JRS: In article , dated Sat, 22 Jan 2005
02:13:42, seen in news:sci.space.policy, Ross Smith
posted :
Dr John Stockton wrote:

JRS: In article , dated Wed, 19 Jan 2005
15:20:10, seen in news:sci.space.policy, Henry Spencer
posted :

The distance
between Titan and Rhea is never *less* than about twice the distance
from the Earth to the Moon.

Contributory, but non-essential.

The effect of tides is proportional to:

m1.r2^4/m2.d^3

where m1 = mass of inducing body
m2 = mass of subject body
r2 = radius of subject body
d = distance between the bodies

Plugging in the numbers ...

But what is your definition of effect? That having the dimensions of
distance, I suppose the effect to be the height of the tide on a
homogeneous liquid body.

The above is equivalent to (P/p) q^3 r where P, p are the mean
densities of the inducing and affected bodies, q is the angular diameter
of the former as seen from the latter, and r is the radius of the
latter.

The displacement being proportional to the radius, and the other
factors, tie in well with Roche considerations - a tide is no more than
a thoroughly failed Roche break-up.

The actual effect is of course different, here on earth; take beach
holidays in the Bay of Naples and the Bay of Fundy to see that; and on
Cowes Roads to get properly confused. I expect the formula is
calculated for tidal deformation of a homogeneous liquid body.

What if the affected body is a solid sphere with a moderately deep but
less dense ocean? The tidal force and the restoring gravitational force
are both proportional to the density of the ocean, so the tidal height
should be independent of ocean density.

That table does not include Phobos & Deimos, nor ISS; their primaries
subtend angles much larger than most of the tabulated cases. But they
are small enough to be rigid.

--
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