Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

The position of a comet with a highly eccentric elliptical orbit ( \(e\) very near 1 ) is measured with respect to a fixed polar axis (sun is at a focus but the polar axis is not an axis of the ellipse) at two times, giving the two points \((4, \pi / 2)\) and \((3, \pi / 4)\) of the orbit. Here distances are measured in astronomical units ( \(1 \mathrm{AU} \approx 93\) million miles). For the part of the orbit near the sun, assume that \(e=1\), so the orbit is given by $$ r=\frac{d}{1+\cos \left(\theta-\theta_{0}\right)} $$ (a) The two points give two conditions for \(d\) and \(\theta_{0}\). Use them to show that \(4.24 \cos \theta_{0}-3.76 \sin \theta_{0}-2=0\). (b) Solve for \(\theta_{0}\) using Newton's Method. (c) How close does the comet get to the sun?

Step-by-Step Solution

Answer

\(\theta_0\) satisfies \(4.24 \cos \theta_0 - 3.76 \sin \theta_0 - 2 = 0\). Use Newton's Method to find \(\theta_0\). Closest distance is \(\frac{d}{2}\).

Step 1: Understanding the Equation and Points

The orbit of the comet is described by the polar equation \( r = \frac{d}{1 + \cos(\theta - \theta_0)} \). We are given two points in polar coordinates \((4, \pi/2)\) and \((3, \pi/4)\). At these angles, the distances \(r\) are 4 and 3 astronomical units respectively.

Step 2: Setting Up Equations for Known Points

Substituting \((r, \theta) = (4, \pi/2)\) and \((3, \pi/4)\) into the equation yields:1. For \( (4, \pi/2) \): \[ 4 = \frac{d}{1 + \cos(\pi/2 - \theta_0)} \] Which simplifies to: \[ d = 4(1 + \cos(\pi/2 - \theta_0)) \]2. For \( (3, \pi/4) \): \[ 3 = \frac{d}{1 + \cos(\pi/4 - \theta_0)} \] Which simplifies to: \[ d = 3(1 + \cos(\pi/4 - \theta_0)) \]

Step 3: Equate the Two Expressions for d

We have two expressions for \(d\):\[ 4(1 + \cos(\pi/2 - \theta_0)) = 3(1 + \cos(\pi/4 - \theta_0)) \]This can be expanded and simplified. Use trigonometry identities: - \( \cos(\pi/2 - \theta_0) = \sin \theta_0 \) - \( \cos(\pi/4 - \theta_0) = \cos \frac{\pi}{4}\cos \theta_0 + \sin \frac{\pi}{4}\sin \theta_0 \)

Step 4: Deriving the Condition for Theta_0

Substitute the trigonometric identities:\[ 4(1 + \sin \theta_0) = 3(1 + \frac{\sqrt{2}}{2} \cos \theta_0 + \frac{\sqrt{2}}{2} \sin \theta_0) \]Expanding and simplifying both sides yields:\[ 4 + 4 \sin \theta_0 = 3 + \frac{3\sqrt{2}}{2} \cos \theta_0 + \frac{3\sqrt{2}}{2} \sin \theta_0 \]Rearranging terms results in the equation:\[ 4.24 \cos \theta_0 - 3.76 \sin \theta_0 - 2 = 0 \]

Step 5: Applying Newton's Method to Solve for Theta_0

Using Newton's Method, iterate to find \(\theta_0\) that satisfies the equation \(4.24 \cos \theta_0 - 3.76 \sin \theta_0 - 2 = 0\).1. Choose an initial guess, \( \theta_0^{(0)}\).2. Compute the function \( f(\theta_0) = 4.24 \cos \theta_0 - 3.76 \sin \theta_0 - 2 \).3. Compute the derivative, \( f'(\theta_0) = -4.24 \sin \theta_0 - 3.76 \cos \theta_0 \).4. Iterate using \( \theta_0^{(n+1)} = \theta_0^{(n)} - \frac{f(\theta_0^{(n)})}{f'(\theta_0^{(n)})} \) until converged.

Step 6: Calculating Closest Distance to the Sun

Once \(\theta_0\) is determined via Newton's Method, compute the closest distance to the sun, which is when \(r\) is minimized, i.e., \(\theta = \theta_0\).Minimize \(r = \frac{d}{1 + \cos(\theta - \theta_0)} \) by setting \(\theta = \theta_0\) which simplifies to \( r = \frac{d}{1 + 1} = \frac{d}{2} \).

Key Concepts

Newton's Method

Newton's Method is a numerical technique used to find approximate solutions to equations. It is particularly useful for solving non-linear equations that are difficult to handle algebraically. In our exercise, we used it to determine the value of \( \theta_0 \) for the comet's orbit equation, given that an analytical solution was not straightforward.
Newton's Method involves using the derivative of a function to iteratively converge on a root. The steps are as follows:

• Choose an initial guess \( \theta_0^{(0)} \).
• Evaluate the function at this guess, \( f(\theta_0) = 4.24 \cos \theta_0 - 3.76 \sin \theta_0 - 2 \).
• Calculate the derivative, \( f'(\theta_0) = -4.24 \sin \theta_0 - 3.76 \cos \theta_0 \).
• Update the guess using \( \theta_0^{(n+1)} = \theta_0^{(n)} - \frac{f(\theta_0^{(n)})}{f'(\theta_0^{(n)})} \).

Repeat these steps until the change between successive \( \theta_0 \) values is small, indicating convergence. This method harnesses the power of calculus to find solutions where simple algebra may fall short.

Trigonometric Identities

Trigonometric identities play a crucial role in simplifying complex equations, particularly in polar coordinates. In this exercise, we use identities to simplify the problem of determining \(d\) and \(\theta_0\) from the given points on the comet's orbit.
Here are the identities we used:

• \( \cos(\pi/2 - \theta_0) = \sin \theta_0 \)
• \( \cos(\pi/4 - \theta_0) = \cos \frac{\pi}{4}\cos \theta_0 + \sin \frac{\pi}{4}\sin \theta_0 \)

By substituting these identities into the orbit equation, we managed to express both known points in terms of \(d\) and \(\theta_0\). This allowed us to set up an equation that we could then solve using Newton's Method.
Understanding and applying trigonometric identities like these is vital in problems involving angles and distances.

Eccentric Elliptical Orbits

Eccentric elliptical orbits describe paths where the shape of the orbit is significantly elongated. The eccentricity \( e \) is a measure of how much the orbit deviates from being circular. In this exercise, the eccentricity is assumed to be very close to 1, indicating a highly elongated path.
For a comet orbiting the sun, this means the comet gets very close to the sun at one point (perihelion) and then travels far away. In polar coordinates, such an orbit is given by the equation \( r = \frac{d}{1 + \cos(\theta - \theta_0)} \). This model assumes a near-parabolic trajectory, reflecting the high eccentricity.

• \(d\) represents the semi-latus rectum of the orbit.
• The angle \( \theta - \theta_0 \) determines the position on the orbit.

Understanding this concept allows us to describe the motion of celestial bodies with non-circular paths, critical for predicting the movement of comets and other astronomical objects.

Astronomical Units

Astronomical Units (AU) are a standard unit of length used in astronomy to describe distances within our solar system. One AU is roughly the mean distance from the Earth to the Sun, approximately 93 million miles or 150 million kilometers.
Using AU simplifies the representation of large distances between celestial objects, making calculations more manageable. In the context of the exercise, distances described in AU allow us to relate the position of the comet to familiar measurements, easing comprehension.

• AU is preferred over kilometers or miles in astronomical calculations due to its convenience and standardization.
• It provides a common frame of reference for comparing the distances of planets, comets, and other bodies in the solar system.

Employing AU helps students and astronomers understand and communicate space distances efficiently, enhancing the clarity of astronomical discourse.