Question

A binary star consist of two stars \(\mathrm{A}(2.2 \mathrm{Ms})\) and \(\mathrm{B}\) (mass \(11 \mathrm{Ms}\) ) where \(\mathrm{Ms}\) is the mass of sun. They are separated by distance \(\mathrm{d}\) and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of \(\operatorname{star} B\). about the centre of mass is \(\\{\mathrm{A}\\} 6\) \(\\{\mathrm{B}\\} \overline{(1 / 4)}\) \(\\{C\\} 12\) \(\\{\mathrm{D}\\}(1 / 2)\)

Short answer

The short answer to the given question is \(\boxed{\text{(A)}\, 6}\).

Step-by-step solution

Find the center of mass

To find the center of mass, we can use the formula: \(\frac{m_A \cdot r_A + m_B \cdot r_B}{m_A + m_B}\), where \(m_A = 2.2 \,\mathrm{Ms}\) and \(m_B = 11 \,\mathrm{Ms}\) are the masses of the stars, and \(r_A\) and \(r_B\) are their respective distances from the center of mass. Since we only need the ratio of the distances, we can set up the equation: \(\frac{r_A}{r_B} = \frac{m_B}{m_A} = \frac{11 \,\mathrm{Ms}}{2.2 \,\mathrm{Ms}} = 5\). This gives us a ratio of distances, with star A being 5 times farther from the center of mass than star B.

Find the individual angular momenta

Next, let's find the angular momenta for star A and star B. Angular momentum (L) is given by the formula: \(L = m \cdot r \cdot v\), where m is the mass, r is the distance from the center of mass, and v is the tangential velocity. Since we only need the ratio of the angular momenta, and both stars have the same angular velocities, we can set up the equation: \(\frac{L_A}{L_B} = \frac{m_A \cdot r_A}{m_B \cdot r_B} = \frac{2.2 \,\mathrm{Ms} \cdot 5}{11 \,\mathrm{Ms}}\).

Calculate the ratio of the total angular momentum to the angular momentum of star B

Now, we find the total angular momentum of the binary star system, which is the sum of the individual angular momenta: \(L_{total} = L_A + L_B\). To find the ratio of the total angular momentum to the angular momentum of star B, we use: \(\frac{L_{total}}{L_B} = \frac{L_A + L_B}{L_B}\). Substituting the values from Step 2, we get: \(\frac{L_{total}}{L_B} = \frac{2.2 \,\mathrm{Ms} \cdot 5 + 11 \,\mathrm{Ms}}{11 \,\mathrm{Ms}}\). Solution: \(\frac{L_{total}}{L_B} = \frac{2.2 \,\mathrm{Ms} \cdot 5 + 11 \,\mathrm{Ms}}{11 \,\mathrm{Ms}} = \boxed{1 + 5} = \boxed{6}\). The correct answer is \(\boxed{\text{(A)}\, 6}\).

Key concepts

Angular Momentum

In a binary star system, angular momentum is a key concept used to understand the dynamics of the stars as they orbit each other. Angular momentum is conserved, which means it remains constant unless acted upon by an external force. For a single star, angular momentum (\(L\)) is calculated using the formula: \[L = m \cdot r \cdot v\]where:

• \(m\) is the mass of the star,
• \(r\) is the distance from the center of mass, and
• \(v\) is the tangential velocity of the star.

In binary systems, the ratio of the stars' angular momentum provides insight into each star's contribution to the total angular momentum. For our given exercise, finding this ratio helps solve the problem more efficiently by understanding the relative motions and contributions of each star.

Center of Mass

The center of mass in a binary star system is a pivotal concept, as both stars revolve around this point. It acts as the "balancing point" of the system. In binary stars, the center of mass (CM) remains stationary unless an external force interferes, meaning the positions of the stars dictate it.To find the CM, use the formula:\[\frac{m_A \cdot r_A + m_B \cdot r_B}{m_A + m_B}\]where:

• \(m_A\) and \(m_B\) are the masses of stars A and B, respectively,
• \(r_A\) and \(r_B\) are their distances from the CM.

In our exercise, \(m_B\) is much larger than \(m_A\), so \(r_A\) is significantly greater than \(r_B\). This tells us star A is further from the CM. Understanding this arrangement is crucial for visualizing the system's dynamics.

Mass Ratio

Mass ratio in binary star systems reveals how the mass of one star compares to the other. This measuring is critical because it affects the dynamics of the system. For example, in this exercise, the mass of star B is given as \(11 \, \mathrm{Ms}\) and star A is \(2.2 \, \mathrm{Ms}\).The mass ratio is calculated simply as:\[\frac{m_B}{m_A} = \frac{11 \, \mathrm{Ms}}{2.2 \, \mathrm{Ms}} = 5\]This means star B has five times the mass of star A. Knowing this ratio enables us to understand many aspects of the system:

• How the center of mass is affected by their masses,
• How each star's angular momentum contributes to the system.

Remember, in binary systems, a larger mass often means a smaller distance to the center of mass, affecting the entire dynamics.

Star Dynamics

Star dynamics in binary systems describe the motion and interaction between the stars as they orbit their common center of mass. The behavior of these stars is driven by gravitational forces, and their dynamics help us understand various stellar interactions. In our scenario, star dynamics can be observed through:

• The rotation about a center of mass, which remains still.
• The conservation of angular momentum, resulting in predictable orbital patterns.
• The ratio of mass and distance influencing each star's path.

Understanding these dynamics allows us to comprehend real stellar behavior, predict positions, and even deduce the presence of unseen objects by perturbations they cause. It's the harmony of these principles that maintains order in binary systems like the one in the exercise.