Question Statement

The orbit of a planet is in elliptical shape with the Sun sitting at one of its two foci. The eccentricity of the ellipse is $0.085$ and the minimum distance of the planet from the Sun is $105$ million miles. What is the maximum distance of the planet from the Sun?

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Background and Explanation

This problem requires understanding the geometry of an ellipse, specifically the relationship between the semi-major axis ($a$), the distance from center to focus ($c$), and the eccentricity ($e=c/a$). For a planet orbiting the Sun at one focus, the closest approach (perihelion) is $a-c$ and the farthest distance (aphelion) is $a+c$.

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Solution

Let the Sun be at focus $F_1$. From the geometry of the ellipse, the planet is at minimum distance from the Sun when it is at vertex $A$ (the vertex closest to $F_1$).

Step 1: Relate the minimum distance to the ellipse parameters

Given that the minimum distance is $105$ million miles: $|A{F}_1|=105$

From the figure, we can express this distance as the difference between the semi-major axis and the focal distance: $|OA|-|O{F}_1|=105$

Since $|OA|=a$ (semi-major axis) and $|O{F}_1|=c$ (distance from center to focus): $\begin{array}{rl}|OA|-|O{F}_1|& =105\\ \Rightarrow a-c& =105\end{array}$

Step 2: Use eccentricity to solve for $a$

Recall that eccentricity $e=\frac{c}{a}$, so $c=ae$. Substituting this into the equation above: $a-ae=105$ $\Rightarrow a(1-e)=105$

Given $e=0.085$: $\begin{array}{rl}a(1-0.085)& =105\\ a(0.915)& =105\\ a& =\frac{105}{0.915}\\ a& \approx 114.75\text{ million miles}\end{array}$

Step 3: Calculate the maximum distance

The planet is at maximum distance from the Sun when it is at vertex $B$ (the vertex farthest from $F_1$). This distance can be calculated as: $|F_{1}B|=|F_1{F}_2|+|F_{2}B|$

Since $|F_1{F}_2|=2c$ (distance between the two foci) and $|F_{2}B|=|F_{1}A|=105$ (by the symmetry of the ellipse, the distance from $F_2$ to the far vertex equals the distance from $F_1$ to the near vertex): $|F_{1}B|=2c+105$

Substituting $c=ae$: $\begin{array}{rl}|F_{1}B|& =2(ae)+105\\ & =2(114.75)(0.085)+105\\ & =19.5075+105\\ & =124.5075\text{ million miles}\end{array}$

Rounding to two decimal places, the maximum distance is approximately $124.51$ million miles (or $124.507$ million miles as calculated).

Alternative verification: The maximum distance can also be calculated directly as $a+c=a(1+e)=114.75(1.085)\approx 124.50$ million miles, confirming our result.

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Key Formulas or Methods Used

• Eccentricity definition: $e=\frac{c}{a}$ where $c$ is the focal distance and $a$ is the semi-major axis
• Minimum distance from focus: $a-c=a(1-e)$ (perihelion distance)
• Maximum distance from focus: $a+c=a(1+e)$ (aphelion distance)
• Relationship between distances: Maximum distance $=2c+\text{minimum distance}$

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Summary of Steps

1. Set up the equation for minimum distance: $a(1-e)=105$
2. Solve for semi-major axis $a$: $a=\frac{105}{1-0.085}\approx 114.75$ million miles
3. Calculate focal distance $c$: $c=ae\approx 114.75\times 0.085\approx 9.75$ million miles
4. Find maximum distance: Use $|F_{1}B|=2c+105$ or $a(1+e)\approx 124.51$ million miles