Gravitation Part 2

Concept:For a satellite orbiting very close to Earth's surface, the time period $T$ depends only on Earth's density $\rho$ and is given by $T = 2\pi \sqrt{\frac{R_e}{g}}$, which is derived from Kepler's law and the relation $g = \frac{GM_e}{R_e^2}$.Explanation:We start from the equilibrium of gravitational force and centripetal force: $\frac{GM_e m}{r^2} = \frac{m v^2}{r}$This gives $v = \sqrt{\frac{GM_e}{r}}$Time period $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{GM_e}}$For a satellite very close to Earth, $r \approx R_e$, so $T \approx 2\pi \sqrt{\frac{R_e^3}{GM_e}}$Assuming Earth is a uniform sphere, $M_e = \frac{4}{3}\pi R_e^3 \rho$Substituting: $T = 2\pi \sqrt{\frac{R_e^3}{G \cdot \frac{4}{3}\pi R_e^3 \rho}} = 2\pi \sqrt{\frac{3}{4\pi G \rho}} = \sqrt{\frac{3\pi}{G \rho}}$Since $3, \pi, G$ are constants, $T$ depends only on density $\rho$. Thus Statement I is true.Also, using $g = \frac{GM_e}{R_e^2}$, we get $GM_e = g R_e^2$Substitute into $T = 2\pi \sqrt{\frac{R_e^3}{GM_e}}$ gives $T = 2\pi \sqrt{\frac{R_e^3}{g R_e^2}} = 2\pi \sqrt{\frac{R_e}{g}}$, confirming Statement II.Both statements are correct.Answer:Option C: Both Statement I and Statement II are true.