Answer:
The asteroid requires 5.14 years to make one revolution around the Sun.
Step-by-step 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:
(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
. 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:
![a = 2.98(1.50x10^(8)Km)](https://img.qammunity.org/2020/formulas/physics/college/az8urafsmjqao1ebxnxzw36inhijj7a7v5.png)
![a = 4.47x10^(8)Km](https://img.qammunity.org/2020/formulas/physics/college/jym3pigc389x5rwyt7mhq35nd95n0eyfkd.png)
That distance will be expressed in terms of astronomical units:
⇒
![2.98AU](https://img.qammunity.org/2020/formulas/physics/college/u3a1r9a712fcz5srvem8quufigydtyzeqo.png)
Finally, from equation 1 the period T can be isolated:
![T = 5.14AU](https://img.qammunity.org/2020/formulas/physics/college/lcxp29krea9wkinb5hkzxsh5ax1nmdr30p.png)
Then, the period can be expressed in years:
![5.14AU.(1yr)/(1AU) ⇒ 5.14 yr](https://img.qammunity.org/2020/formulas/physics/college/rrba7ojdjsiky9opw9x3f204c2u37obn9a.png)
![T = 5.14 yr](https://img.qammunity.org/2020/formulas/physics/college/w65d6pgocxic7u5dcyfbdoolqog2w10j5j.png)
Hence, the asteroid requires 5.14 years to make one revolution around the Sun.