We can now separate the ring for the system of Uranus from the
Uranian nebula, reduced as we have seen to 4,580,000,000 miles in
diameter, volume of 50,303,255,814^{18} cubic miles, and density
of 104,184,535,721 times less than water. Referring to Table II.,
we find the volume of the whole system of Uranus to have been
25,876,388,977,690 cubic miles at the density of water, but we have to
multiply this volume by the new density of 104,184,535,721 times less
than water in order to bring it to the same density as the nebula,
which will make the volume of his system to be 2,695,918,851^{15}
cubic miles at that density. Then, subtracting this volume from
50,303,255,814^{18}, we find that the nebula has been reduced to
50,300,559,895,149^{15} cubic miles in volume.
Then the diameter of the orbit of Uranus being 3,566,766,000 miles, its
circumference will be 11,205,352,065 miles, so that dividing the volume
2,695,918,851^{15} of his system by this length of circumference, the
area of the cross section of the ring would be 240,592,061,166,666
square miles. If we now suppose the diameter of the nebula, after
abandoning the ring for the whole system of Uranus, to have been
2,672,000,000 miles--dimension derived from nearly the half-distance
between the orbits of Uranus and Saturn--we find that the breadth of
the ring would be 954,000,000 miles, which would be the difference
between the radii of the Uranian and Saturnian nebulæ, respectively
2,290,000,000 miles, and 1,336,000,000 miles; so that if we divide the
area of cross section of Uranus' ring or 240,592,070,232,288 square
miles by this breadth we find the thickness of the ring to have been
252,193 miles. But the density of the inner edge of the ring would be
5·036 times more dense than the outer edge, for the same reason as in
the case of the Neptunian ring, which would make the average thickness
to have been about 100,553 miles.
SATURNIAN NEBULA.
We have seen that the volume of the nebula after the separation of
the ring for Uranus' system would be 50,300,559,859,149^{15} cubic
miles, but as we have reduced the diameter of the Saturnian nebula to
2,672,000,000 miles, its volume would also be reduced, or condensed to
9,988,700^{21} cubic miles, so that dividing the larger volume by the
smaller we find that its density must have been increased 5·036 fold.
Then dividing 104,184,535,721 by 5·036, we see that the density would
be reduced, or increased rather, to 20,689,000,000 times less than that
of water. This can be easily found to be 26,750,876 times less than the
density of air, and the air-thermometer would show that the absolute
temperature of the Saturnian nebula must have been 0·000010242° or
-273·99998976°.
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