Question

From Kepler's third law, the square of the period of revolution ($T$) is proportional to the cube of the semi-major axis ($R$). If a planet has $T_{p = 27T_{e}$, then:

• A. $R_{p} = 3R_{e}$
• B. $R_{p} = 9R_{e}$
• C. $R_{p} = 27R_{e}$
• D. $R_{p} = R_{e}$

Hint:

$R \propto T^{2/3}$. If time increases by a factor of 27, radius increases by $27^{2/3} = 9$.

Answer 1

The Correct Option is B

Step 1: Concept
Kepler's Third Law: $T^{2} \propto R^{3}$.
Step 2: Analysis
$\frac{T_{e}^{2}}{T_{p}^{2}} = \frac{R_{e}^{3}}{R_{p}^{3}}$. Given $T_{p} = 27T_{e}$, then $\frac{T_{e}^{2}}{(27T_{e})^{2}} = \frac{R_{e}^{3}}{R_{p}^{3}}$. $\frac{R_{p}}{R_{e}} = (27^{2})^{1/3} = (3^{6})^{1/3} = 3^{2}$.
Step 3: Conclusion
Therefore, $R_{p} = 9R_{e}$.
Final Answer: (B)