Revolving Body

PROPOSITION 12 Problem 12

Suppose a body to move in an hyperbola. It is required to find the law of the centripetal force tending to the focus of that shape.

Let:

• CA, CB are the semi-axes of the hyperbola
• PG, KD are conjugate diameters
• PF is perpendicular to the diameter KD
• Qv is an ordinate to the diameter GP

Draw SP cutting the diameter DK in E, and the ordinate Qv in x, and complete the parallelogram QRPx. It is evident that EP is equal to the semi-transverse axis AC; for drawing HI, from the other focus H of the hyperbola, parallel to EC, because CS, CH are equal, ES, EI will be also equal; so that EP is the half difference of PS, PI; that is (because of the parallels IH, PR, and the equal angles IPR, HPZ), of PS, PH, the difference of which is equal to the whole axis 2AC. Draw QT perpendicular to SP; and putting L for the principal latus rectum of the hyperbola (that is, for …}, we shall have L × QR to L × Pv as QR to Pv, or Px to Pv, that is (because of the similar triangles Pxv, PEC), as PE to PC, or AC to PC.

L × Pv will be to Gv × Pv as L to Gv; and (by the properties of the conic sections) the rectangle GvP is to Qv² as PC² to CD²; and by (Cor. 2, Lem. VII.), Qv² to Qx² the points Q and P coinciding, becomes a ratio of equality; and Qx² or Qv² is to QT² as EP² to PF², that is, as CA² to PF², or (by Lem. XII.) as CD² to CB²: and, compounding all those ratios together, we shall have L × QR to QT² as AC × L × PC² × CD², or 2CB² × PC² × CD² to PC × Gv × CD² × CB², or as 2PC to Gv. But the points P and Q coinciding, 2PC and Gv are equal.

Therefore the quantities L × QR and QT², proportional to them, will be also equal.

Let those equals be drawn into …

We shall have L × SP² equal to …

Therefore (by Cor. I and 5, Prop. VI.) the centripetal force is reciprocally as L × SP², that is, reciprocally in the duplicate ratio of the distance SP. Q.E.I.

The same otherwise.

Find out the force tending from the centre C of the hyperbola. This will be proportional to the distance CP. But from thence (by Cor. 3, Prop. VII.) the force tending to the focus S will be as … that is, because PE is given reciprocally as SP². Q.E.I.

The same way may it be demonstrated, that the body having its centripetal changed into a centrifugal force, will move in the conjugate hyperbola.