Question

For the following exercises, determine the polar equation form of the orbit given the length of the major axis and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units \((A U)\). Jupiter: length of major axis \(=10.408\), eccentricity \(=0.0484\)

Short answer

The polar equation of Jupiter's orbit is \( r(\theta) = \frac{5.191}{1 + 0.0484 \cos\theta} \).

Step-by-step solution

Identify the Polar Equation of an Elliptical Orbit

The polar equation for an ellipse in the form of an orbit is given by \( r(\theta) = \frac{p}{1 + e \cos\theta} \), where \( p \) is the semi-latus rectum and \( e \) is the eccentricity.

Calculate the Semi-Major Axis

The semi-major axis \( a \) is found by dividing the length of the major axis by 2. Thus, \( a = \frac{10.408}{2} = 5.204 \) AU.

Calculate the Semi-Latus Rectum

The semi-latus rectum \( p \) can be calculated using the formula \( p = a(1 - e^2) \). Substituting \( a = 5.204 \) and \( e = 0.0484 \), we get \( p = 5.204(1 - (0.0484)^2) \).

Solve for Semi-Latus Rectum (p)

Calculate \( (0.0484)^2 = 0.00234256 \). Then \( 1 - 0.00234256 = 0.99765744 \). Therefore, \( p = 5.204 \times 0.99765744 = 5.191 \) AU.

Write the Polar Equation

Substitute \( p = 5.191 \) and \( e = 0.0484 \) into the polar equation: \( r(\theta) = \frac{5.191}{1 + 0.0484 \cos\theta} \).

Key concepts

Ellipse Orbit

In celestial mechanics, many orbits, such as those of planets and comets, are in the shape of an ellipse. An ellipse is a kind of elongated circle that has two focuses, with the central body being located at one of these foci. When we talk about elliptical orbits in the context of polar coordinates, the position of a point on an ellipse can be described by a polar equation.
In the case of Jupiter's orbit, the polar equation used is:

• \( r(\theta) = \frac{p}{1 + e \cos\theta} \)

where \( r \) is the radius, \( \theta \) is the angle, \( p \) is the semi-latus rectum, and \( e \) is the orbit's eccentricity. The semi-latus rectum represents one dimension of the ellipse related to its shape, and substituting different values in \( \theta \) allows us to track the path of the orbit on a two-dimensional plane.
This allows us to visualize how far the orbiting body is from the central focus at various points in its orbit, making it an essential tool for astronomers studying planetary orbits.

Astronomical Units

Astronomical Units (AU) are a standard unit of measure used in astronomy to describe large distances, particularly within our solar system. One AU is approximately equal to the average distance from the Earth to the Sun, which is about 93 million miles or 150 million kilometers. Using AUs simplifies the expression of distances that would otherwise involve very large numbers. For example, instead of saying "93 million miles," astronomers can simply say "1 AU."
In the problem with Jupiter, expressing the length of the major axis, which is 10.408 AU, becomes much more manageable than using miles or kilometers. This allows for easier calculations when dealing with planetary distances and makes comparison between different celestial objects more straightforward.
By using AU, scientists can express and calculate distances in a manner that reduces complexity while maintaining accuracy. This standardization plays a crucial role in orbital calculations and enhances communication in the astronomical community.

Eccentricity Calculation

Eccentricity is a mathematical term used to describe the shape of an ellipse. In orbits, it denotes how much an orbit deviates from being circular. A circle has an eccentricity of 0, while more elongated ellipses have values approaching 1. The eccentricity for orbits is important because it helps define the shape and characteristics of an orbit.
To calculate the eccentricity of a planet like Jupiter, we use the formula:

• \( e = \frac{c}{a} \)

where \( c \) is the distance from the center to a focus, and \( a \) is the semi-major axis. However, in the exercise, the eccentricity is given as 0.0484.
Another equation involving eccentricity is used to find the semi-latus rectum \( p \), through:

• \( p = a(1 - e^2) \)

By providing the eccentricity and the length of the major axis, we can find other elements of the ellipse, such as the semi-latus rectum, which helps describe the shape and size of the orbit more accurately.