### AST 353 Astrophysics — Problem Set 1

```AST 353 Astrophysics — Problem Set 1
Prof. Volker Bromm — TA: Aaron Smith∗
I.
SIMPLE STELLAR MODEL
Assume a star has a radius of R = 3R and a quadratic density profile:
r 2 = ρc 1 − x2 .
ρ(r) = ρc 1 −
R
Here, ρc = 20 g/cm3 is the central density. We have defined a new variable x = r/R to make future
integrals easier, so the differential is dr = Rdx.
(a) Mass
To find the total mass M of the star we first find the mass coordinate m(r), which is the mass
interior to a shell of radius r:
Z r
m(r) ≡ 4π
r02 ρ(r0 )dr0
Z0 x
= 4π
(Rx0 )2 ρ(x0 )(Rdx0 )
0
Z x
3
= 4πρc R
x02 − x04 dx0
0
1 5
3 1 3
m(x) = 4πρc R
x − x
.
3
5
The total mass M is the mass inside a radius R, or where x = 1:
M = m(R) = 4πρc R
3
1 1
−
3 5
=
8π
ρc R3 .
15
We can find the solution in terms of solar masses by using the following information: R = 3R ,
R = 7 × 1010 cm, ρc = 20 g/cm3 , and M = 2 × 1033 g. Together this gives:
M=
8π
(20 g/cm3 ) (3 · 7 × 1010 cm)3 = 3.1 × 1035 g = 155 M .
15
(b) Average density and free-fall time
Recall how averages are found:
R
Z
ρ dV
1 R
M
hρi = R
=
4πr02 ρ(r0 )dr0 =
=
V 0
V
dV
∗
[email protected]
8π
3
15 ρc R
4π 3
3 R
2
= ρc = 8 g/cm3 .
5
2
The free-fall time τff is then (using G = 6.67 × 10−8 cm3 /s2 /g)
1
1
1
= 1369 s = 22.8 min ≈
τff ≈ p
=q
hour .
2
2
3
Ghρi
−8
3
(6.67 × 10 cm /s /g) · (8 g/cm )
(c) Pressure
For this we need the equation of hydrostatic equilibrium:
dP
Gρ(r)m(r)
= −ρg = −
.
dr
r2
Both ρ(r) and m(r) depend on the radius!!! Finally we can think of this as a simple ODE with
our boundary condition P (R) = 0 and the pressure is found by integration:
Z R
Z r
0)
r
Gρ(r0 )m(r0 ) 0
dP
(r
0
dr
=
dr .
P (r) = P (r) − P (R) = P (r)R =
dr0
r02
r
R
In the final equality I switched the limits of integration because of the minus sign in the HSE
equation. The integration is easier in terms of x = r/R:
Z 1
G
1 05
02
3 1 03
P (x) =
ρ
x
−
x
(Rdx0 )
1
−
x
4πρ
R
c
2 x02 c
R
3
5
x
Z
4πGρ2c R2 1
=
1 − x02 5x0 − 3x03 dx0
15
x
Z
4πGρ2c R2 1
=
5x0 − 8x03 + 3x05 dx0
15
x
2
2
4πGρc R
5 2
1 6
4
P (x) =
1 − x + 2x − x
.
15
2
2
(d) Central pressure
The central pressure is simply the pressure at r = x = 0
2
2
4πGρ2c R2
4π
Pc =
=
(6.67×10−8 cm3 /s2 /g) 20 g/cm3
3 · 7 × 1010 cm = 9.86 × 1017 dyne/cm2 .
15
15
We may always check our answer by the OoMA estimate given in class
Pc ≈
−4
GM 2
= (6.67 × 10−8 cm3 /s2 /g) (3.1 × 1035 g)2 3 · 7 × 1010 cm
= 3.3 × 1018 dyne/cm2 .
4
R
These two methods are the same order of magnitude (∼ 1018 dyne/cm2 ) so we know we have the
(e) Total gravitational potential energy
The gravitational potential energy is given by
Z M
Gm(r)
Epot = −
dm .
r
0
3
However, the mass m(x) in terms of M is
m(x) = M x
3
5 3 2
− x
2 2
and the mass differential dm is given in terms of M is
dm = 4πr2 ρ(r)dr = 4πR3 x2 ρ(x)dx =
15
M x2 x2 − 1 dx ,
2
so the total gravitational potential energy is given by
Z M
Gm(r)
Epot = −
dm
r
0
Z 1
G
15
3 2
3 5
2
2
=−
Mx
− x
M x x − 1 dx
2 2
2
0 Rx
Z 45 08
GM 2 1 75 04
06
dx
x − 30x + x
=−
R
4
4
0
= −
5 GM 2
.
7 R
These quantities can be evaluated to give the energy in ergs
|Epot | =
5 GM 2
5 (6.67 × 10−8 cm3 /s2 /g) (3.1 × 1035 g)2
=
= 2 × 1052 erg .
7 R
7
(3 · 7 × 1010 cm)
II.
PARTICLE KINETIC ENERGIES
From Special Relativity we know a particle’s total energy is related to its momentum p and
rest mass m0 by
2 = p2 c2 + m20 c4 .
If we define the kinetic energy as
kin = − m0 c2
√
then using a Taylor expansion ( 1 + x ≈ 1 + x/2 if x is small) and using the defining property for
NR particles that pc m0 c2 then the kinetic energy reduces to the NR version:
kin = − m0 c2
q
= p2 c2 + m20 c4 − m0 c2
s
pc 2
2
− m0 c2
= m0 c 1 +
m0 c2
"
#
1
pc 2
2
≈ m0 c 1 +
− m0 c2
2 m0 c2
m0 c2 p2 c2
=
2
m20 c4
p2
1
2
=
.
or m0 v
2m0
2
```