From: ronkean@juno.com
Date: Mon Sep 20 1999 - 09:19:16 MDT
On Sun, 19 Sep 1999 22:30:57 -0700 Spike Jones <spike66@ibm.net> writes:
Ron Kean wrote:
> > Tidal force is proportional to the mass of the body causing the
> force and inversely proportional to the cube of the distance to that
> body....
>
> The distance to the moon varies
> about 3 percent between moon overhead and moon underfoot, whereas the
> distance to the sun, noon vs midnight, only varies only about one
> part in ten thousand, or a hundredth of a percent. Turns out the tides
> are more complicated than that.
>
Your point about the size of the earth being relevant to tidal force is
well taken since the actual force produced by the tidal effect in any
locality is a stretching force which is observed acting on two separated
test masses which tends to pull them further apart. That stretching
force is indeed (nearly) proportional to the distance separating the two
masses. The further apart the two masses along the line of action of the
tidal field, the greater the force which tries to pull them away from
each other. The tidal effect is differential gravity, which is why it is
inversely proportional to the cube of the distance from the massive body
causing the effect, whereas gravity itself is inversely proportional to
the square of the distance from that body.
The height of the tides is also influenced by the earth's gravity. The
earth has a density of 5.5 and the tidal range on the open ocean is about
2 feet. If the earth had a much smaller density, say .01, then the tides
would be much higher, over 1000 feet, I think.
Ron Kean
.
.
.
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