Solution 8

Derivation

Let the masses of the stars be $m_1$ and $m_2$. The spaceship of mass $m$ is located on the line joining the stars at distance $r_1$ from $m_1$ and $r_2$ from $m_2$.

1. Angular Velocity of the System

The two stars rotate about their common Center of Mass (COM) due to their mutual gravitational attraction. Let $R = r_1 + r_2$ be the separation between the stars.

$$ \omega^2 = \frac{G(m_1 + m_2)}{R^3} = \frac{G(m_1 + m_2)}{(r_1 + r_2)^3} \quad \text{…(i)} $$

2. Force Equation for the Spaceship

The spaceship moves in a circular orbit with the same angular velocity $\omega$. The net gravitational force provides the centripetal force.

The radius of the spaceship’s orbit ($r’$) is the distance from the spaceship to the COM. Let the origin be at $m_1$.

• Position of COM: $x_{cm} = \frac{m_2 R}{m_1 + m_2}$
• Position of Spaceship: $r_1$
• Radius of orbit ($r’$): $r_1 – x_{cm} = r_1 – \frac{m_2(r_1+r_2)}{m_1+m_2} = \frac{m_1 r_1 – m_2 r_2}{m_1+m_2}$

The net force on the spaceship (assuming $m_1$ pull is dominant for direction):

$$ F_{net} = F_{g1} – F_{g2} = \frac{Gm_1 m}{r_1^2} – \frac{Gm_2 m}{r_2^2} $$

Equating to centripetal force $m \omega^2 r’$:

$$ \frac{Gm_1}{r_1^2} – \frac{Gm_2}{r_2^2} = \omega^2 \left( \frac{m_1 r_1 – m_2 r_2}{m_1+m_2} \right) $$

Substituting $\omega^2$ from eq (i):

$$ \frac{Gm_1}{r_1^2} – \frac{Gm_2}{r_2^2} = \left[ \frac{G(m_1+m_2)}{(r_1+r_2)^3} \right] \left( \frac{m_1 r_1 – m_2 r_2}{m_1+m_2} \right) $$

Canceling $G$ and $(m_1+m_2)$:

$$ \frac{m_1}{r_1^2} – \frac{m_2}{r_2^2} = \frac{m_1 r_1 – m_2 r_2}{(r_1+r_2)^3} $$

3. Rearranging for Mass Ratio

Group $m_1$ and $m_2$ terms:

$$ m_1 \left( \frac{1}{r_1^2} – \frac{r_1}{(r_1+r_2)^3} \right) = m_2 \left( \frac{1}{r_2^2} – \frac{r_2}{(r_1+r_2)^3} \right) $$ $$ m_1 \left( \frac{(r_1+r_2)^3 – r_1^3}{r_1^2(r_1+r_2)^3} \right) = m_2 \left( \frac{(r_1+r_2)^3 – r_2^3}{r_2^2(r_1+r_2)^3} \right) $$ $$ \frac{m_1}{m_2} = \frac{r_1^2}{r_2^2} \left[ \frac{(r_1+r_2)^3 – r_2^3}{(r_1+r_2)^3 – r_1^3} \right] $$