Download ASTRONOMY 120: GALAXIES AND THE UNIVERSE HOMEWORK

ASTRONOMY 120: GALAXIES AND THE UNIVERSE
HOMEWORK # 3 SOLUTIONS
FALL 2016
1. Main Sequence Lifetime (60 Points)
Calculate the following for a main sequence star with T=10,000K, R=2.0R(sun),
M=3.0M(sun). Assume that the proton-proton chain is the only nuclear reaction. Assume that 74% of the original mass of the star is in the form of
Hydrogen, 25% in the form of Helium, and 1% in heavier elements. Assume
that only 30% of the Hydrogen is in the core and can undergo nuclear fusion.
1.a. What is the total number of Hydrogen nuclei in the core of this star? We
know that the star has mass of M = 3.0 M , where M = 2.0 · 1030 kg, and is composed
of 74% H, and only 30% of this H is in the core. We can then express the mass of the
hydrogen present in core of this star by:
MH
= 0.30 · 0.74 · 3.0 M
If the proton mass is mp , then the number of H nuclei in the core is:
MH
mp
0.30 · 0.74 · 3.0 M
=
1.67 · 10−27 kg
0.30 · 0.74 · 3.0 · 1.989 · 1030 kg
=
1.67 · 10−27 kg
56
= 8.0 · 10
n =
1.b. How much energy is available in each Hydrogen nucleus from nuclear
fusion through the proton-proton chain in the core of this star (in Joules)?
In the process of nuclear fusion through p-p chain, only 0.7% of hydrogen mass entering
the reaction gets converted to energy (derivation in class and discussion sections). Hence,
using the Einstein’s formula for equivalence of energy and mass, energy available in each
H nucleus is:
EH
= 0.007mp c2
= 0.007 · 1.67 · 10−27 kg · (3.0 · 108 m s−1 )2
= 1.05 · 10−12 J
1.c. What is the total amount of energy available for nuclear fusion through
the proton-proton chain in the core of this star (in Joules)? Simply multiply the
number of H atoms in the core (from part a) by the amount of energy released per each
H atom entering the reaction (from part b):
ET OT
= n × EH
= 8.0 · 1056 · 1.05 · 10−12 J
= 8.4 · 1044 J
Date: Fall 2016.
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ASTRONOMY 120: GALAXIES AND THE UNIVERSE HOMEWORK # 3 SOLUTIONS FALL 2016
1.d. What is the luminosity of this star. Give the answer in both solar units and
in Joule/sec. Since we know stellar temperature T and radius R, we can use relation:
L = 4πR2 σT 4
to solve for luminosity. Setting up the ratio of stellar to solar luminosity immediately gives
us the answer in solar units:
4πR2 σT 4
R∗ 2 T∗ 4
L
=
=
L
R
T
4πR2 σT4
4
10000 K
= 2.02 ·
= 35.4
5800 K
L = 35.4 L
To get the result in J s−1 , we have to multiply previous result by numerical value of solar
luminosity L = 3.9 · 1026 J s−1 :
L = 35.4 · 3.9 · 1026 J s−1
= 1.38 · 1028 J s−1
1.e. Given the luminosity of this star, what is its main sequence lifetime. i.e.,
how long can the energy supply (Hydrogen nuclei in core) last (in years)? [DO
NOT USE THE MASS-LIFETIME RELATION TO SOLVE THIS PART!].
ET OT
τms =
L∗
8.35 × 1044 [J]
=
1.36 × 1028 [J s−1 ]
= 6.12 × 1016 [s] ≈ 2 × 109 [years]
1.f. How much mass will the star lose over this main sequence lifetime through
nuclear fusion (in solar masses)? Assuming that all the mass lost from nuclear fusion
was converted to energy:
ET OT
mconverted
= mconverted c2
ET OT
=
c2
8.35 × 1044 [J]
=
(3.0 × 108 [m s−1 ])2
= 9.28 × 1027 [kg]
= 4.66 × 10−3 M
1.g. What will the elemental composition of the star be at the end of its main
sequence lifetime? i.e. what fraction of total mass of the star is in the form of
Hydrogen, Helium, and heavier elements.
Initially we have:
Hydrogen: 0.74 × (3.0 M ) = 2.22 M
Helium: 0.25 × (3.0 M ) = 0.75 M
heavy elements: 0.01 × (3.0 M ) = 0.03 M
At the end of the star’s main sequence lifetime we have:
Hydrogen: 0.74 × (3.0 M ) - 0.3 × 0.74 × (3.0 M ) = 1.554 M
Helium: 0.25 × (3.0 M ) + 0.3 × 0.74 × (3.0 M ) × 0.993 = 1.411 M
ASTRONOMY 120: GALAXIES AND THE UNIVERSE
HOMEWORK # 3 SOLUTIONS
FALL 2016 3
heavy elements: 0.01 × (3.0 M ) = 0.03 M
The total mass of the star is now M = 2.995M (the remaining 0.005M were converted
into energy), of which 51.886% (=1.554/2.995) is Hydrogen, 47.112% (=1.411/2.995) is
Helium and 1.002% (0.03/2.995) are heavy elements. Note that the use of four decimal
places was used here to illustrate a slight change in the percentage of heavy elements (the
mass of heavy elements stays the same, the mass of the star decreases).
2. Red Giant Sun (20 Points)
When the Sun becomes a red giant, its luminosity will be about 2000 times
greater than it is today. Calculate how long our Sun will be a red giant,
assuming:
• that this luminosity is caused only by the fusing of hydrogen,
• that only 40% of the Sun’s original supply of Hydrogen is fused during the red
giant phase,
• and that 74% of the original mass of the star is in the form of Hydrogen.
There are a few different ways you could conceptualize this problem, so your procedure
might be correct even if it doesn’t look exactly like this one. The length of the Sun’s red
giant phase will be dictated by two factors: how much hydrogen fuel will be available to
power it, and how quickly the sun will use up that fuel.
First, calculate the amount of hydrogen that the sun can fuse, as specified by the
problem:
(.74 M mass hydrogen to begin with) × (.40 of that burned during red giant phase) =
30
kg
=(.296 M ) × 2×10
1 M
=5.92 × 1029 kg of hydrogen will be fused during red giant phase
Next, find the total amount of energy that will be released by burning this amount of
hydrogen, given that 0.007 of the original hydrogen mass gets converted to energy:
E = 0.007 × m × c2
E = 0.007 × (5.92 × 1029 kg) × (3 × 108 (m/s))2
E = 3.73 × 1044 J
Now, calculate the rate at which the red giant will be using up the available energy. A
star’s luminosity is the amount of energy it gives off per second, so
L = 2000L
L = 2000 × (3.9 × 1026 (J/s))
L = 7.8 × 1029 (J/s)
Almost there! Divide the amount of energy available by the rate at which it’s being used
to get the hydrogen-fusing lifetime of the sun as a red giant:
3.73 × 1044 J
= 4.8 × 1014 s, or
7.8 × 1029 (J/s)
1 year
= 4.8 × 1014 s ×
= 1.5 × 107 years
3.1 × 107 s
τRG =
4
ASTRONOMY 120: GALAXIES AND THE UNIVERSE HOMEWORK # 3 SOLUTIONS FALL 2016
3. Lifetime of stars (20 Points)
The temperature and brightnesses of the stars in three star clusters are
measured and plotted in an HR diagram. One of the clusters has a main
sequence turnoff at a stellar mass of 0.7M , whereas the others have main
sequence turnoff at a stellar masses of 1.4and 2.1 M . What are the ages of
the 3 star clusters? (Use the mass-lifetime relation for stars, and the lifetime
of our Sun as a reference. Give the answers in years.)
The mass-lifetime relation goes as: τ ∝ M −2.5
And we will assume the lifetime of the Sun, τ , is 1010 years. So, using the Sun as a
reference:
τ∗
τ
Star Cluster 1 :
τ1
τ
τ1
τ
τ1
τ1
Star Cluster 2 :
τ2
τ
τ2
τ
τ2
τ2
Star Cluster 3 :
τ3
τ
τ3
τ
τ3
τ3
−2.5
M∗
M
0.7M
M
=
=
−2.5
= 0.7−2.5
= 2.4τ
= 2.4 × 1010 yrs
=
1.4M
M
−2.5
= 1.4−2.5
= 0.43τ
= 4.3 × 109 yrs
=
2.1M
M
−2.5
= 2.1−2.5
= 0.15τ
= 1.5 × 109 yrs
By assuming τ = 10Gyrs, we find that Star Cluster 1 is 24 Gyrs, Star Cluster 2 is 4.3
Gyrs, and Star Cluster 3 is 1.5 Gyrs old.