Finding the orbits from the focus given

Proposition 94 Theorem 48

If two similar mediums be separated from each other by a space terminated on both sides by parallel planes, and a body in its passage through that space be attracted or impelled perpendicularly towards either of those mediums, and not agitated or hindered by any other force ; and the attraction be every where the same at equal distances either plane, taken towards the same hand of the plane, that the sine of incidence upon either plane will be to the sine of emcr from gence from the other plane in a given ratio.

Case 1

Let:

• Aa and Bb be 2 parallel planes
• the body light on the first plane Aa in the direction of the line GH, and in its whole passage through the intermediate space let it be attracted or impelled towards the medium of incidence, and by that action let it be made to describe a curve line HI

Let it emerge in the direction of the line IK. Let there be erected IM perpendicular to Bb the plane of emergence, and meeting the line of incidence GH prolonged in M, and the plane of incidence Aa in R; and let the line of emergence KI be produced and meet HM in L.

Around the centre L, with the interval LI, let a circle be described cutting both HM in P and Q, and MI produced in N; and, first, if the attraction or impulse be supposed uniform, the curve HI (by what Galileo has demonstrated) be a parabola, whose property is that of a rectangle under its given latus rectum and the line IM is equal to the square of HM; and moreover the line HM will be bisected in L. Whence if to MI there be let fall the perpendicular LO, MO, OR will be equal: and adding the equal lines ON, OI, the wholes MN, IR will be equal also.

Therefore since IR is given, MN is also given, and the rectangle NMI is to the rectangle under the latus rectum and IM, that is, to HM² in a given ratio. But the rectangle NMI is equal to the rectangle PMQ, that is, to the difference of the squares ML², and PL² or LI²; and HM² hath a given ratio to its fourth part ML²; therefore the ratio of ML² - LI² to ML² is given, and by conversion the ratio of LI² to ML², and its subduplicate, the ratio of LI to ML. But in every triangle, as LMI, the sines of the angles are proportional to the opposite sides. Therefore the ratio of the sine of the angle of incidence LMR to the sine of the angle of emergence LIR is given. Q.E.D.

Case 2

Let now the body pass successively through several spaces terminated with parallel planes AabB, BbcC, &c.

Let it be acted on by a force which is uniform in each of them separately, but different in the different spaces; and by what was just demonstrated, the sine of the angle of incidence on the first plane Aa is to the sine of emergence from the second plane Bb in a given ratio.

This sine of incidence upon the second plane Bb will be to the sine of emergence from the third plane Cc in a given ratio; and this sine to the sine of emergence from the fourth plane Dd in a given ratio; and so on in infinitum; and, by equality, the sine of incidence on the first plane to the sine of emergence from the last plane in a given ratio.

Let now the intervals of the planes be diminished, and their number be infinitely increased, so that the action of attraction or impulse, exerted according to any assigned law, may become continual, and the ratio of the sine of incidence on the first plane to the sine of emergence from the last plane being all along given, will be given then also. Q.E.D.