Draw IC meeting KE in N, and on IK let fall the perpendicular NT; and let the interval DE or IN, between the circumferences of the circles be very ſmall; and imagine the bodies in D and I to have equal velocities, Then becauſe the diſtances CD and CI are equal, the centripetal forces in D and I will be alſo equal. Let thoſe forces be expreſs'd by the equal lineolæ DE and IN; and let the force IN (by cor 2. of the laws of motion) be reſolved into two others, NT and IT. Then the force NT acting in the direction of the line NT perpendicular to the path ITK of the body, will not at all affect or change the velocity of the body in that path, but only draw it aſide from a rectilinear courſe, and make it deflect perpetually from the tangent of the orbit, and proceed in the curvilinear path ITKk. That whole force therefore will be ſpent in producing this effect; but the other force IT; acting in the direction of the courſe of the body, will be all employed in accelerating it; and in the leaſt given time will produce an acceleration proportional to it itſelf. Therefore the accelerations of the bodies in D and I produced in equal times, are as the lines DE, IT; (if we take the firſt ratios of the naſcent lines DE, IN, IK, IT, NT); and in unequal times as thoſe lines and the times conjunctly. But the times in which DE and IK are deſcribed, are, by reaſon of the equal velocities (in D and I) as the ſpaces deſcribed DE and IK, and therefore the accelerations in the courſe of the bodies through the lines DE and IK are as DE and IT; and DE and IK conjunctly; that is, as the ſquare of DE to the rectangle IT into IK. But the rectangle IT x IK is equal to the ſquare of IN; that
