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(C) From Kepler's third law,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{r^3}{GM}}$.Since the satellite is revolving close to the surface of

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$\sqrt{\frac{3 \pi}{\rho G}}$

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(C) From Kepler's third law,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{r^3}{GM}}$.Since the satellite is revolving close to the surface of the planet,we take the orbital radius $r = R$.Thus,$T = 2 \pi \sqrt{\frac{R^3}{GM}} \dots (i)$.The mass $M$ of the planet can be expressed in terms of its density $ho$ and radius $R$ as $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3} \pi R^3 ho \dots (ii)$.Substituting the value of $M$ from equation $(ii)$ into equation $(i)$:$T = 2 \pi \sqrt{\frac{R^3}{G 	imes \frac{4}{3} \pi R^3 ho}}$$T = 2 \pi \sqrt{\frac{3}{4 \pi G ho}}$$T = 2 \pi 	imes \frac{1}{2} \sqrt{\frac{3}{\pi G ho}}$$T = \sqrt{\frac{4 \pi^2 	imes 3}{4 \pi G ho}} = \sqrt{\frac{3 \pi}{ho G}}$.

$A$ 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 ($G=$ universal constant of gravitation).

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