Earth is revolving around the sun if the distance of the Earth from the Sun is reduced to 1/4th of the present distance then the present day length reduced by

Options

A. `1/4`
B. `1/2`
C. `1/8`
D. `1/6`

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, this can be expressed as:

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
Let the current distance of the Earth from the Sun be \( r \) and the current orbital period (length of the year) be \( T \).

2. **Apply Kepler's Third Law**:
According to Kepler's Third Law:
\[ T^2 \propto r^3 \]
This can be expressed as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

3. **Determine the New Distance**:
If the distance of the Earth from the Sun is reduced to \( \frac{1}{4} \) of the present distance, the new distance \( r' \) can be expressed as:
\[ r' = \frac{r}{4} \]

4. **Calculate the New Orbital Period**:
Using Kepler's law for the new distance:
\[ T'^2 = k \cdot (r')^3 \]
Substituting \( r' \):
\[ T'^2 = k \cdot \left(\frac{r}{4}\right)^3 \]
\[ T'^2 = k \cdot \frac{r^3}{64} \]
Since \( T^2 = k \cdot r^3 \), we can substitute this into our equation:
\[ T'^2 = \frac{T^2}{64} \]

5. **Taking the Square Root**:
To find the new period \( T' \):
\[ T' = \sqrt{\frac{T^2}{64}} \]
\[ T' = \frac{T}{8} \]

6. **Conclusion**:
The new orbital period \( T' \) is \( \frac{T}{8} \), meaning the length of the year is reduced to \( \frac{1}{8} \) of the original period.

### Final Answer:
The present day length is reduced by a factor of \( 8 \).

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