Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 4, Problem 31QAP
To determine
The two ways through which atmosphere interfere with astronomical observations.
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Earth's daylight surface disk absorbs about 1047 W per m2 from the Sun. Using 6400 km for the Earth's radius, how much of this radiative power is emitted by each square meter of the spherical Earth? (Compare ratio of disk area to spherical surface area)
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Earth's daylight surface disk absorbs about 1036 W per m2 from the Sun. Using 6400 km for the Earth's radius, how much of this radiative power is emitted by each square meter of the spherical Earth? Hint: Compare the ratio of the disk area to the spherical surface area.
Chapter 4 Solutions
Understanding Our Universe
Ch. 4.1 - Prob. 4.1CYUCh. 4.2 - Prob. 4.2CYUCh. 4.3 - Prob. 4.3CYUCh. 4 - Prob. 1QAPCh. 4 - Prob. 2QAPCh. 4 - Prob. 3QAPCh. 4 - Prob. 4QAPCh. 4 - Prob. 5QAPCh. 4 - Prob. 6QAPCh. 4 - Prob. 7QAP
Ch. 4 - Prob. 8QAPCh. 4 - Prob. 9QAPCh. 4 - Prob. 10QAPCh. 4 - Prob. 11QAPCh. 4 - Prob. 12QAPCh. 4 - Prob. 13QAPCh. 4 - Prob. 14QAPCh. 4 - Prob. 15QAPCh. 4 - Prob. 16QAPCh. 4 - Prob. 17QAPCh. 4 - Prob. 18QAPCh. 4 - Prob. 19QAPCh. 4 - Prob. 20QAPCh. 4 - Prob. 21QAPCh. 4 - Prob. 22QAPCh. 4 - Prob. 23QAPCh. 4 - Prob. 24QAPCh. 4 - Prob. 25QAPCh. 4 - Prob. 26QAPCh. 4 - Prob. 27QAPCh. 4 - Prob. 28QAPCh. 4 - Prob. 29QAPCh. 4 - Prob. 30QAPCh. 4 - Prob. 31QAPCh. 4 - Prob. 32QAPCh. 4 - Prob. 33QAPCh. 4 - Prob. 34QAPCh. 4 - Prob. 35QAPCh. 4 - Prob. 36QAPCh. 4 - Prob. 37QAPCh. 4 - Prob. 38QAPCh. 4 - Prob. 39QAPCh. 4 - Prob. 40QAPCh. 4 - Prob. 41QAPCh. 4 - Prob. 42QAPCh. 4 - Prob. 43QAPCh. 4 - Prob. 44QAPCh. 4 - Prob. 45QAP
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- How are Radio waves created in outer space?arrow_forwardWhat is the atmospheric temperature IN KELVIN of a planet 4 AU from the sun, with an albedo of 0.5, and an optical depth of 1? Round your answer to the nearest degree.arrow_forwardGive several reasons Mercury would be a particularly unpleasant place to build an astronomical observatory.arrow_forward
- Why is it difficult to observe at infrared wavelengths? What do astronomers do to address this difficulty?arrow_forwardIn thinking about satellite altimetry, when the distance between a satellite and the ocean surface increases, what is happening with the ocean depth? Is it getting shallower or deeper?arrow_forwardThe surface temperatures of both Venus and Earth are warmer than would be expectedon the basis of their respective distances from the Sun. Why is this so?arrow_forward
- How does the frequency of a particular spectral line observed in sunlight compare with the frequency of that line observed from a source on Earth?arrow_forwardWithout the greenhouse effect, Earth's average surface temperature would be about 250 K. With the greenhouse effect, it is some 40 K higher. Use this information and Stefan's law to calculate the fraction of infrared radiation leaving Earth's surface that is absorbed by greenhouse gases in the atmospherearrow_forwardDiscuss why atoms can escape from the exosphere but not the lower parts of the atmosphere of a planetarrow_forward
- Which of the following is least reasonable regarding the "water hole"? Group of answer choices It consists of frequencies which are greater than the frequencies of atmospheric emissions. It relates to the natural frequencies of vibration of hydroxyl (OH) and hydrogen (H), respectively. It occurs in that part of the electromagnetic spectrum where the galactic "noise" from stars and interstellar clouds is minimized. It is considered the "electromagnetic oasis" for interstellar communication. It corresponds to wavelengths in the 18-21 cm range.arrow_forwardEarths daylight surface disk absorbs about 1045 W per m2 from the Sun. Using 6400 km for the Earths radius, how much of this radiative power is emitted by each square meter of the spherical Earth ?arrow_forwardKepler’s First Law: Elliptical Planetary Orbits: The solar system major planet in the most elliptical solar orbit is little Mercury, which is the closest planet to the Sun. At Perihelion, Mercury’s distance from the Sun (Rp) is 0.31 AU. At Aphelion, Mercury’s distance from the Sun (Ra) is 0.47 AU. The intensity of Sunlight (I) that a planet receives from the Sun is inversely proportional to the square of that planet’s distance from the Sun (R). in other words, I = Constant / R2. Calculate how much more intense the Sunlight received by Mercury is at perihelion (p) than at aphelion (a): Rp2 = Ra2 = Ip / Ia = Ra2 / Rp2 =arrow_forward
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