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:
Show 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.
The angular momentum of a planet in orbit is calculated as \( L = m v R \), where \( m \) is mass, \( v \) is velocity, and \( R \) is orbital radius. Orbital velocity is defined by \( v = \sqrt{\frac{GM}{R}} \). For planet A, angular momentum is \( L_A = m_A v_A R_A = m_A \sqrt{\frac{GM}{R_A}} R_A = m_A \sqrt{GM R_A} \). Similarly, 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 \( R_B = 2R_A \) and \( m_B = \sqrt{2} m_A \), the ratio of angular momenta is computed as \( \frac{L_B}{L_A} = \frac{m_B \sqrt{GM R_B}}{m_A \sqrt{GM R_A}} = \frac{\sqrt{2} m_A \sqrt{GM (2R_A)}}{m_A \sqrt{GM R_A}} = 8 \). Therefore, the ratio \( \frac{L_B}{L_A} \) is 8.
