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The Sitnikov Problem is a ....

Configuration

Figure 1. Configuration of the Sitnikov-Problem

The Sitnikov Problem has a name of the russian mathematician Kirill Aleksandrovich Sitnikov (* 1926). This is an form of three-body problem. It describs the moution of three primaries resulting from gravitational interaction.

The System consist of 2 primaries (for example stars) with the equal masses ($m_{1}=m_{2}=m_{pr}$). The third body have a mass ($m_{3}$), $m_{3}$ is much less, then the mass of primsries. This body is confined to a motion perpendicular to the instantaneous plane of motion of the two prinaries, which are always equally far away from the barycenter of the system (see Fig.1). The point of origin is in centre of mass of primaries. In such system the motion of third body is one-dimensional - it moves just along $z$-Axis.

• General equation of motion

First of all it is necessary to define the energy of the system $E$:

$E={\frac {m_{3}}{2}}\left({\frac {dz}{dt}}\right)^{2}-{\frac {2Gm_{3}m_{pr}}{r}},$

Lets write the second Newton's law:

$\left({\frac {d^{2}z}{dt^{2}}}\right)=-{\frac {2Gm_{pr}z}{r^{3}}},$

From Fig 1. follows:

$r^{2}=a^{2}+z^{2}\,$

Now equation of motion is

$\left({\frac {d^{2}z}{dt^{2}}}\right)+{\frac {2Gm_{pr}z}{(z^{2}+a^{2})^{3/2}}}=0,$

or

$\left({\frac {d^{2}z}{dt^{2}}}\right)+{\frac {z}{(z^{2}+a^{2})^{3/2}}}=0,$

here $2Gm_{pr}=1$.

Circular Case

Datei:Sitnikov Problem CircularCase COLOR.jpg

Figure 2. Motion of body in circular case of Sitnikov Problem

• )Equations of motion

In the case, when the primaries move on circular orbits, consider that $a=1/2$(distance unit is a diameter of the orbit). Final equation of motion with boundary condition in this case

$\left\{{\begin{array}{lll}\left({\frac {d^{2}z}{dt^{2}}}\right)+{\frac {z}{(z^{2}+1/4)^{3/2}}}=0\\{\frac {dz}{dt}}|_{t=0}=0\\z|_{t=0}=z_{ini}.\\\end{array}}\right.$

The last two equations are boundary condition. They signify, that in start time the velocity of body was zero, and z-coordinate was $z_{initial}$. It is possible to use the numerical methods for calculating $z$. For example standart Runge - Kutta method 4th order. The results are plotted in the Fig.2.

It is possible to calculate the period of motion started from principies of theoretical mechanics (see for example ref.1 in the section Literature):

${\frac {p}{4}}={\frac {1}{\sqrt {2}}}\int _{z_{ini}}^{0}{\frac {dz}{\sqrt {-{\frac {1}{\sqrt {1/4+z_{ini}^{2}}}}+{\frac {1}{\sqrt {1/4+z^{2}}}}}}}$

The numerical calculating of this solution give the next periods of the oscillations:

if $z_{initial}=0.2$ then period $p=2.42$

if $z_{initial}=0.7$ then period $p=4.27$

if $z_{initial}=1.3$ then period $p=8.07$

This Results confirm the accuracy of Figure 2.

Elliptic Case

• )Trajectory

Datei:Sitnikov Problem EllipticCase ZTcord.JPG

Figure 3. Motion of body in elliptic case of Sitnikov Problem (Z,T coordinates)

Datei:Sitnikov Problem EllipticCase ZZDOTcord.JPG

Figure 4. Motion of body in elliptic case of Sitnikov Problem (Z,Z_DOT coordinates)

The differential equation of Sitnikov-Problem in elliptic case another. Now it is necessary to take into consideration the dependence of $a(t)$.

$\left({\frac {d^{2}z}{dt^{2}}}\right)+{\frac {z}{(z^{2}+a^{2}(t))^{3/2}}}=0,$

Note that, from well known results in the two-body problem (see ref.1 in the section Litersture):

$a(t)={\frac {1}{2}}(1-e\cdot \cos(t))+O(e^{2})$

Here $O(e^{2})$ is all parts of $a(t)$, that have order of value $e^{2}$. It means that for example if eccentricity e=0.1, then part $O(e^{2})$ have order 0.01. In first approximation it is possible to neglect this part. The results of numerical calculating are plotted in Fig.3-4.

• )The surfaces of section

Datei:Sitnikov Problem EllipticCase SOS.JPG

Figure 5. The surfaces of section in elliptic case of Sitnikov Problem

Sitnikov_Problem_EllipticCase_SOS

Now, when trajectories were calculated in elliptic case, it is possible to plot the surfaces of section (SOS). The phase space have in our case three dimensions: time, velocity and distance from center mass. When one of this dimensions is $const$,then we have just two dimensions and get SOS. SOS is just one of vivid ways to represent numerical results. For example here SOS are in coordinates $(z,z_{dot})$.

In the Fig.5 are some examples of SOS of motion. The area $(z\in (2-,2),z_{dot}\in (-2,2))$ was divided into $21\times 21$ points. At every point was the trajectory calculated (values of $z$ and $z_{dot}$ for each time point). Then were kipped from this trajectory just points, when the primaries are in perihelion. To the finite arrays, that plotted here, were added just finite trajectories (i.e. $z_{max}<10$ is serve as a criterion for finite trajectory).

• ) Finite motion as a whole

Datei:Sitnikov Problem EllipticCase FinitVolum.JPG

Figure 6. The Finite Volume in elliptic case of Sitnikov Problem

Here area $(z\in (2-,2),z_{dot}\in (-2,2))$ was divided into $50\times 50$ points. For every point was calculated trajectory. Next must be checked the max value of $z$. When like previous $z_{max}<10$, it is possible to assume, that the motion is bounded and add at the plot the black point in this place. Fig.6 represents some results.

Figure 7. The Finite Volume in elliptic case of Sitnikov Problem(as whole)

For getting of eccentricity dependence of area of finite motion, it is necessary to count the black points and disjoint at $50\times 50$(total number of points) for different eccentricities. It is plotted in the Fig.7.

Applications

Allthough it is almost impossible, that three celestial bodies can make Sitnikov-Configuration, the Sitnikov-Problem was studied already relativ long time, because it is possible to discover in this fairly simple System all properties of a chaotic system. Such system is excellent serve for generalized research of chaotic effects in dynamical systems.

Literarure

1. Landau, Livshitc: Theoretical Meachanics, Moskwa, "Nauka", 1988.