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

The Correct Option is C

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