C05-TOPIC- Stellar model 2 - Linear density

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Stellar model 2: Linear density model
REFERENCE: Bowers and Deeming
The next simplest form a star of mass M and radius R could take is that
of a linear density star in hydrostatic equilibrium. To analyze such a
model we need the mass-continuity equation, and the equation of
hydrostatic equilibrium, given here, along with a reasonable boundary
condition:
dm
[Eq. 1]
= (r)4r2,
dr
dP = - Gm(r)(r) .
r2
dr
[Eq. 2]
[Eq. 3]
P(R) = 0,
For our linear density model we have
[Eq. 4]
(r) = c 1 – r .
R
Note that Eq. 4 tells us that (0) = c and (R) = 0.
becomes
r
dm = 4c 1 –
r2 dr.
R
Eq. 1 then
[Eq. 5]
Integrating Eq. 5 from 0 to an arbitrary value of r we obtain
r
r3
m(r) =  4c r2 –
dr
R
0
m(r) = 4c
r3
r4
–
.
3
4R
[Eq. 6]
Eq. 6 tells us how much of the star's total mass is enclosed in a
sphere of radius r  R. Recall that this was just the mass that caused
a gravitational attraction toward the center of the star, from Newton's
shell theorem. In interesting use of Eq. 6 is to find the relationship
between the central density c and the average density <>:
We know
that the total mass is given by
M = m(R) = 13 cR3.
But it is also given by
4
M = 3 <>R3,
so that
[Eq. 7]
c = 4<>,
The linear density model predicts four times the central density as the
constant density model.
Now we can solve Eq. 2, the equation of
hydrostatic equilibrium. We will integrate from an arbitrary r to the
surface, so we can apply our boundary condition Eq. 3.
R
r3
r4
P(R) - P(r) = -G4c
–
3
4R
r
R
r2
r
0 - P(r)= -G4c2
–
3
4R
r
c 1 – 1 dr2
R
1 –
1
R
r
dr
P(r) = G4c2
R2
R3
R3 .
R4 .
–
–
+
6
9R
12R
16R2
r2
r3
r3 .
r4 .
+
6
9R
12R
16R2
5R2
r2
r3
r3 .
r4 .
144 - 6 + 9R + 12R - 16R2 .
P(r) = G4c2
[Eq. 8]
At the center of the star the pressure P(0)  Pc will be the highest,
and Eq. 8 can be written
5
[Eq. 9]
Pc = 36 Gc2R2.
Eq. 9 can be rewritten in terms of Pc:
P(r) = Gc2
5R2
2r2
7r3
r4
+
- 2
36
3
9R
4R
5
24r2
28r3
9r4
= 36 Gc2R2 1 +
5R2
5R3
5R4
= Pc 1 where
24r2
28r3
9r4
+
5R2
5R3
5R4
,
[Eq. 10]
5
Pc = 36
Gc2R2.
Note that even with a simple linear model the formulas for m,  and P
become complicated. For our purposes, the constant density model will
be used, but this example shows you how “corrective” numerical
coefficients arise from more rigorous analysis.
We will not do this
analysis (although we could).
Instead, we will freely borrow
coefficients determined by other people’s work when needed.
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