Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Question
Chapter 16, Problem 57Q
To determine
The reason for the Sun to appear smaller in the image taken with visible light as compared to the case when the image was taken with the ultraviolet light.
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Students have asked these similar questions
Mercury's orbit ranges from 46 to 70 million km from the Sun, while Earth orbits at about 150 million km.
a. The Sun has a 30-arc-minute diameter viewed from Earth; what range of sizes does it have when viewed from Mercury?
When Mercury is 46 million km from the Sun, the Sun has a diameter of
When Mercury is 70 million km from the Sun, the Sun has a diameter of
arc-minutes.
arc-minutes.
b. At Mercury's orbital extremes, how many times stronger is the Sun's radiation on Mercury than on Earth?
At 46 million km, the Sun's radiation is
times stronger than on Earth.
At 70 million km, the Sun's radiation is
times stronger than on Earth.
Using a 8-m reflector telescope on Mars, what is the maximum distance we could measure using stellar parallax?
Calculate the total amount of radiative energy per second intercepted by Mars from the Sun using the flux of radiation from the Sun at Mars' orbital radius.
Flux of radiation from the Sun at Mars' orbital radius is 597 W m-2.
The luminosity of the Sun Ls = 3.8×1026 W.
Mars orbits at a distance of 2.25×1011 m (1.5 AU) from the Sun.
Note: Consider carefully the cross-sectional area Mars presents to the outwards flow of radiative energy when answering this question.
Chapter 16 Solutions
Universe: Stars And Galaxies
Ch. 16 - Prob. 1QCh. 16 - Prob. 2QCh. 16 - Prob. 3QCh. 16 - Prob. 4QCh. 16 - Prob. 5QCh. 16 - Prob. 6QCh. 16 - Prob. 7QCh. 16 - Prob. 8QCh. 16 - Prob. 9QCh. 16 - Prob. 10Q
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- From a distance of 800 km above the surface of the Moon, what is the angular diameter of an astronaut in a spacesuit who has a linear diameter of 0.8 m viewed from above? (won Use the small-angle formula, angular diameter (in arc seconds) linear diameter 2.06 x 105 distance arc seconds The unaided human eye has a resolution of about 100 arc seconds in bright lighting conditions. Could someone looking out the command module window have seen the astronauts on the Moon? Yes Noarrow_forwardWhat are the information we can get from knowledge of 1st brillouin zone and why the 1st brillouin zone is more important that the other higher number brillouin zone ?arrow_forwardNext you will (1) convert your measurement of the semi-major axis from arcseconds to AU, (2) convert your measurement of the period from days to years, and (3) calculate the mass of the planet using Newton's form of Kepler's Third Law. Use Stellarium to find the distance to the planet when Skynet took any of your images, in AU. Answer: 4.322 AU Use this equation to determine a conversion factor from 1 arcsecond to AU at the planet's distance. You will need to convert ? = 1 arcsecond to degrees first. Answer: 2.096e-5 AU (2 x 3.14 x 4.322 x (.000278/360) = 2.096e-5) Next, use this number to convert your measurement of the moon's orbital semi-major axis from arcseconds to AU. A) Calculate a in AU. B) Convert your measurement of the moon's orbital period from days to years. C) By Newton's form of Kepler's third law, calculate the mass of the planet. D) Finally, convert the planet's mass to Earth masses: 1 solar mass = 333,000 Earth masses.arrow_forward
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