Jeff Morris wrote:

"Nav" wrote in message
...


Donal wrote:


"Peter S/Y Anicula" wrote in message
.dk...


The moon are a lot closer than the sun. Therefore the gravitational
force of the moon varies more over the earth's surface. It is the
variation in the gravitational force and not the force in itself that
creates the tides.

The moons pull on a water-molecule directly under the moon is larger
than
on a molecule on the far side of the earth, actually it is larger than
"the average pull on the whole earth", and here the moon pulls away

from the earth.

On the far side of the earth (seen from the moon) the gravitation from
the moon is less than average and at this point the moon pulls toward
the earth.


On the far side the tide is "high", ... just the same as at the near side.
If the moon's gravity was pulling the water, then you would expect LW to be
opposite the moon.


Quite so. The tidal gravity force is in the direction of the moon. This
is the potential energy in the system. So, there must be another force
present. The moon has kinetic energy in it's orbital velocity. From
Newton's first law:

F=m r omega^2

It is the difference in the two forces (and the resulting energy minima)
that causes two tides. Simple no? Why invoke something new like
"differential gravity"? Could it be to avoid saying inertial force?


Before I thought you were just arguing philosophically how much we should credit
centrifugal force, but now it appears you haven't really looked at the math at
all. The reason why "differential gravity" is invoked is because it represents
the differing pull of the Moon on differing parts of the Earth. Although this
force is all obviously towards the Moon, when you subtract off the centrifugal
force this is what is left. It is this differing pull that causes the two
tides.


Again you say that gravity causes the two tides -but I say that is not
correct. It is the DIFFERENCE between the centripetal term and the
gravity that causes the tide (How many times do I have to say this?).
You may try to malign me by saying I've not looked at the math but I
have -much closer than you have I think. Here's something for you to try
during coffee:

1) How big is Differential gravity? (1e-6 m/s/s ? It's on the web)
2) How big is the difference in centrifigal acceleration on each side of
the barycenter?

Centrifugal acceleration = r omega^2
The moon orbits the earth every 28 days so omega = 2 pi/ (28 * 24 * 60 *
60).

The barycenter is about at 3/4 r (1/4 r under the earth surface) so the
difference in r from one side to the other makes the imabalance.

The difference in centrifugal acceleration is therefo

1.5 * r * (2 pi/ 28 * 24 * 60 * 60)^2

r is 6.4e^6 m

Get out your calculator and work it out for yourself. It's easy and I
think you'll be surprised at the answer.

Or is my equation for centrifugal acceleration wrong?



You can handwave the centrifugal force causes the outward bulge, but
mathematically, the idealized shape of the Earth is caused specially by the
differential forces. Trying to explain it all by "inertia" is just making it
simple for young children, it doesn't really explain what's going on.


But I never tried to explain it all by intertial forces, Jeff. I always
said it was the _difference_ between inertial forces and gravity. I'd
say it's you who is trying to explain it all by gravity instead!

What this shows is that without land masses to block tidal flow the
tides would be much bigger than they are. That differential gravity
produces the "correct" answer really just shows how it is not the
correct sole explanation (if it had predicted higher tides that would be
expected). Perhaps you would like to think about that approximation used
in the gravity term that "allows" it to cancel the centripetal term. Is
it correct to use this when we are dealing with very very small
fractions of the total acceleration??

Cheers