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Homework 7 Solutions

# 5k r 365 4 g 9m 2 r 3 7 5 10 f 8 g 365 13 12

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Unformatted text preview: bstituting this in for ρc: 2.5K R = 3.65 4πG 9M 2πR3 7 5 × 10 f (η) 8πG = 3.65 −1/3 1/2 1/2 9 2π −1/3 M −1/3 moving the R1/2 over to the left side and squaring gives: R = (3.65) = R0 25 × 107 f (η) 8πG 9 2π η2 1+η+ 1 + η R1/2 −1/3 M −1/3 where the constant R0 can be solved numerically: R0 = 3.5 × 1017 M −1/3 m kg1/3 = 2.8 × 10 1/3 M M 7 m (c) [15 pts] Using your result from part (b), show that η can be written as η = 3×10 −9 M M Tc 4/3 R R0 2 and evaluate this for (i) the Sun and (ii) a 0.1 M, 0.1 R low-­‐mass star with a core temperature of 5x106 K. Start with our definition of η evaluated in the core: ηc = 8 × 10−4 Tc 2/3 ρc substitute in the average density/core density relationship for n=1.5 polytrope: η = 8 × 10 −4 Tc −2/3 9M 2πR3 = 6.3 × 10−4 Tc M −2/3 R2 Now normalize the mass term to solar masses, and the radius term to R0, usi...
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