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.
SOLAR NEBULA.
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
132,376,310,975,609,756^6
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°.
Public-domain text, read in full here on John Shaqi.
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