Final Answer:
The asteroid orbits the sun at a semi-major axis of 3 astronomical units (AU).
Step-by-step explanation:
An astronomical unit (AU) is the average distance between the Earth and the Sun, roughly 149.6 million kilometers. A semi-major axis defines an ellipse's size, half the major axis. For this asteroid, its semi-major axis being 3 AU means it orbits the sun at a distance three times farther than Earth does.
To compute the actual distance, we multiply 3 AU by the average Earth-Sun distance. Therefore, the asteroid's average distance from the sun is approximately 3 * 149.6 million kilometers, which equals 448.8 million kilometers.
This distance impacts the asteroid's orbital period. Kepler's third law of planetary motion states that the square of a planet's orbital period is proportional to the cube of the semi-major axis. Given that the Earth's orbital period is one year (365.25 days), we can use the formula to estimate the asteroid's orbital period.
The ratio of the asteroid's semi-major axis cubed to Earth's semi-major axis cubed is 3^3:1^3, which simplifies to 27:1. Therefore, the square of the asteroid's orbital period compared to Earth's is also 27:1. Consequently, the asteroid's orbital period can be estimated by squaring Earth's orbital period: 365.25 days * 27, resulting in approximately 9865.25 days for the asteroid's orbital period.