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A planet of radius R has a time period of revolution T. Find time period of a planet of radius 9 R ?
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Verified Answer
The correct answer is:
$27 T$
By Kepler's third law of planetory motion,
Given,
$\begin{aligned}
\frac{T_1}{T_2} & =\left(\frac{a_1}{a_2}\right)^{3 / 2} \\
T_1 & =T, a_1=R, T_2=?, a_2=9 R \\
\frac{T}{T_2} & =\left(\frac{R}{9 R}\right)^{3 / 2}, \frac{T}{T_2}=\left[\left(\frac{1}{3}\right)^2\right]^{3 / 2} \\
\frac{T}{T_2} & =\frac{1}{27} \Rightarrow T_2=27 \mathrm{~T}
\end{aligned}$
Given,
$\begin{aligned}
\frac{T_1}{T_2} & =\left(\frac{a_1}{a_2}\right)^{3 / 2} \\
T_1 & =T, a_1=R, T_2=?, a_2=9 R \\
\frac{T}{T_2} & =\left(\frac{R}{9 R}\right)^{3 / 2}, \frac{T}{T_2}=\left[\left(\frac{1}{3}\right)^2\right]^{3 / 2} \\
\frac{T}{T_2} & =\frac{1}{27} \Rightarrow T_2=27 \mathrm{~T}
\end{aligned}$
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