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A geostationary satellite is orbiting the earth at a height of $5 R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2 R$ from the surface of the earth is
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Verified Answer
The correct answer is:
$6 \sqrt{2}$
From Keplar third's law
$T^2 \propto r^3$
Hence, $T_1^2 \propto r_1^3$
So,
$\begin{aligned}
& T_2^2 \propto r_2^3 \\
& \frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3} \\
&=\frac{(3 R)^3}{(6 R)^3}
\end{aligned}$
$\begin{aligned}
& \text {or } \frac{T_2^2}{T_1^2}=\frac{1}{8} \\
& T_2^2=\frac{1}{8} T_1^2 \\
& T_2=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{~h}
\end{aligned}$
$T^2 \propto r^3$
Hence, $T_1^2 \propto r_1^3$
So,
$\begin{aligned}
& T_2^2 \propto r_2^3 \\
& \frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3} \\
&=\frac{(3 R)^3}{(6 R)^3}
\end{aligned}$
$\begin{aligned}
& \text {or } \frac{T_2^2}{T_1^2}=\frac{1}{8} \\
& T_2^2=\frac{1}{8} T_1^2 \\
& T_2=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{~h}
\end{aligned}$
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