Answer:
The asteroid requires 5.14 years to make one revolution around the Sun.
Explanation:
Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:
[tex]T^{2} = a^{3}[/tex] (1)
Where T is the period of revolution and a is the semi-major axis.
In the other hand, the distance between the Earth and the Sun has a value of [tex]1.50x10^{8} Km[/tex]. That value can be known as well as an astronomical unit (1AU).
But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.
For the special case of the asteroid the distance will be:
[tex]a = 2.98(1.50x10^{8}Km)[/tex]
[tex]a = 4.47x10^{8}Km[/tex]
That distance will be expressed in terms of astronomical units:
[tex]4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}[/tex] ⇒ [tex]2.98AU[/tex]
Finally, from equation 1 the period T can be isolated:
[tex]T = \sqrt{a^{3}}[/tex]
[tex]T = \sqrt{(2.98)^{3}}[/tex]
[tex]T = \sqrt{26.463592}[/tex]
[tex]T = 5.14AU[/tex]
Then, the period can be expressed in years:
[tex]5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr[/tex]
[tex]T = 5.14 yr[/tex]
Hence, the asteroid requires 5.14 years to make one revolution around the Sun.