MHT CET · Physics · Gravitation
A satellite is orbiting just above the surface of the planet of density ' \(\rho\) ' with periodic time ' \(T\) '. The quantity \(\mathrm{T}^2 \rho\) is equal to ( \(\mathrm{G}=\) universal gravitational constant)
- A \(\frac{4 \pi^2}{\mathrm{G}}\)
- B \(\frac{3 \pi^2}{\mathrm{G}}\)
- C \(\frac{3 \pi}{\mathrm{G}}\)
- D \(\frac{\pi}{G}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \pi}{\mathrm{G}}\)
Step-by-step Solution
Detailed explanation
Time period of a nearby satellite is given by,
\(\begin{aligned}
& T=2 \pi \sqrt{\frac{R}{g}} \\
\therefore & T^2=4 \pi^2 \times \frac{R}{g} \\
& \text { But, } g=\frac{4}{3} \pi \rho G R . \\
\therefore & T^2=\frac{4 \pi^2 R}{\frac{4}{3} \pi \rho G R} \\
\therefore & T^2 \rho=\frac{3 \pi}{G}
\end{aligned}\)
\(\begin{aligned}
& T=2 \pi \sqrt{\frac{R}{g}} \\
\therefore & T^2=4 \pi^2 \times \frac{R}{g} \\
& \text { But, } g=\frac{4}{3} \pi \rho G R . \\
\therefore & T^2=\frac{4 \pi^2 R}{\frac{4}{3} \pi \rho G R} \\
\therefore & T^2 \rho=\frac{3 \pi}{G}
\end{aligned}\)
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