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frankreed@HistoricalAtlas.net wrote:
> **Above I mentioned the tidal period. So there's no confusion, I'm not
> talking about ocean tides due to the variation in the gravitational forces
> of the Sun and the Moon over the surface of the Earth, but rather the local
> "tidal" acceleration in the vicinity of any point on Earth due to the
> variation in the gravitational force of the Earth itself. For an example,
> imagine an air table with a disc on it ("air hockey" with a floating puck).
> If I level the table perfectly at its center, and if the table is perfectly
> flat, then the floating disc, if released near one edge, will "fall" towards
> the center because that's the only place where the gravity vector is exactly
> perpendicular to the table --everywhere else, the vector is slightly tilted,
> pointing towards the center of the Earth, so a small component of the
> gravity vector is pointing towards the center of the table. The amount is
> directly proportional to the distance from the center of the table. The disc
> will execute simple harmonic motion [natural frequency
> omega=sqrt(g/R)=sqrt(GM/R^3)], as if it's attached to the center of the
> table by a perfect spring, gliding back and forth across the table with a
> period of 84 minutes (ignoring air resistance). Give it a slight sideways
> motion and the disc will travel in a circular or elliptical path around the
> center of the table with that same 84 minute period (no matter how big the
> circle). That's the "tidal field" of the Earth.
>
I challenged several of my smart friends with this problem.   After they
found the solution, one of them asked if the axis of the puck's
back-and-forth motion will rotate because it is in fact a Foucault
pendulum.   I suspect it would.   But another asked about Coriolis
forces on the back-and-forth motion and I didn't have an answer.   Would
the Coriolis force just be another manifestation of the Foucault effect?

Lu Abel

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