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.
Given that you have the period \( T = 250 \) days, we need to convert this into years first. There are approximately 365.25 days in a year, so:
\[ T = \frac{250 \text{ days}}{365.25 \text{ days/year}} \approx 0.684 \text{ years} \]
Now, we can apply Kepler's Third Law. To find \( a \), we rearrange the equation:
\[ a^3 = \frac{T^2}{k} \]
Assuming \( k = 1 \) (which is valid for our solar system's planets when using AU and years):
\[ a^3 = T^2 \] \[ a^3 = (0.684)^2 \approx 0.467 \] \[ a \approx \sqrt[3]{0.467} \approx 0.78 \text{ AU} \]
1 AU is approximately equal to \( 149,597,870.7 \) km. Thus,
\[ a \approx 0.78 \times 149,597,870.7 \text{ km} \approx 116,878,905 \text{ km} \]
This value is much larger than to the options provided (10 km, 25 km, 35 km, and 40 km). Kepler's Third Law and the sizes involved typically apply to celestial orbits and their distances in AU rather than kilometers for these options.
Therefore, it appears that the question may be asking for a conceptual understanding rather than a numerical solution matching the provided answers. If you meant to compare semi-major axis values generally, it seems there are no correct options in terms of kilometers for an orbiting object with a period of 250 days based on Kepler's Third Law.
To conclude, if you are looking for something specific or further clarification, please provide additional context!
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