Gravitational force goes down as r^2, so the effect on the sunward side should be slightly different from that on the other side. This is, of course, neglecting the interaction of the water with the irregular topography of the earth (cf Bay of Fundy) and rotational effects, which are larger. Actual tide tables are very local and wildly different in both amplitude and timing over a distance of tens of miles or less. Far from symmetrical. --R -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, February 11, 2014 5:35 PM To: math-fun Subject: Re: [math-fun] atomic clocks & gravitational time dilation This explanation from the text Mike sugggests is the same explanation I've long heard from reliable sources: ----- . Water on the side of Earth facing the moon is pulled hardest by the moon's gravity. This causes a bulge of water on that side of Earth. That bulge is a high tide. . Earth itself is pulled harder by the moon's gravity than is the ocean on the side of Earth opposite the moon. As a result, there is bulge of water on the opposite side of Earth. This creates another high tide. . With water bulging on two sides of Earth, there's less water left in between. This creates low tides on the other two sides of the planet. ----- But I'm curious: * Are the tides on the sides of the earth nearest and farthest from the moon symmetrical? * If so, why (since the reasons given for those tides are different) ? --Dan On 2014-02-11, at 2:17 PM, Mike Stay wrote:
On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper@gmail.com> wrote:
PS, has anyone ever found a correct explanation of tides in a middle or high school text?
http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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