Satellite is revolving around earth. If it's radius of orbit is increased to 4 times of the radius of geostationary statellite, what will become its time period ?

Options

A. 8 days
B. 4 days
C. 2 days
D. 16 days

Explanation

To solve the problem of finding the new time period of a satellite when its radius of orbit is increased to four times that of a geostationary satellite, we can follow these steps:

### Step 1: Understand the relationship between time period and radius
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. Mathematically, this can be expressed as:
\[ T^2 \propto R^3 \]
This implies:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]

### Step 2: Identify the initial conditions
Let:
- \( T_1 \) = Time period of the geostationary satellite
- \( R_1 \) = Radius of the geostationary orbit
- \( R_2 \) = 4 times the radius of the geostationary satellite, so \( R_2 = 4R_1 \)

### Step 3: Substitute the values into the equation
Using the relationship from Step 1:
\[ \frac{T_2^2}{T_1^2} = \frac{(4R_1)^3}{R_1^3} \]

### Step 4: Simplify the equation
Calculating the right side:
\[ \frac{(4R_1)^3}{R_1^3} = \frac{64R_1^3}{R_1^3} = 64 \]
Thus, we have:
\[ \frac{T_2^2}{T_1^2} = 64 \]

### Step 5: Solve for \( T_2 \)
Taking the square root of both sides:
\[ \frac{T_2}{T_1} = 8 \]
This implies:
\[ T_2 = 8T_1 \]

### Step 6: Conclusion
If the time period of the geostationary satellite \( T_1 \) is 1 day (which is the standard time period for geostationary satellites), then:
\[ T_2 = 8 \times 1 \text{ day} = 8 \text{ days} \]

Thus, the new time period of the satellite when its radius of orbit is increased to four times that of the geostationary satellite will be **8 days**.