Complete the parallelogram S Q R U, and let fall O P_{1}, O P_{2}, O T perpendiculars from O on S Q, S R, S U respectively. Since S Q is the velocity of the particle when at A_{1} the moment of momentum is _m_ O P_{1} × S Q; when the particle is at A_{2} the moment of momentum is _m_ O P_{2} × S R. Whence the difference of the moments of momentum at A_{1} and A_{2} is _m_ (O P_{2} × S R - O P_{1} × S Q) = 2 _m_ (O S R - O S Q) = 2 _m_ O S U = _m_ O T × S U = _m_ O T. Q R = F _t_ × O T. But in the limit S coincides with A_{1} and A_{2}, and we see that the gain in moment of momentum is _t_ times the moment of the force around O. Hence we deduce the following fundamental theorem, in which, by the expression acceleration of moment of momentum, we mean the rate at which the moment of momentum increases:— _If a particle under the action of force describes a plane orbit, then the acceleration of the moment of momentum around any point in the plane is equal to the moment of the force around the point._ If the force is constantly directed to a fixed point, then the moment of the force about this point is always zero. Hence the acceleration of the moment of momentum around this point is zero, and the moment of momentum is constant. Thus we have Kepler’s law of the description of equal areas in equal times, and we learn that the velocity is inversely proportional to the perpendicular on the tangent. § 11. IF TWO OR MORE FORCES ACT ON A POINT, THEN THE ACCELERATION OF THE MOMENT OF MOMENTUM, DUE TO THE RESULTANT OF THESE FORCES, IS EQUAL TO THE ALGEBRAIC SUM OF THE MOMENTS OF MOMENTUM DUE TO THE ACTION OF THE SEVERAL COMPONENTS. Let A D, Fig. 61, be a force, and A C and A B its two components. Then, since O A D = O A B + O A C, we see that the moment of A D around O is equal to the sum of the moments of its components. Hence we easily infer that if a force be resolved into several components the moment of that force around a point is equal to the algebraical sum of the moments of its several components. The acceleration of the moment of momentum around O, due to the resultant of a number of forces, is equal to the moment of that resultant around O. But, as we have just shown, this is equal to the sum of the moments of the separate forces, and hence the theorem is proved. § 12. IF ANY NUMBER OF PARTICLES BE MOVING IN A PLANE, AND IF THEY ARE NOT SUBJECTED TO ANY FORCES SAVE THOSE WHICH ARISE FROM THEIR MUTUAL ACTIONS, THEN THE ALGEBRAIC SUM OF THEIR MOMENTS OF MOMENTUM ROUND ANY POINT IS CONSTANT.