Carrying the comparison a little further, we find that as water is 773·395 times more dense than air, and 11,173·184 times more dense than hydrogen, the density of the nebula could not have been more than 1/403,000,000th part that of air, and 1/27,894,734th that of hydrogen. But, confining the comparison to air, as it suits our purpose better, we see that it would take 403,000,000 cubic feet of the nebula to be equal in mass to 1 cubic foot of air at atmospheric pressure; and that were we to expand this cubic foot of air to this number of times its volume, the space occupied by it would be as nearly in the state of absolute vacuum as could be imagined, far beyond what could be produced by any human means. Now, were heat a material, imponderable substance, as it was at one time supposed to be, we could conceive of its being piled up in any place in space in any desired quantity; but it has been demonstrated not only not to be a substance at all, but that its very existence cannot be detected or made manifest, unless it is introduced by some known means—friction, hammering, combustion—into a real material substance. Therefore, we must conclude that if it existed at all in the nebula, it must have been in a degree corresponding to the tenuity of the medium, and the air thermometer will tell us what the temperature must have been if we only choose to apply it.
Applying, then, this theory of the air thermometer, if we divide[B] 274° by 403,000,000—the number of times the density of the nebula was less than that of air—we get 0·00000068°, as the absolute temperature of the nebula, something very different to excessive heat, incandescence, firemist, or any other name that has been given to its supposed state. Furthermore, as a cubic foot of air weighs 565·04 grains, 403,000,000 divided by 565·04, which is equal to 713,223, would be the number of cubic feet of the space occupied by the nebula, corresponding to each grain of matter in the whole solar system, which would be equal to a cube of very nearly 90 feet to the side. And as the only means by which the nebula could acquire heat would be by collision with each other of the particles of matter of which it was composed; to conceive that two particles weighing 1 grain each, butting each other from an average distance of 90 feet, could not only bring themselves, but all the space corresponding to both of them—which would be 1,426,446 cubic feet, of what?—up to the heat of incandescence, or excessive heat of any kind, is a thing which passes the wit of man. Consequently, neither by primitive piling up, nor by collisions among the particles, could there be any heat in the nebula at the dimensions we have specified, beyond what we have measured above.
Some people believe, at least they seem to say so, that meteors or meteorites colliding would knock gas out of each other, sufficient to fill up the empty space around them, and become incandescent, and so pile up heat in nebulæ sufficient to supply suns for any number of millions of years of expenditure. But they forget that gas is not a nothing. It possesses substance, matter, of some kind, however tenuous. Therefore, if the meteors knock matter out of each other in the form of gas, they must end by becoming gas themselves, and we come back to what we have said above; we have two grains, in weight, of gas abutting each other at an average distance of 30 yards, instead of two grains of granite or anything else, and things are not much improved thereby. And if we compare 30 yards with M. Faye's 3000, where are we?
The next thing to deal with is the formation of the planets.
Separation of Ring for Neptune.
When the nebula was 6,600,000,000 miles in diameter its volume would be 150,532,847,22218[C] cubic miles, and we have just seen that its density must have been 311,754,100,720 times less than that of water, or 403,000,000 less than air, and its temperature 0·00000068° above absolute zero. On the other hand, we find from [Table II]. that the volume of Neptune and his satellite is 29,107,964,680,925 cubic miles at the density of water. Multiplying, therefore, this volume by 311,754,100,720 we get 9,074,53018 cubic miles as the volume of the ring for the formation of Neptune's system at the same density as the nebula. Then, subtracting this volume from 150,532,847,22218, there remain 150,523,772,69218 cubic miles as the volume to which the nebula was reduced by the abandonment of the ring out of which Neptune and his satellite were formed.
Then the mean diameter of the orbit of Neptune being 5,588,000,000 miles, its circumference or length will be 17,555,261,000 miles, and if we divide the volume of his system as stated above, by this length, we get 516,912,620,000,000 square miles as the area of the cross section of the ring, which is equal to the area of a square of 22,735,123 miles to the side. Again, if we divide the circumference of the orbit by this length of side, we find that it is 1/772·165th part of it, and therefore about 28 minutes of arc. Also if we divide the diameter of the orbit by an arc of 22,735,123 miles in length, we find that it bears the proportion of 1 to 246 to the diameter of the orbit. Thus the cross section of the ring would bear the same ratio to its diameter that a ring of 1 foot square would bear to a globe of 246 feet in diameter. Here we find it difficult to believe that by rotating a ball of 246 feet in diameter of cosmic matter, meteorites, or brickbats, we could detach from it, mechanically, by centrifugal force a ring of 1 foot square, and the same difficulty presents itself to us with respect to the nebula. We cannot conceive how a ring of that form could be separated by centrifugal force from a rotating nebula, and have therefore to suppose it to have had some different form, and to apply for that to the example of Saturn's rings—just the same as Laplace no doubt did. We cannot tell how the idea originated that the ring should be of the form we were looking for—perhaps it was naturally—but it seems to have been very general, and in some cases to have led to misconceptions. It is not difficult to show how a Saturnian or flat ring could be formed, but we shall have a better opportunity hereafter of doing so. We must try, nevertheless, to form some notion, however crude it may be, of what might be the thickness of a flat ring of the cross section and volume we have found for Neptune.
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