
across the Earth. This would predict a tidal force 65 times stronger than
differential gravity. If that isn't minimalizing differential gravity, I don't know
what is.


Nothing bogus abot F=m r omaga^2. It's high school physics.


It is quite easy to explain the tides without Centrifugal force - I mentioned the
approach in one of my first posts. There is a gravitational force pulling the Earth
and Moon together. This creates an acceleration such that the Earth is in free fall,
and no net force is felt at the Earth's center. However, there is a larger force on
the Moon side, and a smaller force on the far side. Since the average force has been
accounted for it must be subtracted from these two forces, and the smaller force ends
up being the same magnitude as the larger, but in the opposite direction. Hence, two
tidal bulges, both caused solely by gravity.


Jeff, stop a moment and put what you just said into an equation.
According to you subtraction of a constant from both sides of an
inequality makes an equality?




You, however, have claimed that the variation of the centrifugal force from the near
side to the far creates a force that dominates the tides. Your formula predicts a
force 65 times greater than the differential formula, which would seem to create tides
100 feet or more.

That's because you've forgetten that the tiode is due to the difference
between forces -how many times do I have to repeat this????


Your explanation that land masses dampen these tides runs counter
to common experience. And, your formula predicts the Sun's contribution is only 1% of
the Moon's, which is clearly not the case.

No it doen't. It's your miscomprehension about the force balance.

As to your claim that the rotation around the barycenter "powers" the tides, well,
that's not true.

Jeff, don't be hysterical. Of course it's true. Furthermore, the
barycenter is not the same distance from all points on earth so the
centrifugal force varies across the earth.

Here is the equation for the force on a mass m of water on the far side
for a non-rotating earth (to keep it clear):

GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2


On the near side:

GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2

M is the moon mass, E the earth mass, s the distance from the center of
the earth to the barycenter, omega the angular velocity of the
earth-moon pair G the gravitational constant.


Now look very carefully at the three terms in each equation. The first
two are gravity, the third centrifugal. Two terms are different in both
cases. The gravity term is smaller on the far side and the centrifugal
term is bigger. On the near side the gavity is bigger and the
cenbtrifugal is smaller. That is the proof of my argument.

Cheers