Leonhard Euler’s early lunar theories 1725–1752

Abstract

The analysis of unpublished manuscripts and of the published textbook on mechanics written between about 1730 and 1744 by Euler reveals the invention, application, and establishment of important physical and mathematical principles and procedures. They became central ingredients of an “embryonic” lunar theory that he developed in 1744/1745. The increasing use of equations of motion, although still parametrized by length, became a standard procedure. The principle of the transference of forces was established to set up such equations. Trigonometric series expansions together with the method of undetermined coefficients were introduced to solve these equations approximatively. These insights constitute the milestones achieved in this phase of research, which thus may be characterized as “developing the methods”. The documents reveal the problems Euler was confronted with when setting up the equations of motion. They show why and where he was forced to introduce trigonometric functions and their series expansions into lunar theory. Furthermore, they prove Euler’s early recognition and formulation of the variability of the orbital elements by differential equations, which he previously anticipated with the concept of the osculating ellipse. One may conclude that by 1744 almost all components needed for a technically mature and successful lunar theory were available to Euler.

Appendices

Appendix A: The contents of Ms 167 relevant for the development of Euler’s lunar theory

In the “Introductio” of Ms 167 Euler defines the concept of “scala celeritatum”, which he adopted from Jacob Hermann’s Phoronomia (cf. Hermann 1716, p. 54), a work that Euler studied carefully and “in which the whole science [of mechanics] has been enriched by the treatment of so many selected topics to be found within”, as Euler mentioned in the preface to his “Mechanica” (cf. Euler 1736, Prefatio). The concept “scala” means the graphical representation of values or shortly “graph” or “run of the curve”. It is not to be confused with the concept of a function in the modern analytical sense, although Euler already uses the term “functio” in Ms 167 (cf., e.g., §§ 69, 82, 92–96, 106) and later in his “Mechanica” (cf. Euler 1736), expressing the dependence of the value of a parameter from another one. In that time (about 1730) the term “scala” meant the point-to-point correlation between the values of a certain parameter (force, velocity, time) and the values on which they depend (distances). In a diagram, the latter constitutes the values of the abscissa, the former of the ordinate, thus forming a discrete steplike representation (which is the meaning of the word “scala”), which was considered to be a continuously progressing curve and expressed—in this sense for our context—the “progression of the velocity as a function of the distance covered by the considered body”. Later on, in Chapter I of Section I, Euler also introduces the concepts of “scala potentiarum” and “scala temporum” (cf. Ms 167, §§ 62, 64). He adopted the former from the Phoronomia as well. Hermann introduced not only the term “scala potentiarum”, but other “scalae”, e.g., “scala solicitationum centralium” (cf. Hermann 1716, p. 28 and p. 52). Euler defined the concepts of “scala celeritatum” and “scala temporum” in his “Mechanica” (cf. Euler 1736, Prop. 5), he missed, however, to define the term “scala potentiarum” and used it without explanation in the solution of Proposition 41 (cf. Euler 1736, Prop. 41). The term “scala temporum” does not appear in Hermann’s Phoronomia, but it already occurs in the letters of Johann I Bernoulli to Jacob Hermann written on December 21, 1715, and on May 20, 1716, from which one may conclude that Euler became acquainted with it from them as well. According to the way Euler uses the “scala potentiarum” in Ms 167, one may suppose that it does not represent just one but any curve in the sense of “variational curves”, which is why the term “scala” sometimes may be associated with a whole family or series of curves. The significance of the concept of “scala potentiarum” for the development of lunar theory concerns the fact that Euler learned to find and to determine different “scalas of forces” associated with different given initial conditions or properties defined by the problem. This actually involves not only variational principles, which were still to be developed at that time, but has—by the initial conditions—implications on the choice of appropriate reference frames as well. I will briefly address this topic below.

Euler opens the first part of his treatise, entitled “De motu a potentiis producto” (On the motion produced by forces), with definitions of absolute and relative forces, which are of equal importance with respect to the development of reference frames:

Potentiæ quarum actiones non pendent a celeritate corporis patientis, vocentur absolutæ, Quæ vero aliter agunt, si corpus alia feratur celeritate relativæ vocentur.

(Forces whose actions do not depend on the velocity of the body experiencing them are called absolute forces, whereas those that act in a different way when the body moves with a different velocity are called relative.)

Euler has used first the terms “potentiae purae” (pure forces) and “potentiae impurae” (impure forces) for absolute and relative forces, respectively, which corroborates the conjecture that these concepts are his own neologisms with respect to the meaning defined and used by him (cf. Fig. 6).

Fig. 6

Euler’s sketch in Ms 167 to illustrate the concepts of absolute and relative forces

Dato autem effectu potentiæ absolutæ in corpus quiescens sequenti modo invenietur effectus in motum. Valeat potentia quædam corpus $A$ quiescens tempusculo infinite parvo ex $A$ in $P$ transferre. Nunc vero habeat corpus $A$ celeritatem $z$ secundum directionem AB quæritur quomodo eadem potentia eodem tempusculo motum corporis perturbabit.

(The effect of an absolute force on a body at rest being given, the effect on the moving body will be determined as follows: Let some force be such that it sends a body $A$ at rest in an infinitely small time from $A$ to $P$. Now let the body $A$ have the velocity $z$ in the direction AB. How will the same force now disturb the motion of the body during the same moment of time?)

Quia potentia æque in motum corpus ac in quiescens agere ponitur, concipiatur corpus $A$ motum suum amisisse seu id super plano positum, quod motum habet aequalem et contrarium ei quem habet corpus transferetur id igitur in $P$. Restituo autem motu perveniat id interea in $B$ motu ante concepto. Quamobrem post hoc tempusculum non in $P$ sed in $M$ reperietur, ducta recta PM parallela et æquali rectæ AB. Quoniam enim planum in partes contrarias motum concipiebatur, ut corpus in locum debitum restituatur, oportet plano motum ei, quem ante habere ponebatur, contrarium tribuere, hoc modo punctum $P$ in $M$ transferetur. Quamobrem corpus $A$ interea diagonalem AM descripsisse putandum est. A potentia ergo angulo BAM a sua semita deflectere coactum est, et celeritatem acquisivit, quae se habet ad pristinam ut AM ad AB.

(Since the force is assumed to act equally on a moving body as on one at rest, let us suppose the body $A$ to have lost its motion or to be set on a plane which has a uniform motion opposite to that of the body. Thus it will be carried to $P$. The motion being restored, let it arrive at $B$ by the motion conceived before. It will therefore be found after this moment of time not at $P$ but at $M$, where the line PM is drawn parallel and equal to AB. For since the plane was supposed to move in the opposite direction, in order to return the body to its due place, one needs to attribute to the plane a motion opposite to that which it was supposed to have before: thus the point $P$ will be carried to $M$. The body $A$ must therefore be thought to have covered in the meantime the diagonal AM. Consequently it has been compelled by the force to deviate from its path by the angle BAM and has obtained a velocity that is to the original one as AM to AB.)

In the first section, entitled “De motu a potentiis in punctum liberum agentibus producto” (On the motion produced by forces acting in a free pointlike body) Euler treats uniform rectilinear motions. To be able to solve non-uniform rectilinear motions, he states what we now call “Newton’s second law” or “equation of motion”:

Porro si tempora sunt inæqualia, quia tum incrementa celeritatum sunt ut tempora, habebimus hanc legem incrementum celeritatis esse directe ut tempusculum et potentiam atque ut corpus ipsum inverse.

(If the time intervals are not equal, the following law holds (since then the increments of velocity are proportional to the times): the increment of velocity is directly proportional to the time element and to the force, and inversely proportional to the body itself [i.e., its mass].)(emphasis added)

Euler adopted this law most probably from the Phoronomia, because Hermann presented it in the same context as Euler did, namely when dealing with the transition from uniform to non-uniform motions (cf. Hermann 1716, pp. 55–57).

Euler continues his investigations on rectilinear motions of pointlike bodies in Chapter I of this section, entitled “De motu puncti a potentiis absolutis tracti rectilineo” (On the rectilinear motion of pointlike bodies due to absolute forces). Here he formulates the law of motion in Leibnizian notation:

$$\begin{aligned} \mathrm{d}z = mp\,\text{ d }t : A, \end{aligned}$$

(1)

where $z$ denotes the body’s velocity, $A$ its mass, $p$ the force acting on it, and $m$ a proportionality constant. He substitutes the time element $\text{ d }t=n\,\mathrm{d}x:\mathrm{d}z$, $n$ being a constant, to obtain

$$\begin{aligned} \mathrm{d}z = mnp\,\mathrm{d}x : Az \quad \mathrm{or} \quad Az\,\mathrm{d}z = mnp\,\mathrm{d}x. \end{aligned}$$

(2)

Its integration gives

$$\begin{aligned} Azz = 2mnpx, \end{aligned}$$

(3)

setting the integration constant equal zero. Here we can observe the very origin of the factor 2 occurring in Euler’s equation of motion, which he maintained over many decades of years by choosing appropriate units. Therefore

$$\begin{aligned} z = \sqrt{\frac{2mnpx}{A}}. \end{aligned}$$

(4)

According to Euler, it is more convenient to express the quantity of velocity by the corresponding height or altitude of free fall, which is a distance and therefore easier to measure than the velocity. In the sequel Euler defines and determines the units and constants in Rhinelandian (Prussian) feet, thus obtaining $n=\frac{1}{250}$, which corresponds to the value of gravity on Earth at an altitude of 15,625 Rhinelandian feet.

The next topic concerns the motion of a body which is attracted in any multiple ratio of its distance from the center. In this context Euler introduces the terms “scala potentiarum” (progression of forces) and “scala temporum” (progression of time) and applies these concepts to solve problems closely related to what we call today “calculus of variations” such as (cf. Ms 167, § 80, and Fig. 7):

Fig. 7

Euler’s setch in Ms 167 to illustrate the concepts of “scala” or function

Inveniamus nunc omnes possibiles scalas potentiarum, quæ faciant, ut corpus dictum spatium $AC$ vel eodem tempore percurrat, vel ut in $C$ eundem celeritas gradum assequatur.

(Find now all possible progressions of forces which cause the body either to cover a given space $AC$ in a given time or to reach in $C$ with a given degree of velocity.)

Euler reformulated this kind of “variational” problem in his “Mechanica” (cf. Euler 1736, Prop. 47 and 48). In Ms 167, he also treats the problem of finding a special “scala potentiarum” that causes a falling body to move in a certain way according to a given time schedule (cf. Ms 167, §§  88–102, and Fig. 8).

Fig. 8

Euler’s sketch in Ms 167 to illustrate the construction of a “scala potentiarum” $(B)(D)$, which is equivalent to a constant force BD

Chapter III deals with the curved motion of a pointlike body which is driven by absolute forces. Euler decomposes the force acting on the body into the tangential and normal components with respect to the point of the curve the body is currently located in and notes, that the former can change only the body’s speed, the latter only its direction (cf. Ms 167, §§ 176–177). He defines the curvature radius (“radius osculi”) or curvature (“radius curvedinis”), which he wants to determine now using the normal force: Let (cf. Fig. 9) $Mr$ and $\mathrm{d}s=M \mu$ be the normal and tangential line elements associated with the element $Mm$ of the curve defining the curvature radius $r$ in $M$, and $p$ be the normal force component (assuming the ratio of the normal to the gravitational force as $p$ to $1$). The altitude $v$ that corresponds to the body’s velocity is equal to $p \cdot Mr$ (cf. Euler 1736, Prop. 25, Coroll. 5). This velocity corresponds to a distance $2Mr$ covered by the body during the time element $\frac{2Mr}{\sqrt{Mr \cdot p}}$. This time element is equal to that one used by the body to cover the distance $M \mu$, which is $\frac{\mathrm{d}s}{\surd v}$. By equating these two time elements Euler derives the equation

$$\begin{aligned} Mr = \frac{p\,\mathrm{d}s^2}{4v}. \end{aligned}$$

(5)

Using the equivalence $Mr:Mm=Mm:2r$ and assuming $Mm \approx M \mu =\mathrm{d}s$, Euler solves this for $Mr$ and substitutes it in Eq. (5), obtaining (cf. Ms 167, §178)

$$\begin{aligned} r = \frac{2v}{p}, \end{aligned}$$

(6)

which agrees with the result of Ms 180 (cf. Verdun 2012, Appendix B, Eq. (1), considering that $p:1=N:A$).

Fig. 9

Euler’s sketch in Ms 167 explaining the curvature radius

Let $P$ be the force acting on the body, $\mathrm{d}s=\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}$ the line element of its curved path that it covers in the time element (cf. Ms 167, §179, and Fig. 10). Then the normal and tangential components are given by $\frac{P\,\mathrm{d}x}{\mathrm{d}s}$ and $\frac{P\,\mathrm{d}y}{\mathrm{d}s}$, respectively. Again using the “law” relating the increment of velocity with tangential force (cf. Euler 1736, Prop. 25, Coroll. 5), Euler obtains

$$\begin{aligned} \mathrm{d}v = \frac{-P\,\mathrm{d}y}{\mathrm{d}s}\mathrm{d}s = -P\,\mathrm{d}y. \end{aligned}$$

(7)

He substitutes the normal force $\frac{P\,\mathrm{d}x}{\mathrm{d}s}$ into Eq. (6) and equates the result with the formula for the curvature radius $r=\frac{\mathrm{d}s\,\mathrm{d}y}{\mathrm{dd}x}$ derived from differential geometry, obtaining

$$\begin{aligned} P\,\mathrm{d}x\,\mathrm{d}y = 2v\,\mathrm{dd}x. \end{aligned}$$

(8)

The substitution of Eq. (7) into the last one gives

$$\begin{aligned} \frac{\mathrm{d}v}{v} + \frac{2\,\mathrm{dd}x}{\mathrm{d}x} = 0, \end{aligned}$$

(9)

whose integral is

$$\begin{aligned} v\,\mathrm{d}x^2 = C\,\mathrm{d}s^2, \end{aligned}$$

(10)

where $C$ denotes the integration constant. If $x=0$ then $v=b$ and $\mathrm{d}y:\mathrm{d}s=f:1$, therefore $C=bg^2$, where $g=\sqrt{1-f\!f}$. From Eq. (10) Euler obtains for the altitude

$$\begin{aligned} v = \frac{bg^2\,\mathrm{d}s^2}{\mathrm{d}x^2} \end{aligned}$$

(11)

and for the corresponding velocity

$$\begin{aligned} \surd v = \sqrt{\frac{bg^2\,\mathrm{d}s^2}{\mathrm{d}x^2}}. \end{aligned}$$

(12)

Fig. 10

Euler’s sketch in Ms 167 explaining the tangential and normal force components

Euler considers first parabolic and circular orbits using this result (cf. Ms 167, §§ 183–189), before he treats the general central force problem (cf. Ms 167, §190f, and Fig. 11). Let BMA be the curve described by a body $M$ due to a central force located in $C$. Let the distance between $M$ and $C$ be $y$ and the central force be $P$, supposing the gravitational force $=1$. Let further the length of the perpendicular CT to the tangent MT be $p$ and $MT=\sqrt{yy-pp}=t$. The normal and tangential components of the central force are thus given by $\frac{Pp}{y}$ and $\frac{Pt}{y}$, respectively. Let $v$ be the altitude corresponding to the velocity of $M$ and $v+\mathrm{d}v$ that of $m$. Let $r$ be the curvature radius in $M$. Due to $mr=\mathrm{d}y$ and the equivalence of the triangles CMT and $Mmr$, it follows (cf. Ms 167, §191) $Mr=\frac{p\,\mathrm{d}y}{t}$ and $Mm=\frac{y\,\mathrm{d}y}{t}$. Therefore, due to the tangential force which acts in reverse direction to the motion of $M$, the increment of its velocity becomes

$$\begin{aligned} \mathrm{d}v = - \frac{Pt}{y} \frac{y\,\mathrm{d}y}{t} = - P\,\mathrm{d}y. \end{aligned}$$

(13)

According to Eq. (6), the curvature radius is given by $r=\frac{2vy}{Pp}$. On the other hand, it is also defined by $r=\frac{y\,\mathrm{d}y}{\mathrm{d}p}$. Therefore,

$$\begin{aligned} \frac{y\,\mathrm{d}y}{\mathrm{d}p} = \frac{2vy}{Pp} \quad \mathrm{or} \quad Pp\,\mathrm{d}y = 2v\,\mathrm{d}p. \end{aligned}$$

(14)

The substitution of Eq. (13) into Eq. (14) gives $ppv=C$. Euler sets $p=f$ and $v=b$ to obtain $C=bf\!f$ and thus $ppv=bf\!f$ and $v=\frac{bf\!f}{pp}$. Assuming $bf\!f$ being constant, this result gives the velocity $\surd v$ of the body $M$ defined by the parameter $p$ (cf. Ms 167, §192). Let $T$ be the time needed by the body to cover the arc BM, which is given by the known velocity of the body. The time element $\text{ d }T$ needed to cover the arc $Mm$ is defined by

$$\begin{aligned} \frac{Mm}{\surd v} = \frac{py\,\mathrm{d}y}{tf \surd b} = \frac{y \cdot Mr}{f \surd b}, \end{aligned}$$

(15)

whereby the relations $Mm=\frac{y\,\mathrm{d}y}{t}$, $Mr=\frac{p\,\mathrm{d}y}{t}$, and $v=\frac{bf\!f}{pp}$ were used. Therefore,

$$\begin{aligned} \text{ d }T = \frac{2 \cdot [\mathrm{area}] MCm}{f \surd b}, \end{aligned}$$

(16)

and the time needed to cover the arc BM is thus given by

$$\begin{aligned} dT = \frac{2 \cdot \mathrm{area} \textit{BCM}}{f \surd b}. \end{aligned}$$

(17)

Euler concludes that the time intervals are proportional to the areas covered by the associated radius vectors, which is Kepler’s area law (cf. Ms 167, §193). He substitutes $v=\frac{bf\!f}{pp}$ into Eq. (14) to obtain Keill’s theorem (cf. Ms 167, §194):

$$\begin{aligned} P = \frac{2bf\!f\,\mathrm{d}p}{p^3\,\mathrm{d}y}. \end{aligned}$$

(18)

He derived this theorem already in Ms 398, fol. 26v (see also Euler 1736, Prop. 74, Coroll. 4, § 592; Keill 1708). Euler admits that the application of this theorem is so difficult if the body’s curve, either algebraic or transcendent, can not be reduced to an equation which is general in a way that the curve may be expressed by an equation interrelating orthogonal coordinates (cf. Ms 167, §195):

Fig. 11

Euler’s sketch in Ms 167 used to derive the force components

Aequatio, quæ hoc modo ad curvam invenitur a corpore prolubitu projecto et a vi data $P$ ubique sollicitato [descriptam], est inter distantiam corporis a centro $y$ et inter perpendiculum ad tangentem ex centro $p$. Cum vero ut plurimum difficile sit hinc judicare, qualis sit curva utrum algebraica an transcendens, reducam æquationem hanc generalem ad æquationem inter coordinatas orthogonales quemadmodum curvæ exprimi solent.

(This equation found in this way for the curve described by a body that is propelled in any manner and acted anywhere by a given force $P$, depends on the distance $y$ between the body and the center, and on the perpendicular $p$ from the center to the tangent. But because it is extremely difficult to assess from it what kind of curve – algebraic or transcendent – it is, I will reduce here this general equation to an equation in terms of rectangular coordinates in the same way like curves usually are described.)

We skip Euler’s application of this result to special cases, which is not relevant for the development of his lunar theory, and turn to his investigations on the motion of a body $M$ being subject of two force centers $C$ and $D$. Let the ratio between the central force in $C$ acting on $M$ and the gravitational force as the ratio between the distances CM and the constant straight line $a$, and let the ratio between the central force in $D$ acting on $M$ and the gravitational force as the ratio between the distances $DM$ and the constant straight line $b$. Therefore $\frac{CM}{AM}=a$ and $\frac{DM}{BM}=b$, assuming the gravitational force being $=1$. Let $F$ be the common center of mass of the attracting bodies $C$ and $D$ with masses $\frac{1}{a}$ and $\frac{1}{b}$, respectively, where CF : DF $=$ $a:b$. These two force centers thus act as one single force center being their common center of mass (cf. Ms 167, §253, and Fig. 12).

Fig. 12

Reconstruction of the missing figure in Ms 167 to illustrate two simultaneously acting force centers $C$ and $D$

Let $2\,$ ME be the force acting on $M$ by $F$. Using the law of sines, Euler derives (cf. Ms 167, §254)

$$\begin{aligned} \sin \textit{CMF} : \sin \textit{CMD} = \textit{MB} : 2\,\textit{ME} = (\textit{CF} \cdot \textit{MD}) : (\textit{CD} \cdot \textit{MF}), \end{aligned}$$

(19)

therefore

$$\begin{aligned} \textit{ME} : \textit{MF} = (\textit{CD} \cdot \textit{MB}) : 2\,(\textit{CF} \cdot \textit{MD}) \end{aligned}$$

(20)

and

$$\begin{aligned} (\textit{CF} + \textit{DF}) (\textit{CD}) : \textit{CF} = (a+b) : a. \end{aligned}$$

(21)

He concludes, that

$$\begin{aligned} \textit{ME} : \textit{MF} = (a+b) \textit{MB} : (2a \cdot \textit{MD}) \end{aligned}$$

(22)

and

$$\begin{aligned} 2\,\textit{ME} = \frac{(a+b)\textit{MF} \cdot \textit{MB}}{a \cdot \textit{MD}} = \frac{(a+b)\textit{MF}}{ab} = \textit{MF} \left( \frac{1}{a} + \frac{1}{b} \right) . \end{aligned}$$

(23)

Later, Euler considers the more general case of two acting force centers, where the centripetal forces are no longer proportional to the distances and where one unique common center of force is not determinable (cf. Ms 167, §256, and Fig. 13). However, he investigates only the coplanar case, where the whole orbit BM of $M$ is situated in a plane together with the force centers $C$ and $D$, which attract $M$ with any ratio of the distance (cf. Ms 167, §257). Let $v$ be the altitude corresponding to the velocity of $M$ in the point $M$ of its curve, and $v+\mathrm{d}v$ its velocity in the point $m$, which is situated infinitesimally close to $M$. Let this line element $Mm$ of the curve be $\mathrm{d}s$. Euler denotes CM $=$ y, DM $=$ Y, CT $=p$, DV $=$ $\pi$, MT $=\sqrt{yy-pp}=q$, and MV $=\sqrt{YY- \pi \pi }=\rho$. Let further the curvature radius in $M$ be $=r$ and the distance CD between the two force centers $c$. From the geometry of the two perpendiculars CT and $DV$ to the tangent line TV through $M$ it is easily seen that TV $=q-\rho$ and $DV-CT=\pi -p$. Therefore (cf. Ms 167, §258)

$$\begin{aligned} cc = qq - 2q \rho + \rho \rho + pp - 2p \pi + \pi \pi = yy + YY -2q \rho - 2 \rho \pi . \end{aligned}$$

(24)

Due to the equivalence of the triangles CMT and DVM with their infinitesimal small counterparts it follows $\mathrm{d}s = \frac{y\,\mathrm{d}y}{q}$ and $\mathrm{d}s = \frac{Y\,\mathrm{d}Y}{\rho }$, respectively, and therefore

$$\begin{aligned} \mathrm{d}s = \frac{y\,\mathrm{d}y}{q} = \frac{Y\,\mathrm{d}Y}{\rho }. \end{aligned}$$

(25)

From § 191 we know that the curvature radii are defined either by $r=\frac{y\,\mathrm{d}y}{\mathrm{d}p}$ or by $r=\frac{Y\,\mathrm{d}Y}{\mathrm{d}\pi }$, giving the relation

$$\begin{aligned} \frac{y\,\mathrm{d}y}{\mathrm{d}p} = \frac{Y\,\mathrm{d}Y}{\mathrm{d}\pi }. \end{aligned}$$

(26)

By differentiation of Eq. (24) Euler obtains

$$\begin{aligned} y\,\mathrm{d}y + Y\,\mathrm{d}Y = q\,\mathrm{d}\rho + \rho \,\mathrm{d}q + p\,\mathrm{d}\pi + \pi \,\mathrm{d}p. \end{aligned}$$

(27)

Eqn. (24) may also be written as $\rho = q - \sqrt{cc-(p - \pi )^2}$, and from the combination of the Eqs. (25) and (26) one obtains $\rho = \frac{q\,\mathrm{d}\pi }{\mathrm{d}p}$. Therefore

$$\begin{aligned} q = \frac{\mathrm{d}p \sqrt{cc - (p - \pi )^2}}{\mathrm{d}p - \mathrm{d}\pi } \quad \mathrm{and} \quad \rho = \frac{\mathrm{d}\pi \sqrt{cc - (p - \pi )^2}}{\mathrm{d}p - \mathrm{d}\pi }. \end{aligned}$$

(28)

This result for $q$, substituted into $y=\sqrt{pp+qq}$, gives (cf. Ms 167, §259)

$$\begin{aligned} y = \frac{\sqrt{cc\,\mathrm{d}p^2+pp(\mathrm{d}p-\mathrm{d}\pi )^2-(p-\pi )^2\,\mathrm{d}p^2}}{\mathrm{d}p-\mathrm{d}\pi } \end{aligned}$$

(29)

and analogously

$$\begin{aligned} Y = \frac{\sqrt{cc\,\mathrm{d}\pi ^2+\pi \pi (\mathrm{d}p-\mathrm{d}\pi )^2-(p-\pi )^2\,\mathrm{d}\pi ^2}}{\mathrm{d}p-\mathrm{d}\pi }, \end{aligned}$$

(30)

representing the two distances of the body $M$ from the two force centers $C$ and $D$ used to determine the resulting force acting on $M$. Euler denotes these two forces with $P$ and $Q$, respectively, and decomposes them into the normal and tangential components (cf. Ms 167, §260). Let $\frac{Pp}{y}$ and $\frac{Q \pi }{Y}$ be the normal components of $P$ and $Q$, and its sum $\frac{Pp}{y} + \frac{Q \pi }{Y}$ be their resulting normal force. According to § 178 and assuming the gravitational force $=1$, this normal force becomes

$$\begin{aligned} \frac{Pp}{y} + \frac{Q \pi }{Y} = \frac{2v}{r}, \end{aligned}$$

(31)

from which it follows

$$\begin{aligned} v = \frac{Ppr}{2y} + \frac{Q \pi r}{2Y}. \end{aligned}$$

(32)

Substituting Eq. (26) into this result, Euler obtains

$$\begin{aligned} v = \frac{Pp\,\mathrm{d}y}{2\,\mathrm{d}p} + \frac{Q \pi \,\mathrm{d}Y}{2\,\mathrm{d}\pi }. \end{aligned}$$

(33)

Let $\frac{Pq}{y}$ and $\frac{Q \rho }{Y}$ be the tangential components of $P$ and $Q$, respectively, and its sum $\frac{Pq}{y} + \frac{Q \rho }{Y}$ their resulting tangential force (cf. Ms 167, §261). Using the result derived in § 176, the increment (taken negative against the force direction) of the body’s velocity becomes

$$\begin{aligned} \mathrm{d}v = - \frac{Pq\,\mathrm{d}s}{y} - \frac{Q \rho \,\mathrm{d}s}{Y}. \end{aligned}$$

(34)

Combining this result with Eq. (25) gives

$$\begin{aligned} \mathrm{d}v = - P\,\mathrm{d}y - Q\,\mathrm{d}Y, \end{aligned}$$

(35)

whose integral is

$$\begin{aligned} v = C - \int P\,\mathrm{d}y - \int Q\,\mathrm{d}Y, \end{aligned}$$

(36)

where $C$ denotes the integration constant, which has to be determined using the given initial velocity in a given location of the body’s orbit. Euler derived this result already in Ms 180, fol. 3r (cf. Verdun 2012). Equating the Eqs. (33) and (36) allows him to determine the two forces $P$ and $Q$. Without demonstration Euler presents the result (cf. Ms 167, §262):

$$\begin{aligned} P = \frac{\pi \,\mathrm{d}p\,\mathrm{d}v + 2v\,\mathrm{d}p\,\mathrm{d}\pi }{\mathrm{d}y\,(p\,\mathrm{d}\pi - \pi \,\mathrm{d}p)} \quad \mathrm{and} \quad Q = \frac{p\,\mathrm{d}\pi \,\mathrm{d}v + 2v\,\mathrm{d}\pi \,\mathrm{d}p}{\mathrm{d}Y\,(\pi \,\mathrm{d}p - p\,\mathrm{d}\pi )}. \end{aligned}$$

(37)

In Ms 180 Euler left the integro-differential equation leading to this result still unsolved, which provides evidence for the conjecture, that Ms 167 must have been written some time later. Euler comments this result as follows:

Fig. 13

Euler’s sketch in Ms 167 to illustrate multiple acting force centers

Ex his apparet si detur curva quæcunque et corporis in ea moti celeritas in singulis locis, insuper duo quæcunque puncta, inveniri posse vires ad ea puncta tendentes, quæ faciant ut corpus libere eam curvam describat, et in singulis locis celeritates habeat datas.

(Thus it is clear: if any curve and the velocity of a body moving on it is given for each point, and in addition any two points [of the curve are given], then it is possible to find forces directed on to these points, which cause the body to follow this curve freely and to assume the given velocity in each point.)

However, there still remains the problem to determine the nature of the curve supposing the two central forces $P$ and $Q$ are given. For that purpose the parameters $p$, $q$, $\pi$, and $\rho$ of $y$ and $Y$, which define $\mathrm{d}s$ and $r$, have to be determined. From Eq. (25) it follows that

$$\begin{aligned} q=\frac{y\,\mathrm{d}y}{\mathrm{d}s} \quad \mathrm{and} \quad \rho = \frac{Y\,\mathrm{d}Y}{\mathrm{d}s}. \end{aligned}$$

(38)

This substituted in $q=\sqrt{yy-pp}$ and $\rho =\sqrt{YY-\pi \pi }$ gives (cf. Ms 167, §263)

$$\begin{aligned} p = \frac{y \sqrt{\mathrm{d}s^2-\mathrm{d}y^2}}{\mathrm{d}s} \quad \mathrm{and} \quad \pi = \frac{Y \sqrt{\mathrm{d}s^2-\mathrm{d}Y^2}}{\mathrm{d}s}. \end{aligned}$$

(39)

From the equations $r=\frac{y\,\mathrm{d}y}{\mathrm{d}p}$ and $r=\frac{Y\,\mathrm{d}Y}{\mathrm{d}\pi }$ Euler has the relations

$$\begin{aligned} \mathrm{d}p = \frac{y\,\mathrm{d}y}{r} \quad \mathrm{and} \quad \pi = \frac{Y\,\mathrm{d}Y}{r}. \end{aligned}$$

(40)

This substituted in the first derivatives of $q=\sqrt{yy-pp}$ and $\rho =\sqrt{YY-\pi \pi }$ and considering Eq. (25) gives

$$\begin{aligned} \mathrm{d}p = \mathrm{d}s - \frac{y \sqrt{\mathrm{d}s^2-\mathrm{d}y^2}}{r} \quad \mathrm{and} \quad \mathrm{d}\rho = \mathrm{d}s - \frac{Y \sqrt{\mathrm{d}s^2-\mathrm{d}Y^2}}{r}. \end{aligned}$$

(41)

Substituting Eqs. (38) and (39) into $cc=(q-\rho )^2-(p-\pi )^2$, Euler obtains (cf. Ms 167, §264)

$$\begin{aligned} \mathrm{d}s^2 = \frac{4yY (cc\,\mathrm{d}y\,\mathrm{d}Y + yY\,\mathrm{d}y^2 + yY\,\mathrm{d}Y^2 - yy\,\mathrm{d}y\,\mathrm{d}Y - YY\,\mathrm{d}y\,\mathrm{d}Y)}{2ccyy + 2ccYY + 2y^2Y^2 - c^4 - y^4 - Y^4}. \end{aligned}$$

(42)

If this element of the body’s curve is known, one can find the associated osculting radius $r$ in terms of $y$ and $Y$ together with its differentials, using $r=\frac{y\,\mathrm{d}y}{\mathrm{d}p}$ and Eq. (39). Assuming $\mathrm{d}s$ to be constant, the first derivative of $p$ in Eq. (39) gives

$$\begin{aligned} \mathrm{d}p = \frac{\mathrm{d}s^2\,\mathrm{d}y - \mathrm{d}y^3 - y\,\mathrm{d}y\,\mathrm{dd}y}{\mathrm{d}s \sqrt{\mathrm{d}s^2 - \mathrm{d}y^2}}. \end{aligned}$$

(43)

Therefore, the resulting curvature radius becomes

$$\begin{aligned} r = \frac{y\,\mathrm{d}s \sqrt{\mathrm{d}s^2-\mathrm{d}y^2}}{\mathrm{d}s^2-\mathrm{d}y^2-y\,\mathrm{dd}y} = \frac{Y\,\mathrm{d}s \sqrt{\mathrm{d}s^2-\mathrm{d}Y^2}}{\mathrm{d}s^2-\mathrm{d}Y^2-Y\,\mathrm{dd}Y}. \end{aligned}$$

(44)

It may also be expressed in terms of the acting central forces $P$ and $Q$ by equating (36) and (32) (cf. Ms 167, §265):

$$\begin{aligned} 2C-2\int P\,\mathrm{d}y - 2\int Q\,\mathrm{d}Y = \frac{Ppr}{y} + \frac{Q \pi r}{Y}, \end{aligned}$$

(45)

from which the nature of the curve can be described using Eq. (39):

$$\begin{aligned} 2C-2\int P\,\mathrm{d}y - 2\int Q\,\mathrm{d}Y = \frac{Pp \sqrt{\mathrm{d}s^2-\mathrm{d}y^2}+Qr \sqrt{\mathrm{d}s^2-\mathrm{d}Y^2}}{\mathrm{d}s} \end{aligned}$$

(46)

and thus the resulting curvature radius becomes

$$\begin{aligned} r = \frac{2C\,\mathrm{d}s - 2\,\mathrm{d}s \int P\,\mathrm{d}y - 2\,\mathrm{d}s \int Q\,\mathrm{d}Y}{P \sqrt{\mathrm{d}s^2-\mathrm{d}y^2} + Q \sqrt{\mathrm{d}s^2-\mathrm{d}Y^2}}. \end{aligned}$$

(47)

If the curve that the body describes and one of the central forces, say $P$, are given, then the other central force $Q$ can be determined as well (cf. Ms 167, §266). Let be $\frac{pr}{2y}=x$ and $\frac{\pi r}{2Y}=z$. Therefore, $v=Px+Qz$ and

$$\begin{aligned} \mathrm{d}v = P\,\mathrm{d}x + x\,\mathrm{d}P + Q\,\mathrm{d}z + z\,\mathrm{d}Q = - P\,\mathrm{d}y - Q\,\mathrm{d}Y \end{aligned}$$

(48)

or

$$\begin{aligned} \mathrm{d}Q + Q \frac{\mathrm{d}Y+\mathrm{d}z}{z}+\frac{P\,\mathrm{d}y + x\,\mathrm{d}P + P\,\mathrm{d}x}{z} = 0. \end{aligned}$$

(49)

Multiplying this by $\mathrm{e}^{\int \frac{\mathrm{d}Y+\mathrm{d}z}{z}}$ Euler obtains

$$\begin{aligned} \mathrm{e}^{\int \frac{\mathrm{d}Y+\mathrm{d}z}{z}}Q = D - \int \frac{\mathrm{e}^{\int \frac{\mathrm{d}Y+\mathrm{d}z}{z}} (P\,\mathrm{d}y+x\,\mathrm{d}P+P\,\mathrm{d}x)}{z}, \end{aligned}$$

(50)

where the integration constant $D$ can be determined from the initial conditions.

Finally, Euler treats the case where one of the force centers may be considered as infinitely far away from the other (cf. Ms 167, §267, and Figs. 14, 15). (This situation may, in fact, be assumed for first order approximations in the case of the Earth-Moon-Sun system, where the Sun is regarded as infinitely further away from the Moon than the Earth.) Supposing the force center $D$ situated in infinity, or $c=\infty$. Therefore MD is parallel to CD. Let CP be a perpendicular to CD and denote PM $=z$. Thus $Y=c+z$ and $\mathrm{d}Y=\mathrm{d}z$ holds. Let $Q$ be a function of $z$. From these assumptions, substituted into Eqs. (42) and (44), Euler obtains

$$\begin{aligned} \mathrm{d}s = \frac{\sqrt{yy\,\mathrm{d}y^2 + yy\,\mathrm{d}z^2 - 2yz\,\mathrm{d}y\,\mathrm{d}z}}{\sqrt{yy-zz}} \end{aligned}$$

(51)

and

$$\begin{aligned} r = \frac{\mathrm{d}s \sqrt{\mathrm{d}s^2-\mathrm{d}z^2}}{-\mathrm{dd}z}, \end{aligned}$$

(52)

where $\mathrm{d}z$ is supposed to be constant. The other equation for the curvature radius, Eq. (47), becomes

$$\begin{aligned} r = \frac{2C\,\mathrm{d}s - 2\,\mathrm{d}s \int P\,\mathrm{d}y - 2\,\mathrm{d}s \int Q\,\mathrm{d}z}{P \sqrt{\mathrm{d}s^2-\mathrm{d}y^2} + Q \sqrt{\mathrm{d}s^2-\mathrm{d}z^2}}. \end{aligned}$$

(53)

Equating these results (Eqs. 52 and 53) gives

$$\begin{aligned} \frac{2\,\mathrm{dd}z}{\sqrt{\mathrm{d}s^2-\mathrm{d}z^2}} = \frac{P\sqrt{\mathrm{d}s^2-\mathrm{d}y^2}+Q\sqrt{\mathrm{d}s^2-\mathrm{d}z^2}}{\int P\,\mathrm{d}y + \int Q\,\mathrm{d}z - C}. \end{aligned}$$

(54)

The second possible situation concerns the case where the force center $C$ is also regarded as infinitely far away from the body $M$. Euler considers this case as well, which is, however, of minor importance for the development of his lunar theory and thus may be skipped here.

Fig. 14

Euler’s sketch in Ms 167 illustrating the first case where the force center is infinitely far away

Fig. 15

Euler’s sketch in Ms 167 illustrating the second case where the force center is infinitely far away

Appendix B: The content of Ms 271

In this small treatise entitled “De Motu Lunæ in Ellipsin” Euler develops a formula to construct lunar tables based on the Moon’s elliptic motion represented by its orbital velocity $v$ at any point of the trajectory. He uses two physical principles, one of them implicitly, the other one explicitly. The first one concerns the equation of motion and the balance of centrifugal and centripetal (gravitational) force $F$ allowing him to determine the Moon’s velocity $v$ by Huygens’ centrifugal formula (cf. Euler 1736, §630; see also Bomie 1708):

$$\begin{aligned} F \propto \frac{v^2}{r}, \end{aligned}$$

(55)

where $r$ designates the curvature radius or radius of the osculating circle at each point of the trajectory. This formula may be reconstructed using Euler’s “Mechanica” which he completed already in 1734. From the equation of motion, given by (cf. §157)

$$\begin{aligned} c\,\mathrm{d}c = \frac{np \,\mathrm{d}s}{A}, \end{aligned}$$

(56)

where $c$ is the velocity of the body with mass $A$, $\mathrm{d}s$ the line element covered by the body, $p$ is the force acting on this body and $n$ is a constant, Euler derives the formula (cf. §163)

$$\begin{aligned} npr \mathrm{d}x = Ac^2\,\mathrm{d}s, \end{aligned}$$

(57)

where $r$ is the curvature radius and $\mathrm{d}x$ the line element covered by the body perpendicular to $r$, regarding $\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^2$. This is exactly Huygens formula, because $\mathrm{d}s=c \mathrm{d}t$ and

$$\begin{aligned} np \frac{\mathrm{d}x}{\mathrm{d}s} = \frac{Ac^2}{r}. \end{aligned}$$

(58)

If $p$ is collinear with $r$ as it is the case in central force motions, then $\mathrm{d}y=0$ and $\mathrm{d}x=\mathrm{d}s$, so that this relation becomes (cf. §165)

$$\begin{aligned} r = \frac{Ac^2}{np}. \end{aligned}$$

(59)

Euler always expressed the velocity $c$ by its corresponding height of fall $v$, so that $v=c^2$ (cf. §202). This implies $n=\frac{1}{2}$ (cf. §206) and thus $r=\frac{2Av}{p}$. From this formula, Euler extracts (cf. §552)

$$\begin{aligned} v = \frac{r}{2} \frac{p}{A}. \end{aligned}$$

(60)

Although this derivation is not included in Euler’s treatise Ms 271, it seems to be part of the standard repertoire of physical principles belonging at Euler’s disposal in that time (cf. Varignon 1707a, §XXIII, Corol. 2, p. 198). The second principle makes sure that the first one holds only if the center of force (i.e., the center of the Earth) is at rest with respect to inertial space, or in other words, that the Moon’s motion is described by this approach in an earth fixed reference frame. This condition requires a principle which had never been used before in that context and which was developed as innovative element and applied here for the first time by Euler: the principle of the transference of forces. It became a standard method already in Euler’s “Mechanica” (cf. Euler 1736, Prop. 97, §795). This principle, applied to Euler’s determination of the Moon’s motion, means that the force component acting on the Earth by the Sun has to be transferred in contrary direction to the Moon. Evidence for its first appearance here is given by the peculiarity that Euler had to state more precisely the way how to apply the inverse force by some deletions and by inserting a note referring to another marginal note.

Let (see Fig. 16) $C$, $M$, and $N$ be the centers of the Earth, Moon, and Sun, respectively, which are assumed to be situated in one and the same ecliptic plane. AMBA is the Moon’s elliptical orbit coplanar with the ecliptic. AB is the apsidal line, $A$ the apogee, and $B$ the perigee. The Moon is situated in $M$. The straight line GMTR is the tangent in $M$. MP is the perpendicular through $M$ on the apsidal line AB, CO the perpendicular through $C$ on the straight line joining $M$ and $N$, $TC$ and $RN$ are the perpendicular lines through $C$ and $N$ on the tangent line GMTR, respectively. Euler denotes the masses of Earth and Sun, respectively, by $A$ and $S$, and defines the lengths of the major and minor axes of the Moon’s orbit $=2a$ and $=2c$, respectively, and the length of the major axis of the Sun’s orbit $=2f$. He sets the Earth’s global radius $=1$ and denotes the distances CM $=$ y and CN $=$ z, and designates the curvature or curvature radius in $M$ with $r$. He first determines the resulting normal components to the tangent line GMTR of the gravitational forces acting on the Moon and the Earth. The gravitational forces acting on the Moon by the Earth and the Sun are given by $A:yy$ and $S:{\textit{MN}}^2,$ respectively, the one acting on the Earth by the Sun is given by $S:zz$. The normal components of the first two forces are given by $A \cdot {\textit{CT}}: y^3$ and $S \cdot$ $N\!R:M\!N^3$. The normal component of the latter is given by $S \cdot {\textit{FG}}:{\textit{MF}} \cdot zz$, where the relation $M\!F:F\!G=C\!N:N\!R-CT$ is defined by the equivalence of the correspondent triangles. This normal component becomes thus $=S \cdot ({\textit{NR}}-{\textit{CT}}\,\,):z^3$. Now Euler applies his principle of the transference of forces when summing up these normal components to the resulting centripetal forces acting on the Moon, giving $=A \cdot {\textit{CT}}:y^3+S \cdot {\textit{NR}}:{\textit{MN}}^3-S({\textit{NR}}-{\textit{CT}}\,\,):z^3$. Note that the latter force actually is the normal component of the gravitational force acting on the Earth by the Sun, applied in the reverse direction on the Moon, thus making sure that its motion refers to the Earth resting in inertial space.

In the next step Euler determines the orbital velocity $v$ of the Moon as a function of the curvature radius $r$ in $M$. Using the first principle (cf. eq. 60) described in words by

Fig. 16

The figure of Ms 271 explaining the geometry in the System Earth ($C$), Moon ($M$), and Sun ($N$), AMBA being the elliptic lunar orbit with $C$ in one focus

“Est vero vis gravitatis $=A$ et cum sit vis normalis ad vim gravitatis ut altitudes velocitatem generans ad dimidium radii osculi [...]”,

Euler derives

$$\begin{aligned} v=r \cdot CT:2y^3 + S \cdot r \cdot \textit{NR}: 2A \cdot \textit{MN}^3 - Sr(\textit{NR}-\textit{CT }):2Az^3. \end{aligned}$$

(61)

Without giving any derivation Euler substitutes

$$\begin{aligned} {\textit{CT}} = c \surd y : (a-y) \end{aligned}$$

(62)

and the curvature radius (cf. Varignon 1707b)

$$\begin{aligned} r=(2ay-yy)^{\frac{3}{2}}:ac, \end{aligned}$$

(63)

and sets approximately ${\textit{MN}}={\textit{CN}}+{\textit{MO}}=z+{\textit{MO}}$, thus yielding

$$\begin{aligned} v= \frac{2a-y}{2ay} + \frac{Syy(2a-y)}{2Aaz^3} - 3S \cdot r \cdot {\textit{NR}} \cdot {\textit{MO}}:2Az^4. \end{aligned}$$

(64)

Finally, Euler substitutes the product ${\textit{NR}} \cdot {\textit{MO}}$ by trigonometric functions, resulting in

$$\begin{aligned} v&= (2a-y):2ay + Syy(2a-y):2Aaz^3 \nonumber \\&- 3Syr \left( \sin {\textit{CMT}} + \sin [{\textit{CMT}} + 2{\textit{MCN}}] \right) :4Az^3, \end{aligned}$$

(65)

and the mass ratio $S:A$ by $S:A=2f^3:(8795)^2=2f^3 \cdot n$, leading to

$$\begin{aligned} v&= (2a-y):2ay + f^3yy(2a-y):naz^3 \nonumber \\&- 3f^3yr \left( \sin {\textit{CMT}} + \sin [{\textit{CMT}}+2{\textit{MCN}}]\right) :2nz^3. \end{aligned}$$

(66)

Euler remarks that the first term depends only on the Earth’s attraction, but that the second and third depend on the Sun’s gravitational force. In addition, the second term depends on the Moon’s position in its orbit and on the Earth’s distance from the Sun, while the third term depends on the lunar aspect, namely on the angle between Moon and Sun. He further notes that in $v$ the excess of the terms depending on the Sun over the terms not depending on it is always reciprocal to the cube of the Sun’s distance from the Earth. This is why the Moon’s velocity produced by the Sun is much smaller than that generated by the Earth. The difference in the lunar time period due to the Earth’s and the Sun’s action is thus reciprocal to the cube of the Earth’s distance from the Sun, as well.

Due to the fact that $z \gg y$, Euler approximates $z=f$, resulting in

$$\begin{aligned} v&= (2a-y):2ay + yy(2a-y):na \nonumber \\&- 3yr \cdot \left( \sin {\textit{CMT}} + \sin [{\textit{CMT}}+2{\textit{MCN}}]\right) :2n. \end{aligned}$$

(67)

If the lunar orbit is considered as circle with radius $a$, his velocity is then given by

$$\begin{aligned} v = 1: 2a - aa:2n - 3aa \cdot \cos 2{\textit{MCN}} : 2n. \end{aligned}$$

(68)

Euler derives some corollaries for the elliptic orbit case, defined by the distance $\sqrt{aa-cc} = b$ of the focus from the center of the ellipse, the distance of the apogee $=a+b$ and the perigee $=a-b$. If the Moon is located in the transverse axis or apsidal line, there are two cases: In apogee, $y=a+b$, $r=cc:a$, and $\sin {\textit{CMT}} = 1$, resulting in

$$\begin{aligned} v=(a-b):2a(a+b)-cc(a+b):2an - 3cc(a+b) \cos 2{\textit{MCN}} :2an.\quad \quad \end{aligned}$$

(69)

In perigee, $y=a-b$ and $r=cc:a$, leading to

$$\begin{aligned} v=(a+b):2a(a-b)-cc(a-b):2an -3cc(a-b) \cos 2{\textit{MCN}} : 2na.\quad \quad \end{aligned}$$

(70)

If oppositions or conjunctions happen to be in the apogee (upper sign) and in the perigee (lower sign), then

$$\begin{aligned} v=(a \mp b):2a(a \pm b) - 2cc(a \pm b):na, \end{aligned}$$

(71)

but if quadratures happen to be in the apogee (upper sign) and in the perigee (lower sign), then

$$\begin{aligned} v=(a \mp b):2a(a \pm b) + cc(a \pm b):an. \end{aligned}$$

(72)

If the Moon is located in the conjugate axis, then $y=a$, $r=aa:c$, and $\sin {\textit{CMT}} = c:a$, yielding

$$\begin{aligned} v=1:2a - aa:2n - 3aa \left( c \cos 2{\textit{MCN}} + b \sin 2{\textit{MCN}}\right) :2cn. \end{aligned}$$

(73)

If oppositions or conjunctions happen to be in the conjugate axis, the sine and cosine of the angle $2{\textit{MCN}}$ become $\sin 2{\textit{MCN}} = 0$ and $\cos 2{\textit{MCN}} = 1$, respectively, and therefore

$$\begin{aligned} v=1:2a - 2aa:n. \end{aligned}$$

(74)

But if quadratures happen to be in the conjugate axis, then $\sin 2{\textit{MCN}} = 0$ and $\cos 2{\textit{MCN}} = -1$, therefore

$$\begin{aligned} v=1:aa + aa:n. \end{aligned}$$

(75)

This is an important result derived by Euler, proving that in syzygies the Moon’s velocity is always smaller, in the quadratures always larger than it’s mean motion. Moreover, every time when the Moon in its orbit is in conjunction or opposition, then—due to $\angle {\textit{MCN}}=0^{\circ }$ resp. $\angle {\textit{MCN}}=180^{\circ }$—its velocity becomes

$$\begin{aligned} v=(2a-y):2ay + yy(2a-y):na - 3yr \sin {\textit{CMT}} : n. \end{aligned}$$

(76)

But $\sin {\textit{CMT}} = {\textit{CT}}:{\textit{CM}} = c: \surd (2ay-yy)$, so that in syzygies

$$\begin{aligned} v=(2a-y):2ay - 2yy(2a-y):na, \end{aligned}$$

(77)

and in quadratures

$$\begin{aligned} v=(2a-y):2ay+yy(2a-y):na. \end{aligned}$$

(78)

Euler reduces the general equation (67) by using the relations $\sin {\textit{CMT}} = c: \sqrt{2ay-yy}$ and $r \sin {\textit{CMT}} = (2ay-yy):a$, as well as the addition theorem, gaining

$$\begin{aligned} v&= (2a-y):2ay - yy(2a-y):2na \nonumber \\&- 3yy(2a-y) \cos 2MCN : 2na \nonumber \\&- 3yy (2a-y) \sqrt{2ay-yy-cc} \cdot \sin 2M\!C\!N : 2nac. \end{aligned}$$

(79)

Now he tries to make this formula more handy for the calculation of lunar tables. Using the geometry of the ellipse and introducing the ratio $T:t$ of the periods of revolution (i.e., the mean motions of the sidereal periods) of Sun and Moon, respectively, by $\overline{T-t}:T= 559:605$, he is able to substitute

$$\begin{aligned} \sin 2M\!C\!N = \sin (2D - 1{,}118 \cdot AC\!M : 605), \end{aligned}$$

(80)

where $D$ designates the Sun’s elongation from the apsidal line ${\textit{AD}}$ as reference line of these motions. Furthermore, he substitutes $\cos {\textit{ACM}} = q$, $y=cc:(a-bq)$, $cc:b=g$, and $a:b=K$, yielding $y=g:(K-q)$ and $2a-y=2a-g:(K-q)$. He sets $\sin {\textit{ACM}} =p = \sqrt{1-yy}$, having $\sqrt{2ay-yy-cc}=$ and $bcp:(a-bq)=cp:(K-q)$. Finally, Euler determines the Moon’s angular velocity (“velocitates angulares lunæ”) $u$ expressed in terms of equivalent altitude of fall corresponding to the circular motion in $M$ and defined by the curvature radius $r$ in that point:

$$\begin{aligned} u&= \frac{cc\,v}{y^3(2a-y)} \nonumber \\&= cc:2ay^4-cc:2nay-3cc \cos 2M\!C\!N : 2nay \nonumber \\&- 3c \sqrt{2ay-yy-cc} \cdot \sin 2M\!C\!N : 2nay \nonumber \\&= (K-q)^4:2g^3K - (K-q):2nK - 3(K-q) \cos 2M\!C\!N : 2nK \nonumber \\&- 3p \sin 2MCN : 2nK \nonumber \\&= (K-q)^4:2g^3K-(K-q):2nK \nonumber \\&- 3 \cos 2MCN :2n + 3 \cos (2M\!C\!N+AC\!M) :2nK \nonumber \\&= (K-q)^4:2g^3K-(K-q):2nK -3 \cos (2D-1118 \cdot AC\!M:605):2n \nonumber \\&+ 3 \cos (2D-513 \cdot ACM:605):2n. \end{aligned}$$

(81)

In this result Euler inserts the values $a=60$ and $b=3$, yielding $c=\surd 3{,}591$, $g=1{,}197$, and $K=20$. In order to gain practical numbers for $u$, he scales $u$,

$$\begin{aligned} 1{,}000{,}000{,}000{,}000\,u&= (1{,}000{,}000)^2 \Big ((20-q)^4:40 \cdot 1197^3 - (20-q):40n \nonumber \\&- 3 \cos (2D-1{,}118 \cdot ACM : 605):2n \nonumber \\&- 3 \cos (2D - 513\,ACM : 605):40n \Big ), \end{aligned}$$

(82)

so that the final result becomes

$$\begin{aligned} 1{,}000{,}000 \surd u&= 3.817949\,(20-q)^2 - \frac{42.3261}{(20-q)} \nonumber \\&- \frac{2{,}539.566}{(20-q)^2} \cos \left( 2D - \frac{1{,}118 \cdot {\textit{ACM}}}{605}\right) \nonumber \\&+ \frac{126.9783}{(20-q)^2} \cos \left( 2D - \frac{513\,{\textit{ACM}}}{605}\right) . \end{aligned}$$

(83)

Euler remarks that the parameter $q$ must successively be substituted with the cosine of the angles $1^{\circ }, 2^{\circ }, 3^{\circ }, 4^{\circ }$ etc. In this way one may construct lunar tables depending on the angle $ACM$ as table argument and on the angle $D$ between the Sun and the lunar apogee, measured from the apogee, which may be varied by, e.g, $30^{\circ }$ or $45^{\circ }$, represented by different tables, as well.

Appendix C: The content of Ms 273

This manuscript fragment consists of six Propositions, each of them followed by Corollaries. Unfortunately, all thirteen figures to which Euler refers in the text margins are missing.

Proposition I: Given the mass or active quantity (“quantitas activa”) of a body being at rest, find the motion of another body rotating around the former.

This proposition concerns the determination of the orbital (Keplerian) elements of the two-body-problem. Let $M$ be the mass of the Earth, $r$ its radius, $S$ the mass of the central body (i.e., the Sun) located in one focus of the elliptic orbit (cf. Fig. 17, which was adopted from Ms 397, fol. 121v). Let $A$ be the aphelion and $A\!S=f$ the distance between $A$ and $S$. Let $K$ and $v$ be the altitudes corresponding to the velocities of the body in $A$ and in $M$, respectively. Let further be $S\!M=y$ and $Mn=\mathrm{d}y$, where $n$ denotes a point infinitesimally close to $M$ on the curve, and $S\!P=p$, where $P$ is the intersection between the tangent line through $M$ and the perpendicular through $S$ to this tangent. Let $v+\mathrm{d}v$ be the altitude corresponding to the velocity of the body in $m$ infinitesimally close to the point $M$ on the curve. Due to the dynamic relation

$$\begin{aligned} \mathrm{d}y : \mathrm{d}v = \frac{M}{rr} : \frac{S}{yy}, \end{aligned}$$

(84)

and due to the fact that $\mathrm{d}v$ decreases when $\mathrm{d}y$ increases, Euler starts with the equivalence

$$\begin{aligned} \frac{M\,\mathrm{d}v}{rr} = \frac{-S\,\mathrm{d}y}{yy}, \end{aligned}$$

(85)

whose integral is

$$\begin{aligned} \frac{Mv}{rr} = \frac{S}{y} + A, \end{aligned}$$

(86)

where $A$ denotes the integration constant (not to be confused with the aphelion). If the body $M$ is located in the aphelion, then $v=K$ and $y=f$ holds, so that

$$\begin{aligned} A = \frac{M\!K}{rr} - \frac{S}{f}. \end{aligned}$$

(87)

Substituting Eq. (87) into (86) gives

$$\begin{aligned} \frac{Mv-M\!K}{rr}=\frac{Sf-Sy}{fy} \quad \mathrm{or} \quad v= \frac{Sfrr-Srry}{Mfy}+K. \end{aligned}$$

(88)

Using the relation $K:v=pp: f\!f$, Euler eliminates $v$ in Eq. (88) and solves the result for $p$:

$$\begin{aligned} pp = \frac{Mf^3Ky}{Sfrr-Srry+MfyK}. \end{aligned}$$

(89)

If the body is situated in the perihelion, then $p=y$, and Eq. (89) becomes

$$\begin{aligned} Mf^3K - MfKy^2 - Sfrry + Srry^2 = 0, \end{aligned}$$

(90)

or, after dividing by $f-y$,

$$\begin{aligned} Mf\!fK + MfKy - Srry = 0. \end{aligned}$$

(91)

Solving this equation for the perihelion distance $y \equiv B\!S$ gives

$$\begin{aligned} BS = \frac{Mf\!fK}{Srr-MfK}. \end{aligned}$$

(92)

Thus, the major axis $a \equiv AB$ (“axis transversus”) becomes

$$\begin{aligned} AB = \frac{Sfrr}{Srr - MfK}. \end{aligned}$$

(93)

Substituting this result into Eq. (89) Euler obtains

$$\begin{aligned} pp = \frac{Maf^3Ky}{Safrr-Sfrry} = \frac{Maf\!fKy}{Sarr - Srry}. \end{aligned}$$

(94)

The distance between the two foci is given by $=\frac{Sfrr-2Mf\!fK}{Srr-MfK}$, which Euler uses to determine the length of the minor axis $c$ (“axis conjugatur orbitae”):

$$\begin{aligned} c = 2f \sqrt{\frac{MfK}{Srr-MfK}} = 2f \surd \frac{MaK}{Srr}. \end{aligned}$$

(95)

Therefore

$$\begin{aligned} \frac{Srrcc}{4f\!f}=MaK \end{aligned}$$

(96)

and Eq. (94) becomes

$$\begin{aligned} pp=\frac{ccf\!fy}{4f\!f(a-y)}=\frac{ccy}{4a-4y}. \end{aligned}$$

(97)

The time element used to cover the infinitesimally small distance $Mm$ is given by

$$\begin{aligned} \frac{Mm}{\surd v}= \frac{p \cdot Mm}{f \surd K}. \end{aligned}$$

(98)

Using the relations $p:y=mn:Mm$ or $p \cdot Mm = y \cdot mn$ and thus $mn=\frac{p\,\mathrm{d}y}{\sqrt{yy-pp}}$, Euler obtains for the time element $=\frac{py\,\mathrm{d}y}{f \sqrt{K(\overline{yy-pp}}}$. But

$$\begin{aligned} p=\frac{c}{2} \surd \frac{y}{a-y} \end{aligned}$$

(99)

and

$$\begin{aligned} \sqrt{yy-pp}=\frac{1}{2} \sqrt{\frac{4ayy-4y-ccy}{a-y}}, \end{aligned}$$

(100)

therefore

$$\begin{aligned} \frac{py\,\mathrm{d}y}{f \sqrt{K(\overline{yy-pp})}} = \frac{cy\,\mathrm{d}y}{f \sqrt{4aKy-4Kyy-ccK}}, \end{aligned}$$

(101)

whose integral is the time of revolution. The time element is defined by the uniform straight line motion of the body from $m$ to $n$, and thus given by $\frac{y \cdot mn}{f \cdot \surd K}$. The infinitesimally small area covered in this time is $y \cdot mn = 2\,MSm$. Let the total aera of the orbit’s ellipsis be $A$. The time of revolution (“tempus periodicum”) $T$ is then defined by

$$\begin{aligned} T=\frac{2A}{f \surd K}. \end{aligned}$$

(102)

Let $1:\pi$ be the ratio between the radius of a circle and its perimeter. ⁴ Hence the area of a circle with radius $a$ becomes $\frac{aa \pi }{8}$. The ratio $a:c$ between the major and minor axes of the ellipse may also be expressed by the equivalent fractions $\frac{aa \pi }{8} : \frac{ac \pi }{8}$, from which Euler infers to the area of the ellipsis $A=\frac{ac \pi }{8}$. This substituted into Eq. (102) gives

$$\begin{aligned} T=\frac{\pi ac}{4f \surd K}. \end{aligned}$$

(103)

Substituting Eq. (96) into this result, one obtains

$$\begin{aligned} T=\frac{\pi a \surd Ma}{2r \surd S}. \end{aligned}$$

(104)

When asking for the absolute time of revolution, the variables $a$ and $r$ in this formula have to be expressed in Rhinelandian (Prussian) feet (cf. Ms 167, Section A above), so that the time of revolution in seconds of time is given by

$$\begin{aligned} T=\frac{\pi a \surd 10\,Ma}{50r \cdot \surd S}. \end{aligned}$$

(105)

This proposition is followed by nine corollaries. In the first one Euler solves Eq. (93) for $K$ and substitutes the result into Eq. (88), which gives

$$\begin{aligned} v = \frac{Srr(a-y)}{May}. \end{aligned}$$

(106)

Naming the moving body $M$ by “the mobile”, Euler concludes from this result, that the square of the mobile’s velocity is proportional to the distance $MF$ from the non-attracting focus $F$ and reciprocal to the distance $MS$ from the focus $S$, being the center of force. He notes in a second corollary, that $v=0$ if $y=a$. In the third corollary he solves Eq. (93) for $f$ and substitutes the result into Eq. (96), obtaining thus

$$\begin{aligned} c = \frac{2ar \surd MSaK}{Srr - MaK}. \end{aligned}$$

(107)

In the forth corollary Euler denotes the time of revolution by $T$, which is—according to Eq. (104)—proportional to $\frac{a \surd a}{\surd S}$. He interprets this result in the fifth corollary for the cases where several bodies move around one and the same center and around multiple centers. In the remaining corollaries he evaluates Eq. (105) numerically for the motions of the Earth and of Venus, and determines the mass ratio $S:M$ between the Sun and Earth, obtaining 1,053,531:1. Euler adopted the time of revolution for Venus, “$224$.d.$17$.h.$44^{\prime }.55^{\prime \prime }.$” (cf. Ms 273, fol. 2v), most probably from the 1726 edition of Gregory’s book, from which he possessed a copy in his own library (cf. Gregory 1726, p. IV from the “Editoris præfatio”).

Proposition II: Let $S$ be a [central] body attracting in any way, and $M$ a mobile thus describing any curve located in any point $M$; determine the ellipsis, in which the mobile proceeds, if the formers force immediately decreases according to the squared ratio of the distances.

Let $v$ be the altitude [of free fall] corresponding to the velocity of $M$ and $SM=y$. Let $P$ be the force acting on $M$ by $S$, and $a$ be the major axis (“axis principalis ellipsis”), of which the arc element $Mm$ is an infinitesimally small part. Let the attracting central force $P$ be $=\frac{S}{yy}$ and $c$ be the minor axis of the ellipsis. From Eq. (106) of Corollary I it follows

$$\begin{aligned} v = \frac{Prry(a-y)}{Ma}. \end{aligned}$$

(108)

Euler reformulates Eq. (96) using the relation $Kf\!f = vpp$ and obtains

$$\begin{aligned} Srrcc=4Maf\!fK. \end{aligned}$$

(109)

He substitutes this result into Eq. (93), so that the major axis becomes

$$\begin{aligned} a = \frac{Prryy}{Prry - Mv}. \end{aligned}$$

(110)

When $y$ changes into $y+\mathrm{d}y$, the major axis increases by the element

$$\begin{aligned} \frac{PPr^4yy\,\mathrm{d}y + PMrryy\,\mathrm{d}v-2PMrrvy\,\mathrm{d}y-Mrryyv\,\mathrm{d}P}{(Prry-Mv)^2}. \end{aligned}$$

(111)

Euler inserts the relation $Kf\!f = vpp$ into Eq. (109) and obtains the parameter of the orbit (“parametro orbitæ”):

$$\begin{aligned} \frac{cc}{a} = \frac{4Mvpp}{Prryy}. \end{aligned}$$

(112)

If $y$ increases, then this parameter will also increase by the element (as first derivative of $\frac{4Mvpp}{Prryy}$, where $vpp$ is assumed to be constant)

$$\begin{aligned} \frac{-4Mrrvppyy\,\mathrm{d}P-8MPrrvppy\,\mathrm{d}y}{PPr^4y^4}. \end{aligned}$$

(113)

Euler concludes from this result, that the apsidal line will move according to the order of the signs (“in consequentia”), if the major axis $a$ increases by this element. Analogously, if this element decreases, the major axis decreases as well. But if the major axis either increases or decreases, this element decreases or increases, respectively, and the apsidal line will retrograde against the order of the signs (“in antecedentia”). This is why a motion of the apsidal line can be recognized in any case.

Fig. 17

Euler’s sketch of Ms 397 explaining the central force motion of the body $M$ around the central body $S$, $A$ and $B$ being the apo- and pericenter, respectively

This proposition is followed by seven corollaries. In the first one Euler inserts the relation $\mathrm{d}y:\mathrm{d}v=\frac{M}{rr}:P$ or $\mathrm{d}v = \frac{-Prr\,\mathrm{d}y}{M}$ into Eq. (111) and obtains

$$\begin{aligned} \frac{-2PMrrvy\,\mathrm{d}y-Mrryyv\,\mathrm{d}P}{(Prry-Mv)^2}. \end{aligned}$$

(114)

This element increases the major axis when $y$ increases its element $\mathrm{d}y$. In the second corollary he relates the increments of the parameter and the major axis to each other. The ratio of these increments is as $\frac{4pp}{PPr^4y^4}$ to $\frac{1}{(Prry-Mv)^2}$, which is as $4pp(Prry-Mv)^2$ to $PPr^4y^4$ or as $4pp$ to $aa$. In the third corollary he introduces—probably for the first time in his works on celestial mechanics—a “generalized law of gravitation”, defining

$$\begin{aligned} P=\frac{N}{y^n}, \end{aligned}$$

(115)

where $N$ denotes an arbitrary constant of proportionality. The first derivative of Eq. (115) is given by

$$\begin{aligned} \mathrm{d}P=\frac{-Nn\,\mathrm{d}y}{y^{n+1}}. \end{aligned}$$

(116)

Euler substitutes these Eqs. (115) and (116) into the increment of the major axis (Eq. 114) and of the parameter (Eq. 113), obtaining

$$\begin{aligned} \frac{\overline{n-2} \cdot MNrry\,\mathrm{d}y}{y^{n-1}(Prry-Mv)^2} = \frac{\overline{n-2} \cdot PMrrvy\,\mathrm{d}y}{(Prry-Mv)^2}= \frac{\overline{n-2} \cdot Maav\,\mathrm{d}y}{Pr^2y^3} \end{aligned}$$

(117)

and

$$\begin{aligned} \frac{\overline{n-2} \cdot 4Mppv\,\mathrm{d}y}{Pr^2y^3}, \end{aligned}$$

(118)

respectively. He is thus able to infer important consequences from these results, which are also relevant for the motion of the lunar apses. He concludes in the forth and fifth corollary that, if $n>2$ and $y$ increases or decreases, then the major axis will increase or decrease as well, respectively. If the force $P$ decreases in a ratio larger than the square of the distances, then the major axis of the ellipsis increases with increasing distances of the mobile from the focus, and decreases with decreasing distances from it. The contrary is the case if $n<2$. If therefore the central force $P$ decreases in a superior ratio than in the squared, then—so Euler continues in corollary VI—the apsidal line moves according to the order of the signs (“in consequentia”). In the other case, it moves against the signs (“in antecedentia”). In corollary VII he determines the ratio between the orbit’s velocity, i.e., the velocity of the apsidal line, and the angular velocity of the mobile. Let ${\textit{PMQ}}$ be the tangent line through the mobile $M$, ${\textit{SP}}$ and ${\textit{FQ}}$ perpendiculars through the foci $S$ and $F$ to this tangent. Let $m$ be a point of the ellipsis infinitesimally close to $M$ achieved by the body while proceeding in his motion on the curve and while the major axis is moving from $F$ to $f$ by the element $Ff$. Thus the apsidal line progresses around $S$ by the angle $FSf$. Let $Fg$ and ${\textit{MR}}$ be the perpendiculars through $F$ and $M$ to the apse ${\textit{SF}}$. Therefore, the ratio between the angular motion of the body and the progressive motion of the apsidal line is given by

$$\begin{aligned} \frac{Mn}{{\textit{SM}}} : \frac{Fg}{{\textit{SF}}} = \frac{{\textit{PS}} \cdot mn}{{\textit{PM}} \cdot {\textit{SM}}} : \frac{{\textit{MR}} \cdot Ff}{{\textit{FM}} \cdot {\textit{SF}}}. \end{aligned}$$

(119)

Let ${\textit{SM}}+{\textit{FM}}=a$ be the length of the major axis, ${\textit{SM}}=y$, ${\textit{SP}}=p$, ${\textit{PM}}=q$, and thus ${\textit{FM}}=a-y$. From Eq. (117) Euler concludes

$$\begin{aligned} Ff = \frac{\overline{n-2} \cdot Maav\,\mathrm{d}y}{Pr^2y^3} = \frac{\overline{n-2}\cdot aab\,\mathrm{d}y}{4p^2y}, \end{aligned}$$

(120)

where $b$ denotes the parameter defined by $b=\frac{4Mvpp}{Prryy}$. He expresses the ratio of Eq. (119) by

$$\begin{aligned} \frac{Mn}{{\textit{SM}}} : \frac{Fg}{{\textit{SF}}}&= \frac{{\textit{PS}}}{{\textit{PM}} \cdot {\textit{SM}}} : \frac{\overline{n-2}\cdot aab \cdot {\textit{MR}}}{4\,SP^2 \cdot {\textit{PM}} \cdot {\textit{FM}} \cdot {\textit{SF}}} \nonumber \\&= 4\,PS^2 \cdot {\textit{SF}} \cdot {\textit{FM}} : \overline{n-2} \cdot aab \cdot {\textit{SM}} \cdot {\textit{MR}}. \end{aligned}$$

(121)

Using the relation SF : $SP=2MQ$  : MR, the ratio between the angular velocity of the mobile and the angular velocity of the apsidal line becomes (regarding $mn=\mathrm{d}y$)

$$\begin{aligned}&4 \,{\textit{PS}}^3 \cdot {\textit{SF}}^2 \cdot {\textit{FM}} : \overline{n-2} \cdot 2aab \cdot {\textit{SM}} \cdot {\textit{MQ}} \cdot {\textit{SP}} \nonumber \\&\quad = 2 {\textit{PS}}^2 \cdot {\textit{SF}}^2 \cdot {\textit{FM}} : \overline{n-2} \cdot aab \cdot {\textit{SM}} \cdot {\textit{MQ}}. \end{aligned}$$

(122)

Considering the fact that

$$\begin{aligned} ab = \frac{4pp(a-y)}{y} = \frac{4{\textit{SP}}^2 \cdot {\textit{FM}}}{{\textit{SM}}}, \end{aligned}$$

(123)

the ratio of Eq. (122) finally becomes

$$\begin{aligned}&= 2{\textit{PS}}^2 \cdot {\textit{SF}}^2 \cdot {\textit{FM}} : \overline{n-2} \cdot 4a\,SP^2 \cdot {\textit{FM}} \cdot {\textit{MQ}} \nonumber \\&= {\textit{SF}}^2 : \overline{2n-4} \cdot a{\textit{MQ}} = SF^2 : \overline{2n-4} \cdot {\textit{FM}} \cdot {\textit{PQ}}. \end{aligned}$$

(124)

Proposition III: One asks for the force of the Sun causing the perturbation of the Moon’s motion.

In order to determine the Moon’s motion in an adequate manner, it has to be described with respect to an immobile location. Thus suppose the Earth to be immobile, and therefore consider the total motion impressed to the system of bodies always to be equal and inverse directed to the Earth’s motion. The forces acting on the Earth have to be considered as acting in reverse direction on the whole system. Actually, Euler neglects the Moon’s mass and does not consider the force acting on the Sun by the Earth. Let $T$ be the Earth, ${\textit{ALB}}$ a part of the Moon’s orbit, $S$ the Sun, $L$ the Moon, and ${\textit{LT}}$, ${\textit{LS}}$ its distances from the Earth and Sun, respectively. Let ${\textit{TS}}$ be the distance between the Earth and the Sun. Let $M$ and $S$ be the masses of the Earth and the Sun, respectively. The forces acting on the Moon $L$ by the Earth $T$ and the Sun $S$ are given by $\frac{M}{{\textit{TL}}^2}$ and $\frac{S}{SL^2}$, respectively. The force acting on the Earth by the Sun is given by $\frac{S}{TS^2}$. To consider the Earth to be at rest means that this force $\frac{S}{T\!S^2}$ has to be applied to the Moon in the reverse sense along the direction parallel to TS (Euler wrote erroneously ${\textit{TL}}$). The force, with which the Moon is attracted to the Sun along the direction ${\textit{LS}}$ has to be decomposed into the two directions ${\textit{LT}}$ and ${\textit{TS}}$. Therefore, the forces acting on the Moon by the Sun along ${\textit{LT}}$ and along ${\textit{TS}}$ (Euler wrote erroneously ${\textit{TL}}$) become $\frac{S \cdot {\textit{TL}}}{({\textit{LS}})^3}$ and $\frac{S \cdot {\textit{TS}}}{{\textit{LS}}^3}$, respectively. Taking into account all these force components, the resulting forces acting on the Moon along ${\textit{LT}}$ and ${\textit{TS}}$ becomes

$$\begin{aligned} \frac{M}{{\textit{TL}}^2} + \frac{S \cdot {\textit{TL}}}{{\textit{LS}}^3} = \frac{M \cdot {\textit{LS}}^3 + S \cdot {\textit{TL}}^3}{{\textit{TL}}^2 \cdot {\textit{LS}}^3} \end{aligned}$$

(125)

and

$$\begin{aligned} \frac{S \cdot {\textit{TS}}}{{\textit{LS}}^3} - \frac{S}{{\textit{TS}}^2} = \frac{S \cdot {\textit{TS}}^3 - S \cdot {\textit{LS}}^3}{{\textit{TS}}^2 \cdot {\textit{LS}}^3}. \end{aligned}$$

(126)

Due to Corollary IX of Proposition I, the ratio of the masses between the Sun and the Earth is $S:M=1{,}053{,}531:1$ or $S=1{,}053{,}531 \cdot M$. Euler substitutes this relation into Eq. (125) and obtains the force acting on the Moon along the Earth’s direction $=\frac{M}{TL^2 \cdot LS^3} ( LS^3 + 1{,}053{,}531 \cdot {\textit{TL}}^3 )$. He knew approximately the ratio between the distances as ${\textit{TS}}:{\textit{LT}}=572:1$. As an approximation he assumes ${\textit{LS}} \approx {\textit{TS}}$ and obtains for this force

$$\begin{aligned} \frac{M}{{\textit{TL}}^2} \left( 1 + \frac{1{,}053{,}531}{(572)^2}\right) = \frac{M}{{\textit{TL}}^2} \left( \frac{178}{177}\right) . \end{aligned}$$

(127)

The ratio between the force acting on the Moon by the Earth when disregarding the Sun’s action and the force when regarding the Sun’s action is as 177–178.

Euler adds four corollaries to this proposition which, however, are of minor importance and thus will be skipped here, because they concern only numerical estimations and comparisons of the forces determined for different characteristic points of the lunar orbit.

Proposition IV: Suppose the force acting on the Moon by the Sun is decomposed into two components: one component which attracts the Moon towards the Earth, and one which attracts the Moon along the direction parallel to the conjugation line between the Sun’s and the Earth’s center. One asks for the perturbation of the Moon’s motion caused by the force component attracting towards the Earth.

Let the Moon be attracted towards the Earth by the force $\frac{M \cdot L\!S^3 + S \cdot T\!L^3}{T\!L^2 \cdot L\!S^3}$ according to Eq. (125). Euler denotes LT=y and LS=z. Thus this force becomes

$$\begin{aligned} P=\frac{Mz^3+Sy^3}{y^2z^3}. \end{aligned}$$

(128)

If $y$ goes over into $y+\mathrm{d}y$ and $z$ into $z+\mathrm{d}z$, the increment of the major axis—let us call it $\mathrm{d}a$—may be determined using Corollary I of Proposition II. The first derivative of Eq. (128) is given by

$$\begin{aligned} \mathrm{d}P \!=\! \frac{Sz^3y^4\,\mathrm{d}y \!-\! 2Mz^6y\,\mathrm{d}y \!-\! 3Sy^5z^2\,\mathrm{d}z}{y^4z^6} \!=\! \frac{Szy^3\,\mathrm{d}y\!-\!2Mz^4\,\mathrm{d}y \!-\!3Sy^4\,\mathrm{d}z}{y^3z^4}. \end{aligned}$$

(129)

This result may now be substituted into Eq. (114) to determine the variation of the major axis. In addition, Eq. (108) may be written as

$$\begin{aligned} Prry - Mv = \frac{Prryy}{a}, \end{aligned}$$

(130)

so that the increment or variation of the major axis $\mathrm{d}a$ becomes

$$\begin{aligned}{}[\mathrm{d}a] = \frac{3Saay - 3Sayy}{Mz^4+Sy^3z}(y\,\mathrm{d}z - z\,\mathrm{d}y). \end{aligned}$$

(131)

Euler denotes the major axis of the orbit by $a$, and the orbit’s parameter by $b$. Due to Corollary II of Proposition II the ratio between the increment of the parameter (let us call it d$b$) and the increment of the major axis $\mathrm{d}a$ is $4TP^2$ to $aa$, i.e., $by$ to $aa-ay$. He substitutes this relation into Eq. (131) and obtains for the increment of the parameter

$$\begin{aligned}{}[db] = \frac{3Sby^2}{Mz^4+Sy^3z}(y\,\mathrm{d}z - z\,\mathrm{d}y). \end{aligned}$$

(132)

To find out how the apsidal line moves when $y$ increases by the element $\mathrm{d}y$, one has to investigate the power of $y$ defining the decrease of the (gravitational) force. Let $n$ be this exponent. According to Eq. (117) the increment of the major axis becomes

$$\begin{aligned}{}[\mathrm{d}a] \!=\! \frac{\overline{n\!-\!2} \cdot Maav\,\mathrm{d}y}{Pr^2y^3} \!=\! \frac{\overline{n\!-\!2} \cdot \overline{aa\!-\!ay} \cdot \mathrm{d}y}{y^2} \!=\! \frac{3Sy \cdot (aa\!-\!ay)}{Mz^4\!+\!Sy^3z}(y\,\mathrm{d}z \!-\! z\,\mathrm{d}y),\nonumber \\ \end{aligned}$$

(133)

from which it follows

$$\begin{aligned} \overline{n-2} = \frac{3Sy^3(y\,\mathrm{d}z - z\,\mathrm{d}y)}{z\,\mathrm{d}y (Mz^3+Sy^3)}. \end{aligned}$$

(134)

In Proposition II, Corollary VII, Eq. (124), it was already shown that the ratio between the angular velocity of the Moon and the angular velocity of the apsidal line is given by $S\!F^2$ to $\overline{2n-4} \cdot a \cdot {\textit{MQ}}$. The distance between the foci of the ellipse is given by $S\!F^2=aa-ab$. From the geometry of Fig. 18 Euler concludes that

$$\begin{aligned} {\textit{MQ}}&= \frac{{\textit{PM}} \cdot {\textit{MF}}}{{\textit{SM}}} = \frac{\overline{a-y} \cdot {\textit{PM}}}{y} \nonumber \\&= \frac{\overline{a-y} \cdot \sqrt{4ayy-4y^3-aby}}{y \sqrt{4a-4y}} \nonumber \\&= \frac{\sqrt{(2ay-2yy)^2} - aby \cdot \overline{a-y}}{2y}. \end{aligned}$$

(135)

Therefore,

$$\begin{aligned} {\textit{SF}}^2 : \overline{n-2} \cdot aMQ&= aa-ab : \frac{\overline{n-2} \cdot a}{y} \sqrt{(2ay-2y^2)-by\, \overline{a-y}} \nonumber \\&= ay-by : \overline{n-2} \cdot \sqrt{(2ay-2yy)^2 - aby \cdot \overline{a-y}}.\nonumber \\ \end{aligned}$$

(136)

Using these results, $(n-2)$ can be substituted to obtain the ratio between the Moon’s angular velocity and the angular velocity of the apsidal line, representing—according to Euler—the solution of the problem:

$$\begin{aligned} \overline{a-b} \cdot z\,\mathrm{d}y (Mz^3+Sy^3) : 3Sy^2 (y\,\mathrm{d}z - z\,\mathrm{d}y) \sqrt{(2ay-2yy)^2 - aby \cdot \overline{a-y}}.\nonumber \\ \end{aligned}$$

(137)

Fig. 18

Reconstruction of Fig. 2 in Ms 273

This proposition is followed by eleven corollaries and one scholium. In Corollaries I–III Euler discusses the location and motion of the major axis according to the result achieved in Eq. (131) for special cases, depending on the values of $\mathrm{d}y$ and $\mathrm{d}z$. He uses the results obtained from these conclusions in Corollaries IV–VI to quantify them, assuming the ratio between the masses of Sun and Earth as $S=1{,}053{,}531\,M$, which was derived in Corollary IX of Proposition I. In addition, he supposes $z=572\,y$ as a result gained in Corollary IV of Proposition III, and uses the force ratio $177:178$ as main result of that Proposition. He approximates the distance between the Earth $T$ and the empty focus $F$ of the lunar orbit by $FT=5$ in units of the Earth’s radius. The numerical results are not an important issue for the reconstruction of the development of Euler’s lunar theory and may be skipped here. In the remaining four corollaries Euler analyzes the consequences which may be drawn from the ratio between the Moon’s angular velocity in any place of its orbit and the angular velocity of the lunar apses, using and interpreting Eq. (137), again in terms of the values of $\mathrm{d}y$, $\mathrm{d}z$, $y$, and $z$. For this purpose he considers the infinitesimally small angular sector defined by

$$\begin{aligned} MSm = \frac{{\textit{SP}} \cdot \mathrm{d}y}{{\textit{MP}} \cdot {\textit{SM}}}, \end{aligned}$$

(138)

The first factor of the denominator is given by

$$\begin{aligned} {\textit{MP}} = \frac{\sqrt{\overline{2ay-2yy}^2 -aby \cdot \overline{a-y}}}{2a-2y}, \end{aligned}$$

(139)

and of the nominator by

$$\begin{aligned} {\textit{SP}} = \frac{\surd aby}{2 \sqrt{a-y}}. \end{aligned}$$

(140)

He substitutes the last two equations into Eq. (138) and obtains

$$\begin{aligned} MSm = \frac{\mathrm{d}y\,\sqrt{aby \cdot \overline{a-y}}}{y \sqrt{\overline{2ay-2yy}^2 - aby \cdot \overline{a-y}}}. \end{aligned}$$

(141)

Using the relation (137), the angular element covered by the apsidal line—let us call it $\mathrm{d}\omega$—thus becomes

$$\begin{aligned}{}[\mathrm{d}\omega ] = \frac{3Sy(y\,\mathrm{d}z-z\,\mathrm{d}y)\surd aby \cdot \overline{a-y}}{\overline{a-b} \cdot z (Mz^3 + Sy^3)}, \end{aligned}$$

(142)

whose integral yields the motion of the lunar apse. To prepare this integration, Euler reformulates Eq. (142) to obtain

$$\begin{aligned}{}[\mathrm{d}\omega ] = \frac{3Syz \sqrt{aby \cdot \overline{a-y}}}{\overline{a-b}(Mz^3 + Sy^3)} \left( \frac{y\,\mathrm{d}z - z\,\mathrm{d}y}{zz}\right) . \end{aligned}$$

(143)

Assuming the first factor to be constant, the integration gives

$$\begin{aligned}{}[\omega ] =\frac{3Syz \sqrt{aby \cdot \overline{a-y}}}{\overline{a-y}(Mz^3 + Sy^3)} \left( \frac{B\!S}{e} - \frac{y}{z}\right) , \end{aligned}$$

(144)

where $\frac{BS}{e}$ is the constant of integration with $e$ denoting the distance between Sun and Moon, measured when the latter is in perigee. If the direction of the apsidal line happens to be in quadrature, so he concludes from Eq. (144), then $[\omega ]=41^{\prime }$ per month. The apsidal line thus regresses with $82^{\prime }=1^{\circ }22^{\prime }$ per anomalistic month against the series of the signs (“contra signorum seriem”). Euler does not comment on this result with regard to that one obtained by Newton (cf. Newton 1687, Lib. I, Prop. XLV, Coroll. II). In the Scholion that finishes this proposition, he states:

Vis hæc de qua egi in nodos nullum habet influxum sed eos immutatos et immotos sinit. directior [sic!] enim ejus in ipso plano orbitæ lunæ sita est, efficitque ut Luna in eadem semper plano immoto moveatur. Motus autem nodorum ut et motus lineæ absidum in consequentia debetur alteri vi partiali derivatæ a vi solis, agenti secundum parallelas rectæ Solem et terram jungentis.

(This force I have mentioned has no influence on the nodes, but leaves them unchanged and immobile. Its direction coincides with the plane of the Moon’s orbit and causes the Moon to move always in one and the same immobile plane. Consequently, the motion of the nodes and the motion of the apsidal line is due to the other force component, which comes from the Sun’s force acting along the parallel to the straight line conjoining the Sun and the Earth.)

Proposition V: Suppose the Moon moving in an ellipse and another force is acting on it, which may be derived from the Sun acting along the straight line conjoining the centers of the Sun and Earth; find the perturbation of the Moon’s orbit due to this force.

Let $S$ be the Sun, $T$ the Earth, and $L$ the Moon. Let CT be the line conjoining the Earth and the quadrature points of the lunar orbit. Let further $N$ be the intersection between the lines CT and ${\textit{LS}}$. The force repelling the Moon along the line $C\!N$ is, according to Eq. (126) of Proposition III, given by

$$\begin{aligned} \frac{S \cdot ({\textit{LS}}^3-{\textit{TS}}^3)}{{\textit{TS}}^2 \cdot {\textit{LS}}^3}. \end{aligned}$$

(145)

Denoting ${\textit{LS}}-{\textit{TS}}={\textit{LN}}$, Euler approximates ${\textit{LS}}^3-{\textit{TS}}^3$ by $3 \cdot {\textit{TS}}^2 \cdot {\textit{LN}}$, so that the force becomes

$$\begin{aligned} \frac{3 S \cdot {\textit{LN}}}{\textit{LS}^3}. \end{aligned}$$

(146)

Because $LS$ may be considered constant with respect to $LN$, the force acting on the Moon varies everywhere with its distance from the straight line conjoining the quadratures. And consequently, the Moon is steadily departing from its orbit when moving on the upper side of the quadratures, which is why the orbit becomes prolonged in opposition; in the same way it is pulled away from the line conjoining the quadratures and attracted towards the Sun in the lower part of the orbit, which is why the orbit is also prolonged in conjunction. These prolongations cause the apsidal line to change its position and to retrograde continuously; however, its progression dominates the retrogression easily, as will be shown later on. From the fact that all points of the orbit are pushed away from the line conjoining the quadratures and attracted along the straight line conjoining the centers of Earth and Sun, one may conclude that the line conjoining the nodes has to change, as well as the inclination of the Moon’s orbit with respect to the ecliptic. The angle of inclination must decrease when the Moon is located in the syzygies, because the Moon is then retracted by the Earth and its distance increases while the arc measuring the angle of inclination remains unchanged. This proposition is followed by eleven corollaries which are skipped here.

Proposition VI: Due to the Sun’s force the Moon is everywhere repelled away from the straight line connecting the quadratures; find how far the Moon is at any point dislocated from its orbit.

Suppose this force to act on a third body (which we would call today a “massless test body”) and investigate the effect which is produced on it. Let us begin by assuming, for the sake of simplicity and facility, that the apsidal line coincides with the syzygies (cf. Fig. 19). While the Moon proceeds from $C$ to $A$, the Sun’s force begins to repel the Moon away from the straight line CD due to the transference of the force acting on the Earth by the Sun to the Moon as explained in Proposition III. The resulting forces act along the parallels to the apsidal line and vanish in the line CD due to the Earth considered at rest by the transference of forces. Euler assumes that one of these forces act not on the Moon but on another body $a$ and attract it along the straight line $aP$. Suppose the Moon has arrived at $L$ and the body at $P$, and while the Moon proceeds to $l$, the body—starting from $P$—arrives at $p$. Euler denotes ${\textit{LT}}=y$ and $aP=x$. Let $w$ be the altitude (i.e., the height of fall) corresponding to the velocity of the body in $P$. Then $Pp=\mathrm{d}x$. Let $z$ be the distance between the Sun and the Moon; from $L$, $l$ take the perpendiculars ${\textit{LQ}}$, $lq$ to the (major) axis, defining thus ${\textit{TQ}}=t$ and $Qq=\text{ d }t$. The force moving the body in $P$ is given by Eq. (145). Starting again with Eq. (84) and using the same meaning for the symbols $M$ and $r$ as at the beginning of his treatise, Euler obtains

$$\begin{aligned} \frac{M}{rr} : \mathrm{d}x = \frac{3St}{z^3} : dw \quad \mathrm{or} \quad 3Srrt\,\mathrm{d}x = Mz^3\,dw. \end{aligned}$$

(147)

He assumes that the distances $Pp$ and $Ll$ are covered simultaneously, hence

$$\begin{aligned} Ll : Pp = \surd v : \surd w \quad \mathrm{or} \quad Ll^2 : \mathrm{d}x^2 = v : w. \end{aligned}$$

(148)

Euler denotes $LR=q$ (not to confuse with the point $p$ on the major axis), and concludes from the equivalence of triangles that $Ll=\frac{y\,\mathrm{d}y}{q}$. Therefore,

$$\begin{aligned} v\,\mathrm{d}x^2=\frac{wyy\,\mathrm{d}y^2}{qq} \quad \mathrm{or} \quad vqq\,\mathrm{d}x^2=wyy\,\mathrm{d}y^2. \end{aligned}$$

(149)

Let $a$ and $c$ be the length of the major and minor axes (not to confuse with the test body $a$), $b$ the parameter of the ellipse or “latus rectum”, and $f$ the distance between the foci. From properties of the geometry of the ellipsis, he derives

$$\begin{aligned} TQ = \frac{2ay-aa+f\!f}{2f} \equiv t \quad \mathrm{and} \quad LQ = \surd \frac{4accy-4ccy^2-c^4}{4aa-4cc}. \end{aligned}$$

(150)

Euler rewrites Eq. (100) as

$$\begin{aligned} q = \sqrt{\frac{4ayy-4y^3-ccy}{4a-4y}} \end{aligned}$$

(151)

to obtain for Eq. (149) the equation

$$\begin{aligned} v\,\mathrm{d}x^2 = \frac{(4a-4y)wy^2\,\mathrm{d}y}{4ayy-4y^3-ccy} = \frac{Srr \cdot (a-y)\,\mathrm{d}x^2}{May}, \end{aligned}$$

(152)

and consequently

$$\begin{aligned} \frac{Srr\,\mathrm{d}x^2}{Ma} = \frac{4wyy\,\mathrm{d}y^2}{4ay-4yy-cc} \quad \mathrm{or} \quad r\,\mathrm{d}x \sqrt{\frac{S}{Ma}} = \frac{2y\,\mathrm{d}y \surd w}{\sqrt{4ay-4yy-cc}}. \end{aligned}$$

(153)

Euler then substitutes Eq. (150) for $t$ into Eq. (147) and solves the result for $x$ to obtain

$$\begin{aligned} \mathrm{d}x&= \frac{2Mfz^3\,dw}{3Srr(2ay-cc)} \nonumber \\&= \frac{2Mz^3\,dw \sqrt{aa-ab}}{3Srr(2ay-ab)}\nonumber \\&= \frac{2y\,\mathrm{d}y \surd Maw}{r \surd S \cdot (4ay-4yy-ab)}, \end{aligned}$$

(154)

from which he concludes

$$\begin{aligned} \frac{2Mz^3\,dw \sqrt{a-b}}{3Sr \cdot \surd Mw} = \frac{2y\,\mathrm{d}y (2ay-ab)}{\surd S \cdot \overline{4ay-4yy-4b}}. \end{aligned}$$

(155)

He integrates this equation, assuming the denominator as constant, and obtains

$$\begin{aligned} \frac{2z^3 \surd Mw \cdot \overline{a-b}}{3r \surd S} = \frac{\frac{2}{3}ay^3-\frac{1}{2}aby^2-\frac{2}{3}a{\textit{CT}}^3 +\frac{1}{2}ab \cdot {\textit{CT}}^2}{\sqrt{aa-ab}}, \end{aligned}$$

(156)

which is, when setting ${\textit{CT}}=\frac{b}{2}$, equal to

$$\begin{aligned} \frac{2z^3 \cdot \overline{a\!-\!b}}{3r} \cdot \surd \frac{Mw}{Sa} \!=\! \frac{2}{3}y^3 \!-\! \frac{1}{2}by^2 \!-\! \frac{1}{12}b^3 \!+\! \frac{1}{8}b^3 \!=\! \frac{2}{3}y^3 \!-\! \frac{1}{2}by^2 \!+\! \frac{1}{24}b^3. \end{aligned}$$

(157)

He derives from Eq. (153)

$$\begin{aligned} \surd \frac{Mw}{Sa} = \frac{r\,\mathrm{d}x}{y\,\mathrm{d}y} \surd \frac{a-b}{a}, \end{aligned}$$

(158)

which may be substituted into Eq. (157) to obtain the differential equation for $\mathrm{d}x$,

$$\begin{aligned} \frac{2z^3\,\mathrm{d}x(a-b)^{\frac{3}{2}}}{3y\,\mathrm{d}y \surd a} = \frac{2}{3}y^3-\frac{1}{2}by^2+\frac{1}{24}b^3, \end{aligned}$$

(159)

the integral of which is

$$\begin{aligned} \frac{2z^3x(a-b)^{\frac{3}{2}}}{3\surd a} = \frac{2}{15}y^5-\frac{1}{8}by^4+\frac{1}{48}b^3y^2-\frac{1}{640}b^5 \end{aligned}$$

(160)

and therefore

$$\begin{aligned} x = \frac{3 \surd a}{2z^3 \cdot (a-b)^{\frac{3}{2}}} \left( \frac{2}{15}y^5 - \frac{1}{8}by^4 + \frac{1}{48}b^3y^2 - \frac{1}{640}b^5\right) . \end{aligned}$$

(161)

Euler sets $y=\frac{1}{2}b+s$, where $s$ denotes the exzess over $\frac{1}{2}b$, thus obtaining

$$\begin{aligned} \frac{2}{15}y^5-\frac{1}{8}by^4+ \frac{1}{48}b^3y^2-\frac{1}{640}b^5=\frac{1}{12}bbs^3, \end{aligned}$$

(162)

which may be substituted into Eq. (161) to obtain the rectangular distance of the dislocated position of $p$ from the line CD:

$$\begin{aligned} x = \frac{bbs^3 \surd a}{8z^3 (a-b)^{\frac{3}{2}}}, \end{aligned}$$

(163)

where $s$ contains the accumulation of all “$\mathrm{d}y$”, taken either as positive or negative. If it is introduced by $Ll \cdot \mathrm{d}y$, it has always to be taken as affirmative due to $Ll$.

Fig. 19

Reconstruction of Fig. 11 in Ms 273

Euler provides an alternative solution based upon the distance $LQ=t$. He reformulates the right hand side of Eq. (155) using Eq. (150): from the first one he derives

$$\begin{aligned} 2y = \frac{ac+f \sqrt{cc-4tt}}{c}, \end{aligned}$$

(164)

and from the second one

$$\begin{aligned} \sqrt{4ay-4yy-cc} = \frac{2ft}{c}. \end{aligned}$$

(165)

The first derivative of Eq. (164) gives

$$\begin{aligned} \mathrm{d}y = \frac{-2ft\,\text{ d }t}{c \sqrt{cc-4tt}}. \end{aligned}$$

(166)

Euler solves Eq. (165) for $y$ to obtain

$$\begin{aligned} 2ay-ab = \frac{f\!fc+af \sqrt{cc-4tt}}{c}. \end{aligned}$$

(167)

All these results are substituted into Eq. (155), which becomes

$$\begin{aligned} \frac{2Mz^3\,\mathrm{d}w \sqrt{a-b}}{3Sr \surd Mw} = \frac{- f\,\text{ d }t (fc+a \sqrt{cc-4tt})(ac+f \sqrt{cc-4tt})}{cc \surd S \cdot \overline{cc-4tt}}. \end{aligned}$$

(168)

He introduces $p$ and sets

$$\begin{aligned} \sqrt{cc-4tt}=p \quad \mathrm{or} \quad 2t=\sqrt{cc-pp}, \end{aligned}$$

(169)

from which he gains the first derivative

$$\begin{aligned} \text{ d }t = \frac{-p\,\mathrm{d}p}{2 \sqrt{cc-pp}}. \end{aligned}$$

(170)

He substitutes Eqs. (169) and (171) into Eq. (168), which becomes

$$\begin{aligned} \frac{2ccz^3\,\mathrm{d}w \surd M \cdot \overline{a\!-\!b}}{3fr \surd Sw}&= \frac{p\,\mathrm{d}p (fc\!+\!ap)(ac\!+\!fp)}{2p \sqrt{cc-pp}} \nonumber \\&= \frac{accf\,\mathrm{d}p \!+\! aacp\,\mathrm{d}p \!+\! cf\!fp\,\mathrm{d}p \!+\! afpp\,\mathrm{d}p}{2 \sqrt{cc-pp}}. \end{aligned}$$

(171)

The integration of the left hand side gives

$$\begin{aligned} \frac{4ccz^3 \surd Mw \cdot \overline{a-b}}{3fr \surd S}. \end{aligned}$$

(172)

The integral of the term $\frac{\overline{aa+f\!f} \cdot cp\,\mathrm{d}p}{2 \sqrt{cc-pp}}$ is $= \frac{-c(aa+f\!f)}{2} \sqrt{cc-pp}$. Euler prepares the remaining term for integration by the reformulation

$$\begin{aligned} af \cdot \frac{cc\,\mathrm{d}p + pp\,\mathrm{d}p}{2 \sqrt{cc-pp}}&= af \cdot \frac{2cc\,\mathrm{d}p^2 - cc\,\mathrm{d}p + pp\,\mathrm{d}p}{2 \sqrt{cc-pp}}\nonumber \\&= af \left( \frac{cc\,\mathrm{d}p}{\sqrt{cc-pp}} - \frac{\mathrm{d}p}{2} \sqrt{cc-pp}\right) . \end{aligned}$$

(173)

The integration is straightforward if the ellipse is approximated by a circle:

$$\begin{aligned} \int af \left( \frac{cc\,\mathrm{d}p}{\sqrt{cc-pp}} \!-\! \frac{\mathrm{d}p}{2} \sqrt{cc\!-\!pp}\right) \!=\! af \left( \frac{3}{2}{\textit{DCM}} \!-\! \frac{1}{4}p \sqrt{cc\!-\!pp}\right) , \end{aligned}$$

(174)

where ${\textit{DCM}}$ denotes the sector composed of the straight lines ${\textit{DC}}$ and CM, which inclose the arc MD of the semi-circle ${\textit{ADB}}$. Euler obtains thus the result

$$\begin{aligned} \frac{4ccz^3 \sqrt{Mw \cdot \overline{a-b}}}{3fr \surd S} = \frac{3}{2}af \cdot DCM \!-\! \overline{\frac{1}{4}afp \!-\! \frac{1}{2}aac - \frac{1}{2}f\!fc}\, \sqrt{c^2-p^2} + \mathrm{Const.}.\nonumber \\ \end{aligned}$$

(175)

Euler tries to “construct” this result in such a way that the areas involved in Eq. (175) can easily be computed, which is equivalent with an appropriate series expansion of this result. He substitutes Eqs. (164)–(166) into Eq. (153) and obtains

$$\begin{aligned} r\,\mathrm{d}x \surd \frac{S}{Maw} = \frac{-\text{ d }t (ac+f \sqrt{cc-4tt}}{c \sqrt{cc-4tt}} \quad \mathrm{or} \quad \surd \frac{Mw}{S} = \frac{-cr\,\mathrm{d}x \sqrt{cc-4tt}}{\text{ d }t (ac+f \sqrt{cc-4tt}) \surd a}.\nonumber \\ \end{aligned}$$

(176)

Using this result, the term (172) becomes

$$\begin{aligned} \frac{4ccz^3 \surd Mw \cdot \overline{a-b}}{3fr \surd S} = \frac{-4c^3z^3 \,\mathrm{d}x \cdot \sqrt{cc-4tt} \cdot \overline{a-b}}{3f\,\mathrm{d}t (ac+f \sqrt{cc-4tt}) \surd a} \end{aligned}$$

(177)

Setting $\sqrt{cc-4tt}=p$ and substituting Eq. (171), this term is ready for integration by series expansion:

$$\begin{aligned} \frac{8c^3z^3\,\mathrm{d}x \sqrt{cc-pp}}{3a\,\mathrm{d}p (ac+fp)} = \int \frac{accf\,\mathrm{d}p+aacp\,\mathrm{d}p+cf\!fp\,\mathrm{d}p+afpp\,\mathrm{d}p}{2 \sqrt{cc-pp}} \end{aligned}$$

(178)

Euler develops the integrand into a series that he can integrate term by term:

$$ \begin{aligned}&\frac{1}{2} \int accf\,\mathrm{d}p + aacp\,\mathrm{d}p + cf\!fp\,\mathrm{d}p + afpp\,\mathrm{d}p \cdot \nonumber \\&\quad \left( \frac{1}{c} + \frac{pp}{2c^3} + \frac{1 \cdot 3 \cdot p^4}{2^2 \cdot 1 \cdot 2 c^5} + \frac{1 \cdot 3 \cdot 5 \cdot p^6}{2^3 \cdot 1 \cdot 2 \cdot 3 \cdot c^4} \; \mathrm{ \& c}\right) \nonumber \\&\quad \quad = acfp + \frac{afp^3}{3 \cdot 2 \cdot 1 \cdot c} + \frac{1 \cdot 3 \cdot afp^5}{5 \cdot 2^2 \cdot 1 \cdot 2 \cdot c^3} + \frac{1 \cdot 3 \cdot 5 \cdot af p^7}{7 \cdot 2^3 \cdot 1 \cdot 2 \cdot 3 \cdot c^5}\; \mathrm{ \& c} \nonumber \\&\quad \qquad + \frac{afp^3}{3 \cdot c} + \frac{afp^5}{5 \cdot 2 \cdot 1 \cdot c^3} + \frac{af \cdot 1 \cdot 3 \cdot p^7}{7 \cdot 2^2 \cdot 1 \cdot 2 \cdot c^5}\; \mathrm{ \& c} \left( +aac + f\!fc\right) \frac{pp}{2c} \nonumber \\&\quad \qquad + \frac{cp^4}{4 \cdot 2 \cdot 1 \cdot c^3} + \frac{1 \cdot 3 \cdot p^6}{6 \cdot 2^2 \cdot 1 \cdot 2 \cdot c^5} + \frac{1 \cdot 3 \cdot 5 \cdot p^8}{8 \cdot 2^3 \cdot 1 \cdot 2 \cdot 3 \cdot c^7} \; \mathrm{ \& c} - \mathrm{Const.}\nonumber \\ \end{aligned}$$

(179)

where the value of the integration constant has to be determined in such a way that the whole series vanishes if the Moon is located in the quadratures or if $y=\frac{1}{2}b$ or if $p=\frac{-cf}{a}$ is subtracted. The resulting series becomes thus

$$ \begin{aligned}&\frac{8c^3z^3\,\mathrm{d}x \sqrt{cc-pp}}{3a\,\mathrm{d}p (ac+fp)} = acfp \nonumber \\&\quad + \frac{3afp^3}{5 \cdot 2^1 \cdot 1 \cdot c} + \frac{7 \cdot afp^5}{5 \cdot 2^2 \cdot 1 \cdot 1 \cdot c^3} + \frac{33 \cdot afp^7}{7 \cdot 2^3 \cdot 1 \cdot 2 \cdot 3 \cdot c^5} + \mathrm{ \& c} \nonumber \\&\quad + (aa+f\!f) \cdot \frac{pp}{2} + \frac{1 \cdot p^4}{4 \cdot 2 \cdot 1 \cdot c^2} + \frac{1 \cdot 3 \cdot p^6}{5 \cdot 2^2 \cdot 1 \cdot 2 \cdot c^4} \; \mathrm{ \& c} \nonumber \\&\quad + ccf\!f + \frac{3ccf^4}{3 \cdot 2 \cdot 1 \cdot aa} + \frac{7ccf^6}{5 \cdot 2^2 \cdot 1 \cdot 2 \cdot a^4} \; \mathrm{ \& c} \nonumber \\&\quad - (aa+f\!f) \frac{ccf\!f}{2aa} + \frac{1 \cdot ccf^4}{4 \cdot 2^1 \cdot 1 \cdot a^4} + \frac{1 \cdot 3 \cdot ccf^6}{6 \cdot 2^2 \cdot 1 \cdot 2 \cdot a^6}\; \mathrm{ \& c} \end{aligned}$$

(180)

and therefore

$$ \begin{aligned}&\frac{8c^3z^3 \,\mathrm{d}x \sqrt{cc-pp}}{3a\,\mathrm{d}p (ac+fp)} = acfp \nonumber \\&\quad + \frac{3afp^3}{3 \cdot 2^1 \cdot 1 \cdot c} + \frac{7afp^5}{5 \cdot 2^2 \cdot 1 \cdot 2 \cdot c^3} + \frac{33afp^7}{7 \cdot 2^3 \cdot 1 \cdot 2 \cdot 3 \cdot c^5} \; \mathrm{ \& c} \nonumber \\&\quad + (aa+f\!f) \frac{pp}{2} + \frac{p^4}{4 \cdot 2 \cdot 1 \cdot c^2} + \frac{1 \cdot 3 \cdot p^6}{5 \cdot 2^2 \cdot 1 \cdot 2 \cdot c^4}\; \mathrm{ \& c} \nonumber \\&\quad + \frac{ccf\!f}{2} - \frac{1 \cdot cc \cdot f^4}{4 \cdot 2 \cdot 1 \cdot a^2} - \frac{ccf^6}{80 \cdot a^4} - \frac{3ccf^5}{896 \cdot a^6}\; \mathrm{ \& c} \end{aligned}$$

(181)

Before Euler proceeds with the investigations of this equation he considers the case where the Moon is orbiting in a circle, i.e., if $f=0$ and/or $a=b=c$. In this case

$$\begin{aligned} \frac{8a^3z^3\,\mathrm{d}x \sqrt{aa-pp}}{3a^3\,\mathrm{d}p} = \frac{8z^3\,\mathrm{d}x \sqrt{aa-pp}}{3\,\mathrm{d}p} = - \frac{a^3}{2} \sqrt{aa-pp} + \mathrm{const.} \end{aligned}$$

(182)

If this constant is set to be $=\frac{a^4}{2}$, then

$$\begin{aligned} 16z^3\,\mathrm{d}x = \frac{3a^4\,\mathrm{d}p}{\sqrt{aa-pp}} - 3a^3\,\mathrm{d}p, \end{aligned}$$

(183)

and consequently

$$\begin{aligned} 16z^3x = 3a^4 \int \frac{\mathrm{d}p}{\sqrt{aa-pp}} - 3a^3p. \end{aligned}$$

(184)

Let $a$ be the circle’s radius and $\sin p = A$ in degree. Eqn. (184) becomes

$$\begin{aligned} 16z^3x = \frac{71Aa^4}{1356}-3a^3p, \end{aligned}$$

(185)

hence

$$\begin{aligned} x= \frac{71Aa^4}{16 \cdot 1356 \cdot z^3} - \frac{3a^3p}{16z^3}. \end{aligned}$$

(186)

But if $z=572a$, then

$$\begin{aligned} x = \frac{71Aa}{16 \cdot 1356 \cdot (572)^3} - \frac{3p}{16 \cdot (572)^3} = \frac{71Aa - 4068p}{16 \cdot 1356 \cdot (572)^3}. \end{aligned}$$

(187)

At this point the manuscript ends abruptly.

Appendix D: The content of Ms 276

This unfinished manuscript is well prepared for publication. Euler formulated the paragraphs carefully. The first three introductory paragraphs reveal that Euler became fully aware of the difficulty involved in developing an accurate lunar theory, which is only possible by approximation. The use of series expansions of trigonometric functions appears here probably for the first time and thus represents a further innovation. Due to the importance of Euler’s statements, which reflect his insight and new approach, I will not only paraphrase but translate them in full length:

“That the motion of the Moon—however perturbed—does agree very well with the Newtonian hypothesis of attraction is proved more than sufficiently both by observation and by the conclusions drawn from this hypothesis. Even if this principle of attraction is burdened with such difficulties, that one has to keep it strictly away from a rational way of philosophy, its usefulness—particularly for astronomy—is anyway considerable when considered as phenomenon; and without its help no important things could have been achieved in the theory of celestial motions until now. Observations make clear that planets and comets move in the same way as if they are attracted by the Sun and by each other in the ratio given by Newton. From this, two completely different questions emerge, which should by no means be confused with each other: the first one concerns physics and demands a mechanical reason for this phenomenon consisting of the mutual attraction of the celestial bodies. The second question, however, concerns the determination of the motions caused by this force of attraction in order to complete the theory of astronomy itself, and to calculate and predict the particular phenomena most accurately.⁵

Concerning the second question, by which astronomy has achieved its biggest progress, it has been accomplished so far by its first discoverer Newton that hardly something remains to be added, in particular with regard to the primary planets, the theory of which is no longer fraught with difficulties, apart probably from the tables of the motion of Saturn which need some correction when it is staying near conjunction with Jupiter, because then one observes that its motion is perceptibly perturbed. The motion of the Moon, however, caused by the twofold force of the Sun’s and the Earth’s attraction, is so much difficult to determine, that nothing else than approximations could be done. By appraising the question very carefully Newton showed after all not only that his theory matches with all irregularities of the Moon, but he had completed the lunar theory itself prodigiously, even if the complexity of the driving forces does not allow an analytic approach to the calculation.⁶

If we consider the question in its own rights, which demands the [determination of the] motion of a body like the Moon caused by a twofold driving force, the power of the analysis known until now seems to be insufficient to describe such a motion with confident rules; and it would not have been possible even now to accomplish anything going in that direction, if one would not have called the approximation for help. It is convenient that the force emerging from the Sun is less of a multiple than the force of the Earth driving the Moon, and this is why it becomes possible to treat initially the Moon’s motion in such a way, as if it is driven only by the Earth, but then one had to investigate the deviations from this quite regular motion occurring from the Sun’s force; this would not have been possible if the forces of Earth and Sun would not come close to the ratio of unity. Furthermore, the orbit of the Moon’s motion makes the approximation easier and more accurate at the same time, because it does not much deviate from a circle. Using these resources I tried to find out how much could be achieved by means of calculations.”⁷

In the following paragraphs 4 and 5 Euler treats first the two-body-problem of the Earth-Moon system, then (in the remaining) the three-body-problem of the Sun-Earth-Moon system. He assumes the lunar orbit to be coplanar with the ecliptic. Let $A$ be the perigee, $T$ the Earth, and $AT=a$. Let $b$ be the velocity corresponding to the altitude (=hight of free fall) of the Moon at perigee directed along the straight line $AT$. The Earth is considered to be at rest, while the Sun is revolving around it contrary to its annual motion. Let the Moon be moving from $A$ to any position $L$, thus describing the arc AL or the angle ${\textit{ATL}}=x$ during the time interval $T$. Euler denotes the distance between the Earth and the Moon by ${\textit{LT}}=y$. Let ${\textit{LN}}$ be the tangent through $L$ and $T\!P=p$ the perpendicular from $T$ to ${\textit{LN}}$ intersecting the tangent in $P$. Let $v$ be the velocity corresponding to the altitude of the Moon in $L$, and let $lp=\mathrm{d}y$ be an infinitesimally small radial element of the distance between Earth and Moon. Due to

$$\begin{aligned} {\textit{LP}} = \surd (yy-pp), \end{aligned}$$

(188)

and due to the equivalence of the triangles ${\textit{TPL}}$ and $Lpl$ (do not confuse the point $p$ with the distance $p={\textit{TP}}$), the distances $Lp$ and $Ll$ covered in the time element $\text{ d }t$ become

$$\begin{aligned} Lp = \frac{p\,\mathrm{d}y}{\surd (yy-pp)} \quad \mathrm{and} \quad Ll = \frac{y\,\mathrm{d}y}{\surd (yy-pp)}. \end{aligned}$$

(189)

This time element dT itself, and the corresponding angular element $\mathrm{d}x$, are given by

$$\begin{aligned} \mathrm{d}T = \frac{y\,\mathrm{d}y}{\surd v (yy-pp)} \quad \mathrm{and} \quad \mathrm{d}x = \frac{p\,\mathrm{d}y}{y \surd (yy-pp)}. \end{aligned}$$

(190)

Euler sets both the accelerative gravitational force on the Earth’s surface and the Earth’s radius $=1$. The force acting on the Moon by the Earth along the direction $LT$ then is $=\frac{1}{yy}$, which he decomposes into the tangential and normal components. Thus the former becomes $=\frac{\surd (yy-pp)}{y^3}$, and the latter $=\frac{p}{y^3}$. To determine the Moon’s motion driven only by the Earth’s gravitational force, Euler applies the equation of motion as given in his “Mechanica” (cf. Euler 1736, Def. 15, Corol. 5, § 207: $\frac{\mathrm{d}v}{\mathrm{d}s}=\frac{p}{A}$, and § 209: $r=\frac{2Av}{p}$, where $A=1$), considering that the tangential component is a retarding force:

$$\begin{aligned} \frac{\mathrm{d}v}{Ll} = \frac{\mathrm{d}v \, \surd (yy-pp)}{y\,\mathrm{d}y} = - \frac{\surd (yy-pp)}{y^3} \end{aligned}$$

(191)

and

$$\begin{aligned} \frac{2v}{\mathrm{rad.~Osc.}} = \frac{2v\,\mathrm{d}p}{y\,\mathrm{d}y} = \frac{p}{y^3}. \end{aligned}$$

(192)

But Eq. (191) simply is $\mathrm{d}v = \frac{\mathrm{d}y}{yy}$, which can easily be integrated:

$$\begin{aligned} v = b - \frac{1}{a} + \frac{1}{y}. \end{aligned}$$

(193)

This result, inserted into Eq. (192), gives

$$\begin{aligned} 2 \left( b - \frac{1}{a} + \frac{1}{y}\right) \frac{\mathrm{d}p}{p} = \frac{\mathrm{d}y}{yy} \quad \mathrm{or} \quad \frac{2\,\mathrm{d}p}{p} = \frac{\mathrm{d}y}{y (1+by-\frac{y}{a})}, \end{aligned}$$

(194)

which can be integrated as well:

$$\begin{aligned} 2\,\ell p = \ell \frac{y}{1+by-\frac{y}{a}} + C, \end{aligned}$$

(195)

where $\ell$ denotes the logarithm to the base 10 and $C$ the constant of integration. Euler determines this constant by setting $y=a$ and $p=a$, so that $C$ becomes

$$\begin{aligned} C=2\,\ell a - \ell \frac{a}{ab} = 2\,\ell a + \ell b. \end{aligned}$$

(196)

This constant, inserted into Eq. (195), gives

$$\begin{aligned} 2\,\ell p = \ell \frac{aaby}{1+by-\frac{y}{a}}, \end{aligned}$$

(197)

and therefore

$$\begin{aligned} pp = \frac{aaby}{1+by-\frac{y}{a}}. \end{aligned}$$

(198)

This equation contains the nature of the Moon’s orbit. Denoting the latus rectum tentatively with $CT=c$, Euler derives the relation

$$\begin{aligned} b=\frac{1}{a}-\frac{1}{c} \end{aligned}$$

(199)

and substitutes it into Eq. (198), obtaining

$$\begin{aligned} pp=\frac{a(c-a)y}{c-y}, \end{aligned}$$

(200)

which represents the equation for an ellipse with major axis $c$. Because the Moon’s orbit does not much differ from a circle, one may approximately use the value $c=2a$, otherwise $a=\frac{1}{2}c - \frac{c}{2n}$, where $n$ is a big number. Likewise, if $y$ does not much deviates from $\frac{1}{2}c$, Euler finds

$$\begin{aligned} v = \frac{1}{c} + \frac{\omega }{c}, \end{aligned}$$

(201)

where $\omega$ is a very small value, and thus

$$\begin{aligned} pp = \frac{(nn-1)cc}{4nn(1+\omega )}. \end{aligned}$$

(202)

With these results the two-body-problem is solved, and Euler passes on to the three-body-problem considering also the Sun’s force.

Let the ratio between the forces of the Sun and the Earth be as $m$ to $1$ (at equal distances). Let the distance between the Sun $S$ and the Earth $T$ be $ST=h$, and the angle ${\textit{ATS}}=k$. This angular argument has to be considered as variable. The Earth is attracted towards the Sun along the direction ${\textit{TS}}$ by the force $=\frac{m}{hh}$, which has to be transferred with inverse direction to the Moon and parallel to the direction ${\textit{TS}}$, if the Earth has to be considered at rest. This statement seemed to be quite self-evident for Euler at that time when writing this manuscript:

Attrahetur ergo terra ad solem in directione ${\textit{TS}}$ vi $=\frac{m}{hh}$ quæ quia terram quiescentum spectamus transferri debet in lunam contrarie, ita ut luna sollicitata considerari debeat in directione ${\textit{LM}}$ parallela directioni ${\textit{TS}}$ vi $=\frac{m}{hh}$.

It proves that the principle of the transference of forces had been fully established by him at that time. The procedure following that statement underlines this conjecture. The force acting on the Moon by the Sun along the direction $LS$ is given by $\frac{m}{LS^2}$. Let the geocentric angular distance of the Moon from the Sun be the argument ${\textit{LTS}}=k-x$. Euler derives the distance between the Sun and the Moon from the geometry of the system:

$$\begin{aligned} {\textit{LS}}^2=hh+yy-2hy \cos \mathrm{A} (k-x), \end{aligned}$$

(203)

where $A$ has no mathematical meaning, but has the task to indicate that $k-x$ is the Argument or Arc (“Arcus”). Euler decomposes the force $\frac{m}{LS^2}$ into components acting along the directions ${\textit{LT}}$ and ${\textit{LV}}$, and denotes it with the same symbols: ${\textit{LT}}=\frac{my}{{\textit{LS}}^3}$ and ${\textit{LV}}=\frac{mh}{{\textit{LS}}^3}$. The direction ${\textit{LV}}$ is identical with ${\textit{LM}}$ mentioned in the quotation above. In the following calculations he uses only ${\textit{LM}}$ instead of ${\textit{LV}}$ for the direction parallel to ${\textit{TS}}$. To consider the Earth at rest means, that the force component acting on the Moon along the direction ${\textit{LM}}$ becomes $\frac{m}{hh}-\frac{mh}{{\textit{LS}}^3}$. He expands $\frac{1}{{\textit{LS}}^3}$ into a trigonometric series, probably for the first time in the context of lunar theory:

$$\begin{aligned} \frac{1}{{\textit{LS}}^3}&= \left( hh-2hy \cos \mathrm{A} (k-x) + yy\right) ^{-\frac{3}{2}} \nonumber \\&= \frac{1}{h^3} + \frac{3y \cos \mathrm{A} (k-x)}{h^4} - \frac{3yy}{2h^5} + \frac{15yy( \cos \mathrm{A} (k-x) )^2}{2k^5} \; \mathrm{etc.} \nonumber \\&= \frac{1}{h^3} + \frac{3y \cos \mathrm{A} (k-x)}{h^4} + \frac{9yy}{4h^5} + \frac{15yy \cos \mathrm{A} 2(k-x)}{4h^5}. \end{aligned}$$

(204)

We see Euler here introducing approximations by series expansions into celestial mechanics. Using this result, the forces acting on the Moon along the directions LT and LM become, respectively,

$$\begin{aligned} = \frac{my}{h^3} + \frac{3myy \cos \mathrm{A} (k-x)}{h^4} + \frac{9my^3}{4h^5} + \frac{15my^3 \cos \mathrm{A} 2(k-y)}{4h^5} + \mathrm{etc.} \end{aligned}$$

and

$$\begin{aligned} = -\frac{3my \cos \mathrm{A} (k-x)}{h^3} - \frac{9myy}{4h^4} - \frac{15myy \cos \mathrm{A} 2(k-x)}{4h^4} \; [+] \; \mathrm{etc.} \end{aligned}$$

(205)

Due to the parallelism of the lines ${\textit{TS}}$ and ${\textit{LM}}$, the angle $k-x$ is also defined by ${\textit{MTL}}=k-x$. Furthermore, the sine and cosine of the angle ${\textit{TLP}}$ is given by

$$\begin{aligned} \begin{aligned}{c} \sin {\textit{TLP}}&= \frac{p}{y} \\ \cos {\textit{TLP}}&= \frac{\surd (yy-pp)}{y}, \end{aligned} \end{aligned}$$

(206)

respectively, and the sine and cosine of the angle ${\textit{MLN}}={\textit{MLT}}-{\textit{PLT}}$ is given by

$$\begin{aligned} \begin{aligned}{c} \sin {\textit{MLN}}&= \frac{\surd (yy-pp)}{y} \sin \mathrm{A} (k-x) - \frac{p}{y} \cos \mathrm{A} (k-x) \\ \cos {\textit{MLN}}&= \frac{p}{y} \sin \mathrm{A} (k-x) + \frac{\surd (yy-pp)}{y} \cos \mathrm{A} (k-x), \end{aligned} \end{aligned}$$

(207)

respectively. From the force acting along ${\textit{LM}}$ results the normal force

$$\begin{aligned} \frac{3m \surd (yy-pp)}{2h^3} \sin \mathrm{A} 2(k-x) - \frac{3mp}{2h^3} - \frac{3mp}{2h^3} \cos \mathrm{A} 2(k-x), \end{aligned}$$

(208)

neglecting small terms, and the accelerating tangential force becomes

$$\begin{aligned} \frac{3mp}{2h^3} \sin \mathrm{A} 2(k\!-\!x) \!+\! \frac{3m \surd (yy\!-\!pp)}{2h^3} \!+\! \frac{3m \surd (yy\!-\!pp)}{2h^3} \cos \mathrm{A} 2(k\!-\!x). \end{aligned}$$

(209)

And from the force acting along $LT$ results the normal force

$$\begin{aligned} \frac{mp}{h^3} \end{aligned}$$

(210)

and the accelerating tangential force

$$\begin{aligned} -\frac{m \surd (yy-pp)}{h^3}. \end{aligned}$$

(211)

But the force acting on the Moon by the Earth produces the normal component

$$\begin{aligned} \frac{p}{y^3} \end{aligned}$$

(212)

and the accelerating tangential component

$$\begin{aligned} -\frac{\surd (yy-pp)}{y^3}. \end{aligned}$$

(213)

The resulting tangential force accelerating the Moon thus is given by the combination of Eqs. (191), (209), (211), and (213):

$$\begin{aligned} \frac{\mathrm{d}v \surd (yy\!-\!pp)}{y\,\mathrm{d}y}&= \!-\!\frac{\surd (yy\!-\!pp)}{y^3} \!+\! \frac{m \surd (yy\!-\!pp)}{2h^3} \nonumber \\&+\! \frac{3mp}{2h^3} \sin \mathrm{A} 2(k\!-\!x) \!+\! \frac{3m \surd (yy\!-\!pp)}{2h^3} \cos \mathrm{A} 2(k\!-\!x), \end{aligned}$$

(214)

from which Euler concludes

$$\begin{aligned} \mathrm{d}v \!=\! -\frac{\mathrm{d}y}{yy} \!+\! \frac{my\,\mathrm{d}y}{2h^3}\!+\! \frac{3mpy\,\mathrm{d}y \sin \mathrm{A} 2(k\!-\!x)}{2h^3 \surd (yy\!-\!pp)} \!+\! \frac{3my\,\mathrm{d}y \cos \mathrm{A} 2(k\!-\!x)}{2h^3}. \end{aligned}$$

(215)

The resulting normal force acting on the Moon is given by the combination of Eqs. (192), (208), (210), and (212):

$$\begin{aligned} \frac{2v\,\mathrm{d}p}{y\,\mathrm{d}y}&= \frac{p}{y^3} - \frac{mp}{2h^3} \nonumber \\&+ \frac{3m \surd (yy-pp)}{2h^3} \sin \mathrm{A} 2(k-x) - \frac{3mp}{2h^3} \cos \mathrm{A} 2(k-x). \end{aligned}$$

(216)

Euler substitutes Eq. (190) into Eq. (215) and obtains

$$\begin{aligned} \mathrm{d}v \!=\! - \frac{\mathrm{d}y}{yy} \!+\! \frac{my\,\mathrm{d}y}{2h^3} \!+\! \frac{3myy\,\mathrm{d}x \sin A 2(k\!-\!x)}{2h^3} \!+\! \frac{3my\,\mathrm{d}y \cos A 2(k\!-\!x)}{2h^3}, \end{aligned}$$

(217)

which he can integrate immediately:

$$\begin{aligned} v = C + \frac{1}{y} + \frac{myy}{4h^3} + \frac{3myy \cos \mathrm{A} 2(k-x)}{4h^3} \end{aligned}$$

or

$$\begin{aligned} v = \frac{1}{y} - \frac{1}{c} + \frac{myy}{4h^3} + \frac{3myy \cos \mathrm{A} 2(k-x)}{4h^3}. \end{aligned}$$

(218)

“This equation defines the absolute velocity of the Moon at any position in its orbit, which can be determined from the Moon’s distance from the Earth, from the angular separation between Sun and Moon given by $k-x$, and from the Earth’s distance from the Sun, whose variation does not affect the velocity considerably.”

On the remaining folios of this manuscript Euler tries to prepare the equation of motion (216) due to the normal force for the integration. For that purpose he has first to figure out the items $p$, $v$, $2v\,\mathrm{d}p$, etc. If the Moon’s motion is driven only by the Earth, then Eq. (200) holds, as shown above, representing the unperturbed Keplerian motion. Using the abbreviation $a(c-a)=bb$, this equation becomes

$$\begin{aligned} p=b \surd \frac{y}{c-y}. \end{aligned}$$

(219)

But due to the additional action of the Sun, this parameter needs to be corrected. The results obtained above gives Euler the idea of a general ansatz:

$$\begin{aligned} p = b \surd \frac{y}{c-y} + \frac{mP}{h^3} + \frac{mQ \cos \mathrm{A} 2(k-x)}{h^3} + \frac{mR \sin \mathrm{A} 2(k-x)}{h^3}, \end{aligned}$$

(220)

where $P$, $Q$, and $R$ are coefficients to be determined. We observe here Euler’s very early use of the method of undetermined coefficients for solving differential equations in lunar theory. The first derivative of Eq. (220) is given by

$$\begin{aligned} \mathrm{d}p&= \frac{bc\,\mathrm{d}y}{2(c-y) \surd (cy-yy)} + \frac{m\,\mathrm{d}P}{h^3} \nonumber \\&+ \frac{m\,\mathrm{d}Q \cos \mathrm{A} 2(k-x)}{h^3} + \frac{2mQp\,\mathrm{d}y}{h^3y \surd (yy-pp)} \sin \mathrm{A} 2(k-x) \nonumber \\&+ \frac{m\,dR \sin \mathrm{A} 2(k-x)}{h^3} - \frac{2mRp\,\mathrm{d}y \cos \mathrm{A} 2(k-x)}{h^3 y \surd (yy-pp)}. \end{aligned}$$

(221)

Using Eq. (219), the distance $PL$ becomes

$$\begin{aligned} \surd (yy-pp) = \frac{\surd (cyy-y^3-bby)}{\surd (c-y)}, \end{aligned}$$

(222)

and therefore, approximately,

$$\begin{aligned} \frac{p}{\surd (yy-pp)} \approx \frac{b}{\surd cy-yy-bb}. \end{aligned}$$

(223)

Euler substitutes Eqs. (219), (220), and (223) into the equation of motion (216) to obtain

$$\begin{aligned} 2v\,\mathrm{d}p&= \frac{b\,\mathrm{d}y \surd y}{yy \surd (c-y)} + \frac{mP\,\mathrm{d}y}{h^3yy} \nonumber \\&+ \frac{mQ\,\mathrm{d}y \cos \mathrm{A} 2(k-x)}{h^3yy} + \frac{mR\,\mathrm{d}y \sin \mathrm{A} 2(k-x)}{h^3yy} \nonumber \\&- \frac{mb}{2h^3} y\,\mathrm{d}y \surd \frac{y}{c-y} + \frac{3m y\,\mathrm{d}y \surd (cyy-y^3-bby)}{2h^3 \surd (c-y)} \sin \mathrm{A} 2(k-x) \nonumber \\&- \frac{3mb y\,\mathrm{d}y \surd y}{2h^3 \surd (c-y)} \cos \mathrm{A} 2(k-x) - \frac{2mRp\,\mathrm{d}y \cos \mathrm{A} 2(k-x)}{h^3 y \surd (yy-pp)}. \end{aligned}$$

(224)

On the other hand, the product $2v\,\mathrm{d}p$ is also given by Eqs. (218) and (221), resulting in

$$\begin{aligned} 2v\,\mathrm{d}p&= \frac{b\,\mathrm{d}y}{y \surd (cy-yy)} + \frac{2m(c-y)\,\mathrm{d}P}{h^3cy} + \frac{2m(c-y)\,\mathrm{d}Q \cos \mathrm{A} 2(k-x)}{h^3cy} \nonumber \\&+ \frac{4mb(c-y)Q\,\mathrm{d}y \sin \mathrm{A} 2(k-x)}{h^3cyy \surd (cy-yy-bb)} + \frac{mbcy\,\mathrm{d}y \surd y}{4h^3(c-y)^{\frac{3}{2}}} \nonumber \\&- \frac{4mb(c-y)R\,\mathrm{d}y \cos \mathrm{A} 2(k-x)}{h^3cyy \surd (cy-yy-bb)} + \frac{2m(c-y)\, \mathrm{d}R \sin \mathrm{A} 2(k-x)}{h^3cy} \nonumber \nonumber \nonumber \\&+ \frac{3mbcyy\,\mathrm{d}y \cos \mathrm{A} 2(k-x)}{4h^3(c-y)^{\frac{3}{2}} \surd y}. \end{aligned}$$

(225)

At this point (§ 10), the fragment of the manuscript ends. It is quite possible that Euler determined the coefficients $P$, $Q$, and $R$ on the lost folios. It is therefore not clear whether he already applied the method of undetermined coefficients in full length and whether he was already able to solve the differential equation approximately. Evidence for the former assumption is given by the way of his approach presented above.