What is the average sidereal period for an asteroid moving around the Sun in the asteroid belt, according to Kepler's law?

Question

Levi Anderson · Accepted Answer

At a = 2.2, T = 3.26 years
At a = 3.2, T = 5.72 years
The value in the middle of these two, so a reasonable average, is 4.5 years.

According to Kepler's Third Law, the squared period of an orbiting object equals the cubed distance of the object it is orbiting, or: #T^2=a^3# To describe an object in orbit around the Sun in terms of years (for T) and astronomical units (for a), we can go one step further and use the Earth's period and distance around the Sun to set up a ratio (T and a of the Earth are both equal to 1 when expressed in years and astronomical units). This means that the objects in the belt will take less time if they are closer to the 2.2 distance and more time if they are closer to the 3.2 distance. Let's calculate the range of the asteroid belt's distance from the Sun. It spans a range between 2.2 and 3.2 AU. Let's solve for T first before moving on. #T=a^(3/2)# T = 3.26 years at a = 2.2 and 5.72 years at a = 3.2. Since the question only asks for an average sidereal period, we can just take the middle value of 4.49, or about 4.5 years, from either end of our range. Kep3laws.htm, https://tutor.hix.ai https://tutor.hix.ai