Problem 87 - A Graphical Approach to Precalculus with Limits

Question

Solve each application. The orbit of Mars around the sun is an ellipse with equation $$ \frac{x^{2}}{5013}+\frac{y^{2}}{4970}=1 $$ where \(x\) and \(y\) are measured in millions of miles. Approximate the eccentricity \(e\) of this ellipse.

Step-by-Step Solution

Answer

The eccentricity \( e \) is approximately 0.0927.

Step 1: Identify the form of the ellipse equation

The given equation \( \frac{x^{2}}{5013} + \frac{y^{2}}{4970} = 1 \) is in the standard form of the equation of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 5013 \) and \( b^2 = 4970 \). Identify \( a \) and \( b \) by taking the square roots: \( a = \sqrt{5013} \) and \( b = \sqrt{4970} \).

Step 2: Determine which parameter is the semi-major axis

Compare \( a \) and \( b \). Normally, the semi-major axis is larger. If \( a > b \), then \( a \) is the semi-major axis. For this ellipse, \( a = \sqrt{5013} \approx 70.8 \) and \( b = \sqrt{4970} \approx 70.5 \). Since \( a > b \), \( a \) is the semi-major axis.

Step 3: Calculate the eccentricity

The eccentricity \( e \) of an ellipse is given by the formula \( e = \sqrt{1 - \frac{b^2}{a^2}} \). Substitute \( a^2 = 5013 \) and \( b^2 = 4970 \) into the formula to find \( e \).

Step 4: Perform the calculations

Compute \( \frac{b^2}{a^2} = \frac{4970}{5013} \approx 0.9914 \). Then, find \( e = \sqrt{1 - 0.9914} = \sqrt{0.0086} \approx 0.0927 \).

Step 5: Interpret the Result

The eccentricity value \( e = 0.0927 \) indicates that the orbit of Mars is nearly circular but slightly elongated, as \( e \) is close to 0.

Key Concepts

Ellipse Equation

An ellipse is a fascinating shape in mathematics, often described by its unique equation. The standard form of an ellipse equation looks like this: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a \) and \( b \) represent the lengths of the semi-major and semi-minor axes, respectively. This equation helps determine the shape and orientation of the ellipse in a coordinate plane.

In our example with Mars, the given equation is \( \frac{x^{2}}{5013} + \frac{y^{2}}{4970} = 1 \). By comparing this to the standard form, you can identify that \( a^2 = 5013 \) and \( b^2 = 4970 \). Remember that these expressions are squared, so we take the square root to find \( a \) and \( b \) themselves:

• \( a = \sqrt{5013} \)
• \( b = \sqrt{4970} \)

Understanding the ellipse equation allows us to visualize the elliptical orbit Mars follows around the sun.

Semi-Major Axis

The semi-major axis of an ellipse is crucial as it represents the longer radius of the ellipse. It stretches from the center to the farthest point on its edge. In contrast, the semi-minor axis is shorter.

Determining which is which in equations can sometimes be tricky. In our Mars orbit example, calculate both \( a \) and \( b \) by taking their square roots. You will find:

• \( a = \sqrt{5013} \approx 70.8 \)
• \( b = \sqrt{4970} \approx 70.5 \)

Since \( a > b \), \( a \) is the semi-major axis (the longest one), indicating that \( a \)'s direction has the broader reach. The semi-major axis gives an idea of how stretched the ellipse is compared to a perfect circle, where both axes would be equal.

Ellipse Eccentricity Calculation

Eccentricity is a pivotal concept when discussing ellipses, as it measures how elongated the shape is. An eccentricity of 0 corresponds to a perfect circle, while values closer to 1 suggest a more stretched figure.

To calculate the eccentricity \( e \) of an ellipse, use the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
In Mars' orbit problem, substitute \( a^2 = 5013 \) and \( b^2 = 4970 \) into this formula:

• Calculate \( \frac{b^2}{a^2} = \frac{4970}{5013} \approx 0.9914 \)
• Then, \( e = \sqrt{1 - 0.9914} = \sqrt{0.0086} \approx 0.0927 \)

Thus, the eccentricity \( e = 0.0927 \) reveals that Mars' orbit is almost circular. This small eccentricity indicates just a slight elongation, making the ellipse barely discernible from a perfect circle.