Question

If an asteroid is 4 AU from the Sun, what is the period of revolution around the Sun for the asteroid?

Answer 1

Answer: 8 years

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”:

$T^{2}\propto a^{3}$ (1)

In other words: this law states a relation between the orbital period $T$ of a body (moon, planet, satellite, comet, asteroid) orbiting a greater body in space (the Sun, for example) with the size $a$ of its orbit.  

However, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$), equation (1) becomes:

$T^{2}=a^{3}$ (2)

This means that now both sides of the equation are equal.

Knowing $a=4 AU$ and isolating $T$ from (2):

$T=\sqrt{a^{3}}$ (3)

$T=\sqrt{(4 AU)^{3}}$ (4)

Finally:

$T= 8 years$This is the period of the asteroid