Proposition 29

PROPOSITION 33. PROBLEM

274. With the attraction from the centre C (Fig. 26) to be in some ratio of a multiple of the distances, the body at D now has a given speed ; the point A is required on the line CD produced, from which the descent of the body towards C begins, so that it has acquired this speed when it arrives at D. [p. 110]

SOLUTION

With n denoting as above the exponent of the ratio of multiplication, in what shall be the centripetal force, and f the distance at which the centripetal force is equal to the force of gravity; let CD = b, the speed at D corresponding to the height h, and the distance CA sought which is put equal to q.

Since here therefore q denotes that same distance as a [i. e. CA] in the above proposition, and b likewise that of y [i. e. CP], and h likewise here represents v, this equation is formed: n … = above.]

From which …

Moreover in the particular case, when n = -1, there is obtained … henceand , where e is the number, the logarithm of which is unity. Q. E. I.

Corollary 1

275. If the centripetal force varies directly as the distance, then n = 1 and )2( 2 fhbq += . Which is always a finite quantity, but only if b, f and h are such.

Likewise it arises, provided n + 1 is a positive number. And also in the case n = - 1 the distance q is never infinite.

Corollary 2

276. But if n + 1 is a negative number, for example – m, since n = - m – 1, then …, for which the height is infinite when h is given by m … and if h is a quantity greater than this, [p. 111] q is negative, or rather infinitely greater, or even imaginary. From which it is understood from these cases that only by falling from infinity is the body able to acquire as much speed at D.

Corollary 3

277. By keeping n + 1 equal to the negative number – m and the distance h for the point A at infinity will be m …= .

The distance from the centre C at which the body, falling from the infinite distance, will have the speed h , which is equal to m mh

Corollary 4

278. If the centripetal force is inversely proportional to the square of the distance, then m = 1. On account of which bhf ….

When, therefore b f h 2 = , the distance AC, i. e. q, shall be infinitely great.

Corollary 5

279. If this problem is joined with the preceding one, the motion of the body can be easily determined, since it begins to descent to C from B with the speed h . From the preceding indeed the descent of the body is observed to be made from A, since it begins to descend in the part from D with the speed h , CP is called equal to y, and the speed, that the body has at P, corresponding to the height v, is )1( ..

But )1(11 nnn hfnbq ++= ++ . Hence it becomes

…

Corollary 6

280. From this expression for v, when the descent begins from D with the speed corresponding to the height h, it does not differ from that which is produced, if the descent were made from rest, except that this is quantity is always a distance greater than h.

Scholium

281. Since the time delay, in which the distance AC or any part of this is completed for any hypothetical centripetal force (Fig. 25), that is easily known from the known speeds. Generally the time for any letter n cannot be shown in a finite number of terms, clearly the time to traverse AP is found to equal …

which quantity generally neither can be integrated nor reduced to the quadrature of any known curve. But yet in various cases of n itself, it can be expressed neatly enough , on account of which from the general case set out, particular special cases will be examined in the following propositions.

PROPOSITION 34. PROBLEM.

282. If the centripetal force is in proportion to the distance from the centre C (Fig. 27) and the body falls from A as far as C, [p. 113] it is required to determine the time in which the body completes any part of this distance.

SOLUTION

With AC = a, and the distance from the centre C, in which the centripetal force is equal to the force of gravity , equal to f, some part of the distance CP = y and the speed at P corresponding to the height v. Therefore the time, in which the distance CP is completed, is equal to ∫ v dy ; with the fraction 250 1 ignored, since this can be used for the known time in seconds and can be added as desired. Truly from Prop. 32 (264) by making n = 1 .hence, …

From which the time to travel through PC …

Upon AC the quadrant of a circle is constructed AME, and to this the lines CE and PM as axis. From which is made, as agreed, the arc EM = ∫ − )( 22 ya ady . On account of which the time to traverse PC becomes = a fEM 2. . The time therefore of the total descent through AC will be a fAME 2. . Hence the time of descent through AP = a fAM 2. .

From these therefore the time of descent through any distance travelled through can become known, and that in seconds, if these expressions are divided by 250 and the length f is shown in thousandth parts of Rhenish feet. Q. E. I.

Corollary 1

283. Let 1 : π denote the ratio of the diameter to the circumference , then it becomes 2AME : a = π : 1 and 2 π=a AME . Hence on account of this, the time of descent through AC is equal to f22 π . [p. 114] Since that does not depend on the height dropped or travelled through a, but whatever amount this shall be, it keeps the same value.

Therefore all bodies, which are released towards this centre, reach that in equal amounts of time.

Scholium

284. This equality of the time follows from the expression for the time f … in which a and y are required to have one dimension. Indeed the amount comes about, the times, in which any distances a are travelled through, must be equal to each other. (46).

Corollary 2

285. If besides there should be another of centre of force of this kind, but with a different effectiveness provided, thus in order that the distance at which the centripetal force is equal to the force of gravity is F, the times of the descents shall be to each other as Ff ad . But the effectiveness of each are themselves in this case in the inverse ratio of the distances f : F; indeed these are as the forces, which these forces exercise at equal distances. Wherefore the times of descent to the different centres of force are in the inverse ratio of the square roots of their effectiveness. Which ratio indeed holds in all similar centres of force in place, if the distances traversed are equal to each other, as will be taught in what follows.