[Paddlewise] The Tide/Moon Debate [long]

From: Dave Kruger <dkruger_at_pacifier.com>
Date: Wed, 22 Dec 1999 11:14:32 -0800
'Been lurkin' on this one, watching and listening as the heavy artillery and
mortars flash overhead.  Background:  I taught a "Practical Physics" course for
three years some 2.5 decades ago, carrying into it the usual head-in-the-clouds
attitude of the university-educated, and a couple years of college physics on my
transcript.  Figured it would be *simple* to teach physics at that level.  After
all, it's just "introductory" stuff, right?

Wrong!  One of the most humbling experiences of my life!  

Simple physics is simple to understand as long as it deals with the frame of
reference we are accustomed to:  the "fixed" one we think we are in.  Tides are
the consequence of the forces and dynamics of a *rotating* three-body system,
*in fre fall*:  earth-moon-sun, with the sun a lesser player than the
moon/earth, but not at all a minor player.  The other ingredient which is
massively different is that the earth (in this case) is large enough in diameter
that the attractive force from the moon to a piece of the earth on the near-moon
side is greater than the similar-in-origin attractive force to a piece of the
earth on the distal-side.

Here is a way to understand the origin or tides:  imagine yourself floating
freely in space *with no orbital motion,* feet towards the sun, speeding up as
you get closer and closer.  You experience a very slightly greater force at your
feet ('cause they are a little closer to the sun), and a slightly lesser force
at the top of your head, which would tend to elongate you a very tiny amount. 
You would not notice this difference, although it is the origin of tides.  You
would feel like you feel when you are in free-fall here on earth, pretty much.

Now, reconstruct the same scenario, but add orbital motion, so that you are in
constant-radius orbit.  How would you feel?  Answer:  *the same,* because the
force vectors operate at right angles to the constant speed of orbital motion. 
Why do you feel the same?  Because you are still in "free fall."  The only
difference is that as you "fall" toward the sun, you also move at right angles
to a line connecting you and the sun.  When you are in *stable* orbit, these
effects match each other, so you never get closer to the sun.  (No, Virginia,
there is *not* another force operating on you to oppose the sun's gravitational
force.  What would cause that force?)  Notice that we have nothing in our
ordinary lives here on earth like this scenario, which is why tides are hard to
understand.  Here come the tides.

Now *coat* your body with a nice thick layer of a nonvolatile liquid, maybe Karo
syrup, and think about how the syrup responds to the gravitational force the sun
exerts on it.  [Why nonvolatile?  So it does not evaporate into space.  On
earth, water is "nonvolatile" in the sense that earth's own gravitation keeps it
from escaping.  Unless you are the size of the earth, water on your skin would
just flash off into the vacuum of space.]

Pretty much, the liquid experiences the same stuff, *except* there is a *tiny*
bit greater force/unit mass on a chunk of syrup on the side *near* the sun,
compared to the comparable-in-origin force on a chunk of syrup *distal* to the
sun.  So what happens?  Does the syrup nearer the sun fly off and leave your
bod?  Does the syrup farther from the sun get "left behind?"  No, and here's why
not:  Karo syrup (and water, to a lesser extent) is a visco-elastic material
(meaning:  it sticks to itself, but will flow a little bit when stretched).  So
the syrup nearer the sun begins to elongate a little bit toward the sun, *until
a compensating restorative force operates within the syrup.*  Net effect, there
is a "bulge" of syrup on your Sun side!  That was your feet.

Now what about the side distal to the sun (your head)?  Well, it's similar
there, except a distal syrup chunk needs a force to *pull* it back to you.  That
force is produced when the syrup on the outside has bulged enough to generate a
visco-elastic restoring force.

So there you are, flying through space, with Karo syrup bulges toward the sun,
and away from the sun, happy as a clam in pig-doop!  Except, how come no
"tides?"  Well, all you have to add to the scenario now is rotation of your bod,
around an axis at right angles to the plane of the orbit.  As you rotate around
your bellybutton, the bulges stay in the same position, relative to the line
connecting you and the sun, but the piece of you under each bulge changes. 
Rotate yourself every 24 hours, and you will have two high tides (the bulges),
and two low tides (the anti-bulges -- where the syrup is thinned a bit from the
bulge-making process).

Well, if you imagine your bod as the earth, you now have a picture of how *the
Sun* generates tidal bulges.  Notice I left the moon out.  How does the moon
change things?  Conceptually, not much, except that as the moon rotates around
the earth, *the location of the bulges from the moon changes,* in concert with
its orbital position.  So, when the moon is on the *same* line as the earth-Sun,
its bulges reinforce the bulges caused by the sun, and we get bigger high tides
and lower low tides (spring tides).  Notice this happens whether the moon is
*outside* the earth or *inside* the earth -- a consequence of the "free fall"
which the earth-moon system is in.  Likewise, when the moon is at an orbital
position at right angles to the sun-earth line, the moon-induced tidal bulges
tend to offset the sun's tidal effect, and we get smaller tides (neap tides).

Turns out that the bulges generated by the moon are bigger than those generated
by the sun, so for most of the earth, the principal "high" and "low" tidal
effects follow the moon as we rotate under it, with the sun's tidal effects
being superimposed on those of the moon.

I have not considered any of the secondary effects, specifically friction and
inertia, which "delay" the tidal bulges to a little past when the moon is
directly overhead (or, straight through the center of the earth).  A good
oceanography text will tell you 'way more than you want to know!

Finally, re:  the off-topic suggestion that the extraordinary "lineup" (and
close approach of the moon) we are experiencing today will "cause" earthquakes: 
the tidal effects described above "do" distort the earth's crust.  However, to
the best of my knowledge, as Chuck Holst mentioned, the scientific jury is out
on that.  And, as Jackie said, it is a form of hollering "fire!" in church to
insist that there is a greater chance of catastrophic earthquake *because* of
this lineup.  Minor chance?  Maybe, maybe, maybe.  But, not worth worrying
about.

-- 
Dave Kruger
Astoria, OR
some-time physics teacher

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Received on Wed Dec 22 1999 - 11:17:03 PST

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