Solution:
According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).
According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).
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