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 $4\sqrt{2}$times more massive than planet A. The ratio$\left(\frac{L_B}{L_A}\right)$ of angular momentum$\left({\mathrm{L}}_{\mathrm{B}}\right)$ of planet B to that of planet A $\left({\mathrm{L}}_A\right)$ is closest to integer __________.

Step 1: Angular Momentum of a Planet in Circular Orbit

The angular momentum $L$ of a planet of mass $m$ moving in a circular orbit of radius $R$ around a star is given by: $$L = m v R$$

where:

$m$ is the mass of the planet,

$v$ is the orbital speed,

$R$ is the radius of the orbit.

Step 2: Orbital Speed from Kepler’s Laws

For a planet in a circular orbit, the centripetal force is provided by gravitational attraction:

$$\frac{G M m}{R^2} = \frac{m v^2}{R}$$

Solving for $v$: $$v = \sqrt{\frac{G M}{R}}$$

Thus, the angular momentum of a planet in orbit is:

$$L = m R \sqrt{\frac{G M}{R}}$$

 $$L = m \sqrt{G M R}$$ 

Step 3: Ratio of Angular Momenta

For Planet A:

$$L_A = m_A \sqrt{G M R_A}$$

For Planet B:

$$L_B = m_B \sqrt{G M R_B}$$

Given that:

$R_B = 2 R_A$,

$m_B = 4\sqrt{2} \times m_A$.

Substituting:

$$L_B = (4\sqrt{2} m_A) \sqrt{G M (2 R_A)}$$

 $$L_B = 4\sqrt{2} m_A \cdot \sqrt{2} \cdot \sqrt{G M R_A}$$ 

$$L_B = 4 \times 2 \times m_A \sqrt{G M R_A}$$ 

$$L_B = 8 L_A$$ 

Final Answer

$$\mathbf{8}$$