THE CURVILINEAR MOTION OF FREE POINTS ACTED ON BY ABSOLUTE FORCES OF ANY KIND.

PROPOSITION 86. PROBLEM.

707. To find the law of the force continually pulling downwards which can be constructed along the lines MP (Fig. 62) parallel to each other, in order that the body moves along a given curve AM, and to determine the speed of the body at individual points M. [p. 292]

SOLUTION. The normal AP is drawn through the line MP, AP is called x, and PM y. The curve AM = s and the radius os dsdy osculation MR at M = r, then r = ddx with ds taken as constant. Again the speed of the body at M corresponds to the height v, and the body force at M pulling along MP is put as P. From these established, these two equations are obtained : dv = − Pdy and Pr dx = 2vds (557), from which, if v is known, then P is itself immediately apparent. Therefore P is eliminated, and v is to be found from this dsdy equation : rdxdv = −2vdsdy , which with r replaced by its value ddx , becomes this 2 ddx , which integrated gives: l C − l v = l dx . Moreover, the speed is equation: − dv = 2dx 2 v ds known at the point A, and that corresponds to the height [v = ] c, and if the cosine of the angle MAP or the value of dx , with the point M falling at A, is equal to λ. Therefore this ds equation hence arises : l C = l c + 2l λ and consequently v = λ cds2 , 2 2 dx hence the speed of the body becomes known at the individual points M. Moreover the force P acting can be found to be : 3 ds , and the force is Or, if dx taken as constant, then in which case r = − dxddyEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 410 Finally the time, in which the arc AM is traversed, is equal to : Q.E.I. [Note that ds taken as constant enables us to write d ( dx / ds ) = ∫ dx / ds = log( dx / ds ). Also recall that ∫ ddx dx r=− ( 1+ y’ 2 )3 / 2 ds 3 ds 3 . ]

− 2 3 = − y" ddydx ( ddy / dx )dx Corollary 1. 708. Therefore whatever the force shall be acting, the body always progresses horizontally uniformly on account of the time to travel along AM to be proportional to AP itself, that has been observed above (579). [p. 293] Corollary 2. 709. If a perpendicular is sent from R to the line MP produced RS, and from S another perpendicular ST is sent to MR , and finally a third perpendicular TV is sent from T to MS, 3 then MV = rdx3 . Whereby the force P acting has the ratio to the force of gravity [which ds equals one] as 2λ2c to MV, or P is inversely as MV. [For there is a set of nested similar triangles, where dx/ds = cosθ = cos (TSR). Thus, MV = MT cos θ = MS cos 2 θ = MR cos3 θ . ] Corollary 3. 710. If the angle at A is right, then λ = 0. In which case the body must ascent straight up. But only if λ is made indefinitely small and c infinitely large, in order that 2λ2c has a finite value, is the body able to move by making use of this curve. Example 1. 711. Let the curve AM be a circle, the diameter of which is put on the axis AP, and the radius is equal to a. Thus r = a and ds : dx = a : y . On this account the force P becomes : P = 2λ a3 c and v = λ a2 c 2 y 2 2 2 y Therefore the force pulling downwards on M varies inversely as the cube of the upright MP and the speed varies inversely as this applied line itself [i. e. the y-coordinate]. Truly the height generating the speed at the maximum point of the periphery, where it becomes y = a, is equal to λ2c.EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 411 Example 2. 712. Let the curve AMC (Fig. 63) be a parabola, the axis of which CB is vertical and the parameter is equal to a. The horizontal line MQ is drawn and on putting CQ = t and MQ = z, then z 2 = at . Truly also dx = − dz and dy = − dt [p. 294] and as before, ( dt 2 + dz 2 ) = ds . On account of which, v = λ cds2 and P = 2λ cddt . Let the speed at the 2 2 2 dz 2 dz maximum point C correspond to the height b, and this is given, since ds = dz at C, , with dz taken as constant. From the equation by λ2c = b , and thus v = bds2 and P = 2bddt 2 2 z 2 dz dz 2 2 dz 2 2 zdz = at we have dt = a and ddt = a and ds 2 = dz 2 ( 1 + 4 z2 ) . a Consequently there is found : and P = 4ab . From which it is apparent that the force pulling downwards is constant, which is effective as the body progresses along the parabola. Whereby therefore, if this force is taken as the force of gravity equal to 1, it gives rise to b = a4 , equal to the distance of the focus from the vertex. Which are in agreement with what was found above (564 and onwards.) [This corresponds to our normal presentation of the parabola, y 2 = 4ax , where the focus lies on the semi-latus rectum AB, and CB is taken as equal to a; if here the vertex to focus distance CB is equal to b, while Euler’s a is equal to the length of the latus rectum, 2AB. ] Example 3. 713. Let MAN (Fig. 64) be a hyperbola with centre C described having the vertical axis CP. The semi-transverse axis AC = a and the semi- conjugate axis = e and CP = t , PM = z, and as above the height corresponding to the speed that the body has at A is equal to b, is as we have done 2 for the above parabola : v = bds2 and P = 2bddt 2 dz dz with dz taken as constant. Truly from the nature of the hyperbola, we have a 2 z 2 = −a 2e2 + e 2t 2 , from which there arises on differentiation :EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. 2 a zdz dt = 2 and [on differentiating and substituting these last two equations : ] et page 412 4 Consequently, the force is given by : P = 2a2 3b , or the force pulling the body downwards et everywhere at M is inversely proportional to the cube of the distance ML of the point M from the horizontal LC drawn through the centre C. Again we have : And we have besides : PROPOSITION 87. [p. 295] PROBLEM. 714. For the given curve AMB (Fig. 65) together with the centre of attraction C, to find the law of the centripetal force, which must act, in order that the body is free to move along that curve, and to find the speed of the body at any position M. SOLUTION. Since the curve AMB is given together with the point C, the equation is sought between the distance MC of any point of the curve M from the centre C and the perpendicular CT, that is sent from C to the tangent MT. Whereby with CM = y and CT = p the equation between p and y can be obtained. Now let the speed of the body at the point A correspond to the height c and the perpendicular from C sent to the tangent at A be equal to h. Truly of these unknown quantities, let the height corresponding to the speed at M be equal to v and the centripetal force at M is equal to P. With these put in place, we have 2 v = ch2 (589) and P = p then we have P = 2ch 2 dp (592). Or with the radius of osculation at M put equal to r p 3 dy 2ch 2 y (592). Q.E.I. p 3r Corollary 1. 715. Also the time, in which the body completes some arc AM, is equal to 2 ACM , or is h c proportional to the area ACM (588).EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 413 Corollary 2. 2 716. Since ch is a constant quantity [i. e. the angular momentum of the unit mass], then the centripetal force at some point M is proportional to this value [p. 296] this : y . Therefore the speed p 3r dp or to p 3 dy v is proportional to the reciprocal of the perpendicular CT sent to the tangent MT (589). [You may wish to remind yourself about r using the sketch supplied here, as in Ch. 5 Part a : note that Mmr and MTT’ are incremental triangles, and hence the lines TC, MR, and mR are treated as parallel. We hence have the similar triangles Mmr and CMT : giving dy / ds = MT / y and dy / dx = MT / p . Likewise, triangles MTT’ and RmM give : dp / MT = ds / r ; hence. r = MT .ds / dp = ydy / dp. Note that the first elemental triangle diminishes to a point, keeping its angles intact, while the second diminishes to a finite line segment which stays constant.] Example 1. 717. Let the given curve be an ellipse and the centre of force C placed in the centre of the ellipse. The semi-transverse axis is called a and the semi-conjugate axis b; from the ab . Therefore we have : nature of the ellipse it follows that p = 2 2 2 ( a +b − y ) dp = abydy ( a +b − y ) 2 2 2 3 2 and thus dp y

. On this account the centripetal force produced p 3 dy a 2b 2 is given by : P= 2ch 2 y , a 2b 2 which is therefore proportional to the distance from the centre of the ellipse.EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 414 Example 2. 718. Let the given curve again be an ellipse, but with the centre of force C placed in the other focus. The transverse axis of this is put equal to A and the latus rectum equal to L, ALy and from the nature of the ellipse we find that 4 pp = A− y . Hence of differentiating, we have 8 pdp = dp dy A2 Ldy A 2 L2 y 2 4 . Truly since 16 p

2 2 , then 3 = 2p Ly 2 ( A− y ) ( A− y ) and consequently 2 P = 4ch2 . Ly Therefore the centre of force is inversely proportional to the square of the distance of the body from the centre of force C. Example 3. 719. Let the curve be a logarithmic spiral and the centre of force C is placed at the centre of this; hence p = ny and 2 dp = 21 3 , and thus P = 2ch 3 2 3 . p dy n y n y Whereby the centripetal force varies inversely as the cube of the distance of the body from the centre. [p. 297] PROPOSITION 88. THEOREM. 720. The force pulling towards C (Fig. 66), that is put in place in order that the body moves along the given curve AM, has the ratio to the force pulling towards another centre of force c, which is put in place in order that the body can move around the same curve with the same periodic time, as the cube of the line cV from c to the tangent TM drawn parallel to the line CM, is to the volume formed from the line cM multiplied by the square of the line CM. DEMONSTRATION. Let the speed of the body at a given point A, with the body rotating around the centre of force C, correspond to the height c and the perpendicular sent from C to the tangent at A is equal to h. But when the body is moving around the centre of force c, let the speed at A correspond to the height γ and the perpendicular sent from the centre c to the tangent at A is equal to θ. Moreover since the periodic times around each centre of force are equal, then h c = θ γ or ch 2 = γθ 2 (715). From the centre C and from c perpendiculars CT and ct are again sent to the tangent at M, and the radius of osculation at M is equal to r. With these in place, the centripetal force at M pulling towards C, that weEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 415 2 ch . CM 2 call P, is equal to and the centripetal force at M pulling towards the centre c, that r .CT 3 2γθ 2 .cM (714). On account of this, since ch 2 = γθ 2 then we call Π, is equal to r .ct 3 Moreover with the line drawn cV parallel to the line CM, as triangles TCM and tcV are similar, we have CT : ct = CM : cV . Hence because of this, Q.E.D. [p. 298] Corollary 1. 721. The speed of the body at the same point M, while it is being attracted to the centre of forces C, to the speed, while it is being attracted to the other centre c, vary inversely as CT to ct or directly as cV to CM. This is a consequence of ch 2 = γθ 2 . Corollary 2. 722. If the periodic times are not equal, but are to each other as T to t, then T : t = 1 : 1 or ch 2 : γθ 2 = t 2 : T 2 . Consequently it follows that the ratio of the h c θ γ forces : Or the forces P and Π are in the ratio composed from the ratio assigned for equal periods in the theorem, and inversely as the square of the periodic times. Corollary 3. 723. In this same case, in which the periodic times are unequal, the speed of the body at M, with the centre of forces placed at C, to the speed at M, with the centre of forces placed at c, are in the reciprocal ration composed from the ratio of the perpendiculars CT and ct and in the ratio of the period times T : t. Scholium 1. 724. Newton deduced this Proposition in Book I, prop. VII, coroll. 3 of the Princ., and that was used to find the centripetal force acting at some point, from the known force acting at some other centre. Here we show the use of this in the following single example. [p. 299] Example. 725. Let the given curve be the circle AMc (Fig. 67), and one centre of forces is put at the centre of the circle C. Therefore the force P pulling everywhere towards C is constant and is called g. From this is sought the force acting towards a centre of forces c situated on the periphery and making Π, in order that the body is movingEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 416 around the circle in the same periodic time. Therefore the perpendicular cV is sent from c to the tangent MV, which from the nature of the circle is likewise parallel to the line CM. On account of which, g : Π = cV 3 : cM .CM 2 and thus The line AM is drawn, the triangles cVM and cMA are similar, as the angles cMV = 2 cAM and since cV : cM = cM : cA . Therefore cV = 2cM , from which the force becomes CM : Therefore this force Π varies inversely as the fifth power of the distance Mc of the body from the centre of forces c, now as was found above (692). Corollary 4. 726. Let the speed of the body rotating on the periphery of the circle around the centre C correspond to the height c and the speed of the body at M rotating around the centre of forces c correspond to the height v. Hence : 4 (721) or v = 4c .CM4 . cM On account of which the speed of the body rotating around the centre c is everywhere reciprocally as the square of the distance of that from c. [p. 300] [On account of cons. of ang. mom.] Corollary 5. 727. For, with the centre of force present at the centre C of the circle, then we have 2c (592). On this account, [from above] by h = y = p = r = radius CM, and P = g = CM putting Π = f5 f5 f5 , g

and c

. Thus cM 5 8CM 5 16CM 5 v= f5 . 4cM 4 Scholium 2. 728. In these propositions we have put in place the curve that the body describes, which is given completely with the equation for that curve. But there are also the cases, in which the curve itself is not given that describes the motion, but rather the motion itself must first be found by examining certain conditions, so that the law of the centripetal force can then be found. And here these propositions are concerned, and which have been treated everywhere, with the motion of bodies in moving orbits, concerning which therefore we treat in the following proposition. [We note that the fifth power situation with the centre of force lying on the orbit would not be a physically realizable situation.]EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 417 PROPOSITION 89. PROBLEM. 729. If the orbit ( A)(M)(B) (Fig. 65) is revolving around some centre of forces C, it is required to define the centripetal force always pulling towards C, which is put in place, in order that the body is moving in this moveable orbit. SOLUTION. While the orbit comes from the situation (A)(M)(B) to the situation AMB, the body meanwhile is put to have come from (A) to M, thus as the body in the orbit meanwhile describes the angle (A)C(M) = ACM, but actually the angle (A)CM = (A)C(M) + (A)CA [p. 301] has been described. Initially with the body present at (A), the true speed at this point is not that which it has in the orbit, but corresponds to the altitude c; and the line C(A) is perpendicular both to the orbit as well as to the true curve on which the body moves, and which is equal to a. Again the speed of the body at M , as much as it is moving in the orbit, corresponds to the height u and the true speed of the body at M corresponds to the height v. But the angular speed in the orbit to the true angular speed around C, while the body is moving at M, is as 1 to w. Therefore in the orbit considered stationary, the element (M)(m) is described with the speed u . The distance C(M) is put equal to CM = y and the perpendicular C( T ) = CT = p is sent from C to the tangent to the orbit at (M) or M, and an equation exists between p and y on account of the given orbit. Now while the body describes the element Mm in the orbit, the orbit itself progresses around C with the orbital angular motion through the angle mCμ , and on this account the body is found, not indeed at m, but instead at μ, by taking Cμ = Cm , and meanwhile the element Mμ is agreed to have been described by that speed which corresponds to the height v. Hence we have : and with the small [circular] arc Mv described with centre C (on account of the given angular motions about C in the orbit and in fact in the ratio 1 : w ) for which the ratio is : [Thus, the first proportionality describes the ratio of the displacements of the body in the orbit reference frame to the absolute reference frame in which the orbit itself rotates; theEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 418 second proportionality considers the corresponding angular speed by taking the corresponding arcs of a circle Mn and Mv, the first relative to the orbit reference frame and the second relative to the absolute reference frame .] ydy pdy Truly, with the tangent put in place : MT = ( y 2 − p 2 ) = q , Mm = q and Mn = q [In a small displacement along the orbital curve, CM corresponds to y and Cm corresponds to y + dy; hence mn is equal to dy in the elemental triangle mMn. The curvature remains the same on Mm, and the centre of curvature lies along the normal to the curve and the tangent at M, i. e. on a line parallel to CT (but not passing through C unless the orbit is a circular one); hence, the angle TCM is a measure of the angle between the direction of MC or y and the direction of the normal to the curve at M, which stays constant along Mm. Hence the angle Mmn is the complement of this angle, and hence the angle mMn is equal to the angle TCM. It follows that the triangles Mnm and MTC are similar; hence the ratio dy/Mm = TM/MC = q/y follows. Meanwhile, the arc Mn is traced out by the same elemental angle MCn or MCm, for which the constant radius is y, and the arc length Mn is to dy (the cotangent of the angle mMn or MCT) as p is to q as above. We have met this result several times already, the last occurrence in (716) above.] On which account we have : Mμ = ydy v q u and Mv = wpdy q [The multiples of the distance and angle gone through in the same time on the true curve. Hence, νμ 2 + Mν 2 = Mμ 2 or 1 + Mν2 = 2 dy Mμ 2 , leading to the next result. ] dy 2 From which as μν = mn = dy there is produced 1 + w2 p 2 vy 2

or q2 uq 2 Because Mμ is the element [p. 302] of the true curve that the body describes, the perpendicular CΘ is sent from C to this element produced, and the ratio becomes [on using arguments similar to the above for the equality of the angles] : Mμ : Mν = CM : CΘ , thus we have : 2 Truly from this perpendicular, the true speed of the body is known; for v = a2 cv2 (589) w up 2 [Recall that since for any curve, v = ch2 , and that the speed v = h p c ; here p = CΘ , and p 2 thus v = a2 cv2 and v cancels out.] w up With these values of u and v put in place, then which for the sake of brevity we call π. Moreover from this known value π the centripetal force P can itself be found, which comes about, in order that the body can move in thisEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 419 given orbit, and that the orbit in turn can be moved in this way. For it is given by P = 2a 3cdπ (592) [De Moivre’s Theorem]. But since π 2 = 2 π dy w2 p 2 y 2 , then q 2 + w2 p 2 2 [as q + p2 = y2 : ] and thus or Consequently we have : Q.E.I. Corollary 1. mμ 730. With the radius put equal to 1 then CM = ( w−1 ) pdy is the element of the angle qy (A)CA, that the orbit has completed, while the body travels through the arc (A)(M). For this reason the angle (A)CA= ∫ ( w−1 ) pdy . And ( w − 1 ) : 1 is as the angular speed of the qy orbital to the angular speed of the body, while it is at M in its orbit. Corollary 2. 731. The speed of the body in the orbit, which is as u , is inversely proportional to wp itself. Therefore, unless w is constant, this cannot happen, as the body in the stationary orbit may be moved by this force in this way, attracted to the centre C. [p. 303] [One would need to add another force as a cause for w to change.] Corollary 3. 732. Therefore with w constant, i. e. with the ratio of the angular motion of the body to the angular motion in the orbit always the same, also the speed of the body in the orbit u is inversely proportional to the perpendicular C(T) to the tangent. And the centripetal force attracting and being effective towards C , because the body is moving in a stationary orbit, is equal to 2 a 2 cdp . For the speed with respect to the orbit, that the body w 2 p 3 dy has at (A), corresponds to the height γ , is γ : c = 1 : w (by hypothesis) and c = w2γ , from which the centripetal force acting towards C, as the body in the stationary orbit is moved , is equal to 2a 2γ dp , as indeed has been found from the above treatment (591). p 3 dyEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 420 Corollary 4. 733. Therefore the angle (A)CA in this hypothesis, where w is put constant, which is completed by the orbit, while the body traverses the arc (A)(M), is equal to Hence for one complete revolution of the body in the orbit, the orbit itself rotates about C by the angle ( w − 1).360 degrees. Corollary 5. 734. Moreover the force, which it effects, as the body moving in this orbit proportionately to the angular motion in the orbit itself, is [p. 304] or is equal to Whereby the difference between the centripetal force for the stationary orbit and the force for the moving orbit is the latter part which is inversely proportional to the cube of the distance of the body from the centre of the forces C. Corollary 6. 735. If we put w = 1, then w − 1 = 0 and then there is no motion of the orbit, in which case also the centripetal force is equal to 2a 2γ dp with the other term vanishing. Likewise p 3 dy it comes about, if w = −1 or w − 1 = −2 , in which case the orbit in the preceding moves with twice the speed that the body itself moves in the orbit. [From the factor w – 1 in (734)] But the true curve, that in this motion is described by the body, does not differ from the orbit, except that it is in the opposite sense. Corollary 7. 736. If w > 1, then as a consequence the orbit is moving; and the greater this motion becomes, the greater also becomes the centripetal force. But if w < 1, with the orbit pulling in the opposite direction, and the centripetal force is less, as w2 − 1 is negative. Corollary 8. 737. If w = 0, then this makes c = 0, and the body is moving along a straight line, since in this case the angular motion of the orbit is equal and opposite to the angular motion of the body in its orbit.EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 421 Corollary 9. 738. If w is negative, surely equal to − n , then the body moves on the same curve as if [p. 305] w = + n , only with this distinction, that the body proceeds in the opposite sense. And on this account the centripetal force retains the same value, whether w takes positive or negative values. Likewise the same result holds generally, if w is a variable quantity. Example. 739. Let the curve (A)(M)(B) be an ellipse and the centre of forces C of this one of the focal points. The latus rectum of this is put equal to L and the transverse axis (A)(B) = A; the distance a [from the focus as origin] is given by : and [Note that the use of the eccentricity e had not yet been developed for conic sections, and conics were specified by their width and the length of the latus rectum or focal chord. In modern terminology, with the semi-latus rectum l = a( 1 − e 2 ) and b 2 = a 2 ( 1 − e 2 ) , and 2 the pedal equation of an ellipse is given by b 2 = 2ra − 1 , see Lockwood p. 21. Hence, p p 2 = 2ba −rr = 2 a ( 1− e )r ( A / 2 )( L / 2 ) y = 2alr

, in Euler’ s notation. The first result is 2a − r a −r A− y 2 2 easily found from the ‘constant length of string principle’ used to draw an ellipse, applied to the rt. triangle formed by the semi-latus rectum FP = L/2, the interfocal distance FF’, and the distance F’P, where FP + F’P = A.] If besides, w is constant; the force operating in order that the body travels in a rotating ellipse is equal to: dp 2 a 2γ ( w2 −1 ) 4 a 2γ + (734). [For, on differentiating the 4 p2 equation Ly 2 y3 above, we have dy = 2 p3 .] Truly the angle (A)CA, that the orbit completes, while the Ly 2 body travels through the arc (A)(M), is equal to ( w − 1 )( A )C( M ) (733). The equation for the curve itself that the body describes, of which the element is Mμ , is found by finding the equation between CM = y and CΘ = π . Moreover, we have and which values substituted in the equation π = the curve itself described: wpy ( qq + w 2 p 2 ) give this [pedal] equation forEULER’S MECHANICA VOL. 1. Chapter Five (part c). page 422 Translated and annotated by Ian Bruce. Scholium 1. 740. The curves themselves, which bodies describe acted on by centripetal forces of this kind, otherwise are most difficult to be recognised and their form from this consideration does not mean that in any way they can be determined. [p. 306] Therefore investigations of centripetal forces of this kind have the maximum use for curves generated by some given force, from which in turn, from the given centripetal forces, properties of the curves themselves can become known. For so complex expressions of the forces acting there occur in the motions of heavenly bodies, so that none of the orbits of these can be determined, except perhaps these forces can be understood in some such case, for which the orbits can be determined after the centripetal force has been found. Scholium 2. 741. If a body is taken to be moving in a moving orbit of this kind, the motion of this body and the distance at some time from the centre C can be determined. And just as often as the body in the orbit arrives at the points(A) and (B), so the distance from C is a minimum or a maximum. Whereby when the rotary motion of the line (A)(B), which is called the line of the apses, is given, it is possible to define when the distance from the centre C is a maximum or a minimum. Newton has explored this problem in the Principia, Book I, in the whole of Section IX, and that theory applies the motion of the apses in the determination of the moon’s orbit. But this examination is applied with less accuracy to the moon, since the lunar force is not acting at any fixed point, as we have put here, but is always exerted from some variable point. Therefore we will give the work, in order that, after we have explained relevant matters here, we will offer other more suitable propositions, which can be transferred to the motion of the moon. [p. 307] PROPOSITION 90. PROBLEM. 742. For the known curve, that a body describes acting under some central force V, to determine the curve, that the body describes acted on the centripetal force V + C3 , with y y denoting the distance MC of the body from the centre of forces C. [See also L. Euler Commentationem 232 (the Enestrom index) : De motu corporum coelestium a viribus quibuscunque perturbato, Novi comment. acam. se. Petrop. (1752/53); Leonardi Euleri Opera omnia, series II, vol. 21. P. St.] SOLUTION. With the centripetal force V + C3 acting, the speed of the body, which is projected at y (A) along a direction normal to the direction of the radius C(A), corresponding to the height c, and C(A) is put equal to a. Moreover with the force V acting, (A)(M)(B) is theEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 423 orbit, in which the body moves projected at (A) along the same direction, but with a speed corresponding to the height γ . Now from the preceding proposition it has been shown that a force of the form V + C3 is to be acting, in order that the body in the same orbit y (A)(M)(B), but moving around the centre C in a given ratio to the angular motion of the orbit. Therefore the ratio becomes w − 1 to 1, as the orbital angular motion to the angular motion of the body in that orbit, while it is at M, and also it is given that c = w2γ (732) and the perpendicular from C sent to the tangent of the orbit at M is called CT = p. Hence making the centripetal force equal to 2 a 2γdp , in order that the body moves in the p 3 dy stationary orbit (A)(M)(B), and thus as therefore we are able to construct the equation on account of the given curve (A)(M)(B) (by hypothesis). Moreover the force acting V, [p. 308] in order that the body in the same orbit can move in the manner described, is given by : (734). On account of which we have : From which there is produced : and Hence: and, [as c = w2γ ], Therefore the curve, that the body at (A) describes with the speed of projection corresponding to the height 2a c −2 C acted on by the force V, the force V + C3 acting on 2 2a y the body, in order that the body at (A) projected with the speed c is moving in the same mobile orbit thus, in order that the angular motion of the orbit to the angular motion of the body in this orbit, shall be as 2 a 2 c − ( 2 a 2 c −C ) ( 2a 2 c −C ) to 1. Moreover the motion of the body in that orbit is the same, as that which it has in a stationary orbit acted on only by the force V and projected with a speed at (A) corresponding to the height 2a c −2 C , which 2 2a motion by hypothesis is known. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 424 Corollary 1. 743. Therefore while the body reaches (B) from (A) in the orbit, or rotates around the centre C by an angle of 180, meanwhile the orbit itself has turned about C through an angle of Corollary 2. 744. Therefore if the line (A)(B) is the apse line, the point (A) is the closer point, and (B) truly the greater of the apses, as they are called in Astronomy; therefore the body arrives at the larger from the smaller apse in an absolute motion around C of [p. 309] degrees. Corollary 3. 745. The time, in which the body arrives at M from (A) in the moving orbit, is equal to the time, in which in the stationary orbit it reaches (M) from (A). Therefore the angle (A)CM has to the angle (A)C (M) the ratio w to 1, i. e. as Example. 746. Let the force V be inversely proportional to the square of the distance from the ff centre or V = yy , the curve (A)(M)(B) is an ellipse, in the focus of which is put the centre of the forces C. Let the transverse axis of this ellipse be (A)(B) = A and the latus rectum = L, then [as above] a = 12 A − 12 ( A2 − AL ) = ( A)C and ( B)C = 12 A + 12 ( A2 − AL ) . On ALy account of 4 pp = A− y then we have : Hence there becomes : 4a 2γ = Lff = 4a 2c − 2C , thus 4a cff− 2C and c = 2 Lff + 2C . 4a 2 Which is the height corresponding to the speed of the body at (A) for the orbit to be moving under the central force ff

• C3 . Truly the orbital angular motion is to the angular y2 y motion of the body in the orbit as :EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 425 to 1. And the body arrives at the further apse from the nearer one, after the angular motion has completed an angle of : degrees. [p. 310] PROPOSITION 91. PROBLEM.

747. If the figure that the body describes on being acted on by some central force does differ much from a circle, to determine the motion of the apses. SOLUTION. The motion of this is to be compared with the motion of the body in a moving ellipse with a small eccentricity, of which one or other focus is placed at the centre of the forces. Therefore in this stationary orbit the body is moving acted on by a centripetal force inversely proportional to the square of the distance. Truly the body is moving in the same orbit if the centripetal force is equal to ffy + C (746). With the preceding denominators y3 kept in place, and putting y = a + z ; where z is extremely small with respect to a, since the curve described by the body is put as nearly circular. Whereby the former centripetal aff + C + ffz , and as 2a is approximately equal to the latus rectum L. Now y3 the centripetal force acting is put equal to P3 , in which P is some function of y. The y force is equal to value a + z is put in place of y in P, and P can be changed into E + Fz by rejecting terms in which z has a dimension greater than one, on account of z being so small. Therefore this formula has to be compared with aff + C + ffz : F = ff or f = F and aF + C = E or C = E − aF . With these substituted, the body acted on by this centripetal force P3 comes from the smaller to the larger apse, with the angle y 2C ⎞⎟ degrees [p. 311] for the absolute angular motion (746). Or with 2a put in 180 ⎛⎜1 + Lff ⎝ ⎠ E degrees. If place of L and F in place of ff and C = E − aF , this angle becomes 180 aF indeed the orbit does not disagree much from being circular. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Five (part c). page 426 Translated and annotated by Ian Bruce. Corollary 1.
748. Truly the apsidal line (A)(B), while the body rotates through an angle of 360 degrees about C, is moved by the orbital angular motion through an angle E − aF .360 E degrees. For the angular motion of the orbital is put proportional to the angular motion of the body on account of the centripetal force ff

• C3 y2 y (734). Corollary 2.

749. Since E is such a function of a, just as P is of y, Fz is the increment of E from the increase of a by the element z. Whereby on putting z = da then Fda = dE , and likewise the angle, by which the body arrives at the larger apse from the smaller one, is equal to Eda degrees. 180 adE Corollary 3. Eda in place of a and P
750. Since E is such a function of a, as P is of y, y can be put in adE in place of E. On account of which by the presence of the centripetal force P3 , the body y pdy moves from the small apse to the large apse in an absolute angle of 180 ydp degrees. And if y remains in this expression, a can be written in place of y, clearly with a small discrepancy. [p. 312] Corollary 4. dy
751. If we set da

dE or y > dP , then the ellipse by its own motion expresses the true a E P dy Eda > 1 , etc]. For if < dP , then the line motion of the body by moving forwards [as adE y P dy or P = αy , in which case the centripetal of the apses moves backwards. But if y = dP P force is inversely proportional to the square of the distance, and the apse line remains at rest, or the body, after the angular motion has completed 180 degrees, has gone in turn from the nearer apse to the further apse. Corollary 5.

752. Moreover for a given angle, in which the body goes from one apse to the other and back again, which is 360μ degrees, we have μ 2 = ydP and likewise P μμ = αy or Pdy 1 P = ( αy ) μμ .EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 427 P Therefore the centripetal force 3 which makes the apsidal line have such a motion, is y y 1−3 μ 2 μ2 . [P is thus the function of y required in the proposition]. Corollary 6. Pdy
753. If it happens, that ydP or dP is made negative, then the motion of the apses is imaginary. From which it is understood that the body is never able to proceed from one apse to the other, but continually either recedes further from the centre or approaches the centre, or evidently it is in a closed orbit that does not change. Corollary 7.
754. If the centripetal force is proportional to the power of the distance y n , then P = y n +3 . Whereby ydP = n 1+3 , [p. 313] and the body goes from the closer to the further Pdy apse while going through the absolute angle about C of 180 ( n+3 ) degrees; and to go from the greater or smaller apse to return to the same by going through an angle of 360 degrees. ( n+3 ) Corollary 8.
755. If ( n + 3 ) is a rational number and m the smallest whole number, from which m makes some whole number, then after completing m revolutions about the ( n+3 ) ( n+3 ) centre C the body has fallen on the same point and the curve described by the body has completed as many whole turns before it is restored and closed. But if n + 3 is not a perfect square, then the curve can never be restored to its starting point, but the body indefinitely travels around the centre C, and neither does it at any time revert to the same path. Example 1.
756. If the centre of forces attracts in the inverse cube of the distances, then n + 3 = 0. Hence in this hypothesis, the body cannot travel from one apse to the other except by completing an infinite number of revolutions. And if the centripetal force decreases in a greater ratio than the third power of the distances, the curve clearly does not have two apses, but will either go to infinity or terminate as a logarithmic spiral towards the centre. [p. 314] Example 2.
757. If the centripetal force is inversely proportional to the square of the distance, then n + 3 = 1. Whereby while the body by the absolute angular motion completes an angle of 180 degrees it travels from one apse to the other, and the curve after any revolutionEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 428 returns to its initial condition. For the body it travelling in an ellipse, in either focus of which is placed the centre of forces, and the transverse axis of this ellipse is the line of the apses. Example 3.
758. If the centripetal force varies inversely as the distance, then n + 3 = 2. Therefore the body arrives at one apse from the other after the orbit has turned through an angle of 180 2 or 127 degrees and 17’. Truly the orbit never returns on itself on account of the irrational 2 . Example 4.
759. If the centripetal force is constant at all distances, then n = 0. In this case the body goes from one apse to the other while the angular motion carries the orbit through an angle of 180 degrees, i. e. 103 degrees and 55’ approximately. 3 Example 5.
760. If the centripetal force varies directly as the distance of the body from the centre, in which case the body is agreed to be moving in an ellipse, in the centre of which the centre of force has been placed. (631). Therefore with the apses standing apart by an angle of 90 degrees. The same can truly be deduced from this rule. for on account of n = 1 the angle is 180 = 90 . [p. 315] n +3 Scholium 1.
761. Therefore as often as the body is projected around the centre of force with such a velocity, so that it almost revolves in a circle, with the help of this proposition the true curve that the body describes can be determined, which is not possible by considering the centripetal force alone. For which it is evident on greater contemplation that there are more uses of this kind for these orbits, by determining other more difficult orbits from these that can easily be defined. Newton has set out the same proposition in Sect. IX , Prop. 45. Scholium 2.
762. Now we have shown above that a body acting under a hypothetical centripetal force varying inversely as the cube of the distance falling to the centre, arrives in a finite time and does not to escape from that point, but as it were, to be annihilated suddenly (675 and 676). The same also prevails if the body falls to the centre along a straight line. And in a similar manner, if the centripetal force decreases in a ratio greater than the inverse cube of the distance, the body at once arrives at the centre, where it vanishes and neither progresses further than the centre nor returns. For whatever you please, it would be absurd for the curve that the body describes, projected with a certain velocity, to have two apses (756). Moreover as often as the centripetal force decreases in a ratio less than the cube, as in the simple ratio of the distance or greater than that, the body recedes, after it arrives at the centre, [p. 316] along the same line by which it approached; for this isEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 429 evident for the inverse square of the distance (655) and for the simple ratio, from which it is apparent (271) that the body cannot progress beyond the centre. But if n + 1 > 0, the body falling towards the centre along a straight line has a finite speed, by which beyond the centre it can progress along the same straight line, as long as it was losing speed (273). Therefore in this way we have satisfied the above desire (272), in which it was necessary to define the linear motion of the falling body, when it arrived at the centre.EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 430 CAPUT QUINTUM DE MOTU CURVILINEO PUNCTI LIBERI A QUIBUSCUNQUE PONTENTIIS ABSOLUTIS SOLLICITATI . PROPOSITIO 86. PROBLEMA.
763. Invenire legem vis perpetuo deorsum secundum rectas MP (Fig. 62) inter se parallelas tendentis, quae faciat, ut corpus in data curva AM moveatur, atque determinare corporis in singulus locis M celeritatem. [p. 292] SOLUTIO. Per rectas MP ducatur normalis AP, et vocetur AP x et PM y. Ponatur curva AM = s et radius osculi MR in M = r, dsdy erit r = ddx sumto ds pro constante. Porro debita sit corporis in M celeritas altitudini v, et vis corpus in M trahens secundum MP ponatur P. Ex his habebuntur duae istae aequationes dv = − Pdy et Pr dx = 2vds (557), ex quibus, si cognita esset v, statim apparet quantitas ipsius P. Eliminatur igitur P ad v inveniendum ex hac aequatione dsdy rdxdv = −2vdsdy , quae posito loco r eius valore ddx abit 2 ddx , quae integrata dat l C − l v = l dx . Cognita autem sit celeritas in in hanc − dv = 2dx 2 v ds puncto A, eaque debeatur altitudini c, atque sit consinus anguli MAP seu valor ipsius dx incidente puncto M in A = λ. Hinc ergo erit l C = l c + 2l λ et consequenter ds v = λ cds2 , 2 2 dx unde corporis in singulis locis M celeritas innotescit. Vis autem sollicitans P reperietur 3 ds , erit Sive sumto dx pro constante, quo casu est r = − dxddyEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 431 Tempus denique, quo arcus AM percurritur, erit Q.E.I. Corollarium 1.
764. Quaecunque ergo sit vis sollicitans, corpus perpetuo aequabiliter secundum horizontem progreditur ob tempus per AM proportionale ipsi AP, uti iam supra est observatum (579). [p. 293] Corollarium 2.
765. Si ex R in rectam MP productam demittatur perpendicularis RS et ex S in MR quoque perpendicularum ST atque denique tertium perpendiculum TV ex T in MS, erit MV = rdx3 . Quare vis sollicitans P se habebit ad vim gravitatis ut 2λ2c ad MV, seu P est 3 ds reciproce ut MV. Corollarium 3.
766. Si angulus ad A est rectus, fiet λ = 0. Quo casu corpus directe sursum ascendere debebit. At si tantum sit λ infinite parvum atque c infinite magnum, ita ut 2λ2c finitum habeat valorem, corpus in huiusmodi curva utique moveri poterit. Exemplum 1.
767. Sit curva AM circulus, cuius diameter posita sit in axe AP, et radius = a. Erit itaque r = a et ds : dx = a : y . Hanc ob rem fiet P = 2λ a3 c et v = λ a2 c 2 y 2 2 2 y Vis ergo corpus in M deorum trahens est reciproce ut cubis applicatae MP et celeritas reciproce ut haec ipsa applicata. Altitudo vero generans celeritatem in summo peripheriae puncto, ubi fit y = a, est = λ2c. Exemplum 2.
768. Sit curva AMC (Fig. 63) parabola, cuius axis CB est verticalis et parameter = a. Ducatur horizontalis MQ et ponatur CQ = t et MQ = z, erit z 2 = at . Praeterea vero erit dx = − dz et dy = −dt [p. 294] et ut ante ( dt 2 + dz 2 ) = ds . Quocirca fiet v = λ cds2 and P = 2λ cddt . Debita sit celeritas in puncto summo C altitudini b, eritque, 2 2 dz 2 2 dzEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 432 2 2 bds bddt ob ds = dz in C, λ c = b ideoque v = 2 et P = , sumto dz pro constante. Ex dz dz 2 2 2 et ddt = 2dz atque ds 2 = dz 2 ( 1 + 4 z2 ) . aequatione vero z 2 = at fit dt = 2 zdz a a a 2 Consequenter invenitur et P = 4ab . Ex quo apparet potentiam deorsum tendentem, quae efficit, ut corpus in hac parabola progrediatur, esse constantem. Quare igitur, si aequalis sit gravitati = 1, fiet b = a4 = distantiae foci a vertice. Quae conveniunt cum supra inventis (564 et sqq.) Exemplum 3.
769. Sit curva MAN (Fig. 64) hyperbola centro C descripta habens axem CP verticalem. Ponatur semiaxis transversus AC = a et semiaxis coniugatus = e atque CP = t ac PM = z, sitque insuper altitudo celeritati, quam corpus in A habet, debita = b, erit ut supra pro parabola fecimus, sumto dz constante. Est vero ex natura hyperbola a 2 z 2 = −a 2e 2 + e 2t 2 , ex qua fit dt = a zdz et 2 2 et Consequenter erit 4 P = 2a2 3b , et seu potentia corpus deorsum trahens ubique in M proportionalis est reciproce cubo distantiae ML puncti M ab horizontali LC per centrum C ducta. Porro erit Atque praeterea habebiturEULER’S MECHANICA VOL. 1. Chapter Five (part c). page 433 Translated and annotated by Ian Bruce. [p. 295] PROPOSITIO 87. PROBLEMA.
770. Data curva AMB (Fig. 65) una cum centro virium C invenire legem vis centripetae, quae faciat, ut corpus in hac curva libere moveatur, ut et celeritatem corporis in loco quavis M. SOLUTIO. Quia curva AMB una cum puncto C est dat, quaeratur aequatio inter distantiam MC cuiusque curvae puncti M a centro C et perpendiculum CT, quod ex C in tangentem MT demittitur. Quare posita CM = y et CT = p habebitur inter p et y aequatio. Iam sit corporis in dato loco A celeritas debita altitudini c atque perpendiculum ex C in tangentem in A demissum = h. Eorum vero, quae sunt incognita, vocetur altitudo debita celeritati in M = v et vis 2 centripeta in M = P. His positis erit v = ch2 (589) atque P = p radio osculi in M = r erit P = 2ch 2 dp (592). Vel posito p 3 dy 2 2ch y (592). Q.E.I. p 3r Corollarium 1.
771. Tempus etiam, quo corpus quemvis arcum AM absolvit, erit = 2 ACM seu erit h c proportionale areae ACM (588). Corollarium 2. 2
772. Quia ch est quantitas constans, erit vis centripeta in puncto quovis M proportionalis [p. 296] huic valori dp y seu huic 3 . Celeritas vero 3 p dy pr v proportionalis est reciproce perpendiculo CT in tangentem MT demisso (589). Exemplum 1.
773. Sit curva data ellipsis et centrum virium C in ipso eius centro positum. Vocetur eius ab . semiaxis transversus a et semiaxis coniugatus b; erit ex natura ellipsis p = 2 2 2 ( a +b − y ) Habebitur ergo dp = abydy ( a 2 +b 2 − y 2 ) 3 2 ideoque P= dp y = 2 2 . Quocirca prodibit vis centripeta 3 p dy a b 2ch 2 y , a 2b 2 quae igitur proportionalis est distantiae corporis a centro.EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 434 Exemplum 2.
774. Sit curva data iterum ellipsis, at centrum virium C in eius alterutro foco positum. Ponatur eius axis transversus = A et latus rectum = L, eritque ex natura ellipsis ALy 4 pp = A− y . Differentiando ergo fit 8 pdp = dp dy = 2 2 p3 Ly A2 Ldy A 2 L2 y 2 4 = , erit 2 . Quia vero est 16 p ( A− y ) ( A − y )2 et consequenter 2 P = 4ch2 . Ly Vis igitur centripeta reciproce erit proportionalis quadrato distantiae corporis a centro virium C. Exemplum 3.
775. Sit curva spiralis logarithmica et centrum virium C in eius centro positum; erit p = ny et 2 dp = 21 3 , ideoque P = 2ch 2 3 . 3 p dy n y n y Quare vis centripeta erit reciproce ut cubus distantiae corporis a centro. [p. 297] PROPOSITIO 88. THEOREMA.
776. Vis tendens ad centrum C (Fig. 66), quae facit, ut corpus in data curva AM moveatur, se habet ad vim tendentem ad aliud centriud c, quae facit, ut corpus in eadem curva et eodem tempore periodico moveatur, ut cubis rectae cV ex c ad tangentem TM parallele rectae CM ductae ad solidum ex recta cM in quadratu rectae CM. DEMONSTRATIO. Sit corporis celeritas in dato puncto A, cum corpus circa centrum virium C revolvitur, debita altitudini c et perpendiculum ex C in tangentem in A demissum = h. At cum corpus circa centrum virium c movetur, sit celeritas in A debita altitudini γ et perpendiculum ex centro c in tangentem in A demissum = θ. Quia autem tempora periodica circa utrumque virium centrum sunt aequalia, erit h c = θ γ seu ch 2 = γθ 2 (715). Ex centro C et c porro in tangentem in M demittantur perpendicula CT et ct, sitque radius osculi in M = r. His positis erit vis centripeta in M ad centrum C tendens, 2 atque vis centripeta in M ad centrum c tendens, quam quam vocemus P, = 2ch .CM 3 r .CT vocemus Π, = 2γθ .cM (714). Quamobrem ob ch 2 = γθ 2 erit r .ct 3 2EULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 435 Ducta autem cV parallela rectae CM erit, ob triangula TCM et tcV similia, CT : ct = CM : cV . Hance ob rem erit Q.E.D. [p. 298] Corollarium 1.
777. In eodem puncto M erit celeritas corporis, dum ad virium centrum C attrahitur, ad celeritatem, dum ad alterum centrum c attrahitur, reciproce ut CT ad ct sive directe ut cV ad CM. Sequitur hoc ex eo, quod est ch 2 = γθ 2 . Corollarium 2.
778. Si tempora periodica non sint aequalia, sed sint inter se ut T ad t, erit T : t = 1 : 1 seu ch 2 : γθ 2 = t 2 : T 2 . Consequenter erit h c θ γ Seu vires P et Π erunt in ratione composita ex ratione in theoremate assignata et inversa duplicata temporum periodicorum. Corollarium 3.
779. Eodem hoc casu, quo tempora periodica sunt inaequalia, erit celeritas in M, centro virium in C posito, ad celeritatem in M, centro virium in c posito, in ratione reciproca composita ex ratione perpendiculorum CT et ct et ratione temporum periodicorum T : t. Scholion 1.
780. Propositionem hanc Neutonus deduxit ex Lib. I. prop. VII in coroll. 3 Princ. eaque utitur ad inveniendam vim centripetam tendentem ad punctum quodcunque ex cognita vi ad aliud quodpiam centrum trahente. Nos hic usum eius in unico exemplo sequente ostendemus.

Exemplum. 725. Sit curva data circulus AMc (Fig. 67), alterumque centrum virium positum sit in ipso circuli centro C. Vis igitur centripeta P ad C tendens ubique erit constans dicaturque g. Ex hac quaeratur vis ad centrum virium c in peripheria situm tendens Π faciensque, ut corpus eodem tempore periodico in circulo moveatur. Demittatur ergo ex c in tangentem MV perpendiculum cV, quod ex natura circuli simul parallelum erit rectae CM. Quamobrem erit g : Π = cV 3 : cM .CM 2 ideoque Ducatur recta AM, erunt triangula cVM, cMA ob ang.cMV = cAM similia et propterea 2 cV : cM = cM : cA . Habetur ergo cV = 2cM , ex quo fit CM Est igitur haec vis Π reciproce ut potestas quinta distantiae Mc corporis a centro virium c, ut iam supra (692) est inventum. Corollarium 4. 726. Sit celeritas corporis in peripheria circuli circa centrum C revolventis debita altitudini c et celeritas corporis in M circa centrum virium c revolventis debita altitudini v. Eritque 4 (721) seu v = 4c .CM4 . cM Quamobrem celeritas corporis circa centrum c revolventis erit ubique reciproce ut quadratum distantiae eius ab c. [p. 300] Corollarium 5. 727. Quia, centro virium in centro circuli C existente, est h = y = p = r = radio CM, erit 2c (592). Hanc ob rem fiet, posito Π = P = g = CM v= f5 f5 f5 , g

et c

. Ideoque cM 5 8CM 5 16CM 5 f5 . 4cM 4 Scholion 2. 728. In his propositionibus posuimus curvam, quam corpus describit, absolute esse datam et aequationem pro ea haberi. Sed dantur etiam casus, quibus curva ipsa, quam corpus describit, non datur, sed ex certis conditionibus ad ipsum motum spectantibus ante debet inveniri, quam lex vis centripetae potest determinari. Hucque pertinent, quae passimEULER’S MECHANICA VOL. 1. Chapter Five (part c). Translated and annotated by Ian Bruce. page 437 tradita sunt de motu corporum in orbibus mobilibus, qua de re igitur in sequenti propositione tractabimus.

PROPOSITIO 89. PROBLEMA.

729. Si orbita ( A)(M)(B) (Fig. 65) utcunque revolvatur circa centrum virium C, oportet definiri vim centripetam perpetuo ad C tendentem, quae faciat, ut corpus in hac orbita mobili moveatur. SOLUTIO. Dum orbita ex situ (A)(M)(B) in situm AMB pervenit, ponatur corpus interea ex (A) in M pervenisse, ita ut corpus interea in orbita angulum (A)C(M) = ACM, revera autem angulum (A)CM = (A)C(M) + (A)CA [p. 301] descripserit. Existente initio corpore in (A) sit eius celeritas vera, non ea, quam in orbita habet, debita altitudini c, et recta C(A), quae tam in orbitam quam in veram curvam, in qua corpus movetur, sit perpendicularis, = a. Porro sit celeritas corporis in M , quatenus in orbita movetur, debita altitudini u et vera corporis celeritas in M debita altitudine v. At celeritas angularis in orbita sit ad veram celeritatem angularem circa C, dum corpus in M versatur, ut 1 ad w. In orbita igitur tanquam immobili spectata elementum (M)(m) celeritate u describetur. Ponatur distantia C(M) = CM – y et perpendiculum in tangentem orbitae in (M) vel M ex C demissum C( T ) = CT = p , habebiturque ob orbitam datam aequatio inter p et y. Iam dum corpus in orbita elementum Mm describit, progrediatur ipsa orbita motu angulari circa C per angulum = mCμ , et hanc ob rem corpus reipsa non in m, sed in μ reperietur, sumto Cμ = Cm , atque idcirco interea elementum Mμ descripsisse censendum est, id quod fecit celeritate debita altitudini v. Erit itaque atque centro C descripto arculo Mv (ob datam motuum angularium circa C in orbita et revera rationem 1 : w ) erit Est vero, posita tangente MT = ( y 2 − p 2 ) = q ,EULER’S MECHANICA VOL. 1. Chapter Five (part c). page 438 Translated and annotated by Ian Bruce. Quocirca habebitur Ex quo ob μν = mn = dy prodibit 1 + w2 p 2 v2 y2 = 2 seu 2 q uq Quia Mμ est elementum [p. 302] verae curvae, quam corpus describit, demittatur in hoc productum ex C perpendiculum CΘ , eritque Mμ : Mν = CM : CΘ , unde fit 2 Ex hoc vero perpendiculo cognoscitur vera corporis celeritas ; erit enim v = a2 cv2 (589) w up ideoque His loco u et v positis valoribus erit quam brevitatis gratia vocemus π. Ex hac autem π cognita innotescit ipsa vis centripeta P, quae facit, ut corpus in hac data orbita hocque modo mobili moveatur. Namque erit P = 2a 3cdπ (592). At ob π 2 = 2 π dy w2 p 2 y 2 , erit q 2 + w2 p 2 ideoque seu Consequenter habebitur Q.E.I.

Corollarium 1. mμ 730. Posito radio = 1 est CM = ( w−1 ) pdy elementum anguli (A)CA, quem orbita confecit, qy dum corpus arcum (A)(M) percurrit. Hanc ob rem erit ang. (A)CA= ∫ ( w−1 ) pdy . Estque qy ( w − 1 ) : 1 ut celeritas angularis orbitae ad celeritatem angularem corporis, dum est in M, ipsa orbita.EULER’S MECHANICA VOL. 1. Chapter Five (part c). page 439 Translated and annotated by Ian Bruce. Corollarium 2. 731. Cereritas corporis in orbita, quae est ut u , reciproce proportionalis est ipsi wp. Ergo, nisi w sit constans, fieri non potest, ut corpus hoc modo in orbita quiescente moveatur attractum ad centrum C. [p. 303] Corollarium 3. 732. Posito igitur w constante, i. e. ratione motus angularis corporis ad motum angularem orbitae perpetuo eadem, erit etiam celeritas corporis in orbita u reciproce proportionalis perpendiculo C(T) in tangentem. Atque vis centripeta ad C tendens atque efficiens, ut corpus hac ratione in orbita quiescente moveatur, erit = 2 a 2 cdp . Sit enim w 2 p 3 dy celeritas respectu orbitae, quam corpus in (A) habet, debita altitudini γ , erit γ : c = 1 : w (p. hyp.) atque c = w2γ , ex quo vis centripeta ad C tendens faciensque, ut corpus in orbita quiescente moveatur, erit = 2 a 2γdp , ut etiam ex supra traditis invenitur p 3 dy (591).

Corollarium 4.

733. Angulus igitur (A)CA in hac hypothesi, qua w ponitur constans, qui ab orbita absolvitur, dum corpus arcum (A)(M) percurrit, erit Ergo una tota corporis in orbita revolutione ipsa orbita circa C gyrabitur angulo ( w − 1 )360 graduum. Corollarium 5.
734. Vis autem, quae efficit, ut corpus in hac orbita mobili proportionaliter motui anguli in ipsa orbita moveatur, erit [p. 304] Quare differentia inter vim centripetam pro orbita immobili et vim pro orbita mobili reciproce proportionalis est cubo distantiae corporis a centro virium C. Corollarium 6.
735. Si sit w = 1, erit w − 1 = 0 motusque orbitae nullus, quo casu etiam vis centripeta fit = 2 a 2γdp evanescente altero termino. Idem evenit, si w = −1 seu w − 1 = −2 , quo casu p 3 dy orbita in antecedentia movetur duplo velocius, quam ipsum corpus in orbita ingreditur. At vera curva, quae hoc motu a corpore describitur, non differt ab orbita, nisi quod sit inversa.

Corollarium 7.

736. Si w > 1, orbita in consequentia movetur; qui motus quo sit maior, eo maior etiam erit vis centripeta. At si w < 1, orbita in antecedentia tendit, et vis centripeta fit minor, ob w2 − 1 negativum.

Corollarium 8.

737. Si w = 0, fit etiam c = 0, corpusque in recta linea movebitur, quia motus angularis orbitae hoc casu aequalis fit et contrarius motui angulari corporis in orbita.

Corollarium 9.

738. Si w est numerus negativus, nempe = − n , corpus in eadem movebitur curva, [p. 305] ac si esset w = + n , hoc tantum discrimine, quod corpus in contrarius plagas progrediatur. Et hanc ob rem vis centripeta eundem retinet valorem, sive w affirmative sive negative accipiatur. Idem etiam universaliter, si w est quantitas variabilis, obtinet. Exemplum.

739. Sit curva (A)(M)(B) ellipsis et centrum virium C eius alteruter focus. Ponatur eius latus rectum = L et axis transversus (A)(B) = A; erit Sit praeterea w constans; erit vis, quae facit, ut corpus in hac ellipsi mobili moveatur, = 2 a 2γ ( w2 −1 ) 4 a 2γ

(734). Angulus vero (A)CA, quem orbita absolvit, dum corpus in ea Ly 2 y3 arcum (A)(M) percurrit,erit = ( w − 1 )( A )C( M ) (733). Aequatio vero pro ipsa curva, quam corpus describit, cuius elementum est Mμ , habebitur invenienda aequatione inter CM = y et CΘ = π . Est autem qui valores in aequatione π = wpy ( qq + w 2 p 2 ) substituti dabunt aequationem pro ipsa curva descripta hanc

Scholion 1.

740. Curvae ipsae, quas corpora a huiusmodi viribus centripetis sollicitata describunt, difficillime alias cognoscerentur earumque forma hac consideratione non adhibita nequaquam posset determinari. [p. 306]Maximam igitur habent utilitatem huiusmodi virium centripetarum investigationes pro curvis ex datis utcunque generatis, quo reciproce ex viribus centripetis datis ipsae curvae earumque proprietates innotescant. Occurrunt enim in motibus corporum coelestium tam complexae virium ea sollicitantium expressiones, ut omnino eorum orbitae determinari nequaeant, nisi forte illae vires comprehendantur in tali quodam casu, de quo a posteriori vis centripeta est inventa.