Two planets orbit a star in circular paths with radii $R$ and $4R$, respectively. At a specific time, the two

Question

Two planets orbit a star in circular paths with radii $R$ and $4R$, respectively. At a specific time, the two planets and the star are aligned in a straight line. If the orbital period of the planet closest to the star is $T$, what is the minimum time after which the star and the planets will again be aligned in a straight line (at the initial position)?

Accepted Answer

1. **Kepler's Third Law:** According to the Law of Periods, the square of the time period of revolution is proportional to the cube of the semi-major axis (radius for circular orbits). Formula: $T^2 \propto R^3$ or $T \propto R^{3/2}$.
2. **Calculate Period of Outer Planet ($T_2$):**
   Let $T_1 = T$ and $R_1 = R$. Let $T_2$ be the period of the planet at $R_2 = 4R$.
   $$ \frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2} = \left( \frac{4R}{R} \right)^{3/2} = 4^{3/2} = (2^2)^{3/2} = 2^3 = 8 $$
   $$ T_2 = 8T_1 = 8T $$
3. **Alignment Analysis:** 
   - The inner planet completes an orbit in time $T$. The outer planet completes an orbit in time $8T$.
   - The question asks for the time to align again. While they technically align at the synodic period ($S = \frac{T_1 T_2}{T_2 - T_1} = \frac{8}{7}T$), this option is not available. 
   - Checking the available options, at time $t = 8T$, the inner planet completes exactly 8 revolutions and the outer planet completes exactly 1 revolution. Both planets return to their *initial* positions simultaneously, aligning with the star exactly as they started. Thus, $8T$ is the correct answer in the context of the given options.

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