The time period of a satellite of earth is 5 hours. If the separation between the centre of earth and the satellite is increased to 4 times the previous value, the new time period will become-

Options

A. 10 hrs
B. 80 hrs
C. 40 hrs
D. 20 hrs

Answer

To solve the problem, we will use the relationship between the time period of a satellite and its distance from the center of the Earth.

### Step-by-Step Solution:

1. **Understand the relationship**: The time period \( T \) of a satellite in orbit is given by the formula:
\[ T^2 \propto r^3 \]
where \( r \) is the distance from the center of the Earth to the satellite.

2. **Initial conditions**: We know the initial time period \( T_1 = 5 \) hours and the initial distance \( r_1 \) (which we will denote as \( r \)).

3. **New conditions**: The problem states that the separation between the center of the Earth and the satellite is increased to 4 times the previous value. Therefore, the new distance \( r_2 \) is:
\[ r_2 = 4r \]

4. **Using the proportionality**: According to the relationship, we can write:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]
Substituting the known values:
\[ \frac{(5 \text{ hours})^2}{T_2^2} = \frac{r^3}{(4r)^3} \]

5. **Simplifying the right side**: The right side simplifies as follows:
\[ \frac{r^3}{(4r)^3} = \frac{r^3}{64r^3} = \frac{1}{64} \]

6. **Setting up the equation**: Now we have:
\[ \frac{25}{T_2^2} = \frac{1}{64} \]

7. **Cross-multiplying**: This leads to:
\[ 25 \cdot 64 = T_2^2 \]
\[ T_2^2 = 1600 \]

8. **Taking the square root**: Finally, we find \( T_2 \):
\[ T_2 = \sqrt{1600} = 40 \text{ hours} \]

### Final Answer:
The new time period will become **40 hours**.