"Nav" wrote in message ...


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.

Simply differentiating the gravitational force is not enough. But the equation
normally given for differential gravity is dimensionally correct.


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.

That formula is correct. Applying it with a varying "r" is bogus. That's the issue
here, but you haven't noticed it.





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?

I have no idea what you're saying.

If the force at the center is X, and on the near side is X+D, and on the far side is
X-D, then if you subtract the central force, the near side is D and the far side
is -D.






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????


I just tried to follow all your instructions. You challenged me to work the math, I
did, it produced bogus answers. The difference in the centrifugal force that your
equation predicts is 65 times more that the difference in the gravitational force, so
subtracting still won't help much. Why don't you work out the math? And don't forget
to do it for the Sun also.



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.

I'm just reporting the numbers your equation predicts.



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.

This is your basic mistake. Even the web site you pointed to for support explicitly
says this is not true.
Let me repeat it again
http://www.co-ops.nos.noaa.gov/restles3.html

"While space does not permit here, it may be graphically demonstrated that, for such a
case of revolution without rotation as above enumerated, any point on the earth will
describe a circle which will have the same radius as the radius of revolution of the
center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of
the centrifugal force produced by the revolution of the earth and moon around their
common center of mass (G) is the same at point A or B or any other point on or beneath
the earth's surface. Any of these values is also equal to the centrifugal force
produced at the center-of-mass (C) by its revolution around the barycenter. This fact
is indicated in Fig. 2 by the equal lengths of the thin arrows (representing the
centrifugal force Fc) at points A, C, and B, respectively."

it continues with:
"While the effect of this centrifugal force is constant for all positions on the
earth, the effect of the external gravitational force produced by another astronomical
body may be different at different positions on the earth because the magnitude of the
gravitational force exerted varies with the distance of the attracting body."

Go on. Read it, think about it. You were quick to cite this page when you thought it
supported you.



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.

Nope. This is incorrect. The proper equations a

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

On the near side:
GmM/(R-r)^2 + GmE/r^2 - m s omega^2

Further, since GmE/r^2 = m s omega^2, we are left simply with:
Farside: GmM/(R+r)^2
On the near side: GmM/(R-r)^2

This leads to the traditional differential equation 2GmMr/R^3 as shown in
http://mb-soft.com/public/tides.html


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.

Nope. Only one term is different.

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.

The proof fails because of a faulty assumption. Sorry Nav, Centrifugal Force is
constant. You can use it if you chose, but it doesn't really change the math and
isn't particularly interesting. The part of the equation that actually produces the
two tides is the differential gravity.