AST 341 - Homework V - Solutions TA: Marina von Steinkirch, steinkirch@gmail.com State University of New York at Stony Brook November, 010 1 (1 point) Derive the homologous form of the luminosity equation df dx = Dp t ν x. The homologous relations are x = r R, f = L(r) L 0, We know also the following identities: Now, from class we have following immediately p = P (r) P 0, where P 0 = GM 4πR 4 t = T (r), where T 0 = µgm. T 0 RR g ɛ = ɛ 0 ρt ν, µp (r) ρ = R g T (r), D = D 0 ɛ 0 µ ν M ν+ LR ν+3, dl dr = 4πr ρɛ. dp dx = dp dl dr dl dr dx, df dx = df dl dl dr dr dx, = 1 L 4πr ρɛ R = 1 L 4πr ρ ɛ T (r) µ R 1
Plugging the identities into this last equation and evaluating correctly ρ and D gives df dx = Dp t ν x. 11. (1 point) L M 3.84 10 6 J/s 1. 10 34 J/year 1.988 10 30 kg.1 At what rate is the Sun s mass decreasing due to nuclear reactions? Express in solar masses per year. From the previous chapter, example 10.3., we can just use the Einstein s energy-mass relation, E = mc, we have Ṁ = 6.67 10 14 M /year.. Compare you answer with the mass loss rate due to the solar wind. From example 11..1 in our book, loss rate due to the solar wind is giving my Ṁ = 3 10 14 M /year, which is half of what we have found before..3 Assuming that the solar wind mass loss rate remains constant, would either mass loss process significantly affect the total mass of the Sun over its entire mainsequence lifetime? Let us say that Sun will stay in its main-sequence lifetime for 10 10 year. These both quantities of mass loss, 10 14 M /year 10 10 year 10 4 M, are negligible. 3 11.5 (1 point) 3.1 Using Eq. 9.63 ( λ) 1/ = λ c (kt m + v turb ) ln,
and neglecting turbulence, estimate the full width at half-maximum of the Hydrogen Hα absorption line due to random thermal motions in the Sun s photosphere. Assume that the temperature is the Sun s effective temperature. For H α absorption, λ α T k b m 656.80 nm 5777 K 1.38 10 3 J K 1 m H m p 1.67 10 7 kg. Plugging back in 9.63, we have ( λ) 1/ = 0.03558 nm. 3. Using Hα redshift data for solar granulation, estimate the full width at half-maximum when convective turbulent motions are included with thermal motions. From data we have v turb = 0.4kms 1, and we find a very similar result ( λ) 1/ = 0.0356 nm. 3.3 What is the ratio of v turb to kt/m? The ratio is 10 3, meaning that they are very similar. 3.4 Determine the relative change in the full width at half-maximum due to Doppler broadening when turbulence is included. Does turbulence make a significant contribution to ( λ) 1/ in the solar photosphere? which is negligible. δ( Λ) 1/ ( Λ) 1/ 0.1%, 3
4 11.1 (1 point) Calculate the magnetic pressure in the center of the umbra of a large sunspot. Assume that the magnetic field strength is 0. T. Compare your answer with a typical value of 10 4 N.m for the gas pressure at the base of the photosphere. µ 0 4π 10 7 N.A B 0. T The magnetic pressure in the center of the umbra can be calculated by P B = B µ 0 = 1.5916 10 4 N.m, which is the same order of the typical value at the base of the photosphere. 5 1.10 (1 point) Calculate the Jeans length for the dense core of the giant molecular cloud in example 1..1. T 10K ρ 0 3 10 17 kg m 3 µ M H m p 1.67 10 7 kg G 6.67 10 11 N.m kg 1.38 10 3 J K 1 k b The Jeans length is given by R J ( ) 1/ 15k b T, 4πGµm H ρ 0 5 10 15 m 4
6 1.18 (1 point) 6.1 Beginning with Eq. 1.19, d r dt = GM r r, adding a centripetal acceleration term and using conservation of angular momentum, show that the collapse of a cloud will stop in the plane perpendicular to its axis of rotation when the radius reaches r f = ω 0r 4 0 GM r, where M r is the interior mass and ω 0 and r 0 are the original angular velocity and radius of the surface of the cloud, respectively. Assume that the initial radial velocity of the cloud is zero and r f r 0. Adding a centripetal acceleration term we have d r dt = GM r r + ω r = U du dr, we then multiply both size by dr integrate in the limit,resulting GM r r + ω r Making r = r f and putting it back gives = U (r). ω F = GMr r 3 f r f = ω 0r 4 0 GM r. 6. Assume that the original cloud has a mass of 1 M and an initial radius of 0.5 pc. If collapse is halted at 100 AU, find the initial angular velocity of the cloud. ω 0 = GMr r f r 0 =.8 10 16 rad/s. 6.3 What was the original rotation velocity (in m s 1 ) of the edge of the cloud? ω 0 = v 0 r 0 v 0 = 4. m/s. 5
6.4 Assuming that the moment of inertia is I s = 5 Mr, when the collapse begins and I d = 1 Mr when it stops, determine the rotation velocity at 100 AU. Here we can use conservation of momentum, I 0 ω 0 = I f ω f ω f r f = v f = 3.4 km/s. 7 ( points) T Tauri stars collapse quasi-statically, converting gravitation potential energy into heat. If you could measure the radius of a T Tauri star very accurately, how much smaller would it be after a year? How long would it take before you noticed the star fading? T eff R π M π 4500 K 3R 1M This problem is partially worked out at page 46 of the textbook. First, L = 4πR σt 4 = 1.3 10 7 W, and it is related to the gravitational potential as giving L = GM R dr dr = L/R 660 m/ year. GM To notice the fading we should be able to notice the change of magnitude: δm = 10 3.5 =.5 log (L + δl L ), giving δl = 1. 10 4 W. From the luminosity δl = 8πRdRσT 4 = 8 10 0 W/ year, therefore we would need more than thousand years to notice it. 6
8 13.1 (1 point) 8.1 How long does it takes the 5 M star to cross the Hertzsprung gap relative to its main-sequence lifetime? t = 0.46 million of years (Myr), 0.5%. 8. How long does the 5 M star spend on the blueward portion of the horizontal branch relative to its mainsequence lifetime? t = 1.50 million of years (Myr), 1.6%. 8.3 How long the 5 M star spend on the redward portion of the horizontal branch relative to its main-sequence lifetime? t = 6.3 million of years (Myr), 6.8%. 9 13.5 (1 point) Use equation 10.7, T quantum = Z 1Z e 4 µ m 1π ɛ 0 h k, to show that the ignition of the triple alpha process at the tip of the red giant branch ought to occur at more than 10 8 K. Z 1 Z e 1.6 10 14 C First of all, the reduced mass is µ m p 3.34 10 7 kg. Plugging in everything into the equation gives T reaction = 6. 10 8 K. 7