Radius of a geostationary satellite revolving round the earth is 'r'. Then period of revolution of another satellite revolving in an orbit of radius r/2 is

Options

A. 6 hrs
B. `6//sqrt2` hrs
C. `6sqrt2` hrs
D. 12 hrs

Explanation

To find the period of revolution of a satellite revolving in an orbit of radius \( \frac{r}{2} \) (where \( r \) is the radius of a geostationary satellite), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Time Period Formula**:
The time period \( T \) of a satellite in orbit is given by the formula:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where \( r \) is the orbital radius, \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

2. **Time Period of the Geostationary Satellite**:
For a geostationary satellite, the time period \( T_g \) is equal to the time period of the Earth's rotation, which is 24 hours. Thus:
\[ T_g = 2\pi \sqrt{\frac{r^3}{GM}} = 24 \text{ hours} \]

3. **Determine the Time Period for the New Satellite**:
Now, we need to find the time period \( T_s \) for a satellite at radius \( \frac{r}{2} \):
\[ T_s = 2\pi \sqrt{\frac{(r/2)^3}{GM}} \]
Simplifying this:
\[ T_s = 2\pi \sqrt{\frac{r^3/8}{GM}} = 2\pi \sqrt{\frac{r^3}{GM} \cdot \frac{1}{8}} \]
\[ T_s = 2\pi \sqrt{\frac{r^3}{GM}} \cdot \frac{1}{\sqrt{8}} = T_g \cdot \frac{1}{\sqrt{8}} \]

4. **Substituting the Value of \( T_g \)**:
We know \( T_g = 24 \text{ hours} \):
\[ T_s = 24 \cdot \frac{1}{\sqrt{8}} = 24 \cdot \frac{1}{2\sqrt{2}} = \frac{24}{2\sqrt{2}} = \frac{12\sqrt{2}}{2} = 12\sqrt{2} \text{ hours} \]

5. **Final Calculation**:
To express \( 12\sqrt{2} \) in hours, we can approximate \( \sqrt{2} \approx 1.414 \):
\[ T_s \approx 12 \cdot 1.414 \approx 16.97 \text{ hours} \]

Thus, the period of revolution of the satellite revolving in an orbit of radius \( \frac{r}{2} \) is approximately \( 12\sqrt{2} \) hours.