Download ASTR 1020 – Spring 2017 – Prof. Magnani Answer Key – Homework 5

ASTR 1020 – Spring 2017 – Prof. Magnani
Answer Key – Homework 5
Question 1 – Ch. 15; #11
Hydrostatic equilibrium is established between the force from pressure pushing
outwards and the force of gravity pushing inwards.
Question 2 – Ch. 15; #15
Mass is the single most important property of a star as far as determining its
evolution.
Question 3 – Ch. 15; #21
The way to get heat out of an interstellar cloud is for the particles that make up to
cloud to collide with each other. In a collision between two particles (atoms or
molecules or ions), one of the colliding partners loses some kinetic energy and the
other picks it up. However, the gain in energy could be kinetic or it could go into
exciting one of the electrons of this particle to a higher energy state (there are other
excitation channels, but we will ignore them). When the electrons de-excite, the
photons which are emitted in the process leave the cloud carrying energy with
them. In this way, the cloud loses energy and cools. The key to all this is collisions.
If a gas is low-density and it happens to be hot, then there are so few collisions
(because of the low density) that it will take a long time to cool. The opposite is true
for high-density clouds.
Question 4 – Ch. 15; #27
It gains energy as the gas falls on to it, but half of the energy is radiated away and
the other half produces heat and thus more pressure balancing the extra mass and
keeping the protostar in hydrostatic equilibrium.
Question 5 – Ch. 15; #28
From a composition point of view, they are nearly identical. However, because they
form in different fashion, I would not classify brown dwarfs as supergiant planets.
Question 5 – Ch. 15; #29
If the outer layers of star heat up, the H minus density drop (the enegy from other
particles colliding with the H minus easily knocks off the extra electron). With less H
minus, the gas is more transparent, so radiation flows more freely and the star
cools. If the outer layer of a star cools down, the H minus density goes up, the gas
becomes more opaque and the radiation has a tougher time leaving the star so that
the region heats up again.
Question 6 – Ch. 15; #32
When a hydrogen atom is ionized it splits up into a proton and an electron. The
electron moves much faster because it is about 2000 times less massive than a
proton.
Question 6 – Ch. 15; #33
The average number density (i.e., the number of particles per volume) of the
interstellar medium is 0.1 atoms/cm3. The vast majority (about ¾) is made up of
hydrogen with a mass of 1.67 x 10-27 kg, so the average gas mass density is 1.67 x 1028 kg/cm3 (we’re ignoring helium). There is about 100 times less dust than gas by
mass, so the average dust mass density is 1.67 x 10-29 kg/cm3 ,
We want the number of grains per cubic centimer.
If you look at our dust mass density, its units are kg/cm3 . I can get a number
density by taking that quantity and dividing it by the mass of a single dust grain. In
other words:
Mass density of dust / mass of a single dust particle = number of dust particles per
cubic centimer
Check the units:
kg/cm3 / kg/dust particle = dust particles / cm3
If only we knew the mass of a dust grain….but, wait, that’s exactly what they gave us:
1 x 10-17 kg is the typical mass of a single dust particle, so
1.67 x 10-29 kg/cm3 / 1 x 10-17 kg/dust particle = 1.67 x 10-12 dust particles per
cubic centimeter.
Question 7 – Ch. 15; #35
Lambda_max = 0.0029 / T
Plug in 8000 K for T and you get Lambda_max = 3.6 x 10-7 m or 360 nm.
Question 8 – Ch. 15; #37
A B8 main sequence star puts most of its electromagnetic radiation in the UV (just
plug in its surface temperature, 12,300 K, in the equation we used in problem 7). It
is UV radiation that ionized hydrogen. A K0 giant may have the same luminosity,
but most of its light is coming out in the orange part of the spectrum and those
photons aren’t energetic enough to ionize hydrogen.
Basically, it’s the “hardness” of the radiation. In other words it’s the proportion of
the radiation that comes out in the ultraviolet.
Question 9 – Ch. 15; #42
If 10 billion years = 1 day, then
30 million / 10 billion = x /1 day
solve for x: x = 0.003 days = 0.072 hours = 4.32 minutes
The Hayashi track is 1/3 of that time, so about one and a half minutes.
Question 10 – Ch. 15; #45
L = 4π R2 σT4,
T = 1000 K, R = 7 x 107 m, so
L = 3.5 x 1021 J/s
L(brown dwarf) / L(Sun) = 3.5 x 1021 J/s /3.8 x 1026 J/s = 9 x 10-6
If one brown dwarf is only 9 x 10-6 as luminous as the Sun, then you would need 1.1
x 105 of them to match the Sun’s luminosity.