The distances of two planets from the sun are `10^12` m and `10^10` m respectively.The ratio of the time periods of these planets is

Options

A. `1000:1`
B. `100:1`
C. `10:1`
D. `1:1`

Answer

To solve the problem, we will use Kepler's third law of planetary motion, which states that the square of the time period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit around the Sun. The relationship can be expressed as:

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

### Step-by-Step Solution:

1. **Identify the Distances**:
- Let the distance of the first planet from the Sun be \( r_1 = 10^{12} \) m.
- Let the distance of the second planet from the Sun be \( r_2 = 10^{10} \) m.

2. **Apply Kepler's Third Law**:
- According to Kepler's third law, the ratio of the squares of the time periods of two planets is equal to the ratio of the cubes of their distances from the Sun:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

3. **Express the Time Periods**:
- Taking the square root of both sides gives us the ratio of the time periods:
\[ \frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}} = \frac{r_1^{3/2}}{r_2^{3/2}} \]

4. **Substitute the Values**:
- Substitute \( r_1 \) and \( r_2 \) into the equation:
\[ \frac{T_1}{T_2} = \frac{(10^{12})^{3/2}}{(10^{10})^{3/2}} \]

5. **Simplify the Expression**:
- Calculate \( (10^{12})^{3/2} \) and \( (10^{10})^{3/2} \):
\[ (10^{12})^{3/2} = 10^{12 \times \frac{3}{2}} = 10^{18} \]
\[ (10^{10})^{3/2} = 10^{10 \times \frac{3}{2}} = 10^{15} \]
- Now substitute these back into the ratio:
\[ \frac{T_1}{T_2} = \frac{10^{18}}{10^{15}} = 10^{18 - 15} = 10^3 \]

6. **Final Result**:
- Therefore, the ratio of the time periods is:
\[ \frac{T_1}{T_2} = 1000 : 1 \]

### Conclusion:
The ratio of the time periods of the two planets is \( 1000 : 1 \).