Secondly, since the earth fell in with this body in the same part of its orbit, for several years in succession, it must either have remained there while the earth was performing its whole revolution around the sun, or it must itself have had a revolution, as well as the earth. But I have already shown that it could not have remained stationary in that part of space; therefore, _it must have had a revolution around the sun_. Thirdly, its period of revolution must have either been greater than the earth's, equal to it, or less. It could not have been greater, for then the two bodies could not have been together again at the end of the year, since the meteoric body would not have completed its revolution in a year. Its period might obviously be the same as the earth's, for then they might easily come together again after one revolution of each; although their orbits might differ so much in shape as to prevent their being together at any intermediate point. But the period of the body might also be less than that of the earth, provided it were some _aliquot part of a year_, so as to revolve just twice, or three times, for example, while the earth revolves once. Let us suppose that the period is one third of a year. Then, since we have given the periodic times of the two bodies, and the major axis of the orbit of one of them, namely, of the earth, we can, by Kepler's law, find the major axis of the other orbit; for the square of the earth's periodic time 1^2 is to the square of the body's time (1/3)^2 as the cube of the major axis of the earth's orbit is to the cube of the major axis of the orbit in question. Now, the three first terms of this proportion are known, and consequently, it is only to solve a case in the simple rule of three, to find the term required. On making the calculation, it is found, that the supposition of a periodic time of only one third of a year gives an orbit of insufficient length; the whole major axis would not reach from the sun to the earth; and consequently, a body revolving in it could never come near to the earth. On making trial of six months, we obtain an orbit which satisfies the conditions, being such as is represented by the diagram on page 362, Fig. 69', where the outer circle denotes the earth's orbit, the sun being in the centre, and the inner ellipse denotes the path of the meteoric body. The two bodies are together at the top of the figure, being the place of the meteoric body's aphelion on the thirteenth of November, and the figures 10, 20, &c., denote the relative positions of the earth and the body for every ten days, for a period of six months, in which time the body would have returned to its aphelion. [Illustration Fig. 69'.] Such would be the relation of the body that affords the meteoric shower of November, provided its revolution is accomplished in six months; but it is still somewhat uncertain whether the period be half a year or a year; it must be one or the other.