To find the equation of the ellipse for the orbit of the satellite, we need to determine its semi-major axis, semi-minor axis, and the location of its foci. The general equation of an ellipse centered at the origin, with its major axis along the x-axis, is:

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

where:

• $a$ is the semi-major axis.
• $b$ is the semi-minor axis.
• The foci of the ellipse are located at $(\pm c, 0)$, where $c = \sqrt{a^2 - b^2}$.

Step 1: Determine the semi-major axis $a$

The vertices of the ellipse are at the maximum and minimum distances of the satellite from the planet. These distances are given as 300,000 km and 500,000 km. Since the center of the ellipse is at the origin, the semi-major axis $a$ is the distance from the center to one of the vertices.

Given the units in 100,000 km, the distances become:

• Closest distance: $3$ (300,000 km)
• Furthest distance: $5$ (500,000 km)

Since these distances are from the center to each vertex, the semi-major axis $a$ is half the length of the distance between the two vertices:

$a = \frac{5 + 3}{2} = 4$

Step 2: Determine the focal distance $c$

The distance between a focus and the center is $c = \sqrt{a^2 - b^2}$. The closest distance (300,000 km) corresponds to the periapsis (closest approach to the focus), and the farthest distance (500,000 km) corresponds to the apoapsis (farthest distance from the focus).

The distance from the center to a vertex is $a = 4$, and the distance from a focus to a vertex is $a - c = 3$. Thus:

$c = a - 3 = 4 - 3 = 1$

Step 3: Determine the semi-minor axis $b$

We know that $c = 1$ and $a = 4$, so we can find $b$ using the relationship:

$c = \sqrt{a^2 - b^2}$

Substitute the values for $c$ and $a$:

$1 = \sqrt{4^2 - b^2}$

$1 = \sqrt{16 - b^2}$

Square both sides to solve for $b^2$:

$1^2 = 16 - b^2$

$1 = 16 - b^2$

$b^2 = 15$

Thus:

$b = \sqrt{15}$

Step 4: Write the equation of the ellipse

Now that we have $a = 4$ and $b = \sqrt{15}$, we can write the equation of the ellipse:

$\frac{x^2}{4^2} + \frac{y^2}{(\sqrt{15})^2} = 1$

Simplify the equation:

$\frac{x^2}{16} + \frac{y^2}{15} = 1$

Final Answer

The equation of the ellipse, representing the orbit of the satellite around the planet, is:

$\frac{x^2}{16} + \frac{y^2}{15} = 1$

Would you like more details or have any questions about this solution?

Related Questions

1. How do you find the eccentricity of an ellipse given its semi-major and semi-minor axes?
2. What is the significance of the foci in an elliptical orbit in astronomy?
3. How would the equation change if the semi-major axis was along the y-axis instead?
4. What happens to the shape of the ellipse if the distances to the vertices are equal?
5. How do you derive the area of an ellipse given its semi-major and semi-minor axes?

Tip

When working with ellipses, remember that the sum of the distances from any point on the ellipse to the two foci is constant and equal to the length of the major axis.