The volume of the nebula after abandoning the ring for the system of Neptune was found to be 150,523,772,69218 cubic miles at its original density, but during the separation it has been condensed into a sphere of 4,580,000,000 miles in diameter, whose volume would be 50,303,255,81418 cubic miles; so that if we divide the larger of these two volumes by the smaller, we find that the density of the Uranian nebula would be increased 2·9923 times, and therefore it would then be 311,754,100,720 divided by 2·9923, equal to 104,184,535,721 times less dense than water. Furthermore, if we compare it to the density of air, which we can do by dividing this last quantity by 773·395, we find it to have been 134,710,620 times less than that density; and if we apply the air thermometer to it, we shall find that its absolute temperature must have been 274 divided by 134,710,620 = 0·000002034° or -273·9999796.°
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 50,303,255,81418 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 2,695,918,85115 cubic miles at that density. Then, subtracting this volume from 50,303,255,81418, we find that the nebula has been reduced to 50,300,559,895,14915 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 2,695,918,85115 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 50,300,559,859,14915 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 9,988,70021 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°.
We have just seen that the Saturnian nebula has been condensed to 2,672,000,000 miles in diameter, to volume of 9,988,70021 cubic miles, and density of 20,689,000,000 times less than that of water. Then from [Table II]. we get the volume of the whole of the system of Saturn as 154,370,734,774,315 cubic miles at the density of water, and multiplying this by 20,689,000,000 will give 3,193,775,47815 as its volume at the same density as the nebula; and subtracting this from 9,988,70021 we find that the volume of the nebula had been reduced to 9,985,506,224,52215 cubic miles.
Then the diameter of the orbit of Saturn being 1,773,558,000 miles its circumference would be 5,571,809,813 miles in length, and if we divide the volume of his system, viz. 3,193,775,47815 cubic miles, by this length, we find the area of the cross section of the ring to have been 573,202,529,391,503 square miles. Now, supposing the diameter of the nebula, after abandoning the ring, to have contracted to 1,370,800,000 miles and radius consequently of 685,400,000 miles, the breadth of the ring would be 1,336,000,000 less 685,400,000 or 650,600,000 miles; and if we divide the area of the cross section of the ring, that is, 573,202,529,391,503 square miles, by this breadth, we get 881,037 miles for its thickness. But in the same way as before, the inner edge of the ring would be 7·4037 times more dense than the outer edge, which would reduce its average thickness to 238,000 miles.
Jovian Nebula.
The volume of the nebula after separation of the ring for Saturn's system having been 9,985,506,224,52215 cubic miles, this volume has to be condensed into the volume of the Jovian nebula of 1,370,800,000 miles in diameter, which would be 1,348,720,186,33515 cubic miles. Then if we divide the first of these two volumes by the second, we find the density of the Jovian nebula to have been increased 7·4037 fold over the previous one. But the density of the Saturnian nebula was 20,689,000,000 times less than water, dividing which by 7·4037 makes the Jovian nebula to have been 2,794,417,420 times less dense than water. Dividing this by 773·395 we get a density for it of 3,613,182 times less than that of air, which corresponds to the absolute temperature of 0·00007583° or -273·99992417°.