Astro 160: The Physics of Stars

and finally the temperature can be found by usingP = nk b T = 2ρm pk b T ⇒ T(r) = m pP2ρk band substituting the P and ρ from the previous expressions we findT(r) = m pK 3/52k b[ [ 2 15 K−3/5 GMR − 1 ] ]+ Pc2/5R cb) In detailed solar models, the pressure at the base of the convection zone is ≈ 5.2 × 10 13 dyne/cm 2and the density is ρ ≈ 0.175 g cm −3 . Using your solution from a), estimate the radius of the base of theconvection zone R c . Compare this to the correct answer of R c ≈ 0.71R sunIf we solve the density equation for R c we findand plugging in values we find that1= 1 [R c R −1R c= 1.998 × 10 −9 m −1(ρK 3/2 ) 2/3 − P 2/5c]· 5K3/52GM⇒ R c ≈ 5.11 × 10 8 m = 0.72R sunc) In your model, what is the temperature of the sun at 0.99R sun , 0.9R sun , and at the base of the solarconvection zone. This gives you a good sense of how quickly the temperature rises from its surface valueof ≈ 5800 K as one enters the interior of the sun.To find the temperature as a function of radius we would use the temperature equation derived frompart (a). i.eT(r = 0.99R sun ) = m pK 3/5 [− 2 ]252k b 3 K−3/5 GM r + Pc2/5 ≈ 4.1 × 10 4 K66 · R sunT(r = 0.90R sun ) = m pK 3/5 [− 2 ]152k b 3 K−3/5 GM r + Pc2/5 ≈ 5.1 × 10 5 K18 · R sunT(r = 0.72R sun ) = m pK 3/5Pc2/5 ≈ 1.8 × 10 6 K2k bProblem set 4Problem # 1I mentioned in class that there are two ways to estimate the energy carried by convection. The first isthat the energy flux is F c ≈ 1/2ρv 3 c ≡ F c,1 where v c is the characteristic velocity of the convective motions.12