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 Given Variables:**
- Let the time period of the Earth be \( T = t \).
- Let the distance of the Earth from the Sun be \( R = r \).
- The time period of the planet is given as \( T_p = 64t \).

2. **Apply Kepler's Third Law:**
- For the Earth:
\[ T^2 = k \cdot R^3 \implies t^2 = k \cdot r^3 \]
- For the planet:
\[ T_p^2 = k \cdot R_p^3 \implies (64t)^2 = k \cdot R_p^3 \]

3. **Calculate the Time Period of the Planet:**
- Squaring the time period of the planet:
\[ (64t)^2 = 4096t^2 \]
- Thus, we have:
\[ 4096t^2 = k \cdot R_p^3 \]

4. **Set Up the Ratio:**
- Now we can set up the ratio of the equations for the Earth and the planet:
\[ \frac{T_p^2}{T^2} = \frac{R_p^3}{R^3} \]
- Substituting the values:
\[ \frac{4096t^2}{t^2} = \frac{R_p^3}{r^3} \]
- This simplifies to:
\[ 4096 = \frac{R_p^3}{r^3} \]

5. **Solve for \( R_p \):**
- Rearranging gives us:
\[ R_p^3 = 4096r^3 \]
- Taking the cube root of both sides:
\[ R_p = \sqrt[3]{4096} \cdot r \]

6. **Calculate \( \sqrt[3]{4096} \):**
- Since \( 4096 = 16^3 \):
\[ R_p = 16r \]

### Final Answer:
The distance of the planet from the Sun is \( 16 \) times the distance of the Earth from the Sun.