Proposition 32

Table of Contents

PROPOSITION 32. Problem

If the body is attracted by some force to the centre of force C (Fig. 37), to find the curve AM on which the body is carried with a speed equal to the uniform motion towards C.

Solution

Let the minimum speed of the body at A correspond to the height b, the line CA is a tangent to the curve at A, since the body at A ought to move directly to C. Let AC = a and CM = x, the speed at M corresponds to the height v and the centripetal force at M is equal to P; then we have v = b − Pdx , as the integral has to be taken in order that … with x = a it vanishes and makes v = b. Now the speed at M has to be of such a magnitude, since the element Mm is completed in the same increment of time as the element Pp with the speed b .

Hence it becomes b : v = Pp : Mm = MT : MC , hence this equation is produced [for the triangles Mnm and MTC are similar] : The perpendicular CT to the tangent is equal to p; it becomes

…

Or, if the sine of the angle ACM is put equal to t, it dt.

Hence the following equation is produced … becomes [on differentiation]: MC … and x 2 − p 2 = Pdx , etc. ] : …

Of which for each case, if P is given in terms of x, a suitable curve can be constructed. Q.E.I

Corollary 1

If the centripetal force is proportional to some power of the distances, clearly … then we have :

…

With this substituted we have the following equation for the curve AM :

Which equation must thus be integrated, so that with t = 0 it becomes x = a.

Corollary 2

If ∫ ( 1dt−tt ) is thus taken, so that it becomes equal to zero if t = 0, there is produced by integration from that equation: if with the centre C and with the radius BC = 1, the arc of the circle BSs is described. From which it is apparent that the curve AM has an infinite number of rotations before it arrives at C. For with x = 0, BS is made ∝ .

Corollary 3

Therefore the construction of this curve depends in part on the quadrature of the circle and in part on logarithms, if n + 1 is a positive number. But if n + 1 is a negative number, that term which was given by a logarithm, is also reduced to the quadrature of the circle.

Corollary 4

This curve has a turning point [flexus contrarii]where dp = 0. [p. 132] Therefore in order that this point can be found, the equation is taken ; from which by differentiation, and on putting dp = 0, there is produced

Corollary 5

n 294. Therefore in the case in which P = x n , the turning point is at the place where : f Thus, this equation arises: Which on substitution into the integration gives the angle ACM at which the turning point lies.

Scholium 1

But since from the nature of these kinds of curves it is difficult to produce turning points in general, it is considered that the principles are best effected by descending to the level of special cases.

Example 1

If the centripetal force is proportional to the distances or P = xf , in making n = 1.

Therefore with the arc BS put equal to s, the curve sought is expressed by this equation :

From which equation with any distance of the point M from C the angle BCS is found, at which the body is without doubt present at that distance. Now between the distance MC = x and the perpendicular CT = p, there is this equation : The turning point of this curve is where dp = 0, but this is where … or .. since x cannot be greater than a. Hence it becomes : and

Therefore the cosine of the angle that the curve makes at the turning point with the radius CM is equal to [Euler’s original formula has here been corrected by P. S. in the O. O. to give the one presented here] :

Now the equation of the curve converted into a series on putting ( a 2 − x 2 ) = y is this :

Therefore in this series, the beginning of the curve, where x is not much less than a, or y is extremely small, will be Then from this equation it is apparent on making x = 0 that s becomes ∝ , whereby the curve goes round the centre C in an infinite number of spirals, and when the body is near to the centre , the equation becomes : …

From which it follows that near the centre C the curve goes off in a logarithmic spiral.

Example 2

Let n = – 1 or n + 1 = 0, which case it to be extracted from the differential equation.

For it becomes, on account of P = x thus this equation is obtained : .. the integral of which is : The other equation between the perpendicular p and x is given by

From which the turning point is found at that place, where .. or Therefore by taking this expression, there is obtained : Again it is evident, if x is made equal to 0, that s =∝ or the curve makes an infinite pp number of turns around the centre C ; now in this case we have xx = 1 or p = x. Therefore finally the curve goes round in an infinitely small circle. Example 3. 298. Putting n = – 2, in order that the centripetal force is inversely proportional to the square of the distances, the equation becomes Now ∫ 1+ yy expresses the arc, the tangent of which is y or a −x x ; let this arc be equal to dy t, and the equation becomes : is Therefore everywhere with the distance given x the arc s taken and multiplied by 2ab f equal to the difference between the tangent a − x and the corresponding arc with the x radius put equal to 1. If x is put equal to 0, then this makes s =∝ , from which it follows that the curve falls towards the centre C in an infinite spiral. [p. 135] Besides, on account of it follows thatEULER’S MECHANICA VOL. 2. Chapter 2d. page 201 Translated and annotated by Ian Bruce. pp From which it follows if x vanishes that xx = 1 or the curve finally also goes in an infinitely small circle. If ab = ff, then The turning point in this case lies at that place where 2ax = 3 xx , where either x = 0 or x = 23a . But if it is not the case that ab = 4ff, then and the turning point is found by taking [either x = 0 or] x = a 13 . Scholium 2. 299. Moreover from which it is apparent how the infinite spirals can be compared, if the n centripetal force is proportional to some power of the distance, or P = x n , and the f equation between p and x is considered, which is Here two cases are to be distinguished, the one in which n + 1 is a positive number, and the other in which it is negative. If n + 1 is a positive number, on making x = 0 the equation becomes Hence in this case the curve AM goes around the centre C in a logarithmic spiral. pp But if n + 1 is a negative number, on making x = 0 the equation becomes xx = 1 . Therefore in these cases the curve at C becomes an infinitely small circle. The speed of approaching C in these cases becomes infinitely large, and on account of this, unless the curve becomes a circle [p. 136] the body approaches C with an infinite speed, which is contrary to the condition of the problem. Therefore with the curves determined, upon which the body uniformly approaches the centre of forces, we will investigate these curves in which the body is carries around the centre of force with a uniform motion.EULER’S MECHANICA VOL. 2. Chapter 2d. Translated and annotated by Ian Bruce. page 202