# C01-TOPIC- A quick look at the sun ```A quick look at the sun
REFERENCE: None
Treating the sun as a self-gravitating mass M of radius R, its selfforce is just
[Eq. 1]
F  GMM/R2.
Since pressure P = F/A and since A  R2, from Eq. 1 we get
[Eq. 2]
P  GM2/R4.
From the perfect gas law we have
[Eq. 3]
PV  NkT,
where k is Boltzmann's constant, N is the number of particles in the
volume V, and T is the absolute temperature. Since the sun is mostly
protons,
[Eq. 4]
N = M/mp,
where mp is the mass of a single proton. Since V  R3, Eq. 3 and Eq. 4
give us
[Eq. 5]
PR3  MkT/mp.
Substitution of Eq. 2 into Eq. 5 gives us
(GM2/R4)R3  MkT/mp.
which cleans up to
[Eq. 6]
T  GMmp/Rk.
Eq. 6 predicts (fairly well) the internal temperature of the sun. We
find that due to self-gravitation, the internal temperature of the sun
is
[Eq. 7]
T  2.3107 K.
Note the large value V(r) takes on as r
approaches the value ro = 110-15 m, one
femtometer (the width of a nucleus), at
which point the strong nuclear force
can cause fusion.
From the figure, we
see that if the total energy E of the
protons just equals V(ro), the protons
can just make it into the correct range
for fusion. From Eq. 8 we have
energy
In order to get two protons to fuse, the Coulomb potential wall must be
overcome.
A picture of this wall is easily gotten from the Coulomb
potential energy function:
[Eq. 8]
V(r) = e2/4or.
ro
E = V(ro) = e2/4oro = 2.310-13 J.
[Eq. 9]
Since the average kinetic energy of a typical particle in a gas is
given by E = 3kT/2, from Eq. 9 we see that the temperature necessary
for fusion is
[Eq. 10]
T  11010 K.
Comparing Eq. 10 to Eq. 7, it is clear that something is wrong with our
analysis--the self-gravitational temperature does not appear to be high
enough for fusion to occur, by a factor of 1000.
Early work on sun
models came up against the same problem. Studies showed, however, that
the temperature of Eq. 7 was about correct, and fusion was taking
place.
So, how close do the protons actually come to one another at the
temperature provided by self-gravity? Working backward, the distance r
can be found from
[Eq. 11]
E = 3kT/2 = e2/4or
to be
[Eq. 12]
r = 4.810-13 m,
short by a factor of 10 to 100.
How do we resolve this difficulty?
The solution comes about through quantum mechanics and Heisenberg's
uncertainty principle,
pr  h.
[Eq. 13]
Eq. 13 tells us the more precise one measurement is, the less precise
the other is--at least to the extent that the product is  h. To use
Eq. 13 we need to determine p: To this end, at a temperature of T, the
average kinetic energy of a proton is given by mpv2/2 = 2kT/3, so that
[Eq. 14]
p2  3mpkT.
From Eq. 14 we see that at the temperature predicted
gravitation that the average momentum of a proton is
by
self-
p = 1.210-21 kgm/s.
[Eq. 15]
r  5.510-13 m.
[Eq. 16]
From Eq. 12 we see that at the
temperature
provided
by
selfgravitation, the protons can be pushed
to within r = 4.810-13 m of each other.
But
because
of
the
uncertainty
principle, that distance can vary by r
= 5.510-13.
Thus protons can &quot;breech&quot;
the Coulomb wall and come under the
influence of the strong force, allowing
fusion to take place.
We call this
phenomenon tunneling.
energy
From Eq. 13 we have
ro
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