Let us suppose that the final separation of the ring took place somewhere near the half-distance between his orbit and that of Uranus, say, 2,290,000,000 miles from the centre of the nebula, the breadth of the ring would be the difference between the radius of the original nebula, i.e. 33,000,000,000 miles and the above sum, which is 1,010,000,000 miles. Then if we divide the area of the cross section of the ring by this breadth, that is, 516,912,620,000,000 by 1,010,000,000, we find that the thickness would be 511,794 miles; provided the ring did not contract from its outer edge inwards during the process of separation. This could not, of course, be the case, but, as we have no means of finding how much it would contract in that direction, we cannot assign any other breadth for it; and we shall proceed in the same manner in calculating the thicknesses of the rings for all the other planets as we go along. We can, however, make one small approach to greater accuracy. We shall see presently that the density of the ring would be increased threefold at its inner edge as compared with the outer during the process of separation, which would reduce its average thickness to somewhere about 341,196 miles at density of water, of course. The nebula remaining after Neptune's ring we may now call THE URANIAN NEBULA. The volume of the nebula after abandoning the ring for the system of Neptune was found to be 150,523,772,692^{18} 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 50,303,255,814^{18} 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.°