Two planets A and B having masses m_1 and m_2 | vedclass

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(C) For a planet in a circular orbit,the angular momentum $L$ is given by $L = mvr = m(r\omega)r = mr^2\left(\frac{2\pi}{T}\right)$.Thus,$\frac{L}{m}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a planet in a circular orbit,the angular momentum $L$ is given by $L = mvr = m(r\omega)r = mr^2\left(\frac{2\pi}{T}\right)$.Thus,$\frac{L}{m} = \frac{2\pi r^2}{T}$,which implies $\frac{T}{r^2} = \frac{2\pi m}{L}$.For planet $A$: $\frac{T_A}{r_1^2} = \frac{2\pi m_1}{L}$.For planet $B$: $\frac{T_B}{r_2^2} = \frac{2\pi m_2}{3L}$.Dividing the two equations: $\frac{T_A}{T_B} \cdot \left(\frac{r_2}{r_1}\right)^2 = \frac{m_1}{m_2} \cdot 3 \implies \frac{T_A}{T_B} = 3 \left(\frac{m_1}{m_2}\right) \left(\frac{r_1}{r_2}\right)^2$.From Kepler's Third Law,$T^2 \propto r^3$,so $\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{r_1}{r_2}\right)^3$,which means $\left(\frac{r_1}{r_2}\right)^2 = \left(\frac{T_A}{T_B}\right)^{4/3}$.Substituting this into the expression: $\frac{T_A}{T_B} = 3 \left(\frac{m_1}{m_2}\right) \left(\frac{T_A}{T_B}\right)^{4/3}$.Rearranging: $\left(\frac{T_A}{T_B}\right)^{-1/3} = 3 \frac{m_1}{m_2} \implies \frac{T_A}{T_B} = \left(3 \frac{m_1}{m_2}\right)^{-3} = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3$.

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:

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