Download Solution to Assignment #7

Solution to Assignment #7
Due November 8, 2000
Problems to hand in
1) Draw the HRD together with the evolutionary track of a solar mass star. Explain what
happens at every stage, and why this is so. (Comment on the energy source, and the fashion
in which energy gets transported to the surface. Also comment on the evolution of the core
of the star. Explain how the radius, luminosity and temperature change; ands also explain
why radius, luminosity and temperature change in that particular fashion.)
The Evolution of the Sun in the Herzsprung Russel Diagram
As the sun evolves its luminosity increases, its surface temperature drops, and its radius
increases. These are all observable quantities, which can be (conveniently) plotted into the
Herzsprung Russel diagram (HRD). We refer to this evolution as “the evolutionary tracks in the
HRD”. To understand this evolution is more detail, I have marked several stages in the HRD.
Stage 1:
The sun just started nuclear fusion of H à He and arrived on the main sequence (generally this
is referred to as the Zero-Age-Main-Sequence, or just ZAMS)
Stage 1 à 2:
• As a main sequence star it will continue to burn H à He in a region where the temperature
is hot enough (T>15x106 K). This region is called the “core”.
• Energy source: pp-process (H à He; 4 protons turning into one He-ion)
à in the core the fraction of helium increases gradually
(no mixing as is the case for massive stars – see below)
à the core becomes degenerate (explained below)
• Energy transport in core: radiative diffusion
• Energy transport in envelope: radiative diffusion
Stage 2 à 3:
• now have a core of helium
• hydrogen burning to helium continues in a shell surrounding this core
Stage 3 à 4:
• more energy gets produced in the center of the star (see below)
this energy has to be carried from the center to the surface of the star
à energy transport via radiation is no longer efficient enough
à the envelope turns convective
à energy can then be transported more efficiently to the surface
à luminosity increases
Overall: Stages 1à 2 à 3 à 4: as more H à He
à He mass of core increases (to roughly 0.5 the total mass)
à core contracts slowly (central temperature increases, reaction rates increase)
à envelope expands (radius increases)
à surface temperature decreases (color gets redder)
à luminosity increases (when envelope turns convective)
à star moves to the top right hand side of the HRD
(The Sun has now become a red giant and Mercury and Venus will be part of the sun,
and maybe the Earth too. In any case we will be fried, if we have not yet died…)
At Stage 4:
• Sudden onset of Helium to Carbon fusion (Tripple Alpha process, “3α”)
• He-flash = Core explosion (invisible)
• Followed by:
Re-adjustment rather rapid
Helium core burning and Hydrogen shell burning
How did a HE-flash happen?
Core is degenerate
“electrons as close as possible”
Central temperature keeps on rising
Perfect gas law does not apply to degenerate gases
à As temperature increases, the pressure does no longer balance it out
à Temp can increase further
à reaction rates speed up until He fusion starts
à this reaction is even more energetic (energy generation rate is proportional to T40)
à faster reaction à more energy output à higher temp
à “run-away” process
à “core explosion” - “Helium Flash”
Stage 5: Horizontal branch phase
• Comparable to main sequence phase with one main difference:
• Energy Source: HeàC (3α) and HàHe
• Helium core burning
• Hydrogen shell burning
à Energy generation is a lot higher
Stage 6: A brief Variable star phase (only in certain place of HRD)
• RR Lyrae star
• Instability
over-expansion of envelope
à collapse of envelope
à too much collapse
à rebounce
à expansion
à too much expansion
à etc
à radius increases and decreases
à surface temperature increases and decreases
à luminosity increases and decrease
à Period-Luminosity Relationship
Stage 6à7: On the way to a Giant/Supergiant
• Energy Source: some C à O; He à C (3α) and H à He
• Carbon/Oxygen core
• Helium shell burning
• Hydrogen shell burning
• Stages 1à 4 repeat themselves (only faster)
à Carbon mass of core increases
à core contracts slowly (central temperature increases, reaction rates increase)
à envelope expands (radius increases)
à surface temperature decreases (color gets redder)
à luminosity increases
à star moves to the top right hand side of the HRD
à Red Super Giant
(The Sun has now become a red giant and Mercury and Venus will be part of the
sun, and maybe the Earth too. In any case we will be fried, if we have not yet
died…)
Stage 7à8
• as the stat grows, mass loss happens
outer shells get dispersed
He burning shell gets exposed to the surface
get tripple alpha reactions on the surface
since these are uneven, have several short flashes
these are rather explosive
à material is expelled in shells
• see a planetary nebula
• but the central star appears to be hotter
à hotter means a bluer color
à star moves towards blue in HRD
Stage 9 and beyond
• planetary nebula disappears
• star does no longer undergo any nuclear fusion
• star contracts
• star cools (the light emitted is thermal radiation)
à smaller, dimmer, redder means that the star moves towards the bottom right
in the HRD
• star now is a white dwarf
• eventually it will become an even dimmer red dwarf
2) Look at figure 13-6 in your book. Comment on the evolutionary states of Betelgeuse, Rigel,
the Sun and Capella. Draw the same drawing, and then insert the evolutionary track of the
Sun.
Betelgeuse:
Massive Star, in Red Supergiant phase, probably burning he in its core, but it could also be in a
later stage that is burning heavier elements; it may go supernova soon.
Rigel:
Massive star that is either on its way evolving off the main sequence (ie with a hydrogen burning
shell, and a Helium core, or, it may have been past that stage and is in one of its loops of starting
to burn another heavier element.
Optional Problem
3) What is the energy source of the sun? The Sun’s mass, radius and luminosity are: R¤= 7
× 108 m, M¤= 2 × 1030 kg, L¤= 4 × 1026 W. (Hints: next page…)
a) Chemical Burning: The energy given off in coal burning, a typical chemical reaction, is
about 5 × 106 J/kg (that is 5 × 106 Joules per 1 kg of coal). If chemical burning were
the energy source in the Sun, calculate how much energy would be available. Then
calculate how long (in years) the sun would shine at its present luminosity? (Recall the
luminosity is defined as the energy emitted PER SECOND and is measured in Watts or
in Joules per second.)
b) The sun will burn roughly 10% of its entire hydrogen to helium. During this process
0.7% of that mass will get transformed into energy. If the solar energy is supplied via
nuclear burning, calculate how long (in years) the sun would shine at its present
luminosity.
Answer:
Energy Generation and the Age of the Sun
A rather detailed approach
(also answers the question of whether the energy source could be gravitational
energy)
Stars shine – we know that. But they can only shine until all their energy is used up. So
where do they get their energy from? And how long will stars shine? The lifetime of a star
depends on the amount of fuel divided by the consumption rate. (Analogy: if your car has a full
15-gallon gas tank and it consumes 2 gallons/hour on the highway, then your car can travel for
15 gallons/(2 gallons/hour) = 7.5 hours.) The fuel of the sun corresponds to the total amount of
energy that is available, and the consumption rate is how much energy per second it emits. The
energy per second that a star emits is called Luminosity. Thus the lifetime of a star is the energy
available divided by the luminosity; or in terms of a formula:
t age −of − sun =
E
L
Why is this important? We know the luminosity of the sun, and if we could somehow figure
out how much energy is available, we could calculate the sun’s age.
Is the energy source of the sun coal?
It sounds funny, but that’s what people used to believe. We can test this. We can measure
how much energy is produced when burning coal. It turns out that one kilo-gram of coal
produces 5 ×106 Joules. How much energy would the entire sun produce if it was made of
coal? The sun’s total mass is 2 ×1030 kg, so the energy available is:
E = 5 × 10 6
J
× 2 × 10 30 kg = 1037 J
kg
How long does this last if the sun emits energy at the present rate? The Luminosity of the sun is
4 × 1026 W (1Watt = 1 Joule/sec).
t age −of − sun =
E
=
L
10 37 J
4 × 10 26
J
sec
= 2.5 × 1010 sec =
2.5 × 1010 sec
= 800 years
sec
7
3 × 10
yr
So, if the sun received its energy through burning coal, it would radiate for 800 years.
Clearly this is a ridiculously short time-scale. In fact an energy source is needed that provides
energy for roughly 10,000,000 times as long a time.
Is the energy source of the sun gravitational energy?
In the mid 1800’s Kelvin & Helmholz came up with another suggestion. What if the sun is
collapsing, and what if gravitational energy gets transformed into light energy? Again, this can be
calculated. The gravitational energy of a spherical self gravitating body can be shown (using
calculus) to be:
E grav
3 GM 2
=−
5 R
where M is the total mass of all the particles and R the radius of the sphere.
How much of that energy can be transformed into light? As the star collapses, it turns out
that one half of the gravitational energy goes into heating the star, while the other half is radiated
away. This is known as the “Viral Theorem”. In terms of a formula this is:
1
E light = − E grav
2
Thus the total energy reservoir that is available to be radiated away during the lifetime of the sun
would then be:
E grav
3 GM 2
=
10 R
Again, the age of the sun would be given by:
t age −of − sun
3 GM 2
E 10 R
3 GM 2
= =
=
L
L
10 RL
So, if we know the mass, radius and the luminosity of the sun, we ought to be able to calculate
for how long the sun would radiate at its present rate, if it obtained all of its energy from
gravitational energy. Inserting R¤= 7 × 108 m, M¤= 2 × 1030 kg, L¤= 4 × 1026 W or 4 × 1026
kg×m2/sec3 (since 1Watt = 1 kg×m2/sec3) and G = 6.67 × 10-11 m3/(kg×sec2):
t=
3 GM 2
3
=
10 RL
10
m3
2
30
2 × ( 2 × 10 kg )
kg × sec
= 9.5 × 10 14 sec = 3 × 10 7 years
2
kg
×
m
7 × 10 8 m × 4 × 10 26
sec3
6.67 × 10 −11
Is this age reasonable and can life on earth have formed in that time-span? Well, the
Neandertaler’s existed already a million years ago, and fossils are much older than that, implying
that the sun must have existed for a much longer time. Kelvin & Helmholz already knew that in
the mid 1800’s, but no other energy source was known. This puzzle was resolved in the 1930’s
with the discovery of the hydrogen bomb.
However, while gravity is not the main energy source of the sun, gravitation is indeed an
important energy source. Proto-stars receive all their energy from gravitation!! And proto-stellar
lifetimes are much shorter than the main sequence lifetimes.
Is the energy source of the sun nuclear fusion?
We now believe that the energy source of the sun is nuclear fusion of hydrogen to helium.
This procedure requires four hydrogen atoms. It turns out that four hydrogen atoms have more
mass than one helium atom. So where did the extra mass go? The mass somehow got converted
into energy. A few years prior to this discovery, Einstein had already shown this - at least in
theory. His discovery can be summarized as:
E = mc 2
This means that Energy and Mass are “different faces of the same thing” (c is the velocity of
light, and thus a constant), and that one can be transformed into the other.
If “hydrogen to helium burning” is really what is happening in the sun, we ought to be able to
calculate the total amount of energy that the sun could produce. Since we know the mass of the
hydrogen and the helium atoms, we can calculate how much mass has been lost, and how much
energy was produced during this process. The mass of one hydrogen atom is 1.673 × 10-27 kg
and that of one helium atom is 6.645 × 10-27 kg.
4 × mhydrogen = 6.693 × 10 −27 kg
1 × mhelium = 6.645 × 10− 27 kg
Difference = 0.048 × 10 − 27 kg
Thus the energy is:
E = mc2 = 0.048 × 10 −27 kg × ( 3 × 108 m / sec) 2
E = 4.3 × 10 −12 Joules
The fraction of the hydrogen mass that is lost is:
E = 6.3 × 1014 Joules
0.048 × 10 kg
= 0.007
6.693 × 10 −27 kg
−27
fractionlost − mass =
E = mc 2 = 0.007 kg × ( 3 × 10 8 m / sec) 2
which means that only 0.7% of the hydrogen mass has been converted to energy. One kilogram of hydrogen then produces:
How much energy is that? As a reference, to get the same amount of energy, you would have to
burn 20,000 tons of coal; and 20,000 tons corresponds to roughly 4×1011kg of coal. This is
quite a lot of coal! So you can do quite a lot of damage with only one kilo-gram of hydrogen…
Finally, if the sun’s energy source was nuclear burning, how much of an energy reservoir
would it have? Well, you can calculate that. One kilogram produces 6.3 × 1014 Joules, so the
whole sun would produce
E = 6.3 × 1014
J
× 2 × 10 30 kg = 13
. × 10 45 J
kg
However, the sun has only 75% of hydrogen gas, so the total energy available would be:
E = 13
. × 10 45 J × 0.75 = 9.5 × 10 44 J
If a maximum of 9.5 × 1044 Joules are available and the sun emits energy at a constant rate of L
= 4 ×1026 W (or 4 × 1026 J/sec) its life-time would be:
t age −of − sun =
E
9.5 × 1044 J
=
= 2.4 × 1018 sec = 7.5 × 1010 years
J
L
4 × 1026
sec
This time-scale is certainly long enough. Our estimates show that the sun is probably about 5
billion years old at the present time and will probably last for another 5 billion years on the main
sequence. During its main-sequence lifetime it will convert roughly 10% of its hydrogen into
helium. Its initial main sequence luminosity was actually 30% less than its present one, so
compensating for both effects, one gets an estimate of 10 billion years for the main sequence
lifetime of the sun. This sounds about right.