Final answer:
Eris's average distance from the Sun, or its semimajor axis, is approximately 67.78 astronomical units (AU), determined by using Kepler's third law relating the square of the orbital period to the cube of the semimajor axis.
Step-by-step explanation:
The average distance (semimajor axis) from the Sun of an object that orbits it, such as the dwarf planet Eris, can be determined using Kepler's third law. Kepler's law exploits the relationship between the orbital period and the semimajor axis of an orbit. According to Kepler's third law, the square of the orbital period (P) is proportional to the cube of the semimajor axis (a) for any object orbiting the Sun. Essentially, P2 = a3, where P is the orbital period in Earth years and a is the semimajor axis in astronomical units (AU).
For Eris, which has an orbital period of approximately 557 years, we would set up the equation as 5572 = a3. Without doing the calculation here, we would calculate the cube root of 5572 to find the semimajor axis a in AU. It is widely accepted that Eris's semimajor axis is approximately 67.78 AU, which is its average distance from the Sun. It is important to note that because Eris has a highly elliptical orbit, its actual distance from the Sun will vary significantly over its orbit, reaching as far as approximately 96 AU from the Sun.