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A satellite can be in a geostationary orbit around a planet if it is at a distance $R$ from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
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Verified Answer
The correct answer is:
$\frac{R}{4^{1 / 3}}$
We know that, angular velocity,
$$
\begin{aligned}
\omega &=\frac{2 \pi}{T} \\
\Rightarrow \quad T &=\frac{2 \pi}{\omega} \\
\Rightarrow \quad \frac{T_{2}}{T_{1}} &=\frac{\omega_{1}}{\omega_{2}} \\
&=\frac{\omega_{1}}{2 \omega_{1}} \quad\left[\because \omega_{2}=2 \omega_{1}\right]
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{T_{2}}{T_{1}}=\frac{1}{2} \\
\Rightarrow & T_{2}=\frac{T_{1}}{2}
\end{array}
$$
According to Kepler's law,
$$
\begin{aligned}
\Rightarrow \quad & T^{2} \propto R^{3} \\
\Rightarrow \quad \frac{R_{2}}{R_{1}} &=\left(\frac{T_{2}}{T_{1}}\right)^{2 / 3} \\
&=\left(\frac{T_{1} / 2}{T}\right)^{2 / 3} \\
&=\left(\frac{1}{2}\right)^{2 / 3} \\
R_{2} &=\frac{R_{1}}{2^{2 / 3}} \\
&=\frac{R}{4^{1 / 3}}
\end{aligned}
$$
$$
\begin{aligned}
\omega &=\frac{2 \pi}{T} \\
\Rightarrow \quad T &=\frac{2 \pi}{\omega} \\
\Rightarrow \quad \frac{T_{2}}{T_{1}} &=\frac{\omega_{1}}{\omega_{2}} \\
&=\frac{\omega_{1}}{2 \omega_{1}} \quad\left[\because \omega_{2}=2 \omega_{1}\right]
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{T_{2}}{T_{1}}=\frac{1}{2} \\
\Rightarrow & T_{2}=\frac{T_{1}}{2}
\end{array}
$$
According to Kepler's law,
$$
\begin{aligned}
\Rightarrow \quad & T^{2} \propto R^{3} \\
\Rightarrow \quad \frac{R_{2}}{R_{1}} &=\left(\frac{T_{2}}{T_{1}}\right)^{2 / 3} \\
&=\left(\frac{T_{1} / 2}{T}\right)^{2 / 3} \\
&=\left(\frac{1}{2}\right)^{2 / 3} \\
R_{2} &=\frac{R_{1}}{2^{2 / 3}} \\
&=\frac{R}{4^{1 / 3}}
\end{aligned}
$$
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