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:
To resolve this issue, we will utilize principles of angular momentum in circular orbits and Kepler's planetary motion laws.
The angular momentum \(L\) for a planet in a circular orbit is calculated as: \(L = m \cdot v \cdot r\). Here, \(m\) represents the planet's mass, \(v\) its orbital velocity, and \(r\) the orbital radius.
In circular motion, gravitational force provides the necessary centripetal force: \(\frac{G \cdot M \cdot m}{r^2} = \frac{m \cdot v^2}{r}\). This allows us to express the orbital velocity \(v\) as: \(v = \sqrt{\frac{G \cdot M}{r}}\).
Substituting the expression for \(v\) into the angular momentum formula yields: \(L = m \cdot \sqrt{\frac{G \cdot M}{r}} \cdot r = m \cdot \sqrt{G \cdot M \cdot r}\).
We are given the following conditions: Planet A has \(L_A = L\), and Planet B has \(L_B = 3L\). Based on the angular momentum equations derived, we have: \(L_A = m_1 \cdot \sqrt{G \cdot M \cdot r_1} = L\) and \(L_B = m_2 \cdot \sqrt{G \cdot M \cdot r_2} = 3L\).
Kepler's Third Law states that the square of the orbital period \(T\) is proportional to the cube of the orbital radius \(r\): \(T^2 \propto r^3\). Consequently: \(\left(\frac{T_A}{T_B}\right)^2 = \frac{r_1^3}{r_2^3}\). Taking the square root yields: \(\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{\frac{3}{2}}\).
From the expression derived in Step 5, we find: \(\frac{r_2}{r_1} = \left(\frac{3 \cdot m_1}{m_2}\right)^2\). Therefore: \(\left(\frac{r_1}{r_2}\right)^{\frac{3}{2}} = \left(\frac{m_1}{3 \cdot m_2}\right)^3 = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3\).
Conclusion: Our analysis indicates that the ratio of the time periods is \(\frac{T_A}{T_B} = \frac{1}{27} \left(\frac{m_2}{m_1}\right)^3\). The correct answer is therefore: \(\frac{1}{27} \left(\frac{m_2}{m_1}\right)^3\).
