63. If p be the perpendicular from the origin or pole on the tangent to the path,

pds = 284,

the more nearly as ds and SA are more and more diminished;

65. From the first of equations (3) we get

a being supposed to be a function of r only.

If the initial values of v, r be v′, r',

This shews that when the acceleration depends only on the distance from the fixed point, the change of velocity in passing

from any one point to any other is dependent only upon the distances of the two positions under consideration, and is independent of the particular path described. (NEWTON, Prop. 40.)

66. Also from the second of equations (3)

=a. {chord of curvature through the pole};

= cos (inclination of normal to radius vector).

Now if a point be moving from an initial position of rest, its motion being affected by a constant acceleration a, the direction of which is always the same; and if s, be the space described when the point's velocity is v, we shall have (by Art. 40)

$1

is called the space due to the velocity v: and it must be borne in mind that for this case the acceleration is supposed to be continued constant for the requisite time with the magnitude (a) which it has at the instant under consideration. (NEWTON, Prop. 6, Cor. 4).

67. To determine the path we shall find equations (2)

The solution of this equation determines the path.

68. We shall now proceed to discuss the particular cases mentioned above; and

(j) Let the acceleration vary as the distance from the fixed

If we had taken the positive sign above, the only difference would have been that ẞ would have been changed.

69. We must now determine C, B, and h.

If the point be moving initially with a velocity v'in a direction inclined at an angle to the prime radius vector r', then v'r' sinh, which is therefore known.

Also in (a) we have the left-hand side = 2, (Art. 64);

70. The equation (b) shews that the orbit is an ellipse whose center is at the pole, and the angle vector of its apse is B.

of

If the acceleration were in the opposite direction, the sign would be changed, and the equation (b) would represent an hyperbola whose center is at the polet.

*

By an "apse" is meant any point in the curve where the radius vector is perpendicular to the tangent. In the ellipse, for example, when the origin is the center, the extremities of the axes are apses; if the focus were the origin, the extremities of the major axis only are apses.

+ This case of motion may be easily solved in the following manner.

We have the acceleration in a=-λr cos 0 = −λx, and that in y = — \r sin 0 = − λy, supposing its direction to be always towards the origin;

71. It may be convenient to obtain a value for C in terms

of the axes of the ellipse.

... as in Art. 51, ~= A cos (√.t-ẞ), and y = A'cos (√.t-ß'),

N

take the cosine of each side and write C for cos (ẞ'-ẞ).

To determine whether it is an ellipse or an hyperbola we have

If the direction of the acceleration were away from the origin, then we should

And as (A'B+AB')2 – 4BB'. AA', which =(A'B – AB′), is positive, the path is an hyperbola.