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 $$(L_{B})$$ of planet B to that of planet $$A(L_{A})$$ is closest to integer ________.

The angular momentum of a planet in circular orbit is $$L = mv r$$, where $$v$$ is the orbital velocity.

For a circular orbit we have $$\frac{GMm}{R^2} = \frac{mv^2}{R}$$, so $$v = \sqrt{\frac{GM}{R}}$$.

Substituting this into the expression for angular momentum gives $$L = mR\sqrt{\frac{GM}{R}} = m\sqrt{GMR}$$.

We then consider the ratio of the angular momenta of bodies B and A:

$$ \frac{L_B}{L_A} = \frac{m_B\sqrt{GMR_B}}{m_A\sqrt{GMR_A}} = \frac{m_B}{m_A}\sqrt{\frac{R_B}{R_A}} $$

Using the given relations $$m_B = 4\sqrt{2} \cdot m_A$$ and $$R_B = 2R_A$$, we find

$$ \frac{L_B}{L_A} = 4\sqrt{2} \times \sqrt{2} = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8 $$

The answer is 8.