Sterren: Werkcollege 4
Opgaven: 1, 2, 3 (Handout: Basic concepts in stellar physics)
1 Calculate Egr , hP i and P (r) for two different density distributions.
Case (A) the density is independent of the distance r from the centre ρ(r) = ρ = constant. We will first
calculate the depth dependent mass distribution;
Z r
4
2
ρ(r)4πr dr =
m(r) =
πr3 ρ.
3
0
It follows, that;
M
Z
Gm(r)
dm
r
0
Z R
Gm(r)ρ(r)4πr2
= −
dr
r
0 16
= −
π 2 Gρ2 R5
15
3
GM 2
= −
5
R
Egr = −
−Egr
3V 4
=
πGρ2 R2
15
3GM 2
=
20πR4
hP i =
Z
r
Gm(r)ρ(r)
dr
r2
0
Z r
4
2
= −
πGρ
rdr
3
0
2πGρ2 r2
+C
= −
3
With C an integration constant (boundary condition). This constant can be evaluated using P (r =
R) = 0. We find C = 2πGρ2 R2 /3. It then follows that;
2πGρ2
P (r) =
R2 − r 2
3
r 2 3GM 2
=
1−
8πR4
R
P (r) = −
1
Werkcollege Sterren
2
In the last line we made use of ρ = 3M/4πR3 .
Note that the gravitational energy is proportional to GM 2 /R and the the pressure is proportional to
(GM 2 /R)/V ∝ GM 2 /R4 .
Case (B) The density increases towards the centre as ρ(r) = ρc (1 − r/R) We will again first calculate
the depth dependent mass distribution;
Z r
3r
4
2
3
ρ(r)4πr dr = πρc r 1 −
.
m(r) =
3
4R
0
It then follows, that;
Egr =
=
=
=
R
Gm(r)ρ(r)4πr2
−
dr
r
0
Z
3r 16 2 2 R 4
r
− π Gρc
r 1−
1−
dr
3
4R
R
0
16 2 2 5 1 1 1
3
− π Gρc R
− − +
3
5 6 8 28
26 2 2 5
−
π Gρc R
315
Z
Egr
3V
13
=
πGρ2c R2
630
hP i = −
Z
r
Gm(r)ρ(r)
dr
r2
0
Z r 4
r
3r
2
= − πGρc
r 1−
1−
dr
3
R
4R
0
1 2
r3
1 r3
3r4
4
2
= − πGρc
r −
−
+
+C
3
2
4R 3 R 16R2
1
4
7 r
3 r2
2 2
= − πGρc r
−
+
+C
3
2 12 R 16 R2
P (r) = −
5
R2 . It then follows that;
C can again be evaluated using P (r = R) = 0. We find C = 34 πGρ2c 48
5 2
4
1
7 r
3 r2
2
2
P (r) = πGρc
R −r
−
+
3
48
2 12 R 16 R2
2 a. Determine the total energy radiated by the sun during its contraction (Kelvin-Helmholtz) phase, and
determine the time it takes the sun to reach the main sequence (assume that the luminosity of the sun
during this phase is constant and equal to its current value). We will assume that during this phase we
have ρ = hρi for the sun. From exercise 1 it follows that Egr = −(3/5)GM 2 /R. Invoking the virial
theorem one finds;
Egr
3GM 2
Etot =
=−
2
10R
Werkcollege Sterren
3
The energy radiated during the contraction phase, ∆E, then is (Ri Rf = R and Mi = Mf = M );
∆E = −(Etot,f − Etot,i )
≈ −Etot,f
3GM 2
= −
10Rf
3GM2
= −
10R
= 1.14 × 1041 [J].
I.e., half of the energy liberated by the contraction is radiated away, the rest is converted into heat.
The contraction phase to reach the main sequence the lasts for at time, t = tKH ;
∆E
L
= 3.0 × 1014 [s]
= 9.4 × 106 [yr].
tKH =
b. As the sun is contracting toward the main sequence it will lose energy because the gravitational energy
decreases (becomes more negative). From the virial theorem we know that if the contraction proceeds
quasi-statically, and the sun remains near hydrostatic equilibrium, the kinetic energy must increase in
order to maintain equilibrium (2Ekin + Egr = 0). Hence the sun heats up even while loosing energy.
Another way to see this is to note that the binding energy Ebind = −Etot of the sun goes up as it
contracts which implies that Ekin = Ebind must go up too. The latter relation holds in hydrostatic
equilibrium.
3 a. Estimate the age of the Pleiades cluster from the main sequence lifetime of a 6M star. We will
assume that the luminosity of the star obeys L ∝ M 3 . The luminosity of a 6M star is then;
L6M =
6M
M
3
× L = 216L
This means that the 6M star burns its fuel 63 = 216 times as fast as the sun, however it also contains
6 times more fuel (i.e. mass) as compared to sun. Thus the (main sequence) lifetime of the 6M star
will scale with 62 as compared to the (main sequence) lifetime of the sun. In terms of mass we find
that the lifetime t ∝ M −2 . If we assume that the main sequence lifetime of the sun is tMS, = 1×1010
[yr], it then follows that;
−2
6M
× tMS,
=
M
= 6−2 × tMS,
= 2.8 × 108 [yr]
tM S,6M
b. In the estimate above we have assumed that all hydrogen in the star is burned on the main sequence
and that the star consists of hydrogen only. In reality the star starts out with about 70% hydrogen and
will not burn all of it before the nuclear fusion of H to He stops in the core of the star. Hence the age
estimated above will be too large.