PP: V 1 1 22.-If several bodies revolve in circles about the same or differ. ent centres, the centripetal forces are as the radii directly, and the squares of the periodic times reciprocally. Put the same letters as in Art. 8. Then (Art. 4,) P = st 2R ,2R 2R =tv and PP = 7* tt x and PPF = 26tR; 27attR R whence F = PP 23.-Cor. 1. The centripetal forces are as the velocities directly, and the periodic times reciprocally. R R V For (Art. 16,) VO; ) and Fa. PP P 24.-Cor. 2. The centripetal forces aie as the squares of the velocities directly, and the radii reciprocally. R For (Cor. 1,) Fa and FP a V. But (Art. 9,) Paj, therefore FP FR FR v R 25.-Cor. 3. If the centripetal forces are equal, the velocities are as the periodic times; and the radii as the squares of the periodic times, or as the sqnares of the velocities, 26.-Cor. 4.- If the centripetal forces be as the radii, the periodic times will be equal. R For Fa and and if be a given ratio, will be given, as PP' R PP R PP also P. 27.--Cor. 5. If the centripetal forces be reciprocally as the squares of the distances, the squares of the periodical times will be as the cubes of the distances, and the velocities reciprocally as the square roots of the distances. R R3 For, writing then and a given quantity. And P2 and RR R 28. If several bodies revolve in circles about the same or dif. ferent centres, the radii are directly as the centripetal forces, and the squares of the periodic times. 2R For, (Art. 4,) putting the same letters a: before, P = + V 2R 2R and PP = att x and PPF = 211t RaR. F F 29.-Cor. 1. The radii are directly as the velocities and periodic times. For (Art. 17,) PF & V, but PPF A R; therefore PV R. 30.-Cor. 2. The radii are as the squares of the velocities direcủly, and the centripetal forces reciprocally. VV For (Art. 10,) PO, but (Art. 29,) R o PV; therefore RO F: 31.-Cor. 3. If the radii are equal, the centripetal forces are as the squares of the velocities, and reciprocally as the squares of the periodic times; and the velocities reciprocally as the periodic times. 1 1 for F, ) , = For, if R be given, and PPF, and PV, are given quantities, and F F a VV, or P of and Tap PP SCHOLTUM. -The converse of all these propositions and corollaries are equally true; and, what is demonstrated of centripetal forces, is equally true of centrifugal forces, they being equal and contrary. 32.—The quantities of matter in all attracting bodies, having others revolving about them in circles, are us the cubes of the dis. tances directly, and the squares of the periodical times reciprocally. Let M be the quantity of matter in any central attracting body. Then, since it appears, from all astronomical observations, that the squares of the periodical times are as the cubes of the distances, of the planets, and satellites from their respective centres. Therefore (Art. 14,) the centripetal forces will be reciprocally as the squares of , Foc ) , at a given distance, is as the body M; therefore, the absolute force R M of the body M is as And (Art. 22,) since FQ RR put RR R R3 instead of F, and we have RRPP 33.-Cor. 1. Hence, instead of F in any of the foregoing propositions and M their corollaries, one may subsutute which is the force that the attract RR' ing body in C exerts at A. (Fig. 1.) 34.-Cor. 2. The attractive force of any body is as the quantity of matter directly, and the square of the distance reciprocally. II. The Motion of Bodies in all sorts of Curve Lines, M PP M and Mac S 35. The areas which a rerolving body describes by radii drawn to a fixed centre of force, are proportional to the times of descrip. tion, and are all in the same immoveable plane. Let S (fig. 3,) be the centre of force; and let the time be di. vided into very small equal parts. In the first part of time, let the body describe the line AB ; then, if nothing bindered, it would describe BK = AB, in the second part of time; and then the area ASB = BSK. But, in the point B, let the centripetal force act by a single, but strong, impulse, and cause the body to describe the live BC. Draw KC parallel to SB, and complete the parallelogram BKCr, then the triangle SBC = SBK, being between the same parallels; therefore SBC = SBA, and in the same plane. Also, the body moving uniformly would, in another part of time, describe Cm = CB; but at C, at the end of the second part of time, let it be acted on by another impulse, and carried along the line CD; draw mD parallel to CS, and D will be the place of the body after the third part of time ; and the triangle SCD = SCM = SCB, and all in the same plane. After the same manner, let the force act successively at D, E, F, &c. And, making Dn = DC, and Eo = ED, &c. and completing the parallelograms as before, the triangle CSm = CSD = DSn = DSE = ESO = ESF, &c. and all in the same immoveable plane. Therefore, in equal times, equal areas are described ; and, by compounding, the sum of all the areas is as the time of description. Now, let the number of triangles be io. creased, and their breadth diminished ad infinitum, and the centri. petal force will act continually, and the figure ABCDEF, &c. will become a curve; and the areas will be proportional to the times of description. 36.-Cor. 1. If a body describes areas, proportional to the times, about any point, it is urged towards that point by the centripetal force. For a body cannot describe areas, proportional to the times, about two different points or centres, in the same plane. 37.- Cor. 2. The velocity of a body revolving in a corve is reciprocally as the perpendicular to the tangent, in that point of the curve. For, the area of any of these little triangles being given, the base (which represents the velocity,) is reciprocally as the perpendicular. 38.—Cor. 3. The angular velocity at the centre of force is reciprocally as the square of its distance from that centre. Fig. 5. For, if the small triangles CSD and SBA (fig. 4,) be equal, they are de SC X CQ scribed in equal times The area CSD and area SBA P SB X BP -; therefore SCX CQ=SB X BP. But the angle CSD : angle ASB :: CQ:09 :: SC X CQ : SC X cq :: SB X BP: SC X 09 :: area SBA i area Scq :: SB2 : Sc? or SC?. 39.-If a body revolving in any curve VIL, (fig. 5,) be urged by a centripetal force tending towards the centre S, the centripetal force in any point I of the curve will be as where p = perpen pid dicular SP on the tangent at I, and d = the distance SI. For, take the point K infinitely near I, and draw the lines SI, SK; and the tangents IP, Kf; and the perpendiculars SP, Sf. Also draw Km, Kn, parallel to SP, SI, and KN perp. to SI. The triangles ISP, IKN, nKm, are similar ; as also IKm, IPq. Therefore, lq or IP : IK :: qP : Km. And PS : IP :: Km: mn. And IN : IK::mn : nk. And, inultiplying the terms of these three proportions, IP x PS X IN : IK x IP X IK :: P x Km x mn : Km x mn x nk. That is, PS X IN : IK? :: QP: nK = Pq. X IK2 But (Mechan. Prop. 6,) the space nk, through PS X IN which the body is drawn from the tangent, is as the force and square of the time; that is, (Art. 35,) as the force and square of the area (SK, or as the force x SI? x KN?, or because $I KN = twice the triangle ISK = IK * SP; therefore nK is as the nK force x IK x PSTherefore the force at I is as IKx PS Pq X IK? PS X IN X IK? x PS2 PS3 X IN nK NK 40.-Cor. 1. The centripetal force at I is as SI2X KNP SP2 x 1K2 SI XIN 41.-Cor. 2. Hence, the radius of curvature in I is = P9 IK X IP PY SI X IN = (by the similar triangles IPS, INK,) Pq 42.—To find the law of the centripetal force, requisite to make a body move in a given curve line. Let the distance SI = d, the perpendicular SP (upon the tangent at I,) = p; then, from the nature of the curve, find the value of p in terms of d, and substitute it and its Auxion, in the quantity p pa nK nK Or find the value of SI2 X KN2) X ' will give the law of centripetal force, by the last Prop. Ex. 1. If a body revolves in the circumference of a circle; to find the force directed to a given poist S. (Fig. 6.) Pq or as or whence I = 2r Is as mp3 1 Draw SI to the body at I, SP perp. to the tangent PI, SG perp. to the radins CI, Then SP = GI; because SGIP is a parallelogram. . Pot si = d, SP =p, SC = a, CI=1, CD= * ID being perp. to SD. Then, in the obtuse apgle SCI, ŞI? = SC2 + Cl2 + 2SCD, or dd = a + T + 241; dd ---- 44 — ". The triangles SCG and CID are similar, whenee dd CI (r): CD () :: SC (a); CG ; and p =It dd - aa – IT dd + rs aa dd Therefore the force d x 8r3 that is, the force is as jo x dd + ir au And, if « = r, the force is as Ex, 2. If a body revolves in an ellipsis ; to find the force tending 10 the centre C. (Fig. 7.) Let { transverse CV="; conjugate CD = c, draw CI =d, and its semiconjugate CR = b. Then, by the known properties of the ellipsis, kb + dd = rr + cc, whence b = V tr + cc – dd; and, again, 6 or vrt te - -dd crad :::-:p= and Therefore pd crda d Therefore the force is directly as rr + cc c3r3d the distance CI. After the same manner, the force tending to the centre of an hyperbola will -d be found which is a centrifugal force, directly as the distance. Ex. 3. If a body revolves in an ellipsis; to find the law of centripetal force, tending to the focus S. (Fig. 8.) cr ccrr сст Let the semitransverse OV = r, the semiconjugate OD , draw SI = el; and OJ, and its conjugate OK, = b. |