If `S_(1)` is surface satellite and `S_(2)` is geostationary satellite, with time periods `T_(1)` and `T_(2)`, orbital velocities `V_(1)` and `V_(2)`

Options

A. `T_(1)gtT(_(2),V_(1)gtV_(2)`
B. `T_(2)gtT_(2),V_(1)ltV_(2)`
C. `T_(1)ltT_(2),V_(1)ltV_(2)`
D. `T_(1)ltT_(2),V_(1)gtV_(2)`

Answer

To solve the problem regarding the relationship between the time periods and orbital velocities of a surface satellite (S1) and a geostationary satellite (S2), we can follow these steps:

### Step 1: Understand the Definitions
- **Surface Satellite (S1)**: A satellite that orbits very close to the Earth's surface.
- **Geostationary Satellite (S2)**: A satellite that orbits the Earth at a height where it appears to be stationary relative to the Earth's surface, typically at an altitude of about 36,000 km.

### Step 2: Determine the Distances from the Center of the Earth
- Let \( R_1 \) be the distance of the surface satellite from the center of the Earth.
- Let \( R_2 \) be the distance of the geostationary satellite from the center of the Earth.
- Since the geostationary satellite is much higher, we have \( R_2 > R_1 \).

### Step 3: Apply 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 (or radius in circular orbits) of its orbit:
\[ T^2 \propto R^3 \]
This implies:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]
Since \( R_2 > R_1 \), it follows that:
\[ T_2 > T_1 \]

### Step 4: Determine the Orbital Velocities
The orbital velocity \( V \) of a satellite is given by the formula:
\[ V = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

For the surface satellite (S1):
\[ V_1 = \sqrt{\frac{GM}{R_1}} \]
For the geostationary satellite (S2):
\[ V_2 = \sqrt{\frac{GM}{R_2}} \]
Since \( R_2 > R_1 \), it follows that:
\[ V_2 < V_1 \]

### Step 5: Summarize the Relationships
From the above analysis, we conclude:
- The time period of the geostationary satellite is greater than that of the surface satellite:
\[ T_2 > T_1 \]
- The orbital velocity of the geostationary satellite is less than that of the surface satellite:
\[ V_2 < V_1 \]

### Final Answer
Thus, the relationships can be summarized as:
- \( T_1 < T_2 \)
- \( V_1 > V_2 \)