Understanding Earth Year Duration and Distance from the Sun

The question asks how the duration of the Earth's year would change if its distance from the sun were reduced to one-fourth of its current distance. To solve this, we need to use Kepler's laws of planetary motion, specifically Kepler's Third Law.

Kepler's Third Law and Orbital Period

Kepler's Third Law describes the relationship between the orbital period of a planet and its average distance from the sun. It states that the square of the orbital period (\(T\)) is directly proportional to the cube of the average distance (\(r\)) from the sun. Mathematically, this can be written as:

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

This means that the ratio \( \frac{T^2}{r^3} \) is a constant for all planets orbiting the same central body (like the sun). Let's call this constant \(k\):

\[ T^2 = kr^3 \]

or

\[ \frac{T^2}{r^3} = k \]

Calculating the New Earth Year Duration

Let's denote the current duration of the Earth year as \(T_1\) and the current average distance from the sun as \(r_1\). According to the question, the current Earth year duration is 1 year. So, \(T_1 = 1\) year.

If the Earth's distance from the sun is changed, let the new duration of the year be \(T_2\) and the new distance be \(r_2\). The problem states that the new distance is 1/4 of the present distance:

\[ r_2 = \frac{1}{4} r_1 \]

Using Kepler's Third Law, we can set up a proportion comparing the two scenarios:

\[ \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} \]

We want to find \(T_2\), so let's rearrange the equation:

\[ T_2^2 = T_1^2 \left( \frac{r_2^3}{r_1^3} \right) \]

\[ T_2^2 = T_1^2 \left( \frac{r_2}{r_1} \right)^3 \]

Now, substitute the given relationship \( r_2 = \frac{1}{4} r_1 \), which means \( \frac{r_2}{r_1} = \frac{1}{4} \):

\[ T_2^2 = T_1^2 \left( \frac{1}{4} \right)^3 \]

Calculate the cube of \(\frac{1}{4}\):

\[ \left( \frac{1}{4} \right)^3 = \frac{1^3}{4^3} = \frac{1}{64} \]

So, the equation becomes:

\[ T_2^2 = T_1^2 \times \frac{1}{64} \]

Now, take the square root of both sides to find \(T_2\):

\[ T_2 = \sqrt{T_1^2 \times \frac{1}{64}} \]

\[ T_2 = T_1 \times \sqrt{\frac{1}{64}} \]

\[ T_2 = T_1 \times \frac{1}{8} \]

Since the current duration of the Earth year \(T_1\) is 1 year, the new duration \(T_2\) is:

\[ T_2 = 1 \text{ year} \times \frac{1}{8} = \frac{1}{8} \text{ year} \]

Therefore, if the Earth were 1/4 of its present distance from the sun, the duration of the year would be 1/8 of the present duration. This calculation using Kepler's Third Law clearly shows the impact of distance from the sun on the orbital period and thus the Earth year duration.

Conclusion

Based on Kepler's Third Law, reducing the Earth's distance from the sun to one-fourth results in the Earth year duration being one-eighth of the current year. This demonstrates the non-linear relationship between orbital period and distance described by the law.