Proposition 6 Theorem 5

Assume a space with an immovable centre and void of resistance.

A body revolves in any orbit around that center. In the shortest time draws any arc just then nascent.

The versed sine of that arc is supposed to be drawn bisecting the chord, passing through the centre of force.

The centripetal force in the middle of the arc will be as the versed sine directly and the square of the time inversely.

This is because the versed sine in a given time is as the force (by Cor. 4, Prop. 1).

Augmenting the time in any ratio, because the arc will be augmented in the same ratio, the versed sine will be augmented in the duplicate of that ratio (by Cor. 2 and 3, Lem. XI.).

Therefore is as the force and the square of the time.

Subduct on both sides the duplicate ratio of the time, and the force will be as the versed sine directly, and the square of the time inversely. Q.E.D.

The same thing is also easily demonstrated by Corol. 4, Lem. X.

Cor. 1

If a body P revolving about the centre S describes a curve line APQ, which a right line ZPR touches in any point P; and from any other point Q of the curve, QR is drawn parallel to the distance SP, meeting the tangent in R; and QT is drawn perpendicular to the distance SP; the centripetal force will be reciprocally as the solid

…

if the solid be taken of that magnitude which it ultimately acquires when the points P and Q coincide. For QR is equal to the versed sine of double the arc QP, whose middle is P: and double the triangle SQP, or

…

is proportional to the time in which that double arc is described; and therefore may be used for the exponent of the time.

Cor. 2

By a like reasoning, the centripetal force is reciprocally as the solid

…

if SY is a perpendicular from the centre of force on PR the tangent of the orbit. For the rectangles

…

and

…

are equal.

Cor. 3

If the orbit is either a circle, or touches or cuts a circle concentrically, that is, contains with a circle the least angle of contact or section, having the same curvature and the same radius of curvature at the point P.

If PV is a chord of this circle, drawn from the body through the centre of force; the centripetal force will be reciprocally as the solid SY2 x PV. For PV is QP2/QR.

Cor. 4

The same things being supposed, the centripetal force is as the square of the velocity directly, and that chord inversely. For the velocity is reciprocally as the perpendicular SY, by Cor. 1. Prop. I.

Cor. 5

Hence if any curvilinear figure APQ is given, and therein a point S is also given, to which a centripetal force is perpetually directed, that law of centripetal force may be found, by which the body P will be continually drawn back from a rectilinear course, and being detained in the perimeter of that figure, will describe the same by a perpetual revolution.

That is, we are to find, by computation, either the solid

…

or the solid

SY2 x PV reciprocally proportional to this force. Examples of this we shall give in the following Problems.