NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Air Hockey Pucks and Gravity
From: Frank Reed
Date: 2008 Feb 02, 00:45 -0500
From: Frank Reed
Date: 2008 Feb 02, 00:45 -0500
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 -~----------~----~----~----~------~----~------~--~---