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A SIMPLE EXAMPLE OF A CENTRAL ORBIT WITH MORE THAN TWO APSIDAL DISTANCES.*

By DR. F. L. GRIFFIN, Williams College, Williamstown, Mass.

If a central force be a single-valued function of the distance, every rbit is symmetrical with respect to each apsidal line where the radius vector s a maximum or minimum, the apsidal angle is constant, and the orbit has hot more than two apsidal distances.† These conclusions do not extend to orces which are multiple-valued. The purpose of this note is to call attenion to an example of an orbit not having the properties mentioned above; t seems sufficiently simple to serve for class-room illustration.

Let ABCD be an ellipse, and O be the point of intersection of two hormals which make angles of 45° with the major axis. If this ellipse be described as a central orbit under a force directed to O, there will be four psidal lines, OA, OB, OC, and OD, three of the apsidal distances being distinct. There are two distinct apsidal angles, 17 and 1. A similar state

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of affairs exists, if O be any point of the major axis, within the evolute of the ellipse (except the center).

The general case may be treated as follows. Let O be selected as the origin of co-ordinates; and let the center of the ellipse be at (-d, 0), where 0<d<ae2. The equation of the ellipse is then:

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Or, using polar co-ordinates, and denoting by the longitude measured from OB, and by u the reciprocal of the radius vector, [r], the equation becomes

(2)

(3)

or

(1-e) (cos +du)2+sin20=a2 (1—e2)u2,

r cos 0=au±√ (ßu2+r),

where re2, a=d(1—e2), ß·= (d2 — a2e2) (1—e2), and the sign of the radical is to be determined.

At 0=0, u=

2

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1 α

-d'

and, therefore,

(a–d) e2=d(1— e2) ±1/[(d2 — a2 e2) (1—e2)+e2 (a−d)2], or (a-d)e2=d(1−e2)± (ae2 —d).

*Read before the American Mathematical Society, April 27, 1907.

†E. J. Routh, Dynamics of a Particle, pp. 270-272.

Evidently the positive sign must be taken. At 6, however, the reverse is true, for u=

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1
a+d'

and

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which requires the negative sign. Since cos is obviously a continuous function of u in the ellipse, the sign of the radical in (3) can change only for a value of u which makes the radical vanish. There must be such a value taken somewhere in the orbit.

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the values of u for which du/d0=0 are given by Bu2+x=0 and sin 0=0.

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the negative value of u having no meaning. That this value of u (which is real, since d<ae' <ae) is actually taken in the orbit, is seen from the value obtained for ", corresponding to this value of u. Thus

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which is real and less than unity; for d <ae, and hence d3 (1-e) <e2 (a2 e2-d2).

1

There are, then, four apsidal lines, 0=0, 0=101, 0, as in the special case first mentioned, where 1. The apsidal angles are ", and π-01, which are equal only for 0,=; i. e., except for d=0, (when the center of force is at the center of the ellipse), there are two distinct apsidal angles. Evidently the sign of the radical in equation (3) must be positive from 0=-01 to 0=+91, and negative from 0-0, to 0-2-01.

1,

It remains to ascertain whether the radius vector is a maximum 1 1 or minimum at each apse. It is easily seen that u1> -> a-d a+d' from (d-ae)>0, follows e2 (a−d) > (a2e2-d2) (1—e2), or

For,

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Since the radius vector can have maxima and minima only at 0=0, 0=±01, and T, and since r1 [=1/u]<(a-d)<a+d, the radius vector is a maximum at 0-0 and 0=, and is a minimum at 0±0.

=

Since the ellipse is asymmetrical with respect to the lines 0=0,, it follows from the statement made at the outset that the force is a multiplevalued function of the distance. This conclusion may be verified by deriving the law of force. Thus,

(6)

f=h2u2 (u +

d2u do

=

h2y 2 (Br+a2-B) u*
[Bu±a v ́ (Bu2+r)]3,

where h is the constant of areas, and the positive and negative signs are to be given as in (3). Thus, for values of r between r1 and (a–d), the required force is a multiple-valued function.

The force has one striking peculiarity. For u>u, Bu2+r<0, so that the force is imaginary. From this fact it easily follows that the force is such as not to permit a real orbit anywhere within a circle of radius r about the center of force.

DEPARTMENTS.

SOLUTIONS OF PROBLEMS.

ALGEBRA.

NOTE ON PROBLEM 266. While in Washington, D. C., during the latter part of last August, I called on Mr. Theodore L. DeLand of the United States Treasury Department. While there, he called my attention to the usage among practical computers and actuaries in England regarding the finite series. He showed me a number of books in which infinite series were used and in which it was assumed that the series is determined by the terms that are given.

Thus, in the series 1+7+12+21, it is assumed that the series has for its first differences, 5, 9, etc.; for the second differences, 4, 4, 4, etc. If only three terms were given, it would be assumed that the first differences are 5, 5, 5, etc. The series 1+3+7+17+... of which the sum of n terms

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Then, x2 [412 (c-a) (c−b) +4m2 (a−b) (a−c) +4n2 (b−a) (b−c)
- (m+n+1)(m+n−1) (m−n+1)(n+l−m)]

-2x [12 (c-a) (c−b) (a+b+2c)+m2 (a-b) (a−c) (b+c+2a)
+n2 (b-a) (b-c) (a+c+2b)+cl2 (12-m2-n2) + am2 (m2 —n2 - 1o)
+bn2 (n2 - 12 — m2)]+[c1l2+a+m2 + b + n2 + 12 m3 n2
+(a2b2+c2l2) (l2 —m2 — n2) + (b2c2 +a2m2) (m2 — l2 —n2 )
+(a2c2+b2n2) (n2 —l2 — m2)]=0... (2).

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x2 [4abc (a+b+c) - (ab+ac+bc)2]-2xabc (ab+ac+bc)-a2b2c2=0... (3).

From which, x=

abc

.(4).

ab+ac+bc±21 [abc(a+b+c)]***

We have thus found the radii of the circles having centers D and D', when the circles intersect, are tangent, or are non-tangent.

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