The distances of two planets A nad B from the sun are `10^(12)m` and `10^(11)m` and their periodic times are `T_(A)` and `T_(B)` respectively. If their orbits are assumed to be circular, then the ratio of their periodic times `((T_(A))/(T_(B)))` will be

Options

A. 10
B. 100
C. `10sqrt(10)`
D. `10sqrt(3)`

Explanation

To solve the problem, we will use Kepler's Third Law of Planetary Motion, which states that the square of the periodic time (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, it is expressed as:

\[ T^2 \propto r^3 \]

This can also be written as:

\[ \frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3} \]

Where:
- \( T_A \) and \( T_B \) are the periodic times of planets A and B, respectively.
- \( r_A \) and \( r_B \) are the distances of planets A and B from the sun, respectively.

Given:
- \( r_A = 10^{12} \, \text{m} \)
- \( r_B = 10^{11} \, \text{m} \)

Now, we can find the ratio of the periodic times \( \frac{T_A}{T_B} \):

1. **Calculate \( r_A^3 \) and \( r_B^3 \)**:
- \( r_A^3 = (10^{12})^3 = 10^{36} \)
- \( r_B^3 = (10^{11})^3 = 10^{33} \)

2. **Set up the ratio**:
\[ \frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3} = \frac{10^{36}}{10^{33}} = 10^{3} \]

3. **Take the square root to find the ratio of the periodic times**:
\[ \frac{T_A}{T_B} = \sqrt{10^{3}} = 10^{1.5} = 10 \sqrt{10} \]

Thus, the ratio of their periodic times \( \frac{T_A}{T_B} \) is \( 10 \sqrt{10} \).

### Final Answer:
\[ \frac{T_A}{T_B} = 10 \sqrt{10} \]