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Question

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 :

The correct answer is
Statement I is true but Statement II is false

Statement I Analysis: Satellite Time Period and Earth Density

For a satellite orbiting very close to the Earth's surface, the orbital radius $r$ is approximately equal to the Earth's radius $R_e$. The time period $T$ is given by the formula:

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

where $M_e$ is the mass of the Earth and $G$ is the gravitational constant.

We know the acceleration due to gravity at the surface is $g = \frac{GM_e}{R_e^2}$, which implies $GM_e = gR_e^2$. Substituting this into the time period formula for $r \approx R_e$:

$T \approx 2\pi \sqrt{\frac{R_e^3}{gR_e^2}} = 2\pi \sqrt{\frac{R_e}{g}}$

Now, let's express $g$ in terms of Earth's density ($\rho_e$). The mass of the Earth is $M_e = \rho_e \times (\frac{4}{3}\pi R_e^3)$.

So, $g = \frac{G(\rho_e \frac{4}{3}\pi R_e^3)}{R_e^2} = \frac{4}{3}\pi G \rho_e R_e$.

Substituting this expression for $g$ back into the time period formula:

$T \approx 2\pi \sqrt{\frac{R_e}{\frac{4}{3}\pi G \rho_e R_e}} = 2\pi \sqrt{\frac{3}{4\pi G \rho_e}}$

This derivation shows that the time period $T$ is inversely proportional to the square root of the Earth's density ($\rho_e$). Therefore, Statement I is true.

Statement II Analysis: Satellite Time Period Formula

Statement II provides the formula $T = 2\pi \sqrt{\frac{R_e}{g}}$ for a satellite very close to the Earth's surface.

As derived above, this formula is an approximation obtained by setting the orbital radius $r$ equal to the Earth's radius $R_e$ in the general time period equation $T = 2\pi \sqrt{\frac{r^3}{GM_e}}$.

However, for any satellite in orbit, the orbital radius $r$ must be strictly greater than the Earth's radius $R_e$ ($r > R_e$). Even for orbits "very close" to the surface, $r$ is slightly larger than $R_e$. Using $R_e$ directly instead of $r$ makes the formula an approximation, not a strictly accurate representation for an orbit.

Because the formula uses $R_e$ instead of the actual orbital radius $r$, it is considered technically inaccurate. Therefore, Statement II is false.

Conclusion

Based on the analysis, Statement I is true, and Statement II is false.

The correct option is: Statement I is true but Statement II is false.

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