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

To find the equation of the ellipse that describes the orbit of a satellite around a planet, we need to identify key properties of the ellipse:

1. **Vertices**: The points on the ellipse that are closest and farthest from the center.  
2. **Foci**: The two points inside the ellipse for which the sum of distances to any point on the ellipse is constant.
3. **Major Axis**: The longest diameter of the ellipse, which passes through both foci and vertices.
4. **Minor Axis**: The shortest diameter of the ellipse, perpendicular to the major axis.

Given the problem:

- The planet is at one focus of the ellipse.
- The distance of the satellite varies from \(300,000\) km to \(500,000\) km.
- The center of the ellipse is at the origin \((0, 0)\).
- The vertices are along the x-axis.

### Step-by-Step Solution:

1. **Convert units**: Since all units are in \(100,000\) km, the distances become:
   - Closest distance (vertex at left): \(300,000 \, \text{km} = 3 \times 100,000 \, \text{km} = 3\).
   - Farthest distance (vertex at right): \(500,000 \, \text{km} = 5 \times 100,000 \, \text{km} = 5\).

2. **Determine the center and vertices of the ellipse**:
   - The center of the ellipse is at \((0, 0)\).
   - The vertices are at \((-5, 0)\) and \((5, 0)\) along the x-axis.
   - The semi-major axis length \(a\) is the distance from the center to either vertex. So, \(a = 5\).

3. **Calculate the distance to the foci**:
   - The closest distance (perigee) from the center to the satellite is \(3\). This is the semi-major axis \(a\) minus the focal distance \(c\).
   - The farthest distance (apogee) from the center to the satellite is \(5\). This is the semi-major axis \(a\) plus the focal distance \(c\).

   Using these distances:
   \[
   a - c = 3 \quad \text{and} \quad a + c = 5
   \]

   Solving these equations:
   \[
   a = 5, \quad c = 5 - 3 = 2
   \]

4. **Calculate the semi-minor axis \(b\)**:
   - The relationship between the semi-major axis \(a\), semi-minor axis \(b\), and focal distance \(c\) is given by:
   \[
   c^2 = a^2 - b^2
   \]
   Plugging in the known values:
   \[
   2^2 = 5^2 - b^2 \implies 4 = 25 - b^2 \implies b^2 = 21
   \]

5. **Equation of the ellipse**:
   - The standard form of the ellipse equation centered at the origin with its major axis along the x-axis is:
   \[
   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
   \]
   Plugging in \(a^2 = 25\) and \(b^2 = 21\):
   \[
   \frac{x^2}{25} + \frac{y^2}{21} = 1
   \]

### Conclusion

The equation of the ellipse that describes the orbit of the satellite is:
\[
\frac{x^2}{25} + \frac{y^2}{21} = 1
\]

Would you like more details or have any questions? Here are some related questions that might be interesting:

1. How would the equation change if the semi-minor axis was longer than the semi-major axis?
2. How do you determine the coordinates of the foci of this ellipse?
3. What is the significance of the focal distance in determining an ellipse's shape?
4. How does the eccentricity of the ellipse affect the orbit of a satellite?
5. What would be the equation of the ellipse if the orbit was centered at a point other than the origin?

**Tip:** In orbital mechanics, the shape and size of an orbit are often described by the ellipse's semi-major axis and eccentricity, which provide key information about the energy and period of the orbit.

Learn how to find the equation of an ellipse describing the orbit of a satellite around a planet. This problem involves key geometric concepts, including ellipse properties and distance calculations. Suitable for advanced high school students and enthusiasts of orbital mechanics.

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