Re: [math-fun] newton rotating liquid mirror
Thank you, Gene. Perhaps even my poor keyboard is still too fast for my brain. So, the surface of the Earth -- and indeed the surface of an unconstrained liquid (e.g., the ocean?) -- at the South Pole is concave downward. Now consider again our liquid _in a fixed container_ at the South Pole. Unless the Earth spins very fast, the Earth retains a nearly spherical shape, and therefore appears to the container and its contents as if the entire mass of the Earth were concentrated at a point at the Earth's center. So the container at 4000 miles feels a really "flat" parallel force field. It would take a pretty large container to "see" a large angle relative to the earth's center. So it would appear that the surface of our liquid in a container would be concave upwards, after all, unless its radius were quite large. Q: what is the size of such a container at the South Pole which would produce a truly flat liquid surface? At 09:50 AM 3/5/2015, Eugene Salamin via math-fun wrote:
A lake at the South Pole would have the shape of the geoid, the equipotential surface due to gravity and centrifugal force.
Would you expect the lake to somehow have a concave surface?
On 3/5/2015 9:32 PM, Henry Baker wrote:
Thank you, Gene. Perhaps even my poor keyboard is still too fast for my brain.
So, the surface of the Earth -- and indeed the surface of an unconstrained liquid (e.g., the ocean?) -- at the South Pole is concave downward.
Now consider again our liquid _in a fixed container_ at the South Pole. Unless the Earth spins very fast, the Earth retains a nearly spherical shape, and therefore appears to the container and its contents as if the entire mass of the Earth were concentrated at a point at the Earth's center. So the container at 4000 miles feels a really "flat" parallel force field. It would take a pretty large container to "see" a large angle relative to the earth's center. So it would appear that the surface of our liquid in a container would be concave upwards, after all, unless its radius were quite large.
Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere? I think Eugene is right. Brent
Q: what is the size of such a container at the South Pole which would produce a truly flat liquid surface?
At 09:50 AM 3/5/2015, Eugene Salamin via math-fun wrote:
A lake at the South Pole would have the shape of the geoid, the equipotential surface due to gravity and centrifugal force.
Would you expect the lake to somehow have a concave surface?
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No, I wouldn't expect a lake at the south pole to have a concave area in the center, but I would expect a child's round swimming pool at the south pole to have a concave surface. There will be forces against the side of the pool to keep the edge of the water higher than the middle. In the middle of a lake, these forces will be dissipated because the water will move away from the center. At 09:40 PM 3/5/2015, meekerdb wrote: Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere?
So the shape of the surface near the center of the pool is not determined by the local forces on the water at the center of the pool, but by some kind of "spooky action at a distance from the pool walls? And accurate measurement of the water at the center of the pool can determine the size of the pool? Can it also determine the shape of the pool? This sure sounds wrong to me. Suppose the pool is 100 feet in diameter. Consider the circle 50 feet in diameter around the center of the pool. No water is flowing in or out of that circle. So if we introduce a new boundary there, there will be no change in the height of water inside or outside that circle. Andy On Fri, Mar 6, 2015 at 8:21 AM, Henry Baker <hbaker1@pipeline.com> wrote:
No, I wouldn't expect a lake at the south pole to have a concave area in the center, but I would expect a child's round swimming pool at the south pole to have a concave surface.
There will be forces against the side of the pool to keep the edge of the water higher than the middle. In the middle of a lake, these forces will be dissipated because the water will move away from the center.
At 09:40 PM 3/5/2015, meekerdb wrote: Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere?
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Within a few millimeters of the wall there is surface tension which usually pulls the liquid up (water on glass), but could pull it down (mercury on glass). But then, this has nothing to do with Earth's rotation, and would make a tiny liquid mirror. The gravitational acceleration g = 9.8 m s^-2 acting on a molecule on the surface at distance d from Earth's axis has a component gd/R pulling it inward, where R = 6.4e6 m is Earth's radius. The centrifugal acceleration pulling it outward is ω^2 d where ω is the angular velocity. In order to flatten the convex surface, these forces must be equal. This requires ω = √(g/R) = 0.00124 s^-1. The rotation period P = 2π/ω = 5080 s = 1.4 hours. You need to spin faster than this to have a concave surface. -- Gene From: Henry Baker <hbaker1@pipeline.com> To: math-fun <math-fun@mailman.xmission.com> Cc: meekerdb <meekerdb@verizon.net> Sent: Friday, March 6, 2015 5:21 AM Subject: Re: [math-fun] newton rotating liquid mirror No, I wouldn't expect a lake at the south pole to have a concave area in the center, but I would expect a child's round swimming pool at the south pole to have a concave surface. There will be forces against the side of the pool to keep the edge of the water higher than the middle. In the middle of a lake, these forces will be dissipated because the water will move away from the center. At 09:40 PM 3/5/2015, meekerdb wrote: Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere?
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meekerdb