Available online throug
ISSN 2229 – 5046
International Journal of Mathematical Archive- 9(3), March-2018 157
STABILITY OF THE EQUILIBRIUM POSITION
OF THE CENTRE OF MASS OF AN EXTENSIBLE CABLE CONNECTED SATELLITES SYSTEM
IN THE ELLIPTIC ORBIT UNDER THE INFLUENCE OF PERTURBATIVE FORCES
VIJAY KUMAR
Assistant Professor, Dept. of Mathematics, Mangalayatan University, Aligarh, India.
(Received On: 09-01-18; Revised & Accepted On: 17-02-18)
ABSTRACT:
D
uring the motion of a cable connected satellites system, at least one equilibrium point exists when perturbative forces like Air-resistance, shadow of the earth due to solar radiation radiation pressure, magnetic force and oblateness of the earth act simultaneously. We have obtained two equilibrium points out of which only one is found to be stable in case of perturbative forces like shadow of the earth due to solar radiation pressure, magnetic force and oblateness of the earth act simultaneously on the motion of two extensible cable-connected satellites. Lyapunov’s theorem has been used to examine the stability of the equilibrium points.Keywords: Stability, perturbative forces, equilibirium Point, satellites, Elliptic orbit.
1. INTRODUCTION
The present paper is devoted to examine the stability of the equilibrium points of the centre of mass of a system of two satellites connected by a light, flexible and extensible string under the influence of shadow of the earth due to solar radiation pressure, magnetic force and oblateness of the earth acting simultaneously in elliptic orbit. Beletsky; V.V[1] is the pioneer worker in this field. It is the generalisation of the works done by Beletsky[1], Singh[2], Sinha[3], Singh[4] and Das et. al [7].
2. EQUATION OF MOTION
The equation of motion of one of the two satellites moving along a keplerian elliptic orbit in Nechvill’s coordinates can be obtained by exploiting Lagrange’s equation of motion of first kind in the form:
(
)
3 4 0
1
4
cos
" 2 ' 3
Bx
cos
cos
1
l
D
i
x
y
x
A
x
r
α
ρ
ρ
ν α
λ ρ
ρ
ρ
ρ
−
−
−
+ Ψ
∈
−
= −
−
−
(
)
y
r
l
A
By
x
y
−
−
=
−
∈
Ψ
−
+
+
ρ
ρ
λ
α
ν
ρ
ρ
1 3cos
sin
α 41
0'
2
"
(1)
where,
ρ
=;
cos
1
1
ν
e
+
2 2
y
x
r
=
+
v
=
True anomaly of the centre of mass of the system
3
1 2
1 2
B
B
P
A
m
m
µ
=
−
= Solar pressure parameter2 2
3
P
k
B
=
= Oblateness force parameter.1 2
1 2
E
Q
Q
C
m
m
p
µ
µ
=
−
= Magnetic force parameter.1
Ψ
= Shadow function parameter.Corresponding Author: Vijay Kumar
the Elliptic Orbit Under the Influence of Perturbative Forces / IJMA- 9(3), March-2018.
(
)
3
1 2
0
0 1 2
;
mod
m
m
p
Natural length of the string
m m
ulus of elasticity
α
λ
λ
µ
λ
+
=
=
=
Where
µ
denotes the product of gravitational constant and mass of the earth. Here B1 and B2 are the absolute values of forces due to the direct solar pressure exerted on the masses m1 and m2 andα
be the angular separation of solar position vector projected on the orbital plane from the orbit perigee. Here∈
is the inclination of the osculating plane of the orbit of the centre of mass of the system with the plane of ecliptic andp
and e are focal parameter and eccentricity of the earth.Here, dashes denote the differentiation with respect to the true anomaly v of the elliptic orbit of the centre of mass .
The condition of constraint is given by 2
2 2 0
2
x
y
ρ
+
≤
(2)
To find the Jacobian integral of the problem, the averaged values of the secular terms due to periodic terms presents in
the equations of motion (1) can be deduced as given follow :
2 0
1
1
1 ;
2
dv
π
π
∫
ρ
=
2 2
0 2
1
1
1
2
2
e
dv
π
π
∫
ρ
=
+
(
)
2
0 1
2 2
1
1
;
2
1
dv
e
π
ρ
π
∫
=
−
(
)
2
2 3
0 5
2 2
1
2
2
2 1
e
dv
e
π
ρ
π
+
∫
=
−
;(
)
2
2 4
0 7
2 2
1
2
3
2
2 1
e
dv
e
π
ρ
π
+
∫
=
−
;(
)
(
)
0 1
1 1
2
2 3 3 3
0 1 1 1
2 5 2 2
1
1
cos
cos(
)
cos
cos(
)
cos
cos(
)
2
2
2
cos
cos
sin
2 1
A
v
dv
A
v
dv
A
v
dv
e
A
e
θ π
π
θ θ
ρ ψ
α
ρ ψ
α
ρ ψ
α
π
π
α
θ
π
Ψ =− Ψ =
∫
∈
−
=
∫
∈
−
+ ∫
∈
−
−
+
∈
=
−
and
(
)
(
)
0 1
1 1
2
2 3 3 3
0 1 1 1
2 5 2 2
1
1
cos
sin(
)
cos
sin(
)
cos
sin(
)
2
2
2
cos
sin
sin
2 1
A
v
dv
A
v
dv
A
v
dv
e
A
e
θ π
π
θ θ
ρ ψ
α
ρ ψ
α
ρ ψ
α
π
π
α
θ
π
Ψ =− Ψ =
∫
∈
−
=
∫
∈
−
+ ∫
∈
−
+
∈
=
−
(3)
where
θ
is taken to be constant.Using (3) in (1), we get
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 2 2
1 5 7 5
2 2 2 2 2 2 2 2
2
2 3
2
3
cos
cos
sin
" 2 '
4
cos
1
2 1
2 1
2
1
o
e
e
e
x
A
x
y
Bx
x
D
i
e
e
e
r
e
α
α
θ
λ
π
+
∈
+
+
−
−
−
−
= −
−
−
−
−
−
−
(
)
(
)
(
)
(
)
(
)
(
)
2 2 2
5 7 5
2 2 2 2 2 2
2
cos
sin
sin
2 3
2
" 2 '
2 1
2 1
2
1
o
e
A
e
e
and
y
x
By
y
e
e
r
e
α
α
θ
λ
π
+
∈
+
+
+
+
−
= −
−
−
−
−
(4)
© 2018, IJMA. All Rights Reserved 159 The condition of constraint given by (2) takes the form
2 0 2 2
2
2
1
+
≤
+
y
e
x
(5)
From equations of motion (4) it follows that the true anomaly v does not appear explicitly in the equations of motion, so there must exist Jacobian integral for the problem.
Multiplying first and second equations of (4) by 2x’ and 2y’ respectively and adding and integrating, we get the Jacobian integral in the form.
(
)
(
)
(
)
(
)
(
)
(
)(
)
(
)
(
)(
)
(
)
2 2
2
2 2 2 2
1 5 5
2 2 2 2 2 2
1
2 2 2 2 2 2 2
7 5
2 2 2 2
2
cos
cos
sin
2
cos
sin
sin
3
4
1
1
1
2
3
2
cos
2 1
1
o
e
Ax
e
Ay
x
x
y
Bx
By
e
e
e
e
x
y
e
x
y
Dx
i
h
e
e
α α
α
θ
α
θ
π
π
λ
λ
+
∈
+
∈
′
+
′′
−
−
+
−
−
−
−
−
+
+
+
+
+
−
+
=
−
−
(6)
where, h is the constant of integration.
The curve of zero velocity is obtained by putting
x
′
2+
y
′
2=
o
in (6) as(
)
(
)
(
)
(
)
(
)
(
)(
)
(
)
(
)(
)
(
)
2 2
2
2 2
1 5 5
2 2 2 2 2 2
1
2 2 2 2 2 2 2
7 5
2 2 2 2
2
cos
cos
sin
2
cos
sin
sin
3
4
1
1
1
2
3
2
cos
2 1
1
o
e
Ax
e
Ay
x
Bx
By
e
e
e
e
x
y
e
x
y
Dx
i
h
o
e
e
α α
α
θ
α
θ
π
π
λ
λ
+
∈
+
∈
+
−
+
+
−
−
−
+
+
+
+
−
+
−
+ =
−
−
(7)
Hence we conclude that the satellite by mass m1 will move inside the boundaries of different curves of zero velocity represented by (7) of (6) for different values of Jacobian constant h.
3. EQUILIBRIUM POSITION OF THE SYSTEM
We have obtained the system of equations (4) of the particle of mass m1 of the system in rotating frame of reference. It has been assumed that the system is moving with effective constants and hence the string connecting the two satellites of masses m1 and m2 will remain always tight
The equilibrium positions of the system are given by the constant values of the coordinates in the rotating frame of reference Now, let x = x0 and y = y0 give the equilibrium position where x0 and y0 are constants.
y
y
and
x
x
′
=
=
′′
′
=
=
′′
∴
0
0
Thus, equations given by (4) take the following form:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 2 2
0
0 0
1 5 7 5
2 2 2 2 2 2 2 2
2
2 3
2
3
cos
cos
sin
4
cos
1
2 1
2 1
2
1
o
o
e
e
e
x
A
Bx
D
i
x
e
e
e
r
e
α
α
θ
λ
π
+
+
+
−
∈
−
−
+
= −
−
−
−
−
−
(
)
(
)
(
)
(
)
(
)
(
)
2 2 2
5 7 5
2 2 2 2 2 2
2
cos
sin
sin
2
3
2
2 1
2 1
2
1
o
o o
o
e
A
e
e
By
y
e
e
r
e
α
α
θ
λ
π
+
∈
+
+
−
= −
−
−
−
−
(8)
where
r
0=
x
02+
y
02the Elliptic Orbit Under the Influence of Perturbative Forces / IJMA- 9(3), March-2018.
Therefore, for our further investigation, it has been assumed that α=0
This means that the sun rays is in the line of the perigee of the elliptical orbit of the centre of mass of the system.
Putting α =0, the equations of motion given by (8) take the form:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 2 2
1 0
1 5 7 5
2 2 2 2 2 2 2 2
2
sin
2 3
2
3
4
1
2 1
2 1
2
1
o o o
e
A
e
e
B x
x
e
e
e
r
e
α
θ
λ
+
+
+
−
+
−
= −
−
−
−
−
−
(
)
(
)
(
)
(
)
2 2
7 5
2 2 2 2
2 3
2
2 1
2
1
o
o o
o
e
e
and By
y
e
r
e
α
λ
+
+
= −
−
−
−
(9)
From (9), we get the equilibrium points as
[
]
(
)
(
)
(
)
(
) (
)
(
)
5
2 2 2
1
1 1
3
2 2 2 2
2
sin
2 1
cos
, 0
, 0
2
3
2 1
3 4
1
o
e
A
e
D
i
a
e
e
B
e
α
α
λ
θ
λ
+
+
−
−
=
+
−
−
+
−
(10)
(
)
(
)
(
)
(
)
5
2 2 2
1
1 1 1
2
2 2 2
2 1 cos 2 sin
[ , ] ,
2 1 3 5 1
e D i e A
b c
e B e
θ
− − +
=
− + −
(
)(
)
(
)
(
)
(
)
(
)
(
)
2 2
2 2 2
1
7 1
2 2 2 2 2 2
2 1 cos 2 sin
2 3 2 1 2 1 3 5 1
o e e D i e A
e B e e B e
α
α
λ θ
λ
+ − − +
−
+ + − − + −
(11)
Now, Now, it can be easily seen that the equilibrium point given by (10) only gives a meaningful value of the Hook’s
modulus of elasticity.
4. STABILITY OF THE SYSTEM
We shall study the stability of the equilibrium position given by (10) of the system in the sense of Liapunov. For this,
let us assume that there are small variation in the coordinate at the given equilibrium point [a1, 0]. let
η
1andη
2besmall variation in x and y-coordinates respectively for the given position of equilibrium.
1 1
+
η
=
∴
x
a
andy
=
η
21
'
=
η′
∴
x
andy
'
=
η′
22
"
=
η
′′
x
andy
"
=
η
2′′
(12)Putting the values of x, y, x’, y’, x” and y” from (12) in (4), we get on putting and, as
1
cos
A
A ∈=
π
α
=
0
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 2 2
1 '' '
1 2 2 1 1 5 7 5 1
2 2 2 2 2 2
2
sin
2
3
2
3
2
4
cos
(
)
1
2 1
2 1
2
1
o
e
A
e
e
B
a
D
i
a
e
e
αe
r
e
θ
η
−
η
−
+
+
η
−
+
+
= −
λ
+
−
+
+
η
−
−
−
−
(
)
(
)
(
)
(
)
2 2
" '
2 1 2 7 5 2
2 2 2 2
2 3
2
2
2 1
2
1
o
e
e
and
B
e
r
e
α
η
−
η
+
η
= −
λ
+
−
+
η
−
−
(13)
© 2018, IJMA. All Rights Reserved 161 To obtain Jacobian integral of the equations of motion (13), we multiply first equation of (13) by 2(a1+n1)’ and second
equation of (13) by
η
2' and add them together, we get after integrating(
)
(
)
(
)
(
)
(
)
(
)
{
(
)
}
(
)
(
)
{
(
)
}
2
2 1 1
2 2 2
1 2 2 1 1 1 1 2 5
2 2
2 2 1
2 2 0 2 2 2
1 1 2 1 1 2 1
7 5
2 2 2 2
2
sin
3
4
2 (
) cosi
1
2 1
2 3
2
2 1
1
e
a
A
B
a
D a
B
e
e
e
e
l
a
a
h
e
e
α α
η
θ
η
η
η
η
η
λ
λ
η
η
η
η
+
+
′
+
′
−
+
+
+
+
+
−
−
−
+
+
+
+
+
−
+
+
=
−
−
(14)
where h1 is the constant of integration.
To test the stability in the sense of Liapunov, we take Jacobian integral (14) as Liapunov’s function
V
(
η
1′
,
η
2′
,
η
1,
η
2)
and is obtained by expanding the terms of (14) as:(
η
1′
,
η
2′
,
η
1,
η
2)
V
=(
)
(
)
2
2 2 2
1 2 1 7 2
2 2
2
3
3
4
1
2 1
e
B
e
e
α
λ
η
′
+
η
′
+
η
+
−
−
−
−
(
)
(
)
(
)
(
)
7 5
2 2
2 2
0 2
2
2 2
1
2 3
2
2 1
2
1
e
e
B
e
a
e
α α
λ
λ
η
+
+
+
+
−
−
−
2 2
2
1 1
1 2 1 5 1 7 5
2 2 2 2 2 2
(2 3 )
(2
)
6
(2
)
8
sin
2
cos
1
(1
)
(1
)
(1
)
o
e a
e
a
e
Ba
A
D
i
e
e
e
e
α α
λ
λ
η
−
+
θ
+
+
+
−
−
+
+
−
−
−
−
−
2 2 2
2 2
2 1 1 1 1 1
1 2 5 7 5 1
2 2 2 2 2 2
(2
)
(2 3 )
3
(2
)
sin
4
2
cos
1
(1
)
2(1
)
(1
)
o
e
a
e a
a
a
e A
Ba
Da
i
e
e
e
e
α α
λ
+
λ
+
+
θ
+ −
−
−
+
−
+
−
−
−
−
+
0
(
3
)
=
h
1
(15) where 0 (3) stand for the third and higher order terms in the small quantities
η
1and
η
2.Now, by Lyapunov theorem on stability it follows that the only criterion for the given equilibrium position (a1, 0) to be stable is that v defined by (15) must be positive definite and for this the following conditions must be satisfied:
(i)
2 2
2
1 1
1 1
1 5 7 5
2 2 2 2 2 2 2 2
(2 3 )
(2
)
6
(2
)
8
sin
2
cos
0
(1
)
(1
)
(1
)
(1
)
o
e a
e
a
e
Ba
A
D
i
e
e
e
e
α α
λ
λ
θ
+
+
−
−
−
+
+
−
+
=
−
−
−
−
(ii)
0
)
1
(
3
4
)
1
(
2
)
3
2
(
2 1 2 2
7 2
2
>
−
−
−
−
+
e
B
e
e
α
λ
(iii)
0
)
1
(
2
)
2
(
)
1
(
2
)
3
2
(
2 5 2 1
2 0
2 7 2
2
>
−
+
−
−
+
+
e
a
e
e
e
B
λ
αλ
α
(16)
Condition (i): First of all it is to be noted that
(
)
(
)
(
)
(
) (
)
(
)
5
2 2 2
1
1 1
3
2 2 2 2
2
sin
2 1
cos
2
3
2 1
3 4
1
o
e
A
e
D
i
a
e
e
B
e
α
α
λ
θ
λ
+
+
−
−
=
+
−
−
+
−
the Elliptic Orbit Under the Influence of Perturbative Forces / IJMA- 9(3), March-2018.
Putting the values a1 on the L.H.S of (i) weget
L.H.S of (i)=
(
) (
)
7(
) (
)
73 3
2 2 2 2 2 2 2 2
1 1
7 7
2 2 2 2
2
3
6 1
8 (1
)
2
3
6 1
8 (1
)
0
(1
)
(1
)
e
e
B
e
a
e
e
B
e
a
e
e
α α
λ
+
−
−
−
−
λ
+
−
−
−
−
−
=
−
−
Hence the first condition is identically satisfied.
Condition (ii):
7
2 2 2 2 3
2
7 1 7
2 2 2 2 2 2
(2
3 ) 8 (1
)
6(1
)
(2
3 )
3
4
2(1
)
(1
)
2(1
)
e
B
e
e
e
B
ve
e
e
e
α α
λ
λ
+
+
−
−
−
−
−
−
=
= +
−
−
−
Since the denomirator of a1 is positive.
Condition (iii):
2 2 2
2 2
1 0
0
7 5 7
2 2 2 2 2 2
1 1
2 4
1 0 1 0 0
1 0
7 2 2 1
(2
3 )
(2
)(1
)
(2
3 )
(2
)
2(1
)
2 (1
)
2 (1
)
2(
)
(3
)
0 if
2 (1
)
e a
e
e
e
e
B
B
e
a
e
a
e
a
e
a
e
B
a
a
e
α
α α
α
λ
λ
λ
λ
+
−
+
−
+
+
+
−
= +
−
−
−
−
+
+
+
= +
>
>
−
1 0
Thus the third condition is identically satisfied if
a
>
.
CONCLUSION
we conclude that the equilibrium position [a1, 0] of the system is stable in the sense of Liapunov if a1 > l0.
REFERENCES
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