Discussion
A satellite revolves around a planet in a circular orbit of radius R with the time period \(t\) . If the orbital radius is 3R, then its time period of revolution is
Answer: C
The new period is \(\sqrt{27}t\).
By Kepler's third law \(T^{2} \propto R^{3}\), so \(T \propto R^{3/2}\):
\[ \begin{aligned} \frac{T'}{t} &= \left(\frac{3R}{R}\right)^{3/2} \\ &= 3^{3/2} \\ &= \sqrt{27} \end{aligned} \]
So \(T' = \sqrt{27}t = 3\sqrt{3}t\), about 5.2 times longer. \(\sqrt{3}t\) uses \(T \propto R^{1/2}\), and \(t\) would mean the period is independent of the radius.
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