Question

A space ship sent to study a binary star system is set into an orbit in order to remain always collinear with the stars. If distances of the spaceship from the stars are $r_1$ and $r_2$, find ratio of masses of the stars. Is this orbit a stable one?

Answer 1

Ratio of masses: $\frac{M_1}{M_2} = \frac{r_1^2}{r_2^2} \frac{(r_1+r_2)^3 - r_2^3}{(r_1+r_2)^3 - r_1^3}$ , Orbit is unstable.

Answer 2

The problem involves a spaceship in a collinear orbit with a binary star system. To find the ratio of the masses of the stars, we analyze the gravitational forces and centripetal acceleration acting on the spaceship.

Assumptions:

• The spaceship is at the Lagrangian point $L_1$, located between the two stars.
• The distance between the stars is $D = r_1 + r_2$, where $r_1$ and $r_2$ are the distances from the spaceship to each star, respectively.

Derivation:

1. Gravitational Forces: The gravitational forces exerted by the two stars on the spaceship are in opposite directions.

2. Centripetal Force: The net gravitational force provides the centripetal force required for the spaceship to maintain its orbit around the center of mass of the binary system.

3. Equation Setup: $\frac{GM_1}{r_1^2} - \frac{GM_2}{r_2^2} = m \omega^2 (x_s - x_{CM})$ where:

   • $M_1$ and $M_2$ are the masses of the stars.
   • $G$ is the gravitational constant.
   • $m$ is the mass of the spaceship.
   • $\omega$ is the angular velocity of the binary system.
   • $x_s$ is the position of the spaceship from $M_1$, so $x_s = r_1$.
   • $x_{CM}$ is the position of the center of mass from $M_1$, given by $x_{CM} = \frac{M_2 D}{M_1+M_2} = \frac{M_2 (r_1+r_2)}{M_1+M_2}$.

4. Simplifying: $x_s - x_{CM} = r_1 - \frac{M_2 (r_1+r_2)}{M_1+M_2} = \frac{M_1 r_1 - M_2 r_2}{M_1+M_2}$.

5. Substituting: $\frac{GM_1}{r_1^2} - \frac{GM_2}{r_2^2} = m \frac{G(M_1+M_2)}{(r_1+r_2)^3} \frac{M_1 r_1 - M_2 r_2}{M_1+M_2}$ $\frac{M_1}{r_1^2} - \frac{M_2}{r_2^2} = \frac{M_1 r_1 - M_2 r_2}{(r_1+r_2)^3}$

6. Solving for Mass Ratio: $M_1 \left(\frac{1}{r_1^2} - \frac{r_1}{(r_1+r_2)^3}\right) = M_2 \left(\frac{1}{r_2^2} - \frac{r_2}{(r_1+r_2)^3}\right)$ $\frac{M_1}{M_2} = \frac{\frac{1}{r_2^2} - \frac{r_2}{(r_1+r_2)^3}}{\frac{1}{r_1^2} - \frac{r_1}{(r_1+r_2)^3}} = \frac{r_1^2}{r_2^2} \frac{(r_1+r_2)^3 - r_2^3}{(r_1+r_2)^3 - r_1^3}$

Stability of Orbit:

The collinear Lagrangian points ($L_1, L_2, L_3$) are unstable. Any small displacement from these points will cause the spaceship to drift away, requiring active station-keeping to maintain the orbit.

Final Answer:

The ratio of masses of the stars is: $\frac{M_1}{M_2} = \frac{r_1^2}{r_2^2} \frac{(r_1+r_2)^3 - r_2^3}{(r_1+r_2)^3 - r_1^3}$

The orbit is unstable.