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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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 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_forwardDiscuss why atoms can escape from the exosphere but not the lower parts of the atmosphere of a planetarrow_forwardWhich 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_forward
- A pixel on a NOAA-19 image has channel 4 brightness temperature of 285.5 K and a channel 5 brightness temperature of 285.0 K. If the satellite is 70 degrees above the horizon at the time that the measurement was taken during daylight, what is the sea surface (15) temperature in degrees Celsius of the pixel?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_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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