Problem 39 - Algebra and Trigonometry

Question

Orbit of a Satellite A satellite is in an elliptical orbit round the earth with the center of the earth at one focus. The height of the satellite above the earth varies between 140 \(\mathrm{mi}\) and 440 \(\mathrm{mi}\) . Assume the earth is a sphere with radius 3960 \(\mathrm{mi}\) . Find an equation for the path of the satellite with the origin at the center of the earth.

Step-by-Step Solution

Answer

The equation of the satellite's path is \(\frac{x^2}{18062500} + \frac{y^2}{17868900} = 1\).

Step 1: Determine the Semi-Major Axis

The semi-major axis, \(a\), is the average of the minimum and maximum distances from the center of the Earth to the satellite. The minimum distance is \(3960 + 140 = 4100\, \text{mi}\) and the maximum distance is \(3960 + 440 = 4400\, \text{mi}\). Thus, \(a = \frac{4100 + 4400}{2} = 4250\, \text{mi}\).

Step 2: Calculate the Semi-Minor Axis Using the Foci Property

Given that the semi-major axis is \(a = 4250\, \text{mi}\) and the Earth is at one of the foci, we determine \(c\) using \(c = 4400 - 3960 = 440\, \text{mi}\) (the maximum minus Earth's radius since this is the longest focus-to-satellite distance). Now, use the relationship \(b^2 = a^2 - c^2\): \[b^2 = 4250^2 - 440^2\]. Calculate this to find \(b\).

Step 3: Simplify the Semi-Minor Axis Calculation

We have \(b^2 = 4250^2 - 440^2 = 18062500 - 193600 = 17868900\). Thus, \(b = \sqrt{17868900} \approx 4228.8\, \text{mi}\).

Step 4: Formulate the Equation of the Ellipse

The standard equation for an ellipse centered at the origin with a horizontal major axis is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Substitute \(a = 4250\) and \(b = 4228.8\) to get the equation \[\frac{x^2}{4250^2} + \frac{y^2}{4228.8^2} = 1\].

Step 5: Simplify and Present the Final Equation

Calculate \(4250^2 = 18062500\) and \(4228.8^2 \approx 17868900\). Hence, the equation of the satellite's path is \[\frac{x^2}{18062500} + \frac{y^2}{17868900} = 1\].

Key Concepts

Semi-Major Axis

In an elliptical orbit, the semi-major axis is a crucial element as it represents half the longest dimension of the ellipse. In the context of orbital mechanics, it plays a significant role. For our satellite, this axis determines the average distance from the center of the Earth, around which the satellite revolves.

The calculation of the semi-major axis involves taking the average of the minimum and maximum distances of the satellite from the center of the Earth. These distances are derived from the Earth's center to the satellite at its closest and farthest points.

• The minimum distance reached by the satellite is the sum of Earth's radius and the satellite's closest approach, calculated as: 3960 miles + 140 miles = 4100 miles.
• The maximum distance is Earth's radius plus the satellite's farthest orbit point, which is: 3960 miles + 440 miles = 4400 miles.

To find the semi-major axis, average these two distances: \[ a = \frac{4100 + 4400}{2} = 4250 \text{ miles} \]
This axis not only sets a scale for the ellipse but also is pivotal in determining the orbit period and energy dynamics.

Semi-Minor Axis

The semi-minor axis is the second primary measure of an ellipse's size, and it stretches along the shortest dimension of the ellipse. In satellite orbits, it complements the semi-major axis and helps in shaping the orbit's specific characteristics.

To find the semi-minor axis, we utilize the relationship between the semi-major axis, the focal distance, and the semi-minor axis itself \( b^2 = a^2 - c^2 \). Here, \(c\) represents the focal distance, which is the distance from the Earth's center to the satellite at its furthest point, taking into account the focus of the ellipse at Earth's center.

• The focal distance, \(c\), was determined to be the longest line from focus to satellite minus Earth's radius: 440 miles (4400 miles - 3960 miles).

The equation to calculate the semi-minor axis becomes: \[ b^2 = 4250^2 - 440^2 \] Calculating it gives:
1. \( 4250^2 = 18062500 \)
2. \( 440^2 = 193600 \)
Thus, \( b^2 = 18062500 - 193600 = 17868900 \)
And the semi-minor axis is approximately:
\[ b = \sqrt{17868900} \approx 4228.8 \text{ miles} \]
This dimension completes the size and set-up of the elliptical orbit, allowing it to adjust to the gravitational influences of the planet and keep its elliptical shape.

Equation of an Ellipse

The equation of an ellipse mathematically represents the path of the satellite's orbit in our scenario. It relates various points on the elliptical path to the axes' lengths, providing a formula to describe the orbit in a standard coordinate system.

For an ellipse centered at the origin with the major axis along the x-axis, the equation is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting the known semi-major and semi-minor axes values, we formulate the satellite's orbit equation.

• With \( a = 4250 \) miles and \( b = 4228.8 \) miles, substitute into the ellipse equation:

\[ \frac{x^2}{4250^2} + \frac{y^2}{4228.8^2} = 1 \]
The equation simplifies with these calculated squares:

• \(4250^2 = 18062500\)
• \(4228.8^2 \approx 17868900\)

Thus the final form of the ellipse's equation:\[ \frac{x^2}{18062500} + \frac{y^2}{17868900} = 1 \]
This expression provides a precise mathematical description of the satellite's elliptical path, showcasing how these parameters control and allow prediction of the orbital mechanics in play.