Zero-Velocity Curves

Zero-Velocity Curves

Calculation of a particle's motion in the gravitational field of many other particles is a n-body problem that cannot be solved in closed algebraic form. But, if the particle's mass is infinitesimally small as compared to two other objects (stars), then the particle's motion in the gravitational field of the larger objects can be solved in closed form. The motion of this infinitesimal particle is governed by the equations of the restricted three-body problem. The mass of the infinitesimal particle, m, is negligible compared to the masses of the other two objects, $M_1$ and $M_2$. The (x,y,z) coordinate of the particle m is the instantaneous position of the particle in the rotating frame of reference of the system. The origin of this coordinate system is the center of mass. The x-axis defines the line between the centers of the two larger masses. Of these two massive bodies, the more massive one is located on the negative x-axis, while the smaller mass is located on the positive x-axis. The orthogonal x and y axes define the orbital plane of the system. The z-axis defines the axis of orbital revolution. $M_1$ and $M_2$ orbit around the center of mass in circular orbits. The motion of m is restricted to the orbital plane. Therefore, the infinitesimal particle can only move in the (x,y) plane of the system and not in the z direction. Figure 4.1 displays the coordinate system used in the restricted three-body problem.

Figure 4.1: Coordinate system used in the restricted three-body problem. The origin is the center of mass. The system revolves counterclockwise around the center of mass in X-Y coordinate plane. The motion of an infinitesimal particle is restricted to this orbital plane.

With the above assumptions, the two differential equations that describe the motion of m are given as

$$\frac{d^2x}{dt^2} - 2 \frac{dy}{dt} = \frac{\partial U}{\partial x},$$ (4.1)

$$\frac{d^2y}{dt^2} + 2 \frac{dx}{dt} = \frac{\partial U}{\partial y},$$ (4.2)

where U describes the effective gravitational potential acting on particle m. This effective gravitational potential is defined as

$$U=\frac{1}{2} (x^2 + y^2) + \frac{1-\mu}{r_1} + \frac{\mu}{r_2},$$ (4.3)

which includes the gravitational potential arising from $M_1$ and $M_2$ on mass m, as well as centrifugal forces due to rotation around the center of mass. The variables $r_1$ and $r_2$ measure the distance of particle m from $M_1$ and $M_2$, respectively. These values are measured in fractional units of distance between the two massive objects. The variable $\mu$ defines the reduced mass of the system. These variables are defined as

$$r^{2}_{1} = (x+\mu)^2 + y^2,$$ (4.4)

$$r^{2}_{2}= (x+\mu-1)^2 + y^2,$$ (4.5)

and

$$\mu=\frac{M_2}{M_1+M_2}.$$ (4.6)

Equations 4.1 and 4.2 can be specifically solved in closed algebraic form if Eqn. 4.1 is multiplied by $2\frac{dx}{dt}$ and Eqn. 4.2 is multiplied by $2\frac{dy}{dt}$. Then adding together these two altered equations yields

$$2\frac{dx}{dt}\frac{d^2x}{dt^2} + 2\frac{dy}{dt}\frac{d^2y}{... ... U}{\partial x} + 2\frac{dy}{dt}\frac{\partial U}{\partial y}.$$ (4.7)

Integrating Eqn. 4.7 yields

$$\frac{d}{dt}\left( \frac{dx}{dt} \right)^2 + \frac{d}{dt}\left( \frac{dy}{dt} \right)^2 = 2\frac{d}{dt}({U}),$$ (4.8)

$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 2U - C,$$ (4.9)

or similarly,

$$V^2 = 2U-C.$$ (4.10)

Equations 4.9 and 4.10 are named the Jacobian integral, where C is the Jacobian constant. If the value of C is known, then the velocity of particle m can be calculated in the rotating frame of the system. If the values of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ equal zero for a given value of C, then the position of the particle within the system will trace out a surface, called a zero-velocity curve. Setting the velocity components in Eqn. 4.9 to zero yields the following simplified equation

$$C = x^2 + y^2 + \frac{2(1-\mu)}{r_1} + \frac{2\mu}{r_2}.$$ (4.11)

Refer to Fig. 1.1 for a schematic diagram of zero-velocity curves around the stellar components for KU Cygni.

Joseph Louis Lagrange was the first mathematician to tackle the restricted three-body problem and to solve the differential equations in closed form. Lagrange also calculated five equilibrium points within the rotating system. At these Lagrangian points, the net force (gravitational and centrifugal) exerted on particle m by the other objects is zero; thus, particle m remains at rest in the rotating frame of the system at these locations. The gravitational potential describes the gravitational field at a certain point in the system. For the gravitational force to be negligible at any point in the system, the gradient of the potential must equal zero. Taking the gradient of the U, given in Eqn. 4.3, yields the following expression,

$$\hat{i}\left[x-(1-\mu)\left(\frac{x+\mu}{r^{3}_{1}}\right) -... ..._{1}}\right) - \mu\left(\frac{y}{r^{3}_{2}}\right)\right] = 0.$$ (4.12)

If Eqn. 4.12 must equal zero for gravitational equilibrium, each expression within the brackets of Eqn. 4.12 must equal zero. In Eqn. 4.12 the second expression within the brackets will equal zero, if the value of y is zero. Therefore, if y equals zero, the following equation must be satisfied,

$$x - (1-\mu)\frac{x+\mu}{\mid x+\mu \mid^3} - \mu\frac{x+\mu-1}{\mid x + \mu - 1 \mid^3} = 0.$$ (4.13)

Lagrange calculated three points along the x-axis that satisfy Eqn. 4.13. One point is located between $M_1$ and $M_2$, called the inner Lagrangian point or L1 point. The other two points are located on either side of $M_1$ and $M_2$ on the x-axis, called the outer Lagrangian points or the L3 and L2 points, respectively. These three points satisfy the straight line solution of Eqn. 4.9. If the value of $y\neq 0$, then another solution to Eqn. 4.12, would require that

$$1 - \frac{1-\mu}{r^{3}_{1}} - \frac{\mu}{r^{3}_{2}} = 0.$$ (4.14)

The above equation can only be satisfied if $r_1=r_2=1$. Two points satisfy the equilateral triangle solution of Eqn. 4.14. These two points are named the L4 and L5 points. Figure 4.2 displays the positions of the Lagrangian points for KU Cygni for $\mu=0.11085$, where the mass ratio (q=0.125) was taken from Olson, Etzel & Dewey (1995). Slight external forces exerted on a particle located at L4 or L5 points will not greatly perturb the orbit of a particle at this position. But, external forces exerted on a particle located at the L1, L2, or L3 points will dramatically perturb and alter its orbit. These perturbations become very important in close binary systems.

Figure 4.2: Location of Lagrangian points for KU Cygni. The closed surfaces around M1 and M2 are the critical Roche lobes around the more massive and less massive stars, respectively. The center of mass (CM) is labeled. The system orbits around the CM in counterclockwise fashion.