ASTR 680Problem Set 4Due Thursday, April 51. Fermi energy.Dr. Sane has been trying his hand at particle physics, and has reported that in his model thereis a previously unsuspected particle that he calls a “saneon” with a mass-energy of 8 GeV. Heasserts that the inner 0.5 M¯sphere of slowly rotating neutron stars has a high enough numberdensity of neutrons that it is energetically favorable to convert them to saneons (that is, thetotal energetic cost of adding another neutron exceeds 8 GeV). Based on this idea, Dr. Sane hasapplied for a physics faculty position at the University of Maryland. Drew Baden, chair of thephysics department, has been dubious about Dr. Sane’s previous ideas but is excited about thisone. What report do you give to Professor Baden? Note that the Fermi momentum at a neutronnumber density nnis pF= (3h3nn/(8π))1/3, where h = 6.63 × 10−27cm2g s−1.2. Pulsar planets.The amazing stability of pulsars has been used for many purposes, one of which was the firstdiscovery of planets outside the Solar System. To see how this was possible, consider a pulsarwith moment of inertia I = 1045g cm2, mass M = 1.5 M¯, radius R = 106cm, surface magneticfield B = 109G, and spin frequency 500 Hz (hence angular frequency Ω = 2π × 500 rad s−1). Towithin a factor of 5, compute the minimum mass of a planet in a circular orbit of radius 1 AUaround this pulsar such that the Doppler shifts it produces on the pulsar are at least a factor of 10times larger than the steady spindown of Ω over the course of one planetary orbit. Assume thatthe system is edge on to us.For this problem, please assume an orthogonal rotator, meaning that the energy lost inspindown is˙E = d(IΩ2/2)/dt = −B2R6Ω4/(6c3).If poor unwanted Pluto were to move to this system, could we detect it in a 1 AU orbit? Forcomparison, the smallest mass planets detected by observation of normal stars are at least severaltimes the mass of the Earth.3. Neutron star accretion from the interstellar medium.The spinning magnetic field of a neutron star presents a barrier to accreting matter. The matter,however, exerts a torque that slows down the neutron star rotation, so that eventually the mattercan accrete.(a) Suppose that the rate at which matter flows towards the star,˙M, is the Bondi-Hoyle rate thatwe have used before. If the neutron star has mass M and dipolar magnetic moment µ, this impliesan Alfv´en radius rA. If the orbital angular velocity Ωorbat rAis less than the spin angular velocityΩs, the matter is flung off, thus slowing the star down. Assume the speed at which the matteris thrown off is ΩsrA, so that the specific angular momentum after being thrown away is Ωsr2A.Write the differential equation for the evolution of Ωs, assuming the star has moment of inertia I.(b) Begin at the time when the star has a rotation period of P = 10 seconds. To within a factorof 3, compute how long it will take before the star can accrete from the interstellar medium. Forthis, we assume that the speed of the star relative to the ISM is v = 107cm s−1and the density atinfinity is ρ∞= 2 × 10−25g cm−3(these are needed for the Bondi-Hoyle accretion rate); and thatthe stellar mass is M = 1.5 M¯, its moment of inertia is I = 1045g cm2, and its dipole moment isµ = 1030G cm3.4. Helium core flash.Let’s get some insight into thermonuclear flashes in a slightly different context: the evolution of amassive star. Given the equations below (which indicate the energy generation rate per time intriple-alpha fusion of helium, plus how the temperature adjusts to that energy), write a code tofollow the temperature with time.I’ll need from you (a) a plot of the log temperature as a function of time (in units of days),and (b) an e-mail copy of your code, which has to be able to compile and run on the astromachines. Note that the temperature rises very suddenly during the flash, so you need to resolvethe rise; you don’t have to have a tiny time step the whole way, though.The details:Fix the density at ρ = 2 × 105g cm−3and assume the composition is always pure helium. Theinitial temperature is T = 1.5 × 108K. The energy generation rate for the triple-alpha reaction is²3α=5.1 × 108ρ2Y3T39e−4.4027/T9erg g−1s−1. (1)where T9≡ T/109K and Y = 1 is the helium mass fraction. For a given time step, in which somenumber of ergs per gram is produced, the change in temperature is determined by the sum of thespecific heats of the helium and the electrons. That is:dTdt=1cV(He) + cV(e)²3α. (2)Here cV(He) = 3.1 × 107erg g−1K−1is the specific heat for helium andcV(e) = 0.346(T/1 K) erg g−1K−1(3)is the specific heat for the electrons (these are not general formulae; the coefficients are specific tothis