Given below are two statements:
Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II: The time period of revolution of the satellite is $$T= 2\pi \sqrt{\frac{R_{e}}{g}}$$ (for satellite very close to the earth surface), where $$R_{e}$$ radius of earth and g acceleration due to gravity. 

In the light of the above statements , choose the correct answer from the options given below:

Statement 1:

To find the dependence on density ($$\rho$$), we substitute Earth's mass ($$M_e$$) using the volume of a sphere:

$$M_e = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R_e^3 \rho$$

$$T = 2\pi \sqrt{\frac{R_e^3}{G \left( \frac{4}{3} \pi R_e^3 \rho \right)}}$$

$$T = 2\pi \sqrt{\frac{3}{4\pi G \rho}} = \sqrt{\frac{3\pi}{G \rho}}$$

The expression shows that the time period $$T$$ is inversely proportional to the square root of the Earth's density. Therefore, Statement I is true.

Statement 2:

The orbital time period $$T$$ for a satellite at a radius $$r$$ is given by: $$T = 2\pi \sqrt{\frac{r^3}{GM_e}}$$

For a satellite orbiting very close to the Earth's surface, the orbital radius $$r$$ is approximately equal to the Earth's radius $$R_e$$.

$$T = 2\pi \sqrt{\frac{R_e^3}{GM_e}}$$

Using $$GM_e = g R_e^2$$: $$T = 2\pi \sqrt{\frac{R_e^3}{g R_e^2}} = 2\pi \sqrt{\frac{R_e}{g}}$$

Thus, Statement II is true.