A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.

Options

A. `3sqrt(2)h`
B. `1.5sqrt(2)h`
C. `6sqrt(2)h`
D. `12sqrt(2)h`

Explanation

To solve the problem, we need to find the time period of a satellite that is orbiting at a height of 2.5R above the Earth's surface, given that a geostationary satellite orbits at a height of 6R above the surface.

### Step-by-step Solution:

1. **Identify the height of the geostationary satellite (Hg)**:
\[ H_g = 6R \]
where \( R \) is the radius of the Earth.

2. **Identify the height of the satellite for which we need to find the time period (Hs)**:
\[ H_s = 2.5R \]

3. **Calculate the distance from the center of the Earth to the geostationary satellite (Rg)**:
\[ R_g = R + H_g = R + 6R = 7R \]

4. **Calculate the distance from the center of the Earth to the satellite (Rs)**:
\[ R_s = R + H_s = R + 2.5R = 3.5R \]

5. **Use Kepler's Third Law**:
According to Kepler's Third Law, the square of the time period (T) of a satellite is directly proportional to the cube of the semi-major axis (R) of its orbit:
\[ T^2 \propto R^3 \]
This can be expressed as:
\[ \frac{T_g^2}{T_s^2} = \frac{R_g^3}{R_s^3} \]

6. **Substituting the known values**:
The time period of the geostationary satellite \( T_g \) is 24 hours:
\[ \frac{(24 \text{ hours})^2}{T_s^2} = \frac{(7R)^3}{(3.5R)^3} \]

7. **Simplifying the right side**:
\[ \frac{(7R)^3}{(3.5R)^3} = \frac{343R^3}{42.875R^3} = \frac{343}{42.875} = 8 \]

8. **Setting up the equation**:
\[ \frac{576}{T_s^2} = 8 \]

9. **Solving for \( T_s^2 \)**:
\[ T_s^2 = \frac{576}{8} = 72 \]

10. **Taking the square root to find \( T_s \)**:
\[ T_s = \sqrt{72} = 6\sqrt{2} \text{ hours} \]

### Final Answer:
The time period of the satellite at a height of 2.5R above the surface of the Earth is \( 6\sqrt{2} \) hours.