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

1 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.

Explanation:

Kepler's Third law states that the period of an orbiting object squared is equal to its distance from what it is orbiting from cubed, or:

#T^2=a^3#

We can go one step further and use the period and distance of the Earth 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), so that we can describe an object in orbit around the Sun in terms of years (for T) and astronomical units (for a).

So how far away is the asteroid belt from the Sun? It spans a range between 2.2 and 3.2 AU. So the objects in the belt will take a shorter time if closer to the 2.2 distance and longer if closer to the 3.2 distance. Let's calculate the range.

Before we do that, let's first solve for T:

#T=a^(3/2)#

At a = 2.2, T = 3.26 years
At a = 3.2, T = 5.72 years

The question asks for an average sidereal period, so let's just find the middle value of the two ends of our range, which is 4.49 or roughly 4.5 years.

http://www-spof.gsfc.nasa.gov/stargaze/Kep3laws.htm

http://theplanets.org/asteroid-belt/