Page:Newton's Principia (1846).djvu/472

Cor. 2. But their orbits will be so near to parabolas, that parabolas may be used for them without sensible error.

Cor. 3. And, therefore, by Cor. 7, Prop. XVI, Book 1, the velocity of every comet will always be to the velocity of any planet, supposed to be revolved at the same distance in a circle about the sun, nearly in the subduplicate proportion of double the distance of the planet from the centre of the sun to the distance of the comet from the sun's centre, very nearly. Let us suppose the radius of the orbis magnus, or the greatest semidiameter of the ellipsis which the earth describes, to consist of 100000000 parts; and then the earth by its mean diurnal motion will describe 1720212 of those parts, and 71675½ by its horary motion. And therefore the comet, at the same mean distance of the earth from the sun, with a velocity which is to the velocity of the earth as $\scriptstyle {\sqrt {2}}$ to 1, would by its diurnal motion describe 2432747 parts, and 101364½ parts by its horary motion. But at greater or less distances both the diurnal and horary motion will be to this diurnal and horary motion in the reciprocal subduplicate proportion of the distances, and is therefore given.

Cor. 4. Wherefore if the latus rectum of the parabola is quadruple of the radius of the orbis magnus, and the square of that radius is supposed to consist of 100000000 parts, the area which the comet will daily describe by a radius drawn to the sun will be 1216373½ parts, and the horary area will be 50682¼ parts. But, if the latus rectum is greater or less in any proportion, the diurnal and horary area will be less or greater in the subduplicate of the same proportion reciprocally.

LEMMA V.

To find a curve line of the parabolic kind which shall pass through any given number of points.

Let those points be A, B, C, D, E, F, &c., and from the same to any right line HN, given in position, let fall as many perpendiculars AH, BI, CK, DL, EM, FN, &c.

$b$		$2b$		$3b$		$4b$		$5b$
	$c$		$2c$		$3c$		$4c$
		$d$		$2d$		$3d$
			$e$		$2e$
				$f$

Case 1. If HI, IK, KL, &c., the intervals of the points H, I, K, L, M, N, &c., are equal, take b, 2b, 3b, 4b, 5b, &c., the first differences of the perpendiculars AH, BI, CK, &c.; their second differences c, 2c, 3c, 4c, &c.; their third, d, 2d, 3d, &c., that is to say, so as AH - BI may be = b, BI