The time period of earth is taken as T and its distance from sun as R. What will be the distance of a certain planet from sun whose time period is 64 times that of earth ?

Explanation

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 mathematically as:

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

### Step-by-Step Solution:

1. **Identify the known values**:
- Let the time period of the Earth be \( T \) and its distance from the Sun be \( R \).
- The time period of the planet is given as \( T' = 64T \).

2. **Apply Kepler's Third Law**:
- According to Kepler's Third Law:
\[ \frac{T^2}{R^3} = \frac{T'^2}{R'^3} \]
where \( R' \) is the distance of the planet from the Sun that we need to find.

3. **Substitute the known values**:
- Substitute \( T' = 64T \) into the equation:
\[ \frac{T^2}{R^3} = \frac{(64T)^2}{R'^3} \]

4. **Simplify the equation**:
- This simplifies to:
\[ \frac{T^2}{R^3} = \frac{4096T^2}{R'^3} \]
- Cancel \( T^2 \) from both sides (assuming \( T \neq 0 \)):
\[ \frac{1}{R^3} = \frac{4096}{R'^3} \]

5. **Cross-multiply to find \( R' \)**:
- Rearranging gives:
\[ R'^3 = 4096R^3 \]

6. **Take the cube root**:
- Taking the cube root of both sides:
\[ R' = R \cdot \sqrt[3]{4096} \]

7. **Calculate \( \sqrt[3]{4096} \)**:
- Since \( 4096 = 16^3 \), we find:
\[ \sqrt[3]{4096} = 16 \]

8. **Final expression for \( R' \)**:
- Therefore, we have:
\[ R' = 16R \]

### Conclusion:
The distance of the planet from the Sun is \( 16R \).