The ratio of distance of two satellites from the centre of earth is `1:4`. The ratio of their time periods of rotation will be

Options

A. `1:4`
B. `4:1`
C. `1:8`
D. `8:1`

Explanation

To solve the problem, we need to use Kepler's third law of planetary motion, which relates the time period of a satellite to its distance from the center of the Earth.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let the distance of the first satellite from the center of the Earth be \( r_1 \).
- Let the distance of the second satellite from the center of the Earth be \( r_2 \).
- According to the problem, the ratio of the distances is given as:
\[ \frac{r_1}{r_2} = \frac{1}{4} \]
- This implies that \( r_2 = 4r_1 \).

2. **Using Kepler's Third Law**:
- Kepler's third law states that the square of the time period \( T \) of a satellite is directly proportional to the cube of the semi-major axis (or distance from the center of the Earth for circular orbits):
\[ T^2 \propto r^3 \]
- This can be expressed as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

3. **Finding the Time Periods**:
- For the first satellite:
\[ T_1^2 = k \cdot r_1^3 \]
- For the second satellite:
\[ T_2^2 = k \cdot r_2^3 \]
- Substituting \( r_2 = 4r_1 \) into the equation for \( T_2^2 \):
\[ T_2^2 = k \cdot (4r_1)^3 = k \cdot 64r_1^3 \]

4. **Finding the Ratio of Time Periods**:
- Now we can find the ratio of the squares of the time periods:
\[ \frac{T_1^2}{T_2^2} = \frac{k \cdot r_1^3}{k \cdot 64r_1^3} = \frac{1}{64} \]
- Taking the square root to find the ratio of the time periods:
\[ \frac{T_1}{T_2} = \frac{1}{8} \]

5. **Conclusion**:
- The ratio of the time periods of the two satellites is:
\[ \frac{T_1}{T_2} = \frac{1}{8} \]

### Final Answer:
The ratio of their time periods of rotation will be \( 1:8 \).