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Explanation: [tex]$\square$[/tex]

Because Kepler's third law is a direct variation, the quotient [tex]$\frac{t^3}{d^3}$[/tex] will equal the same constant of variation, [tex]$k$[/tex], for all combinations of time, [tex]$t$[/tex], in years and distance from the Sun, [tex]$d$[/tex], in millions of miles.

Julie's strategy is to use the equation [tex]$\frac{t^3}{d^3}=k$[/tex] twice. First, she substitutes the values of [tex]$d$[/tex] and [tex]$t$[/tex] known for Earth and evaluates [tex]$\frac{1^3}{93^3}$[/tex] to find the decimal value of [tex]$k$[/tex]. Then, she substitutes that decimal value and 1,488 into the equation and solves for [tex]$t$[/tex]. Julie's strategy is valid, but the decimal value of [tex]$k$[/tex] is less than 0.000002 and doesn't repeat in the first 85 digits. As a result, she may have issues because of rounding.

Mona's strategy uses the transitive property of equality to set two ratios that equal the same value of [tex]$k$[/tex] equal to each other. She can solve the proportion [tex]$\frac{1^3}{93^3}=\frac{t^3}{1,488^3}$[/tex] by cross-multiplying and won't have to solve for [tex]$k$[/tex]. As a result, she is less likely to have issues related to rounding.

Question:

Type the correct answer in the box. Use numerals instead of words.

It will take the asteroid [tex]$\square$[/tex] years, in Earth years, to complete one orbit of the Sun.



Answer :

GinnyAnswer
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.

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