A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution ?

Options

A. 25
B. 50
C. 100
D. 20

Answer

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

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For the first planet, the time period \( T_1 = 200 \) days and the distance from the Sun \( r_1 = r \).
- For the second planet, the distance from the Sun \( r_2 = \frac{r}{4} \).

2. **Apply Kepler's Third Law:**
- According to Kepler's Third Law:
\[ T_1^2 \propto r_1^3 \quad \text{(1)} \]
\[ T_2^2 \propto r_2^3 \quad \text{(2)} \]

3. **Express the Proportionality in Terms of a Constant:**
- We can write:
\[ T_1^2 = k \cdot r_1^3 \]
\[ T_2^2 = k \cdot r_2^3 \]
- Where \( k \) is a constant.

4. **Set Up the Ratio of the Two Equations:**
- Dividing equation (2) by equation (1):
\[ \frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} \]

5. **Substitute the Values of \( r_1 \) and \( r_2 \):**
- Substitute \( r_1 = r \) and \( r_2 = \frac{r}{4} \):
\[ \frac{T_2^2}{T_1^2} = \frac{\left(\frac{r}{4}\right)^3}{r^3} \]
- Simplifying the right-hand side:
\[ \frac{T_2^2}{T_1^2} = \frac{\frac{r^3}{64}}{r^3} = \frac{1}{64} \]

6. **Express \( T_2 \) in Terms of \( T_1 \):**
- Rearranging gives:
\[ T_2^2 = \frac{T_1^2}{64} \]
- Taking the square root:
\[ T_2 = \frac{T_1}{8} \]

7. **Substitute the Value of \( T_1 \):**
- Since \( T_1 = 200 \) days:
\[ T_2 = \frac{200}{8} = 25 \text{ days} \]

### Final Answer:
The time taken by the second planet to complete one revolution around the Sun is **25 days**.

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