Proposition 102

Table of Contents

PROPOSITION 102. PROBLEM

If a body is acted on at individual points M (Fig. 79) by two forces, the first in the direction MA, and the other, the direction of which is MQ along the normal sent from M to the plane APQ, then it is required to determine the motion of the body M and its orbit.

SOLUTION

With MP drawn, which is the normal to AP, the force MA is resolved into forces acting along Mf parallel to AP and the other along MP. Truly the force along MP can be resolved into forces acting along [p. 357] MQ and Mg. Therefore with AP = x, PQ = y and QM = z put as before, and the force pulling along MA equal to V and the force along MQ equal to W and with the resolution of the forces put in place and with a comparison made with Prop. 100 (823) it is found that and

From the equations of the same proposition there are produced : From these it is found : and hence on integration : or This value substituted in place of v gives these equations : From which the described curve can be determined. Q.E.I.

Corollary 1

The time in which the body comes as far as M, is equal to Moreover since then that time is given by : which becomes known from the quadrature of the given curve in the plane APQ.

Corollary 2

If we put y = px and z = qx, the following equations are produced

Which are helpful in making the curve known.

Example

If the force V is proportional to the distance MA, and the force W is proportional to the perpendicular MQ, on putting and thus Whereby from the equations of the preceding corollary, these equations become : The integral of this equation is x 2 = C − 2a 3 fdx 2 , x 4 dp 2 [Euler has not indicated how he solved this equation. However, if we set dx then dp = dx and ddp = ddx − dx dt = − dx dt ; on substituting these into the above …

equation, we find that : x dx = 4a ft dx − 2a fxtdt . This can be rearranged to give the integrals …

and which gives the above result on integration, and xdx = 4a3 f t dx 5 − 2a f 4 treating t and x as independent variables,] from which there arises [on setting C = c2, ] and the integral of this : for the equation of the projected curve described in the plane APQ, which therefore is an ellipse, the centre of which has been put at A. From the value of dp found, again from which with the values of dp and ddp substituted into the other equation, there arises

On putting q = e ∫ rdx , there is produced

On setting r = x2

u , there comes into being : ( c2 − x2 ) Which on putting t = ( c2 − x2 ) or x = c , the equation is transformed into : x ( 1+tt ) of which we will show the integral later. In order to know the plane, in which the elements Mm and mμ (Fig. 76) are situated, on account of : [p. 359] with h in place of f + g it is found that : and the tangent of the angle of the plane Mmμ with the plane APQ is equal to xxdp , is given by : a a Then the time, which the body takes to arrive at M, since it is equal to ∫ Thus this time is proportional to the angle, of which the sine is the abscissa x, with the total sine taken equal to c, or of which the sine is cx , if the total sine is taken as equal to

From which it is evident that the motion of the body projected on the plane APQ makes equal angles around A in equal times, and the time of one revolution is proportional to f . Corollary 3.
For the projection of the curve described in the plane APQ makes a circle, if n = 0 and − a 2af = cc ; the centre of which is at A and the radius = c. Therefore we have y = ( c 2 − x 2 ) . And z is given from this equation: gddz gxdz − hzdx = cc − xx , which extends to dx the case of the preceding example equally widely, even if only this particular case is considered.

Corollary 4

In order to discover the value of z from the equation dx = …

z = e ∫ rdx . From which being done, this differential equation of the first degree is produced: Put r = u and there is produced :

With gh or f +g = m 2 this equation arises : g in which the indeterminates can now be separated from each other in turn.

Corollary 5

Truly, .. and ..

Therefore with the constant added and with the given numbers, we have then: and hence [These integrals are easily shown to be true, on differentiation.]

Since truly lz = rdx and r =

Corollary 6

, then we have ( c − x2 ) 2EULER’S MECHANICA VOL. 1. Chapter Five (part e). On putting ∫ Translated and annotated by Ian Bruce. dx = s , then there is ( c2 − x2 ) and 2 ms or on putting e −1 = t ,so that ds = dt2mt−1 , then we have and hence it becomes :

Corollary 7

Now from the value of z we have .. and ..

Again putting y = ( c 2 − x 2 ) there will be

Corollary 8

Hence from these there is found : and tang. PQR =

In a similar manner, from these the angle of inclination between the plane is found in which the body is moving and the plane APQ. [p. 362]

Scholium

The application to finding the value of z and the inclination of the orbit is very difficult on account of the imaginary quantities occurring in turn. On this account we are unwilling to tarry longer with the intersections of the curve described by the body with the plane APQ, to be determined. Moreover since this is the great question of the moment in astronomy in finding the motion of the nodes, the following proposition has been designated to this business, in which, we investigate when the body by its own motion shall arrive in the plane APQ. For the body completes part of its motion above the plane, and part below; and whether it is above or below, the body is always drawn from the other towards this plane by a force W in the direct ratio of the distance from this plane. Truly the point in the plane APQ, through which the body passes from the upper part to the lower part, is called the descending node, and the point in which it reverts to the upper, is called the ascending node.