The orbit of geostationary satellite is circular, the time period of satellite depeds on (i) mass of the satellite, (ii) mass of earth, (iii) radius of the orbit and (iv) height of the satellite from the surface of the earth

Options

A. 1. (i) only
B. 2. (i) and (ii)
C. 3. (i),(ii) and (iii)
D. 4. (ii),(iii) and (iv)

Explanation

To solve the problem regarding the time period of a geostationary satellite, we will analyze the factors that influence it step by step.

### Step-by-Step Solution:

1. **Understanding the Orbit**:
A geostationary satellite orbits the Earth in a circular path. The satellite remains stationary relative to a point on the Earth's surface, meaning it has the same angular velocity as the Earth.

2. **Defining Variables**:
- Let \( M_e \) be the mass of the Earth.
- Let \( m \) be the mass of the satellite (which will cancel out later).
- Let \( r \) be the distance from the center of the Earth to the satellite. This distance is equal to the radius of the Earth (\( R_e \)) plus the height (\( h \)) of the satellite from the Earth's surface:
\[ r = R_e + h \]

3. **Setting Up the Forces**:
For a satellite in circular orbit, the gravitational force provides the necessary centripetal force to keep the satellite in orbit. Thus, we can set the gravitational force equal to the centripetal force:
\[ \frac{G M_e m}{r^2} = \frac{m v^2}{r} \]
where \( G \) is the gravitational constant and \( v \) is the orbital speed of the satellite.

4. **Cancelling Mass**:
Since the mass of the satellite \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{G M_e}{r^2} = \frac{v^2}{r} \]

5. **Solving for Velocity**:
Rearranging the equation gives us the orbital speed:
\[ v^2 = \frac{G M_e}{r} \]

6. **Finding the Time Period**:
The time period \( T \) of the satellite is related to the orbital radius and speed:
\[ T = \frac{2 \pi r}{v} \]
Substituting for \( v \) gives:
\[ T = \frac{2 \pi r}{\sqrt{\frac{G M_e}{r}}} = 2 \pi \sqrt{\frac{r^3}{G M_e}} \]

7. **Identifying Dependencies**:
From the final expression for the time period \( T \):
\[ T = 2 \pi \sqrt{\frac{r^3}{G M_e}} \]
We can see that:
- \( T \) depends on \( M_e \) (mass of the Earth).
- \( T \) depends on \( r \) (which includes the height of the satellite from the surface of the Earth).

8. **Conclusion**:
Therefore, the time period of a geostationary satellite depends on:
- (ii) Mass of the Earth
- (iii) Radius of the orbit (which includes the height of the satellite from the Earth's surface).

The correct answer is that the time period depends on the mass of the Earth, the radius of the orbit, and the height of the satellite from the surface of the Earth.

### Final Answer:
The time period of a geostationary satellite depends on:
- (ii) Mass of the Earth
- (iii) Radius of the orbit
- (iv) Height of the satellite from the surface of the Earth

Thus, the correct options are (ii), (iii), and (iv).