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R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.92 (Online Publication: Dec., 2013) 

 

 

BIBECHANA 
A Multidisciplinary Journal of Science, Technology and Mathematics 

ISSN 2091-0762 (online) 
Journal homepage: http://nepjol.info/index.php/BIBECHANA 

 

Solutions of the Sitnikov's circular restricted three body problems 

when primaries are oblate spheroid 
 

 
R.R.Thapa

1*
, M.R. Hassan

2
 

1Dept. of Mathematics, P.G.Campus, Biratnagar, Morang, Nepal. 
2
Dept. of Mathematics, S.M. College, Bhagalpur, Under T.M.B.University, Bhagalpur  

*
Email: Email: thaparajuram@yahoo.com    

 

Article history: Received 25 September, 2013; Accepted 26 November, 2013 

 

Abstract 

The solutions of  sitnikov's circular restricted three body problem has been tried to obtain by using 

Lindstedt poincare method if the primaries are oblate spheroid. 

Keywords: Lindstedt poincare method, Oblate spheroid, Perturbation, Primaries, Sitnikov's circular 

restricted three body problem.  

 

1. Introduction 

 
MacMillan [1] has studied an integrable case in restricted problem of three bodies by imposing further 

restrictions on the restricted three body problem by supposing the two finite bodies of equal masses 

and an infinitesimal body be moving in their common axis of revolution. 

Sitnikov [2] motion in three body problem has two primaries of equal masses and these two moving in 

(i) circular orbits and (ii) elliptic orbits around their center of mass. The third body (infinitesimal 

mass) is moving along a line which is perpendicular to the plane of primaries and passing through the 

center of mass of the primaries. 

Further Hagel [3] have studied a higher order perturbation analysis of the Sitnikov problem and 

investigated the low amplitude bounded oscillatory solutions in full range of primary eccentricities – 

0.99 < e < 0.99. They have found that near integrals in a polynomial form can be obtained for 
sufficiently small oscillation amplitudes in the entire interval of eccentricities. In addition they 

derived a relation for the non-linear frequency of the oscillatory solution as function of e and T0. 

Faruque [4] has studied a new analytic expression for the position of the infinitesimal body in the 
elliptic Sitnikov problem. This solution is valid for small bounded oscillations in cases of moderate 

primary eccentricities. The final solution to the equation with non-linear force included is obtained 

through first the use of a courant and Synder transformation followed by Lindstedt-poincare 
perturbation method and again an application of courant and Synder transformation. 

We have studied the Sitnikov’s circular restricted problem of the bodies when both the primaries are 

oblate spheroids and moving in circular orbits around their centre of mass. The Sitnikov’s problem is 

a special case of the restricted three body problem when both the primaries are of equal masses (m1 = 

m2 = 1/2) moving in circular orbits or elliptic orbits under Newtonion force of attraction and the third  

 



 

 

R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.93 (Online Publication: Dec., 2013) 

 

body of mass m3 (m3 is much less then the primaries) moves along the line perpendicular to the plane 

of motion of the primaries and passes through the center of mass of the primaries. 

2. Equations of motion  

Here the system consists of two oblate primaries with equal masses (m1 = m2 = 1/2). The third body of 
mass m3 is much less than the masses of the primaries. We know that both the primaries in the 

Sitnikov’s problem move on the circumference of the same circle if the primaries are spheroid in 

shape. But if the primaries are oblate spheroid in shape then due to their oblateness the primaries will 

not be equidistant from their center of mass. To keep the primaries equidistant from the centre of mass 

of the primaries, the following conditions are to be imposed : 

(i)  Principal axes of the oblate bodies should be parallel to the synodic axes. 

(ii)  The masses of both the primaries should be equal. 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let ai, bi, ci (i = 1, 2) be the semi axes of the primaries. When ai = bi, as the primaries are oblate, then 

the moment of inertia of the oblate spheroid m1 and m2 about the principal axes of body are  

 
 

 

 

 

 

m1 
a/2 

a/2 
k̂  î  

ĵ  

X 

p (0, 0, z) 

p1(a/2, 0, 0) 

p2(-a/2, 0, 0) 

m2 

r2 r1 

Y 

Z 

O(0, 0, 0) 



 

 

R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.94 (Online Publication: Dec., 2013) 

 

2 2 2 2

1 2

2 2

1 2

2 2 2

1 2

5 5

5

2

5 5

b c a c
A A

a c
B B

a b a
C C

+ +
= = = 


+ 

= = 

+

= = = 


   (1) 

Thus the potential between two bodies m1 and m2 is given by 

( )

( )

2 2 21 2 1 1 1 1
1 1 1 1 1 13

2 2 22 2 2 2
2 2 2 2 2 23

3

2 2

3

2 2

+ + − = + − + +  

+ + + − + +  

Gmm Gm A B C
v Al B m C n

r r

Gm A B C
A l B m C n

r

   (2) 

1 2 1 2

3

( )Gm m GA m m
v

r r

+
⇒ − = +   (3) 

where 

2 2

10

a c
A

−
=  = oblatenless of the primaries 

If op1 = r1 and op2 = r2 and 1̂u  and 2û  are the unit vectors along r1 and r2, then the equation 

of motion of m1 can be written as 

2

1 1 2 23 5 3 5

1 4 1 4
ˆ ˆ 0

2 2

A A
n ru r u

r r r r

   ⇒ − + + − + =   
   

  (4) 

Similarly from equation of motion of m2, we get  

2

1 1 2 23 5 3 5

1 4 1 4
ˆ ˆ 0

2 2

A A
ru n r u

r r r r

   − + + − + + =   
   

  (5) 

1 1 2 2
ˆ ˆ 0ru r u⇒ + =   (6) 

As Mccuskey [5] the equations (4), (5) and (6) have the non-trivial solutions if the 2 × 2 
determinant obtained from any two of the above three equations is zero. i.e. 

2

3 5

1 8A
n

r r
⇒ = +  (7) 

Now following Suraj et.al.[6], we can find the equation of motion of the third body as  

2 2

2 2 2 3/ 2 2 2 5/ 2 2 2 7 / 2

9 15

( ) ( ) ( )

d z z Az Az

dt z r z r z r

−
= − +

+ + +
 

2
8 16

1 1
3 3

A A
b

 ⇒ + = + = 
 

, say 

 
 



 

 

R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.95 (Online Publication: Dec., 2013) 

2 2

2 2 3/ 2 2 5/ 2 2 7 / 2

9 15
0

( ) ( ) ( )

d z z Az Az

dt z b z b z b
⇒ + + − =

+ + +
 (8) 

3. Solutions by Lindsted-Poincare Method 

The equation of motion of third body is  

( ) ( ) ( )
2 2

3/ 2 5/ 2 7 / 22 2 2 2

9 15
0+ + − =

+ + +

d z z Az Az

dt z b z b z b
 

2
2 3

2 3/ 2 5/ 2 7 / 2 5/ 2 7 / 2

4
5

9/ 2 7 / 2 9/ 2

1 9 15 3 45

2 2

105 15 315
... 0

2 8 8

   ⇒ + + − − +   
   

 + + + − = 
 

d z A A A
z z z

dt b b b b b

Az
z

b b b

 

( )
2

2 3

2

4 5

3 5
1 15 1

2 3

105 15 784
11 ... 0

2 4 3

d z A
A z Az z

dt

A A
z z

 ⇒ + + − − + 
 

 + + − + = 
 

 (9) 

Since, there is no effect of independent perturbating terms i.e. 15Az
2
, 

4105

2

Az
 and hence they may 

be omitted from the equation (9) 

We get, ( )
2

3

2

3 5
1 1 ... 0

2 3

d z A
A z z

dt

 + + − + + = 
 

 

2
2 3

02
0

d z
z z

dt
η ε⇒ + − =  (10) 

Where, 
2

0

3 5
1  and (1 )

2 3
A Aη ε= + = +  

Let us write, 

2 3

0 1 2 3( ) ( ) ( ) ( ) ( ) ...

& ,  where  is some parameter

z t z t z t z t z t

t

d

dt

ε ε ε

τ η η
τ

η

= + + + +

=

=

 

Let us write 
2 3

0 1 2 3 ...η η εη ε η ε η= + + + +  

From (10) 

 
 

 

 

 

 



 

 

R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.96 (Online Publication: Dec., 2013) 

 

( ) ( )

( ) ( )

2
2 2 3

02

2
2

2 3 2 3

0 1 2 3 0 1 2 32

3
2 2 3 2 3

0 0 1 2 3 0 1 2 3

2 22
2 2 20 01
0 0 0 0 1 1 0 02 2 2

2
2 2 22

0 0 2 12

0

... ...

... ... 0

2

2

+ − =

⇒ + + + + + + + +

+ + + + + − + + + + =

   
⇒ + + + + −   

   

+ + − +

d z
z z

d

d
z z z z

d

z z z z z z z z

d z d zd z
z z

d d d

d z
z

d

η η ε
τ

η εη ε η ε η ε ε ε
τ

η ε ε ε ε ε ε ε

η ε η η ηη α
τ τ τ

ε η η α η
τ

( )

( )

( )

22
2 01

1 0 1 2 02 2

2 2 2
3 2 2 23 2 1

0 0 3 2 1 0 1 2 02 2 2

2

0
1 2 3 0 2

2

2 2

2 ... 0 (11)

 
+ + 

 


+ + − + + +




+ + + =



d zd z

d d

d z d z d z
z

d d d

d z

d

η η η η
τ τ

ε η η α ηη η η η
τ τ τ

ηη η η
τ  

Hence (11) is identity in ε so the coefficients of 2 3 4, , , ,...ε ε ε ε  must be zero. 

2

0
02

0
d z

z
dτ

∴ + =  (12) 

22
2 301
0 1 1 0 02 2

2 0
d zd z

z z
d d

η ηη
τ τ

 
+ + − = 

 
 (13) 

( )
22 2

2 2 202 1
0 2 1 2 0 1 0 0 12 2 2

2 2 3 0
d zd z d z

z z z
d d d

η η η η ηη
τ τ τ

 
+ + + + − = 

 
 (14) 

( ) ( )

( )

2 2 2
2 23 0 1
0 3 1 2 3 0 1 2 02 2 2

2
2 22

1 0 0 2 0 12

2 2

2 3 0 (15)

 
+ + + + + 

 

+ − + =

d z d z d z
z

d d d

d z
z z z z

d

η ηη η η η η η
τ τ τ

ηη
τ

 

The general solution of the equation (12) is 

 z0 = c1 cos τ + c2 sin τ 

z0 = c cos τ (16)    

From (16),  

22
2 301
0 1 1 0 02 2

2 0
 

+ + − = 
 

d zd z
z z

d d
η ηη

τ τ
 



 

 

R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.97 (Online Publication: Dec., 2013) 

To avoid the secular terms, equating the coefficient of cos τ to zero,  

we get 

( )

2

1

3

8 1

c

A
η

−
=

+
  (17) 

2 3

1
12 2

0

cos3
4

d z c
z

d
τ

τ η
+ =      (18) 

The general solution of the equation (18) is 

           

3

1 3 4 2

0

0 1

3

3 4 2

0

3

3 4 2

0

cos sin cos3
32

cos cos sin cos3
32

3
sin sin cos sin 3

32

= + −

∴ = +

= + + −

= − − + +ɺ

c
z c c

z z z

c
z c c c

c
z c c c

τ τ τ
η

ε

ε
τ ε τ ε τ τ

η

ε
τ ε τ ε τ τ

η

 

3 3

2 2

0 0

cos cos cos3
32 32

c c
z c

ε
τ ε τ τ

η η
∴ = + −  (19) 

or,    ( )
3

1 2

0

cos cos3
32

c
z τ τ

η
= −  (20) 

        
( )

( )
3

1 cos cos3
32 1

c
z

A
τ τ= −

+
  (21) 

   
( )

( )
2 3

1

2
9cos3 cos

32 1
= −

+
d z c

d A
τ τ

τ
 

Now from (14) 

 
( )

22 2
2 2 202 1
0 2 1 2 0 1 0 0 12 2 2

4

2 3/ 2

2 2 3 0

21

256(1 )

d zd z d z
z z z

d d d

c

A

η η η η ηη
τ τ τ

η

 
+ + + + − = 

 

−
=

+

 

 
 

 



 

 

 R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.98 (Online Publication: Dec., 2013) 

 

 

( )

5 5

2 5 6 4 4

0 0

2

0 1 2

3

2

0

5 5
2

5 6 4 4

0 0

3
and cos sin cos 3 cos 5

128 1024

since, 

cos cos cos 3
32

3 cos 5
cos sin cos 3

128 1024

= + − +

= + +

= + −

 
+ + − + 

 

c c
z c c

z z z z

c
z c

c c
c c

τ τ τ τ
η η

ε ε

ε
τ τ τ

η

τ
ε τ τ τ

η η

 

 , 

( )

( ) ( )

3

5 5 2 5 2
2

2 4 4

0 0

3 2 5

2 4

0 0

cos cos cos3
32(1 )

23 3 cos5
cos cos3

1024(1 ) 128 1024

cos cos 3cos3 23cos 24cos3 cos5
32 1024

c
z c

A

c c c

A

c c
z c

ε
τ τ τ

ε ε τ
ε τ τ

η η

ε ε
τ τ τ τ τ τ

η η

∴ = + −
+

+ − +
+

= + − + − +

 

Now following, Raju Ram Thapa et.al.[6],  

 

( ) ( )
3 2 5

2 4

0 0

7 7
3

6 6

0 0

cos cos cos3 23cos 24cos3 cos5
32 1024

547 297 1
cos cos3 3cos5 cos7

32768 2048 8 16

c c
z c

c c

ε ε
τ τ τ τ τ τ

η η

ε τ τ τ τ
η η

= + − + − +

 − + + + −  
  

 

2

0

3 5
But, 1  and 1

2 3
A Aη ε  = + = + 

 
 

 

( )

( )

( )

3

2

5

2

3 7

3

7

3

3 5
1

2 3
cos cos cos3

32(1 )

9 5
1

4 3
23cos 24 cos3 cos5

1024(1 )

27 5 547
1 cos

8 3 32768 1

297 1
cos3 3cos5 cos 7

8 165
2048 1

3

A c

z c
A

A c

A

c
A

A

c

A

τ τ τ

τ τ τ

τ

τ τ τ

 + 
 = + −

+

 + 
 + − +

+

 + +  
  +


− + + −   +  

  

 

 

 

 

 

  

 

 

 



 

 

R. R. Thapa and M.R. Hassan / BIBECHANA 10 (2014) 92-99 : BMHSS, p.99 (Online Publication: Dec., 2013) 

 

( ) ( )

( ) ( )

5
3

7

3 5

3 9
cos cos cos 3 23cos 24 cos 3 cos 5

64 4096

27 547 297 3 1
cos cos 3 cos 5 cos 7

8 32768 16384 2048 32768

1 3
cos cos 3 23cos 24 cos 3 cos 5

32 1024

27 547 297
cos

4 32768 1638

c
z c t t t c t t t

c
t t t t

A t t c t t t c

t

η η η η η η

η η η η

η η η η η

η

= + − + − +

 + − + −  

+ − + − +

+ − 73 1
cos 3 cos 5 cos 7 ...

4 2048 32768
t t t cη η η

 + − +    

 

4. Conclusion 

The solution Z of Sitnikov's circular restricted three body problem depends on constant 'C', 

independent variable τ  & oblateness parameter A. The third body moves around the equilibrium 

point. 

Acknowledgement 

We are thankful to Prof. Dr. Jaishanker Jha, Professor and Former Head, University Department of 
Mathematics, T.M.B. University Bhagalpur, India, for his kind co-operation and help during the 

research. 

Reference 

1.  W.D. Macmillan, Astronomical Journal, 27(1913) 11. 

2.  K.A. Stinikov,  Dokl.Akad.Nauk, USSR,  133(1960) 303-306. 
3.  J. A. Hagel, Celestial mechanics and Dynamical Astronomy,  56 (1993) 81. 

4.  S.B. Faraque,  Celes. Mech & Dyn. Astron., 87(2003) 353-369. 

5.  S.W. Mccuskey,  Introduction to Cel. Mechanics, Addison-Wesley publishing company, inc., 1963 
6. M.S. Suraj, and M.R. Hassan, Solution of Sitnikov restricted four body problem when all the 

primaries are oblate bodies: Circular case, proceeding of PAS, 50 (1)(2013) 61-79.  

7.  R.R. Thapa and M.R. Hassan, Solutions of Sitnikov's restricted three body problem when the 

primaries are sources of radiation. Modern trends in science and Technology Devendra Adhikari, 

Shiva Kumar Rai and Kul Prasad Limbu (editors). Nepal Physical Society (Eastern Chapter) and 

Research Council of Science and Technology (Publishers), (2013)97-111.