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.
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°.