Page:Encyclopædia Britannica, Ninth Edition, v. 15.djvu/739

Rh MECHANICS 707 of gravity at different points of the path, by remembering that, as shown in 49, the path of an unresisted pro jectile is an ellipse, one of whose foci is at the centre of the earth. The following, among many other analogous propositions, are easily proved. (1) The locus of the second foci of the paths of all pro jectiles leaving a given point with a given speed, in a vertical plane, is a circle. (2) The direction of projection, for the greatest range on a given line passing through the point of projection, bisects the angle between the vertical and the line. (3) Any other point on the line which can be reached at all can be reached by two different paths, and the directions of projection for these are equally inclined to the direction which gives the maximum range. (4) If a projectile meet the line at right angles, the point which it strikes is the vertex of the other path by which it may be reached. (5) The envelop of all possible paths in a vertical plane is an ellipse, one of whose foci is the centre of the earth, and the other the point of projection. To prove these propositions, let E (fig. 46) be the earth s centre, P the point of projection, A the point which the projectile would reach if fired vertically upwards. With centre E, and radius EA, describe a circle in the common plane of projection. This, the circle of zero velocity, cor responds to the common direc trix of the parabolic paths in the ordinary theory If F bo the second focus of any path, we must have EP + PF con stant, because the axis major depends on the speed, not the direction, of projection. Hence (I) the locus of F is the circle AFO, centre P. Again, since, if F be the focus of the path which meets PR in Q, we must have FQ = QS, it is obvious that the greatest range P/ is to be found by the condition Oq = qs. O is therefore the second focus of this trajectory, and therefore (2) the direction of projection for the greatest range on PR bisects the angle APR. If QF = QF = QS, F and F are the second foci of the two paths by which Q may be reached; and, as L F PO= L FPO, we see the truth of (3). If Q be a point reached by the projectile when moving in a direction perpendicular to PR, we must evidently have L PQF = L PQF = L SQR = L EQP; i.e., EQ passes through F. When this is the case, the ellipse whose second focus is F evidently meets PR, at right angles ; and that whose second focus is F has (4) its vertex at Q. The locus of q is evidently the envelop of all the trajectories. Now Hence or (5) the envelop is an ellipse, whose foci are E and P, and which passes through A. 142. One of the most important problems in this branch of our subject is that of planetary motion, which forms a good typical example of the processes to be employed in the treatment of central orbits. One or two definitions, and a general property of central orbits, must be premised. 143, Def. An &quot;apse&quot; is a point, in a central orbit, at which the path is perpendicular to the radius-vector along which the central force acts. Pic. 46. The length of the radius-vector is therefore, at such a point, generally a maximum or a minimum. This radius vector, drawn to an apse, is called an &quot; apsidal line.&quot; A central orbit is symmetrical about every apsidal line. The simplest proof of this theorem depends upon the general principle of &quot; reversibility,&quot; which holds in all con servative sj^stems. In fact if, at any instant, the velocity- vector of a particle, moving under the action of a conserva tive system of forces, be reversed, the particle will simply retrace its previous path. For if we suppose a smooth tube, in the form of the previous path, to be employed to guide it back, the speed at every point will be of the same magnitude as before. The curvature also of the path will be the same ; and thus the normal component of the applied forces will balance the so-called centrifugal force, i.e., will suffice to produce the requisite curvature, so that there will be no pressure on the tube, and it is not required. Hence since, at an apse, the velocity is perpendicular to the radius-vector, the two halves of the orbit on opposite sides of the apsidal line are similar and equal. Hence, however many apses there may be, there can be at most only two apsidal distances. For the property of symmetry about each apsidal line shows that, if there be more apses than one, the firs.t, third, fifth, &c., must have their apsidal distances equal, as also must the second, fourth, sixth, &c. If there be one apse only, it may correspond either to a minimum or to a maximum value of the radius-vector ; but, if there be more than one. they must be maxima and minima alternately. 144. We now proceed to the gravitation case already promised. We will take, first, the direct problem as in 49, where the force is assigned and the orbit is to be found. A particle is projected from a given point in a given direction and Plane- with a given speed, and moves under the action of a central attrac- tary tion varying inversely as the square of the distance ; to determine motion. the orbit. We have P=|im 2, and therefore by the last part of 47 +-*- where h is double the constant area, or .. = 0, the integral of which is M-~= or, as it is usually written, u= To{ This gives by differentiation a). (2). Let R be the distance of the point of projection from the centre, and /3 the angle, and V the speed, of projection ; then, when 6 = 0, 1, a /ldu u=- T , cot/3=- - ---) it u de/ e= o Hence, by (1) and by (2) From these and But wherefore and h&quot; -rr - l=ecosa , (ill 7/ 2 != -esma. tan a = - T (3), . (4), tan a = 1-e 2 M - VRsiinS 1 V 2 It M/ . . (3 ), (4 )-