ASTR 1010 – Spring 2012 – Prof. Magnani
Answer Key – Homework 4
Question 1 – Ch. 4; #2
A light year is a measure of distance. It is the distance a photon travels in one year
at a speed of 2.99792 x 108 m/s. This comes out to about 9.5 trillion km (9.5 x 1012
km).
Question 2 – Ch. 4; #3
Light manifests both particle and wave-properties. When you do an experiment that
showcases light’s particle properties, then it behaves as particles (photons). When
you do one that showcases light’s wave properties, then it behaves like a wave
phenomenon.
Question 3 – Ch. 4; #10
Protons, Neutrons (both types in nucleus), and Electrons
Question 4 – Ch. 4; #11
The Heisenberg Uncertainty Principle states that you can only simultaneously know
the position and momentum of a particle to a level that is approximately the product
of the position of the particle and its momentum. If you wish to improve the
precision with which you know one of these quantities past that point, then you will
know the other quantity less precisely. This is not a measurement issue, it is a
statement that, at the subatomic level, nature assumes a granular, pixel-light
behavior rather than a continuous, smooth behavior.
Question 5 – Ch. 4; #14
The ground state of an atom is the lowest possible energy state of the atom. Any
electron in that state cannot achieve a lower energy because, by definition, there are
no lower energy states. You can get out of the ground state by gaining energy via
collisions or absorbing photons, but then the atom would be in an excited state.
Question 6 – Ch. 4; #17
A redshifted object shows spectral lines shifted towards the lower frequency/longer
wavelength part of the spectrum (called the “red” part of the spectrum even if you
are not in the optical regime). These objects are moving away from you according to
the Doppler formula. A blueshifted objects shows spectral lines shifted towards the
higher frequency/shorter wavelength part of the spectrum. These objects are
moving towards the observer.
Question 7 – Ch. 4; #20
The temperature of an object is a measure of the motion of the atoms or molecules
comprising the object. The relationship is particularly straightforward in the case of
a gas where the temperature is directly related to the kinetic energy.
Question 8 – Ch.4; #22
Brightness in the book is what we called apparent brightness or flux in class. This is
how bright something looks to an observer and so this quantity can change with
distance. The apparent brightness of a source of light drops as one over the distance
squared (in mathematical terms: 1/d2). Luminosity is the intrinsic energy output of
something and is independent of distance. Think of a 100 Watt lightbulb (a Watt is a
Joule per second). The 100 Watts indicates the luminosity of the bulb and doesn’t
change whether the bulb is near or far. But if the bulb is near, the apparent
brightness is a large number; if it is far, then the apparent brightness decreases to a
low value. In terms of flux,
F = L/(4 pi r2)
where r is the distance of the source.
Question 9 – Ch. 4; #25
The nearer planet to the Sun could have a very high albedo so that a lot of the
sunlight hitting it would be reflected and not absorbed. If some of the energy is not
absorbed, then it cannot be used to heat the planet and raise its surface
temperature.
Question 10 – Ch. 4; #27
 = c/f = 3.0 x 108 / 7.9 x 105 = 380 m
 = c/f = 3.0 x 108 / 9.83 x 107 = 3.1 m
Question 11 – Ch. 4; #30
Frequency = 440 Hz  wavelength of c/f = 0.773 m
c = 340 m/s
v = 25 m/s
[lambda_obs – lambda_o]/lambda_o = v/c
[lambda_obs – 0.773]/0.773 = -25/340
lambda_obs = 0.716 m (remember, the object is approaching you so it will be shifted
to shorter wavelengths! That’s why we need the minus sign for the velocity)
Convert now back to frequency: 475 Hz
Question 12 – Ch. 4; #33
Brightness decreases as 1/d2 , so if Neptune is 30 times the distance of the Earth
from the Sun, the Sun will be lower in brightness at Neptune by a factor of 1/302 =
1/900. Thus, the Sun looks fainter by a factor of 900 from Neptune compared to
how it looks from Earth.
Question 13 – Ch.4; #38
Calculate the Sun’s luminosity:
Lsun = 4 pi Rsun2 sigma Tsun4 = 4 pi (6.96 x 108 m)2(5.67 x 10-8)(5780 K)4 = 3.85 x 1026
J/s
Notice that I had to convert the Sun’s radius to meters.
Question 14 – Ch. 4; #42
Albedo = 0.1
So T = 279 (0.9/1)0.25 = 272 K
Albedo = 0.9
So T = 279 (0.1/1)0.25 = 157 K
Question 15 – Ch. 4; #43
The situation here is we have a planet with the same albedo as Mercury but 4 times
closer to the Sun. Because the temperature drops as 1/d0.5, we see that 1/40.5 = 2,
so this planet should be twice the surface temperature of Mercury or 900 degrees.
Question 16 – Ch.4; #45
We can use the formula that we used in problem 43, as long as we convert the
distance (97.7 AU) to meters, but we can also use the formula
T = 279[(1 - a)0.25]dAU-0.5
For a = 0.8, T = 19 K