The Mathematical Principles of Natural Philosophy, Book III: Of the System of the World
Proposition XLI. Problem XXI
From three observations given to determine the orbit of a comet moving in a parabola.
This being a Problem of very great difficulty, I tried many methods of resolving it; and several of these Problems, the composition whereof I have given in the first Book, tended to this purpose. But afterwards I contrived the following solution, which is something more simple.
‸Select three observations distant one from another by intervals of time nearly equal; but let that interval of time in which the comet moves more slowly be somewhat greater than the other; so, to wit, that the difference of the times may be to the sum of the times as the sum of the times to about 600 days; or that the point E may fall upon M nearly, and may err therefrom rather towards I than towards A. If such direct observations are not at hand, a new place of the comet must be found, by Lem. VI.
‸Let S represent the sun; T, t, , three places of the earth in the orbis magnus; TA, tB, , three observed longitudes of the comet; V the time between the first observation and the second; W the time between the second and the third; X the length which in the whole time V + W the comet might describe with that velocity which it hath in the mean distance of the earth from the sun, which length is to be found by Cor. 3, Prop. XL, Book III; and tV a perpendicular upon the chord. In the mean observed longitude tB take at pleasure the point B, for the place of the comet in the plane of the ecliptic; and from thence, towards the sun S, draw the line BE, which may be to the perpendicular tV as the content under SB and St2 to the cube of the hypothenuse of the right angled triangle, whose sides are SB, and the tangent of the latitude of the comet in the second observation to the radius tB. And through the point E (by Lemma VII) draw the right line AEC, whose parts AE and EC, terminating in the right lines TA and, may be one to the other as the times V and W: then A and C will be nearly the places of the comet in the plane of the ecliptic in the first and third observations, if B was its place rightly assumed in the second.
Upon AC, bisected in I, erect the perpendicular Ii. Through B draw the obscure line Bi parallel to AC. Join the obscure line Si, cutting AC in, and complete the parallelogram iI. Take equal to; and through the sun S draw the obscure line equal to. Then, cancelling the letters A, E, C, I, from the point B towards the point, draw the new obscure line BE, which may be to the former BE in the duplicate proportion of the distance BS to the quantity. And through the point E draw again the right line AEC by the same rule as before; that is, so as its parts AE and EC may be one to the other as the times V and W between the observations. Thus A and C will be the places of the comet more accurately.
Upon AC, bisected in I, erect the perpendiculars AM, CN, IO, of which AM and CN may be the tangents of the latitudes in the first and third observations, to the radii TA and. Join MN, cutting IO in O. Draw the rectangular parallelogram, as before. In IA produced take ID equal to. Then in MN, towards N, take MP, which may be to the above found length X in the subduplicate proportion of the mean distance of the earth from the sun (or of the semi-diameter of the orbis magnus) to the distance OD. If the point P fall upon the point N; A, B, and C, will be three places of the comet, through which its orbit is to be described in the plane of the ecliptic. But if the point P falls not upon the point N, in the right line AC take CG equal to NP, so as the points G and P may lie on the same side of the line NC.
By the same method as the points E, A, C, G, were found from the assumed point B, from other points b and assumed at pleasure, find out the new points e, a, c, g; and, , , . Then through G, g, and, draw the circumference of a circle, cutting the right line in Z: and Z will be one place of the comet in the plane of the ecliptic. And in AC, ac, , making AF, af, , equal respectively to CG, cg, ; through the points F, f, and, draw the circumference of a circle, cutting the right line AT in X; and the point X will be another place of the comet in the plane of the ecliptic. And at the points X and Z, erecting the tangents of the latitudes of the comet to the radii TX and, two places of the comet in its own orbit will be determined. Lastly, if (by Prop. XIX., Book I) to the focus S a parabola is described passing through those two places, this parabola will be the orbit of the comet. Q.E.I.
The demonstration of this construction follows from the preceding Lemmas, because the right line AC is cut in E in the proportion of the times, by Lem. VII, as it ought to be, by Lem. VIII.; and BE, by Lem. XI., is a portion of the right line BS or in the plane of the ecliptic, intercepted between the arc ABC and the chord AEC; and MP (by Cor. Lem. X.) is the length of the chord of that arc, which the comet should describe in its proper orbit between the first and third observation, and therefore is equal to MN, providing B is a true place of the comet in the plane of the ecliptic.
But it will be convenient to assume the points B, b, , not at random, but nearly true. If the angle AQt, at which the projection of the orbit in the plane of the ecliptic cuts the right line tB, is rudely known, at that angle with Bt draw the obscure line AC, which may be to in the subduplicate proportion of SQ to St; and, drawing the right line SEB so as its part EB may be equal to the length Vt, the point B will be determined, which we are to use for the first time. Then, cancelling the right line AC, and drawing anew AC according to the preceding construction, and, moreover, finding the length MP, in tB take the point b, by this rule, that, if TA and intersect each other in Y, the distance Yb may be to the distance YB in a proportion compounded of the proportion of MP to MN, and the subduplicate proportion of SB to Sb. And by the same method you may find the third point, if you please to repeat the operation the third time; but if this method is followed, two operations generally will be sufficient; for if the distance Bb happens to be very small, after the points F, f, and G, g, are found, draw the right lines Ff and Gg, and they will cut TA and in the points required, X and Z.