Question

Two planets, A and B are orbiting a common star in circular orbits of radii \( R_A \) and \( R_B \), respectively, with \( R_B = 2R_A \). The planet B is \( \sqrt{2} \) times more massive than planet A. The ratio \( \frac{L_B}{L_A} \) of angular momentum (\( L \)) of planet B to that of planet A (\( L_A \)) is closest to integer:

Hint:

The angular momentum of a planet in orbit depends on both its mass and the radius of its orbit. Use the relationship \( L = m \cdot v \cdot R \) and account for the velocity from gravitational force for such problems.

Answer 1

Correct Answer: 8

The angular momentum of a planet in orbit is given by: \[ L = m v R \] Where \( m \) is the mass, \( v \) is the velocity, and \( R \) is the radius of the orbit. The velocity \( v \) of a planet in orbit can be expressed as: \[ v = \sqrt{\frac{GM}{R}} \] For planet A: \[ L_A = m_A v_A R_A = m_A \sqrt{\frac{GM}{R_A}} R_A = m_A \sqrt{GM R_A} \] For planet B: \[ L_B = m_B v_B R_B = m_B \sqrt{\frac{GM}{R_B}} R_B = m_B \sqrt{GM R_B} \] Given that \( R_B = 2R_A \) and \( m_B = \sqrt{2} m_A \), the ratio of angular momentum is: \[ \frac{L_B}{L_A} = \frac{m_B \sqrt{GM R_B}}{m_A \sqrt{GM R_A}} = \frac{\sqrt{2} m_A \sqrt{GM (2R_A)}}{m_A \sqrt{GM R_A}} = 8 \] Thus, the correct answer is , \( \frac{L_B}{L_A} = 8 \).