MHT CET · Physics · Gravitation
A satellite is revolving around a planet in a circular orbit close to its surface. Let ' \(\rho\) ' be the mean density and ' \(R\) ' be the radius of the planet. Then the period of the satellite is ( \(\mathrm{G}=\) universal constant of gravitation)
- A \(\sqrt{\frac{4 \pi}{\rho G}}\)
- B \(\sqrt{\frac{\pi}{\rho G}}\)
- C \(\sqrt{\frac{3 \pi}{\rho G}}\)
- D \(\sqrt{\frac{2 \pi}{\rho G}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{3 \pi}{\rho G}}\)
Step-by-step Solution
Detailed explanation
From Kepler's third law,
\(\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \\
& \therefore \quad \mathrm{~T} =2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}}}
\end{aligned}\)
As the satellite is very close to the planet, \(r=R\).
\(\therefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}}...(i)\)
\(\begin{aligned}
& \text {We know, } \text {Mass } =\text { Volume } \times \text { Density }(\rho) \\
& =\frac{4}{3} \pi \mathrm{R}^3 \times \rho...(ii)
\end{aligned}\)
Putting (ii) into (i)
\(T=2 \pi \sqrt{\frac{R^3}{G \times \frac{4}{3} \pi R^3 \rho}}=\sqrt{\frac{3 \pi}{\rho G}}\)
\(\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \\
& \therefore \quad \mathrm{~T} =2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}}}
\end{aligned}\)
As the satellite is very close to the planet, \(r=R\).
\(\therefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}}...(i)\)
\(\begin{aligned}
& \text {We know, } \text {Mass } =\text { Volume } \times \text { Density }(\rho) \\
& =\frac{4}{3} \pi \mathrm{R}^3 \times \rho...(ii)
\end{aligned}\)
Putting (ii) into (i)
\(T=2 \pi \sqrt{\frac{R^3}{G \times \frac{4}{3} \pi R^3 \rho}}=\sqrt{\frac{3 \pi}{\rho G}}\)
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