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

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[edit source]

Our Sun is an approximate black body with a peak wavelength at approximately 500nm. If λ is the peak wavelength, then the absolute temperature (i.e., Kelvins) is related to λ by λT = 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/λ. If λ 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 λ is the peak wavelength, then the absolute temperature (i.e., Kelvins) is related to λ by λT = 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/λ. If λ 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 λ is the peak wavelength, then the absolute temperature (i.e., Kelvins) is related to λ by λT = 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/λ. If λ changed from 500 to 100 it got smaller by a factor of 5. Therefore T got larger by a factor of 5.

## Question 9[edit source]

The distance to a star in parsecs is related to a planet's parallax angle, θ, by the formula, d = r/θ, where d is measured in parsecs, r is the radius of the planet's orbit in AU, and θ 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, θ, by the formula, d = r/θ, where r is the radius of the planet's orbit, and θ 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, θ, by the formula, d = r/θ, where r is the radius of the planet's orbit, and θ 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