Question

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 the angular momentum of A is $L$ and that of B is $3L$, the ratio of time periods $\frac{T_A}{T_B}$ is:

• A. $\left(\frac{r_2}{r_1}\right)^{\frac{3}{2}}$
• B. $\left(\frac{r_1}{r_2}\right)^3$
• C. $\frac{1}{27} \left(\frac{m_2}{m_1}\right)^3$
• D. $27 \left(\frac{m_1}{m_2}\right)^3$

Answer 1

Answer: (C)

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

Answer 2

Solution: For a circular orbit:

$\pi r^2 \propto \frac{L}{m}.$

For planet A:
$\pi r_1^2 \cdot T_A \propto \frac{L}{2m_1}.$

For planet B:
$\pi r_2^2 \cdot T_B \propto \frac{3L}{2m_2}.$

Taking the ratio of time periods:
$\frac{T_A}{T_B} = \frac{m_2}{m_1} \cdot \left(\frac{r_1}{r_2}\right)^2.$

Squaring both sides:
$\left(\frac{T_A}{T_B}\right)^2 = \frac{m_2^2}{m_1^2} \cdot \left(\frac{r_1}{r_2}\right)^4.$

Taking the cube root:
$\frac{T_A}{T_B} = \frac{1}{27} \cdot \left(\frac{m_2}{m_1}\right)^3.$

Final Answer: $\frac{1}{27} \cdot \left(\frac{m_2}{m_1}\right)^3$.