On Fri, 17 Sep 1999 22:58:33 -0700 Spike Jones <spike66@ibm.net> writes:
> > Ron Kean wrote:
> > Tides on earth are caused by both the sun and moon, though the
> moon's> > effect dominates. If the moon were removed, tides would not
> entirely> > stop.
>
> Not entirely but almost. The delta between high tide and low would
> be on the order of 1 cm. spike
>
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. Plugging in the mean distances and the masses for the sun and moon, I get the result that the tidal force due to the moon is 2.176 times that due to the sun, or, to put it another way, the sun's effect is 46% of the effect of the moon. Of course that ratio varies somewhat as the distances vary somewhat during the month and year.
Tidal ranges are minimized when the sun and moon are about 90 degrees apart in the sky (quarter moon), and maximized when the sun and moon are lined up with the earth (full moon, new moon). The minimum is called 'neap tide', the maximum 'spring tide".
Interestingly, for spherical bodies of the same density, the tidal effect is simply proportional to the cube of the apparent angular size of the body. The sun and moon are just about the same size as seen in our sky, so based on that approximation the sun would be about 46% as dense as the moon. The actual ratio is 42.2%, since the apparent diameter of the sun at mean distance is really 2.9% greater than the apparent diameter of the moon at mean distance. (Check: 1.029 cubed times 42.2% = 46%). The tidal effect of Jupiter on the Earth, at closest approach, is .00128% that of the sun.
Ron Kean
.
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