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From: lew@ihlpa.UUCP (Lew Mammel, Jr.)
Newsgroups: net.astro
Subject: Estimating tidal influences
Message-ID: <182@ihlpa.UUCP>
Date: Fri, 12-Apr-85 18:18:51 EST
Article-I.D.: ihlpa.182
Posted: Fri Apr 12 18:18:51 1985
Date-Received: Sat, 13-Apr-85 06:43:28 EST
Distribution: net
Organization: AT&T Bell Laboratories
Lines: 18

It's been noted that the sun's tidal force (on the earth) is less than
the moon's. Here's an interesting way to see this.

The magnitude of the tidal force is proportional to m/r^3, where m is the
mass of the tide-producing body and r is its distance from earth. If we
divide top and bottom by R^3, the radius of the body cubed, we see that:

	F ~ rho * theta^3

where rho is the density of the body and theta is its apparent angular
diameter.  Since the sun and moon have equal apparent diameters, the
ratio of their tidal forces on the earth is just the ratio of their
densities.

This also gives an easy way to estimate the tidal influences of the
planets on the earth. (remember the Jupiter effect?)

	Lew Mammel, Jr. ihnp4!ihlpa!lew