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Astronomy 15 - Problem Set Number 5 - Due Wed. May 8 Be sure to have your paper topic selections ready to hand in on Wednesday May 8. Start rooting around for a topic now! The first four problems are from Shu’s book; they’re embedded in a magnificent chapter on stellar structure as applied to the sun. 1) Shu problem 5.11 2) Shu problem 5.12 3) Shu problem 5.13, with the following extension. In Shu 5.11 you computed how long it takes a photon to leak out of the sun. This is comparable to the time it would take all the radiant energy inside the sun to escape. As you’ve just shown, there’s much more energy than this present in the plasma, in the form of the kinetic energy of the particles. Imagine that the radiant energy leaks out, then is replaced by transferal from the plasma energy, then it leaks out again, etc. (This would actually occur as a continuous process, but never mind; it’s a completely artificial ‘thought experiment’, or Gedankenexperiment, anyway.) This process would continue until the supply of plasma energy was exhausted, which would require a time time = Radiation leak time × ratio of plasma to radiant energy. Compute this time by multiplying your answers from (5.11) and (5.13). 4) Shu problem 5.14. Compare your answer to that in my extension of the previous problem. It should agree to order of magnitude. 5) Use the virial theorem to estimate the internal temperature of the sun. To do this, note that: • The average kinetic energy per particle in the sun is (3/2)kT ; • The gravitational binding energy of mass M and radius R is of order GM 2 /R; and • The number of particles in the sun is about 2× Avogadro’s number × the mass of the sun in grams. The factor of two arises because the sun is almost fully ionized, so there are two particles per atom of hydrogen (and atomic hydrogen has 1 gram per mole). 6) This problem calls for crude estimates of some of the most important timescales in stellar structure and evolution. There are a lot of words here, but not much work, as usual. a) The shortest timescale is the amount of time it takes for the star to respond to disturbances in it pressure equilibrium. Suppose, for instance, a cosmic giant comes along and 1 give the star a kick. The star’s surface will slosh around on some characteristic timescale, which is the amount of time it takes for gravity to bring the surface back. Or, suppose that all the internal pressure holding the star up were suddenly to fail (this just about happens under some special circumstances). The amount of time it would take for the star to collapse would be of the same order of magnitude. This time is called the dynamical timescale of the star. It is of the same order as the time it would take a satellite to orbit the star once, just grazing its surface. Take the dynamical timescale, τd , to be the same as the time required for such a low orbit. Show that τd is independent of the size of a body, being dependent only on the average density. That is, show that to order of magnitude 1 τd = √ , Gρ where ρ is the average density and G is Newton’s universal gravitational constant. Estimate the dynamical timescale for the sun (1 gm cm−3 ), for a white dwarf (106 gm cm−3 ), and for a neutron star (1014 gm cm−3 ). Also, find the dynamical timescale for a body with mean density 10−30 gm cm−3 and convert this last answer to years. (This has cosmological implications!) b) The next shortest timescale is called the thermal or Kelvin-Helmholtz timescale. (This part is pro forma, since you just calculated this in (3) and (4) earlier.) It is the time it would take for all the thermal energy of the star to leak out at its present level of power output, which (by the virial theorem) is about the same as the time it would take for the star to radiate away all its gravitational binding energy. All this assumes that the star shines at constant brightness, which may be unrealistic but which is adequate for orderof-magnitude purposes. You don’t have to calculate this; just repeat your answers from the earlier problems! c) The longest timescale – usually – is called the nuclear timescale, which is the amount of time it would take the star to burn the available nuclear fuel at its present luminosity. Generally speaking, a star will undergo major changes if it burns 1/10 of the available nuclear fuel. Let’s calculate this timescale for the sun. First note that the conversion of hydrogen to helium – the reaction which powers the sun – results in the release of 0.007 of the rest energy of the original fuel. That is, 0.7 per cent of the original matter is converted into energy, or in other words if one takes a mass m of hydrogen and turns it into helium, one releases an energy E = 0.007mc2 . Compute how long (in years) 0.1 M of hydrogen would last if it were burned at a rate giving 1 L of luminosity. 2 7) The following is a list of the weights several types of nuclei, expressed in terms of the weight of the hydrogen atom (which is not standard). The difference between these weights and the weights of collections of hydrogen atoms having the same number of baryons (e. g., the difference between the weight of 12 hydrogen atoms and the weight of 12 C) is the nuclear binding energy of the nucleus. Compute the binding energy per nucleon (that is, per particle in the nucleus) of the nuclei below. Express your answers in MeV (the mass of the hydrogen atom is about 938 MeV). Plot the binding energy per nucleon as a function of atomic weight from these data; you might want to break the horizontal scale into two parts to avoid squishing the light elements all into the left-hand side. (Example: Carbon 12 has a binding energy of 12 − 11.90683 = 0.09317 in units of the hydrogen mass; thus its total binding energy is 0.09317 × 938 MeV = 87.39 MeV, and its binding energy per nucleon is 87.395 MeV = 7.28 MeV nucleon−1 12 nucleons Table of Atomic Masses. Isotope Mass (H = 1) 1 H He 7 Li 9 Be 11 B 12 C 14 N 16 O 20 Ne 24 Mg 28 Si 32 S 40 Ca 56 Fe 82 Kr 140 Ce 208 Pb 235 U 1.00000 3.97152 6.96154 8.94222 10.92385 11.90683 13.89435 15.87071 19.83721 23.79881 27.75971 31.72383 39.65231 55.50061 81.27750 138.81904 206.36102 233.22402 4 3