Astro 162 – Planetary Astrophysics – Solutions to Set 9Problem 1. The Solar CoreIs the core of the Sun degenerate (or is the pressure there instead given by the ideal gaslaw)? Justify your answer quantitatively.We can either look up the core temperature and core number density of the Sun(completely valid decision for this problem), or we can try to estimate our way there.Let’s try the latter ap proach first. I remember hydrogen fusion kicks in at about T ∼107K, so I will use that as the central temperature. To estimate the number density,one might just say ρ/µ, where ρ ∼ M/V, µ ∼ 2 × 10−24g, and Vis the volume of theSun. This naive procedu re gives ρ ∼ 1 g cm−3. Sadly, I also remember that this is toolow an estimate; the central density is actually a few orders of magnitude higher thanthe mean density.To estimate the true core density, I will employ the following rus e. I remember fromtaking Astro-7-esque classes in the past that I can estimate the central pressure prettyaccurately by a brazen use of hydrostatic equilibrium:PR∼ gρ ∼GMR2MR3. (1)ThereforeP ∼GM2R4. (2)Let’s assume that the core gas is ideal and not degenerate. Now I understand the wholepoint of the problem is to decide this issue; but we can assume it’s ideal and then checkto see if our assumption is correct, or at least close to being correct. Then P = ρkT/µ,and thereforeρµ∼GM2R4kT(3)where we have plugged in for P above. Plugging in M = M, R = R, and T = 107K,I getnp∼ρµ∼ 8 × 1024cm−3= ne(4)1where npand neare the numb er densities of protons and of electrons, respectively.Our estimates above are not too far off from the truth, where the truth is given bydetailed calculations of stellar structure:T ∼ 1.6 × 107K (5)np= ne∼ 8 × 1025cm−3(6)This npimplies that ρ ∼ 160 g cm−3. This central density for the Sun is probably worthremembering.Now we finally check to see whether such a temperature and density corresponds toa degenerate gas. First we check to see whether the electrons are degenerate. Calculatethe de Broglie wavelength of each electron,λe=hpe(7)where the momentum of the electron, pe, is given byp22me∼32kT . (8)Thenλe=h√3kT me∼ 2 × 10−9cm.(9)Compare this to the mean spacing between electrons ,n−1/3e∼ 2 × 10−9cm. (10)The fact that λe∼ n−1/3eimplies that the electron gas is actually mildly degenerate.So you shouldn’t think of degeneracy as only afflicting “compact obj ects” like whitedwarfs, brown dwarfs, and neutron stars. The electrons in the core of the Su n are prettytightly packed. Note that had we used our crude estimates, we would have arrived atpretty much the same answer (the 1/3 power in n−1/3esaves us).2What about the pr otons? We can subs titute mpfor meeverywhere above to findthat λp n−1/3pand therefore the proton gas is not degenerate (i.e., the proton gasbehaves like an ideal gas).Problem 2. Liquid GiantsThis problem is adapted from problem 11.2 of Landstreet. Mostly I have exchanged hisSI units for more sensible units. You can read the wording of his problem for hints, ifyou need them.In Saturn, the temperature is about 110 K at a pressure of 0.5 bar (1 bar = 106dyne cm−2).Below this level the atmosphere is convecting and the temperature increases with depth sat the quasi-adiabatic rate of dT/ds ≈ 0.7 K km−1. Take the atmosphere to be composedpurely of molec ular hydrogen.Gas liquifies once its density approaches that of liquid, of order ∼1 g cm−3. Liquifica-tion also requires that the tempe rature be just right, but let us ignore the temperaturedependence for this problem and just focus on the density dependence.ESTIMATE the depth, s, at which molecular hydrogen liquifies. Use hydrostatic equilib-rium, and assume that the gravitational acceleration g is constant (it will be, roughly, aslong as you don’t wander too close to the center of the planet).Thereby argue that the gas giants, Jupiter and Saturn, are more appropriately called“liquid giants.”Hydrostatic equilibriu m statesdPds= gρ (11)where s increases downwards, and g is assumed to be a positive constant. Insert theideal gas law into the left-hand-side to writedPds=kµTdρds+ ρdTds. (12)Now T is a known function based on the information given in th e problem:T = T0+dTdss (13)where T0= 110 K is the temperature at the reference level s = 0 (where th e pressure,P0= 0.5 bar). Let’s write this more conven iently as3T = T0+ βs (14)where β ≡ dT/ds = 0.7 K k m−1= 7 × 10−6(cgs) is a known constant. Plug everythin ginto (12):dPds=kµ(T0+ βs)dρds+kµβρ (15)Insert into (11) and re-arrange:(T0+ βs)dρds=µgk− βρ (16)This is a separable equation,dρρ=µgk− βdsT0+ βs, (17)whose solu tion isρ = C (T0+ βs)µgkβ−1(18)(after exponentiating both sides). Solve for the constant of integration C using theboundary condition at s = 0. At s = 0, ρ = ρ0= µP0/(kT0) = 1.1 × 10−4g cm−3, whereµ = 3.4 × 10−24g. The exponent in equation (18) equalsµgkβ− 1 ≈ 2.5 (19)where g = GMS/R2S= 103[cgs] is the gravitational acceleration near the Saturniansurface (coin cidentally nearly the same as that on Earth). Therefore, putting it alltogether,C ≈ρ0T2.50≈ 8.7 × 10−10[cgs] (20)Now all the constants are known. Finally, use (18) to solve for the depth s at whichρ = ρliquid∼ 1 g cm−3:4sliquid≈1β"ρliquidC12.5− T0#≈ 6000 km (21)Thu s , we need to travel about6000 km downwards to attain liquid conditions. Sincethe radius of Saturn is about 60000 km, we can conclude (modulo the dependence of thephase diagram on temperature) that most of Saturn is liquid .Problem 3. The Incredible Invariant Radius(a) Combine the equation for hydrostatic equilibrium,dPdr= gρ , (22)Poisson’s equation for gravity,∇2φ = 4πGρ , (23)and an approximate equation of state for liquid hydrogen,P = Kρ2, (24)to derive a single second-order differential equation for density as a function of radius,ρ(r). Here φ is the gravitational potential, g = −∇φ is the gravitational acceleration,P is pressure, and K = 2.7 × 1012[cgs] is a constant appropriate for liquid hydrogen.Assume spherical symmetry (so work i n spherical coordinates). Notice that I am askingyou to use P ∝ ρ2, which is close, but not qui te equal to what we derived in class,P ∝ ρ5/3. The latter law is really only appropriate for a pure degenerate gas of freeelectrons, and masses of order a J upiter only achieve