Question

If a planet at a distance $r$ from the sun takes 200 days to complete one revolution, what will be the orbital period of a planet at a distance $\frac{r}{4}$ from the sun?

• A. 50 days
• B. 25 days
• C. 100 days
• D. 12.5 days

Answer 1

Answer: (B)

25 days

Answer 2

According to Kepler's Third Law of Planetary Motion, the square of the orbital period ($T$) is proportional to the cube of the semi-major axis of the orbit ($r$). Mathematically, $T^2 \propto r^3$. Let $T_1$ and $r_1$ be the period and distance for the first planet, and $T_2$ and $r_2$ for the second planet. Given: $T_1 = 200$ days, $r_1 = r$, and $r_2 = \frac{r}{4}$. We can write the relationship as: $\frac{T_2^2}{T_1^2} = \left(\frac{r_2}{r_1}\right)^3$ Substituting the given values: $\frac{T_2^2}{(200 \text{ days})^2} = \left(\frac{r/4}{r}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$ $T_2^2 = \frac{(200 \text{ days})^2}{64}$ $T_2 = \sqrt{\frac{(200 \text{ days})^2}{64}} = \frac{200 \text{ days}}{8} = 25 \text{ days}$