The volume of the last nebula after the separation of the ring for the Asteroids was found to have been 215,634,760,890,416,6209 cubic miles, which had to be condensed into the volume of the Martian nebula of 402,000,000 miles in diameter, which would give a volume of 34,015,582,677,165,3549 cubic miles. Dividing then, the larger of these volumes by the smaller, we find that the density of the Martian nebula had been increased 6·339 times by the condensation. But we found the density of the Asteroidal nebula to have been 447,218,905 times less dense than water, dividing which by 6·339 makes the Martian nebula to have been 70,547,110 times less dense than water. This divided again by 773·395 makes it 91,259 times less dense than air, and consequently its absolute temperature to have been 0·00300243° or -273·99699757°.
From the Martian nebula of 402,000,000 miles in diameter, volume 34,015,582,677,165,3549 cubic miles, and density 70,547,110 times less than water, we have to deduct the volume of his ring, which by [Table II]., was estimated at 160,728,460,000 cubic miles at density of water. Multiplying this by 70,547,110 we find its volume to be 11,338,927,1549 cubic miles at the same density as the nebula, deducting which from its whole volume we get 34,015,571,338,237,209 cubic miles as the volume after the separation of the ring.
For finding the dimensions of the ring we have 283,300,000 miles as the mean diameter of the orbit of Mars, which makes its circumference 890,015,280 miles in length. Then dividing the volume of the ring 11,338,927,1549 cubic miles by this length, the area of its cross-section comes to be 12,740,148,859 square miles, which, divided by the breadth of 83,690,000 miles—that is one-half of the difference between the diameters of the Martian and Earth nebula, respectively 402,000,000 and 234,620,000 miles—makes the thickness of the ring to have been 152 miles. But as before, the inner having become through condensation, 5·0302 times more dense than the outer edge, the average thickness would be 61 miles.
Earth Nebula.
As the volume of the nebula was 34,015,571,338,237,2009 cubic miles after the separation of the ring for Mars, we have to condense it into the volume of the earth nebula, which at 234,620,000 miles in diameter would be 6,762,303,076,923,0319 cubic miles. Dividing the larger of these volumes by the smaller we find that the density of the nebula has been increased 5·0302 times, as employed above. But we found the density of the Martian nebula to have been 70,547,110 times less than that of water, dividing which by 5·0302 makes the earth nebula to have been 14,024,781 times less dense than water. Dividing this again by 773·395 we find it to have been 18,134 times less dense than air, and 274° divided by this density of air—the same as in all the respective cases—gives 0·0151097° as the absolute temperature of the nebula and corresponds to -273·9848903°.
From the earth nebula 234,620,000 miles in diameter, 6,762,303,076,923,0319 cubic miles in volume, and 14,024,781 times less dense than water, we have to subtract the volume of the ring of the earth's system, which, in [Table II]., appears as 1,489,310,236,000 cubic miles at density of water. Multiplying this by 14,024,781 we find it to have been 20,887,249,5539 cubic miles at the same density as the nebula. And subtracting this quantity from 6,762,303,076,923,0319, we get 6,762,282,189,673,4789 cubic miles for the volume of the previous nebula after the separation of the ring for the system of the earth.
For finding the dimensions of the ring we have 185,930,000 miles for the mean diameter of the Earth's orbit, which makes the circumference 584,117,688 miles in length, and dividing the volume of the ring for the system, which was found to be 20,887,249,5539 cubic miles, by this length, the area of its cross section comes to be 35,760,344,109 square miles, which divided by the breadth of 37,205,000 miles—that is one-half of the difference between the diameters of the Earth and Venus nebulæ, respectively 234,620,000 and 160,210,000 miles—makes the thickness of the ring to have been 961 miles. But the inner will presently be seen to have been 3·141 times more dense than the outer edge when its separation was completed, so that the average thickness would be 612 miles.
Venus Nebula.
As the volume of the nebula was 6,762,282,189,673,4789 cubic miles after the separation of the ring for the system of the Earth, we have to condense it into the volume of the Venus nebula, which at 160,210,000 miles in diameter would be 2,153,120,792,079,2089 cubic miles. Then dividing the larger of these two volumes by the smaller, we find that the density of the Venus nebula had been increased to 3·141 times what that of the Earth nebula was. But we found the density of that nebula to have been 14,024,781 times less than that of water, dividing which by 3·141 makes the Venus nebula to have been 4,465,512 times less dense than water. Dividing this again by 773·395 we find it to have been 5,774 times less dense than air, which would make its absolute temperature to have been 0·04745486°, which corresponds to -273·9525459°.
From the Venus nebula of 160,210,000 miles in diameter, volume 2,153,120,792,079,207,9216 cubic miles, and density 4,465,512 times less than that of water, we have now to deduct the volume of her ring, which by [Table II]. is 1,131,960,000,000 cubic miles at the density of water. Multiplying this volume by 4,465,512 we find the volume of the ring to have been 5,054,780,604,6516 cubic miles at the same density as the nebula, and subtracting this amount from 2,153,120,792,079,207,9216 we get 2,153,115,737,298,6036 cubic miles for the volume to be condensed into the nebula following.