u60COR. 3. If the centripetal force of the particles be reciprocally as the cube of the distance of the corpuscle attracted by it, and DN be made as , the force with which the corpuscle is attracted by the whole sphere will be as the area ANB.

COR. 4. And universally if the centripetal force tending to the several particles of the sphere be supposed to be reciprocally as the quantity V; and DN be made as ; the force with which a corpuscle is attracted by the whole sphere will be as the area ANB.

PROPOSITION LXXXI. PROBLEM XLI.

The things remaining as above, it is required to measure the area ANB.

From the point P let there be drawn the right line PH touching the sphere in H; and to the axis PAB, letting fall the perpendicular HI, [Pg 229]bisect PI in L; and (by Prop. XII, Book II, Elem.) PE2 is equal to PS2 + SE2 + 2PSD. But because the triangles SPH, SHI are alike, SE2 or SH2 is equal to the rectangle PSI, Therefore PE2 is equal to the rectangle contained under PS and PS + SI + 2SD; that is, under PS and 2LS + 2SD; that is, under PS and 2LD. Moreover DE2 is equal to SE2 - SD2, or SE2 - LS2 + 2SLD - LD2, that is, 2SLD - LD2 - ALB. For LS2 - SE2 or LS2 - SA2 (by Prop. VI, Book II, Elem.) is equal to the rectangle ALB. Therefore if instead of DE2 we write 2SLD - LD2 - ALB, the quantity , which (by Cor. 4 of the foregoing Prop.) is as the length of the ordinate DN, will now resolve itself into three parts ; where if instead of V we write the inverse ratio of the centripetal force, and instead of PE the mean proportional between PS and 2LD, those three parts will become ordinates to so many curve lines, whose areas are discovered by the common methods. Q.E.D.

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.

In a sphere described about the centre S with the interval SA, if there be taken SI, SA, SP continually proportional; I say, that the attraction of a corpuscle within the sphere in any place I is to its attraction without the sphere in the place P in a ratio compounded of the subduplicate ratio of IS, PS, the distances from the centre, and the subduplicate ratio of the centripetal forces tending to the centre in those places P and I.

As if the centripetal forces of the particles of the sphere be reciprocally as the distances of the corpuscle attracted by them; the force with which the corpuscle situate in I is attracted by the entire sphere will be to the force with which it is attracted in P in a ratio compounded of the subduplicate ratio of the distance SI to the distance SP, and the subduplicate ratio of the centripetal force in the place I arising from any particle in the centre to the centripetal force in the place P arising from the same particle in the centre; that is, in the subduplicate ratio of the distances SI, SP to each other reciprocally. These two subduplicate ratios compose the ratio of equality, and therefore the attractions in I and P produced by the whole sphere are equal. By the like calculation, if the forces of the particles of the sphere are reciprocally in a duplicate ratio of the distances, it will be found that the attraction in I is to the attraction in P as the distance SP to the semi-diameter SA of the sphere. If those forces are reciprocally in a triplicate ratio of the distances, the attractions in I and P will be to each other as SP2 to SA2; if in a quadruplicate ratio, as SP3 to SA3. Therefore since the attraction in P was found in this last case to be reciprocally as PS3 × PI, the attraction in I will be reciprocally as SA3 × PI, that is, because SA3 is given reciprocally as PI. And the progression is the same in infinitum. The demonstration of this Theorem is as follows:

The things remaining as above constructed, and a corpuscle being in any[Pg 232] place P, the ordinate DN was found to be as . Therefore if be drawn, that ordinate for any other place of the corpuscle, as I, will become (mutatis mutandis) as . Suppose the centripetal forces flowing from any point of the sphere, as E, to be to each other at the distances and as to (where the number n denotes the index of the powers of PE and IE), and those ordinates will become as and whose ratio to each other is as to . Because SI, SE, SP are in continued proportion, the triangles SPE, SEI are alike; and thence IE is to PE as IS to SE or SA. For the ratio of IE to PE write the ratio of IS to SA; and the ratio of the ordinates becomes that of to . But the ratio of PS to SA is subduplicate of that of the distances PS, SI; and the ratio of to (because IE is to PE as IS to SA) is subduplicate of that of the forces at the distances PS, IS. Therefore the ordinates, and consequently the areas which the ordinates describe, and the attractions proportional to them, are in a ratio compounded of those subduplicate ratios. Q.E.D.

PROPOSITION LXXXIII. PROBLEM XLII.

To find the force with which a corpuscle placed in the centre of a sphere is attracted towards any segment of that sphere whatsoever.

Let P be a body in the centre of that sphere and RBSD a segment thereof contained under the plane RDS, and the sphærical superficies RBS. Let DB be cut in F by a sphærical superficies EFG described from the centre P, and let the segment be divided into the parts BREFGS, FEDG. Let us suppose that segment to be not a purely mathematical but a physical superficies, having some, but a perfectly inconsiderable thickness. Let that thickness be called O, and (by what Archimedes has demonstrated) that superficies will be as . Let us suppose besides the attractive forces of the particles of the sphere to be reciprocally as that power of the distances, of which n is index; and the force with which the superficies EFG attracts the body P will be (by Prop. LXXIX) as , that is, as . Let the perpendicular FN drawn into[Pg 233] O be proportional to this quantity; and the curvilinear area BDI, which the ordinate FN, drawn through the length DB with a continued motion will describe, will be as the whole force with which the whole segment RBSD attracts the body P. Q.E.I.