To solve these problems, we'll use the properties of ellipses. An ellipse with its center at the origin and aligned with the coordinate axes can be described by the equation:

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

where $a$ is the semi-major axis (half of the longest diameter) and $b$ is the semi-minor axis (half of the shortest diameter). The foci are located at $\pm c$ along the major axis, where $c = \sqrt{a^2 - b^2}$.

Problem 1: Satellite Orbit

• Closest distance to the planet (periapsis): 300,000 km
• Farthest distance from the planet (apoapsis): 500,000 km

The distance from the center of the ellipse to the vertices is the semi-major axis $a$, and the total distance is 800,000 km. Hence:

$a = \frac{500,000 + 300,000}{2} = 400,000 \, \text{km}$

The distance between the foci (which is twice the distance from the center to a focus, $2c$) is the difference between the apoapsis and periapsis distances:

$2c = 500,000 - 300,000 = 200,000 \, \text{km}$

Thus,

$c = \frac{200,000}{2} = 100,000 \, \text{km}$

Now, using the relationship $c = \sqrt{a^2 - b^2}$:

$100,000 = \sqrt{400,000^2 - b^2}$

$(100,000)^2 = (400,000)^2 - b^2$

$10^8 = 16 \times 10^8 - b^2$

$b^2 = 16 \times 10^8 - 10^8 = 15 \times 10^8$

$b = \sqrt{15} \times 10^4 \approx 387,298 \, \text{km}$

Since units are in 100,000 km:

• $a = 4$
• $b = \sqrt{15} \approx 3.87$

The equation of the ellipse is:

$\frac{x^2}{4^2} + \frac{y^2}{(\sqrt{15})^2} = 1 \implies \frac{x^2}{16} + \frac{y^2}{15} = 1$

Problem 2: Planet Orbit

• Closest distance to the star: 420 million km
• Farthest distance from the star: 580 million km

The total length of the major axis is:

$a = \frac{580 + 420}{2} = 500 \, \text{million km}$

The distance between the foci (distance from closest to farthest point) is:

$2c = 580 - 420 = 160 \, \text{million km}$

So,

$c = \frac{160}{2} = 80 \, \text{million km}$

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

$80 = \sqrt{500^2 - b^2}$

$80^2 = 500^2 - b^2$

$6400 = 250000 - b^2$

$b^2 = 250000 - 6400 = 243600$

$b = \sqrt{243600} \approx 493.56 \, \text{million km}$

The equation of the ellipse is:

$\frac{x^2}{500^2} + \frac{y^2}{493.56^2} = 1 \implies \frac{x^2}{250000} + \frac{y^2}{243600} = 1$

Questions to Explore Further:

1. How does changing the distance between the foci affect the shape of an ellipse?
2. What is the significance of the semi-major and semi-minor axes in defining an orbit?
3. Can the equation of an ellipse be derived if only one of the distances (closest or farthest) is known?
4. How are ellipses used to model other celestial orbits, such as those of comets?
5. How does the eccentricity of an ellipse relate to the orbit's shape and the distance between foci?

Tip:

In problems involving orbits, distances from the central focus to the orbiting object at closest and farthest points provide direct information about the lengths of the semi-major axis and the focal distance, which are key to determining the equation of the orbit.