To determine how long it will take the asteroid to complete one orbit of the Sun, we can use Kepler's Third Law, which states that the ratio of the cube of the period of orbit ([tex]\(t^3\)[/tex]) to the cube of the average distance from the Sun ([tex]\(d^3\)[/tex]) is the same for all objects orbiting the Sun. This can be expressed as:
[tex]\[
\frac{t_1^3}{d_1^3} = \frac{t_2^3}{d_2^3}
\][/tex]
For Earth:
- [tex]\(t_{\text{Earth}} = 1\)[/tex] year
- [tex]\(d_{\text{Earth}} = 93\)[/tex] million miles
For the asteroid:
- [tex]\(d_{\text{asteroid}} = 1,488\)[/tex] million miles
We need to find [tex]\(t_{\text{asteroid}}\)[/tex], the period of the asteroid's orbit in Earth years.
First, set up the equation using the known values:
[tex]\[
\frac{1^3}{93^3} = \frac{t_{\text{asteroid}}^3}{1,488^3}
\][/tex]
Now, cross multiply to solve for [tex]\(t_{\text{asteroid}}^3\)[/tex]:
[tex]\[
t_{\text{asteroid}}^3 = \left(\frac{1,488^3}{93^3}\right)
\][/tex]
Calculate the ratio:
[tex]\[
\left(\frac{1,488}{93}\right)^3 = 16^3 = 4,096
\][/tex]
So:
[tex]\[
t_{\text{asteroid}}^3 = 4,096
\][/tex]
To find [tex]\(t_{\text{asteroid}}\)[/tex], take the cube root of both sides:
[tex]\[
t_{\text{asteroid}} = \sqrt[3]{4,096} = 16
\][/tex]
Therefore, it will take the asteroid approximately 16 years, in Earth years, to complete one orbit of the Sun.
