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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.

Short Answer

Expert verified
The eccentricity of the orbit is approximately 0.0926.

Step by step solution

01

Identify the general form of the ellipse

The given equation of the ellipse is \( \frac{x^2}{5013} + \frac{y^2}{4970} = 1 \). This is in the standard form of an ellipse equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
02

Recognize semi-major and semi-minor axes

In an ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the terms \(a^2\) and \(b^2\) represent the squared lengths of the semi-major and semi-minor axes, respectively. Here, \(a^2 = 5013\) and \(b^2 = 4970\). Since \(5013 > 4970\), \(a^2 = 5013\) is the semi-major axis.
03

Calculate the semi-major and semi-minor axes

Calculate \(a\) and \(b\) by taking the square roots: \( a = \sqrt{5013} \) and \( b = \sqrt{4970} \).
04

Find the focus of the ellipse

The foci of the ellipse are located at a distance \( c \) from the center, where \( c = \sqrt{a^2 - b^2} \). Substituting the values, \( c = \sqrt{5013 - 4970} = \sqrt{43} \).
05

Compute the eccentricity

The eccentricity \( e \) of an ellipse is given by the formula \( e = \frac{c}{a} \). Substituting the known values, \( e = \frac{\sqrt{43}}{\sqrt{5013}} \). Calculate \( e \).
06

Simplify the eccentricity calculation

First, calculate \( \sqrt{43} \approx 6.56 \) and \( \sqrt{5013} \approx 70.79 \). Thus, \( e = \frac{6.56}{70.79} \approx 0.0926 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ellipse Equation
An ellipse is a geometric shape that looks like a stretched circle. To describe an ellipse mathematically, we use the ellipse equation. The standard form of the ellipse equation is: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This represents the set of all points \((x, y)\) that form the shape of the ellipse. Here, \(a^2\) and \(b^2\) denote the squares of the lengths along the principal axes of the ellipse. The values of \(a\) and \(b\) determine how the ellipse is stretched or compressed along the x-axis and y-axis, respectively. In the orbit of Mars, the corresponding equation \( \frac{x^2}{5013} + \frac{y^2}{4970} = 1 \) shows the unique shape and orientation of Mars' orbit around the sun.
Semi-Major Axis
The semi-major axis of an ellipse is the longest radius of the ellipse. It extends from the center to the farthest point on the ellipse's boundary. Mathematically, the semi-major axis is denoted by \(a\) and in the standard ellipse equation, it corresponds to \(a^2\), the larger denominator in \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). For Mars’ orbit, we determined that \(a^2 = 5013\). To find \(a\), we simply take the square root: \(a = \sqrt{5013}\). This length reflects how stretched the ellipse is along its longest axis. The semi-major axis is crucial because it helps astronomers understand the orbit's shape and size.
Semi-Minor Axis
The semi-minor axis is the shorter radius of an ellipse and spans from the center to the shortest edge of the ellipse. It contrasts with the semi-major axis by being the shortest path to the ellipse's boundary. It is represented by \(b\) in the standard equation. For Mars’ orbital ellipse, we calculate \(b\) from \(b^2 = 4970\) as \(b = \sqrt{4970}\). This axis is essential to understanding how 'tall' or 'compressed' the ellipse appears. Comparing it to the semi-major axis provides insight into the eccentricity, or how much an ellipse deviates from being a perfect circle — a concept that's important when considering the eccentricity of planetary orbits.
Focus of Ellipse
Every ellipse has two points inside it called foci (singular: focus). These points are vital because they are used to define the ellipse. Basically, for any point on the ellipse, the total distance to both foci is constant. For our specific ellipse equation, the focus is derived from the values of \(a\) and \(b\). We calculate \(c\), the distance from the center to a focus, using the formula: \( c = \sqrt{a^2 - b^2} \). Substituting the numbers for Mars’ orbit, we find \( c = \sqrt{5013 - 4970} = \sqrt{43} \). Knowing the foci's distances helps in finding the eccentricity \(e\), which is \( e = \frac{c}{a} \). This measurement tells us how "stretched" the ellipse is compared to a circle, with Mars’ orbit eccentricity calculating approximately as 0.0926.

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Most popular questions from this chapter

Find an equation of a parabola that satisfies the given conditions. Focus \((2,1) ;\) directrix \(x=-1\)

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