Two satellites `A` and `B` of masses `m_(1)` and `m_(2)(m_(1)=2m_(2))` are moving in circular orbits of radii `r_(1)` and `r_(2)(r_(1)=4r_(2))`, respectively, around the earth. If their periods are `T_(A)` and `T_(B)`, then the ratio `T_(A)//T_(B)` is

Options

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

Explanation

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the period of revolution of a planet (or satellite) is directly proportional to the cube of the semi-major axis of its orbit (in this case, the radius of the circular orbit).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have two satellites, A and B.
- Masses: \( m_1 = 2m_2 \)
- Radii of their orbits: \( r_1 = 4r_2 \)
- Periods of revolution: \( T_A \) for satellite A and \( T_B \) for satellite B.

2. **Applying Kepler's Third Law:**
- According to Kepler's Third Law, we can express the relationship between the periods and the radii as:
\[ T^2 \propto r^3 \]
- For satellite A:
\[ T_A^2 = k \cdot r_1^3 \]
- For satellite B:
\[ T_B^2 = k \cdot r_2^3 \]
- Here, \( k \) is a constant of proportionality.

3. **Finding the Ratio of the Periods:**
- To find the ratio \( \frac{T_A^2}{T_B^2} \), we can divide the two equations:
\[ \frac{T_A^2}{T_B^2} = \frac{r_1^3}{r_2^3} \]

4. **Substituting the Relationship Between Radii:**
- We know \( r_1 = 4r_2 \). Substituting this into the equation gives:
\[ \frac{T_A^2}{T_B^2} = \frac{(4r_2)^3}{r_2^3} \]
- Simplifying this:
\[ \frac{T_A^2}{T_B^2} = \frac{64r_2^3}{r_2^3} = 64 \]

5. **Taking the Square Root:**
- Now, taking the square root of both sides to find the ratio of the periods:
\[ \frac{T_A}{T_B} = \sqrt{64} = 8 \]

6. **Final Result:**
- Therefore, the ratio of the periods \( \frac{T_A}{T_B} \) is \( 8 \).

### Conclusion:
The final answer is:
\[ \frac{T_A}{T_B} = 8 \]