EXAMPLE 1. If the centripetal force tending to the several particles of the sphere be reciprocally as the distance; instead of V write PE the distance, then 2PS × LD for PE2; and DN will become as . Suppose DN equal to its double ; and 2SL the given part of the ordinate drawn into the length AB will describe the rectangular area 2SL × AB; and the indefinite part LD, drawn perpendicularly into the same length with a continued motion, in such sort as in its motion one way or another it may either by increasing or decreasing remain always equal to the length LD, will describe the area , that is, the area SL × AB; which taken from the former area 2SL × AB, leaves the area SL × AB. But the third part , drawn after the same manner with a continued motion perpendicularly into the same length, will describe the area of an hyperbola, which subducted from the area SL × AB will leave ANB the area sought. Whence arises this construction of the Problem. At the points, L, A, B, erect the perpendiculars Ll, Aa, Bb; making Aa equal to LB, and Bb equal to LA. Making Ll and LB asymptotes, describe through the points a, b,[Pg 230] the hyperbolic curve ab. And the chord ba being drawn, will inclose the area aba equal to the area sought ANB. EXAMPLE 2. If the centripetal force tending to the several particles of the sphere be reciprocally as the cube of the distance, or (which is the same thing) as that cube applied to any given plane; write for V, and 2PS × LD for PE2; and DN will become as that is (because PS, AS, SI are continually proportional), as . If we draw then these three parts into the length AB, the first will generate the area of an hyperbola; the second the area ; the third the area , that is, . From the first subduct the sum of the second and third, and there will remain ANB, the area sought. Whence arises this construction of the problem. At the points L, A, S, B, erect the perpendiculars Ll Aa, Ss, Bb, of which suppose Ss equal to SI; and through the point s, to the asymptotes Ll, LB, describe the hyperbola asb meeting the perpendiculars Aa, Bb, in a and b; and the rectangle 2ASI, subducted from the hyperbolic area AasbB, will leave ANB the area sought. EXAMPLE 3. If the centripetal force tending to the several particles of the spheres decrease in a quadruplicate ratio of the distance from the particles; write for V, then for PE, and DN will become as . These three parts drawn into the length AB, produce so many areas, viz. into ; into ; and into . And these after due reduction come forth , [Pg 231], and . And these by subducting the last from the first, become . Therefore the entire force with which the corpuscle P is attracted towards the centre of the sphere is as , that is, reciprocally as Q.E.I. By the same method one may determine the attraction of a corpuscle situate within the sphere, but more expeditiously by the following Theorem. PROPOSITION LXXXII. THEOREM XLI.