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hwk05ans
Ast 4001, October 2015
Answers for problem 3 in homework set 5 -- plus another hint for puzzle 1
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1. -- A fallacy- puzzle, no calculations are needed.
Here’s a fraudulent derivation of hydrostatic equilibrium in a star. Consider a thin shell inside it.
If subscripts 1 and 2 denote the inner and outer boundaries
of the thin shell, then the inner pressure force pushing the
shell outward is area  pressure = 4  r 12 P 1 , while
the outer pressure force pushing it inward is 4  r 22 P 2 .
The net outward force due to pressure is the difference
between 4  r 12 P 1 and 4  r 22 P 2 .
In the limit as  r = r 2  r 1 becomes infinitesimally
small, this difference is
(net pressure force) =  4  { d ( r 2 P ) / d r }  r .
Meanwhile the shell is also pulled inward by gravity. The volume of the thin shell is 4  r 2  r ,
so its mass is 4 r 2  r . (Here  is mass density as usual.) If g is the local gravitational
acceleration, then gravity exerts a radial force  g m =  4 r 2  g  r on the shell.
Summing the pressure and gravity forces, we see that static equilibrium requires
4  { d ( r 2 P ) / d r }  r + 4 r 2  g  r = 0 ,
which simplifies to
d( r2 P ) / d r
  r 2  g .
This looks credible but it’s wrong ! The correct formula is d P / d r    g . Anyone
who tries to use our erroneous result will get poor models for stellar interiors.
So: Where’s the fallacy in the above derivation ?? If you don’t see it immediately, make
a serious effort to figure it out before asking someone who already knows.
Strong hint: Imagine a classical Roman arch in a stone building or an arch bridge. What
holds the keystone up? That's the stone at the middle of the arch, with nothing directly
below it. Draw a sketch of the vector forces there. The top of a classical dome works
the same way, of course.
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4001hwk05ans - p2
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2. (Practice only; useful for problem 3.) Calculate the gas constant  for a fully ionized
gas which is 72% hydrogen and 28% helium by mass.
Answer:
  1.36  10 8 c.g.s. units.
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3. Let’s estimate the total binding energy of the Sun, based on very little information and
a few simple assumptions.
(a) Assume that the density distribution inside the Sun is  ( r ) = { 1  r / R }  0 .
(Why? Because it's the simplest formula that has  ( R ) = 0, increases
monotonically toward the center, and gives a finite value at r = 0.)
Deduce a formula for m ( x ) where x = r / R . Evaluate the central density  0
that is consistent with the Sun’s known mass M and radius R .
Solution:
m(r) =
total

4 r 2  ( r ) d r = ( 4   0 r 3 / 3 ) ( 1  3 r / 4 R ) ,
M = m(R) =
 0 R 3 / 3
 1.047  0 R 3 .
To make this consistent with the Sun's mass and radius, we need  0  5.66 g cm .
(In reality this is a serious underestimate, because the genuine Sun is much more
centrally condensed. The center of the Sun is denser than almost anyone would
expect when first looking at the problem.)
(b) Do another integral to find a formula for the total gravitational energy E G , a negative
quantity, in terms of total mass M and outer radius R . Then use the Virial Theorem
to convert this to a formula for the Sun’s total energy including both gravitational and
thermal energy. Assume that it’s a perfect monatomic gas so U = 1.5 P .
Solution: Tedious algebra, ugh...
EG
=   { G m ( r ) / r } d m =   { G m ( r ) / r } 4 r 2  ( r ) d r
= 4 G

m ( r )  ( r ) r dr .
The expressions are slightly simpler if we use dimensionless variable x = r / R .
Then  ( r ) = ( 1  x )  0 , and part (a) above told us that
m ( r ) = ( 4   0 R 3 / 3 ) ( x 3  0.75 x 4 ) .
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4001hwk05ans - p3
----- problem 3b, continued -----
Substituting r = R x and recalling part (a), we have
E G =  4 G R 2

m  x d x =  4 G R 2

16   G  02 R 5 

m  x dx
= 16 G  02 R 5
( x 3  0.75 x 4 ) ( 1  x ) x d x
=
( x 4  1.75 x 5 + 0.75 x 6 ) d x ,


with integration limits x = 0 and 1.
The dimensionless definite integral is ( 1 / 5  7 / 24 + 3 / 28 )  0.01548
so the total is E G  0.815 G  02 R 5 .
But we should express this in terms of M and R , not  0 and R . Substituting
 0 = 3 M /  R 3 from part (a) , we find
EG
= 0.743 G M 2 R .
With a perfect, non-relativistic, monatomic gas, the virial theorem says that
thermal energy =  E G / 2 , so
total energy = gravitational + thermal = E G / 2  0.371 G M 2 R .
With the Sun's mass and radius, this amounts to about  1.4 × 10 48 ergs .
(Perhaps the main relevance of this calculation is that we can use it to guess a
rudimentary Kelvin-Helmholtz timescale, roughly 10 million years with these
numbers.)
(c) Integrate the hydrostatic equation to estimate P ( 0 ), the pressure at the center of the Sun.
Let’s interpret this in terms of temperature. Work out a reasonable gas constant for the
ideal gas law, P =   T . (Problem 2 above.) Use these results to estimate the
temperature at the center of the Sun in this idealized model.
Solution:
Integrate the hydrostatic equation, remembering that P = 0 at r = R .
P 0 =  { G m ( r ) / r 2 }  (r) d r
=  4  G  02 R 2 / 3 )
 (1
[ integrated from r = 0 to R, or x = 0 to 1 ]
 0.75 x ) ( 1  x ) x d x
=  4  G  02 R 2 / 3 )  0.1042

0.436 G  02 R 2 .
As in part (b) , express this in terms of M and R by substituting  0 = 3 M /  R 3 :
P 0  0.398 G M 2 / R 4 .
For the Sun this turns out to be about 4.5 × 10 15 c.g.s. units.
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4001hwk05ans - p4
----- end of problem 3(c) -----
Recalling the central density that we estimated in part (a) , now we can find the
temperature:
 T0 = P 0 /  0  8.0 × 10 14 cm2 s .
With   1.36 × 10 8 c.g.s. units estimated in problem 2, we find
T0  6 million degrees K.
This turns out to be far too low, because the real Sun is more centrally condensed
than our simple model. But at least T0 is of the right order of magnitude, helping
to suggest the idea of nuclear reactions.
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