* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Solutions - faculty.ucmerced.edu
Survey
Document related concepts
International Ultraviolet Explorer wikipedia , lookup
Outer space wikipedia , lookup
Formation and evolution of the Solar System wikipedia , lookup
Drake equation wikipedia , lookup
Dyson sphere wikipedia , lookup
Aquarius (constellation) wikipedia , lookup
Corvus (constellation) wikipedia , lookup
Andromeda Galaxy wikipedia , lookup
Negative mass wikipedia , lookup
Structure formation wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Observational astronomy wikipedia , lookup
Stellar evolution wikipedia , lookup
H II region wikipedia , lookup
Timeline of astronomy wikipedia , lookup
Transcript
Physics 161 Homework 7 - Solutions Wednesday November 16, 2011 Make sure your name is on every page, and please box your final answer. Because we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish them. The homework is due at the beginning of class on Wednesday, November 23rd. Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions during the discussion session or during office hours. 1. (a) Assuming that the surface temperature of the Sun is T = 5800 K, use Stefan’s law to determine the rest mass lost per second to radiation by the Sun. Take the Sun’s radius to be R = 7.0 × 108 meters. (b) What fraction of the Sun’s rest mass is lost each year from electromagnetic radiation? Take the Sun’s rest mass to be M = 2.0 × 1030 kg. ———————————————————————————————————— Solution (a) The intensity of radiation (in watts per square meter) from the Sun is I = P/A = σT 4 , where σ is the Stefan-Boltzman constant and T is the temperature. Thus 2 σT 4 , where R is the radius of the Sun. Now, the the power is P = AI = 4πR power is P = Ė, and since the energy is coming from the rest mass, E = mc2 , then Ė = ṁc2 . Thus, the rate of mass loss is 2 σ 4 4πR dm = T . 2 dt c So, plugging in the constants gives 2 4πR σ 4 4π (7 × 108 )2 (5.67 × 10−8 ) dm T = = (5800)4 = 4.39 × 109 kg/sec. 2 8 dt c2 (3 × 10 ) (b) So, in one year the Sun burns through ∆m = dm · ∆t = 4.39 × 109 × 3.15 × 107 = 1.38 × 1017 kg. dt This is a fraction ∆m 1.38 × 1017 = = 6.9 × 10−14 , 30 M 2 × 10 meaning that the Sun loses roughly 10−13 % of its mass per year. 2. Emission lines of hydrogen, Hβ (n = 4 → 2 and λrest = 4861Å) are observed in the spectrum of a spiral galaxy at redshift z = 0.9. The galaxy disk is inclined by 45◦ to the line of sight (if the inclination was 0◦ then we’d see the galaxy from a top view, while if it was 90◦ , we’d see it edge-on). The Hβ wavelength of lines from one side of the galaxy are shifted to the blue by 5Å relative to the emission line from the center of the galaxy, and to the red by 5Å on the other side. (a) What is the galaxy’s rotation speed? (b) Find the age of the Universe when the light was emitted from this galaxy. ———————————————————————————————————— Solution (a) There are two Doppler shifts in this problem, one from the redshift z, and the other from the rotational motion of the galaxy. The redshift from the galaxy shifts the wavelength λobs = 1 + z = 1.9. λemit Thus, the observed wavelength to the center of the galaxy is λobs = (1 + z) λemit = 1.9λemit = 9236 Å. Next, the rotation of the galaxy adds an additional redshift, depending on the rotational velocity. This velocity also depends on the angle of orientation. If the angle was zero, giving a top view of the galaxy, then we wouldn’t measure any rotational velocity, since it would be rotating perpendicular to the line of sight. If the angle was 90◦ , then we’d see the galaxy edge-on, and would measure the full rotational velocity. Thus, we see that we care about the sine component of the velocity, v sin θ. Now, the Doppler effect in the light emitted from the rotational velocity is (taking the specific case of the edge that is moving away from us), r c + v sin θ λ0obs = , λobs c − v sin θ where λ0obs is the wavelength seen at the edge of the galaxy, and λobs is the wavelength at the center of the galaxy ((1 + z) λemit ). Since the rotation velocity is much smaller that light we can expand r 2 λ0obs v v 1 + sin θ = 1 + sin θ. ≈ λobs c c So, λ0obs − λobs ∆λ v = = sin θ, λobs λobs c where ∆λ is the difference in wavelength between the edge and center of the galaxy (5 Å). Now, λobs = (1 + z) λemit , and so, all together the velocity is v= ∆λ c . sin θ (1 + z) λemit Plugging in the numbers gives v = 230 km/s (which shows that we were justified in assuming v small in our derivation). (b) The redshift is defined in terms of the scale factor as 1 a (t) = . a0 1+z The galaxy emitted light in the matter-dominated era, and so a ∼ t2/3 . Thus, a (t) = a0 2/3 t 1 , = t0 1+z where t0 is the current age of the Universe. Solving for the emission time, t we find t = t0 (1 + z)−3/2 . Taking t0 = 13.7 billion years, then t= t0 (1 + z)3/2 = 13.7 × 109 (1 + 0.9)3/2 = 5.23 billion years. 3. We have seen that, for a star composed of a classical, nonrelativistic, ideal gas, Etotal = tot tot Eth + Egrav = −Eth , and therefore the star is bound. Repeat the derivation, but for a classical relativistic gas of particles. Recall that the equation of state of a relativistic total , and therefore Etotal = gas is P = 31 EVth . Show that, in this case, Egrav = −Eth tot Eth + Egrav = 0; i.e., the star is marginally bound. As a result, stars dominated by radiation pressure are unstable. ———————————————————————————————————— Solution The pressure of a relativistic gas is P = Eth /3V , which means that hP i = 1 tot E . 3V th Furthermore, we’ve seen that 1 Egrav , 3 V tot tot = −Eth . Since Etot = Eth + Egrav = −Egrav + Egrav = 0. hP i = − which tells us that Egrav 4. Because of the destabilizing influence of radiation pressure, the most massive stars that can form are those in which the radiation pressure and the nonrelativistic kinetic pressure are approximately equal. Estimate the mass of the most massive stars, as follows: (a) Assume that the gravitational binding energy of a star of mass M and radius R is Egrav ∼ GN M 2 /R. Use the virial theorem hP i = − to show that P ∼ 4π 34 1/3 1 Egrav , 3 V GN M 2/3 ρ4/3 , where ρ is the typical density. (b) Show that if the radiation pressure, Prad = 13 aT 4 , equals the kinetic pressure, then the total pressure is 1/3 4/3 3 kB ρ , P =2 a m̄ where m̄ = mH /2 is the mean mass. (c) Equate the expressions for the pressure in parts (a) and (b), to obtain an expression for the maximal mass of a star. Find its value, in solar masses, assuming a fully ionized hydrogen composition. ———————————————————————————————————— Solution (a) The pressure P ∼ GN M 2 . 3RV Now, 1/V = ρ/M , while R = (3M/4πρ)1/3 , and so 1/3 1/3 GN M ρ 4πρ 4π P ∼ = GN M 2/3 ρ4/3 . 3 3M 34 (b) The total pressure is P = Pgrav + Prad = pressures are equal then ρkB T m̄ + 31 aT 4 . If the kinetic and radiation ρkB T 1 = aT 4 ⇒ T = m̄ 3 3ρkB m̄a 1/3 Then, the total pressure is 1 2 2 P = Pg + aT 4 = aT 4 = a 3 3 3 3ρkB m̄a 4/3 1/3 4/3 3 kB ρ =2 . a m̄ (c) Equating the two expressions gives 4π 34 1/3 GN M 2/3 4/3 ρ 1/3 4/3 3 kB ρ ∼2 . a m̄ 4 This gives for the mass, after plugging in for a = 8π 5 kB /15h3 c3 , s 3 14580 hc = 2.27 × 1032 kg, M∼ π 6 m4H GN which is M ∼ 113M . So, the largest stars have a mass of roughly 100 solar masses. 5. Suppose a star of total mass M and radius R has a density profile r ρ = ρC 1 − , R where ρC is the central density (assume constant). (a) Find M (r). (b) Express the total mass M in terms of R and ρC . (c) Solve for the pressure profile, P (r), with the boundary condition P (R) = 0. ———————————————————————————————————— Solution (a) We can determine M (r) via the mass continuity equation, r3 dM (r) 2 2 = 4πr ρ (r) = 4πρC r − . dr R Separating and integrating noting that M (r = 0) = 0, we find Z M (r) Z dM (r) = 4πρC 0 0 r r3 2 dr r − . R Integrating both sides gives 4π 3r 3 ρC r 1 − M (r) = . 3 4R (b) Finding the total mass is very simple. We can either repeat the above derivation in part (a), integrating to r = R, or just plug in r = R in to our result for M (r). Then, 3 π 4π 3 ρC R 1 − = ρC R3 . M = M (R) = 3 4 3 (c) Now, to determine the pressure, P (r), we use the hydrostatic equilibrium equation, dP (r) GN M (r) ρ (r) . =− dr r2 The density ρ (r) was given, and we have just found M (r), so we find dP (r) dr N 2 3 = − 4πG ρC r 1 − 3r2 3r × 4R r 3r − 4R R N 2 = − 4πG ρC r 1 − 3 2 4πGN 2 = − 3 ρC r − 7r 4R 1 − Rr 2 + 3r R2 3r3 + 4R2 . Now, since we know that P (R) = 0, we integrate from r = r to r = R, Z 0 Z 4πGN 2 R 7r2 3r3 dP (r) = − ρC dr r − + 3 4R 4R2 P (r) r Integrating gives 4πGN 2 ρC −P (r) = − 3 5 2 r2 7 r3 3 r4 R − + − 48 2 12 R 16 R2 , or, P (r) = πGN ρ2C R2 5 2 r 2 7 r 3 1 r 4 − + − . 36 3 R 9 R 4 R Notice that the pressure at the center of the star (in this model) is P (0) = 5πGN 2 2 ρ R . 36 C