© J R Stockton, ≥ 2015-02-23

(Chapitres I et II)

by Joseph-Louis Lagrange.

(Essay on the Three Body Problem)

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- About the Translation :-
- "Essay on the Three-Body Problem"
by J-L Lagrange (
*typeset algebra*) :-- Notice
- Chapter I - General Formulae for the Solution of the Three-Body Problem - Recapitulation
- Chapter II - Solution of the Three-Body Problem in Different Cases
- Remainder (not translated) :-
- Chapter III - Modification of the formulae of the previous chapter, for the case where one supposes that one of the three bodies is distant from the other two
- Chapter IV - On the Theory of the Moon
- Editor's Note (part translated)

- Collaborators
- Questions
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- Code

- Gravity 4 : The Lagrange Points - References
- Gravity 5 : A Lagrange Movie
- Gravity 6 : Constant-Pattern Motion
- The Lagrange Points Reconsidered
- The History of the Lagrange Points
- Imaged extracts about the Paris Academy prize
- About Euler on Lagrange Points
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- For Testing MathJax Code Snippets

STOP PRESS -
http://gallica.bnf.fr/ark:/12148/bpt6k5401485r.image.swf has, at the
end, a c.18 copy of the 1772 prizewinning papers of Euler and Lagrange.
Compare for errata!

My introductory Lagrange Point page is Gravity 4 : The Lagrange Points.

Lagrange shared an Academy of Paris prize with Leonhard Euler, but
the *Essai sur le Problème des Trois Corps* was not a joint
work. Euler was awarded his share for his *Nouvelles recherches sur le
vrai mouvement de la Lune*, which has nothing to do with the Lagrange
Points.

**Requirements : **
*Perfectionner les méthodes sur lesquelles est fondée
la théorie de la Lune, fixer par ce moyen celles des
équations de cette planète qui sont encore incertaines et
examiner en particulier si on peut rendre raison par cette
théorie de l'équation séculaire du mouvement de la
Lune.*

To perfect the methods on which the theory of the Moon is founded, thereby settling those equations of that planet which are still uncertain and in particular to examine whether we can explain by this theory the secular equation of motion of the Moon.

Joseph-Louis Lagrange, for his *Essai sur le Problème des
Trois Corps* (Essay on the Three-Body Problem), should be credited
for providing the general theory of what are now called the Lagrange Points L1, L2, L3, L4, and L5. It
explains the behaviour of bodies which were later observed to remain
near L4 and L5. However, Leonhard Euler had, in *Considerationes de motu corporum
coelestium*, already briefly described L1 and L2.

Few Web authors appear to be familiar with the actual contents (as
opposed to the reputed contents) of Lagrange's *Essai*, perhaps
because it is in French. Because I could not find an English version on
the Web, I felt it desirable to translate the *Essai* into English
and HTML.

Chapters I & II of Lagrange's *Essai* are on the general
three-mass problem and then its two constant-pattern (self-similar)
solutions. From that, the subsequent derivation of the five Points, for
one body having negligible mass, which Lagrange did not describe,
would have been trivial.

**Question** : Did Lagrange write anything else relevant to the
L-points? A search of the indexes of *Œuvres de Lagrange* at
Gallica has found nothing.

For ease of reading, I have tried to translate Lagrange's late 18th century French mathematical language into English late 20th century mathematical language. The parts of the text which seem to be of greatest importance are highlighted (pages 229, 230, 272, 292).

To see the original Algebra, open in parallel a copy of *Essai sur
le Problème des Trois Corps*, for which see References below. JavaScript is used to expand the
source notation for algebra. If it is not available, Algebra will be as
represented in the source file, by "words" beginning with "##".

There is now a button which will (but not in IE8) replace the
substitute equations by real ones typeset by MathJax. *Currently, any
double-click on this page will initiate typesetting.* It should be
possible, in time, to treat maths which is inline with the text in a
similar fashion. MathJax is written in JavaScript.

I have read, but do not expect to translate, Chapters III and IV.

The translated pages, pp. 229-292, being Chapters I and II, have been checked/corrected by a native Francophone.

These *Œuvres* of Lagrange were edited by
J.-A. Serret; his "Note de l'Éditeur" that follows Chapter IV in
pp. 324-331 may well be significant for the Points, which could call for
its translation here - eventually.

I have found at Gallica, on 2012-08-19, "Recueil des pieces qui ont remporte les prix de l'Academie Royale des Sciences", Tome 9 (1764-1772), published in 1777. It contains the papers for which Euler and Lagrange shared the 1772 Prize. It is a little hard to read, particularly some of the mathematics, but it is authoritative.

My prime source has been the copy at the University of Liege (ULg) web site. That I believe to be identical in content to the copy in Tome 6 of the "Œuvres de Lagrange" at Gallica, edited in the second half of the 19th century by J.-A. Serret. The Liege copy is easier to use. The printing is good. The pagination differs. The known errors differ.

I use the pagination of Liege and the "Œuvres de Lagrange".

Known errors are noted on a background like this. Typos seen in the c.19 "Œuvres" version have been checked against the c.18 "Prize" version.

Button | For doing the typesetting | |
---|---|---|

Notice | 229-231 | Overall introduction |

Ch. I | 231-267 | General Formulae for the Solution of the Three-Body Problem. |

I | 231-232 | First equations |

II | 232-234 | |

III | 234-235 | |

IV | 236-238 | |

V | 238-240 | |

VI | 241-242 | |

VII | 242-242 | |

VIII | 243-243 | |

IX | 243-246 | |

X | 246-247 | |

XI | 248-249 | |

XII | 249-251 | |

XIII | 251-252 | |

XIV | 252-253 | |

XV | 254-254 | |

XVI | 255-255 | |

XVII | 255-259 | |

XVIII | 260-261 | |

XIX | 261-264 | |

XX | 264-266 | |

XXI | 266-267 | |

XXII | 267-271 | Recapitulation - Summary of Chapter I |

Ch. II | 272-292 | Solution of the Three-Body Problem in Different Cases. |

XXIII | 272-272 | Introduction |

XXIV | 272-273 | Constant-size cases |

XXV | 273-275 | Equilateral circular case |

XXVI | 275-277 | Collinear circular case |

XXVII | 277-279 | A possibility shown to be uninteresting |

XXVIII | 279-280 | On collinear speeds |

XXIX | 280-283 | Constant-ratio cases |

XXX | 283-287 | Equilateral constant-ratio case |

XXXI | 288-290 | Collinear constant-ratio case |

XXXII | 290-292 | Conic section |

XXXIII | 292-292 | Conclusion : exactly soluble in two cases |

Ch. III | 293-303 |
Modification of the formulae of the previous chapter,
for the case where one supposes that one of the three bodies is distant from the other two. |

Ch. IV | 303-324 |
On the theory of the Moon. |

Ed-Note | 324-331 | Editor's Note |

I judge, from the correspondence between Lagrange and D'Alembert, that this work was at least well-in-hand in the summer of 1770.

The button should cause the equation-substitutes to be replaced
throughout, after a while, by MathJax "typeset" versions of the
equations in the original PDF - demonstration :-
##
$$ \sin^2\phi + \cos^2\phi = 1 $$
A progress indication appears at bottom left of the window.
For me, Chrome is faster than Firefox, and Opera is slow.

Tome 6, pp. 229-292;

Joseph-Louis, Comte Lagrange (1736-1813)

AD 1772.

229

ESSAY

ON

THE THREE-BODY PROBLEM

Juvat integros accedere fontes. |

LUCR. |

I love stumbling upon yet unfound sources. LUCRETIUS, De Rerum Natura, I, v. 927 Tr. EjH. |

It is pleasant to handle an untouched subject.
Tr. Henry Fielding? |

(Prize of the Royal Academy of Sciences of Paris, volume IX, 1772.)

These researches contain a method of solving the three body problem, different from all those which have been given before. It consists of not using in the determination of the orbit of each body any other elements than the distances between the three bodies, that is to say, the triangle formed by these bodies at each instant. For that, it is necessary first to find the equations which determine those distances as a function of time ; then, by supposing the distances known, it is necessary to deduce from them the relative movement of the bodies with respect to some fixed plane. One will see, in the first Chapter, what I have done to fulfil these two objects, of which the second particularly requires a delicate and rather complicated analysis. At the end of this Chapter, I collect

230

the main formulae that I have found, which will contain the solution to the three-body problem taken in full generality.

the main formulae that I have found, which will contain the solution to the three-body problem taken in full generality.

The second Chapter is intended to examine how and in which cases the three bodies can move so that their distances shall always be constant, or shall at least keep constant ratios between themselves. I find that these conditions can only be satisfied in two cases : one, when the three bodies are arranged in one straight line, and the other, when they form an equilateral triangle ; then each of the three bodies describes around the two others circles or conic sections, as if there were only two bodies. This research is really no more than for pure curiosity ; but I have thought that it would not be out of place in a work which is principally concerned with the three-body problem, viewed in its full extent.

In the third Chapter, I suppose that the distance of one of the three bodies from the other two is very large, and I apply the general solution of the first Chapter to that situation, which is, as one knows, that of the Earth, the Moon, and the Sun.

Finally, in the fourth Chapter, I treat in particular the Theory of the Moon ; I give the formulae which contain that Theory, and I show, by a small sample calculation, how one must use these formulae to deduce the irregularities of the movement of the Moon around the Earth.

Lack of time and other essential occupations have not allowed me to embark on the topic in all the detail necessary to respond in a proper manner to the principal points of the question proposed by the Academy : so I was at first reluctant to present these researches for the Contest, and I was only persuaded by the hope that this illustrious company would perhaps find my method of solving the three-body problem worthy of some attention, as much by its novelty and its singularity as by the considerable difficulties of calculation which it contains.

If the Academy deigns to honour my work by its vote, it will be a powerful motive for me to work on perfecting it, and I do not despair

231

of being able to get by my method a Theory of the Moon as complete as one can ask for in the present imperfect state of Analysis.

of being able to get by my method a Theory of the Moon as complete as one can ask for in the present imperfect state of Analysis.

FIRST CHAPTER.

GENERAL FORMULAE FOR THE SOLUTION OF THE THREE-BODY PROBLEM

Let A, B, C be the masses of the three bodies which are
mutually attracted in direct ratio of their masses and in inverse ratio
of the square of the distances ; and name *x, y, z* as the
rectangular coordinates of the orbit of body B around body A, *x', y',
z'* the rectangular coordinates of the orbit of body C around the
same body A, coordinates which one supposes always parallel to three
fixed and mutually perpendicular lines ; finally let * r, r', r''*
be the distances between the bodies A and B, A and C, B and C, so that
one has
##.
$$ r = \sqrt { x^2+y^2+z^2 }, ~~~~
r' = \sqrt { x'^2+y'^2+z'^2 }, ~~~~
r'' = \sqrt { (x'-x)^2+(y'-y)^2+(z'-z)^2 } . $$
One will have, as one knows, by taking the element of time *dt*
constant, the following six equations
##A3;
$$
\left\{ \begin{array}{l}
{d^2x \over dt^2} + ( {{A+B} \over r^3} + {C \over r''^3} ) x + C ( {1 \over r'^3} - {1 \over r''^3} ) x' = 0, \\
\tag{A} {d^2y \over dt^2} + ( {{A+B} \over r^3} + {C \over r''^3} ) y + C ( {1 \over r'^3} - {1 \over r''^3} ) y' = 0, \\
{d^2z \over dt^2} + ( {{A+B} \over r^3} + {C \over r''^3} ) z + C ( {1 \over r'^3} - {1 \over r''^3} ) z' = 0;
\end{array} \right. $$
##B3;
$$
\left\{ \begin{array}{l}
{d^2x' \over dt^2} + ( {{A+C} \over r'^3} + {B \over r''^3} ) x' + B ( {1 \over r^3} - {1 \over r''^3} ) x = 0, \\
\tag{B} {d^2y' \over dt^2} + ( {{A+C} \over r'^3} + {B \over r''^3} ) y' + B ( {1 \over r^3} - {1 \over r''^3} ) y = 0, \\
{d^2z' \over dt^2} + ( {{A+C} \over r'^3} + {B \over r''^3} ) z' + B ( {1 \over r^3} - {1 \over r''^3} ) z = 0;
\end{array} \right. $$

232

with the aid of which one will be able to determine the relative orbits of bodies B and C around body A.

with the aid of which one will be able to determine the relative orbits of bodies B and C around body A.

If one also does
##,
$$ x'-x = x'', ~~~~ y'-y = y'', ~~~~ z'-z =z'', $$
such that *x'', y'', z''* are the rectangular coordinates of the
orbit of body C around B, one will have
##,
$$ r'' = \sqrt{x''^2+y''^2+z''^2}, $$
and, subtracting respectively the first three equations from the
last three, one will have these three
##C3,
$$
\left\{ \begin{array}{l}
{d^2x'' \over dt^2}
+ ( {{B+C} \over r''^3} + {A \over r^3} ) x''
+ A ( {1 \over r'^3} - {1 \over r^3} ) x' = 0, \\
\tag{C} {d^2y'' \over dt^2}
+ ( {{B+C} \over r''^3} + {A \over r^3} ) y''
+ A ( {1 \over r'^3} - {1 \over r^3} ) y' = 0, \\
{d^2z'' \over dt^2}
+ ( {{B+C} \over r''^3} + {A \over r^3} ) z''
+ A ( {1 \over r'^3} - {1 \over r^3} ) z' = 0,
\end{array} \right. $$
which will express the relative movement of body C around body B.

It is good to notice the analogy which exists between these nine
equations (A), (B), (C); it is that equations (A) become equations (B)
by changing just B to C, *x* to *x'*, *y* to *y'*,
*z* to *z'*, *r* to *r'*, and reciprocally ; and
that likewise these equations become equations (C) by changing A to C,
*x* to *x''*, *y* to *y''*, *z* to *z''*,
*r* to *r''*, and *vice versa* ; and the same analogy
will hold in all of the formulae which we shall find in the
following.

Multiply the first of the equations (A) by *y* and the
second by *x*, and then subtract one from the other, to get
##.
$$ { { yd^2x - xd^2y } \over dt^2 }
+ C ( {1 \over r'^3} - {1 \over r''^3} ) (yx'-xy') = 0 .$$

233

Combining similarly the first two of equations (B) and the first two of equations (C), one gets these two ##, ##. $$ { { y'd^2x' - x'd^2y' } \over dt^2 } + B ( {1 \over r^3} - {1 \over r''^3} ) (xy'-yx') = 0 , \\ { { y''d^2x'' - x''d^2y'' } \over dt^2 } + A ( {1 \over r'^3} - {1 \over r^3} ) (x'y''-y'x'') = 0 .$$

But ##; $$ x'' = x' - x, ~~~~ y'' = y' - y ~; $$ so ##; $$ x'y'' - y'x'' = xy' -yx' ~; $$ thus, in adding together the three preceding equations, after having divided the first by C, the second by B and the third by A, one gets this ##. $$ { {yd^2x-xd^2y} \over Cdt^2} + { {y'd^2x'-x'd^2y'} \over Bdt^2} + { {y''d^2x''-x''d^2y''} \over Adt^2} = 0. $$

One will find in the same manner these two further equations ##, ##, $$ { {zd^2x-xd^2z} \over Cdt^2} + { {z'd^2x'-x'd^2z'} \over Bdt^2} + { {z''d^2x''-x''d^2z''} \over Adt^2} = 0, \\ { {zd^2y-yd^2z} \over Cdt^2} + { {z'd^2y'-y'd^2z'} \over Bdt^2} + { {z''d^2y''-y''d^2z''} \over Adt^2} = 0, $$

Thus one will get, on integrating
##D3,
$$
\left\{ \begin{array}{l}
{ {ydx-xdy} \over Cdt} + { {y'dx'-x'dy'} \over Bdt} + { {y''dx''-x''dy''} \over Adt} = a, \\
\tag{D} { {zdx-xdz} \over Cdt} + { {z'dx'-x'dz'} \over Bdt} + { {z''dx''-x''dz''} \over Adt} = b, \\
{ {zdy-ydz} \over Cdt} + { {z'dy'-y'dz'} \over Bdt} + { {z''dy''-y''dz''} \over Adt} = c,
\end{array} \right. $$
*a, b, c* being arbitrary constants.

Moreover, if one multiplies the first of equations (A) by
*dx*/C, the first of equations (B) by *dx'*/C, and the first
of equations (C) by

234

*dx''*/A, and then one adds them together, one gets,
because *x'' = x'-x*,
##.
$$ \frac{dxd^2x}{Cdt^2} +
\frac{dx'd^2x'}{Bdt^2} +
\frac{dx''d^2x''}{Adt^2} +
(A+B+C) \left(
\frac{xdx}{Cr^3} +
\frac{x'dx'}{Br'^3} +
\frac{x''dx''}{Ar''^3} \right) = 0 . $$

one gets an integrable equation of which the integral will be
##E1
$$ \tag{E} {\Tiny
\frac{dx^2+dy^2+dz^2}{Cdt^2} +
\frac{dx'^2+dy'^2+dz'^2}{Bdt^2} +
\frac{dx''^2+dy''^2+dz''^2}{Adt^2} - 2 (A+B+C)
\left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) = f . } $$
*f* being an arbitrary constant.

Similarly, one finds ##, ##. $$ \frac{dyd^2y}{Cdt^2} + \frac{dy'd^2y'}{Bdt^2} + \frac{dy''d^2y''}{Adt^2} + (A+B+C) \left( \frac{ydy}{Cr^3} + \frac{y'dy'}{Br'^3} + \frac{y''dy''}{Ar''^3} \right) = 0 , \\ \frac{dzd^2z}{Cdt^2} + \frac{dz'd^2z'}{Bdt^2} + \frac{dz''d^2z''}{Adt^2} + (A+B+C) \left( \frac{zdz}{Cr^3} + \frac{z'dz'}{Br'^3} + \frac{z''dz''}{Ar''^3} \right) = 0 . $$

Then, adding together these three equations and putting

r dr | in place of | x dx + y dy + z dz, |

r' dr' | in place of | x' dx' + y' dy' + z' dz', |

r'' dr'' | in place of | x'' dx'' + y'' dy'' + z'' dz'', |

Those are the only exact integrals which one has been able to
find so far ; so, as there are in all six variables *x, y, z, x', y',
z'*, it is clear that, if one can find two further integrals, the
problem will be reduced to first differences ; but one scarcely dares
hope to reach that in the present state of imperfection of Analysis.

Let us suppose, for brevity, ##, ##, ##, $$ u^2 = \frac{dx^2 + dy^2 + dz^2}{dt^2} , \\ u'^2 = \frac{dx'^2 + dy'^2 + dz'^2}{dt^2} , \\ u''^2 = \frac{dx''^2 + dy''^2 + dz''^2}{dt^2} , $$

235

so that*u, u', u''* express the relative speeds of bodies
B, C around A, and of C around B ; it is clear that one will have
##,
##,
##.
$$
\frac{d^2(r^2)}{2dt^2} =
\frac{xd^2x + yd^2y + zd^2z}{dt^2} + u^2 , \\
\frac{d^2(r'^2)}{2dt^2} =
\frac{x'd^2x' + y'd^2y' + z'd^2z'}{dt^2} + u'^2 , \\
\frac{d^2(r''^2)}{2dt^2} =
\frac{x''d^2x'' + y''d^2y'' + z''d^2z''}{dt^2} + u''^2 . $$

so that

So, putting in these equations instead of ##, ... $$ \frac{d^2x}{dt^2}, ~~~~ \frac{d^2y}{dt^2}, ~~~~ \frac{d^2z}{dt^2}, ~~~~ \frac{d^2x'}{dt^2}, ~ ... $$ their values drawn from equations (A), (B), (C), and observing that ##, $$ x^2 + y^2 + z^2 = r^2, ~~~~ x'^2 + y'^2 + z'^2 = r'^2, ~~~~ x''^2 + y''^2 + z''^2 = r''^2, $$ and ##, ##, $$ xx' + yy' + zz' = \frac{r^2 + r'^2 -r''^2}2 , \\ x'x'' + y'y'' + z'z'' = x'^2 + y'^2 + z'^2 - (x'x + y'y + z'z) = \frac{r'^2 + r''^2 - r^2}2 , $$ one will have, after having moved all of the terms to the same side, ##F3. $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} + \left(\frac{A+B}{r^3} + \frac C{r''^3} \right)r^2 + \frac C2 \left( \frac 1{r'^3} - \frac 1{r''^3} \right) (r^2 + r'^2 - r''^2) - u^2 = 0 , \\ \tag{F} \frac{d^2(r'^2)}{2dt^2} + \left(\frac{A+C}{r'^3} + \frac B{r''^3} \right)r'^2 + \frac B2 \left( \frac 1{r^3} - \frac 1{r''^3} \right) (r^2 + r'^2 - r''^2) - u'^2 = 0 , \\ \frac{d^2(r''^2)}{2dt^2} + \left(\frac{B+C}{r''^3} + \frac A{r^3} \right)r''^2 + \frac A2 \left( \frac 1{r'^3} - \frac 1{r^3} \right) (r'^2 + r''^2 - r^2) - u''^2 = 0 . \end{array} \right. $$

Then, if one can have the values of *u ^{2},
u'^{2}, u''^{2}* expressed in terms
of

236

Thus one has, by differentiating the values of *u ^{2}*,

Let, for brevity, ##, ##, ##, $$ dR = \frac{2rdr}{r''^3} + 2\left( \frac1{r'^3} - \frac1{r''^3} \right) dV , \\ dR' = \frac{2r'dr'}{r''^3} + 2\left( \frac1{r^3} - \frac1{r''^3} \right) dV' , \\ dR'' = \frac{2r''dr''}{r^3} + 2\left( \frac1{r'^3} - \frac1{r^3} \right) dV'' , $$

237

and one will get ##, ##, $$ u^2 = \frac{2(A+B)}{r} - CR , \\ u'^2 = \frac{2(A+C)}{r'} - BR' , \\ u''^2 = \frac{2(B+C)}{r''} - AR'' ; $$ ##; so that equations (F) become ##G3; $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} - \frac{A+B}{r} + C\left[\frac{r^2}{r''^3} + \frac12\left(\frac1{r'^3}-\frac1{r''^3}\right)(r^2+r'^2-r''^2) + R\right] = 0 , \\ \tag{G} \frac{d^2(r'^2)}{2dt^2} - \frac{A+C}{r'} + B\left[\frac{r'^2}{r''^3} + \frac12\left(\frac1{r^3}-\frac1{r''^3}\right)(r^2+r'^2-r''^2) + R'\right] = 0 , \\ \frac{d^2(r''^2)}{2dt^2} - \frac{B+C}{r''} + A\left[\frac{r''^2}{r^3} + \frac12\left(\frac1{r'^3}-\frac1{r^3}\right)(r'^2+r''^2-r^2) + R''\right] = 0 ; \end{array} \right. $$ and no more is needed than to find the values of ##. $$ dV, ~~~ dV', ~~~~ dV'' . $$

and one will get ##, ##, $$ u^2 = \frac{2(A+B)}{r} - CR , \\ u'^2 = \frac{2(A+C)}{r'} - BR' , \\ u''^2 = \frac{2(B+C)}{r''} - AR'' ; $$ ##; so that equations (F) become ##G3; $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} - \frac{A+B}{r} + C\left[\frac{r^2}{r''^3} + \frac12\left(\frac1{r'^3}-\frac1{r''^3}\right)(r^2+r'^2-r''^2) + R\right] = 0 , \\ \tag{G} \frac{d^2(r'^2)}{2dt^2} - \frac{A+C}{r'} + B\left[\frac{r'^2}{r''^3} + \frac12\left(\frac1{r^3}-\frac1{r''^3}\right)(r^2+r'^2-r''^2) + R'\right] = 0 , \\ \frac{d^2(r''^2)}{2dt^2} - \frac{B+C}{r''} + A\left[\frac{r''^2}{r^3} + \frac12\left(\frac1{r'^3}-\frac1{r^3}\right)(r'^2+r''^2-r^2) + R''\right] = 0 ; \end{array} \right. $$ and no more is needed than to find the values of ##. $$ dV, ~~~ dV', ~~~~ dV'' . $$

For that, I make ##, $$ d\rho = x'dx + y'dy + z'dz - xdx' -ydy' - zdz' , $$ and as one has ##, $$ xx' + yy' + zz' = \frac{r^2 + r'^2 - r''^2}2 , $$ one will get, by differentiation, ##; $$ xdx' + ydy' + zdz' + x'dx + y'dy + z'dz = rdr + r'dr' - r''dr'' ; $$ then ##, ##, $$ dV = \frac{rdr + r'dr' - r''dr'' + d\rho}2 , \\ dV' = \frac{rdr + r'dr' - r''dr'' - d\rho}2 , $$ and then ##, $$ dV'' = r'dr' - dV , $$

238

to get ##. $$ dV'' = \frac{r'dr' + r''dr'' - rdr - d\rho}2 . $$

to get ##. $$ dV'' = \frac{r'dr' + r''dr'' - rdr - d\rho}2 . $$

All that remains to be done is to get the value of
*dρ* ; to get that, I differentiate, and I have
##;
$$ d^2\rho = x'd^2x + y'd^2y + z'd^2z - xd^2x' - yd^2y' - zd^2z' ; $$
I substitute in the place of *d ^{2}x, d^{2}y,
d^{2}z*, ... the values drawn from equations (A) and (B),
and making other suitable substitutions, I find
##,
$$
{\small \frac{d^2\rho}{dt^2} = - \frac12
\left(\frac{A+B}{r^3} + \frac C{r''^3} - \frac{A+C}{r'^3} - \frac B{r''^3} \right)
(r^2 + r'^2 -r''^2)
- C \left(\frac 1{r'^3} - \frac 1{r''^3} \right) r'^2
+ B \left(\frac 1{r^3} - \frac 1{r''^3} \right) r^2 , } $$
or else
##.
$$
{\small \frac{2d^2\rho}{dt^2} +
A \left( \frac1{r^3} - \frac1{r'^3} \right)(r^2 + r'^2 - r''^2) +
B \left( \frac1{r^3} - \frac1{r''^3} \right)(r'^2 - r^2 - r''^2) +
C \left( \frac1{r''^3} - \frac1{r'^3} \right)(r^2 - r'^2 - r''^2)
= 0 . } $$

Let us suppose, to put our formulae in a simpler form,
##,
##,
##,
$$
{r'^2 + r''^2 -r^2 \over 2} = p , ~~~~
{r^2 + r''^2 -r'^2 \over 2} = p' , ~~~~
{r^2 + r'^2 -r''^2 \over 2} = p'' , \\
{1 \over r'^3} - {1 \over r''^3} = q , ~~~~
{1 \over r^3} - {1 \over r''^3} = q' , ~~~~
{1 \over r'^3} - {1 \over r^3} = q'' = q - q' , \\
{ u'^2 + u''^2 - u^2 \over 2 } = \nu , ~~~~
{ u^2 + u''^2 - u'^2 \over 2 } = \nu' , ~~~~
{ u^2 + u'^2 - u''^2 \over 2 } = \nu'' , $$
and one will get first, for the determination of *dρ*, this
equation
##H1.
$$ \tag{H} \frac{d^2\rho}{dt^2} + Cpq - Bp'q' - Ap''q'' = 0 . $$

One has next ##, $$ dV = \frac{dp'' + d\rho}2 , ~~~~ dV' = \frac{dp'' - d\rho}2 , ~~~~ dV'' = \frac{dp - d\rho}2 , $$

239

from which ##, ##, ##. $$ dR = {2rdr \over r''^3} + q(dp'' + d\rho) , \\ dR' = {2r'dr' \over r''^3} + q'(dp'' - d\rho) , \\ dR'' = {2r''dr'' \over r^3} + q''(dp - d\rho) . $$

from which ##, ##, ##. $$ dR = {2rdr \over r''^3} + q(dp'' + d\rho) , \\ dR' = {2r'dr' \over r''^3} + q'(dp'' - d\rho) , \\ dR'' = {2r''dr'' \over r^3} + q''(dp - d\rho) . $$

But ##; $$ \frac1{r''^3} = \frac1{r^3} - q', ~~~~ 2rdr = dp' + dp'' ; $$ so ##; $$ \frac{2rdr}{r''^3} = \frac{2dr}{r^2} - q'(dp' + dp'') ; $$ one will find likewise ##, ##; $$ \frac{2r'dr'}{r''^3} = \frac{2dr'}{r'^2} - q(dp + dp'') , \\ \frac{2r''dr''}{r^3} = \frac{2dr''}{r''^2} + q'(dp' + dp) ; $$ so that in substituting these values, and making for simplicity ##I3, $$ \left\{ \begin{array}{l} dQ = q'dp' - q''dp'' - qd\rho , \\ \tag{I} dQ' = q dp + q'' dp'' + q'd\rho , \\ dQ'' = -qdp -q'dp + q''d\rho , \end{array} \right. $$ one gets ##, $$ R = - \frac2r - Q, ~~~~ R' = - { 2 \over r'} - Q' , ~~~~ R'' = - { 2 \over r''} - Q '' , $$ and from that ##J3. $$ \left\{ \begin{array}{l} u^2 = \frac{2(A+B+C)}r + CQ , \\ \tag{J} u'^2 = \frac{2(A+B+C)}{r'} + BQ' , \\ u''^2 = \frac{2(A+B+C)}{r''} + AQ'' . \end{array} \right. $$

240

Now one gets
##,
##;
$$
\frac{r^2}{r''^3} = \frac1r - q'(p'+p'') , \\
\frac12 \left(\frac1{r'^3} - \frac1{r''^3} \right)
(r^2 + r'^2 -r''^2) = qp'' ; $$
so, adding those two equations, and putting *q''* in place of
*q-q'*, one gets
##;
$$
\frac{r^2}{r''^3} +
\frac12 \left(\frac1{r'^3} - \frac1{r''^3} \right)
(r^2 + r'^2 -r''^2) = \frac1r - p'q' + p''q'' ; $$
one finds likewise
##,
##;
$$
\frac{r'^2}{r''^3} +
\frac12 \left(\frac1{r^3} - \frac1{r''^3} \right)
(r^2 + r'^2 -r''^2) = \frac1{r'} - pq - p''q'' , \\
\frac{r''^2}{r^3} +
\frac12 \left(\frac1{r'^3} - \frac1{r^3} \right)
(r'^2 + r''^2 -r^2) = \frac1{r''} + pq + p'q' ; $$
so, making all these substitutions in equations (G) or (F) of the
preceding Articles, they become these
##K3.
$$
\left\{ \begin{array}{l}
\frac{d^2(r^2)}{2dt^2} - {A+B+C \over r} - C(p'q'-p''q''+Q) = 0 , \\
\tag{K} \frac{d^2(r'^2)}{2dt^2} - {A+B+C \over r'} - B(pq+p''q''+Q') = 0 , \\
\frac{d^2(r''^2)}{2dt^2} - {A+B+C \over r''} - A(-pq-p'q'+Q'') = 0 . \\
\end{array} \right. $$

Thus one will be able to, with the aid of these three
equations, determine the three radii *r, r', r''* in terms of
*t*, which will give for each instant the relative position of the
bodies among themselves.

It is good to note that, if one divides the first of these
equations by C, the second by B and the third by A, and that next one
adds them together, one gets (because *d*Q + *d*Q' +
*d*Q'' = 0, and in consequence Q + Q' + Q'' = const.) this
##L1,
$$
\tag{L} {\small
\frac{d^2(r^2)}{2Cdt^2} +
\frac{d^2(r'^2)}{2Bdt^2} +
\frac{d^2(r''^2)}{2Adt^2} -
(A+B+C) \left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) =
\text{const.} , } $$
which can take the place of any one of the three equations (K).

241

One can also put the same equations (K) in another form as follows.

I multiply the first of these equations by
*d(r ^{2})*, and I integrate it next to have
##,
$$ \frac{d(r^2)^2}{4dt^2} - 2(A+B+C)r
- C\int (p'q'-p''q'')d(r^2) - C\int Qd(r^2) + L = 0 , $$
L being an arbitrary constant.

Then
##;
$$ \int Qd(r^2) = Qr^2 - \int r^2dQ ; $$
but
##;
$$ dQ = q'dp' - q''dp'' - q d\rho ; $$
moreover, because *r ^{2}=p'+p''*, one gets
##
##;
$$ \begin{aligned}
& (p'q'-p''q'')d(r^2) - r^2(q'dp'-q''dp'') ~~~~~~~~~~ \\
& ~~ = - (p'q'-p''q'')(dp'+dp'') + (p'+p'')(q'dp'-q''dp'')
= q(p''dp'-p'dp'') ;
\end{aligned} $$
such that if one makes, for brevity,
##,
$$ dP = q(p''dp' - p'dp'' -r^2d\rho) , $$
one gets, by omitting the constant L which can be supposed to be
contained in P, and dividing the whole equation by

Making likewise ##, ##, $$ dP' = q'(p''dp - pdp'' + r'^2d\rho) , \\ dP'' = q''(pdp' - p'dp' + r''^2d\rho) , $$

242

one will find by operations resembling those preceding ##, ##. $$ \frac{dr'^2}{dt^2} - \frac{2(A+B+C)}{r'} + B\left(\frac{P'}{r'^2} - Q'\right) = 0 , \\ \frac{dr''^2}{dt^2} - \frac{2(A+B+C)}{r''^2} + A\left(\frac{P''}{r''^2} - Q''\right) = 0 . $$

one will find by operations resembling those preceding ##, ##. $$ \frac{dr'^2}{dt^2} - \frac{2(A+B+C)}{r'} + B\left(\frac{P'}{r'^2} - Q'\right) = 0 , \\ \frac{dr''^2}{dt^2} - \frac{2(A+B+C)}{r''^2} + A\left(\frac{P''}{r''^2} - Q''\right) = 0 . $$

And, if one reduces these equations respectively by equations
(K) found above, then next one divides the remaining equations by *r,
r', r''*, one gets these three
##M3.
$$
\left\{ \begin{array}{l}
\frac{d^2r}{dt^2} + \frac{A+B+C}{r^2} -
C\left(\frac{p'q'-p''q''}r + \frac P{r^3}\right) = 0 , \\
\tag{M} \frac{d^2r'}{dt^2} + \frac{A+B+C}{r'^2} -
B\left(\frac{pq+p''q''}{r'} + \frac{P'}{r'^3}\right) = 0 , \\
\frac{d^2r''}{dt^2} + \frac{A+B+C}{r''^2} -
A\left(\frac{-pq-p'q'}{r''} + \frac {P''}{r''^3}\right) = 0 .
\end{array} \right. $$

We have thus reduced the six initial equations (A), (B) which
contain the solution to the three-body problem taken in full generality
to three other equations between the three distances *r, r', r''*
and the time *t*. It is true that these reductions each contain two
integration signs (which is evident on substituting the values of Q, Q',
Q'', or of P, P', P'' and of *dρ*). and that in this regard
they are less simple than the initial equations ; but, on the other
hand, they have the advantage of not containing any radical, which seems
to me to be of great importance in these types of problems.

Let us suppose then that one has determined by equations (K)
or (M) the three variables *r, r', r''* in terms
of *t* ; one still only knows from that the relative
positions of the bodies, that is to say, the triangle that the three
bodies form at each instant ; therefore it remains to be seen how
one can determine next the orbit itself of each body, that is to say,
the six variables *x, y, z, x', y', z'*.

243

In order to do this, we remark firstly that by knowing *r, r',
r''* one will know also *u, u', u''*, and *dV, dV', dV''*
by the formulae of Article V. So that one will have {by putting
*p''* in place of (*r ^{2} + r'^{2} -
r''^{2}*)/2 and

However, by considering the quantities *x, y, z, x', y', z',
dx, dy, dz, dx', dy', dz'* as so far unknown, it is clear that the
preceding equations do not suffice to determine them, because one would
have twelve unknowns, and only ten equations ; but, if one includes with
these equations the three equations (D) of article II, one will then get
one equation more than there are unknowns, and the difficulty will
consist only in solving these equations.

I see, with regard to the equations of the previous article, that they can only take the place of nine equations, because, in eliminating some of the unknowns, it happens that others also disappear,

244

so that one arrives by this means at one equation where there enter only the quantities*r*^{2}, r'^{2}, p'',
... . In order to prove it as easily as possible, I take first the three
equations
##,
##,
##,
$$
xdx + ydy + zdz = rdr , \\
x'dx + y'dy + z'dz = dV , \\
dx'dx + dy'dy + dz'dz = \nu''dt^2 , $$
and I get from them by the ordinary rules of elimination the values of
*dx, dy, dz* ; I have, in doing that, for brevity.
##,
##,
##,
##,
$$
\alpha = y'dz'-z'dy',~~~~ \alpha' = zdy'-ydz',~~~~ \alpha'' = yz'-y'z, \\
\beta = z'dx'-x'dz',~~~~ \beta' = xdz'-zdx',~~~~ \beta'' = zx'-z'x, \\
\gamma = x'dy'-y'dx',~~~~ \gamma' = ydx'-xdy',~~~~ \gamma'' = xy'-yx', \\
\delta = x(y'dz'-z'dy') - y(x'dz'-z'dx') + z(x'dy'-y'dx') , $$
I have, I say,
##,
##,
##.
$$
dx = { \alpha rdr + \alpha' dV + \alpha''\nu''dt^2 \over \delta } , \\
dy = { \beta rdr + \beta' dV + \beta'' \nu''dt^2 \over \delta } , \\
dz = { \gamma rdr + \gamma' dV + \gamma''\nu''dt^2 \over \delta } . $$

so that one arrives by this means at one equation where there enter only the quantities

However, I remark that one has ##, ##, ##, ##, $$ \begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= (x'^2+y'^2+z'^2)(dx'^2+dy'^2+dz'^2) - (x'dx'+y'dy'+z'dz')^2 \\ &= r'^2u'^2dt^2 - (r'dr')^2 , \\ \alpha'^2 + \beta'^2 + \gamma'^2 &= (x^2+y^2+z^2)(dx'^2+dy'^2+dz'^2) - (xdx'+ydy'+zdz')^2 \\ &= r^2u'^2dt^2 - dV'^2 , \\ \alpha''^2 + \beta''^2 + \gamma''^2 &= (x^2+y^2+z^2)(x'^2+y'^2+z'^2) - (xx'+yy'+zz')^2 \\ &= r^2r'^2 - p''^2 , \\ \alpha\alpha' + \beta\beta' + \gamma\gamma' &= (x'dx'+y'dy'+z'dz')(xdx'+ydy'+zdz') \\ & ~~~~~~ - (xx'+yy'+zz')(dx'^2+dy'^2+dz'^2) \\ &= r'dr'dV' - p''u'^2dt^2 , \end{aligned} $$

245

## ## ##, ## ## ##; $$ \begin{aligned} \alpha\alpha'' + \beta\beta'' + \gamma\gamma'' &= (x'dx'+y'dy'+z'dz')(xx'+yy'+zz') \\ & ~~~ - (xdx'+ydy'+zdz')(x'^2+y'^2+z'^2) \\ &= p''r'dr' - r'^2dV' , \\ \alpha'\alpha'' + \beta'\beta'' + \gamma'\gamma'' &= (xdx'+ydy'+zdz')(xx'+yy'+zz') \\ & ~~~ - (x'dx'+y'dy'+z'dz')(x^2+y^2+z^2) \\ &= p''dV' - r^2r'dr' ; \end{aligned} $$ such that, if one squares the three preceding equations, and then adds them together, one gets, after multiplying by δ^{2},
and making suitable substitutions,
##
##
##,
$$ \begin{aligned}
\delta^2(dx^2+dy^2+dz^2) =
& (rdr^2)[r'^2u'^2dt^2-(r'dr')^2] + dV^2(r^2u'^2dt^2 - dV'^2) \\
& + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\
& + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') .
\end{aligned} $$
Similarly, if one takes the three equations
##,
##,
##,
$$
xx + yy + zz = r^2 , \\
x'x + y'y + z'z = p'' , \\
xdx' + ydy' + zdz' = dV' , $$
and one takes from them the values of *x*, *y* and *z*,
it is easy to see that one gets for *x*, *y*, *z*
the same expressions that one has found higher up
for *dx*, *dy*, *dz*, by changing there only
*rdr* to *r*^{2},
*dV* to *p''*,
*v''dt*^{2} to *dV'*;
then, doing the same operations and the same substitutions as above,
one gets another equation
##
##
##,
$$ \begin{aligned}
\delta^2(x^2+y^2+z^2) =
& r^4 [r'^2u'^2dt^2 - (r'dr')^2] + p''^2(r^2u'^2dt^2 - dV'^2) \\
& + dV'^2(r^2r'^2-p''^2) +2r^2p''(r'dr'dV' -p''u'^2dt^2) \\
& + 2r^2dV'(p''r'dr'-r'^2dV') + 2p''dV'(p''dV'-r^2r'dr') .
\end{aligned} $$
Now one has
##
$$ dx^2 + dy^2 + dz^2 = u^2dt^2, ~~~\text{and} ~~~~ x^2+y^2+z^2 = r^2 $$

## ## ##, ## ## ##; $$ \begin{aligned} \alpha\alpha'' + \beta\beta'' + \gamma\gamma'' &= (x'dx'+y'dy'+z'dz')(xx'+yy'+zz') \\ & ~~~ - (xdx'+ydy'+zdz')(x'^2+y'^2+z'^2) \\ &= p''r'dr' - r'^2dV' , \\ \alpha'\alpha'' + \beta'\beta'' + \gamma'\gamma'' &= (xdx'+ydy'+zdz')(xx'+yy'+zz') \\ & ~~~ - (x'dx'+y'dy'+z'dz')(x^2+y^2+z^2) \\ &= p''dV' - r^2r'dr' ; \end{aligned} $$ such that, if one squares the three preceding equations, and then adds them together, one gets, after multiplying by δ

246

then one gets the following two equations ## ## ##,

## ## ##, $$ \begin{aligned} \delta^2u^2dt^2 & = (rdr)^2(r'^2u'^2dt^2 - r'^2dr'^2) + dV^2(r^2u'^2dt^2-dV'^2) \\ & + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\ & + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') , \\ \delta^2r^2 = & r^4(r'^2u'^2dt^2-r'^2dr'^2) + p''^2(r^2u'^2dt^2-dV'^2) \\ & + dV'^2(r^2r'^2-p''^2) + 2r^2p''(r'dr'dV'-p''u'^2dt^2) \\ & + 2r^2dV'(p''r'dr'-r'^2dV') + 2p''dV'(p''dV'-r^2r'dr') . \end{aligned} $$

then one gets the following two equations ## ## ##,

## ## ##, $$ \begin{aligned} \delta^2u^2dt^2 & = (rdr)^2(r'^2u'^2dt^2 - r'^2dr'^2) + dV^2(r^2u'^2dt^2-dV'^2) \\ & + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\ & + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') , \\ \delta^2r^2 = & r^4(r'^2u'^2dt^2-r'^2dr'^2) + p''^2(r^2u'^2dt^2-dV'^2) \\ & + dV'^2(r^2r'^2-p''^2) + 2r^2p''(r'dr'dV'-p''u'^2dt^2) \\ & + 2r^2dV'(p''r'dr'-r'^2dV') + 2p''dV'(p''dV'-r^2r'dr') . \end{aligned} $$

From which, eliminating
δ^{2}, one gets an equation in only the known quantities
*r ^{2}, r'^{2}*, ... .

If one takes from the last equation the value of
δ^{2}, one gets, by reducing and removing that which goes
out.
##;
$$ \delta^2 =
r'^2(r^2u'^2dt^2-r^2dr'^2-dV'^2) + 2p''r'dr'dV' - p''^2u'^2dt^2 ; $$
and, this value of δ^{2} being substituted in the other
equation, one gets
##
##
##
##;
$$ \begin{aligned}
r'^2 & (r^2u'^2dt^2-r^2dr'^2-dV'^2)u^2dt^2 +
(2p''r'dr'dV'-p''^2u'^2dt^2)u^2dt^2 \\
= & (rdr)^2(r'^2u'^2dt^2-r'^2dr'^2) + dV^2(r^2u'^2dt^2-dV'^2) \\
& + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\
& + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') ;
\end{aligned} $$
or else, by arranging the terms,
##
##
##
##.
$$ \begin{aligned}
& (r^2r'^2-p''^2)(u^2u'^2-\nu''^2)dt^4 + (rdr.r'dr'-dVdV')^2 \\
& ~~ - [r^2(r'dr')^2-2p''r'dr'dV' + r'^2dV'^2]u^2dt^2 \\
& ~~ - [r'^2(rdr)^2 - 2p''rdrdV + r^2dV^2]u'^2dt^2 \\
& ~~ - 2[p''(rdr.r'dr'+dVdV')-r^2r'dr'dV-r'^2rdrdV']\nu''dt^2 = 0 .
\end{aligned} $$

Now (Article V), ##; $$ dV = {dp''+d\rho \over 2} , ~~~~ \text{and} ~~~~ dV'' = {dp''-d\rho \over 2} ; $$

247

moreover one has, by the formulae of the same Article, ##, $$ r^2 = p'+p'', ~~~~ r'^2 = p+p'', ~~~~ r''^2 = p+p' , $$ and likewise ##; $$ u^2 = \nu'+\nu'', ~~~~ u'^2 = \nu+\nu'', ~~~~ u''^2 = \nu+\nu' ; $$ so, if one makes the substitutions, and one supposes for greater simplicity ##, ##, ##, $$ \Sigma = p\left(\frac{2rdr}{dt}\right)^2 + p'\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp'}{dt}\right)^2 - 2\left(p''\frac{dp'}{dt} - p'\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma' = p'\left(\frac{2r'dr'}{dt}\right)^2 + p\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp}{dt}\right)^2 + 2\left(p''\frac{dp}{dt} - p\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r'^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma'' = p''\left(\frac{2r''dr''}{dt}\right)^2 + p'\left(\frac{dp}{dt}\right)^2 + p\left(\frac{dp'}{dt}\right)^2 + 2\left(p\frac{dp'}{dt} - p'\frac{dp}{dt}\right)\frac{d\rho}{dt} + r''^2\left(\frac{d\rho}{dt}\right)^2 , $$ the following equation becomes, after having been multiplied by 16/*dt*^{4},
##N1.
$$ \tag{N} {\Tiny
16(pp'+pp''+p'p'')(\nu\nu'+\nu\nu''+\nu'\nu'') -
4(\Sigma\nu + \Sigma'\nu' + \Sigma''\nu'') + \left(
\frac{dpdp'+dpdp''+dp'dp''+d\rho^2}{dt^2} \right) ^2 = 0 . } $$

moreover one has, by the formulae of the same Article, ##, $$ r^2 = p'+p'', ~~~~ r'^2 = p+p'', ~~~~ r''^2 = p+p' , $$ and likewise ##; $$ u^2 = \nu'+\nu'', ~~~~ u'^2 = \nu+\nu'', ~~~~ u''^2 = \nu+\nu' ; $$ so, if one makes the substitutions, and one supposes for greater simplicity ##, ##, ##, $$ \Sigma = p\left(\frac{2rdr}{dt}\right)^2 + p'\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp'}{dt}\right)^2 - 2\left(p''\frac{dp'}{dt} - p'\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma' = p'\left(\frac{2r'dr'}{dt}\right)^2 + p\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp}{dt}\right)^2 + 2\left(p''\frac{dp}{dt} - p\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r'^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma'' = p''\left(\frac{2r''dr''}{dt}\right)^2 + p'\left(\frac{dp}{dt}\right)^2 + p\left(\frac{dp'}{dt}\right)^2 + 2\left(p\frac{dp'}{dt} - p'\frac{dp}{dt}\right)\frac{d\rho}{dt} + r''^2\left(\frac{d\rho}{dt}\right)^2 , $$ the following equation becomes, after having been multiplied by 16/

It is necessary then that this equation be valid at the same
time as the three equations (K) of Article V ; so that, as it
contains nothing more than the same variables as equations (K), and that
is of an order lower by one unit than that, one can regard it as an
integral of those equations (K), but a particular integral because it
contains no new constant ; thus, if one integrates the equations
(K) by adding to them the necessary constants, those constants must be
such that they themselves satisfy equation (N). So that, if one does not
want to use that last equation instead of one of the equations (K), it
is nonetheless necessary to have regard to it in the determination of
the constants ; but for that it will suffice to suppose everywhere
*t*=0.

Later we will always make use of that equation to determine the
constant which must enter in the value of *dρ/dt*, resulting
from the integration of equation (B) of Article V.

248

Let us now take again the equations (D) of Article II, and
making, for brevity,
##,
##,
##,
$$
\lambda = ydx-xdy, ~~~~ \lambda' = y'dx'-x'dy', ~~~~ \lambda'' = y''dx''-x''dy'', \\
\mu = zdx-xdz, ~~~~ \mu' = z'dx'-x'dz', ~~~~ \mu'' = z''dx''-x''dz'', \\
\nu = zdy-ydz, ~~~~ \nu' = z'dy'-y'dz', ~~~~ \nu'' = z''dy''-y''dz'', $$
one gets, after having multiplied by *dt*,
##O3.
$$
\left\{ \begin{array}{l}
\frac{\lambda}C + \frac{\lambda'}B + \frac{\lambda''}A = adt , \\
\tag{O} \frac{\mu}C + \frac{\mu'}B + \frac{\mu''}A = bdt , \\
\frac{\nu}C + \frac{\nu'}B + \frac{\nu''}A = cdt .
\end{array} \right. $$

Now I find, as higher up,
##,
$$ {\small \lambda^2+\mu^2+\nu^2 = (x^2+y^2+z^2)(dx^2+dy^2+dz^2)
- (xdx+ydy+zdz)^2 = r^2u^2dt^2 - (rdr)^2 , } $$
and by analogy
##,
##;
$$ \lambda'^2+\mu'^2+\nu'^2 = r'^2u'^2dt^2 - (r'dr')^2 , \\
\lambda''^2+\mu''^2+\nu''^2 = r''^2u''^2dt^2 - (r''dr')^2 ; $$
*Above, $(r''dr')^2$ to be $(r''dr'')^2$ ??
Yes, c.18 p.29 has $(r''dr'')^2$ clearly enough.*

I find likewise
##
##,
$$
{\Tiny \begin{aligned} \lambda\lambda'+\mu\mu'+\nu\nu'
& = (xx'+yy'+zz')(dxdx'+dydy'+dzdz') -
(x'dx+y'dy+z'dz)(xdx'+ydy'+zdz') \\
& = p''\nu''dt^2 - dVdV' = p''\nu''dt^2
- \left(\frac{dp''}2\right)^2 + \left(\frac{d\rho}2\right)^2 ,
\end{aligned} } $$
and by analogy
##,
##.
$$
\lambda\lambda''+\mu\mu''+\nu\nu'' = p'\nu'dt^2
- \left(\frac{dp'}2\right)^2 + \left(\frac{d\rho}2\right)^2 , \\
\lambda'\lambda''+\mu'\mu''+\nu'\nu'' = p\nu dt^2
- \left(\frac{dp}2\right)^2 + \left(\frac{d\rho}2\right)^2 . $$

249

Then, if one makes, for greater simplicity,
##,
##,
##,
$$ \Pi = r^2u^2 - \left(\frac{rdr}{dt} \right)^2 , \\
\Pi' = r'^2u'^2 - \left(\frac{r'dr'}{dt} \right)^2 , \\
\Pi'' = r''^2u''^2 - \left(\frac{r''dr''}{dt} \right)^2 , $$
and
##,
##,
##,
$$ \Psi = p\nu -
\left(\frac{dp}{2dt} \right)^2 + \left(\frac{d\rho}{2dt} \right)^2 , \\
\Psi' = p'\nu' -
\left(\frac{dp'}{2dt} \right)^2 + \left(\frac{d\rho}{2dt} \right)^2 , \\
\Psi'' = p''\nu'' -
\left(\frac{dp''}{2dt} \right)^2 + \left(\frac{d\rho}{2dt} \right)^2 , $$
such that one has
##,
##,
##,
$$ \lambda^2+\mu^2+\nu^2 = \Pi dt^2, ~~~~
\lambda'\lambda''+\mu'\mu''+\nu'\nu'' = \Psi dt^2 , \\
\lambda'^2+\mu'^2+\nu'^2 = \Pi' dt^2, ~~~~
\lambda\lambda''+\mu\mu''+\nu\nu'' = \Psi' dt^2 , \\
\lambda''^2+\mu''^2+\nu''^2 = \Pi'' dt^2, ~~~~
\lambda\lambda'+\mu\mu'+\nu\nu' = \Psi'' dt^2 , $$
one gets, on squaring the three equations (O) and adding them
together,
##P1,
$$ \tag{P} {\Pi \over C^2} + {\Pi' \over B^2} + {\Pi'' \over A^2} +
{2\Psi \over AB} + {2\Psi' \over AC} + {2\Psi'' \over BC} =
a^2+b^2+c^2 , $$
an equation which is also, as one sees, of an order lower by one unit
than the equations (K); and as it contains the arbitrary constant
*a ^{2} + b^{2} + c^{2}* which is not found
in equations (K), one can regard it as a complete integral of
these same equations.

One might believe that equation (E) which we have found in
Article II might thus, by substituting in it the values of *u, u',
u''*, give a new integral, but it is easy to see that it will only
result

250

in an identical equation, because the equation that it concerns reduces first to ##; $$ \frac{u^2}C + \frac{u'^2}B + \frac{u''^2}A - 2(A+B+C) \left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) = f ; $$ and, putting for*u, u',* and *u''* their values taken from
formulae (J), one gets, by omitting that which goes out,
##,
$$ Q+Q'+Q'' = f , $$
which contains no new condition, because the quantities Q, Q', Q'' are
already themselves such that *d*Q+*d*Q'+*d*Q''=0 (Article
V).

in an identical equation, because the equation that it concerns reduces first to ##; $$ \frac{u^2}C + \frac{u'^2}B + \frac{u''^2}A - 2(A+B+C) \left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) = f ; $$ and, putting for

Besides, if one combines the equation
##
$$ Q+Q'+Q'' = f $$
with the equations (N) and (P), after having there substituted the
values of *u, u'* and *u''*, one can, by means of these three
equations, determine the three quantities Q, Q', Q'', which will
consequently contain only the finished variables *r, r', r''* and
their first derivatives *dr, dr', dr''* with the quantity
*dρ/dt*; thus, substituting these values in equations (K), one
will have three equations of the second order in the variables *r,
r'*, and *r''*, in which there is no more than to substitute the
value of *dρ/dt*. Thus, if with the help of one of these
equations one eliminates the quantity *dρ/dt* from the two
others, one first gets two equations purely of the second order between
the variables *r, r', r''* and *t* ; next, if one
differentiates the value of *dρ/dt* and one puts the value of
*d ^{2}ρ/dt^{2}* in equation (R), one gets a
third equation in the same variables, which is only of the third order.
Then one gets, by that means, for the determination of the variables

We believe meanwhile that it is even simpler and more convenient

251

for the calculation to substitute in equations (K) the values of Q, Q', and Q'' taken from equations (J) ; because, although the resulting equations can mount to orders higher than the second, they have always the great advantage that the variables are found to be less mixed together, and that the analogy which holds there will greatly facilitate their resolution.

for the calculation to substitute in equations (K) the values of Q, Q', and Q'' taken from equations (J) ; because, although the resulting equations can mount to orders higher than the second, they have always the great advantage that the variables are found to be less mixed together, and that the analogy which holds there will greatly facilitate their resolution.

Of the ten equations of Article VIII there only remain thus no
more than nine, and of the three equations (D) of Article II, or (O) of
Article XI, there only remain no more than two ; such that one has
only in total eleven equations for the determination of the six
variables *x, y, z, x', y', z'* and their differentials *dx,
dy,*, ... ; from where one sees that it is impossible to
determine these variables directly and by the operations of Algebra
alone ; but one can obtain them in the end by means of an
integration, as one will see.

I suppose that one would wish to know the values of *x, y,
z* ; one has first the equation
##Q1.
$$ \tag{Q} x^2+y^2+z^2 = r^2 . $$

Next, multiplying the three equations (O) of article XI respectively by λ, μ, ν, and adding them together, one gets ##; $$ \frac{\lambda^2+\mu^2+\nu^2}C + \frac{\lambda\lambda'+\mu\mu'+\nu\nu'}B + \frac{\lambda\lambda''+\mu\mu''+\nu\nu''}A = (a\lambda+b\mu+c\nu)dt ; $$ or then, on making the substitutions of the same Article, ##R1. $$ \tag{R} \left( \frac\Pi C + \frac{\Psi''}B + \frac{\Psi'}A \right) dt = a(ydx-xdy) + b(zdx-xdz) + c(zdy-ydz) . $$

Then multiplying the same equations (O) respectively by *z, -y,
x,* and adding them together, one gets
##.
$$ \frac{\lambda z - \mu y + \nu x }C +
\frac{\lambda' z - \mu' y + \nu' x }B +
\frac{\lambda'' z - \mu'' y + \nu'' x }A =
(az-by+cx)dt . $$
Now it is easy to see that one has
λ*z*-μ*y*+ν*x*=0, and that

252

-λ'*z*+μ'*y*-ν'*x* is the same
quantity that we have designated higher up by δ (Article IX) ;
then, because one has already found (Article X)
##,
$$ \delta^2 = (r^2r'^2-p''^2)u'^2dt^2
-r^2r'^2dr'^2 + 2p''r'dr'dV' - r^2dV'^2 , $$
one gets, on making the substitutions of the same Article X,
##,
$$ (\lambda'z-\mu'y+\nu'x)^2 = \left[(pp'+pp''+p'p'')u'^2
- \frac{\Sigma'}4 \right] dt^2 , $$
and by analogy
##.
$$ (\lambda''z-\mu''y+\nu''x)^2 = \left[(pp'+pp''+p'p'')u''^2
- \frac{\Sigma''}4 \right] dt^2 . $$
Such that the equation above becomes
##S1.
$$
\tag{S} \frac{\frac12\sqrt{4(pp'+pp''+p'p'')u'^2-\Sigma'}}B +
\frac{\frac12\sqrt{4(pp'+pp''+p'p'')u''^2-\Sigma''}}A =
az-by+cx . $$

-λ'

Thus one will have three equations (Q), (R), and (S), with the
aid of which one can easily determine the values of *x, y, z*,
as soon as one knows those of *r, r'* and *r''*.

One can find similar formulae for the determination of *x', y',
z'* ; and likewise, without making a new calculation, it will suffice
to change in the previous B to C and C to B, to put accents on the
letters which have no accent and to remove the accent from those which
have one, without altering those which have two accents. It is only
necessary to observe that the quantity *dρ* does not change its
value, but only its sign, when one changes among themselves the masses
A, B, C and the accented letters, which is seen clearly from equation
(H) of Article XIV.

Suppose, for brevity. ##, ##, $$ \begin{aligned} T & = \frac{\Pi}C + \frac{\Psi''}B + \frac{\Psi'}A , \\ Z & = \frac{\sqrt{4(pp'+pp''+p'p'')u'^2-\Sigma'}}{2B} + \frac{\sqrt{4(pp'+pp''+p'p'')u''^2-\Sigma''}}{2A} , \end{aligned} $$

253

and one will have these three equations ##, ##, ##. $$ \begin{aligned} &x^2+y^2+z^2 = r^2 , \\ &az-by+cx = Z , \\ &a(ydx-xdy) + b(zdx-xdz) + c(zdy-ydz) = T dt . \end{aligned} $$

and one will have these three equations ##, ##, ##. $$ \begin{aligned} &x^2+y^2+z^2 = r^2 , \\ &az-by+cx = Z , \\ &a(ydx-xdy) + b(zdx-xdz) + c(zdy-ydz) = T dt . \end{aligned} $$

As the constants *a, b, c* are arbitrary (Article H), and
depend only on the position of the plane of projection of the orbits of
bodies B and C around body A, it is easy to see that one can take this
plane in a manner such that one has *b*=0 and *c*=0 ; because
for that it suffices that one has *b*=0 and *c*=0 at the
beginning of the motion, that is to say, when *t*=0.

Let us suppose then *b*=0 and *c*=0, one gets
##;
$$ az = Z , ~~~~ a(ydx-xdy) = Tdt ; $$
so
##,
$$ z = \frac Za , $$
and, because *x ^{2} + y^{2} + z^{2} =
r^{2}*, one gets
##;
$$ x^2+y^2 = r^2 - \frac{Z^2}{a^2} ; $$
so
##;
$$\frac{ydx-xdy}{x^2+y^2} = \frac{aTdt}{a^2r^2-Z^2} ; $$
so that on making
##,
$$ d\varphi = \frac{aTdt}{a^2r^2-Z^2} , $$
one gets
##,
$$\frac yx = \operatorname{tang} \varphi , $$
and from that
##.
$$ z = \frac Za,
~~~~ y = \sqrt{r^2-\frac{Z^2}{a^2}}\sin\varphi ,
~~~~ x = \sqrt{r^2-\frac{Z^2}{a^2}}\cos\varphi . $$

254

But if one does not want to keep to the supposition of
*b*=0 and *c*=0, which necessitates taking the plane of
projection in a specified manner, here is how one can determine the
quantities *x, y, z* with the maximum generality.

Let
##,
$$ lz -my +nx = X, ~~~~ \lambda z - \mu y + \nu x = Y , $$
*l, m, n, λ, μ, ν* being undetermined coefficients,
and X, Y two new variables ; one gets
##;
$$ YdX-XdY = (m\nu-n\mu)(ydx-xdy) +
(n\lambda-l\nu)(zdx-xdz) + (l\mu-m\lambda)(zdy-ydz) ; $$
then, making
##,
$$ m\nu-n\mu = ka, ~~~~ n\lambda-l\nu = kb, ~~~~ l\mu-m\lambda = kc , $$
one gets the equation (Article XIV)
##.
$$ YdX-XdY = kTdt . $$

Let us suppose that now one has
##,
$$ f(X^2+Y^2) + gZ^2 = x^2+y^2+z^2 , $$
substituting the values of X, Y, Z with
*x, y* and *z*, and comparing next the terms which contain the
same powers of *x, y* and *z*, one gets these six equations
##,
##,
##,
$$ \begin{aligned}
&f(l^2~+\lambda^2) + ga^2 = 1, ~~~~ f(lm + \lambda\mu) + gab = 0 , \\
&f(m^2+\mu ^2) + gb^2 = 1, ~~~ f(ln + \lambda\nu) + gac = 0 , \\
&f(n^2+\nu ^2) + gc^2 = 1, ~~~~ f(mn + \mu \nu) + gbc = 0 ,
\end{aligned} $$
which, being combined with the three preceding, will serve to determine
the nine unknowns *l, m, n, λ, μ, ν, f, g, k*.

255

That done, one has then, because
*x ^{2} + y^{2} + z^{2} = r^{2}*,
the equation
##,
$$ f(X^2+Y^2) + gZ^2 = r^2 , $$
from where
##;
$$ f(X^2+Y^2) = r^2 - gZ^2 ; $$
so
##.
$$ \frac{YdX-XdY}{X^2+Y^2} = \frac{fkTdt}{r^2-gZ^2} . $$
So, if one does
##,
$$ d\varphi = \frac{fkTdt}{r^2-gZ^2} , $$
one gets
##.
$$ Y = \sqrt\frac{r^2-gZ^2}f\sin\varphi, ~~~~
X = \sqrt\frac{r^2-gZ^2}f\cos\varphi . $$

Thus one knows the three quantities X, Y, Z, with the aid of which
one can determine *x, y, z*.

For that, one takes the three equations
##,
$$ lz - my + nx = X, ~~~~
\lambda z - \mu y + \nu x = Y, ~~~~ az - by + cx = Z , $$
and one adds them together after having multiplied them respectively :
1 by *fl, fλ, ga*; 2 by *fm, fμ, gb*;
3 by *fn, fν, gc*; one immediately gets,
by virtue of the equations of the preceding article,
##.
$$
z = f(l X + \lambda y) + gaZ, ~~~~
y = - f(m X + \mu Y) - gbZ, ~~~~
x = f(n X + \nu Y) + gcZ . $$

Now, as one has supposed
##,
$$ x^2 + y^2 + z^2 = f(X^2+Y^2) + gZ^2, $$
one gets, on substituting the values of *x, y, z* which one has
just found, and

256

comparing the homogeneous terms, ##, ##, ##, $$ \array{ f(l^2+m^2+n^2) = 1 , & l\lambda + m\mu + n\nu = 0 , \\ f(\lambda^2+\mu^2+\nu^2) = 1 , & la + mb + nc = 0 , \\ g(a^2+b^2+c^2) = 1 , & \lambda a + \mu b + \nu c = 0 , ~ } $$ and these equations must be identical with the six that have been found above (Article XV), and can in consequence be used in place of those for the determination of the unknowns*l, m,* ... .

comparing the homogeneous terms, ##, ##, ##, $$ \array{ f(l^2+m^2+n^2) = 1 , & l\lambda + m\mu + n\nu = 0 , \\ f(\lambda^2+\mu^2+\nu^2) = 1 , & la + mb + nc = 0 , \\ g(a^2+b^2+c^2) = 1 , & \lambda a + \mu b + \nu c = 0 , ~ } $$ and these equations must be identical with the six that have been found above (Article XV), and can in consequence be used in place of those for the determination of the unknowns

Then, as it is necessary to satisfy at the same time to these other
three equations (Article XV)
##,
$$ m\nu - n\mu = ka, ~~~~ n\lambda - l\nu = kb, ~~~~ l\mu - m\lambda = kc, $$
I remark that, if one adds together these last equations, after having
multiplied them respectively : 1 by *l, m, n*; 2 by λ, μ,
ν, one gets these two
##,
$$ k(la + mb + nc) = 0 , ~~~~
k(\lambda a + \mu b + \nu c) = 0, $$
which are in accordance with with the fifth and sixth of the
previous ; thus one can already reduce to a single one the three
equations concerned, and one satisfies that by the
determination of the unknown *k*. So, if one adds together the
squares of these equations, one gets
##
##
$$ \begin{aligned}
k^2(a^2+b^2+c^2) &=
(m\nu-n\mu)^2 + (n\lambda-l\nu)^2 + (l\mu-m\lambda)^2 \\
&= (l^2+m^2+n^2)(\lambda^2 + \mu^2 + \nu^2) -
(l\lambda + m\mu + n\nu)^2 = \frac1{f^2}
\end{aligned} $$
by virtue of the six equations above ; so that one gets
##.
$$ k = \frac 1 {f\sqrt{a^2+b^2+c^2}} . $$

So there is no more than to satisfy the six equations found higher up ; one can do that in several ways because there are more unknowns than equations.

In p.256, is *k* a
Lagrange's Undetermined Multiplier? Any LUMs elsewhere?

Full *Essai* - pp.229-332, PDF images, 4.79MB ;
reveal image of page 256 :

257

One has first
##;
$$ g = \frac 1 {a^2 + b^2 + c^2} ; $$
next, if one eliminates λ from those two equations
##,
$$ l\lambda + m\mu + n\nu = 0, ~~~~ \lambda a + \mu b + \nu c = 0 , $$
one gets
##,
$$ (am-bl)\mu + (an-cl)\nu = 0 , $$
and, eliminating μ, one gets likewise
##,
$$ (bl-am)\lambda + (bn-cm)\nu = 0 , $$
from which I conclude that one gets
##,
$$ \lambda = (cm-bn)\delta, ~~~~
\mu = (an-cl)\delta, ~~~~
\nu = (bl-am)\delta, $$
δ being an unknown that one determines by the equation
##,
$$ f(\lambda^2+\mu^2+\nu^2) = 1 , $$
which gives
##;
$$ f\delta^2[(cm-bn)^2 + (an-cl)^2 + (bl-am)^2] = 1 ; $$
but one has
##,
$$
{\small (cm-bn)^2 + (an-cl)^2 + (bl-am)^2
= (a^2+b^2+c^2)(l^2+m^2+n^2) -(al+bm+cn)^2 = \frac1{fg} , } $$
because of
##
$$ a^2+b^2+c^2 = \frac1g, ~~~~ l^2+m^2+n^2 = \frac1f, ~~~~ al+bm+cn = 0 $$
by the equations above ; so one gets
##,
$$ \frac{\delta^2}g = 1, ~~~~
\delta = \sqrt g = \frac1{\sqrt{(a^2+b^2+c)}} , $$
*ERRATUM : "$+c^2$" ??
c.18 p.41 also had "$+c$"*

258

and it only remains to satisfy these two equations ##. $$ f(l^2+m^2+n^2) = 1, ~~~~ la+mb+nc = 0 . $$

and it only remains to satisfy these two equations ##. $$ f(l^2+m^2+n^2) = 1, ~~~~ la+mb+nc = 0 . $$

Let us suppose, for greater simplicity,
##;
$$ a = h\cos\alpha, ~~~~
b = h\sin\alpha\cos\epsilon, ~~~~
c = h\sin\alpha\sin\epsilon ; $$
one gets
##,
$$ \delta = \sqrt g = \frac1h , $$
such that
##,
$$ h = \sqrt{a^2+b^2+c^2} , $$
and the second of the two preceding equations becomes
##;
$$ l\cos\alpha + \sin\alpha(m\cos\epsilon + n\sin\epsilon) = 0 ; $$
let then
##,
$$ l = \sin\alpha\sin\eta , $$
and one gets, making for greater simplicity *f*=1,
##.
$$ \sin^2\alpha\sin^2\eta + m^2 + n^2 = 1, ~~~~
\cos\alpha\sin\eta + m\cos\epsilon + n\sin\epsilon = 0 . $$

Thus ##; $$ m\cos\epsilon + n\sin\epsilon = -\cos\alpha\sin\eta ; $$ so ## ##; $$ \begin{aligned} m^2 + n^2 - (m\cos\epsilon+n\sin\epsilon)^2 &= (m\sin\epsilon - n\cos\epsilon)^2 \\ &= 1 - \sin^2\alpha\sin^2\eta - \cos^2\alpha\cos^2\eta = 1 - \sin^2\eta = \cos^2\eta ; \end{aligned} $$ and, taking the square root. ##; $$ m\sin\epsilon - n\cos\epsilon = \cos\eta ; $$ such that one gets ##, ##; $$ m = \sin\epsilon\cos\eta - \cos\alpha\cos\epsilon\sin\eta , ~~ \\ n = - \cos\epsilon\cos\eta - \cos\alpha\sin\epsilon\sin\eta ; $$ and from that one finds the values of λ, μ, ν from the preceding formulae.

259

One gets in this manner ##, ##, ##, ##, ##, ##. $$\begin{aligned} l &= \sin\alpha\sin\eta , \\ m &= \sin\epsilon\cos\eta - \cos\alpha\cos\epsilon\sin\eta , \\ n &= - \cos\epsilon\cos\eta - \cos\alpha\sin\epsilon\sin\eta , \\ \lambda &= \sin\alpha\cos\eta , \\ \mu &= - \sin\epsilon\sin\eta - \cos\alpha\cos\epsilon\cos\eta , \\ \nu &= \cos\epsilon\sin\eta - \cos\alpha\sin\epsilon\cos\eta . \end{aligned} $$

If one substitutes these values in the expressions in *x, y,*
and *z* of Article XVI, it is easy to see that the quantities X, Y,
and Z/*h* are nothing else but the rectangular coordinates of the
same curve, which is represented by the coordinates *x, y,* and
*z*, but referred to another plane of projection, of which the
position depends on the angles α, ε and η. Indeed,
if one considers the two planes of coordinates *x, y,* and of
coordinates X, Y, the angle α will be that of the inclination of
the planes, the angle η will be that which the line of intersection
of the planes makes with the axis of abscissas *x*, and the angle
ε will be that which the axis of abscissas X makes with the same
line of intersection. So, as the expression of coordinates X, Y, and
Z/*h* is simpler than that of the coordinates *x, y, z,* it is
clear that the plane of projection to which belong the coordinates X, Y,
and Z/*h* is more suitable than any other plane for there referring
the movements of the three bodies, or rather the movement of two of the
bodies around the third.

One sees then that the position of the plane of projection is not at all unimportant, and that, among all of the possible planes that one can make pass through body A, there is one which must be chosen by preference, because the movements of bodies B and C around A are with respect to that plane the simplest that is possible.

That remark, which seems to me of some importance in the three-body problem, has not previously been made, because nobody as far as I know, has up to the present envisaged this problem in a manner as general as that which we have just done.

260

We will take then, in place of the coordinates *x, y, z,*
these X, Y, Z/*h*, to represent the movement of body B around A ;
and as one has, because of *h* = 1/√*g* (Article XV),
##,
##,
$$ X^2 + Y^2 + \left(\frac Zh\right)^2 = x^2+y^2+z^2 = r^2 , $$
$$ Y = \sqrt{r^2 - \left(\frac Zh\right)^2}\sin\varphi, ~~~~
X = \sqrt{r^2 - \left(\frac Zh\right)^2}\cos\varphi, $$
it is clear that φ will be the angle described by the body B around
A in the plane of projection, that is to say, the longitude of body B in
the same plane ; and that Z/*hr* will be the sine of the
latitude. Thus one has (Article XVI), because of *f=1*,
##.
$$ k = \frac1{\sqrt{a^2+b^2+c^2}} = \frac1h . $$

For body B : | |||

Radius rector of the orbit | . . . | r, | |
---|---|---|---|

Longitude | . . . | ∫ Tdt / (h[r^{2} - (Z/h)^{2}]), | |

Sine of the latitude | . . . | Z/hr. | |

For body C : | |||

Radius rector of the orbit | . . . | r', | |

Longitude | . . . | ∫ T'dt / (h[r'^{2} - (Z'/h)^{2}]), | |

Sine of the latitude | . . . | Z'/hr'. |

The values of T and of Z are given by the formulae of Article XIV,

Lagrange, or more likely the printer, appears not to have known the difference between a vector and a rector.

261

and to have those of T' and Z' one only needs to change in those the accent zero to ' and ' to zero, and then B to C and C to B.

and to have those of T' and Z' one only needs to change in those the accent zero to ' and ' to zero, and then B to C and C to B.

As for the quantity A, it is an arbitrary constant which depends on
the initial movement of the bodies ; but it is necessary to take it
such, that it agrees with equation (P) of Article XI, in which the
second member is
##;
$$ a^2 + b^2 + c^2 = h^2 ; $$
such that it is only necessary to take for *h* the square root
of the value of the first member of that equation when one makes
*t*=0.

The formulae which we have found serve to determine the orbits
of bodies B and C around body A with respect to a fixed plane passing
through that same body ; but One still has to see how one can
determine, by their means, the mutual position of these orbits. For
that, we begin by remarking that if one considers the triangle formed at
each instant by the three bodies A, B, C, and of which the three sides
are *r, r',* and *r'',* and that one names as ζ, ζ',
ζ'' the three angles opposite to these sides, one will have, as one
knows, by elementary geometry,
##,
##,
##.
$$ \cos\zeta = \frac{r'^2 + r''^2 - r^2}{2r'r''} = \frac{p}{r'r''} , \\
\cos\zeta' = \frac{r^2 + r''^2 - r'^2}{2rr''} = \frac{p'}{rr''} , \\
\cos\zeta'' = \frac{r^2 + r'^2 - r''^2}{2r'r''} = \frac{p''}{rr'} . $$

Then one has (Article VIII)
##;
$$ p'' = xx' + yy' + zz' ; $$
so
##,
$$ \cos\zeta'' = \frac{xx'+yy'+zz'}{rr'} , $$
ζ'' being the angle formed at the centre of body A by the radius
rectors *r* and *r'* of the two other bodies B and C.

262

Now imagine two planes passing, one through body A and through the
two infinitely close points in which is found the body B at the
beginning and at the end of the infinitely small time *dt*, and the
other through the same body A, and through the two infinitely close
points where body C is at the beginning and at the end of the same time
*dt* ; these two planes will be those of the orbits of bodies B and
C around A, and will necessarily intersect in a straight line passing
through body A which will thus be the line of nodes of the two orbits.

Let ω be the inclination of these two planes the one to the
other, ξ the distance from body B to the intersection of the two
planes or to the line of nodes, that is to say, the angle included
between the radius *r* and the line of nodes, and ξ' the
distance of body C to the same line of nodes, that is to say, the angle
formed by the radius *r'* and the line of nodes ;
if one imagines a sphere described around A as centre, and that by the
points where the two radii *r, r'* and the line of nodes cross the
surface of that sphere, of which we suppose the radius to be equal to 1,
one takes the arcs of great circles, one gets a spherical triangle of
which the three sides will be ξ, ξ' and ζ'', and of which the
angle opposite to the side ζ'' will be ω ; such that one gets,
from known formulae,
##;
$$ \cos\zeta'' = \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega ; $$
so
##
$$ \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega =
= \frac{xx'+yy'+zz'}{rr'} $$

Let us suppose now that during the time *dt* the body B
describes around A the infinitely small angle *dθ*, and that
body C describes the angle *dθ'*, it is clear that, whereas
the lines *x, y, z, r* grow by their differentials *dx, dy, dz,
dr,* the angle ξ grows by *dθ*, and the angle ω
remains the same, because one supposes that the positions of the planes
of the orbits of bodies B and C is the same at the beginning and at the
end of the instant *dt* ; likewise, in making grow the lines *x',
y', z', r'* by their differentials *dx', dy', dz', dr',* it will
be only the angle ξ' which will vary in growing by *dθ'*.
So, as the preceding equation must be identical to and independent of
the law of movements of bodies B and C, it is clear that one can there

263

vary the quantities*x, y, z, r* and ξ which belong to
body B independently of the quantities *x', y', z', t'* and
*Z'* which belong to body C, and *vice versa* these
independently of those ; from which it follows that by varying first
*x, y, z, r* and ξ, next *x', y', z', r'* and ξ', then
the ones and the others at the same time, one obtains from the equation
concerned the three following
##,
##,
##
##
##.
$$ -- \sin\xi\cos\xi' + \cos\xi\sin\xi'\cos\omega)d\theta =
- \frac{(xx'+yy'+zz')dr}{r^2r'} + \frac{x'dx+y'dy+z'dz}{rr'} , \\
-- \cos\xi\sin\xi' + \sin\xi\cos\xi'\cos\omega)d\theta' =
- \frac{(xx'+yy'+zz')dr'}{rr'^2} + \frac{xdx'+ydy'+zdz'}{rr'} , $$
*Leading "(", instead of minus, twice ?? Yes,
c.18 p.47 has "$(-$".*
$$ \begin{aligned}
& (\sin\xi\sin\xi' + \cos\xi\cos\xi'\cos\omega)d\theta d\theta' \\
& ~~~~ = \frac{(xx'+yy'+zz')drdr'}{r^2r'^2}
- \frac{(x'dx+y'dy+z'dz)dr'}{rr'^2} \\
& ~~~~~~~~ - \frac{(xdx'+ydy'+zdz')dr}{r^2r'}
+ \frac{dxdx'+dydy'+dzdz'}{rr'} .
\end{aligned} $$

vary the quantities

Then, if one makes in all of these equations the substitutions of Article VIII, one gets these four ##, ##, ##, ##. $$ \begin{aligned} \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega &= \frac{p''}{rr'} , \\ - \sin\xi\cos\xi' + \cos\xi\sin\xi'\cos\omega &= - \frac{p''dr+rdV}{r^2r'd\theta} , \\ - \cos\xi\sin\xi' + \sin\xi\cos\xi'\cos\omega &= - \frac{p''dr'+r'dV'}{rr'^2d\theta} , \\ \sin\xi\sin\xi' + \cos\xi\cos\xi'\cos\omega &= \frac{p''drdr'-dV.rdr'-dV'.r'dr+rr'\nu''dt^2}{r^2r'^2d\theta d\theta'} , \end{aligned} $$

Then it is easy to conceive that the square of the small space
that body B travels in time *dt* is equally expressed by
*dx ^{2} + dy^{2} + dz^{2}* and by

264

and likewise ##. $$ d\theta' = \frac{\sqrt{u'^2dt^2-dr'^2}}{r'} = \frac{dt\sqrt{\Pi'}}{r'^2} . $$

and likewise ##. $$ d\theta' = \frac{\sqrt{u'^2dt^2-dr'^2}}{r'} = \frac{dt\sqrt{\Pi'}}{r'^2} . $$

Thus the second members of the four preceding equations will
all be given, as soon as one knows *r, r'* and *r''* in terms
of *t* (cited Article) ; so that one gets four equations in
three unknowns ξ, ξ' and ω, by which one can not only
determine the three unknowns, but also have an equation in the
quantities *r, r', r'', u, u',* ,,, , and this equation will be the
same as that which one has already found higher up (Article X) by a very
different route.

Suppose, for brevity, that the preceding equations are represented as follows ##, ##, ##, ##, $$ \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega = \lambda , \\ \sin\xi\cos\xi' - \cos\xi\sin\xi'\cos\omega = \mu , \\ \cos\xi\sin\xi' - \sin\xi\cos\xi'\cos\omega = \nu , \\ ~~~~ \sin\xi\sin\xi' + \cos\xi\cos\xi'\cos\omega = - \varpi , $$ by making ##, ##; $$ \lambda = \frac{p''}{rr'}, ~~~~ \mu = \frac{p''dr + rdV}{r^2r'd\theta}, ~~~~ \nu = \frac{p''dr'+r'dV'}{rr'^2d\theta'} ; \\ \varpi = \frac{-p''drdr'+dV.rdr'+dV'.r'dr - rr'\nu''dt^2} {r^2r'^2d \theta d\theta'} ; $$ it is easy to reduce those four equations to these two ##, ##, $$ \cos(\xi\pm\xi')(1\mp\cos\omega) = \lambda\pm\varpi , \\ \sin(\xi\pm\xi')(1\mp\cos\omega) = \mu\pm\nu , $$ which, because of the ambiguity of the signs, are really equivalent to four equations. Raising these two equations to the square, and then adding them together, one has ##, $$ (1\mp\cos\omega)^2 = (\lambda\pm\varpi)^2 + (\mu\pm\nu)^2 , $$

265

from where, because of the ambiguity of the signs, one takes ##, $$ - \cos\omega = \lambda\varpi + \mu\nu, ~~~~ 1 + \cos^2\omega = \lambda^2 + \mu^2 + \nu^2 + \varpi^2 ; $$ such that eliminating cosω one gets ##. $$ 1 + (\lambda\varpi + \mu\nu)^2 = \lambda^2 + \mu^2 + \nu^2 + \varpi^2 . $$

from where, because of the ambiguity of the signs, one takes ##, $$ - \cos\omega = \lambda\varpi + \mu\nu, ~~~~ 1 + \cos^2\omega = \lambda^2 + \mu^2 + \nu^2 + \varpi^2 ; $$ such that eliminating cosω one gets ##. $$ 1 + (\lambda\varpi + \mu\nu)^2 = \lambda^2 + \mu^2 + \nu^2 + \varpi^2 . $$

If one substitutes in that equation the values of
λ, μ, ν, ω, as also those of *dθ* and
*dθ'*, one gets an equation which is the same
as equation (N) of Article X ; which can serve to confirm the
validity of our calculations.

The equation ## $$ - \cos\omega = \lambda\varpi + \mu\nu $$ gives ##, $$ \cos\omega = \frac{p''\nu''dt^2 - dVdV'}{r^2r'^2 d\theta d\theta'} = \frac{\Psi''}{\sqrt{\Pi\Pi'}} , $$ which makes known the inclination ω of these two orbits.

Knowing ω, one knows easily ξ and ξ' ; because, by multiplying the two equations, ##, ## $$ \cos(\xi+\xi')(1-\cos\omega) = \lambda + \varpi , \\ \cos(\xi-\xi')(1+\cos\omega) = \lambda - \varpi ~ $$ the one by the other, one gets this ## $$ ½ (\cos2\xi + \cos2\xi')\sin^2\omega = \lambda^2 - \varpi^2 ; $$ and likewise the other two equations ##, $$ \sin(\xi+\xi')(1-\cos\omega) = \mu+\nu, ~~~~ \sin(\xi-\xi')(1+\cos\omega) = \mu-\nu, $$ being multiplied together, give ##, $$ - ½ (\cos2\xi-\cos2\xi')\sin^2\omega = \mu^2 - \nu^2 , $$ from which one takes ##, $$ \cos2\xi = \frac{\lambda^2-\varpi^2-\mu^2+\nu^2}{\sin^2\omega}, ~~~~ \cos2\xi' = \frac{\lambda^2-\varpi^2+\mu^2-\nu^2}{\sin^2\omega}, $$

266

or also, by putting in place of ω^{2} its value
##,
$$ 1 + \cos^2\omega -\lambda^2 -\mu^2 -\nu^2 , $$
taken from the equation found above, one gets, because of
cos^{2}ω=1-sin^{2}ω,
##,
$$ \cos2\xi = 1 + \frac{2(\lambda^2+\nu^2-1)}{\sin^2\omega}, ~~~~
\cos2\xi' = 1 + \frac{2(\lambda^2+\mu^2-1)}{\sin^2\omega}, $$
from which one takes
##;
$$ \sin\xi = \frac{\sqrt{1-\lambda^2-\nu^2}}{\sin\omega}, ~~~~
\sin\xi' = \frac{\sqrt{1-\lambda^2-\mu^2}}{\sin\omega}; $$
that is to say, on substituting the values of λ, μ, ν, and
observing that *r*^{2}dθ^{2} =
u^{2}dt^{2}-dr^{2} and
*r'*^{2}dθ'^{2} =
u'^{2}dt^{2}-dr'^{2},
##,
##,
$$ \sin\xi = \frac
{\sqrt{(r^2r'^2-p''^2)u'^2dt^2-r^2(r'dr')^2+2p''(r'dr')dV'-r'^2dV'^2}}
{rr'^2\sin\omega d\theta'} , $$
$$ \sin\xi' = \frac
{\sqrt{(r^2r'^2-p''^2)u^2dt^2-r'^2(rdr)^2+2p''(rdr)dV-r^2dV^2}}
{r'r^2\sin\omega d\theta} , $$
or else (Article XIII)
##,
##.
$$ \sin\xi = \frac{\sqrt{4(pp'+pp''+p'p'')u'^2 - \Sigma'}}
{2r\sqrt{\Pi'}\sin\omega} , $$
$$ \sin\xi' = \frac{\sqrt{4(pp'+pp''+p'p'')u^2 - \Sigma}}
{2r'\sqrt{\Pi}\sin\omega} . $$

or also, by putting in place of ω

If one wants the three bodies to move in a single plane, one has then ω=0, and in consequence cosω=1 and sinω=0 ; so ##, $$ \Sigma = 4(pp'+pp''+p'p'')u^2, ~~~~ \Sigma' = 4(pp'+pp''+p'p'')u'^2 , $$ and by analogy ##. $$ \Sigma'' = 4(pp'+pp''+p'p'')u''^2 . $$

Therefore the quantities Z and Z' (Article XIV) are zero, and in consequence the movements of the three bodies occur in the same

267

plane as we have taken for the plane of projection (Article XVIII). So, if one substitutes the values of*u*^{2},
u'^{2}, u''^{2} taken from the preceding equations
in equation (P) of Article XI, one gets an equation in *r, r', r''*
and *dr/dt, dr'/dt, dρ/dt* , by which one can determine this
last quantity *dρ/dt* ; substituting next the value of
*dρ/dt* in those of Σ, Σ', Σ'', one gets the
values of *u*^{2}, u'^{2}, u''^{2}
expressed in terms of terms of *r, r', r''*
and *dr/dt, dr'/dt, dr''/dt* only ; thus, putting the values
of *u*^{2}, u'^{2}, u''^{2} in equations
(F) of Article III, one gets at last three equations in *r, r',
r''* and *t*, which will be simple differentials of the second
order, in place of the general equations (K) rising to the fourth order,
when one frees them of the integration signs.

plane as we have taken for the plane of projection (Article XVIII). So, if one substitutes the values of

Finally I believe that, in the present case, these last
equations will always be preferable, because they have the singular
advantage of not containing any radical, which would not be the case in
the equations where one uses the values of *u, u', u''* determined
by the equations above, values which would necessarily contain
square roots.

To sum up what has just been demonstrated in this Chapter, let
there be named : A, B, C the masses of the three bodies ; *r, r',
r''* the distances between the bodies A and B, A and C, B and
C ; and supposing, for brevity,
##,
##,
$$ \array {
p = \frac{r'^2+r''^2-r^2}2 ,&
p' = \frac{r^2+r''^2-r'^2}2 ,&
p'' = \frac{r^2+r'^2-r''^2}2 , \\
q = \frac1{r'^3} - \frac1{r''^3},&
q' = \frac1{r^3} - \frac1{r''^3},&
q'' = \frac1{r'^3} - \frac1{r^3} = q-q' ,
} $$

268

one gets, by taking the element of time*dt* as constant,
##H1;
$$ \tag{H} \frac{d^2\rho}{dt^2} + Cpq - Bp'q' - Ap''q'' = 0 ; $$
##I3;
$$
\left\{ \begin{array}{l}
dQ = q'dp' - q''dp'' - qd\rho , \\
\tag{I} dQ' = qdp + q''dp'' + q'd\rho , \\
dQ'' = - qdp - q'dp' + q''d\rho ;
\end{array} \right. $$
##K3.
$$
\left\{ \begin{array}{l}
\frac{d^2(r^2)}{2dt^2} - \frac{A+B+C}{r} - C(p'q'-p''q''+Q) = 0 , \\
\tag{K} \frac{d^2(r'^2)}{2dt^2} - \frac{A+B+C}{r'} - B(pq+p''q''+Q') = 0 , \\
\frac{d^2(r''^2)}{2dt^2} - \frac{A+B+C}{r''} - A(-pq-p'q'+Q'') = 0 .
\end{array} \right. $$

one gets first
##J3.
$$
\left\{ \begin{array}{l}
u^2 = \frac{2(A+B+C)}{r} + CQ , \\
\tag{J} u'^2 = \frac{2(A+B+C)}{r'} + BQ' , \\
u''^2 = \frac{2(A+B+C)}{r''} + AQ'' ,
\end{array} \right. $$

one gets, by taking the element of time

These equations will serve to determine the values of the
distances *r, r', r''* in terms of
*t* ; after which one will be able to find directly and
without any integration the values of all the other elements, on which
depends the determination of the orbits of bodies B and C around body A.

Indeed, if one names

u, | B, | A, | ||

u', | the speed of body | C, | around | A, |

u'', | C, | B, |

If one names next

- φ the angle run by body B around A in a plane supposed fixed and passing through A, that is to say, the longitude of B,

269

- ψ the angle of latitude of B with respect to the same plane,
- φ' the longitude of C,
- ψ' its latitude,

and one makes, for brevity, ##, ##, ##, ##, ##, ##, ##, ##, ##, ##, $$ \begin{aligned} P &= pp' + pp'' + p'p'' = r^2r'^2-p''^2 = ¼ (2r^2r'^2+2r^2r''^2+2r'^2r''^2-r^4-r'^4-r''^4) \\ &= - ¼(r+r'+r'')(r+r'-r'')(r-r'+r'')(r-r'-r'') , \\ \Sigma &= p\left(\frac{d(r^2)}{dt}\right)^2 + p'\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp'}{dt}\right)^2 - 2 \left(p''\frac{dp'}{dt}-p'\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma' &= p'\left(\frac{d(r'^2)}{dt}\right)^2 + p\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp}{dt}\right)^2 + 2 \left(p''\frac{dp}{dt}-p\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r'^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma'' &= p''\left(\frac{d(r''^2)}{dt}\right)^2 + p'\left(\frac{dp}{dt}\right)^2 + p\left(\frac{dp'}{dt}\right)^2 +2 \left(p\frac{dp'}{dt}-p'\frac{dp}{dt}\right)\frac{d\rho}{dt} + r''^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Pi &= r^2u^2 - \left(\frac{d(r^2)}{2dt}\right)^2 , \\ \Pi' &= r'^2u'^2 - \left(\frac{d(r'^2)}{2dt}\right)^2 , \\ \Pi'' &= r''^2u''^2 - \left(\frac{d(r''^2)}{2dt}\right)^2 , \\ \Psi &= p\nu - \left(\frac{dp}{2dt}\right)^2 + \left(\frac{d\rho}{2dt}\right)^2 , \\ \Psi' &= p'\nu' - \left(\frac{dp'}{2dt}\right)^2 + \left(\frac{d\rho}{2dt}\right)^2 , \\ \Psi'' &= p''\nu'' - \left(\frac{dp''}{2dt}\right)^2 + \left(\frac{d\rho}{2dt}\right)^2 , \end{aligned} $$ by supposing ##, $$ \nu = \frac{u'^2+u''^2-u^2}2 , ~~~~ \nu' = \frac{u^2+u''^2-u'^2}2 , ~~~~ \nu'' = \frac{u^2+u'^2-u''^2}2 , $$ one now gets ##, $$ \sin\psi = \frac{\frac1B\sqrt{4Pu'^2-\Sigma'} + \frac1A\sqrt{4Pu''^2-\Sigma''}}{2hr}, ~~~~ \frac{d\varphi}{dt} = \frac{\frac{\Pi}C + \frac{\Psi'}A + \frac{\Psi''}B}{hr^2\cos^2\psi}, $$

270

and ##. $$ \sin\psi' = \frac{\frac1C\sqrt{4Pu^2-\Sigma} + \frac1A\sqrt{4Pu''^2-\Sigma''}}{2hr'}, ~~~~ \frac{d\varphi'}{dt} = \frac{\frac{\Pi'}B + \frac{\Psi}A + \frac{\Psi''}C}{hr'^2\cos^2\psi'} . $$

and ##. $$ \sin\psi' = \frac{\frac1C\sqrt{4Pu^2-\Sigma} + \frac1A\sqrt{4Pu''^2-\Sigma''}}{2hr'}, ~~~~ \frac{d\varphi'}{dt} = \frac{\frac{\Pi'}B + \frac{\Psi}A + \frac{\Psi''}C}{hr'^2\cos^2\psi'} . $$

It is necessary to remark that these formulae contain two
constants which are not arbitrary, but which must be determined by
particular equations ; they are : one, the constant *h*, and the
other, the constant which can be added to the value of *dρ/dt*
deduced from equation (H) by means of integration.

Here are, then, the equations which will serve to determine these constants ##N1, $$ \tag{N} 16PU - 4(\Sigma\nu +\Sigma'\nu' +\Sigma''\nu'') + \left(\frac{dpdp'+dpdp''+dp'dp''+d\rho^2}{dt^2}\right)^2 = 0 , $$ by supposing ## $$ U = \nu\nu' + \nu\nu'' + \nu'\nu'' = u^2u'^2-\nu''^2 = ¼(2u^2u'^2 + 2u^2u''^2 + 2u'^2u''^2 - u^4 - u'^4 - u''^4) $$ and ##P1. $$ \tag{P} \frac{\Pi}{C^2} + \frac{\Pi'}{B^2} + \frac{\Pi''}{A^2} + 2\left(\frac{\Psi}{AB} + \frac{\Psi'}{AC} + \frac{\Psi''}{BC}\right) = h^2 . $$

One would be able, if one wished, to use these equations
instead of any two of the equations (K) ; but, as they are complicated
enough, it would be better to use them only in the determination of the
constants concerned ; and for that it is clear that one
can suppose everywhere *t*=0.

Then if, for greater simplicity, one supposes that, when *t*=0,
one has *dr/dt*=0, *dr'/dt*=0, and also that the radii *r,
r'* coincide, such that the angle ζ'' between these radii
(Article XIX) is zero, which is always permitted when that angle
is variable, one gets, because *r'' ^{2} = r^{2} +
r'^{2} = 2rr'*cosζ'' (cited Article),
##;
$$ r''^2 = (r'-r)^2, ~~~~ \frac{dp''}{dt} = 0 ; $$
so
##;
$$ \frac{dp}{dt} = 0, ~~~~ \frac{dp'}{dt} = 0, ~~~~ \frac{dp''}{dt} = 0 ; $$

271

such that equation (N) becomes ##; $$ 16PU - 4(r^2\nu+r'^2\nu'+r''^2\nu'')\left(\frac{d\rho}{dt}\right)^2 + \left(\frac{d\rho}{dt}\right)^4 = 0 ; $$ but because*r''*^{2}=(r'-r)^{2} one gets P=0 ; so
also *dρ/dt*=0. Thus it would be necessary to take the value of
*dρ/dt*=0, such that it becomes zero when *t*=0
( * ).

such that equation (N) becomes ##; $$ 16PU - 4(r^2\nu+r'^2\nu'+r''^2\nu'')\left(\frac{d\rho}{dt}\right)^2 + \left(\frac{d\rho}{dt}\right)^4 = 0 ; $$ but because

Equation (P) simplifies as much by the same suppositions, and
becomes
##P'1;
$$ \tag{P'} h^2 =
\frac{r^2u^2}{C^2} +
\frac{r'^2u'^2}{B^2} +
\frac{r''^2u''^2}{A^2} +
2\left(\frac{p\nu}{AB} + \frac{p'\nu'}{AC} + \frac{p''\nu''}{BC}\right) , $$
where it is necessary to take for
*r, r', r'', u, u', u''* the values which correspond to *t*=0.

As for the constants which can enter into the values of Q, Q', and
Q'', they will be entirely arbitrary and only depend on the initial
values of *u, u', u''*, which are at will.

Finally, if one also names

*dθ*the element of angle described by body B around body A in the instant*dt*,*dθ'*the corresponding angle described by body C around A,*ω*the mutual inclination of the orbits of bodies B and C around A,*ξ*the distance of body B from the node of these two orbits*ξ'*the distance of body C from the same node.

( * ) It is necessary to remark that, in the present case, equation (N) is still satisfied if one takes ##. $$ \left(\frac{d\rho}{dt}\right)^2 = 4(r^2\nu+r'^2\nu'+r''^2\nu'') . $$

(Editor's Note)

272

CHAPTER II.

SOLUTION OF THE THREE-BODY PROBLEM IN DIFFERENT CASES

In this Chapter, we shall examine some particular cases, where the three-body problem is greatly simplified and admits of an exact or almost exact solution ; although these cases do not occur in the System of the world, we think nevertheless that they merit the attention of Geometers, because they can cast light on the general solution of the three-body problem.

The first case which presents itself is that where the three
distances *r, r', r''* are constant, so that the triangle formed by
the bodies stays always the same and only changes its position.

One has in this case ##, $$dr = 0, dr' = 0, dr'' = 0,$$ and in consequence also ##; $$dp = 0, dp' = 0, dp'' = 0,$$ so the three equations (K) become ##a3; $$ \left\{ \begin{array}{l} {A+B+C \over r } + C( p'q'-p''q''+Q ) = 0 \\ \tag{a} {A+B+C \over r' } + B( p q +p''q''+Q' ) = 0, \\ {A+B+C \over r''} + A(-p q -p' q' +Q'') = 0; \end{array} \right. $$ from which one sees that the quantities Q, Q', Q'' will be likewise constant,

273

so that one has ##, $$ dQ=0, ~~~~ dQ'=0, ~~~~ dQ''=0, $$ meaning that equations (I) reduce to these ##, $$ qd\rho=0, ~~~~ q'd\rho=0, ~~~~ q''d\rho=0, $$ which give either*q=0, q'=0, q''=0*, or *dρ=0*.
Let us examine these two cases separately.

so that one has ##, $$ dQ=0, ~~~~ dQ'=0, ~~~~ dQ''=0, $$ meaning that equations (I) reduce to these ##, $$ qd\rho=0, ~~~~ q'd\rho=0, ~~~~ q''d\rho=0, $$ which give either

Put first ##; $$ q=0, ~~~~ q'=0, ~~~~ q''=0; $$ so ##; $$ r=r'=r''; $$ such that the triangle formed by the three bodies will be equilateral ; the equations (a) then give ##, $$ CQ = BQ' = AQ'' = - { {A+B+C} \over r } , $$ and, these values being substituted in the formulae (J), one has ##. $$ u^2 = u'^2 = u''^2 = - { {A+B+C} \over r } . $$

Now one has
##;
$$ p = p' = p'' = { r^2 \over 2 } , ~~~~
\nu = \nu' = \nu'' = { u^2 \over 2 } ; $$
so
##;
$$ P = { 3r^4 \over 4 }, ~~~~ U = { 3u^4 \over 4 }; $$
moreover equation (H) will give
*d ^{2}ρ/dt^{2}=0*, consequently
##,
$$ { d\rho \over dt } = \alpha, $$
α being an arbitrary constant which must satisfy equation (N).

So one finds ##; $$ \Sigma = \Sigma' = \Sigma'' = r^2\alpha^2; $$

274

such that the equation that we are discussing becomes ##, $$ 9r^4u^4 - 6 r^2u^2\alpha^2 + \alpha^4 = 0, $$ that is to say, ##, $$ ( 3r^2u^2 - \alpha^2 )^2 = 0, $$ from where ##. $$ \alpha^2 = 3r^2u^2 = 3(A+B+C)r. $$

such that the equation that we are discussing becomes ##, $$ 9r^4u^4 - 6 r^2u^2\alpha^2 + \alpha^4 = 0, $$ that is to say, ##, $$ ( 3r^2u^2 - \alpha^2 )^2 = 0, $$ from where ##. $$ \alpha^2 = 3r^2u^2 = 3(A+B+C)r. $$

So one satisfies all the equations of the Problem ;
so that the value of *r* remains indeterminate ; from
which it follows that the system of the three bodies can move so that
the three bodies always form an ordinary equilateral triangle.

Having found ##, $$ P = { 3r^4 \over 4 }, ~~~~ \Sigma = \Sigma' = \Sigma'' = r^2\alpha^2 = 3r^4u^2, $$ one has ##; $$ 4Pu^2-\Sigma = 0, ~~~~ 4Pu'^2-\Sigma' = 0, ~~~~ 4Pu''^2-\Sigma'' = 0; $$ so ##; $$\sin\psi = 0, ~~~~ \sin\psi' = 0 ; $$ from which one sees that the three bodies will always necessarily be in one given plane.

One finds next
##,
##;
$$ \Pi = \Pi' = \Pi'' = r^2u^2, \\
\Psi = \Psi' = \Psi'' = p\nu + {\alpha^2 \over 4} =
{r^2u^2 \over 4} + {3r^2u^2 \over 4} = r^2u^2; $$
then, because *ψ=0* and *ψ'=0*, one gets
##;
$$ {d\varphi \over dt} = {d\varphi' \over dt} =
( {1 \over C} + {1 \over A} + {1 \over B}) {u^2 \over h} ; $$
but equation (P) will give
##,
$$ h^2 = ({1 \over C^2} + {1 \over B^2} + {1 \over A^2} +
{2 \over AB} +{2 \over AC} +{2 \over BC} ) r^2u^2, $$
or else
##,
$$ h^2 = ({1 \over C} + {1 \over B} + {1 \over A} )^2 ~ r^2u^2. $$

275

consequently ##; $$ h = ({1 \over C} + {1 \over B} + {1 \over A} ) ~ ru; $$ so ##, $$ {d\varphi \over dt} = {d\varphi' \over dt} = {u \over r} = \sqrt { {A+B+C} \over r^3 } . $$

consequently ##; $$ h = ({1 \over C} + {1 \over B} + {1 \over A} ) ~ ru; $$ so ##, $$ {d\varphi \over dt} = {d\varphi' \over dt} = {u \over r} = \sqrt { {A+B+C} \over r^3 } . $$

Thus the bodies B and C just rotate around body A with a constant angular velocity equal to √ (A+B+C)/r³.

Let us examine now the other case, where *dρ/dt=0*
without *q, q', q''* being zeroes, and substitute first in
equations (J) the values of CQ, BQ' and AQ'' drawn from equations (a)
above ; one gets
##b3;
$$
\left\{ \begin{array}{l}
u^2 = {{A+B+C} \over r} - C( p'q'-p''q'') , \\
\tag{b} u'^2 = {{A+B+C} \over r'} - B( pq+p''q'') , \\
u''^2 = {{A+B+C} \over r''} - A(-pq-p'q') ;
\end{array} \right. $$
from which one sees that the relative speeds of the bodies will also
be constant, but not equal among themselves as in the preceding
case.

Then, because it is necessary that *dρ/dt=0*, one
gets then also *d ^{2}o/dt^{2}=0*, and equation (H)
becomes
##c1.
$$ \tag{c} Cpq - Bp'q' - Ap''q'' = 0. $$
Next equation (N) becomes (because of

276

thus, in combining one or other of these equations with the previous equation (c), one will be able to, by their means, determine any two of the three unknowns*r, r', r''*, and the problem will be
solved.

thus, in combining one or other of these equations with the previous equation (c), one will be able to, by their means, determine any two of the three unknowns

Suppose first P=0, one gets
##;
$$ (r+r'+r'')(r+r'-r'')(r-r'+r'')(r-r'-r'') = 0 ; $$
then, because *r, r', r''* are supposed positive, one gets
these equations
##,
$$
r+r'-r'' = 0, ~~~~ \text{or} ~~~~ r-r'+r'' = 0, ~~~~ \text{or} ~~~~ r-r'-r'' = 0, $$
from which one gets
##;
$$
r'' = r+r', ~~~~ \text{or} ~~~~ r' = r+r'', ~~~~ \text{or} ~~~~ r = r'+r''; $$
that is to say, that one of the three distances must be equal to the sum
of the other two, which shows that the bodies must always be arranged in
a straight line.

Let us imagine that the three bodies A, B, C are arranged
successively in the same direction, so that one has
##,
$$ r'' = r' - r, $$
and, taking for more simplicity *r'=mr*, one only needs to
substitute in equation (c) *mr* in place of *r'*, and
*(m-1)r* in place of *r''*; the unknown *r* will go out,
and one will have an equation which will serve to determine *m*.
One finds then
##,
##,
##,
##,
##,
##;
$$ \begin{aligned}
p &= { { m^2 + (m-1)^2 - 1 } \over 2 } r^2 = (m^2-m)r^2, \\
p' &= { { 1 + (m-1)^2 - m^2 } \over 2 } r^2 = (1-m)r^2, \\
p'' &= { { 1 + m^2 - (m-1)^2 } \over 2 } r^2 = mr^2, \\
q &= [ {1 \over m^3} - {1 \over (m-1)^3} ] {1 \over r^3} =
- {{3m^2-3m+1} \over m^3(m-1)^3} {1 \over r^3}, \\
q' &= [ 1 - {1 \over (m-1)^3} ] {1 \over r^3} =
{{m^3-3m^2+3m} \over (m-1)^3} {1 \over r^2}, \\
q'' &= ({1 \over m^3} - 1) {1 \over r^3} =
{{1-m^3} \over m^3} {1 \over r^3};
\end{aligned} $$
*
Should the "$q' =$" line end with "$\frac1{r^3}$" ?
Yes, c.18 p.65 has a fuzzy 3.*

First equation : cf. Heron's Formula.

277

and, these substitutions having been made in equation (c), it becomes, after having been multiplied by*m*^{2}(m-1)^{2}r,
##d1,
$$ \tag{d}
C(-3m^2+3m-1) + Bm^3(m^2-3m+3) - A(m-1)^2(1-m^3) = 0, $$
which being ordered with respect to *m* will mount to the fifth
degree, and will always have in consequence a real root.

and, these substitutions having been made in equation (c), it becomes, after having been multiplied by

It is good to note here that, although we have supposed
*r''=r'-r*, the solution nevertheless contains all the possible
cases, because the distances *r, r', r''*, being taken in one
straight line, can be positive or negative, according to the different
position of the bodies.

Now, because ##, $$ dr=0, ~~~~ dr'=0, ~~~~ dr''=0, ~~~~ d\rho=0, $$ one gets ##; $$ \Sigma=0, ~~~~ \Sigma'=0, ~~~~ \Sigma''=0; $$ so that, as one already has P=0, one gets ##, $$ \sin\psi=0, ~~~~ \sin\psi'=0, $$ which shows that the three bodies must move in a fixed plane.

Let us suppose now the other factor U to be equal to zero, one gets
##.
$$ U = \nu\nu' + \nu\nu'' + \nu'\nu'' = u^2u'^2 - \nu''^2 = 0. $$
Then equations (b) give
##,
$$ {u^2 \over r^2} - {u'^2 \over r'^2} = -
(A+B+C)q'' - {C \over r^2}(p'q'-p''q'') + {B \over r'^2}(pq+p''q''), $$
from which, on multiplying by *r ^{2}r'^{2}* and
putting in the place of

278

but ## $$ Cpq -Bp'q'-Ap''q'' = 0 $$ from equation (c) ; so one has simply ##, $$ r'^2u^2-r^2u'^2 = (-Cq+Bq'-Aq'')P, $$ and one finds likewise by analogy ##, ##, $$ r''^2u^2-r^2u''^2 = (Cq+Bq'-Aq'')P, \\ r''^2u'^2-r'^2u''^2 = (Cq+Bq'+Aq'')P, $$ from which it is easy to get ##, $$ p''u^2 - r^2\nu'' = -CqP, ~~~~ p''u'^2-r'^2\nu'' = - Bq'P, $$ and consequently ##; $$ \nu'' = {{p''u^2+CqP} \over r^2} = {{p''u'^2 +Bq'P} \over r'^2} ; $$ so ##, $$ \nu''^2 = { { p''^2u^2u'^2 + p''P(Bq'u^2 + Cqu'^2) + BCqq'P^2 } \over r^2r'^2 }, $$ and from that, because P =*r*^{2}r'^{2} - p''^{2},
##
##.
$$ \begin{aligned}
U = u^2u'^2-\nu''^2
&= {P \over r^2r'^2} \left[u^2u'^2 - p''(Bq'u^2+Cqu'^2) - BCqq'P\right] \\
&= {P \over r^2r'^2} \left[(u^2-Cqp'')(u'^2-Bq'p'') - BCr^2r'^2qq'\right]
\end{aligned} . $$

but ## $$ Cpq -Bp'q'-Ap''q'' = 0 $$ from equation (c) ; so one has simply ##, $$ r'^2u^2-r^2u'^2 = (-Cq+Bq'-Aq'')P, $$ and one finds likewise by analogy ##, ##, $$ r''^2u^2-r^2u''^2 = (Cq+Bq'-Aq'')P, \\ r''^2u'^2-r'^2u''^2 = (Cq+Bq'+Aq'')P, $$ from which it is easy to get ##, $$ p''u^2 - r^2\nu'' = -CqP, ~~~~ p''u'^2-r'^2\nu'' = - Bq'P, $$ and consequently ##; $$ \nu'' = {{p''u^2+CqP} \over r^2} = {{p''u'^2 +Bq'P} \over r'^2} ; $$ so ##, $$ \nu''^2 = { { p''^2u^2u'^2 + p''P(Bq'u^2 + Cqu'^2) + BCqq'P^2 } \over r^2r'^2 }, $$ and from that, because P =

But the same equations (b) give
##,
##;
$$ u^2 - Cqp'' =
\left(\frac{A+B}{r^3} + \frac{C}{r''^3}\right) ~ r^2 ~, $$
$$ u'^2 - Bq'p'' =
\left(\frac{A+C}{r'^3} + \frac{B}{r''^3}\right) ~ r'^2 ~; $$
so, on substituting those values as well as those of *q* and
*q'*, one gets
##,
$$ U = P \left[
\left(\frac{A+B}{r^3 } + \frac{C}{r''^3}\right)
\left(\frac{A+C}{r'^3} + \frac{B}{r''^3}\right)
- BC
\left(\frac{1}{r'^3} - \frac{1}{r''^3}\right)
\left(\frac{1}{r^3 } - \frac{1}{r''^3}\right)
\right] ~, $$

279

or indeed ##. $$ U = P\left[ \frac{A^2}{r^3r'^3} + \frac{B^2}{r^3r''^3} + \frac{C^2}{r'^3r''^3} + \frac{AB}{r^3} \left( \frac{1}{r'^3} + \frac{1}{r''^3} \right) + \frac{AC}{r'^3} \left( \frac{1}{r^3} + \frac{1}{r''^3} \right) + \frac{BC}{r''^3} \left( \frac{1}{r^3} + \frac{1}{r'^3} \right) \right] . $$ From which one sees that the equation U=0 can only give this P=0, the other factor of U can never become zero, because the radii*r, r', r''* and the masses A, B, C are positive quantities.

or indeed ##. $$ U = P\left[ \frac{A^2}{r^3r'^3} + \frac{B^2}{r^3r''^3} + \frac{C^2}{r'^3r''^3} + \frac{AB}{r^3} \left( \frac{1}{r'^3} + \frac{1}{r''^3} \right) + \frac{AC}{r'^3} \left( \frac{1}{r^3} + \frac{1}{r''^3} \right) + \frac{BC}{r''^3} \left( \frac{1}{r^3} + \frac{1}{r'^3} \right) \right] . $$ From which one sees that the equation U=0 can only give this P=0, the other factor of U can never become zero, because the radii

The equation P=0 being then the only one which can satisfy the case which we are examining, this case will only occur, as we have seen above, when the three bodies are arranged in one straight line, and their distances will be in the relationship expressed by equation (d).

Then we have already found that the three bodies must move in
a fixed plane ; so that, knowing the speed *u* of body B around A,
it is only necessary to divide by *r* to have the angular velocity
of bodies B and C ; but, if one wants to make use of the general
formulae of Article XXII, one remarks that because P=0 one has (Article
XXVII)
##;
$$ \frac{u^2}{r^2} = \frac{u'^2}{r'^2} = \frac{u''^2}{r''^2} ; $$
but equations (b) give
##;
$$
\frac{u^2}C + \frac{u'^2}B + \frac{u''^2}A = (A+B+C)
\left(\frac 1r + \frac 1{r'} + \frac 1{r''} \right) ; $$
so, substituting the previous values of *u' ^{2}* and

280

so also ##. $$ \nu = kp , ~~~~ \nu' = kp' , ~~~~ \nu'' = kp'' . $$ Thus one has (Article XXII) ##, ##; $$ \Pi = kr^4 , ~~~~ \Pi' = kr'^4 , ~~~~ \Pi'' = kr''^4 , \\ \Psi = kp^2 , ~~~~ \Psi' = kp'^2 , ~~~~ \Psi'' = kp''^2 ; $$ but, because P=0, one has ##; $$ p^2 = r'^2r''^2, ~~~~ p'^2 = r^2 r''^2, ~~~~ p''^2 = r^2r'^2 ; $$ so ##; $$ \Psi = kr'^2r''^2, ~~~~ \Psi' = kr^2r''^2, ~~~~ \Psi'' = kr^2r'^2 ; $$ so equation (P) becomes ##, $$ h^2 = k \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right)^2 , $$ from which ##; $$ h = k \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right) \sqrt k ; $$ next, because ψ=0 and ψ'=0, ##. $$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac kh \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right) = \sqrt k . $$

so also ##. $$ \nu = kp , ~~~~ \nu' = kp' , ~~~~ \nu'' = kp'' . $$ Thus one has (Article XXII) ##, ##; $$ \Pi = kr^4 , ~~~~ \Pi' = kr'^4 , ~~~~ \Pi'' = kr''^4 , \\ \Psi = kp^2 , ~~~~ \Psi' = kp'^2 , ~~~~ \Psi'' = kp''^2 ; $$ but, because P=0, one has ##; $$ p^2 = r'^2r''^2, ~~~~ p'^2 = r^2 r''^2, ~~~~ p''^2 = r^2r'^2 ; $$ so ##; $$ \Psi = kr'^2r''^2, ~~~~ \Psi' = kr^2r''^2, ~~~~ \Psi'' = kr^2r'^2 ; $$ so equation (P) becomes ##, $$ h^2 = k \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right)^2 , $$ from which ##; $$ h = k \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right) \sqrt k ; $$ next, because ψ=0 and ψ'=0, ##. $$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac kh \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right) = \sqrt k . $$

We have supposed above that the radii *r, r', r''* were
constant. and we have seen that that can only occur in two cases, namely
: when the three radii are all equal, and when one of them is equal to
the sum of the other two. Let us suppose now that these three radii are
only in a constant ratio among themselves, and let us see how that
condition can occur. So let
##.
$$ r'=mr, ~~~~ r'' = nr , $$

281

*m* and *n* being constant quantities, and one will
first get (Article XXII)
##,
##,
$$ p = \mu r^2, ~~~~ p' = \mu' r^2, ~~~~ p'' = \mu'' r^2, \\
q = \frac{\varpi}{r^3}, ~~~~
q' = \frac{\varpi'}{r^3}, ~~~~
q'' = \frac{\varpi''}{r^3}, $$
by making, for brevity,
##,
##.
$$ \mu = \frac{m^2+n^2-1}2, ~~~~
\mu' = \frac{1+n^2-m^2}2, ~~~~
\mu'' = \frac{1+m^2-n^2}2, \\
\varpi = \frac 1 {m^3} - \frac 1 {n^3}, ~~~~
\varpi' = 1 - \frac 1 {n^3}, ~~~~
\varpi'' = \frac 1 {m^3} - 1 = \varpi - \varpi'. $$
Then equation (H) becomes
##,
$$ \frac{d^2\rho}{dt^2} +
\frac {C\mu\varpi - B\mu'\varpi' - A\mu''\varpi''}r = 0, $$
or else, on making
##,
$$ \lambda = C\mu\varpi - B\mu'\varpi' - A\mu''\varpi'', $$
for brevity,
##,
$$ \frac{d^2\rho}{dt^2} + \frac \lambda r = 0, $$
and integrating
##,
$$ \frac{d\rho}{dt} = \alpha - \lambda \int \frac{dt}r , $$
α being an arbitrary constant equal to the value of
*dρ/dt* when *t*=0.

Then one gets ##, ##, ##; $$ dQ = 2(\mu'\varpi'-\mu''\varpi'')\frac{dr}{r^2} - \varpi \left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3}, \\ dQ' = 2(\mu\varpi +\mu''\varpi'')\frac{dr}{r^2} + \varpi' \left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3}, \\ dQ'' = 2(-\mu\varpi-\mu'\varpi')\frac{dr}{r^2} + \varpi'' \left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3}; $$

282

so, by integrating, ##, ##, ##, $$ Q = - \frac{2(\mu'\varpi'-\mu''\varpi'')}r - \varpi \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} + k, \\ Q' = - \frac{2(\mu\varpi +\mu''\varpi'')}r + \varpi' \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} + k', \\ Q'' = - \frac{2(-\mu\varpi-\mu'\varpi')}r + \varpi'' \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} +k'', $$*k, k'* and *k''* being arbitrary constants.

so, by integrating, ##, ##, ##, $$ Q = - \frac{2(\mu'\varpi'-\mu''\varpi'')}r - \varpi \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} + k, \\ Q' = - \frac{2(\mu\varpi +\mu''\varpi'')}r + \varpi' \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} + k', \\ Q'' = - \frac{2(-\mu\varpi-\mu'\varpi')}r + \varpi'' \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} +k'', $$

Making all those substitutions in equations (K), and next
dividing the second by *m ^{2}* and the third by

Those last two conditions can always be complied with by means
of the indeterminate constants *k, k'* and *k''* ; thus
the difficulty only consists of satisfying those of groups 1° and
2°.

Then, if one makes, for brevity, ##, $$ \delta = \mu\mu' + \mu\mu'' + \mu'\mu'' = - ¼ (1+m+n)(1+m-n)(1-m+n)(1-m-n), $$

283

one will be able to reduce the two equations of group 1° to these, by transformations analogous to those of Article XXVII, ##e1, $$ \tag{e} (-C\varpi + B\varpi' - A\varpi'')\delta + \mu''\lambda = 0 , ~~~~ ( C\varpi + B\varpi' - A\varpi'')\delta - \mu' \lambda = 0 . $$

one will be able to reduce the two equations of group 1° to these, by transformations analogous to those of Article XXVII, ##e1, $$ \tag{e} (-C\varpi + B\varpi' - A\varpi'')\delta + \mu''\lambda = 0 , ~~~~ ( C\varpi + B\varpi' - A\varpi'')\delta - \mu' \lambda = 0 . $$

Then it is only necessary to combine these two equations with those of group 2°, thus ##; $$ C\varpi = - \frac{B\varpi'}{m^2} = - \frac{A\varpi''}{n^2}, ~~ \text{ou bien} ~~ \alpha = 0 ~~ \text{and} ~~ \lambda = 0 ; $$ which makes two cases that we will examine separately.

Let there be first ##; $$ C\varpi = - \frac{B\varpi'}{m^2} = - \frac{A\varpi''}{n^2} ; $$ then ##; $$ \varpi' = - \frac{m^2C\varpi}B, ~~~~ \text{and} ~~~~ \varpi'' = - \frac{n^2C\varpi}A ; $$ but one has (Article XXIX) ##; $$ \varpi - \varpi' - \varpi'' = 0 ; $$ so ##, $$ \varpi \left( 1 + \frac{m^2C}B + \frac{n^2C}A \right) = 0 , $$ viz. ##. $$ \varpi \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right) = 0 . $$ So one can see that the quantity 1/C + *m ^{2}*/B
+

284

which will give ##, $$ n = m = 1 , $$ and consequently ##, $$ r = r' = r'' , $$ That is to say, the distances between the bodies are all equal, as in the case of Article XIV ; but with this difference, that in the present case they can change.

which will give ##, $$ n = m = 1 , $$ and consequently ##, $$ r = r' = r'' , $$ That is to say, the distances between the bodies are all equal, as in the case of Article XIV ; but with this difference, that in the present case they can change.

To know the movement of the bodies in that case, one takes
again the differential equations of article XXIX, which, by making
##,
$$ f = Ck = Bk' = Ak'' , $$
reduce to this single equation
##.
$$ \frac{d^2(r^2)}{2dt^2} - \frac{A+B+C}r - f = 0 , $$
Multiplying by *d(r ^{2})*, and integrating together, one
gets
##,
$$ \left[ \frac{d(r^2)}{2dt} \right]^2 - 2 (A+B+C)r - fr^2 = H , $$
H being an arbitrary constant, and from there
##,
$$ dt = \frac{rdr}{\sqrt{H + 2(A+B+C)r + fr^2)}} , $$
by means of which we get

Now, since ω=0, ω'=0, ω''=0, one gets ##; $$ Q = k, ~~~~ Q'=k', ~~~~ Q'' = k'' ; $$ so (Article XXII) ##. $$ u^2 = u'^2 = u''^2 = \frac{2(A+B+C)}r + f . $$

Moreover, having λ=0, one gets ##, $$ \frac{d\rho}{dt} = \alpha , $$

285

and that constant α must be determined so that it satisfies equation (N) ; one can give to that at*t* whatever
value one wants ; but, making no particular supposition, equation (N)
must be identical with that which we have found above for the
determination of *r*, and their comparison will give the value of
α.

and that constant α must be determined so that it satisfies equation (N) ; one can give to that at

Indeed, because *r=r'=r''*, one gets
##;
$$ p = p' = p'' = \frac{r^2}2 ; $$
so
##;
$$ P = \frac{3r^4}4 ; $$
and likewise, because *u=u'=u''*, one gets
##;
$$ \nu = \nu' = \nu'' = \frac{u^2}2 ; $$
then
##;
$$ U = \frac{3u^4}4 ; $$
together
##;
$$
\Sigma = \Sigma' = \Sigma'' =
\frac{r^2}2 \left( \frac{2rdr}{dt} \right) ^2 +
r^2 \left( \frac{rdr}{dt} \right)^2 + r^2\alpha^2 =
3r^2 \left( \frac{rdr}{dt} \right)^2 + r^2\alpha^2 ; $$
and equation (N) becomes
##,
$$
9r^4u^4
- 6u^2 \left[ 3r^2\left(\frac{rdr}{dt}\right)^2 + r^2\alpha^2 \right] +
\left[ 3\left(\frac{rdr}{dt}\right)^2 + \alpha^2 \right]^2 = 0 , $$
so
##;
$$ \left[ 3r^2u^2 - 3\left(\frac{rdr}{dt}\right)^2 -\alpha^2 \right]^2
= 0 ; $$
from which one gets
##.
$$ 3r^2u^2 - 3\left(\frac{rdr}{dt}\right)^2 -\alpha^2 = 0 ; $$

Then one has already found ##; $$ u^2 = \frac{2(A+B+C)}r + f ; $$

286

so, substituting that value and resolving the equation, it becomes ##. $$ dt = \frac{rdr} {\sqrt{-\frac{\alpha^2}3 + 2(A+B+C)r + fr^2}} . $$

so, substituting that value and resolving the equation, it becomes ##. $$ dt = \frac{rdr} {\sqrt{-\frac{\alpha^2}3 + 2(A+B+C)r + fr^2}} . $$

Comparing then that equation with the previous, one gets ##, $$ H = - \frac{\alpha^2}3 , $$ and consequently ##; $$ \alpha = \sqrt{-3H} ; $$ from which one sees that H must necessarily be a negative quantity.

One gets next ##, $$ \Pi = \Pi' = \Pi'' = r^2u^2 - \left( \frac{rdr}{dt} \right)^2 = \frac{\alpha^2}3 , $$ and ##; $$ \Psi = \Psi' = \Psi'' = \frac{r^2u^2}4 - \left( \frac{rdr}{2dt} \right)^2 + \left( \frac{\alpha}2 \right)^2 = \frac{\alpha^2}{4.3} + \frac{\alpha^2}4 = \frac{\alpha^2}3 ; $$ so equation (P) becomes ##, $$ \frac{\alpha^2}3 \left( \frac1C + \frac1B + \frac1A \right)^2 = h^2 , $$ from which one gets ##. $$ h = \frac{\alpha}{\sqrt3} \left( \frac1C + \frac1B + \frac1A \right) . $$

So, because one has already found ##, $$ \Sigma = \Sigma' = \Sigma'' = 3r^2 \left( \frac{rdr}{dt} \right)^2 + r^2\alpha^2 , $$ and that ##, $$ 3r^2u^2 - 3\left( \frac{rdr}{dt} \right)^2 - \alpha^2 = 0 , $$ one gets ##; $$ \Sigma = \Sigma' = \Sigma'' = 3r^4u^2 ; $$

287

moreover one has ##; $$ 4P = 3r^4, ~~~~ \text{and} ~~~~ u = u' = u'' ; $$ so one has ##; $$ 4Pu^2 - \Sigma = 0, ~~~~ 4Pu'^2 - \Sigma' = 0, ~~~~ 4Pu''^2 - \Sigma'' = 0 ; $$ and in consequence ##, $$ \sin\psi = 0 , ~~~~ \sin\psi' = 0 , $$ which is to say, ##; $$ \psi = 0 , ~~~~ \psi' = 0 , $$ which shows that bodies B and C must move in one fixed plane through body A.*f* is
the major axis and 2α^{2}/3(A+B+C)
the parameter.

moreover one has ##; $$ 4P = 3r^4, ~~~~ \text{and} ~~~~ u = u' = u'' ; $$ so one has ##; $$ 4Pu^2 - \Sigma = 0, ~~~~ 4Pu'^2 - \Sigma' = 0, ~~~~ 4Pu''^2 - \Sigma'' = 0 ; $$ and in consequence ##, $$ \sin\psi = 0 , ~~~~ \sin\psi' = 0 , $$ which is to say, ##; $$ \psi = 0 , ~~~~ \psi' = 0 , $$ which shows that bodies B and C must move in one fixed plane through body A.

Now, if one substitutes in the expressions *dψ/dt* and
*dψ'/dt* the values of H, H', Ψ, Ψ', Ψ'', and of
*h* found above, one gets
##;
$$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac\alpha{r^2\sqrt3} ; $$
and in consequence, in substituting the above value of *dt*,
##,
$$
d\varphi = { dr \over { r \sqrt {
-1 + {6(A+B+C) \over \alpha^2}r + {3f \over \alpha^2}r^2 }}}, $$
words.

So the bodies B and C describe in this case around body A two
similar and equal conic sections, of which the type and the shape depend
on the arbitrary quantities *f* and α, which can be
determined by the equations
##,
$$
\alpha^2 = 3r^2u^2 - 3\left(\frac{rdr}{dt} \right)^2 , ~~~~
f = u^2 - \frac{2(A+B+C)}r , $$
by giving to *u*, *r*, and *dr/dt* the values which fit
at the first instant.

288

It remains to examine the case where α=0 and λ=0; then the supposition that λ=0 first reduces equations (e) to these ##, $$ (-C\varpi + B\varpi' - A\varpi'')\delta = 0 , ~~~~ ( C\varpi + B\varpi' - A\varpi'')\delta = 0 , $$ which lead to ##, $$ \delta = 0 , $$ or else ##, $$ -C\varpi + B\varpi' - A\varpi'' = 0 , ~~~~ \text{and} ~~~~ C\varpi + B\varpi' - A\varpi'' = 0 , $$ which is to say, ##. $$ C\varpi = 0 ~~~~ \text{and} ~~~~ B\varpi' - A\varpi'' = 0 . $$

Then I observe first that these last two equations are unfruitful; because one first has ω=0, then, because ω''=ω-ω', one has ω''=-ω'; so that the equation Bω'-Aω''=0 becomes (B+A)ω'=0, which gives ω'=0; one has then ω=ω'=ω''=0, which returns to the case that we have examined before.

It is necessary then to make δ=0, so that the solution to the problem will be contained in these three equations ##. $$ \delta = 0 , ~~~~ \lambda = 0 , ~~~~ \alpha = 0 . $$

The first will give (Article XXIX)
##;
$$ (1+m+n)(1+m-n)(1-m+n)(1-m-n) = 0 ; $$
so
##,
$$ 1 \pm m \pm n = 0 , $$
and in consequence
##,
$$ r \pm r' \pm r'' = 0 , $$
which is to say, that one of the three distances *r, r', r''* must
be equal to the sum of the other two, and consequently that the three
bodies must be all arranged in a single straight line.

This case is therefore analogous to that of Article XXVI, but it is more

P.288, after "The first will give (Article XXIX)" : cf. Heron's formula for the area of a triangle.

289

general, in that the distances between the bodies can be variable, provided that their ratios are constant.

general, in that the distances between the bodies can be variable, provided that their ratios are constant.

One determines these ratios from the equation λ=0, and
for that one can suppose, as in the cited Article, that the three bodies
A, B, C are initially placed in a single straight line, such that
*r''=r'-r*, which gives *n=m-1* ; one substitutes then
this value of *n* in the expression for λ in Article XXIX,
and one gets an equation in *m* which is the same as equation
(d) of Article XXVI. But it is still necessary to see whether the
condition of α=0 can be satisfied ; and as the constant α
must be determined by equation (N), all reduces to knowing if that
equation can hold good by making α=0, which is to say,
*dρ/dt*=0, because of *dρ/dt* =
α - λ ∫ *dt/r* (Article XXIX) and of
λ=0.

So, by supposing *dρ/dt*=0, and substituting for
*r', r''* and *p, p', p''* their values (cited Article) one
gets
##,
##,
##,
##,
$$ \begin{aligned}
P &= (\mu\mu' + \mu\mu'' + \mu'\mu'')r^4 , \\
\Sigma &= (\mu + \mu'\mu''^2 + \mu''\mu'^2)~r^2
\left( \frac{2rdr}{dt} \right)^2 , \\
\Sigma' &= (\mu'm^4 + \mu\mu''^2 + \mu''\mu^2)~r^2
\left( \frac{2rdr}{dt} \right)^2 , \\
\Sigma'' &= (\mu''n^4 + \mu'\mu^2 + \mu\mu'^2)~r^2
\left( \frac{2rdr}{dt} \right)^2 ,
\end{aligned} $$
##;
$$
\frac{dpdp' + dpdp'' + dp'dp'' + d\rho^2}{dt^2} =
(\mu\mu' + \mu\mu'' + \mu'\mu'') \left(\frac{2rdr}{dt} \right)^2 ; $$
but, by the nature of the quantities μ, μ', μ'', one has
##;
$$ \mu' + \mu'' = 1 , ~~~~ \mu + \mu'' = m^2 , ~~~~ \mu + \mu' = n^2 ; $$
also, one has, by virtue of the equation δ=0,
##;
$$ \mu\mu' + \mu\mu'' + \mu'\mu'' = 0 ; $$
so one will have also
##;
$$
\mu + \mu'\mu''^2 + \mu''\mu'^2 = 0 , ~~~~
\mu'm^4 + \mu\mu''^2 + \mu''\mu^2 = 0 , ~~~~
\mu''n^4 + \mu'\mu^2 + \mu\mu'^2 = 0 ; $$

290

so that all the preceding quantities P. Σ, ... will be zero, and consequently equation (N) is itself verified.

so that all the preceding quantities P. Σ, ... will be zero, and consequently equation (N) is itself verified.

Now it is clear that because of α=0 and λ=0 the three differential equations of Article XXIX will reduce to this ##, $$ \frac{d^2(r^2)}{2dt^2} - \frac Fr - f = 0 , $$ by making, for brevity, ##, ##. $$ F = A + B + C(1 - \mu'\varpi' + \mu''\varpi'') , \\ f = Ck = \frac{Bk'}{m^2} = \frac{Ak''}{n^2} . $$

That equation being then multiplied by
*d(r ^{2})*, and then integrated, will give
##,
$$ \left(\frac{rdr}{dt} \right)^2 - 2Fr - fr^2 = H , $$
H being an arbitrary constant ; from which one gets
##,
$$ dt = \frac{rdr}{\sqrt{H + 2Fr + fr^2}} , $$
by means of which one determines

Also, if in the equations (J) of Article XXII one substitutes the values of Q, Q', Q'' of Article XXIX, one has, by virtue of the equations of group 1° of that Article, ##; $$ u^2 = \frac{2F}r + f , ~~~~ u'^2 = \frac{2m^2F}r + m^2f , ~~~~ u''^2 = \frac{2n^2F}r + n^2f ; $$ so ##, $$ \nu = \frac{2\mu F}r + \mu f , ~~~~ \nu' = \frac{2\mu'F}r + \mu'f , ~~~~ \nu'' = \frac{2\mu''F}r + \mu''f . $$

291

From that one will find ##, ##, ##, ##, ##, ##, $$ \begin{aligned} \Pi &= 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 = - H , \\ \Pi' &= m^4 \left[ 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 \right] = - Hm^4 , \\ \Pi'' &= - Hn^4 , \\ \Psi &= \mu^2 \left[ 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 \right] = - H\mu^2 = - Hm^2n^2 , \\ \Psi' &= - H\mu'^2 = - Hn^2 , \\ \Psi'' &= - H\mu''^2 = - Hm^2 , \end{aligned} $$ because of ##, $$ \mu\mu' + \mu\mu'' + \mu'\mu'' = 0 , $$ and in consequence ##. $$ \mu^2 = m^2n^2 , ~~~~ \mu'^2 = n^2 , ~~~~ \mu''^2 = m^2 . $$

From that one will find ##, ##, ##, ##, ##, ##, $$ \begin{aligned} \Pi &= 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 = - H , \\ \Pi' &= m^4 \left[ 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 \right] = - Hm^4 , \\ \Pi'' &= - Hn^4 , \\ \Psi &= \mu^2 \left[ 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 \right] = - H\mu^2 = - Hm^2n^2 , \\ \Psi' &= - H\mu'^2 = - Hn^2 , \\ \Psi'' &= - H\mu''^2 = - Hm^2 , \end{aligned} $$ because of ##, $$ \mu\mu' + \mu\mu'' + \mu'\mu'' = 0 , $$ and in consequence ##. $$ \mu^2 = m^2n^2 , ~~~~ \mu'^2 = n^2 , ~~~~ \mu''^2 = m^2 . $$

Thus, if one substitutes these values in equation (P), it becomes ##, $$ h^2 = -H \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right)^2 , $$ from which ##, $$ h = \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right) \sqrt{-H} , $$ from which one sees that H must be a negative quantity,

So, because of P=0 and of Σ=0, Σ'=0, Σ''=0, one gets ##, $$ \sin\psi = 0 ~~~~ \text{and} ~~~~ \sin\psi' = 0 , $$ which shows that the two bodies B and C must move on one fixed plane passing through the body A, and one then finds for the angles of rotation ##, $$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac{-H}{hr^2} \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right) = \frac{\sqrt{-H}}{r^2} . $$

292

And, if one substitutes the value of*dt*, found above, one
gets
##,
$$ d\varphi =
{ dr \over r \sqrt { -1 + {2F \over -H} r + {f \over -H} r^2 } }, $$
the polar equation of a conic section, with respect to the focus, in
which 2F/*f* will be the major axis and -2H/F the parameter.

And, if one substitutes the value of

So we come to see that the three body problem is solvable exactly, either with the distances between the three bodies being constant, or with them only remaining in constant ratios, and that there are two cases – when the three distances are all equal, so that the three bodies always form an equilateral triangle, and when one of the distances is equal to the sum or the difference of the other two, so that the three bodies are always arranged in a straight line.

Now, if one supposes that the distances *r. r', r''* are
variable, but in such a manner that their values differ very little from
those that they should have had for one of the preceding cases to apply,
it is clear that the Problem will be soluble roughly, by the known
methods of approximation ; but we will not enter here into that
detail, which will take us too far from our main object.

I acknowledge, moreover, that one could solve the preceding Problems in a more straightforward manner by the ordinary formulae of the Three-Body Problem with vector radii and the angles described by those radii, if one wanted to restrict oneself initially to the hypothesis that the bodies would move in a fixed plane ; but it would not be easy, I think, to achieve the purpose by those formulae, if one supposed, as we have done, that the bodies could move in different planes.

293-303

CHAPTER III.

MODIFICATION OF THE FORMULAE OF THE PREVIOUS CHAPTER, FOR THE CASE WHERE
ONE SUPPOSES THAT ONE OF THE THREE BODIES IS DISTANT FROM THE OTHER
TWO.

XXXIV.

**... ...**

The following partial translation of the
*Editor's Note* has not been independently checked.

324

The first Chapter of the Memoir that one has just
read deserves to be counted among the most important works of the
illustrious Author. The differential equations of the Three-Body
Problem, when one considers, which is permissible, only relative
motions, constitute a system of the *twelfth order*, and the
complete solution requires, in consequence, *twelve* integrations;
the only known integrals are those of the kinetic energies and the three
which furnish the area rule *(i.e. Kepler's 2nd Law. JRS)* : there
remain *eight* to discover.
** ... ...
**

325

** ... **

The method of Lagrange is most remarkable; it shows
that the complete solution of the problem requires only that one knows
at each instant the sides of the triangle formed by the three bodies;
the coordinates of each Body are then determined without any
difficulty. ** ... ... **

326

** ... ... **

327

** ... ... **

328

** ... ... **

329

** ... ... **

330

** ... ... **

331

** ... ... **

(*Editor's Note.*)

-==-0-==-

There is no reference to Euler, which suggests that Lagrange did not think that Euler had contributed to finding the constant-pattern solutions of the three-body problem.

My translation has been reviewed and somewhat corrected by a native French speaker, to whom many thanks.

Are the "undetermined coefficients" in Page 254 related to
"Lagrange's undetermined multipliers"? Or *k* in Page 256?

The Web has (2012-07-19) over a thousand references to "Lagrange's
quintic equation" - can that equation be identified in the
*Essai*? Perhaps Equation (*d*) at the top of p.277?
Also around p.289?

- Lagrange, Joseph-Louis,
*Essai sur le Problème des Trois Corps*: Notice (pp.229-231), Chapter I (pp.231-271), Chapter II (pp.272-292) :-- The
*Essai*alone - pp.229-332, PDF images, 4.79MB - at l'Université de Liège - (#) - Recueil des pièces qui ont remporté les prix de l'Académie royale des sciences, Tome 9 (1764-1772), published 1777, at Gallica. The 1772 prize-winning papers by Euler (p427/697, 40pp.) and by Lagrange (p563/697, 126pp.) are at the end.

- The
- Euler - see References in About Euler on Lagrange Points
- JRS :-
- Google Translate, French to English

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A final $ e^{i\pi} = -1 $ equation