To find the dimensions of the ring we have 71,974,000 miles for the mean diameter of the orbit of Mercury, which makes its circumference 226,113,518 miles in length. Then dividing the volume of his ring, i.e. 110,787,355,300^{ 6} cubic miles, as above, by this length, the area of its cross-section comes to be 489,963,459 square miles. Here we have to determine the breadth of the ring in a new way, that is empirically. Seeing that the breadth of the ring for the earth's system was 37,205,000 and of that for Venus 28,489,000 miles, we shall assume 20,000,000 miles for the breadth of the ring for Mercury. This will make the residuary, now the Solar nebula, to have been 31,616,000 miles in radius and 63,232,000 miles in diameter. Returning now to the area of the cross-section of the ring, that is, 489,963,459 square miles, and dividing it by the assumed breadth 20,000,000 miles, makes the thickness of the ring to have been 25 miles. But, as before, its inner edge having become 4·354 times more dense than the outer one during the process of separation (see below) the average thickness must have been only 11 miles.
Lastly, as the volume of the nebula was 576,026,502,545,978,033^{ 6}
cubic miles after the separation of the ring for Mercury, we have to condense it into the volume of the Solar nebula, which at 63,232,000 miles in diameter would be
cubic miles. Then dividing the first of these two volumes by the second, we find that its density must have been increased 4·3514 fold. But we found that the density of the Mercurian nebula was 1,194,666 times less than that of water, dividing which by 4·3514 makes the Solar nebula to have been 274,546 times less dense than water. Dividing this in turn by 773·395 shows it to have been 355 times less dense than air, and, still further, dividing 274° by this air density makes its absolute temperature to have been 0·7718585° equal to -273·2281415°.
We might conclude our analysis here, but it will be more convenient to carry our calculations a few steps further, to save the additional trouble that might be occasioned by having to return to them later on.
First we shall condense the Solar nebula to 211,911 times less dense than water, and therefore 274 times less dense than air, which we may note will increase its density 1·2956 times. This supposed to be done, its diameter would be 58,002,920 miles, its volume 102,176,129,412^{ 12} cubic miles, and its density 1/274th of an atmosphere--about _one-ninth_ inch of mercury--which would, in consequence, make its absolute mean heat equal to _one degree_ of the ordinary Centigrade scale, or, in another way of expressing it, equal to -273°.