Problem 25 - Calculus Early Transcendentals

Question

The orbit of Mars around the sun is an ellipse with eccen- tricity 0.093 and semimajor axis \(2.28 \times 10^{8} \mathrm{km} .\) Find a polar equation for the orbit.

Step-by-Step Solution

Answer

The polar equation for Mars' orbit is \( r = \frac{2.26042608 \times 10^8}{1 + 0.093 \cos \theta} \).

Step 1: Understanding Elliptical Orbits

The orbit of Mars is an ellipse, which can be described using a polar equation. The two key variables involved are the eccentricity (\(e\)) and the semimajor axis (\(a\)). These define the shape and size of the orbit.

Step 2: Form of a Polar Equation for an Ellipse

The polar equation for an ellipse can be written as \( r = \frac{a(1-e^2)}{1 + e \cos \theta} \), where \( r \) is the distance from the focus to a point on the ellipse, \( e \) is the eccentricity, and \( \theta \) is the angle from the closest approach.

Step 3: Substitute Given Values

We know that the eccentricity \(e\) is 0.093 and the semimajor axis \(a\) is \(2.28 \times 10^{8}\) km. Substitute these values into the polar equation of the ellipse: \[ r = \frac{2.28 \times 10^8 (1 - (0.093)^2)}{1 + 0.093 \cos \theta} \].

Step 4: Simplify the Expression

Calculate \((1 - (0.093)^2)\): \((0.093)^2 = 0.008649\) so \(1 - 0.008649 = 0.991351\). The polar equation becomes \[r = \frac{2.28 \times 10^8 \times 0.991351}{1 + 0.093 \cos \theta} \]. Multiply to simplify further: \(2.28 \times 10^8 \times 0.991351 = 2.26042608 \times 10^8 \).

Step 5: Final Equation

The simplified polar equation for Mars' orbit is \( r = \frac{2.26042608 \times 10^8}{1 + 0.093 \cos \theta} \). This equation represents the orbit of Mars around the Sun in polar coordinates.

Key Concepts

Understanding Ellipses

An ellipse is a shape that looks like a stretched out circle or an oval. It has two main axes: the major axis and the minor axis. The major axis is the longest diameter of the ellipse, while the minor axis is the shortest. In an ellipse:

• The center is the midpoint of both the major and minor axes.
• There are two focal points (foci), and the sum of the distances from any point on the ellipse to the two foci is a constant.

When discussing the orbits of planets, it's important to note that many of them travel in an elliptical path around the Sun. This is why understanding ellipses is crucial for celestial mechanics. The longer the distance between the two foci, the more stretched or eccentric the ellipse is.

Exploring Eccentricity

Eccentricity is a measure of how much an orbit deviates from being a perfect circle. It is a number between 0 and 1, denoted as \(e\).

• If \(e = 0\), the shape is a perfect circle.
• If \(0 < e < 1\), the orbit is an ellipse.
• As \(e\) approaches 1, the ellipse looks more like a flattened shape.

For example, in the provided solution, the eccentricity \(e\) of Mars’ orbit is 0.093, which indicates it is almost circular but still slightly elliptical. Knowing the eccentricity helps us understand how "stretched out" the orbit is. Lower eccentricity means the planet's path is closer to a circle. This value plays a key role in predicting how the planet moves along its orbit, altering the distance from the Sun at different points in its path.

Semimajor Axis and Its Importance

The semimajor axis is half of the longest diameter in the ellipse. It is denoted as \(a\) and is an integral part of defining the size of an ellipse. In planetary orbits:

• The semimajor axis represents the average distance from the planet to the Sun.
• It is one of the elements used in Kepler's laws of planetary motion.

For Mars, the semimajor axis is \(2.28 \times 10^{8}\) km, indicating how far the planet is from the Sun on average. The semimajor axis also helps in determining the orbital period, which is the time it takes for a planet to complete one full orbit around the Sun. Understanding this concept is crucial for astronomers, as it helps them calculate the dynamics of celestial bodies in space.