Equilateral triangle solution of three-body problem for Apollo Earth-Moon libration point mission

Pei, R. Y.

English

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BELLCOMM, INC. 
5UBJECT: A Note on a S o l u t i o n  of DAG December 13,  1965 
t h e  Three Body Problem 
Case- 218 FROM: R.  Y .  P e i  
E 
AESTRACT 
The E q u i l a t e r a l  T r i ang le  S o l u t i o n  of  t he  three 
body problem i s  de r ived  i n  a s imple manner. The geometric 
p r o p e r t i e s  i nhe ren t  i n  such a conf igu ra t ion  are d iscussed .  
The problem i s  p e r t i n e n t  t o  earth-moon l i b r a t i o n  po in t  
miss ions  t h a t  have been suggested f o r  t h e  Apollo Applica- 
t i o n s  Program. 
. 
0 N66 UCCESSION NUMBER) 2 6 2 5 1  ITHRUl 
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I - 
BELLCOMM, INC. 
SUBJECT: A Note on a S o l u t i o n  of 
t h e  Three Body Problem 
Case- 218 
DATE: Decemher 13, 1965 
FROM: R. Y'. Pel .  
MEMORANDUM FOR FILE 
INTRODUCTION 
The E q u i l a t e r a l  Tr iangle  So lu t ions  of the f i n i t e  three 
body problem have been t r a d i t i o n a l l y  based upon the  work o f  
Lagrange i n  h i s  p r i z e  memoir, "Essai s u r  l e  Probleme des T r o i s  
Corps". (') 
e n t i a 1  equat ions  involv ing  t h e  three mutual d i s t a n c e s .  One of  
these equat ions  w a s  of t h e  t h i r d  order ,  and t h e  remaining two 
of  t h e  second o rde r  each, making t h e  whole problem of the 
seventh  o rde r .  The d i f f e r e n t i a l  equat ions  were t h e n  i n t e g r a t e d  
by assuming t h a t  t h e  r a t i o s  o f  t he  mutual d i s t a n c e s  were con- 
s t a n t s .  T h e  p re sen t  no te  a t tempts  t o  de r ive  the same s o l u t i o n  
i n  a s imple r  manner and br ing  out  the geometric p r o p e r t i e s  
i n h e r e n t  i n  such a conf igura t ion .  The problem i s  of i n t e r e s t  
i n  view o f  some earth-moon l i b r a t i o n  po in t  missions tha t  have 
been suggested f o r  Apollo Applicat ions.  
H i s  method requi red  t h e  d e r i v a t i o n  of three differ-  
PROOF 
Assume that there  e x i s t s  a stable p lane  conf igu ra t ion  
i n  which three f i n i t e  bodies revolve  about a given p o i n t .  
I n  t h e  absence o f  any d i s tu rbances ,  t h e  c e n t r i f u g a l  
f o r c e s  are counterbalanced by t h e  mutual a t t r a c t i o n s  among t h e  
three bodies ,  such mutual a t t r a c t i o n s  being equal  and oppos i t e  
f o r  each pair. Figure 1 d e p i c t s  such a sys t em.  
Lema- If the  fo rces  of  a t t r a c t i o n  between three bodies  obey 
Newton's t h i r d  l a w ,  t hen  t h e  three r e s u l t a n t  f o r c e s  are con- 
c u r r e n t ,  r e g a r d l e s s  of  t h e  geometry of  t h e  t r i a n g l e  formed by 
t h e  three bodies  as v e r t i c e s .  The proof i s  given i n  Appendix I. 
( l )Co l l .  Works, V o l .  V I . ,  p .  229. T i s se rand ' s  Mec. Cel. 
Vol. I., Chap. VIII. 
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I n  o r d e r  t o  de r ive  the e q u i l a t e r a l  t r i a n g l e  s o l u t i o n ,  
and 
the  o r i g i n ,  0,  i s  taken  as t h e  barycenter .  
t h e r e f o r e ,  the  v e c t o r  sum of  the  three q u a n t i t i e s  Ml$, M2g 
M must form a c losed  t r i a n g l e ,  as shown i n  Figure 2. 
By d e f i n i t i o n ,  
3 
Appl ica t ion  of t h e  Sine Law now y i e l d s  t h e  fol lowing:  
a : b : c = s i n  a : s i n  0 : s i n  (a+B) 
r : s : c = s i n  (8-m)  : s i n  a : s i n  (a+B-m) 
r : b : x = s i n  8 : s i n  p : s i n  ( a - g )  
s : a : x = s i n  (a+B+e) : s i n  q : s i n  m 
M l r  : M 2 s  . M3x = s i n  q : s i n  p : s i n  (Q+B-m) 
[ s i n  m s i n  ( a - ~ ) ]  
The magnitude o f  t h e  a t t r a c t i v e  f o r c e s ,  according t o  
Newton's law, are also r e l a t e d  i n  t h e  fol lowing manner: 
M1M2 . -  M1M3 2 
C 
2 - - -  a b2 
Equating ( 6 )  and (71, t h e  fol lowing expres s ion  i s  
M.. 
obta ined  f o r  mass r a t i o  1 : 
M3 
From Equation (51, t h e  same mass r a t i o  i s  ab ta ined  
as fo l lows:  
("~i le  same re fe rence  poin t  i s  used i n  a i i  c l a s s i c a l  p roofs  
and fo l lows  immediately from the  absence o f  e x t e r n a l  f o r c e s .  
( 7 )  
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From Equations ( 2 )  and 
- 3 -  
S s i n  e 
c s i n  (!i+B-m) 
- =  
s i n  
X 
( 3 )  , 
x = ,  b s i n  (!i+B-m) s i n  ( a - a )  
s c  s i n  p s i n  a 
S u b s t i t u t i n g  (10) i n t o  ( 9 )  and equa t ing  w i t h  ( 8 ) ,  t h e  
fo l lowing  i s  obta ined:  
- -  b s i n  (a-E) 
c s i n  II 
or, 
b = c  
S i m i l a r l y ,  from Equations ( 6 )  and ( 7 ) ,  
- - -  M1  b2 . s i n  II s i n  m 
M2 a 2 s i n  ( c t - e )  s i n  ( e - m )  
Equat ion ( 5 )  y i e l d s ,  
M1 s s i n  q 
r s i n  p M2 
- = -  
Equations (21, ( 3 )  and ( 4 )  y i e l d ,  
- = -  s a s i n  (a+B+e)  s i n  p 
r b  s i n  e s i n  q 
- a s i n  II s i n  m s i n  p 
b s i n  ( a - E )  s i n  (6-m) s i n  q 
- -  
BELLCOMM, I N C .  - 4 -  
S u b s t i t u t i n g  (14) i n t o  (13 )  and equa t ing  w i t h  ( 1 2 ) ,  t h e  fol lowing 
i s  obta ined:  
*1 b2 s i n  II s i n  m 
M2 a 
- = -  
2 s i n  (a-k) s i n  (B-m) 
E q u a l i t i e s  
Dynamic Eq 
a s i n  II s i n  rn 
b s i n  ( a - E )  s i n  (B-m) 
= -  
a = b  
(11) and (15) prove t h e  p a r t i c u l a r  s o l u t i o n  i n  ques t ion .  
i 1 i b r  i urn 
The e q u i l a t e r a l  t r i a n g l e  s o l u t i o n  i s  s u f f i c i e n t  to i n s u r e  
dynamic equ i l ib r ium provided the  s y s t e m  i s  g iven  a s u i t a b l e  
angu la r  v e l o c i t y  of r o t a t i o n .  T h i s  can a l s o  be shown from t h e  
geometric r e l a t i o n s h i p s  der ived i n  t h i s  no te .  S p e c i f i c a l l y ,  t h e  
fo l lowing  proof w i l l  be  given. 
L e t  Ql, n2 and n be three angu la r  v e l o c i t i e s  of r o t a t i o n  3 
necessary  t o  achieve dynamic equ i l ib r ium of t h e  t h r e e  masses 
under cons ide ra t ion .  It w i l l  be shown t h a t  these three angu la r  
v e l o c i t i e s  are equal .  
I n  o r d e r  t o  achieve dynamic equ i l ib r ium f o r  M1, t h e  
fo l lowing  cond i t ion  has t o  be m e t :  
2 
M Q r -  1 1  
S i m i l a r l y ,  f o r  M2 
M2 Q22 s 
2 x 3 3  
and M3,  w e  have 
( 1 7 )  
(18) 
BELLCOMM, 1NC. - 5 -  
From equa t ions  (16) and (17), w e  o b t a i n  
From equa t ions  ( 5 )  and observing that  a=S, w e  o b t a i n  
'12 1321 s i n  p s i n  m 
s i n  q s i n  k 
- =  
I f 3 1  2 sT2 
Equat ions (31, ( 4 1 ,  ( 6 )  and (15) y i e l d  the  fol lowing:  
= s i n  (a-t) 
s i n  q s i n  m 
S u b s t i t u t i n g  ( 2 1 )  and (22)  i n t o  equat ion  ( 2 0 ) ,  w e  ob ta in :  
S i m i l a r l y ,  w i t h  equa t ions  (16) and (181, w e  may w r i t e ,  
Noting t h a t  a = b = e ,  w e  ob ta in  from equat ion  ( 7 )  
(21 )  
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From equa t ion  ( 3 ) :  
s i n  (a-8) x =  
l? s i n  8 
- 6 -  
S u b s t i t u t i n g  equat ions  (25)  and (26) i n t o  (24) ,  and n o t i n g  equat ion  
(221, w e  ob ta in :  
1 0  13-RY P- t aa 
Attachment 
Appendix I 
2 2 
nl = n3 
- F!. Y .  P e i  
copy t o  
Messrs. S. W.  Fordyce - NASA/MLO 
W. D.  Green. Jr. - NASA/MLA 
T.' A. Keegan - NASA/MA-2 
G .  M. Anderson 
J. 0. Cappe l l a r i ,  Jr. 
P. Gunther 
J .  A. Hornbeck 
B. T. Howard 
H. S. London 
J .  Z .  Menard 
C .  R.  Moster 
V. S .  Mummert 
I. D. Nehama 
T. L. Powers 
I. M. Ross 
T. H. Tnompson 
R.  L. Wagner 
A l l  Members, D iv i s ion  1 0 1  
Department 1 0 2 3  
C e n t r a l  F i l e  
L i b r a r y  
( 2 7 )  
S. E. ROSS - NASA/MTS 
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~ 
APPENDIX I 
I n  t h i s  Appendix i t  is  proven tha t  the  dynamic equi- 
l i b r i u m  of  th ree  bodies  revolv ing  about a given po in t  i n  space 
i s  i n d i f f e r e n t  t o  t h e  shape of t h e  t r i a n g l e  formed by t h e  three 
bodies  as v e r t i c e s ,  as long as the  mutual a t t r a c t i o n s  between 
bodies are equa l  and oppos i t e .  The problem i s  restated as 
fo l lows  : 
L e t  an a r b i t r a r y  p o i n t  0 be chosen wi th in  t h e  t r i a n g l e  
(F igu re  1). L e t  equa l  segments If 1 be marked o f f  from 
t h e  two base v e r t i c e s  as shown. 
from r e s u l t a n t s  which must l i e  along M 0 and M20 respec- 
t i v e l y  t o  a s s u r e  dynamic equi l ibr ium,  t o  y i e l d  F2 and F1 
t h a t  are i n  t h e  d i r e c t i o n s  o f  M1M3 and M M The theorem 
w i l l  be proven i f  i t  can b e  shown t h a t  t h e  angle 8 formed 
by M M 
marked o f f  from M t hen  i t s  s u b t r a c t i o n  from a r e s u l t a n t  
t h a t  l i e s  a long  M 0 w i l l  y i e l d  a f o r c e  equa l  and oppos i t e  
t o  f, i n  t h e  d i r e c t i o n  of  M2M3. 
3 
3 Sub t rac t  f,  and -f 
1 
2 3 '  
and OM is  such t h a t  i f  a segment equa l  t o  If2/  i s  1 3  3 
3' 
3 
PROOF: Applying t h e  Sine law t o  t r i a n g l e s  M10M3, M20M3 
and M10M2, t he  fol lowing i s  obta ined:  
- X  r 
s i n  6 s i n  ( a - 8 )  
S =x 
s i n  ( y - 6 )  s i n  m 
S = -  r 
s i n  (6 -m)  s i n  II 
s i n  e - s i n  ( a - 8 )  s i n  ( s - m )  
s i n  ( y -e )  s i n  8 s i n  m 
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Appendix I - 2 -  
From the  f o r c e  diagrams a t  t h e  base of t h e  t r iangle ,  the 
fo l lowing  may be w r i t t e n :  
- -  a 
s i n  (B-m) s i n  m 
Comparing 'Equations ( A - 1 )  and ( A - 2 ) ,  it follows t h a t  
( A - 2 )  
which proves t h e  theorem. 
t 
F I G U R E  I FORCE DIAGRAM 
F I G U R E  2 MOMENT VECTOR DIAGRAM 


A note on a solution of three body problem

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