6.5 Q-7

Question Statement

An asteroid orbits the sun in an elliptical path with the sun at one focus. The distance of the asteroid from the sun ranges from 17 million miles (closest approach) to 183 million miles (farthest distance). Write the equation of the asteroid’s orbit.

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Background and Explanation

Elliptical orbits follow the standard equation of an ellipse:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$

where:

• $a$ is the semi-major axis (half the total length of the major axis),
• $b$ is the semi-minor axis,
• $c$ is the distance from the center to the focus, and
• $e$ is the eccentricity, given by $e = \frac{c}{a}$.

For this problem:

• The sun is at one focus of the ellipse.
• The closest distance (perihelion) is $a(1-e) = 17$ million miles.
• The farthest distance (aphelion) is $a(1+e) = 183$ million miles.

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Solution

Step 1: Determine the semi-major axis ($a$)

The total distance between the closest and farthest points is the major axis:

$$2a = 183 + 17 = 200 , \text{million miles}.$$

Thus:

$$a = \frac{200}{2} = 100 , \text{million miles}.$$

Step 2: Calculate the eccentricity ($e$)

The distance from the focus to the closest approach is:

$$a(1-e) = 17.$$

Substitute $a = 100$:

$$100(1-e) = 17 \quad \Rightarrow \quad 1-e = \frac{17}{100} \quad \Rightarrow \quad e = 1 - 0.17 = 0.83.$$

Step 3: Calculate the semi-minor axis ($b$)

Using the relation $c = ae$, find $c$:

$$c = 100 \times 0.83 = 83 , \text{million miles}.$$

Now, use $c^2 = a^2 - b^2$ to solve for $b^2$:

$$83^2 = 100^2 - b^2 \quad \Rightarrow \quad 6889 = 10000 - b^2 \quad \Rightarrow \quad b^2 = 3111.$$

Step 4: Write the equation of the ellipse

Substitute $a^2 = 100^2 = 10000$ and $b^2 = 3111$ into the standard equation:

$$\frac{x^2}{10000} + \frac{y^2}{3111} = 1.$$

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Key Formulas or Methods Used

1. Standard Ellipse Equation:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

2. Eccentricity Relation:

$$e = \frac{c}{a}$$

3. Focal Relation:

$$c^2 = a^2 - b^2$$

4. Closest and Farthest Distances:

$$\text{Closest distance: } a(1-e), \quad \text{Farthest distance: } a(1+e).$$

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Summary of Steps

1. Calculate $a = 100$ from the major axis length $2a = 200$.
2. Use the closest distance $a(1-e) = 17$ to find $e = 0.83$.
3. Solve for $b^2 = 3111$ using $c^2 = a^2 - b^2$.
4. Write the ellipse equation as:

$$\frac{x^2}{10000} + \frac{y^2}{3111} = 1.$$

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Final Equation

$$\frac{x^2}{10000} + \frac{y^2}{3111} = 1$$