The value of tanr and tan g' is easily found from the triangles BOG, BOK, and tanr may be reduced to an expression in terms of the sides by making use of the usual formulas, which give sin A, 14 cos A, &c. in terms of the sides, and sina, cosa, &c. in terms of the angles. Whence will result the value of tan r tan g' given above.

SIMON NEWCOMB.

W. P. G. BARTLETT.
TRUMAN HENRY SAFFOrd.

PRIZE ESSAY ON CENTRAL FORCES.

By DAVID TROWBRIDGE, Perry City, Schuyler Co., N. Y.

1. In the following essay I shall take the well-known and fundamental equations of motion

(1) D2x + X = 0, D2y + Y= 0, D2 z + Z= 0,

for granted, as they are to be found in treatises on Mechanics. In these equations x, y, z are the rectangular co-ordinates of the moving body, and X, Y, Z are the forces directed along the co-ordinate axes of x, y, z respectively.

2. Let R represent the central force acting on the moving body, M; let a, ß, y be the angles which its direction makes with the coordinate axes x, y, z respectively; let r equal the distance which

separates the centre of gravity of the body M from the centre of gravity of the system, this latter point being the centre of attraction. We then have, assuming this centre for the origin of coordinates,

Substituting these values in equations (4), clearing of fractions, and transposing into the first member, we obtain

(6) XyYx = 0, Yz-Zy0, Zx - Xz = 0.

Substituting in these equations the values of XYZ, as given by equations (1), we obtain

(7) y D2 x - x D2 y = 0, z D2y—y D2 z = 0, x D2 z — z D2 x — 0.

3. The integrals of (7) give, letting c, c', and e" be the arbitrary constants introduced by integration,

(8) y D, x-x Dy=c, z Dy-y D,z=c, x D1z — z D, x=c". If we multiply each of these equations by the variable which it does not contain, and take their sum, it will reduce to

(9)

c' x + c" y + c z = 0..

This is the equation of a plane passing through the origin of coordinates, or centre of attraction; and hence the co-ordinates of the body M are also the co-ordinates of a plane; and therefore the curve described by M is always found in the same plane, or it is a

plane curve. It will not be necessary, then, to retain z in our discussion of the subject; and by putting it equal to nothing our investigation will be much simplified, and the conclusions none the less general.

4. Putting z and Z equal to nothing in equations (1) and (8), we obtain

y=c.

(10) D2 x+X=0, (11) D2y+Y=0, (12) y D ̧ x — x D ̧y = c. The integral of (12) gives

(13)

f(ydx-xdy) = et + C.

5. Let us now interpret equation (13). In Fig. 1 let PMY represent the curve described by the body M; S, the centre of attraction; SY and SP, the axes of y and x respectively; PSM== the

=

angle which the radius-vector makes with the axis of x, and the other lines as marked in the figure. Then the area PXM=fydx'. But SP-r; and since SP is constant, we have dxdx. Substituting this value in the above equation, we find the area PXM-fydx. The sector PSM is made up of the area PXM and the triangle sector PSM=SMX+PXM=}xy+ This being differentiated and reduced, gives yda—rdy

xy-fydx.

d (sector PSM) = —

We therefore have

=

2

-edt, from (12).

(14) 2 sector PSM -f(ydx-x dy) = ct + C.

Had we chosen the sector S YM, we should have found it equal to +(ydx-xdy), the only difference being in the sign of the expression. In the first case the area is reckoned from the axis SP, and in the second it is reckoned from the axis SY. We therefore

infer that equation (13) is an expression for the area passed over by the radius-vector. This area ought to be equal to nothing when t=0; hence C0. This reduces (13) to

(15)

с

S (ydx-xdy) = ct.

Since c is constant (15) makes known this law; The area passed over by the radius-vector is proportional to the time.

6. This was found to hold true in the motion of the planet Mars, by the celebrated JOHN KEPLER; and he extended it by analogy to all the planets. It has since been called KEPLER'S second law. This law admits of a generalization wholly unknown to its discoverer ; since it is altogether independent of the form of the curve described, and of the nature of the force, either attractive or repulsive. All that is necessary is, that the force be directed along a single line, and that line the one which connects the centres of gravity of the central and moving bodies.

=

7. To determine the constant c, let t 1 in equation (15), and we find c equal to twice the area passed over in the unit of time. 8. By referring to Fig. 1 we easily find

(16)

x = r cos p, y = r sin 9.

Differentiating with respect to the time, t, we obtain

(17) D,x= Dr cosp―r sing D, q, D,y=D,r sing+r cosy D ̧q. Substituting these values in equation (12) it reduces to

Or changing the axis from which the area is reckoned (Art. 5) we have

Since ዋ is the space described at the unit of distance, it follows that D,q, is the velocity at the unit of distance, or it is the angular velocity of the body M. We hence see that (20) makes known this

law, namely, The angular velocity varies inversely as the square of the distance. According to NEWTON, the force of gravity in the solar system varies inversely as the square of the distance; hence, in the case of nature, the force of gravity varies directly as the angular velocity. But equation (20) comprehends more, for it is as general as the law given by (15), from which it was derived.

9. Equation (19) can be used in computing the time when the angle is given, and vice versa, knowing the form of the orbit.

(21)

If we suppose

to be a function of t, as it in general is, we can cff(t) dt (23), and this can be obtained in the form of a

10. To obtain a second integral of equations (10) and (11), multiply the first by 2 dx, and the second by 2 dy, take their sum, and integrate, and we obtain

s2 v2

(24) D+D=D ==C−2f(Xda +Ydy).

Substitute the values of X and Y derived from equations (2), where cos a and cos ẞ correspond to cos q and sing in equations (16), and sin p = 2, and we obtain

cos p

=

y

(25) D ̧x2+D‚y2—C—2ƒ(R.Z.dx+R.%.dy)—C—25 R(xdx+ydy).

But r2 = x2 + y2, and rdr = xdx+ydy, which being substituted in (25) gives

(26)

D1 x2 + D1y2 = C — 2f R dr.

Substituting for D, and Dy their values given by equations (17), we find

(27)

gı2 D1 q2 + D1 p2 = C-2 Rdr.

:

c2

Equations (18) and (19) both give 2 D, q2 = and this value substituted in (27) gives