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

To solve this problem, we need to use the concepts of angular momentum in circular orbits and Kepler's laws of planetary motion.

1. The angular momentum \(L\) of a planet moving in a circular orbit is given by: 
   \(L = m \cdot v \cdot r\) 
   Where \(m\) is the mass of the planet, \(v\) is the orbital velocity, and \(r\) is the radius of the orbit.
2. For circular motion, the centripetal force is provided by gravitational force. Thus: 
   \(\frac{G \cdot M \cdot m}{r^2} = \frac{m \cdot v^2}{r}\) 
   From this, the orbital velocity \(v\) can be expressed as: 
   \(v = \sqrt{\frac{G \cdot M}{r}}\)
3. Substitute the expression for \(v\) into the angular momentum formula: 
   \(L = m \cdot \sqrt{\frac{G \cdot M}{r}} \cdot r = m \cdot \sqrt{G \cdot M \cdot r}\)
4. We are given: 
   - Planet A: \(L_A = L\) 
   - Planet B: \(L_B = 3L\) 
   From the angular momentum expressions, we have: 
   \(L_A = m_1 \cdot \sqrt{G \cdot M \cdot r_1} = L\) 
   \(L_B = m_2 \cdot \sqrt{G \cdot M \cdot r_2} = 3L\)
5. Divide the expressions of angular momentum: 
   \(\frac{m_2 \cdot \sqrt{r_2}}{m_1 \cdot \sqrt{r_1}} = 3\) 
   On rearranging: 
   \(m_2 \cdot \sqrt{r_2} = 3 \cdot m_1 \cdot \sqrt{r_1}\)
6. Kepler's Third Law states that the square of the time period \(T\) of orbit is proportional to the cube of the radius \(r\): 
   \(T^2 \propto r^3\) 
   Therefore: 
   \(\left(\frac{T_A}{T_B}\right)^2 = \frac{r_1^3}{r_2^3}\) 
   Taking the square root: 
   \(\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{\frac{3}{2}}\)
7. From the expression in Step 5: 
   \(\frac{r_2}{r_1} = \left(\frac{3 \cdot m_1}{m_2}\right)^2\) 
   Thus: 
   \(\left(\frac{r_1}{r_2}\right)^{\frac{3}{2}} = \left(\frac{m_1}{3 \cdot m_2}\right)^3 = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3\)

Conclusion: Based on our analysis, the ratio of time periods is given by \(\frac{T_A}{T_B} = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3\). Hence, the correct answer is: \(\frac{1}{27} \left(\frac{m_2}{m_1}\right)^3\).

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