Question

An asteroid has a semi-major axis distance of 3.29 AU, what is its orbital period?

Answer 1

The orbital period of the asteroid is 5.9675 years.

Explanation:

Kepler's third law relates the orbital period `T` to the semi-major axis distance `a` of a body. Newton derived an exact formula:

`T^2=(4 pi^2)/(G(M+m))a^3`

Where `G` is the gravitational constant, `M` is the mass of the Sun and `m` is the mass of the asteroid.

Now the mass of the asteroid is insignificant compared to the mass of the Sun and can be ignored.

If the semi-major axis distance is in AU and the period is in years then:

`(4 pi^2)/(GM)=1`

Therefore:

`T^2=a^3`

For the asteroid `a=3.29AU`. Cubing and taking the square root gives `T=5.9675` years.