Two planets $$A$$ and $$B$$ of equal mass are having their period of revolutions $$T_A$$ and $$T_B$$ such that $$T_A = 2T_B$$. These planets are revolving in the circular orbits of radii $$r_A$$ and $$r_B$$ respectively. Which out of the following would be the correct relationship of their orbits?

We are given two planets A and B with equal mass, revolving in circular orbits with periods $$T_A = 2T_B$$. According to Kepler’s third law, $$T^2 \propto r^3$$, or more precisely $$T^2 = \frac{4\pi^2}{GM}r^3$$, so for the two planets we have $$\frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$.

Substituting $$T_A = 2T_B$$ into this relation gives $$\frac{(2T_B)^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$, which simplifies to $$\frac{4T_B^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$, or $$4 = \frac{r_A^3}{r_B^3}$$. Therefore, $$r_A^3 = 4r_B^3$$. The correct answer is Option C.