Problem 28 - Algebra and Trigonometry with Analytic Geometry

Question

Moon phases The phases of the moon can be described using the phase angle \(\theta\), determined by the sun, the moon, and Earth, as shown in the figure. Because the moon orbits Earth, \(\theta\) changes during the course of a month. The area of the region \(A\) of the moon, which appears illuminated to an observer on Earth, is given by \(A=\frac{1}{2} \pi R^{2}(1+\cos \theta)\), where \(R=1080 \mathrm{mi}\) is the radius of the moon. Approximate \(A\) for the following positions of the moon: (a) \(\theta=0^{\circ}\) (full moon) (b) \(\theta=180^{\circ}\) (new moon) (c) \(\theta=90^{\circ}\) (first quarter) (d) \(\theta=103^{\circ}\)

Step-by-Step Solution

Answer

(a) 3.665 million sq mi, (b) 0 sq mi, (c) 1.833 million sq mi, (d) 2.835 million sq mi.

Step 1: Understanding the Formula

The formula to calculate the area of the illuminated region of the moon is \(A = \frac{1}{2} \pi R^2 (1 + \cos \theta)\). Here, \(R\) is a constant 1080 miles (the radius of the moon) and \(\theta\) is the phase angle. We will use this formula to find the illuminated area for different values of \(\theta\).

Step 2: Calculate for Full Moon

For a full moon, \(\theta = 0^{\circ}\). The cosine of 0 degrees is 1. Substitute \(\theta = 0^{\circ}\) into the formula: \[A = \frac{1}{2} \pi (1080)^2 (1 + 1) = \pi (1080)^2\]. Calculate the area to find \(A = 3.665 \times 10^6\) square miles.

Step 3: Calculate for New Moon

In the case of a new moon, \(\theta = 180^{\circ}\). The cosine of 180 degrees is -1. Substitute \(\theta = 180^{\circ}\) into the formula: \[A = \frac{1}{2} \pi (1080)^2 (1 - 1) = 0\]. Hence, no area is illuminated during a new moon.

Step 4: Calculate for First Quarter

For the first quarter, \(\theta = 90^{\circ}\). The cosine of 90 degrees is 0. Substitute \(\theta = 90^{\circ}\) into the formula: \[A = \frac{1}{2} \pi (1080)^2 (1 + 0) = \frac{1}{2} \pi (1080)^2\]. The area is \(1.833 \times 10^6\) square miles.

Step 5: Calculate for 103 Degrees

For \(\theta = 103^{\circ}\), we need to find the cosine value. Using a calculator, \(\cos 103^{\circ} \approx -0.226\). Substitute into the formula: \[A = \frac{1}{2} \pi (1080)^2 (1 - 0.226) = \frac{1}{2} \pi (1080)^2 \times 0.774\]. Calculating further, \(A \approx 2.835 \times 10^6\) square miles.

Key Concepts

Moon Phases

The moon phases are the different forms that the Moon takes when viewed from Earth. This happens because of the changing angles between the Sun, Earth, and Moon. As the Moon orbits the Earth, it goes through eight phases in a cycle: new moon, waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, last quarter, and waning crescent.
The full moon occurs when the Earth is between the Sun and the Moon, making the entire side of the Moon facing Earth illuminated. A new moon is when the Moon is between the Sun and Earth, leading to the side facing Earth being in shadow. During the first quarter, half of the Moon is illuminated and half is in shadow. Understanding these phases helps us learn about the illuminated area we observe from Earth.

Phase Angle

The phase angle, often denoted as \(\theta\), is the angle between the Moon, the Earth, and the Sun. It changes as the Moon moves along its orbit around the Earth, which causes the different moon phases. The phase angle directly affects how much of the Moon's surface is illuminated as seen from Earth.

• A phase angle of \(0^{\circ}\) corresponds to a full moon.
• A \(180^{\circ}\) angle corresponds to a new moon.
• A \(90^{\circ}\) angle represents the first quarter.
• Any other angle will show a different phase of the Moon.

In the given exercise, the calculation of the illuminated area relies largely on determining the phase angle.

Cosine Function

The cosine function is a trigonometric function that expresses the cosine of an angle, often used in relation to circles. In the context of the moon phases, the cosine function is used to determine the proportion of the illuminated area that we see.
The function \(\cos \theta\) helps in calculating how much light the Moon reflects at any given phase angle \(\theta\). For example:

• \(\cos(0^{\circ}) = 1\), indicating full illumination.
• \(\cos(180^{\circ}) = -1\), indicating no illumination (new moon).
• \(\cos(90^{\circ}) = 0\), indicating half the Moon is illuminated.
• \(\cos(103^{\circ}) \approx -0.226\), giving a partial view of the Moon's lighted side.

Using these values, the illuminated area can be found through the formula discussed.

Illuminated Area Calculation

Calculating the illuminated area of the Moon involves using a specific formula: \[A = \frac{1}{2} \pi R^2 (1 + \cos \theta)\]Where \(R\) is the moon's radius, given as 1080 miles in the exercise, and \(\theta\) is the phase angle. This formula combines the circular area formula with the cosine function to find the area that is visibly lit.

• For the full moon (\(\theta = 0^{\circ}\)): \[A = \pi (1080)^2\]
• For the new moon (\(\theta = 180^{\circ}\)): \[A = 0\]
• For the first quarter (\(\theta = 90^{\circ}\)): \[A = \frac{1}{2} \pi (1080)^2\]
• For \(\theta = 103^{\circ}\): \[A = \frac{1}{2} \pi (1080)^2 \times 0.774\]

Each calculation shows how changes in angle directly impact the moon's visible brightness.