Math Problem Statement

An orbit of a satellite around a planet is an ellipse with the planet at one focus of this ellipse. The distance of the sateihte from this stor varies froth 300,000 km to 500.000 km attained when the satellite is at each of the two vertices Find the equation of this ellipse if its center is at the origin and the vertices are on the r-axis. Assume all units are in 100.000 km

Solution

To find the equation of the ellipse for the satellite's orbit, we start by using the information provided:

  • The distance of the satellite from the planet varies from 300,000 km to 500,000 km.
  • The vertices are on the x-axis.
  • The center of the ellipse is at the origin.
  • The units are in 100,000100,000 km.

Step 1: Convert the distances into units

Given that the distances are in 100,000100,000 km, we convert:

  • The minimum distance (closest distance): 300,000 km=3300,000 \text{ km} = 3 units.
  • The maximum distance (farthest distance): 500,000 km=5500,000 \text{ km} = 5 units.

Step 2: Determine the values of aa, cc, and bb

In an ellipse centered at the origin, with the major axis along the x-axis, the standard form of the equation is:

x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Where:

  • aa is the semi-major axis.
  • cc is the distance from the center to each focus.
  • bb is the semi-minor axis.

Vertices: The vertices are located at (±a,0)(\pm a, 0), and since the vertices are at ±5\pm 5 units: a=5a = 5

Foci: The planet is at one of the foci of the ellipse. The distance from the center to the focus is cc, and this focus lies on the x-axis. The closest distance to the planet (at the other focus) is the periapsis (3 units). So, the distance to the focus cc is the difference between the semi-major axis and the periapsis: c=aperiapsis=53=2c = a - \text{periapsis} = 5 - 3 = 2

Calculate b2b^2 using the relationship c2=a2b2c^2 = a^2 - b^2: c2=a2b222=52b2c^2 = a^2 - b^2 \Rightarrow 2^2 = 5^2 - b^2 4=25b24 = 25 - b^2 b2=21b^2 = 21

Step 3: Write the equation of the ellipse

Now, substitute a2=25a^2 = 25 and b2=21b^2 = 21 into the standard form equation: x225+y221=1\frac{x^2}{25} + \frac{y^2}{21} = 1

This is the equation of the ellipse for the satellite's orbit.


Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

  1. What is the significance of the semi-minor axis in the context of an orbit?
  2. How does the eccentricity of the ellipse relate to the shape of the orbit?
  3. What would the equation of the ellipse be if the planet was not at the origin but at a different point?
  4. How would the equation change if the major axis was along the y-axis instead?
  5. What happens to the orbit if the distance from the satellite to the planet increases uniformly?

Tip: Remember that the foci of an ellipse play a crucial role in defining the eccentricity, which determines how "stretched" the ellipse is. The closer the foci are to each other, the more circular the ellipse.

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Math Problem Analysis

Mathematical Concepts

Geometry
Conic Sections
Ellipses

Formulas

Standard form of ellipse equation

Theorems

Properties of ellipses

Suitable Grade Level

Grades 11-12