Download Energy Source of the Sun

Energy Generation and the Age of the Sun
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:
fractionlost − mass =
0.048 × 10 −27 kg
= 0.007
6.693 × 10 −27 kg
which means that only 0.7% of the hydrogen mass has been converted to energy. One kilo-gram of
hydrogen then produces:
E = mc 2 = 0.007 kg × ( 3 × 10 8 m / sec) 2
E = 6.3 × 1014 Joules
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.