https://www.irjet.net/archives/V6/i6/IRJET-V6I6505.pdf

1. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2386
Characteristic Exponent of Collinear Libration Point in L2 in Photo –
gravitational Restricted Problem of 2+2 Bodies When Bigger Primary
is a Triaxial Rigid Body Perturbed by Coriolis and Centrifugal Forces
Ziaul Hoque1 and Deb Narayan Garain2
1Department of Mathematics, S. P. Mahila College, Dumka, Jharkhand, India, 814101
2P.G. Department of Mathematics, S. K. M. University, Dumka, Jharkhand, India, 814110
--------------------------------------------------------------------***-----------------------------------------------------------------------
Abstract - Stability of collinear libration point 2L in photo – gravitational restricted problem of 2+2 bodies when bigger
primary is a triaxial rigid body perturbed by coriolis and centrifugal forces has been studied in which it is found that 2L is
unstable.
Key Words: Stability, Collinear libration point, Photo-gravitation, Triaxial rigid body, Coriolis force, Centrifugal
force.
1. INTRODUCTION
Equilibrium solutions of restricted problem of 2+2 bodies are derived by Whipple (1984) in which 1M and 2M
are two point masses moving in the circular Keplerian orbit about their centre of mass. He assumed that 21 MM  . Two
minor bodies 1m and 2m  221, Mmm  move in the gravitational fields of primaries  21 andMM . They attract
each other but do not perturb the primaries. He showed the existence of fourteen equilibrium solutions. Six of these
solutions are located about the collinear Lagrangian points of classical restricted problem of three bodies, eight solutions
are found in the neighborhood of triangular Lagrangian points.
Sharma, Taqvi and Bhatnagar (2001) studied the existence and stability of libration points in the restricted three
body problem when primaries are triaxial rigid body and source of radiation. They found five libration points, two
triangular and three collinear they also observed that collinear points are unstable while triangular libration points are
stable for the mass parameter crit0   .
Garain and Chakraborty (2007) derived libration points and examined stability in Robe’s three body restricted
problem when second primary is a triaxial rigid body perturbed by coriolis and centrifugal forces. They found that the
collinear libration points are deviated due to perturbation of centrifugal forces and triaxility of the second primary.
Perturbation of coriolis force and triaxility character of the body play important role for finding the region of stability.
Hoque and Garain (2014) computed collinear libration point 2L . In the case of 2+2 body problem when
perturbation effects act in coriolis and centrifugal forces, small primary is a radiating body and bigger primary as a triaxial
rigid body.
2. EQUATIONS OF MOTION
Whipple’s (1984) equation of motion of restricted problem of 2 + 2 bodies in synodic system be
i
ii
x
T
y2x


  (1)
i
ii
y
T
x2y


  (2)
i
i
z
T
z


 , (i =1, 2) (3)

2. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2387













2
1
3
21
22
2
)1(
2
)(
i
i
ii
ii
i
rrr
yx
T


 ,21,i,,
21
i
21
2





MM
m
MM
M i
 2222
1 )( iiii zyxr   ,
2222
2 )1( iiii zyxr  
2
21
2
)21
2
21
2
)()()( zzyyxxr 
Here we consider 2M , a radiating body and 1M , a triaxial rigid body, we have also considered effect of perturbation in
coriolis and centrifugal forces in this configuration.
Hence force function T reduces to U as follows:





 






 

 5
1
2
21
3
1
213
21
222
1 2
))(1(3
2
)2)(1(
2
)1(
2
)(
i
i
i
i
ii
ii
i
i
r
y
rrr
q
r
yx
U

 (4)
Where,  1,1 q and 2,1i .
Equilibrium points of the system are those points where 2)1,(i,0 









iii
iii
z
U
y
U
x
U
zyx  .
Thus we have,
5
11
121
3
212
3
21
1
3
11
1
1
2
))(2)(1(3)()1())(1(
r
x
r
xx
r
xq
r
x
x










0
2
))()(1(15
7
11
2
1121



r
yx 
(5)
5
11
121
3
212
3
21
1
3
11
1
1
2
)2)(1(3)()1(
r
y
r
yy
r
yq
r
y
y








0
52
2
13
7
11
3
1
5
11
121










r
y
r
y)σμ)(σ(
(6)
0
2
))(1(15
2
)2)(1(3)()1(
7
11
1
2
121
5
11
121
3
212
3
21
1
3
11
1









r
zy
r
z
r
zz
r
zq
r
z  (7)
5
12
221
3
121
3
22
2
3
12
2
2
2
))(2)(1(3)()1())(1(
r
x
r
xx
r
xq
r
x
x










0
2
))()(1(15
7
12
2
2221



r
yx 
(8)
5
12
221
3
121
3
22
2
3
12
2
2
2
)2)(1(3)()1(
r
y
r
yy
r
yq
r
y
y








0
52
2
))(1(3
7
12
3
2
5
12
221










r
y
r
y
(9)
0
2
))(1(15
2
)2)(1(3)()1(
7
12
2
2
221
5
12
221
3
121
3
22
2
3
12
2









r
zy
r
z
r
zz
r
zq
r
z 
(10)
From equations (7) and (10), we have 021  zz .
By inspection, it can be seen that equations (6) and (9) are satisfied when 021  yy .
Now we have to determine 1x and 2x such that the following simplified forms of equations (5) and (8) are satisfied.

3. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2388
3
21
212
3
1
1
3
1
1
1
)(
1
)1())(1(
xx
xx
x
xq
x
x
x















0
2
))()(1(15
2
))(2)(1(3
7
1
2
1121
5
1
121











x
yx
x
x
(11)
And 3
12
122
3
2
2
3
2
2
2
)(
1
)1())(1(
xx
xx
x
xq
x
x
x















0
2
))()(1(15
2
))(2)(1(3
7
2
2
2221
5
2
221











x
yx
x
x
(12)
The solution of equations (11) and (12) can be obtained with the help of power series.
Let )2,1(,
)( 3
2
21


 ii
i


(13)
∴
3
2
21
2
2
3
2
21
1
1
)()(
and







 (14)
∴ )(21
3
2
21
21
12
)(
sayk

 


 (15)
Let 3,2,1,
1
211 for  
iaLx
n
j
j
ji (16)
Where 321 ,, LLL  equilibrium points be in photo-gravitational restricted problem of three bodies when
bigger primary is a triaxial rigid body perturbed by coriolis and centrifugal forces and 1x be the coordinate
of first small body.
3,2,1
1
122 for  
iaLx
n
j
j
ji (17)
Similar to Whipple,
32,1,where
)()(
)1(
3
2
21
2
3
11 


 iLx
xx
i



(18)
And
32,1,where
)()(
)1(
3
2
21
1
3
12 


 iLx
xx
i



(19)
Hoque and Garain (2014) obtained two values  0,0,1x and  0,0,2x of 2L in which
2211212121211 2  ddcbax 
and
2221222222222 2  ddcbax 

4. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2389
Where
3
2
21
3
4
2
22
221
3
2
21
3
1
2
2
221
)(3
)1(
b,
)(
)1(
1











A
B
b
A
aa
3
2
21
3
4
2
21
222
3
2
21
3
1
2
1
222
)(3
)1(
b,
)(
)1(
1











A
B
b
A
aa
3
2
21
3
4
2
21
222
3
2
21
3
4
2
21
222
)(3
)1(
and
)(3
)1(










A
D
dd
A
C
cc
3
2
21
3
4
2
22
221
3
2
21
3
4
2
22
221
)(3
)1(
,
)(3
)1(










A
D
dd
A
C
cc
They obtained the particular values of 1x and 2x for different values of  , 1 and 2 .
Our characteristic equation corresponding to the point  0,0,1x is
  0
1
4)( 2
2
11
24
11111111
11








 yxyyxxyy
xx
UUUU
U
f

 (20)
       










 7
11
2
121
5
11
21
3
2
3
21
3
111
145216221211
r
y
rrr
q
r
U xx 


       313
2
34131
5
21
2
1
5
11
2
1
5
11
21
3
2
3
21
3
111
11










r
yq
r
y
rrr
q
r
U yy 


       







 9
11
4
121
7
11
2
121
5
2
212
2
1105
2
671153
r
y
r
y
r
yy 
       


 




 5
21212
5
21
11
5
11
11
1
31313
11 r
yyxx
r
yxq
r
yx
yx
U


       
7
11
1121
7
11
1121 115
2
2115
r
yx
r
yx  



In this case, 01 y and 02 y .
Therefore the above equations reduce to
 
     
  
  
















 5
1
21
3
21
2
3
1
3
11
2162
1
21211







 xxxx
q
x
U xx
0
1
11


yxU
And
 
     
  
  
















 5
1
21
3
21
2
3
1
3
11 2
3413
1
111







 xxxx
q
x
U yy
For the collinear equilibrium solutions the partial derivatives contained in equation (20) reduce to
    

















 5
1
21
3
21
2
3
1
3
11
2162
1
21211







 xxxx
q
x
U xx

5. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2390
0
1
11


yxU
        



















52
3413
1
11
1
1
21
3
21
2
3
1
3
11
11






 xxxxx
U yy
Let 221121212121
1
11


EDCBA
U yy

2211212121211 2  ddcbax 
and
2221222222222 2  ddcbax 
221121212121
1
11


EDCBA
U yy

Where,
  ,
1
1
1 3
2221
2
3
21
3
21
21
aaaa
A












   ,
3
1
3
1
13
4
2221
22212
4
21
21
3
21
4
21
21
21
aa
bb
a
b
aa
b
B

















   ,
3
1
313
1 4
2221
22212
4
21
21
4
21
21
21
aa
cc
a
c
a
c
C













     
5
21
4
2221
22212
4
21
21
4
21
21
21
166
1
616

















aaa
dd
a
d
a
d
D
and
     
5
21
4
2221
22212
4
21
21
4
21
21
21
2
193
1
313

















aaa
dd
a
d
a
d
E
Here we see that 0and0 1111
 yxxx UU . Therefore equation (20) reduces to
0
11
4)( 11111111 2
11
24






 yyxxyyxx UUUUf

 (20a)
Case I: Let
2
1 and
2
2 be the roots of equation (20a).
Sub case (i): If
2
1 and
2
2 both are real and one of them is positive, let
2
1 is a positive quantity, then square root
of
2
1 must be real and of opposite sign. In this case characteristic roots will be a real number. So in this case
2L is unstable. We may get similar result in the case of
2
2 .
Sub case (ii) If
2
1 and
2
2 both are real and negative, let
2
1 is a negative quantity, then two roots of
2
1 be purely
imaginary. Similarly two roots of
2
2 be purely imaginary. In this case 2L should be stable. Again when
2
1 and
2
2
both are negative quantity then
2
2
2
1  = a positive quantity (20b)
From equation (20a), we have obtained that

6. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2391
11112
1
2
2
2
1
1
yyxx UU

  = a negative quantity  0,0 1111
 yyxx UU (20c)
Since (20b) and (20c) contradict each other. So 2L is unstable.
Now let 23
1
11
11
4 AU
U
yy
xx


Case II: If 023 A i.e., 1111
1
1
4 yyxx UU 

then 0
1
)( 11112
1
2
23
4
 yyxx UUAf

 .
Sub case (i): It is clear that 011
xxU and let 011
yyU then 0
1
11112
1
yyxx UU

so, 0)( f has no change of signs and
as such it has no positive real roots. 0
1
)( 11112
1
2
23
4
 yyxx UUAf

 also has no change of signs and as such it
has no positive real roots i.e.,  f has no negative real roots. So in this case we can say that all roots of   0f are
imaginary.
Sub case (ii): If 011
yyU then similar to Yadav’s case 0
1
11112
1
yyxx UU

.
Let 0,
1
23232
1
1111
 BBUU yyxx

, so, 0)( 23
2
23
4
 BAf  . Here 0)( f has only one change of sign and
as such it has one positive real root. 0)( 23
2
23
4
 BAf  has only one change of sign and as such it has one
positive real root i.e., )(f has one negative real root.
Case III: If 023 A i.e., 1111
1
1
4 yyxx UU 

then let 0, 232323  CCA then
0
1
)( 11112
1
2
23
4
 yyxx UUCf


Sub case (i): Let 011
yyU , then .0
1
11112
1
yyxx UU

Then, 0)( f has two changes of signs and as such it has at most two positive real roots.
0
1
)( 11112
1
2
23
4
 yyxx UUCf

 also has two change of signs and as such it has at most two positive real
roots i.e., )(f has at most two negative real roots.
Sub case (ii): If 011
yyU , then 0
1
11112
1
yyxx UU

.
Let 0D,
1
23232
1
1111
 DUU yyxx

. Here 0)( 23
2
23
4
 DCf  has only one change of sign and as such it has
one positive real root. 0)( 23
2
23
4
 DCf  has only one change of sign and as such it has one positive real
root i.e.,  f has one negative real root.
In both the cases we see that 0)( f has one positive and one negative real root. So the libration point 2L is
unstable.
The characteristic equation corresponding to the point  0,0,2x be

7. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2392
  0
11
4)( 2
2
22
24
2222222222






 yxyyxxyyxx UUUUUf

 (21)
 
     
  
 















 5
2
21
3
21
1
3
2
3
22
22 2162
1
212







 xxxx
q
x
U xx
 
     
  
 















 5
2
21
3
21
1
3
2
3
22
22
2
3413
1
1







 xxxx
q
x
U yy
0
2
22


yxU
    

















 5
2
21
3
21
1
3
2
3
22
2162
1
21222







 xxxx
q
x
U xx
0
2
22


yxU
        


















 5
2
21
3
21
1
3
2
3
22 2
3413
1
11
122






 xxxxx
yx
U
Let 222122222222
2
22


EDCBA
U yy
 2211212121211 2  ddcbax 
and
2221222222222 2  ddcbax 
Where,
  ,
1
1
1 3
2122
1
3
22
3
22
22
aaaa
A












   ,
3
1
3
1
13
4
2122
21221
4
22
22
3
22
4
22
22
22
aa
bb
a
b
aa
b
B

















   ,
3
1
313
1 4
2122
21221
4
22
22
4
22
22
22
aa
cc
a
c
a
c
C













     
5
22
4
2122
21221
4
22
22
4
22
22
22
166
1
616

















aaa
dd
a
d
a
d
D
and
     
5
22
4
2122
21221
4
22
22
4
22
22
22
2
193
1
313

















aaa
dd
a
d
a
d
E
Here we see that 022
xxU and 022
yxU
Therefore, Equation (21) reduces to
0
11
4)( 22222222 2
22
24






 yyxxyyxx UUUUf

 (21a)
Case I: Let
2
1 and
2
2 be the roots of equation (21a).

8. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2393
Sub case (i): If
2
1 and
2
2 both are real and one of them is positive, let
2
1 is a positive quantity, then square root of
2
1
must be real and of opposite sign. In this case characteristic roots will be a real number. So in this case 2L is unstable. We
may get similar result in the case of
2
2 .
Sub case (ii): If
2
1 and
2
2 both are real and negative, let
2
1 is a negative quantity, then two roots of
2
1 be purely
imaginary. Similarly two roots of
2
2 be purely imaginary. In this case 2L should be stable. Again when
2
1 and
2
2 both
are negative quantity then
2
2
2
1  = a positive quantity (21b)
From equation (21a), we have obtained that
22222
2
2
2
2
1
1
yyxx UU

  = a negative quantity  0,0 2222
 yyxx UU (21c)
Since (21b) and (21c) contradict each other. So 2L is unstable.
Now let 24
2
2222
1
4 AUU yyxx 

Case II: If 024 A i.e., 2222
2
1
4 yyxx UU 

.
Sub case (i): Let 022
yyU then 0
1
22222
2
yyxx UU

so, 0
1
)( 22222
2
2
24
4
 yyxx UUAf

 has no change of
signs and as such it has no positive real roots. 0
1
)( 22222
2
2
24
4
 yyxx UUAf

 also has no change of signs
and as such it has no positive real roots i.e., )(f has no negative real roots. So in this case we can say that all roots of
0)( f are imaginary.
Sub case (ii): If 022
yyU then similar to Yadav’s case 0
1
22222
2
yyxx UU

.
Let ,0,
1
24242
2
2222
 BBUU yyxx

so 0)( 24
2
24
4
 BAf  . Here 0)( f has only one change of sign and
as such it has one positive real root. 0)( 24
2
24
4
 BAf  has only one change of sign and as such it has one
positive real root i.e., )(f has one negative real root.
Case III: If 024 A i.e., 2222
2
1
4 yyxx UU 

then let 0, 242424  CCA then
0
1
)( 22222
2
2
24
4
 yyxx UUCf


Sub case (i): Let 022
yyU , then 0
1
22222
2
yyxx UU

. Then, 0)( f has two changes of signs and as such it has at
most two positive real roots. 0
1
)( 22222
2
2
24
4
 yyxx UUCf

 also has two change of signs and as such it has at
most two positive real roots i.e., )(f has at most two negative real roots.
Sub case (ii): If 022
yyU , then 0
1
22222
2
yyxx UU

.

9. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2394
Let .0,
1
24242
2
2222
 DDUU yyxx

Here 0)( 24
2
24
4
 DCf  has only one change of sign and as such it
has one positive real root. 0)( 24
2
24
4
 DCf  has only one change of sign and as such it has one positive
real root i.e., )(f has one negative real root. In both cases we see that 0)( f has one positive and one negative real
root.
3. CONCLUSION
Hence, we get that the libration point 2L is unstable.

10. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2395
Figure 1 Figure 2
Figure 3 Figure 4
Figure 5 Figure 6

11. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2396
Figure 7 Figure 8
Figure 9 Figure 10
Figure 11 Figure 12

12. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 06 | June 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 2397
REFERENCES
[1] Z. Hoque and D. N. Garain: “Computation of 2L in 2+2 body problem when perturbation effects act in coriolis and
centrifugal forces, small primary is a radiating body and bigger primary is a triaxial rigid body” OIIRJ 2014; Vol. - IV,
Issue - I, pp. 92-100.
[2] Sharma, R. K., Taqvi, Z. A. and Bhatnagar, K. B: “Existence and stability of libration points in the restricted three-body
problem when the primaries are triaxial rigid bodies”, Celestial Mechanics and Dynamical Astronomy, 2001; 79, 119-133.
[3] Garain, D. N. and Chakraborty, R: “Libration point and stability in Robe’s three body restricted problem when second
primary is a triaxial rigid body perturbed by coriolis and centrifugal forces”, Proc. Nat. Acad. Sci., India, 2007; 77(A), IV,
pp. 329-332.
[4] Whipple, A. L: “Equilibrium solutions of the restricted problem of 2 + 2 bodies”, Celest. Mech. 1984; 33, 271-294.
BIOGRAPHY
Ziaul Hoque
M.Sc, M.Phil, PhD
Assistant Professor of Mathematics,
S. P. Mahila College, Dumka,
Jharkhand, India, PIN-814101