A body revolved around the sun 27 times faster then the Earth what is the ratio of their radii

Options

A. `1//3`
B. `1//9`
C. `1//27`
D. `1//4`

Explanation

To solve the problem of finding the ratio of the radii of a body that revolves around the Sun 27 times faster than the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( T_1 \) be the time period of the Earth.
- Let \( R_1 \) be the radius of the Earth's orbit.
- Let \( T_2 \) be the time period of the other body.
- Let \( R_2 \) be the radius of the other body's orbit.

2. **Relate the Time Periods:**
- Since the body revolves 27 times faster than the Earth, its time period \( T_2 \) can be expressed as:
\[ T_2 = \frac{T_1}{27} \]

3. **Apply Kepler's Third Law:**
- According to Kepler's third law, the square of the time period is directly proportional to the cube of the radius:
\[ T^2 \propto R^3 \]
- For Earth:
\[ T_1^2 \propto R_1^3 \]
- For the other body:
\[ T_2^2 \propto R_2^3 \]

4. **Set Up the Proportionality:**
- From the above relationships, we can write:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]

5. **Substitute \( T_2 \):**
- Substitute \( T_2 = \frac{T_1}{27} \) into the equation:
\[ \left(\frac{T_1}{27}\right)^2 = \frac{R_2^3}{R_1^3} \]
- This simplifies to:
\[ \frac{T_1^2}{729} = \frac{R_2^3}{R_1^3} \]

6. **Rearrange to Find the Ratio of Radii:**
- Rearranging gives:
\[ \frac{R_2^3}{R_1^3} = \frac{T_1^2}{729} \]
- Taking the cube root of both sides:
\[ \frac{R_2}{R_1} = \left(\frac{T_1^2}{729}\right)^{1/3} \]

7. **Simplify the Expression:**
- Since \( 729 = 27^2 \), we can rewrite:
\[ \frac{R_2}{R_1} = \frac{T_1^{2/3}}{27} \]
- Now, we need to express this in terms of a ratio:
\[ R_2 = \frac{R_1}{9} \]

8. **Final Ratio:**
- Therefore, the ratio of the radii is:
\[ \frac{R_2}{R_1} = \frac{1}{9} \]

### Conclusion:
The ratio of the radii of the body revolving around the Sun to that of the Earth is \( \frac{1}{9} \).