Kepler's Third Law can be expressed mathematically as:
\[ T^2 = k \cdot a^3 \]
where:
- \( T \) is the orbital period (in years),
- \( a \) is the semi-major axis (in astronomical units, AU),
- \( k \) is a constant that depends on the units used. For our purposes, we can use \( k = 1 \) if we are considering orbits in our solar system.
First, we need to convert the period from days to years. Since there are approximately 365.25 days in a year, we can convert:
\[ T = \frac{250 \text{ days}}{365.25 \text{ days/year}} \approx 0.685 \text{ years} \]
Next, we'll use Kepler's Third Law to find \( a \):
\[ T^2 = a^3 \]
Substituting the value of \( T \):
\[ (0.685)^2 = a^3 \]
Calculating \( (0.685)^2 \):
\[ 0.685^2 \approx 0.469225 \]
Now we solve for \( a \):
\[ a^3 \approx 0.469225 \]
Taking the cube root of both sides:
\[ a \approx \sqrt[3]{0.469225} \approx 0.77 \text{ AU} \]
Since 1 AU is approximately 149,597,870.7 km, we can convert \( a \) to kilometers:
\[ a \approx 0.77 \times 149,597,870.7 \text{ km} \approx 115,825,178 \text{ km} \]
This calculated semi-major axis (approximately 115.8 million km) does not correspond to the answer choices provided (10 km, 25 km, 35 km, 40 km).
If we consider the context of the problem, the choices likely suggest a much smaller scale, potentially reaching out for a different planet or celestial body. Can you please clarify the context of this problem?