Final answer:
The orbital period of the asteroid is approximately 3.30 years, and the eccentricity of its orbit around the Sun is 1/3.
Step-by-step explanation:
The question presented is related to an asteroid's orbit around the Sun and requires us to find the orbital period and the eccentricity of the asteroid's orbit. You have already supplied the closest approach to the Sun, known as the perihelion, which is 2 AU (Astronomical Units), and the farthest distance, known as the aphelion, which is 4 AU. To determine the orbital period, we use Kepler's third law which states that the square of the orbital period is proportional to the cube of the semi-major axis of its orbit.
The semi-major axis is the average of the aphelion and perihelion distances. So in this case, the semi-major axis (a) is (2 AU + 4 AU) / 2, which is 3 AU. Now we use Kepler's third law in the form P² = a³ (with P in years and a in AU), to find the orbital period P. Plugging in the values, we get P² = 3³, so P = √27, which is about 3.30 years.
To find the eccentricity (e) of the orbit, we utilize the formula e = (aphelion - perihelion) / (aphelion + perihelion). Substituting the given values leads to e = (4 AU - 2 AU) / (4 AU + 2 AU), which simplifies to e = 2/6, or 1/3. Therefore, the eccentricity of the asteroid's orbit is 1/3.