In space science, the two-body, the three-body and the restricted three-body problems have been studied by different authors starting from Newton to date. MacMillon et al. [1] provided detailed proof of two theorems for the existence of a quadrilateral configuration in the field of four-body configurations. Brumberg [2] invented a permanent solution for a four-body configuration. For the first time, Albouy et al. [3] [4] examined the symmetric central configuration of equal masses. Cors et al. [5] and Corbera et al. [6] examined the cyclic configuration in the Newtonian four-body problem, in which they used the six mutual distances of the particles as their coordinates and demonstrated that the four-point masses constitute a kite lying on a two-dimensional plane.

They have also demonstrated that a line of symmetry must exist in any central configuration with two equal masses on opposite sides. By describing the masses of the central configuration in terms of angle coordinates, Balint et al. [7] extended the work of Cors et al. [5] in three cases (two concave cases and one convex case). They further asserted that the exact analytical solutions of the four-body configuration are represented by the obtained formulas. Deng et al. [8] demonstrated that the diagonals of a cyclic quadrilateral can’t be perpendicular unless the configuration is a kite by using mutual distances as the coordinates. Further, they verified the same theorem in the four-vortex central configuration.

Hassan [9] [10] classified the kite configuration into two categories and proved three theorems on the existence of cyclic kite configurations. In the first theorem, he uniquely expressed the masses of the four particles in terms of a mass parameter $\mu$and the total mass $M$ of the system as $m_1=M\left(1-\mu \right)/2$, $m_2=M\mu$, $m_3=M\left(1-3\mu \right)/2$, $m_4=M\mu$ and their coordinates as $\left(R,0\right)$, $\left(-R/2,\sqrt{3}R/2\right)$, $\left(-R,0\right)$, $\left(-R/2,-\sqrt{3}R/2\right)$ where $R$ is the radius of the common circular orbit of the kite. Hassan [9] [10] calculated the mean motion of the rotating frame lying on the kite’s plane as a function of $\mu$.

Khatun et al. [11] extended the work of Hassan [9] [10] by taking the first body of mass $m_1$ as an oblate spheroid and showed the effects of oblateness on the mean motion of the system. It is found that the mean motion increases with the increase of oblateness. Further, showed that the axial libration points move away from the origin due to an increase in the oblateness parameter $A$. As the analytical existence of non-axial libration points wasn’t possible so, we presently propose Python programming for finding the location of non-axial libration points with the help of points of intersection of contour plots of zero partial derivatives of the potential function of the satellite.

Let at any time $t$, $P\left(x,y\right)$ be the position of the satellite moving in the gravitational field of the four-point masses of the kite configuration, and $P_k\left(x_k,y_k\right),k=1,2,3,4$ be the positions of the four vertices of the kite configuration at which the four bodies of respective masses $\left(1-\mu \right)/2,\mu ,\left(1-3\mu \right)/2,\mu$ be located on the rotating frame $\left(O,XY\right)$ as shown in Figure 1 .

Let

$OP=\rho$, then $\rho =xi+yj$ $\Rightarrow$ ${\rho }^2=x^2+y^2$ and

$\begin{array}{l}{P}_{k}P=\left(x-x_k\right)\hat{i}+\left(y-y_k\right)\hat{j}={\rho }_k,\\ \Rightarrow {\rho }_k^2={\left(x-x_k\right)}^2+{\left(y-y_k\right)}^2.\end{array}$

As $P_k\equiv \left(x_k,y_k\right)$ then

$P_1\equiv \left(\frac{1}{2},0\right),P_2\equiv \left(-\frac{1}{4},\frac{\sqrt{3}}{4}\right),P_3\equiv \left(-\frac{1}{2},0\right),P_4\equiv \left(-\frac{1}{4},-\frac{\sqrt{3}}{4}\right),$

$\begin{array}{l}{\rho }_1^2={\left(x-\frac{1}{2}\right)}^2+y^2,{\rho }_2^2={\left(x+\frac{1}{4}\right)}^2+{\left(y-\frac{\sqrt{3}}{4}\right)}^2,\\ {\rho }_3^2={\left(x+\frac{1}{2}\right)}^2+y^2,{\rho }_4^2={\left(x+\frac{1}{4}\right)}^2+{\left(y+\frac{\sqrt{3}}{4}\right)}^2.\end{array}\}$(1)

In a rotating frame, the equations of motion of the satellite at $P\left(x,y\right)$ can be written as

$\begin{array}{l}\ddot{x}-2n\stackrel{\dot{}}{y}=\frac{\partial \Omega }{\partial x}={\Omega }_x\\ \ddot{y}+2n\stackrel{\dot{}}{x}=\frac{\partial \Omega }{\partial y}={\Omega }_y\end{array}\}$, (2)

where is $n$ the mean motion of the rotating frame and $\Omega$ is the kinetic potential given by

$\Omega =\frac{n^2}{2}\left(x^2+y^2\right)+\frac{1-\mu }{2{\rho }_1}+\frac{\mu }{{\rho }_2}+\frac{1-3\mu }{2{\rho }_3}+\frac{\mu }{{\rho }_4}.$(3)

Let us write the System (2) as

$\begin{array}{l}\ddot{x}-2n\stackrel{\dot{}}{y}=\frac{\partial \Omega }{\partial x}={\Omega }_x=f\left(x,y\right),\\ \ddot{y}+2n\stackrel{\dot{}}{x}=\frac{\partial \Omega }{\partial y}={\Omega }_y=g\left(x,y\right).\end{array}\}$(4)

For libration points $\stackrel{\dot{}}{x}=\stackrel{\dot{}}{y}=\ddot{x}=\ddot{y}=0$ i.e., ${\Omega }_x=0$ and ${\Omega }_y=0$.

The points of intersection of ${\Omega }_x=0$ and ${\Omega }_y=0$ give the locations of libration points of the kite configuration. The contour plots of these equations are given below:

The symbols $L_1,L_2,\cdots ,L_7$ in Figure 2 are the positions of seven libration points indicated through the intersection of ${\Omega }_x=0$ and ${\Omega }_y=0$ for $\mu =0.03$, $n=1.901802$ and those are in Figure 3 for $\mu =0.07$, $n=1.888594$. Here black dots $P_i\left(i=1,2,3,4\right)$ represent four bodies forming a kite and four pink spots $L_j\left(j=4,5,6,7\right)$ represent non-axial libration points.

The symbols $L_1,L_2,\cdots ,L_7$ in Figure 4 are the positions of seven libration points indicated through the intersection of ${\Omega }_x=0$ and ${\Omega }_y=0$ for $\mu =0.09$, $n=1.882361$ and those are in Figure 5 for $\mu =0.11$, $n=1.876288$. Here, black dots $P_i\left(i=1,2,3,4\right)$ represent four bodies forming a kite and four pink spots $L_j\left(j=4,5,6,7\right)$ represent nonaxial libration points.

The symbols $L_1,L_2,\cdots ,L_{13}$ in Figure 6 are the positions of thirteen libration points indicated through the intersection of ${\Omega }_x=0$ and ${\Omega }_y=0$ for $\mu =0.15$, $n=1.821047$ and those are in Figure 7 for $\mu =0.22$, $n=1.624152$. Here black dots $P_i\left(i=1,2,3,4\right)$ represent four bodies forming a kite and ten pink spots $L_j\left(j=4,5,\cdots ,13\right)$ represent non-axial libration points.

The symbols $L_1,L_2,\cdots ,L_{11}$ in Figure 8 are the positions of eleven libration points indicated through the intersection of ${\Omega }_x=0$ and ${\Omega }_y=0$ for $\mu =0.30$, $n=1.364779$ and those are in Figure 9 for $\mu =0.31$, $n=1.328802$. Here black dots $P_i\left(i=1,2,3,4\right)$ represent four bodies forming a kite and eight pink spots $L_j\left(j=4,5,\cdots ,11\right)$ represent non-axial libration points.

For different values of $\mu \in \left(0,3^{-1}\right)$ and corresponding values of $n$, the coordinates $\left(x,y\right)$ of libration points $L_j\left(j=4,5,\cdots ,13\right)$ are given in the 2^nd, 3^rd, 4^th and 5^th columns respectively of the Stability Tables 1-10 .

The motion of a satellite is said to be stable near the libration points when given a very small displacement and small velocity, the satellite oscillates for a considerable time around the points.

Let $\xi ,\eta$ denote the small displacement of the infinitesimal body (artificial satellite) from the libration points $\left(x_0,y_0\right)$, then the variational equations of motion can be easily obtained by substituting $x=x_0+\xi ,y=y_0+\eta$ in Equation (4). Thus, Equation (4) becomes

$\begin{array}{l}\ddot{\xi }-2n\stackrel{\dot{}}{\eta }=f\left(x_0+\xi ,y_0+\eta \right),\\ \ddot{\eta }+2n\stackrel{\dot{}}{\xi }=g\left(x_0+\xi ,y_0+\eta \right).\end{array}\}$(5)

Now applying Taylor’s theorem in the neighbourhood of $\left(x_0,y_0\right)$ in the right-hand side of the above equations, we get

$\begin{array}{l}\ddot{\xi }-2n\stackrel{\dot{}}{\eta }=f\left(x_0,y_0\right)+\left(\xi \frac{\partial }{\partial x}+\eta \frac{\partial }{\partial y}\right)f\left(x_0,y_0\right)+\cdots +\text{higher order term of}\xi \&\eta \\ \text{and}\ddot{\eta }+2n\stackrel{\dot{}}{\xi }=g\left(x_0,y_0\right)+\left(\xi \frac{\partial }{\partial x}+\eta \frac{\partial }{\partial y}\right)g\left(x_0,y_0\right)+\cdots +\text{higher order term of}\xi \&\eta \end{array}$

$\begin{array}{l}\Rightarrow \ddot{\xi }-2n\stackrel{\dot{}}{\eta }=f\left(x_0,y_0\right)+\xi {\left(\frac{\partial f}{\partial x}\right)}_0+\eta {\left(\frac{\partial f}{\partial y}\right)}_0+\cdots +\text{higher order term of}\xi \&\eta \\ \text{and}\ddot{\eta }+2n\stackrel{\dot{}}{\xi }=g\left(x_0,y_0\right)+\xi {\left(\frac{\partial g}{\partial x}\right)}_0+\eta {\left(\frac{\partial g}{\partial y}\right)}_0+\cdots +\text{higher order term of}\xi \&\eta \end{array}\}$(6)

But at the libration points ${\Omega }_x^0=f\left(x_0,y_0\right)={\Omega }_y^0=g\left(x_0,y_0\right)=0$ So, from Equation (6), we get

$\begin{array}{l}\ddot{\xi }-2n\stackrel{\dot{}}{\eta }=\xi \frac{\partial }{\partial x}{\left(\frac{\partial \Omega }{\partial x}\right)}_0+\eta \frac{\partial }{\partial y}{\left(\frac{\partial \Omega }{\partial x}\right)}_0+\cdots +\text{higher order term of}\xi \&\eta \\ \text{and}\ddot{\eta }+2n\stackrel{\dot{}}{\xi }=\xi \frac{\partial }{\partial x}{\left(\frac{\partial \Omega }{\partial y}\right)}_0+\eta \frac{\partial }{\partial y}{\left(\frac{\partial \Omega }{\partial y}\right)}_0+\cdots +\text{higher order term of}\xi \&\eta .\end{array}$

For linear stability, neglecting the higher-order terms of $\xi$ and $\eta$, the variational equations are reduced to

$\begin{array}{l}\ddot{\xi }-2n\stackrel{\dot{}}{\eta }=\xi {\Omega }_{xx}^0+\eta {\Omega }_{yx}^0\text{,}\\ \ddot{\eta }+2n\stackrel{\dot{}}{\xi }=\xi {\Omega }_{xy}^0+\eta {\Omega }_{yy}^0.\end{array}\}$(7)

where ${\Omega }_{xx}^0,{\Omega }_{xy}^0,{\Omega }_{yx}^0,{\Omega }_{yy}^0$ represent the second-order derivatives of $\Omega$ at the libration points $\left(x_0,y_0\right)$. The above system of equations can be extended as

$\begin{array}{l}\stackrel{\dot{}}{\xi }=\xi \cdot 0+\eta \cdot 0+\stackrel{\dot{}}{\xi }\cdot 1+\stackrel{\dot{}}{\eta }\cdot 0,\\ \stackrel{\dot{}}{\eta }=\xi \cdot 0+\eta \cdot 0+\stackrel{\dot{}}{\xi }\cdot 0+\stackrel{\dot{}}{\eta }\cdot 1,\\ \ddot{\xi }=\xi {\Omega }_{xx}^0+\eta {\Omega }_{xy}^0+\stackrel{\dot{}}{\xi }\cdot 0+\stackrel{\dot{}}{\eta }\cdot 2n,\\ \ddot{\eta }=\xi {\Omega }_{xy}^0+\eta {\Omega }_{yy}^0+\stackrel{\dot{}}{\xi }\left(-2n\right)+\stackrel{\dot{}}{\eta }\cdot 0.\end{array}\}$ (8)

The system of Equation (8) can be written in the form of a single matrix equation as

$\stackrel{\dot{}}{X}=TX,$(9)

where $T=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ {\Omega }_{xx}^0& {\Omega }_{yx}^0& 0& 2n\\ {\Omega }_{xy}^0& {\Omega }_{yx}^0& -2n& 0\end{array}\right]$ and $X=\left[\begin{array}{c}\xi \\ \eta \\ \stackrel{\dot{}}{\xi }\\ \stackrel{\dot{}}{\eta }\end{array}\right]$.

If any matrix $X$ satisfy the equation

$TX=\lambda X$,(10)

then $X$ is said to be an eigenvector of the coefficient matrix $T$ and scalar $\lambda$ is its corresponding eigenvalue. If $T$ is thought of as a transformation matrix, then the result of applying $T$ to the particular vector $X$ satisfying Equation (10) is to produce a vector in the same direction as $X$ but of a different magnitude.

Now, Equation (10) can be written as $\left(T-\lambda I\right)X=0$.

The set of four simultaneous linear equations in four unknowns $\xi ,\eta ,\stackrel{\dot{}}{\xi },\stackrel{\dot{}}{\eta }$ will have non-trivial solutions provided the determinant of the characteristic matrix $\left(T-\lambda I\right)$vanishes.

$\text{i}\text{.e}\text{.,}\left|A-\lambda I\right|=0,$(11)

$\Rightarrow {\lambda }^4-\left({\Omega }_{xx}^0+{\Omega }_{yy}^0-4n^2\right){\lambda }^2+2n\left({\Omega }_{yx}^0-{\Omega }_{xy}^0\right)\lambda +\left({\Omega }_{xx}^0{\Omega }_{yy}^0-{\Omega }_{xy}^0{\Omega }_{yx}^0\right)=0,$

$\Rightarrow {\lambda }^4-\left({\Omega }_{xx}^0+{\Omega }_{yy}^0-4n^2\right){\lambda }^2+\left\{{\Omega }_{xx}^0{\Omega }_{yy}^0-{\left({\Omega }_{xy}^0\right)}^2\right\}=0.\left(\text{as}{\Omega }_{xy}^0={\Omega }_{yx}^0\right)$(12)

This equation is called a characteristic equation corresponding to the equations of a matrix $A$. Therefore, Equation (12) can be written as

${\lambda }^4-A{\lambda }^2+B=0$,(13)

where $A={\Omega }_{xx}^0+{\Omega }_{yy}^0-4n^2$ & $B={\Omega }_{xx}^0{\Omega }_{yy}^0-{\left({\Omega }_{xy}^0\right)}^2$.

Equation (13) is biquadratic in $\lambda$, so taking ${\lambda }^2=\wedge$, Equation (13) is reduced to a quadratic equation

${\wedge }^2-A\wedge +B=0.$ (14)

Let ${\wedge }_1$ and ${\wedge }_2$ be the two roots of the characteristic Equation (14), then

${\wedge }_1+{\wedge }_2=A\text{and}{\wedge }_1{\wedge }_2=B.$

But from Equation (13), $\wedge ={\lambda }^2=\frac{A\pm \sqrt{A^2-4B}}{2}$.

Let

${\wedge }_1=\frac{A+\sqrt{A^2-4B}}{2},{\wedge }_2=\frac{A-\sqrt{A^2-4B}}{2}.$

As ${\lambda }^2=\wedge$, so let ${\lambda }_1^2={\wedge }_1$ and ${\lambda }_2^2={\wedge }_2$, then

${\lambda }_1^2=\frac{A+\sqrt{A^2-4B}}{2}\text{and}{\lambda }_2^2=\frac{A-\sqrt{A^2-4B}}{2}.$

$\Rightarrow {\lambda }_1=\pm {\left(\frac{A+\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}\text{and}{\lambda }_2=\pm {\left(\frac{A-\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}.$

Let ${\lambda }_{11}=+{\left(\frac{A+\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}\text{and}{\lambda }_{12}=-{\left(\frac{A+\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}.$

${\lambda }_{21}=+{\left(\frac{A-\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}\text{and}{\lambda }_{22}=-{\left(\frac{A-\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}.$

For simplicity, let us write

${\lambda }_1^{*}=+{\left(\frac{A+\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}\text{and}{\lambda }_2^{*}=-{\left(\frac{A+\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}},$

${\lambda }_3^{*}=+{\left(\frac{A-\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}\text{and}{\lambda }_4^{*}=-{\left(\frac{A-\sqrt{A^2-4B}}{2}\right)}^{\frac{1}{2}}.$

as the four roots of the Characteristic Equation (13).

The criteria for linear stability of non-axial libration points are as follows:

Any libration point $L_j\left(x_j,y_j\right),j=4,5,6,\cdots ,13$ is said to be stable if ${\lambda }_1^2,{\lambda }_2^2$ are negative real and

$\begin{array}{l}\text{(i)}A={\lambda }_1^2+{\lambda }_2^2<0\\ \text{(ii)}B={\lambda }_1^2{\lambda }_2^2>0\\ \text{(iii)}D=A^2-4B\ge 0\end{array}\}$ Refer Szebehely [12] (15)

are satisfied together.

To satisfy the above conditions of stability, we need ${\Omega }_{xx}^0,{\Omega }_{xy}^0$ and ${\Omega }_{xy}^0$. So, differentiating the potential function $\Omega$ partially twice with respect to $x$ and $y$, we get

$\begin{array}{l}{\Omega }_{xx}=n^2-\left[\frac{1-\mu }{2{\rho }_1^3}+\frac{\mu }{{\rho }_2^3}+\frac{1-3\mu }{2{\rho }_3^3}+\frac{\mu }{{\rho }_4^3}\right]\\ +3\left[\frac{\left(1-\mu \right){\left(x-1/2\right)}^2}{2{\rho }_1^5}+\frac{\mu {\left(x+1/4\right)}^2}{{\rho }_2^5}+\frac{\left(1-3\mu \right){\left(x+1/2\right)}^2}{2{\rho }_3^5}+\frac{\mu {\left(x+1/4\right)}^2}{{\rho }_4^5}\right],\end{array}$

$\begin{array}{l}{\Omega }_{yx}=\frac{3\left(1-\mu \right)\left(x-1/2\right)y}{2{\rho }_1^5}+\frac{3\mu \left(x+1/4\right)\left(y-\sqrt{3}/4\right)}{{\rho }_2^5}\\ -\frac{3\left(1-3\mu \right)\left(x+1/2\right)y}{2{\rho }_3^5}+\frac{3\mu \left(x+1/4\right)\left(y+\sqrt{3}/4\right)}{{\rho }_4^5}={\Omega }_{xy},\end{array}$

$\begin{array}{l}{\Omega }_{yy}=n^2-\left[\frac{1-\mu }{2{\rho }_1^3}+\frac{\mu }{{\rho }_2^3}+\frac{1-3\mu }{2{\rho }_3^3}+\frac{\mu }{{\rho }_4^3}\right]\\ +3\left[\frac{\left(1-\mu \right)y^2}{2{\rho }_1^5}+\frac{\mu {\left(y-\sqrt{3}/4\right)}^2}{{\rho }_2^5}+\frac{\left(1-3\mu \right)y^2}{2{\rho }_3^5}+\frac{\mu {\left(y+\sqrt{3}/4\right)}^2}{{\rho }_4^5}\right].\end{array}$

The symbols $L_1,L_2,\cdots ,L_7$ in Figure 10 are the positions of seven libration points indicated through the intersection of ${\Omega }_x=0$ and ${\Omega }_y=0$ for $\mu =0.10$ & $n=1.879308$. Here black dots $P_i\left(i=1,2,3,4\right)$ represent four bodies forming a kite and four pink spots $L_j\left(j=4,5,6,7\right)$ represent non-axial libration points.

The values of ${\Omega }_{xx}^0,{\Omega }_{xy}^0,{\Omega }_{yy}^0,A,B,D$ are given in the 6^th, 7^th, 8^th, 9^th, 10^th and 11^th columns respectively in stability tables from Table 1-10 . The nature of the stability of each non-axial libration point satisfying the conditions of Equation (15), is mentioned in the last column of each Stability table.