Discussion
The period of revolution of the planet \(A\) around the sun is 27 times that of another planet \(B\). If the distance of \(A\) from the sun is \(x\) times greater than that of \(B\) from the sun, then the value of \(x\) is
Answer: C
\(x = \) 9.
By Kepler's third law, \(T^{2} \propto r^{3}\):
\[ \begin{aligned} \left(\frac{r_{A}}{r_{B}}\right)^{3} &= \left(\frac{T_{A}}{T_{B}}\right)^{2} = 27^{2} = 729 \\ \frac{r_{A}}{r_{B}} &= \sqrt[3]{729} = 9 \end{aligned} \]
3 is the cube root of 27, which forgets to square the period ratio first; 8, 4 and 12 do not satisfy the law.
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