Phys 132 Homework 3 Solutions
Professor Crystal Martin
TA: Ellie Hadjiyska
Problem 3: We are asked to find the total energy radiated during Kelvin-Helmholtz contraction
as the Sun evolves onto the main sequence from a temperature of 30,000 K to 6 ×106K and it
remains close to hydrostatic equilibrium. The total energy radiated is
Erad =Etotal(30000 K) −Etotal(6×106K)
Applying the virial theorem, we have that
Etotal =−KE =K E +EGR ⇒KE =−1
2EGR (1)
Since the kinetic energy per particle is 3
2kT , then the kinetic energy within the spherical cloud is
KE =3
2NkT
where the number of particles N=M¯/¯m(since for every Hydrogen atom that ionized we get an
ion and an electron, ¯m= 0.5mH), M¯is the mass of the cloud (mass of the Sun), and mHis mass
of Hydrogen atom. The total energy from Eq. (1) is therefore (Phillips Eq. (1.21))
Etotal =−KE =−3
2
M¯
¯mkT =−3
2·M¯
0.5mH¸kT =−3M¯
mH
kT
This gives us the radiated energy of the cloud during the collapse as
Erad = 3M¯
mH
k[T6×106K−T30000 K]
Erad = 3 1.99 ×1033 g
1.67 ×10−24 g(1.38 ×10−16 ergs K−1)[6 ×106K−30000 K] = 2.95 ×1048 ergs
The Sun’s luminosity is L¯= 3.827 ×1033 ergs/s. To find the time it takes to reach the main
sequence, we assume the luminosity stays constant over the lifetime of the contraction and get
tcontraction =Erad
L¯
=2.95 ×1048 ergs
3.827 ×1033 ergs s−1= 7.70 ×1014 s = 24 Myr