MHT CET · Physics · Gravitation
If \(\rho\) is the density of the planet, the time period of nearby satellite is given by
- A \(\sqrt{\frac{4 \pi}{3 G \rho}}\)
- B \(\sqrt{\frac{4 \pi}{G \rho}}\)
- C \(\sqrt{\frac{3 \pi}{G \rho}}\)
- D \(\sqrt{\frac{\pi}{G \rho}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{3 \pi}{G \rho}}\)
Step-by-step Solution
Detailed explanation
Time period of nearby satellite
\(
\begin{aligned}
T &=2 \pi \sqrt{\frac{r^{3}}{G M}} \\
&=2 \pi \sqrt{\frac{R^{3}}{G M}} \\
&=\frac{2 \pi\left(R^{3}\right)^{1 / 2}}{\left[G \cdot \frac{4}{3} \pi R^{3} \rho\right]^{1 / 2}} \\
&=\sqrt{\frac{3 \pi}{G \rho}}
\end{aligned}
\)
\(
\begin{aligned}
T &=2 \pi \sqrt{\frac{r^{3}}{G M}} \\
&=2 \pi \sqrt{\frac{R^{3}}{G M}} \\
&=\frac{2 \pi\left(R^{3}\right)^{1 / 2}}{\left[G \cdot \frac{4}{3} \pi R^{3} \rho\right]^{1 / 2}} \\
&=\sqrt{\frac{3 \pi}{G \rho}}
\end{aligned}
\)
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