Talk:Astronomy college course/Introduction to stellar measurements/questions


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These questions will be on Test 4 for Astronomy at Lake. The equations can be used to solve problems on the test that do not give you the equation.

Question 7
Our Sun is an approximate black body with a peak wavelength at approximately 500nm. If &lambda; is the peak wavelength, then the absolute temperature (i.e., Kelvins) is related to &lambda; by &lambda;T = k, where k is a constant. An object emits thermal (blackbody) radiation with a peak wavelength of 250nm. How does its temperature compare with the Sun? - 5 times colder than the Sun - 2 times colder than the Sun - 5 times hotter than the Sun - The temperature is the same + 2 times hotter than the Sun

Why: Solve for T = k/&lambda;. If &lambda; changed from 500 to 250 it got smaller by a factor of 2. Therefore T got larger by a factor of 2.

Our Sun is an approximate black body with a peak wavelength at approximately 500nm. If &lambda; is the peak wavelength, then the absolute temperature (i.e., Kelvins) is related to &lambda; by &lambda;T = k, where k is a constant. An object emits thermal (blackbody) radiation with a peak wavelength of 1000nm. How does its temperature compare with the Sun? - 5 times colder than the Sun + 2 times colder than the Sun - 5 times hotter than the Sun - The temperature is the same - 2 times hotter than the Sun

Why: Solve for T = k/&lambda;. If &lambda; changed from 500 to 1000 it got larger by a factor of 2. Therefore T got smaller by a factor of 2.

Our Sun is an approximate black body with a peak wavelength at approximately 500nm. If &lambda; is the peak wavelength, then the absolute temperature (i.e., Kelvins) is related to &lambda; by &lambda;T = k, where k is a constant. An object emits thermal (blackbody) radiation with a peak wavelength of 100nm. How does its temperature compare with the Sun? - 5 times colder than the Sun - 2 times colder than the Sun + 5 times hotter than the Sun - The temperature is the same - 2 times hotter than the Sun

Why: Solve for T = k/&lambda;. If &lambda; changed from 500 to 100 it got smaller by a factor of 5. Therefore T got larger by a factor of 5.

Question 9
The distance to a star in parsecs is related to a planet's parallax angle, &theta;, by the formula, d = r/&theta;, where d is measured in parsecs, r is the radius of the planet's orbit in AU, and &theta; is the parallax angle in arcseconds. An orbiting satellite makes a circular orbit 5 AU from the Sun. It measures a parallax angle of 0.2 of an arcsecond (each way from the average position). What is the star's distance? + a) 25 parsecs - b) 5 parsecs - c) 50 parsecs - d) 1 parsec - e) 10 parsecs

Why: Use d = 5/.2, and multiply top and bottom (numerator and denominator) by 5, using the fact that (5)(0.2) = 1. H

The distance to a star, d, is related to a planet's parallax angle, &theta;, by the formula, d = r/&theta;, where r is the radius of the planet's orbit, and &theta; is the parallax angle measured in radians. An orbiting satellite makes a circular orbit 5 AU from the Sun. It measures a parallax angle of 1 arcsecond (each way from the average position). What is the star's distance? - a) 25 parsecs + b) 5 parsecs - c) 50 parsecs - d) 1 parsec - e) 10 parsecs

Why: d = 5/1 = 5

The distance to a star, d, is related to a planet's parallax angle, &theta;, by the formula, d = r/&theta;, where r is the radius of the planet's orbit, and &theta; is the parallax angle measured in radians. An orbiting satellite makes a circular orbit 5 AU from the Sun. It measures a parallax angle of 0.1 arcsecond (each way from the average position). What is the star's distance? - a) 25 parsecs - b) 5 parsecs + c) 50 parsecs - d) 1 parsec - e) 10 parsecs

Why: Use d = 5/.1, and multiply top and bottom by 10. Use (10)(0.1) = 1


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