#3
dM/dr=4 pi r^2 rho
R dM/dr=dM/dx=4 pi R^3 x^2 rho=4 pi R^3 rho_c x^2(1-x)
M= 4 pi R^3 rho_c (x^3/3-x^4/4)  = 4 pi/3 R^3 rho_c x^3(1-3 x/4)
M(@x=1)=  pi/3 R^3 rho_c
avg rho = M(@x=1)/(4 pi/3 R^3)= (1/4) rho_c


#4 Eq 8.2-12 implies that if we have two stars with the same mass & q:

T_1 R_1/mu_1 = T_2 R_2/mu_2

we'll let 1=Sun, so T_2=4 T_1

Mu for  H: mass of 1 for 2 particles, mass/particle = mu_1 = 1/2
Mu for He: mass of 4 for 3 particles, mass/particle = mu_2 = 4/3

R_2 = (mu_2/mu_1) (T_1/T_2) R_1 = (4/3)/(1/2) (1/4) R_1 = (2/3) R_1


#5
p.184 says 4H->He mass loss = .0479e-27 kg
assume: at start 100% of Sun is H, and we convert 10% of it to He

* ? 0.1*.0479e-27*c^2*(Msun/(4*mp))
    .1279500255372699E+45

    1.3E44 J; 1 J = 1e7 erg

compare flesh (1 W/kg) to Sun:
* ? Lsun/Msun
    .1925088509816543E-03

    2e-4
    so per kg, flesh releases much more heat than the Sun