jeffie, run this by your wife before you respond again.

"Jeff Morris"
Date: 10/15/2004 9:46 PM Eastern Daylight Time
Message-id:


"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.