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 find the ratio of the time periods of two planets revolving around the sun, we can use Kepler's Third Law of planetary motion. This law states that the square of the time period of a planet's orbit is directly proportional to the cube of the semi-major axis (or the distance from the sun for circular orbits).

### Step-by-Step Solution:

1. **Identify the distances of the planets from the sun**:
- Let the distance of Planet 1 from the sun be \( r_1 = 10^{12} \, \text{m} \).
- Let the distance of Planet 2 from the sun be \( r_2 = 10^{10} \, \text{m} \).

2. **Apply Kepler's Third Law**:
- According to Kepler's Third Law, we have:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]
- Here, \( T_1 \) and \( T_2 \) are the time periods of Planet 1 and Planet 2, respectively.

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

4. **Substitute the values of \( r_1 \) and \( r_2 \)**:
- Substitute \( r_1 = 10^{12} \) and \( r_2 = 10^{10} \):
\[ \frac{T_1}{T_2} = \frac{(10^{12})^{3/2}}{(10^{10})^{3/2}} = \frac{10^{18}}{10^{15}} = 10^{3} \]

5. **Calculate the ratio**:
- Thus, we find:
\[ \frac{T_1}{T_2} = 10^3 = 1000 \]

6. **Express the final ratio**:
- Therefore, the ratio of the time periods of the two planets is:
\[ T_1 : T_2 = 1000 : 1 \]

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

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