Two planets $$A$$ and $$B$$ having masses $$m_1$$ and $$m_2$$ move around the sun in circular orbits of $$r_1$$ and $$r_2$$ radii respectively. If angular momentum of $$A$$ is $$L$$ and that of $$B$$ is $$3L$$, the ratio of time period $$\left(\frac{T_A}{T_B}\right)$$ is:

For a planet of mass $$m$$ moving in a circular orbit of radius $$r$$ around the Sun (mass $$M_\odot$$), the necessary centripetal force is provided by gravitation:

$$\frac{G M_\odot m}{r^{2}} = \frac{m v^{2}}{r}$$

Solving for the orbital speed,

$$v = \sqrt{\frac{G M_\odot}{r}} \quad -(1)$$

Angular momentum about the Sun is

$$L = m v r$$

Substituting $$v$$ from $$(1)$$,

$$L = m r \sqrt{\frac{G M_\odot}{r}} = m \sqrt{G M_\odot\, r} \quad -(2)$$

Applying $$(2)$$ to planets $$A$$ and $$B$$:

$$L_A = m_1 \sqrt{G M_\odot\, r_1}$$
$$L_B = m_2 \sqrt{G M_\odot\, r_2}$$

Given $$L_B = 3L_A$$,

$$m_2 \sqrt{G M_\odot\, r_2} = 3 m_1 \sqrt{G M_\odot\, r_1}$$

Canceling the common factor $$\sqrt{G M_\odot}$$,

$$m_2 \sqrt{r_2} = 3 m_1 \sqrt{r_1} \quad -(3)$$

From $$(3)$$,

$$\frac{\sqrt{r_2}}{\sqrt{r_1}} = \frac{3 m_1}{m_2}$$
Squaring both sides: $$\frac{r_2}{r_1} = \frac{9 m_1^{2}}{m_2^{2}} \quad -(4)$$

Kepler’s third law for circular orbits gives the time period:

$$T = 2\pi \sqrt{\frac{r^{3}}{G M_\odot}} \propto r^{3/2} \quad -(5)$$

Therefore,

$$\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} \quad -(6)$$

Insert $$\frac{r_1}{r_2}$$ from $$(4)$$ into $$(6)$$:

$$\frac{T_A}{T_B} = \left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2}$$

Compute the exponent:

$$\left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2} = \left(\frac{m_2}{m_1}\right)^{3} \left(\frac{1}{9}\right)^{3/2}$$

Since $$9 = 3^{2}$$, we have $$9^{3/2} = 3^{3} = 27$$. Thus

$$\frac{T_A}{T_B} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$$

Hence, the correct ratio is $$\frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$$, which corresponds to Option B.