Jeff,
Remember that I first posted that very same sentiment, and even provided a
graphic. I still believe that to be true, but have modified my internal
model, giving allowance for the centrifugal force. I'm not a physicist, but
the way I'm seeing it, there is a middle ground in this discussion. I'm
curious to know if you're discounting centrifugal force as a contributor to
the far bulge.
Scout


"Jeff Morris" wrote in message
...

"Donal" wrote in message
...

"Jeff Morris" wrote in message

I've still got a suspicion that if we expand your equation, we will
find
that the sun has a greater gravitational influence on the earth than
the
moon does.

Yes, its does. The direct gravitational pull of the Sun is enormous,
much
larger than the Moon's. However, the tides are caused by the
difference
in pull
between the near side and the far side.

Sorry! This doesn't make any sense at all. How does the water on the
far
side(of the earth) know that there is a different pull on the other side?

It doesn't "know" anything. Because the Earth and Moon are an "orbiting
pair," as you
say, they are falling towards each other. Because the gravitational field
varies, the
near side falls faster than the middle; and the far side falls slower.
Hence, they
bulge out from the middle. That's actually all that is needed to explain
the tides;
its so simple a lot of people have trouble getting it.



Even if you were correct, then there would be a high tide facing the
moon, a
low tide at right angles to the moon, and a much lower *high* tide
opposite
the moon. The reality is that the HW opposite the moon is only
fractionally
smaller.

Well, you're right that there are low tides at right angles, but the way
the math
works out the far side high tides are virtually the same. The magnitude
of the
differing pull 4000 miles closer to the Moon is about the same as 4000
miles further
out. (Though I'm curious now just how much they do differ from each other
...)


Centrigugal force explains why there is a high tide on the oppisite side
of
the Earth from the moon - if you consider that the two bodies are
rotating
around a common centre.

OK. Centrifugal force is the explanation for children. Its kind of like
the Tooth
Fairy of physics. The problem is that while CF can explain the "outward
force" needed
for the far bulge, its still the differing gravitational force that
defines the size
and shape of the tidal force. And CF isn't needed at all if you can
accept the "free
fall) argument.



Since the Moon is a lot closer, that
difference is more significant. If you remember any calculus, you'd
know
that
differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse
cube of
the Sun's distance becomes a tiny number compared to the Moon's. The
net
result is that the Moon's effect on the tides is 2.2 stronger than the
Sun's.

Tsk...tsk.

The moon only has a stronger effect on tides because the Earth and Moon
are
an orbiting pair.

The Earth and Sun are also an orbiting pair. There is no qualitative
difference, only
quantitative.



Where's Gilligan when you need him?


Gilly is an educated man. I'm sure he agrees with me.