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

Step-by-step 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