Astronomy

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					                                      AST 376/381 Scalo
                         1st homework assignment, due Monday, Feb. 5.

       You will turn in 3 of the 5 multipart questions below. Don’t let the length scare you
off. Each problem’s parts are (usually) very short. If you get stuck, call me or move on. Do
NOT use mathematica, maple, or anything fancy. These are meant to get you thinking
about comparisons and order of magnitude learning.
       I have my own answers to all of these, but except in possibly one case I do not think
they have been discussed in the astronomy literature, so don’t bother looking.

PART A. Energies.
Choose either question 1 or 2. I suggest you spend a little time with both questions,
whichever you decide to work on.

         Part of the problem in reading about an interdisciplinary subject is that by the time you
get to the “serious” level, people in different fields have deeply ingrained conventions about
terminology, and especially units. Often the units are appropriate for the problem at hand, but in
order not to get slowed down having to convert every time you wonder “Is that a lot of energy?”
or to not appreciate a point because you didn’t have time to think about how to convert a
“picometer” (common in molecular biophysics). This question is supposed to help you with
energy units, and set you up for others. Also to make sure you can get to essential mathematical
relations that might be needed.
         You only have to turn in answers to the questions in bold below. A lot of the rest is in
case you are interested, or preliminary information you need to answer the question, or help in
getting started.

Question 1. Energy comparison: Bonds, Thermal.

        1a. Do you have a unit converter on your computer? You will probably need more than
one. You will also need a handy list of trig formulas, derivative and integral tables, etc. Look
around on the internet and see which are most extensive in their coverage or useful for our
purposes, and write down the urls of these sites, and a brief description, as your answer here. For
example, one of the handiest tools I have found is the “Astro-Calculator” which is a pretty good
calculator which also contains the properties of the planets and sun that an astronomer would use
most often. And you can choose cgs or SI units.
         A famous but long out of print book by Zombeck called “Handbook of Space Astronomy
and Astrophysics” had an invaluable collection of tables and plots for all sorts of things you never
thought you’d see in one place (emphasis on high energy physics, but really covered everything).
There is now a new edition, and there is a web site where you can access all the information.
However as with many things, the revision and update and high cost of publication has forced
him to omit some of the best graphs and tables, so I urge you to look for a used Zombeck online
some time and buy it if you intend to have a career in astronomical research.
        The most useful of all astronomy reference works was Allen’s Astrophysical Quantities,
the OLD edition. The new version is a pale committee job, useful, but not like the old version. If
you find its data online, please let me know. I do not think it exists. In that case, find a hard copy
for cheap if you can.

        Now do some useful conversions of units and store the answers in a safe accessible
location. Convert kcal/mol to electron volts (eV), joules, and ergs. (We will usually use eV or
kcal/mol). You will need this conversion in the next part of the problem and throughout the
course. You may need to know that in order to convert energy per mole to energy per constituent
molecule you multiply with Avogodro’s number 6.02x10^23, the number of molecules per mole
(by definition). Memorize the conversion between kilo-Joule and kcal (a number like 4....) for
future reference in your readings.

This would be a good time to memorize the typical energies (eV are most convenient in my
opinion) of covalent bonds, noncovalent bonds, H bonding, the thermal energy per molecule in
the Earth’s atmosphere at 300K, and any other energies you can think of.

         1b. Covalent bonds are about 300 kcal/mol for many organic molecules. Intermolecular
noncovalent forces are about 1 kcal/mol (with a range of a factor of at least 10). Compare these
to the average thermal energy of molecules at room temperature. (Assume room temperature and
the temperature of disks and planetary atmospheres and surfaces are 300K, and that the average
thermal energy of molecules is kT (we neglect the factors of order unity), where k=1.38e-16 cgs
is Boltzmann’s constant. If this is unfamiliar to you, look it up.) What does this suggest about
the nature and importance of the two types of binding energies? You may want to use the
Boltzmann factor exp(-E/kT) for your estimate of the fraction that can stay bound.

        1c. Consider the cases T = 1000K (brown dwarf, close-in giant exoplanet, inner
protoplanetary disk) and T = 10K (interstellar molecular clouds, very outer disks). Are your
conclusions different?

Question 2. Results of a collision between two bodies (more challenging, or at least more
steps)

         2. This question is going to find out how fast particles can be moving relative to each
other in order to not vaporize each other or break each other into fragments, using a simple
energy argument: To break something up, you have to hit it with an energy greater than the
energy that is holding it together, in this case by chemical bonds.
         It looks long only because I am trying to guide you through the process. If it is already
transparent to you, sorry, but remember we have a range of backgrounds in this class, and it
wouldn’t hurt for you to check that your reasoning is correct.
          A “particle” here means anything in the condensed phase: an asteroid, atmospheric
aerosol droplet, grain in a protoplanetary disk. Think about it in whatever way you find most
interesting, or even write it in specific terms (e.g. “Haze particles in Titan’s atmosphere should
have sizes that are controlled by collisional fragmentation processes. Consider a particle of size a
= 10^-4 (a/1micron) cm....” You will find this is quite a handy exercise that you may use again if
you are headed for research. You can also apply it to anything, like throwing a rock against a
wall. Also, remember we only want order of magnitude in the end.

         2a. Take a particle to be a sphere of density s = 3 g/cm^-3 (for water s = 1 g/cm^-3, s=3
is roughly a silicate rock; I will always use “s” for the interior density of any condensed phase
particle). If the size of a particle is a = 1 micron (typical within an order of magnitude for a
variety of environments) and each molecule is MgSiO4 (so mass of each bound molecule is
m_mol = 124 m_H = 2.07E-22 g):
         What is the particle’s mass m_p? (Just use the fact that for any object the mass is the
volume times the mass density s.) Give all answers in general form, e.g. m_p = 4 pi (s^1/2)
(a_mum^1/4) where a_mum^1/4 means “a in microns to the ¼ power.” (That is not the correct
answer in case you wondered.) See if you plug in the radius of the Earth if you get something
like the mass of the Earth (no need to turn in; just to check).
         Many people think the terrestrial planets, and even the cores of Jupiter and Saturn,
formed by collisional growth of these initially microscopic particles. That is a lot of growth!
       How many of these particles would you need to make the mass of the Earth?
(Assume they all have the proper composition and anything else--we are only concerned with
mass here.)

        2b. How many molecules of the silicate MgSiO4, call it N_mol, are in the particle of
size 1 micron? We want to know how much collisional kinetic energy it will take to break every
one of these bonds, i.e. for the particle to completely sublime or vaporize. If the energy holding
the molecules in the solid together is E_bond = 3eV (E_bond/3 eV) for each bond, what is the
total binding energy E_p of the particle? (Compare: The “heat of sublimation” of water is
about 50 kJ/mol.) Give the dependence of your answer in general form, like

                E_p = 7e8 E_bond (a/1micron)^-3 (s/3 g cm^-3)^1/2 erg.

[You should give all your answers, from now on, in a form like this, so that anyone (including
you) can easily see the dependence on parameters and variables, and can recompute your result
for a different set of parameters.]

         2c. When the relative kinetic energy E_pp of two identical colliding particles exceeds
their total binding energy E_p, you expect a smash up. Consider the energy required to break all
the bonds in the particle, i.e. sublime/vaporize it. What particle-particle velocity does this
correspond to for a particle of size a = 10^-4 (a/1 micron) cm? How does the result depend
on particle size? (Give answer as in the E_p example above.) For this exercise just take the
relative kinetic energy to be (1/2) m_p v_pp^2.
         Express the result in a way that is physically simple and transparent, by thinking in terms
of energy per gram instead of total energy.
         You should get an answer that is of order 0.l km/sec, within a factor of ten.
         What if there is about one noncovalent interaction (of various sorts) for every Y
molecules in the particle, and they have a binding energy of 5 kcal/mol? If the collision with
another particle is at far too small a speed to break up the particles, could it still disrupt a
lot of these noncovalent bonds? Remember, we are working toward trying to understand if
particles in different astrophysical situations could be dominated by noncovalent forces, like the
sticky gooey stuff on the Earth. (Assume that there is some way that the collision can transfer the
collisional energy into internal energy of the particle, making all the molecules in the particle
vibrate or move around more (depending on what kind of substance it is)).


         2d. What do you expect the relative velocities of particles to be in different
environments (say planetary atmospheres and protoplanetary disks, assuming 300K again)?
Assume the particle relative velocities are due only to Brownian motion, neglecting other
possibilities which we will discuss in class later. If you are unfamiliar with Brownian motion,
look it up. I will try to cover it in class by Friday. If you are comfortable using formulas whose
origin you don’t understand, use this: The average particle-particle velocity due to Brownian
motion will be given by

                                 v_pp(Brownian) = (m_gas/m_p)^1/2 v_th

where m_gas is the average mass of the gas particles that are colliding with the particle (take
m_gas = m_H, with m_H the mass of a hydrogen atom or proton and  the mean molecular
weight, which you can take as 2.3; this is appropriate for a medium that is mostly molecular
hydrogen). The quantity v_th is the average thermal speed of the gas atoms and molecules that
are hitting the grain; use v_th=(kT/m_H)^1/2, ignoring all the possible factors of order unity in
the coefficient.

          2e. Finally, consider how many bonds you’d have to break in order to shatter the
particle into N_fr pieces, and so determine how much smaller the colllisional velocity can be
yet still break up the particle, even if you can’t vaporize it. Maybe sticking is extremely
difficult because even at very low speeds you still crack up particles rather than coalesce
them during collisions. That would be the end of the dominant theory of planet formation!
          This problem seems long because I have done most of it here. But it is very
instructive if you can stay awake (actually I stayed up all night solving it).
          Let me get you started with some results you can use. It is useful to consider the particle
of mass m_p and number of molecules in the whole particle N_mol to be divided up into N_fr
(<<N_mol) identical cubes with a side of length a_fr. Then the area of a fragment is a_fr^2 and
its volume is a_fr^3.
          Notice that for the whole particle, the average distance between the bound molecules is l
= n_mol^-1/3 where n_mol is the number of molecules per unit volume in the particle (the 1/3
power is what I went over in class last week): n_mol = N_mol/a^3 = (m_p/m_mol)/a^3. Also, the
ratio of particle mass to the mass of an individual molecule (m_p/m_mol) gives you the total
number of molecules in the particle N_mol. Using m_p=s a^3 (forget about spheres in favor of
cubes to keep it easy), this gives a number density of molecules inside the particle of n_mol =
s/m_mol (recall s=3 g/cm^3 and m_mol was the mass of a MgSiO4.
          Now the idea is to find a simple way to estimate how many fewer bonds you have to
break to fragment something, creating N_fr identical, cubical fragments, compared to how many
bonds you have to break to completely vaporize the particle. First we need to find the average
size of the fragments in terms of N_fr and the size of the particle (both of which are considered
given).
          The number of fragments in the particle is N_fr = total volume of particle/volume of a
fragment = a^3/a_fr^3 = (a/a_fr)^3, giving
                           a_fr = a x N_fr^-1/3.
The number of bonds in the whole particle is just (a/l)^3 = a^3 n_mol (because the mean
separation of molecules in the particle is n_mol^-1/3), which equals a^3 s/m_mol.
          The number of bonds you have to break in the fragments is related to their total area,
which is the area of a fragment a_fr^2 times the number of fragments which is given as N_fr,
divided by the area occupied by a single molecule, which is l^2, so there are N_fr (a_fr/a)^2
bonds that have to be broken to produce N_fr fragments.
          Now all you have to do is take the ratio of the number of bonds in the surfaces of
fragments to the number of bonds in the volume of the fragment. The answer (call it X_bond) is
X_bond (<1) = N_fr^1/3(l/a) ~ 10^-4 N_fr^1/3 (a/1micron)^-1. Stop and think about this to
make sure it makes sense--I could have made a mistake easily.
          Now we are almost there because the energy you need is proportional to the number of
bonds you have to break, and the ratio of the critical velocities for doing this is going to be the
square root of the energy you need divided by the mass of the particle (because the energy is (1/2)
m_p v_pp^2--the kinetic energy of the incoming particle doesn’t know how many fragments it
has to make or anything else!
          Here is the answer I get: The answer I get is:

                 v_pp(fragment)/v_pp(sublimation) = (m_mol N_fr / s)^1/6 a^-1/2
        = 0.020 (m_mol/124m_H)^1/6 (s/3 g/cm^3)-1/6 (a/1micron)^-1/2 N_fr^1/6.
The weak 1/6 power dependence of some things comes in because to get the velocity you take the
square root of the binding energy which has some things to the 1/3 power in it.
The reason I think the expression is corrrect is that 1. I checked the units (never forget to do this),
and 2. You can show that the answer is (N_fr/N_mol)^1/6, or that the ratio of energies goes like
(N_fr/N_mol)^1/3. This makes sense to me, or at least it is remarkably simple. And notice that
the very weak dependence on everything means it is almost parameter independent: You can
break the particle into 2 or a million pieces and the answer only changes by about a factor of 10.
        So this is telling us that even for particles moving at 10 percent of the critical velocity for
complete sublimation, you could still have enough energy to break up the particles.

Another thing to notice is that the critical velocity for sublimation didn’t depend on the particle
size because the kinetic energy of the collision and the number of bonds you have to break are
both proportional to the total mass of the particle. So that a^-1/2 is in v_pp. That means that as
the particle size gets larger, the critical velocity decreases. But the Brownian velocity decreases
more strongly, like a^-3/2 (see above). That means that if the smallest grains can avoid
fragmenting collisions or sublimation, it will be even easier for the larger particles as they grow.
There are qualifications, but I hope you see the importance: The hard part of forming planets may
have to do with the very smallest particles at the earliest time--maybe the ability of a given star to
form planets in its debris is a sensitive function of how big the grains happen to be that the disk
inherits from the cloud from which it collapsed--that means that you can’t solve the planet
formation problem without understanding the initial conditions in detail--bad news!
         We will stop there--if you have any thoughts, write them in whatever you turn in.
                  EXTRA CREDIT:
         What would it take for the collision to cause the grains to melt and fuse rather than to
crack up or vaporize? Does it take more or less energy (per particle) to melt and fuse a solid into
a liquid than it does to break its bonds?

        Now take a break and watch simulations of collisions of giant impactors (comets,
asteroids) with your favorite planet, even the Earth, at http://janus.astro.umd.edu/astro/impact/

Everyone must do question 3, since it is not very long and will be beneficial (I hope).
Question 3. Size-related conversions and comparisons.

        We are interested in particles that have a wide range in sizes, from complex molecules
and volcanic emissions as small as 10 A (1 Angstroms = 0.1 nanometer (nm) = 10^-10 m = 10^-8
cm) to dust grains in the ISM and biocellular assemblies of 1 m (you can use “mum” instead of
m to save time; 1 mum = 10^-6 m = 10^-4 cm = 10^4 A = 1000 nm) to the largest cloud
droplets (brown dwarf and planetary atmospheres) and bacterial cells of 100 m (=10^-2 cm =
0.1 mm) to even larger entities (e.g. pre-planetesimals in disks, dispersed matter in oceans, ...).
Yet the sizes of the macroscopic systems we are concerned with are typically much larger, with
sizes measured in km or even pc.
        Part of the problem in doing interdisciplinary work is getting used to the different units
that you will encounter, and their relation to other phenomena. For some of them, like size or
length scales, the conversions have to become second nature. Astronomers (stellar to
extragalactic) like Angstroms or microns for their microscopic units, while planetary scientists
and biochemists like nm and microns. Most scientists measure macroscopic scales in m, while
astronomers and this class will continue to use cm (with the occasional km when convenient).
This exercise is meant to give you practice in that area, as well as having to look up some
information on your own.

         3a. The size unit of preference is often a result of the wavelength range in which certain
sizes of particle phenomena can be observed. As a general rule, particles scatter and absorb light
whose wavelength is roughly their own size.
        i. Using the fact that for a photon E = h = hc/, what wavelengths do you expect to be
associated with electronic transitions in atoms and molecules, given that the energy levels
have a typical spacing of a few electron volts (eV)? Using 5 eV to be definite, give your
answer in A, nm, microns, and cm. What is this wavelength region called (e.g. infrared,
ultraviolet, visible, radio,...)?

       ii. Give the general numerical relation between photon energy E in eV and
wavelength in microns. Memorize it.

         iii. A planetesimal (accumulating ball of rock destined to become a planet) has a size of 1
km in most theories for the origin of our solar system. How many size-doubling collisions did
it have to undergo, starting with 1 micron particles, in order to reach this size? How many
more to reach a size like the Earth?
         Try to derive a formula for how many mass-doubling collisions you need to make a mass
X times larger than the mass of the original particles, or you will have to do it by brute force,
which is something you usually will not be able to do.
         Can you see a possible timescale problem? We only see disks of small particles around
stars during their first 10 million years or so, and various estimates of the Earth’s formation time
are about 30 million years. Don’t try to calculate if there is enough time--that is too difficult a
problem unless you immediately see through it. If you do decide to investigate, you would need
to know that the density in the protosolar disk at 1AU might have been about 10^14 H atoms per
cm^3 (i.e. about 10^-10 g/cm^3) and the particles (grains) comprised about 1 percent of the disk
mass.


         Part C. Solids, liquids, intermolecular forces, surface tension.
Do one of the following two multi-step problems 4 or 5. Both are more oriented toward
originality and independence than the previous problems, and in some cases there will be no hard
and fast answer. Instead this is along the lines of assigning homework and grading it on the basis
of evidence that you have put in a good effort in thinking about the questions.
         Remember, the sentences that are the “real” problem, the thing whose answer you
will turn in, are in bold, the rest is either background, idle speculation, or assistance in
getting started on the problem. In whatever you turn in, feel free to speculate as much as you
want, or to examine any other ideas that you get. This is how research begins--and not having
enough time will not stop you if you are actually interested in the subject.

4. Surface tension and the shapes of macromolecules and aerosols: Can there be raindrops
on Titan? Why isn’t DNA round?

        Self-gravitating objects bigger than, say, the asteroid Ceres, have a nearly spherical shape
because self-gravity is spherically symmetric. But for smaller objects, like the moons of Mars or
rocks you see laying around, there is no reason to expect spherical particles. Yet most people,
with exceptions, model particles from interstellar grains to pre-planetesimals as having some
simple shape, often spherical, “to get an order of magnitude result.” Researchers who try to
improve the situation go as far as fractal grains, without realizing how restrictive that is. But how
irregular could these particles be? How does that extreme irregularity feed back to cloud
formation and climate, to planet formation through strange drag forces, and so on? They could be
wispy with tendrils of porous filaments or cross-linked polymers that “melt” and reform, or any
of dozens of other creatures in the mesoscale zoo. But we know one thing: Raindrops are more or
less spherical, maybe teardrop-shaped. Is the effect that makes them round important for other
kinds of substances?
         In the paper on liquids by Lower, he explains clearly the origin of surface tension in
liquids, water in particular, as due to the asymmetry the H2Os “feel” at the surface, where they
are pulled back into the liquid by imbalanced noncovalent forces, especially H bonding. He then
explains how, by forcing the surface to minimize its area, that is why droplets are round. Let’s
look at this a little more closely.
         4a. Many amorphous solid substances (say ice) form H-bonding networks. Why doesn’t
ice adopt the shape of spheres? I am not referring to the ice-cube scale or larger, but the very
smallest bits of ice, say the ones that are the sizes of raindrops. Many people assume that there
interstellar and disk grains that are ~ 0.1 micron size and made of various kinds of amorphous ice.
Should they be spherical?
We see irregular materials all the time, but maybe they are being deformed on timescales smaller
than the time it takes for surface tension to form spheres. And biomacromolecules are certainly
not round (think DNA, proteins), but lipid bilayers (cell walls) and their relatives (the micelles)
tend to be spherically symmetric, so maybe it is a peculiarity of water again.
         How to approach this? You mostly only need to know that the change in surface area in
going from some form, say a sheet, to a sphere requires work, and the proportionality between the
change in area A and the energy it takes W has a constant of proportionality which is called the
surface tension :
                                    W =  A =  4a^2 erg cm
where I too an area of 4a^2 for the droplet. Surface tension has funny units that are usually
given as milli-Newtons per square meter (really just like a pressure), which translates into 1 N
m^-2 = 1 J m^-1 = 10^7 erg/100 cm/m = 10^5 erg/cm. The surface tension of water is 72.8 mN
m^-2 at 20C and 50 mN m^-2 at 100C. We’ll adopt 70 mN m^-2 = 7.0 e3 erg/cm.
         For an interesting comparison, I find that  for liquid hexane, methanol, and ethanol, are
all in the range 18 to 22, and benzene is at 29. Also notice that  for Mercury is a whopping 472!
I guess we now see why it beads up so well.
         Use all your internet prowess to find the surface tension of some other substances.
Try to include some weird ones, and see if there are any non-liquids mentioned. Methane and
ethane are particularly significant since they are often suggested as the rain and surface liquid on
Titan. If the surface tension of methane drops are much smaller than for water, maybe
Titan’s “rain” can’t even get very far on the road to roundness. Maybe the rain is irregularly
shaped spokey fluid microblobs if it happens at all.
          (A little-known secret is that terrestrial cloud drops can’t actually get round--in theory.
I’ll explain why in class.)
         4b. Return to the question of why solids and other substances don’t seem to form round
objects (or do they?) but instead seem to grow into irregular eerie forms. I’ll copy a few in here.
One is an “IDP” (interplanetary dust particle), another is a laboratory simulation of a grain that
has grown under conditions like those in a pre-planetary disk, and finally a computer simulation
of the same thing. (Sorry, I couldn’t resist putting in a mycoplasma bacterium, number two in the
row.)

.
What Lower’s paper should have shown you was why surface tension makes drops round when 
A is greater than whatever energy is associated with forces that would deform the drop. (On Earth
gravity makes them teardrop-shaped, but there could be many other effects).
         Calculate the energy that is used to make a drop of size 10^-4 (a/1micron) by
surface tension, and compare that with the forces that might disrupt grains in other
circumstances. Use your imagination for “forces.” Are smaller or larger drops favored?
         Generically, for a given “deforming energy” E_def write the relation between the
size of the drop that you can form (it will take too much work to sphericalize a large
particle) and the surface tension.
         What does this imply about the size of droplets in the clouds of Jupiter? Of Titan?
         On Earth that critical size is thought to be about 10^-7 cm, and then the sphericity is
assumed to be maintained through later growth because the accretion of gas is so isotropic that
the grain can grow to (say) 10 micron or even the mm-size raindrops you have experienced. That
sounds good, but actually the drops are supposed to grow by collisions, as the larger and smaller
grains catch up with each other as they fall to the ground.
         How can droplets maintain their spherical shape using surface tension when they
are being smacked around in collisions. Can surface tension really be effective at a scale of
a mm water droplet? If you can’t think of an answer, then answer “I can’t think of anything.”
         One thing to thing about that might be helpful is: why aren’t biological macromolecules
spherical? I think in that case it is basically covalent bond energies dominating any tendency to
sphericalize (but that is a guess). In addition, the formation process itself is polymerization that
can only occur in a chain (think peptide bonds joining amino acids to make secondary structure of
protein):
         Calculate the “surface area” of a DNA molecule that has a length X and a
“diameter” of each strand Y. Look up the number of nucleotides (monomers) in some
typical DNA (the unit will be “Dalton”, but that just means the mass in nucleotides; you can
search under “genome size” for starters) for X (pick a range of genome sizes that you encounter
in your search) and find (like in a picture) or guess the thickness of a strand Y (it is just a few
molecules). Now consider that it is a tightly coiled structure and use the width of the coil
(find a picture!) for Y and recalculate the range of areas. It is not a “liquid” because there
are no “phases” at this level of complex structure, but IF macroscopic properties like surface
tension have meaning in terms of intermolecular forces, you would expect there to be some
anisotropic pull into the interior of the coiled DNA.
         How does the surface area of this coiled DNA compare with the area of a spherical
object with the same mass ? Could the shapes of some of these structures reflect something
like “the best that surface tension can do” given the other constraints? (You may be
interested to know that the outside and inside of DNA are coated with hydrophobic and
hydrophilic molecules, respectively.)
         Later in the course we will see that there is a “helix-coil” transition that occurs for many
polymers and see how it is related to something like the surface tension trying to minimize the
energy of the object. The trick for a living polymer is that it can reverse its uncoiling! This
process, which is central for the goals of this course, is called “denaturation” if you want to get an
early start.

5. Vapor pressure and the ocean: How thick an atmosphere does the Earth or another
planet need to keep a pressure lid on the oceans, before they evaporate?
         Obviously (there was a time when it wasn’t obvious to me) if there were no atmosphere,
there would be no gas whose pressure can balance that of the flux coming from the oceans, which
would then evaporate, no matter what the temperature (this is why there can’t be oceans on Mars-
-the water would just sublime or vaporize because no pressure. In our thick atmosphere (1 bar at
the surface), it only takes a small abundance of H2O gas to balance this flux. But the Earth’s
atmosphere was probably thicker in the past and will be thinner in the future
         In your reading by Lowers on liquids, he explains the significance of the “equilibrium
vapor pressure” of water. For example, judging from the numbers he quotes (he has a few typos,
so I looked them up at one of the web sites you are supposed to find in question 1, in particular
the NIST site has an amazing amount of thermochemical and other data), the vapor pressure of
H2O at 24C = 297K is 24 torr = 0.03 atm = 0.03 x 1.013 bar. 1 bar = 750 Torr mm Hg, 1 Torr =
1333 bar. Planetary scientists always use the bar. Stellar and disk astronomers don’t use the
pressure much.
         I take this 0.03 bar vapor pressure to mean that at that temperature, the flux of water
molecules from the ocean just balances the flux of water molecules condensing from the
atmosphere to the ocean if the partial pressure of H20 is 0.03 bar, compared to the total pressure
of 1 bar. Just when 3% of the atmosphere is H2O, it is saturated, and you start feeling
uncomfortable because your body can’t evaporate any moisture. I am worried that Lowers has
made several mistakes, so the first thing to do is find vapor pressure data for water and verify if
he is indeed correct that you need that much water vapor--it seems like too much to me.
         According to his graph, the vapor pressure increases exponentially with temperature, and
at 60, 80, and 100 C it looks like the vapor pressure is 150, 300, and 750 bar, which is hard to
believe (we are at 1 bar--those numbers imly that at 80K you’d need 300 times the atmosphere of
the earth to contain the oceans, which is obviously not the case. Did he mean Torr? I think he
means millibar all around, so 0.03 mbar = 3x10^-5 bar (sounds more like it) at 24C, 0.15, 0.3, and
0.75 mbar at T=60, 80, 100. This is still scary since it says the atmosphere would have to be 30%
water vapor at 80 C. I am sure it is never that large.
         So your first assignment in this question is to throroughly research the vapor pressure of
water and hopefully find a handy formula for it as a function of temperature (in K please!).
         Now consider what would happen if the Earth’s atmoshere had a smaller column density,
i.e. a smaller mass and so smaller number of particles per unit area ( a number which is about
1000 g/cm^3 for the present Earth’s atmoshere). Think about bringing it down to 100, then 10
g/cm^2.
          Is it valid to say that the surface pressure would decrease, since the pressure has to
balance gravity, you will always have 1 bar pressure at the surface (to balance gravity), and
it would just thin out rapidly to higher altitudes? The vapor pressure curve (and the Clausius-
Clapyron equation that describes it as a Boltzmann like equation p = p_o exp(H/kT) where p_0
is the reference equilibrium pressure and H is the free energy of vaporization) could be simple
enough, and might require some enormous water vapor partial pressure at some temperature, as it
appears in Lowers’ article. Is this a problem if the atmosphere doesn’t contain that much gas to
begin with?
         The answer is no, because all that matters is the H2O pressure, which is coming from the
ocean. So as long as there is enough ocean, the atmosphere could have 100 times its present mass
due to evaporation from the oceans. All would be well except that I don’t know what would
happen to all that water vapor, and all present organsim would die.
          At present a lot of the water vapor is photodissociated and the resulting H escapes the
atmosphere. This suggests that the key will be having an atmosphere so thin that there is no
ozone layer, resulting in the UVC, as wel as UVB, reaching the near-surface where it could
photodissociate the H2O coming off the ocean. Then, no matter how much water the oceans are
putting into the atmosphere, if the UV radiation can break it up faster than it is injected (say by
volcanoes), the H will escape and you will have lost a lot of water vapor and liquid water but
gained some O2 and O3 (produced by photodissiation of product O of H2O). Can you set up this
problem in mathematical turns?
         Another condition might be that there has to be sufficient mass of ocean that it won’t just
dry up before it can give the atmosphere whatever vapor pressure is required. What is the mass
of the ocean, in absolute units, and relative to the earth’s total mass? For a given pressure at
the surface, let’s say 1 bar (you will answer in the first part whether that is today’s atmosphere),
write a condition for the mass of water vapor in the atmosphere (equivalently, its partial
pressure) at which a (possibly huge) equilibrium vapor pressure leaves no ocean, as a
function of ocean mass and average temperature. This should produce a curve that is a
function of temperature, for probably four ocean masses (10, 1, 0.1, 0.01 times the mass of
the Earth’s ocean--you can change these choices when you see which give interesting
answers). Use this opportunity to produce an interesting plot that is nearly publication
quality in terms of presentation (remember, labels on axes and legends must be large
because they are reduced upon reproduction). If you aren’t sure what makes a good graph,
look through some issues of Astrophysical Journal, Monthly Notices, Astronomy and
Astrophysics, in Peridier library or online.
         This problem illustrates a problem associated with global warming--more ocean will
evaporate as it heats up, the greeenhouse due to H2O heats still more, producing more H2O...
I’m sure the chemistry is much more complicatedl, but that is one scenario, in which the
terrestrial planet loses its ocean and possesses a temporary steam atmosphere which is gradually
lost to space as the H2O is photodissociated by the UV radiation from a solar-type star. Can you
see how this problem might be avoided if you lived on a planet orbiting an M-type star? (You
would have other, probably more serious, problems.) So I am asking for a little commentary on
the problem here. Or you can find the error in any of the reasoning above, and explain my
error clearly.

Finally, THE END.

				
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