AST 353 Astrophysics — Problem Set 1 Prof. Volker Bromm — TA: Aaron Smith∗ I. SIMPLE STELLAR MODEL Assume a star has a radius of R = 3R and a quadratic density profile: r 2 = ρc 1 − x2 . ρ(r) = ρc 1 − R Here, ρc = 20 g/cm3 is the central density. We have defined a new variable x = r/R to make future integrals easier, so the differential is dr = Rdx. (a) Mass To find the total mass M of the star we first find the mass coordinate m(r), which is the mass interior to a shell of radius r: Z r m(r) ≡ 4π r02 ρ(r0 )dr0 Z0 x = 4π (Rx0 )2 ρ(x0 )(Rdx0 ) 0 Z x 3 = 4πρc R x02 − x04 dx0 0 1 5 3 1 3 m(x) = 4πρc R x − x . 3 5 The total mass M is the mass inside a radius R, or where x = 1: M = m(R) = 4πρc R 3 1 1 − 3 5 = 8π ρc R3 . 15 We can find the solution in terms of solar masses by using the following information: R = 3R , R = 7 × 1010 cm, ρc = 20 g/cm3 , and M = 2 × 1033 g. Together this gives: M= 8π (20 g/cm3 ) (3 · 7 × 1010 cm)3 = 3.1 × 1035 g = 155 M . 15 (b) Average density and free-fall time Recall how averages are found: R Z ρ dV 1 R M hρi = R = 4πr02 ρ(r0 )dr0 = = V 0 V dV ∗ [email protected] 8π 3 15 ρc R 4π 3 3 R 2 = ρc = 8 g/cm3 . 5 2 The free-fall time τff is then (using G = 6.67 × 10−8 cm3 /s2 /g) 1 1 1 = 1369 s = 22.8 min ≈ τff ≈ p =q hour . 2 2 3 Ghρi −8 3 (6.67 × 10 cm /s /g) · (8 g/cm ) (c) Pressure For this we need the equation of hydrostatic equilibrium: dP Gρ(r)m(r) = −ρg = − . dr r2 Both ρ(r) and m(r) depend on the radius!!! Finally we can think of this as a simple ODE with our boundary condition P (R) = 0 and the pressure is found by integration: Z R Z r 0) r Gρ(r0 )m(r0 ) 0 dP (r 0 dr = dr . P (r) = P (r) − P (R) = P (r)R = dr0 r02 r R In the final equality I switched the limits of integration because of the minus sign in the HSE equation. The integration is easier in terms of x = r/R: Z 1 G 1 05 02 3 1 03 P (x) = ρ x − x (Rdx0 ) 1 − x 4πρ R c 2 x02 c R 3 5 x Z 4πGρ2c R2 1 = 1 − x02 5x0 − 3x03 dx0 15 x Z 4πGρ2c R2 1 = 5x0 − 8x03 + 3x05 dx0 15 x 2 2 4πGρc R 5 2 1 6 4 P (x) = 1 − x + 2x − x . 15 2 2 (d) Central pressure The central pressure is simply the pressure at r = x = 0 2 2 4πGρ2c R2 4π Pc = = (6.67×10−8 cm3 /s2 /g) 20 g/cm3 3 · 7 × 1010 cm = 9.86 × 1017 dyne/cm2 . 15 15 We may always check our answer by the OoMA estimate given in class Pc ≈ −4 GM 2 = (6.67 × 10−8 cm3 /s2 /g) (3.1 × 1035 g)2 3 · 7 × 1010 cm = 3.3 × 1018 dyne/cm2 . 4 R These two methods are the same order of magnitude (∼ 1018 dyne/cm2 ) so we know we have the right answer! (e) Total gravitational potential energy The gravitational potential energy is given by Z M Gm(r) Epot = − dm . r 0 3 However, the mass m(x) in terms of M is m(x) = M x 3 5 3 2 − x 2 2 and the mass differential dm is given in terms of M is dm = 4πr2 ρ(r)dr = 4πR3 x2 ρ(x)dx = 15 M x2 x2 − 1 dx , 2 so the total gravitational potential energy is given by Z M Gm(r) Epot = − dm r 0 Z 1 G 15 3 2 3 5 2 2 =− Mx − x M x x − 1 dx 2 2 2 0 Rx Z 45 08 GM 2 1 75 04 06 dx x − 30x + x =− R 4 4 0 = − 5 GM 2 . 7 R These quantities can be evaluated to give the energy in ergs |Epot | = 5 GM 2 5 (6.67 × 10−8 cm3 /s2 /g) (3.1 × 1035 g)2 = = 2 × 1052 erg . 7 R 7 (3 · 7 × 1010 cm) II. PARTICLE KINETIC ENERGIES From Special Relativity we know a particle’s total energy is related to its momentum p and rest mass m0 by 2 = p2 c2 + m20 c4 . If we define the kinetic energy as kin = − m0 c2 √ then using a Taylor expansion ( 1 + x ≈ 1 + x/2 if x is small) and using the defining property for NR particles that pc m0 c2 then the kinetic energy reduces to the NR version: kin = − m0 c2 q = p2 c2 + m20 c4 − m0 c2 s pc 2 2 − m0 c2 = m0 c 1 + m0 c2 " # 1 pc 2 2 ≈ m0 c 1 + − m0 c2 2 m0 c2 m0 c2 p2 c2 = 2 m20 c4 p2 1 2 = . or m0 v 2m0 2
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