NavList:

   

Reply

Lu, you wrote:
"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?"

Yes, the Coriolis acceleration is responsible for rotating the swinging
plane of the Foucault pendulum and will precess the general elliptical path
of the "air hockey puck". They are one and the same. There's also some
centrifugal force. It's easy to work out the cases for the equator (no
coriolis, no centrifugal) and also at the poles. At the poles, you can
easily think of it from the point of view of an inertial frame. The table
turns underneath the fixed orbit of the puck so the path across the air
table advances about 22 degrees on each pass. In fact, a satellite in low
polar orbit about the Earth will have the same orbital period as the
oscillation period of the puck on the table, and it will precess about 22
degrees in longitude on every orbit, too. Or you can look at it from the
frame of reference attached to the rotating Earth. There the deflection is
due to Coriolis force. Likewise for the orbiting satellite. Same physics -
different coordinates.

Now try setting up that frictionless table on an asteroid ten miles across.
Level the table, and time the oscillation of a disc across the table. Do the
same in the basket of a balloon (hot air) hanging high above the clouds in
the atmosphere of Jupiter. And try it on the surface of the Earth's Moon,
the surface of Mars; name your planet, pick your asteroid. You will find the
same oscillation at almost the same rate. The period of oscillation does
vary --it scales in proportion to the object's mean density, but since
there's not much range in density among the planets, moons, minor planets,
etc. which make up the Solar System (and presumably other Solar Systems),
this same tidal oscillation period of about an hour and a half comes up
again and again, give or take a factor of 3 or less (the range is from about
60 to about 180 minutes).

And just to get a bit of navigation into this, if you were to deploy an
inertial navigation platform on another planet, you would see the same
oscillation of errors at this "local tidal" rate. The motion would be
harmonic in the plane perpendicular to local gravity, anti-harmonic
(exponentially growing, rather than oscillatory) in the direction aligned
with local gravity. On the Earth's Moon, for example, the period would only
be about 30% longer than here on the Earth despite the fact that the Moon is
4 times smaller in diameter and 81 times smaller in mass.

I had forgotten the name for this rate in inertial navigation terminology,
but I came across it just today. It's called the "Schuler frequency". If you
look that up, beware! There are some very screwy non-physical explanations
of this phenomenon. If they start talking about a pendulum as long as the
Earth's radius, jump to the next paragraph. The math is probably fine, but
that physical explanation is double-talk.

 -FER
PS: sorry for being off-topic.


--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to NavList@fer3.com
To , send email to NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---

   

Reply