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An earth's satellite nejar the surface of the earth takes about 90 min per revolution. A satellite orbiting the moon also takes about 90 min per revolution. Then which of the following is true?
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Verified Answer
The correct answer is:
\(\rho_{\mathrm{m}}=\rho_{\mathrm{e}}\)
Hint : 
\(m\left(E_g\right)=m \omega^2 R\)
\(\begin{aligned}
& m \cdot 4 \pi \frac{G \rho R}{3}=m \omega^2 R \\
& \Rightarrow \omega^2 \propto \rho \\
& \Rightarrow T \propto \frac{1}{\sqrt{\rho}}
\end{aligned}\)
If \(\mathrm{T}\) are equal, so will be \(\rho\)
\(m\left(E_g\right)=m \omega^2 R\)
\(\begin{aligned}
& m \cdot 4 \pi \frac{G \rho R}{3}=m \omega^2 R \\
& \Rightarrow \omega^2 \propto \rho \\
& \Rightarrow T \propto \frac{1}{\sqrt{\rho}}
\end{aligned}\)
If \(\mathrm{T}\) are equal, so will be \(\rho\)
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