Question

the planet mercury moves in an elliptical orbit with the sun at one focus. given that mercurys closest approach to the sun is approximately 46 million kilometers and that the eccentricity of mercurys orbit is approximately 0.206, estimate this planets maximum distance from the sun. express your answer as a decimal rounded to two decimal places.

Explanation:

Step1: Recall the formula for the closest and farthest distance in an ellipse

For an ellipse with the sun at one focus, the closest distance \(d_{min}=a - c\) and the farthest distance \(d_{max}=a + c\), where \(a\) is the semi - major axis and \(c\) is the distance from the center to the focus. Also, the eccentricity \(e=\frac{c}{a}\), so \(c = ea\).
Given \(d_{min}=a - c=a(1 - e)\) and \(d_{min} = 46\) million kilometers, \(e=0.206\).
We can solve for \(a\) from \(a(1 - e)=d_{min}\).

$$a=\frac{d_{min}}{1 - e}$$

Substitute \(d_{min}=46\) and \(e = 0.206\) into the formula:

$$a=\frac{46}{1 - 0.206}=\frac{46}{0.794}\approx57.9345$$

Step2: Calculate the maximum distance

Since \(d_{max}=a + c\) and \(c = ea\), then \(d_{max}=a(1 + e)\).
Substitute \(a=\frac{46}{0.794}\) and \(e = 0.206\) into \(d_{max}=a(1 + e)\):

$$d_{max}=\frac{46}{0.794}(1 + 0.206)$$

$$d_{max}=\frac{46\times1.206}{0.794}$$

$$d_{max}=\frac{55.476}{0.794}\approx69.87$$

Answer:

\(69.87\) million kilometers