Question

A binary star system consists of two stars of masses $m_1$ and $m_2$, separated by a fixed distance $a$, rotating about their common center of mass with angular velocity $\omega$. Derive the gravitational potential $V(x, y)$ at an arbitrary point $(x, y)$ in the plane of rotation.

(Assume the center of mass is at the origin and the stars always lie on the x-axis in the rotating frame.)

Answer 1

V(x, y) = -G \left[ \frac{m_1}{\sqrt{\left(x + \frac{m_2 a}{m_1 + m_2}\right)^2 + y^2}} + \frac{m_2}{\sqrt{\left(x - \frac{m_1 a}{m_1 + m_2}\right)^2 + y^2}} \right]

Answer 2

The problem asks us to derive the gravitational potential $V(x, y)$ at an arbitrary point $(x, y)$ in the plane of rotation of a binary star system.

1. Locate the stars:

Let the masses of the two stars be $m_1$ and $m_2$. They are separated by a fixed distance $a$. The center of mass (CM) is at the origin $(0, 0)$, and the stars always lie on the x-axis in the rotating frame.
Let the coordinates of $m_1$ be $(x_1, 0)$ and $m_2$ be $(x_2, 0)$.
Since the CM is at the origin:
$m_1 x_1 + m_2 x_2 = 0 \quad (1)$
The distance between the stars is $a$:
$x_2 - x_1 = a \quad (2)$ (assuming $x_2 > x_1$, i.e., $m_1$ is on the negative x-axis and $m_2$ on the positive x-axis).

From equation (1), $x_1 = -\frac{m_2}{m_1} x_2$.
Substitute this into equation (2):
$x_2 - \left(-\frac{m_2}{m_1} x_2\right) = a$
$x_2 \left(1 + \frac{m_2}{m_1}\right) = a$
$x_2 \left(\frac{m_1 + m_2}{m_1}\right) = a$
$x_2 = \frac{m_1 a}{m_1 + m_2}$

Now, find $x_1$:
$x_1 = a - x_2 = a - \frac{m_1 a}{m_1 + m_2} = \frac{a(m_1 + m_2) - m_1 a}{m_1 + m_2} = \frac{m_2 a}{m_1 + m_2}$
So, the coordinates of $m_1$ are $\left(-\frac{m_2 a}{m_1 + m_2}, 0\right)$ and $m_2$ are $\left(\frac{m_1 a}{m_1 + m_2}, 0\right)$.

2. Derive the gravitational potential:

The gravitational potential $V$ at a point $(x, y)$ due to a point mass $M$ at a distance $r$ is given by $V = -\frac{GM}{r}$.
For a system of multiple masses, the total gravitational potential at a point is the scalar sum of the potentials due to individual masses.

The distance $d_1$ from the point $(x, y)$ to $m_1$ at $\left(-\frac{m_2 a}{m_1 + m_2}, 0\right)$ is:
$d_1 = \sqrt{\left(x - \left(-\frac{m_2 a}{m_1 + m_2}\right)\right)^2 + (y - 0)^2} = \sqrt{\left(x + \frac{m_2 a}{m_1 + m_2}\right)^2 + y^2}$

The distance $d_2$ from the point $(x, y)$ to $m_2$ at $\left(\frac{m_1 a}{m_1 + m_2}, 0\right)$ is:
$d_2 = \sqrt{\left(x - \frac{m_1 a}{m_1 + m_2}\right)^2 + (y - 0)^2} = \sqrt{\left(x - \frac{m_1 a}{m_1 + m_2}\right)^2 + y^2}$

The gravitational potential $V(x, y)$ at the point $(x, y)$ is the sum of the potentials due to $m_1$ and $m_2$:
$V(x, y) = V_1(x, y) + V_2(x, y)$
$V(x, y) = -\frac{G m_1}{d_1} - \frac{G m_2}{d_2}$

Substituting the expressions for $d_1$ and $d_2$:
$V(x, y) = -G \left[ \frac{m_1}{\sqrt{\left(x + \frac{m_2 a}{m_1 + m_2}\right)^2 + y^2}} + \frac{m_2}{\sqrt{\left(x - \frac{m_1 a}{m_1 + m_2}\right)^2 + y^2}} \right]$
This is the gravitational potential at an arbitrary point $(x, y)$ in the plane of rotation. The angular velocity $\omega$ is relevant for the dynamics in the rotating frame (e.g., for the effective potential including centrifugal terms), but not for the gravitational potential itself, which depends only on the masses and their positions.