Homework 7 Solutions

# 3 104 tc 3 10 9 tc m m m m 2 3 m 4 3 r r0 23 r r0 2

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Unformatted text preview: ng the expression above: η = 6.3 × 10−4 Tc = 3 × 10 −9 Tc M M M M 2/3 M 4/3 R R0 −2/3 R R0 2 2 2.8 × 107 M M 1/3 2 Evaluating this for M = 0.1 M, R = 0.1 R and Tc = 5x106 K, we get R0 = 6.0x107 m and η = 0.4 – the interior of this kind of star is partially degenerate! (d) [20 pts] As a brown dwarf collapses, its core density and temperature increase, so η increases. However, at some point the core temperature stops increasing as the radius gets smaller; i.e., dTc/dR < 0 turns to dTc/dR > 0. Using the equations above for R (part b) and η (part c), find the critical value of η at which this transition occurs, and show that Tc,max = 5×10 M M 7 4/3 K [NOTE: this involves solving a transcendental equation of η, which is most easily solved with a computer] Rearranging the result from (c), and substituting our expression for R/R0 from (b), we get: Tc = 3 × 108 η M M 4/3 f (η)−2 The core temperature hits a minimum when dTc dTc dη = =0 dR dη dR = (f (η)−2 − 2ηf (η)−3 df 2 )( η) dη R0 η2 2η(1 + η) − η 2 2η 1+η+ − 2η 1 + = R0 f (η)3 1+η (1 + η)2 ignoting the trivial case where η = 0 (zero-­‐temperature gas) or f(η) = ∞ (requires η = -­‐1, which is not possible) then this derivative is equal to 0 when 2η(1 + η) − η 2 η2 = 2η 1 + 1+η+ 1+η (1 + η)2 after a healthy dose of algebraic manipulation: 2η 3 + 4...
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## This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.

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