The periodic time of a communication satellite is ......... $hours$.

Why can the entire surface of the Earth be viewed with the help of a polar satellite?

Given below are two statements:
Statement $I$: $A$ satellite is moving around the Earth in an orbit very close to the Earth's surface. The time period of revolution of the satellite depends upon the density of the Earth.
Statement $II$: The time period of revolution of the satellite is $T = 2\pi\sqrt{\frac{R_e}{g}}$ (for a satellite very close to the Earth's surface), where $R_e$ is the radius of the Earth and $g$ is the acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below:

The mass density of a spherical galaxy varies as $\frac{K}{r}$ over a large distance $r$ from its centre. In that region,a small star is in a circular orbit of radius $R$. Then the period of revolution,$T$ depends on $R$ as

$A$ satellite of mass $m$ is revolving around the Earth of mass $M$ in an orbit of radius $r$. The angular momentum of the satellite about the centre of the orbit will be:

Using the data given below,find the height at which a communication satellite can reside. $(G=6.67 \times 10^{-11} \text{ N-m}^2 \text{ kg}^{-2}, M=5.98 \times 10^{24} \text{ kg}, R=6.4 \times 10^6 \text{ m})$ (in $\text{ km}$)