Question

The period of revolution of planet A around the sun is 8 times that of planet B. How many times the distance of A from the sun is greater than that of B from the sun?

• A. 5 times
• B. 2 times
• C. 3 times
• D. 6 times

Hint:

Be careful with the wording "greater than"! First, find the total scale factor using Kepler's law: $8 \rightarrow \text{square it} \rightarrow 64 \rightarrow \text{cube root} \rightarrow 4$. Since its total distance is 4 times as large, it is exactly $4 - 1 = 3$ times greater than the baseline value.

Answer 1

The Correct Option is C

Step 1: State Kepler's third law.
The square of an orbital period is proportional to the cube of the mean orbital radius: $T^2 \propto R^3$.
Step 2: Write the ratio form.
For two planets, \[ \left(\frac{R_A}{R_B}\right)^3 = \left(\frac{T_A}{T_B}\right)^2. \]
Step 3: Insert the period ratio.
Given $T_A = 8T_B$, so $\dfrac{T_A}{T_B} = 8$ and \[ \left(\frac{R_A}{R_B}\right)^3 = 8^2 = 64. \]
Step 4: Take the cube root.
\[ \frac{R_A}{R_B} = \sqrt[3]{64} = 4, \] so $R_A = 4R_B$.
Step 5: Read the question wording.
It asks how many times greater $A$'s distance is, so we compare the extra distance to $R_B$: \[ \frac{R_A - R_B}{R_B} = \frac{4R_B - R_B}{R_B} = 3. \]
Step 6: Conclude.
Planet $A$ is $3$ times further out than $B$, option (C). \[ \boxed{3\ \text{times}} \]