Astronomy 106.02
Solutions to Problem Set 4 due Feb. 19, 1997
Chapter 7:
1. Human body temperature is about 310 K (98.6 degrees F). At what wavelength do
humans radiate the most energy? What kind of radiation do we emit?
Use the formula relating temperature to maximum wavelength:
max = 3,000,000 /
T , where T is temperature in Kelvin.
For T = 310 K,
max = 3,000,000 /
T = 3,000,000/310
[ANSWER: max = 9680
nm. This is infrared radiation (See Figure 6-2 and note that 9680 nm = 9.68 X 10-6
m. ]
3. Infrared observations of a star show that it is most intense at a wavelength of
2000 nm. What is the temperature of the star's surface?
Here you use the wavelength-temperature formula again, but solve for the
temperature T.
max = 3,000,000/T
or T = 3,000,000 / 
max
For max = 2000 nm,
T = 3,000,000 / 2000
[ANSWER: T = 1500 Kelvin.]
*5. If one star has a temperature of 6000 K and another has temprature of 7000 K, how much more energy per second will the hoe star radiate from each square meter of its surface?
We can use the formula at the bottom of page 126, which gives E, the energy per second per square meter as a function of temperature, T. Since we are interested in the ratio of the energy radiated at 7000 K to the energy radiated at 6000 K, we can set it up as follows:
E(7000) / E (6000) = ( 
(7000)4 ) /
( (6000)4 ).
Thus the factor of cancels out and we are left with: E(7000) / E (6000) = (7000)4 / (6000)4 = (7000 / 6000)4 = (7/6)4 = (1.17)4 = 1.85.
[ANSWER: The hotter star (7000 K) radiates 1.85 times as much energy per square meter per second as the cooler star (6000 K).]
9. In a laboratory, the Balmer beta line has a wavelength of 486.1 nm. If the line
appears in a star's spectrum at 486.3 nm, what is the star's radial velocity? Is
it approaching or receding?
We use the Doppler shift formula. For this situation, 0=
486.1 nm and since
= 486.3 nm, 
= 486.3 nm - 486.1 nm = 0.2 nm.
Thus Vr = ( / 0
) X c
Vr = (0.2 nm / 486.1 nm)(3.0 X 108 m/s)
[ANSWER: Vr = 123,000 m/s = 123 km/s. Since > 0, the shift is a redshift
which means that the star is receding.]
Chapter 8
1. The radius of the sun is 0.7 million km. What percentage of the radius is taken up by the chromosphere?
The thickness of the chromosphere is given on page 149 as 10,000 km. Comparing this to the diameter of the sun, 0.7 X 106 km, we can get the ratio:
10,000 / 0.7 X 106 = 0.014.
[AMSWER: The thickness of the chromosphere is about 1.4 % of the diameter of the sun.]
2. The smallest detail visible with ground-based solar telescopes is about 1 second
of arc. How large a region does this represent on the sun? (HINT: Use the small-angle
formula.)
The small-angle formula tells us that:
(
/ 206,265") = ( d
/ D )
Here we are solving for d, the distance on the sun.
d = ( / 206,265") D.
= 1".
D = 1.5 X 108 km (from Data File One)
d = (1" / 206,265")(1.5 X 108 km)
[ANSWER: The region d = 730 km across.]
*3. What is the angular diameter of a star like the sun located 5 ly from the earth? Is the Hubble Space Telescope able to detect detail on the surface of such a star?
This is another small angle formula problem. Here we are solving for the angular size, . We convert D, the distance from light years to kilometers: D = (5 ly) ( 9.46 X 1012 km / 1 ly) = 4.73 X 1013 km. The diameter d is the diameter of the sun, d = 1.39 X 106 km.
Putting these numbers into the formula gives:
= (206,265)(d / D ) = (206,265)( 1.39 X 106 km / 4.73 X 1013 km ) = (206,265) ( 2.94 X 10-8 ) = 0.006 seconds of arc.
[ANSWER: The angular diameter of such a star is about 0.006 seconds of arc. The resolution of the Hubble Space Telescope is 0.023 seconds of arc (from Problem Set 3), so the Hubble would not be able to resolve detail on a star at this distance.]