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 = 4a^2 erg cm where I too an area of 4a^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.