A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit is

Options

A. 4.8 hours
B. `48 sqrt(2)` hours
C. 24 hours
D. `24 sqrt(2)` hours

Answer

To find the time period of a satellite in a new orbit that is twice the distance from the center of the Earth as compared to its original geostationary orbit, we can follow these steps:

### Step 1: Understand the relationship between time period and radius
The time period \( T \) of a satellite in orbit is given by Kepler's third law, which states that:

\[ T \propto r^{3/2} \]

where \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Define the original and new distances
Let \( r_1 \) be the radius of the original geostationary orbit and \( r_2 \) be the radius of the new orbit. According to the problem, the new radius is:

\[ r_2 = 2r_1 \]

### Step 3: Set up the ratio of the time periods
Using the proportionality from Step 1, we can express the ratio of the time periods \( T_1 \) (original) and \( T_2 \) (new) as follows:

\[ \frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} \]

### Step 4: Substitute the new radius into the equation
Substituting \( r_2 = 2r_1 \) into the equation gives:

\[ \frac{T_1}{T_2} = \left(\frac{r_1}{2r_1}\right)^{3/2} = \left(\frac{1}{2}\right)^{3/2} \]

### Step 5: Simplify the expression
Calculating \( \left(\frac{1}{2}\right)^{3/2} \):

\[ \left(\frac{1}{2}\right)^{3/2} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]

### Step 6: Rearranging to find \( T_2 \)
Now, we can rearrange the equation to find \( T_2 \):

\[ T_2 = T_1 \cdot 2\sqrt{2} \]

### Step 7: Substitute the value of \( T_1 \)
For a geostationary satellite, the time period \( T_1 \) is 24 hours (or 86400 seconds). Thus:

\[ T_2 = 86400 \cdot 2\sqrt{2} \]

### Step 8: Calculate \( T_2 \)
Calculating \( T_2 \):

\[ T_2 \approx 86400 \cdot 2 \cdot 1.414 \approx 86400 \cdot 2.828 \approx 244,944 \text{ seconds} \]

### Final Answer
The time period in the second orbit is approximately **244,944 seconds**.

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