Q 11-07-070JEE MainJEE Main 2024 (8 Apr, Shift 1)Hard
Two planets $A$ and $B$ having masses $m_1$ and $m_2$ move around the sun in circular orbits of radii $r_1$ and $r_2$ respectively. If angular momentum of $A$ is $L$ and that of $B$ is $3L$, the ratio of time periods $\left(\dfrac{T_A}{T_B}\right)$ is:
Answer: (B) $\dfrac{1}{27}\left(\dfrac{m_2}{m_1}\right)^3$
In a circular orbit $v = \sqrt{GM/r}$, so $L = mvr = m\sqrt{GMr}$.
$$\frac{L_B}{L_A} = \frac{m_2}{m_1}\sqrt{\frac{r_2}{r_1}} = 3 \;\Rightarrow\; \frac{r_1}{r_2} = \frac19\left(\frac{m_2}{m_1}\right)^2$$
By Kepler's law $T \propto r^{3/2}$:
$$\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^3$$
Solution by Sreeraj P, M.Sc Physics