/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Review Conceptual Example 2 befo... [FREE SOLUTION] | 91Ó°ÊÓ

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Review Conceptual Example 2 before attempting to work this problem. The moon has a diameter of \(3.48 \times 10^{6} \mathrm{~m}\) and is a distance of \(3.85 \times 10^{8} \mathrm{~m}\) from the earth. The sun has a diameter of \(1.39 \times 10^{9} \mathrm{~m}\) and is \(1.50 \times 10^{11} \mathrm{~m}\) from the earth. (a) Determine (in radians) the angles subtended by the moon and the sun, as measured by a person standing on the earth. (b) Based on your answers to part (a), decide whether a total eclipse of the sun is really "total." Give your reasoning. (c) Determine the ratio, expressed as a percentage, of the apparent circular area of the moon to the apparent circular area of the sun.

Short Answer

Expert verified
(a) Moon: 9.04x10^-3 rad, Sun: 9.27x10^-3 rad; (b) Not truly total; (c) 95.26%.

Step by step solution

01

Calculating the Angle Subtended by the Moon

To find the angle subtended by an object, use the formula \( \theta = \frac{d}{D} \), where \( d \) is the diameter of the object, and \( D \) is the distance from the observer to the object. For the moon: \( \theta_{\text{moon}} = \frac{3.48 \times 10^6 \text{ m}}{3.85 \times 10^8 \text{ m}} = 9.04 \times 10^{-3} \text{ radians}. \)
02

Calculating the Angle Subtended by the Sun

Similarly, use the same formula for the sun: \( \theta_{\text{sun}} = \frac{1.39 \times 10^9 \text{ m}}{1.50 \times 10^{11} \text{ m}} = 9.27 \times 10^{-3} \text{ radians}. \)
03

Comparing Angles to Determine Eclipse Totality

Compare the angles subtended by the moon and the sun. Since \( \theta_{\text{moon}} = 9.04 \times 10^{-3} \) radians is slightly less than \( \theta_{\text{sun}} = 9.27 \times 10^{-3} \) radians, a total eclipse is not completely total because the moon's angle is smaller.
04

Finding the Ratio of Apparent Circular Areas

The apparent circular area of a celestial body is proportional to the square of the angle it subtends: \( A \propto \theta^2 \). Calculate the ratio: \( \left(\frac{\theta_{\text{moon}}}{\theta_{\text{sun}}}\right)^2 = \left(\frac{9.04 \times 10^{-3}}{9.27 \times 10^{-3}}\right)^2 = 0.9526. \)
05

Expressing the Ratio as a Percentage

Convert the decimal ratio to a percentage to express the apparent area ratio: \( 0.9526 \times 100\% = 95.26\% \). This means the apparent area of the moon is about 95.26% of the sun's apparent area.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eclipse Totality
An eclipse occurs when one celestial body moves into the shadow of another. In particular, a total solar eclipse happens when the moon completely covers the sun, as viewed from Earth. The key to understanding eclipse totality lies in the angles subtended by the moon and the sun. These angles determine whether or not the moon can entirely obscure the sun in the sky.
For a total solar eclipse to occur, the apparent angular size of the moon, as seen from Earth, must be equal to or greater than that of the sun. If the angle subtended by the moon is slightly less than that of the sun, you witness a partial or annular eclipse, where a ring of the sun’s surface, known as the solar corona, is still visible.
Thus, eclipse totality is reached when the moon’s apparent size is large enough to fully obscure the sun, resulting in the brief period of darkness we call totality during an eclipse.
Angular Diameter
The angular diameter is a measure of how large an object appears to be from a given point, expressed in radians. It's determined by the object's actual size and its distance from the observer. The formula to calculate angular diameter: \[ \theta = \frac{d}{D} \] where \( d \) is the diameter of the object and \( D \) is the distance to the object. This measure helps us understand how celestial bodies like the moon and sun appear in the sky. During the solar eclipse calculations, the formula signifies how the moon and sun appear almost the same size from Earth.
For the moon: - Diameter: \(3.48 \times 10^6 \) meters- Distance: \(3.85 \times 10^8 \) meters
The moon's angular diameter \( \theta_{\text{moon}} \) is calculated as \( 9.04 \times 10^{-3} \) radians. For the sun, with a greater diameter and larger distance, the angular diameter \( \theta_{\text{sun}} \) is \( 9.27 \times 10^{-3} \) radians. Observing both values is crucial in understanding the slight variation that prevents a solar eclipse from being totally complete every time.
Apparent Circular Area
The apparent circular area of a celestial object is not its real size but how big it appears to be from Earth. It's directly proportional to the square of the angular diameter. Therefore, as the angle an object subtends becomes larger, its apparent area also becomes larger.
The formula to calculate apparent circular area is proportional to the square of the angular diameter: \[ A \propto \theta^2 \] In our exercise example, when you calculate the apparent circular areas of the moon and sun, their ratio provides insight into how much of the sun the moon can cover. The ratio of the moon's apparent area to the sun's is about 95.26%. This implies that while the moon covers a significant portion of the sun, its slightly smaller angle compared to the sun's makes it not entirely total during an eclipse.
Celestial Distances
Celestial distances refer to the vast spaces between cosmic entities such as the Earth, moon, and sun. These distances are critical in determining how large celestial bodies appear to us, affecting phenomena like eclipses.
For instance, in the given exercise:- The moon is approximately \(3.85 \times 10^8 \) meters away from Earth.- The sun is roughly \(1.50 \times 10^{11} \) meters from Earth.
Though the sun is significantly larger than the moon, it is also much farther away, leading to its reduced angular size as seen from the Earth. The distances help reconcile the observation that the sun and moon appear almost the same size in the sky despite their true size discrepancy. Understanding these celestial distances is essential for explaining how and why eclipses occur, why the eclipse is only sometimes total, and how phenomena like the apparent circular area arise when we observe celestial bodies.

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Most popular questions from this chapter

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The vertical drop of the arrow may be ignored. There is a maximum value for the angular speed \(\omega\) of the propeller beyond which the arrow cannot pass through an open space without being struck by one of the blades. (a) If the arrow is to pass through an open space, does it matter if the arrow is aimed closer to or farther away from the axis (see points \(\mathrm{A}\) and \(\mathrm{B}\) in the drawing, for example)? Explain. (b) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow speed \(v ?\) Why? (c) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow length \(L\) ? Justify your answer.

A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from \(3.00\) to \(5.00\) rev \(/ \mathrm{s}\) while rotating through one-half of a revolution. How much time does this maneuver take?

Two identical dragsters, starting from rest, accelerate side-by- side along a straight track. The wheels on one of the cars roll without slipping, while the wheels on the other slip during part of the time. (a) For which car, he winner or the loser, do the wheels roll without slipping? Why? For the dragster whose wheels roll without slipping, is there (b) a relationship between its linear speed and the angular speed of its wheels, and (c) a relationship between the magnitude of its linear acceleration and the magnitude of the angular acceleration of its wheels? If a relationship exists in either case, what is it? Problem A dragster starts from rest and accelerates down the track. Each tire has a radius of \(0.320 \mathrm{~m}\) and rolls without slipping. At a distance of \(384 \mathrm{~m},\) the angular speed of the wheels is \(288 \mathrm{rad} / \mathrm{s}\). Determine, (a) the linear speed of the dragster and (b) the magnitude of the angular acceleration of its wheels.

Two Formula One racing cars are negotiating a circular turn, and they have the same centripetal acceleration. However, the path of car A has a radius of \(48 \mathrm{~m}\), while that of car \(\mathrm{B}\) is \(36 \mathrm{~m}\). Determine the ratio of the angular speed of car \(\mathrm{A}\) to that of car \(\mathrm{B}\).

Refer to Interactive Solution \(\underline{8.31}\) at to review a model for solving this problem. The take-up reel of a cassette tape has an average radius of \(1.4 \mathrm{~cm} .\) Find the length of tape (in meters) that passes around the reel in \(13 \mathrm{~s}\) when the reel rotates at an average angular speed of \(3.4 \mathrm{rad} / \mathrm{s}\)

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