In a distant solar system, the radius of Planet A's circular orbit is 9 times the radius of Planet B

Question

In a distant solar system, the radius of Planet A's circular orbit is 9 times the radius of Planet B's orbit. If the length of a year on Planet A is 729 Earth days, what is the length of a year on Planet B?

Professor PC Thomas Classes · Answer

(a) : According to Kepler's Third Law, the square of the time period is directly proportional to the cube of the orbital radius, $T^2 \propto R^3$. Therefore, $\left(\frac{T_B}{T_A}\right)^2 = \left(\frac{R_B}{R_A}\right)^3$. We are given $R_A = 9 R_B$, so $\frac{R_B}{R_A} = \frac{1}{9}$. Substituting this gives $\left(\frac{T_B}{729}\right)^2 = \left(\frac{1}{9}\right)^3 = \frac{1}{729}$. This means $\frac{T_B^2}{729^2} = \frac{1}{729}$, which simplifies to $T_B^2 = \frac{729^2}{729} = 729$. Taking the square root, $T_B = \sqrt{729} = 27$ Earth days.