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Question: Answered & Verified by Expert
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
PhysicsGravitationCOMEDKCOMEDK 2020
Options:
  • A $2 R$
  • B $\frac{R}{2}$
  • C $\frac{R}{2^{1 / 3}}$
  • D $\frac{R}{4^{1 / 3}}$
Solution:
1756 Upvotes 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,
$$
\begin{aligned}
\Rightarrow \quad & T^{2} \propto R^{3} \\
\Rightarrow \quad \frac{R_{2}}{R_{1}} &=\left(\frac{T_{2}}{T_{1}}\right)^{2 / 3} \\
&=\left(\frac{T_{1} / 2}{T}\right)^{2 / 3} \\
&=\left(\frac{1}{2}\right)^{2 / 3} \\
R_{2} &=\frac{R_{1}}{2^{2 / 3}} \\
&=\frac{R}{4^{1 / 3}}
\end{aligned}
$$

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