A satellite moves in a circular orbit around the earth at height `(R_(e))//2` from earth's surface where `R_(e)` is the radius of the earth. Calculate its period of revolution. Given `R = 6.38 xx 10^(6) m`.

Explanation

To find the period of revolution of a satellite moving in a circular orbit at a height of \( \frac{R_e}{2} \) from the Earth's surface, we can follow these steps:

### Step 1: Determine the radius of the orbit
The radius of the Earth \( R_e \) is given as \( 6.38 \times 10^6 \) m. The height \( h \) of the satellite above the Earth's surface is \( \frac{R_e}{2} \).

The radius \( r \) of the satellite's orbit is:
\[
r = R_e + h = R_e + \frac{R_e}{2} = R_e \left(1 + \frac{1}{2}\right) = \frac{3R_e}{2}
...