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Homework 4
Astro 111 - Order of Magnitude: Fall 2016
1) Supernova Explosions. Type Ia supernovae are believed to be the thermonuclear explosions
of accreting white dwarfs that have reached the Chandrasekhar limit.
A). Mean pressure inside a white dwarf of mass M and radius R.
To maintain hydrostatic equilibrium, the total outward pressure must balance gravitational pressure, which is
Fg
Eg
3GM 2
GM 2
=
≈
=
(1)
P =
V
4πR4
4πR4
A
Illustrating that using the energy density is roughly equivalent to finding the gravitational force
per unit area.
p
B). Estimate, the speed of sound, vs ∼ P/ρ. Within how much time does the flame traverse the
radius of the white dwarf, assuming R = 104 km, M = 1.4M ? Show that this sound crossing time
is comparable to the free-fall timescale.
For the speed of sound:
vs ≈
3GM 2
4πR4
3M
4πR3
!1/2
=
GM
R
1/2
(2)
The sound crossing time:
3 1/2
r
R
t≈ =
(3)
v
GM
√
which is equivalent to the free-fall timescale: vf f = 1/ Gρ. Plugging in for R and M yields, t ≈ 2
s.
C). Calculate the total energy output of the explosion, assuming each atom of the white dwarf is
converted from carbon to nickel. Compare this energy to the gravitational binding energy of the
white dwarf, to demonstrate that the white dwarf explodes completely, without leaving any remnant.
There was potential for confusion with the wording of this question, since each individual atom
is not converted from carbon to nickel (this would give you a huge ∆m, and the number of nucleons
would increase a lot).
The number of nucleons in the WD is
1.4 · 2 · 1033 g
MW D
=
≈ 1057 nucleons.
(4)
Nnucleons =
mnucleon
1.66 · 10−24 g
The energy per nucleon of carbon is the mass-energy of carbon divided by the number of nucleons
in carbon. Same for nickel. The amount of energy released per nucleon going from carbon to nickel
is the difference in these energies per nucleon:
m
∆mc2
mN i 2
C
=
−
c ≈ 7 × 10−6 ergs
(5)
nucleon
12
59
Then the total energy released in the explosion is
Nnucleons ·
∆mc2
≈ 1052 ergs
nucleon
(6)
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
The gravitational binding energy of the WD is
U≈
GM 2
≈ 5 × 1050 ergs,
R
(7)
so there is no remnant.
2) Binary Stars and Supernovae.
A). Starting with two stars (M1 & M2 ) in a circular orbit, find the maximum mass that can be lost
by one of the stars for the system to remain bound.
We can do this problem two ways: (1) using the CM frame, and (2) using the virial theorem.
The first involves more algebra.
First way:
The limiting case (between bound and unbound) is simply when the total energy of the binary
system, in the C.O.M. frame after the super-nova, is zero.
Figure 1.
Consider the parameters as indicated in Fig. 1, where after the supernova
M1 → M10 = M1 − ∆M . Because the supernova was instantaneous and isotropic, neither velocity
is altered in the explosion. The C.O.M. however, will obtain a velocity! The total energy, in the
new C.O.M. frame will be
1
M 0 M2
Ecom = (M10 v12 + M2 v22 ) − Tcom − G 1
2
r
(8)
To find the kinetic energy of the C.O.M. (Tcom ), we must find its velocity (vcom ). The COM
velocity is vcom ≡ (M10 v1 + M2 v2 )/(M10 + M2 ), where from the pre-initial binary we know that
v1 = −M2 v2 /M1 , and from our definitions, M10 = M1 − ∆M . With a little algebra, we can find
that
∆M M2
vcom = v2
(9)
M1 (M1 + M2 − ∆M )
Note that if ∆M = M1 (i.e. the entire star is destroyed), the center of mass velocity is simply that
of the second star (v2 ), as it must.
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
The rest is just plugging in and doing some ugly algebra, but the end result is that a final energy
Ecom = 0 requires
1
∆M = (M1 + M2 )
(10)
2
If more than half the mass of the binary is lost in the supernova, the system is inevitably unbound.
Second way:
Recall that the virial theorem tells us
Etot = −U/2.
(11)
Consider a general mass distribution (not necessarily a binary). If this system is bound, it will
have
GMi2 1
−
+ Mi vi2 < 0.
(12)
Ri
2
Now consider our system, with M1 and M2 . Say M1 loses the mass. Initially, M2 is moving at
vkeplerian at the initial orbit radius. After M1 loses mass, M2 will be going too fast and so overshoots
into a larger orbit. To be unbound, we want vkeplerian = vesc for this system, so that the final orbit
radius is infinity. Equating the energy after the explosion (LHS) to the virial energy of the unbound
system (RHS), we have
GMf2
GMf2 1
GMi
+ Mf
=−
,
(13)
−
Ri
2
Ri
2Rf
where we assume that the radial separation does not change instantaneously when the mass does.
When the system is unbound, Rf → ∞, so the RHS is 0. Solving gives Mf = Mi /2. So if
1
1
∆M > Mi = (M1 + M2 )
2
2
(14)
then the system is unbound.
B). 1% anisotropy in a single direction, how much velocity could this impart upon the star? (energy
released 1051 ergs.)
1049 ergs is thus bestowed upon the star—lets say a 1M? neutron star.
r
r
2∆E
1049
v=
≈
≈ 108 cm s−1 = 1000 km s−1
m
1033
This is consistent with observations of pulsars and supernova remnants.
(15)
C). Conceptually, what effects would it have on the binary?
If the velocity kick adds (vectorally) with the orbital motion, it will aid in unbinding the system.
If the kick subtracts from the orbital motion, the binary could tighten (one rare possibility is a
Thorne-Zytkow object). Note, though, that even if the kick and orbit vectors are in opposite
directions, the binary can loosen if the kick velocity is more than twice the orbit velocity. What
really matters for whether the binary remains bound/unbound is whether the total energy is still
negative after the kick.
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
3) Neutron Star Spin-Up. Consider a Neutron Star of mass M = 1.4M , radius R = 104
m, and period of 1 s. The neutron star is accreting mass through an accretion disk, at a rate of
Ṁ = 10−9 M yr−1 .
A). Derive a differential equation for Ṗ , the rate at which the neutron star period decreases.
The angular momentum of a neutron star with period P, will be: Lns = Iω = 2πI
P ; where we
have assumed that the moment of inertia is constant due to the slow rate of mass accretion. Rearranging and taking the derivative, we find for the change in period: Ṗ = −2πIL−2
ns L̇. Conservation
of momentum forces the change in momentum of the NS (Lns ) to be the same as that lost by each
particle (Lp ) accreted: L̇p = L̇ns . Theqvelocity of a particle, infinitesimally above the surface of
the NS, in a circular orbit will be v = GM
R . Thus the momentum of an accreted particle will be
given by
√
(16)
Lp = mvR = m GM R
and therefore
√
L̇p = ṁ GM R = L̇ns
(17)
In equation 9, it has additionally been assumed that the acceleration of the keplerian circular
velocity due to increasing neutron star mass, is negligible—again, due to the low accretion rate.
Plugging this in the for change in period of the neutron star yields:
√
ṁ GM R 2
2πI
P ≡ αP 2
(18)
Ṗns = 2 L̇ns =
Lns
2πI
B). What is the maximum spin rate of such a neutron star?
The maximum spin rate occurs when the centrifugal force on a particle at the stars edge matches
the force of gravity. At this point, in effect, the particle enters orbit around the star. By equating
forces, we see that at this point
r
mv 2
GM m
GM
=
→v=
(19)
R
R2
R
Or using P = 2πR/v
r
R3
2π
Pc = 2π
=√
≈ 6 × 10−4 s
(20)
GM
Gρ
C). Solve the equation from part ‘a’ to find how long it takes to reach the break-up velocity (part ‘b’).
We can rearrange equation 10 to be
dp
(21)
αP 2
≈ 3×10−12 . This can be integrated from initial period (P0 ) to the final (critial)
dt =
where α ≡ ṁ
period (Pc ):
√
GM R
2πI
1
t=
α
1
1
−
P0 Pc
≈ 5 × 1014 s ≈ 107 yrs
(22)
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
Multiplying this resultant time with the mass accretion rate shows that the neutron-star will only
increase its mass by about 1%—validating our previous assumptions.
4) Statistics and Stellar Remnants. ≈ 1011 stars, initial mass function dN/dM ∝ M −2.35 in
the range M ∈ 0.4 − 100M .
A). Find the fraction of stars formed with M > 8M . How many NS and BHs should there be in
the galaxy, with roughly how much mass?
Given the initial mass distribution function, dN ∝ M −2.35 dM , the fraction of stars with mass
greater than 8 solar masses is simply:
R 100 −2.35
M
dM
0.043
≈ 0.017
(23)
f8 = R8100
≈
−2.35
2.56
M
dM
0.4
where the proportionality constants (or normalization constants) have been canceled by taking
the ratio [this is equivalent to multiplying the numerator by a constant which is equal to k =
1/(the integral over all masses) ≈ 0.39].
The total number of stars above 8 solar masses, is thus N8 = 1011 f8 ≈ 2 × 109 stars.
The total mass of these stars can be found by performing the integral over M dN :
Z 100
M × M −2.35 dM ≈ 3 × 1010 M
(24)
M8 = 1011 × 0.39
8
Which also suggests that the average mass of one of these objects would be about M̄ ≈ 16M .
B). Assume that every stellar core collapse distributed 0.05M of iron into the interstellar medium.
If the MW started with 5 × 1010 M of gas, what is the mean interstellar mass abundance of iron
in the Galaxy?
From the previous problem, we can expect the number of supernovae to have been about 2 × 109 ,
distributing a total of MF e ≈ 108 M . Comparing this with the total initial amount of gas, yields
a primordial mass abundance of ZFe ≈ 0.002. This abundance is only slightly greater than that of
the sun—suggesting that the sun formed after most of these supernovae had already occurred.
C). How many pars of stars in the MW were formed in which both companions were more massive
than 8M ?
The probability that a given star is both in a binary system and more massive than 8M is
1
2 f8 . The probability that both stars in the binary were formed greater than 8 solar masses is thus
about ( f28 )2 ; suggesting there may be about 7 × 106 NS-NS, NS-BH and BH-BH binaries in the
galaxy.
5) Neutrinos and Opacity.
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
A). M = 1.4M and a radius R =10km, estimate the mean nucleon density
If we assume the mass of a neutron star is entirely neutrons/protons (nucleons), we have
Np =
1033 g
M?
≈ −24 = 1057
Mp
10 g
(25)
and the density is thus n ≈ 1039 cm−3
B). σνn = 10−42 cm2 ; mean free path assuming the density you found in ‘A?’
l = (nσ)−1 ≈ 103 cm
(26)
C). How many seconds does it take a neutrino to emerge in a random walk
This is the diffusion time-scale1, which we found in class to be
R2
(106 )2 cm2
≈ 3
= 0.1 s
lc
10 cm 1010 cm s−1
The actual solution is about 10s, because the radius of interest is about 10 times larger.
tD =
(27)
D). Twelve electron anti-neutrinos were detected by a spherical tank with 3 kton of water. σνw =
10−43 cm2 how many neutrinos are passing through your body a second?
Water has a mass of 18 grams per mole (and thus 18 nucleons per molecule), along with a density
of 1 g/cm3 ; thus the nucleon number density is
!
1 g/cm3
(6 × 1023 molecules/mol)(18 nucleons/molecule) ≈ 1 × 1024
(28)
18 g/mol
And the mean-free path is about 1019 cm (which is about 10 light-years). We can assume that the
probability of interaction of a neutrino over some distance x will be roughly proportional to x/l
(you can derive this rigorously by taylor expanding the Beer-Lambert law: I = I0 e−x/l ). 3 kton is
about 3 × 109 g, and therefore about 3 × 109 cm3 of water. The size of the detector is then about
x ≈ V 1/3 ≈ 2000 cm.
The number of neutrinos which interact with water nucleons (N ) can thus be related to the total
number of neutrinos which passed through the detector (N0 ), as
19 10
x
→
N0 ≈ 12
≈ 6 × 1016
(29)
N = N0
l
2000
This is the number of neutrinos passing through the entire tank, in an equivalent time period (about
10 seconds actually, or 0.1 s using the result of part ’C’). A 7 × 104 g person (with the same density
as water) is 4 × 104 times smaller than the tank, so finally: during the supernova, there were about
1013 (or 1011 for a 10s dispersion) neutrinos passing through your body per second.
1Because the neutrinos are highly relativistic, we can simply use the speed of light in place of their actual velocity.
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
E). Number density of nucleons in lead is 1025 cm−3 , what is the mean free path?
The difference between the cross-sections of neutrinos with water and with neutron-star matter
is only an order of magnitude, thus either σνw = 10−43 cm2 or σνn = 10−42 cm2 are good estimates.
Thus,
l = (nσ)−1 ≈ 1018 cm ≈ 1 light-year
(30)
To have a good chance of interacting with a single neutrino, one would need one light-year of
solid lead.
Problem set 4
Gravitational Focusing
6) Gravitational Focusing and Accretion.
One
of the
most important
concepts
in pairwise accretion is the “gravitation
A). Find the ratio of
effective
cross-section,
to physical
cross-section
deflection of a small body by a massive body. This deflection works to increase t
Consider a star moving
a dust cloud
anyisparticulate
region),
with some velocity
v∞
of thethrough
two bodies.
Your(or
goal
to reproduce
a calculation
first done
by Safrono
relative to the material
far
away.
If
the
star
were
massless,
it’s
effective
area
would
just
be
its
cross
no
notes,
no
discussion
with
anyone.
Really.
Consider
a
test
particle
approach
sectional area πR2 , and it would accrete all of the material along a cylinder of that size. With
radius R. The impact parameter is b and the velocity at infinite distance is v∞ (i.
gravity, however, additional material will be pulled in; we can describe this by an effective area
the to
two
would
approach
distance
of b of each
other and with re
πb2 , where b is referred
as bodies
the ‘impact
parameter,’
See within
Fig. 2. a
The
impact parameter
describes
2
absencetoofthat
gravity,
cross-section
for collision
would
be πR
. We would like
the distance, perpendicular
of the the
motion,
which determines
the border
between
particles
that will and will not
be accreted. increases due to the effects of gravity.
cross-section
a. What Focusing.
is the escape
velocity
from
large
body?
(you’ll need this for late
Figure 2. Gravitational
Impact
parameter
‘b,’the
which
is the
maximum
perpendicular distance away at which particles will be accreted. Figure from Geoff
a diagram
ofEvolution
a grazingof collision
Blake’s Ge/Ay b.
133:Draw
The Formation
and
Planetary between
Systems. the two bodies. (A grazing
point mass had just slightly more energy it would not collide with the ma
where
Conservation of angularattention
momentumtostates
thatthe two bodies touch.
b v∞just
= Rdrew,
vimp use the principle of conservation
(31)
c. For the case you
of angular mom
relating
b, vthe
, R, andis vjust
the velocity
of the it
object
impact. (Hint: W
∞ particle
imp ,barely
for an impact velocity vimp
. Because
being accreted,
is justatbarely
bound—i.e. its energy is zero
(E < 0 is bound,
> 0 is unbound).
momentum
in the E‘before’
picture, shown above, and the total angular mo
and
1 impact,
M m set these equal)
2GM
2
2
2
2
mvimp − G
R
=0
→
vimp =
R
= vesc
(32)
d. Write an energy conservation equation relating v∞ , vimp , R, and M .
e. What is the maximum value for b for which an impact will occur? Write
b2 = R2 (1 + f ) where f is called the focusing or Safronov parameter, wh
terms of the escape velocity.
f. (you can start collaborating again at this point) How large does an object ne
Homework 4
Astro 111 - Order of Magnitude: Fall 2016
Therefore,
2
b2
vesc
=
2
R2
v∞
(33)
B). In a dense globular cluster (106 stars, radius about 10 pc), dispersion velocity?
If we apply the virial theorem, we see that
r
Mm
GM
2
mv ≈ G
→
v=
(34)
d
d
where d is the radius of the cluster, and M = N M? is the total mass of the cluster. If we plug in
the given values, we see that
r
10−7 106 1033
≈ 106 cm s−1
(35)
v≈
1019
C). Rate of collisions?
Here we put it all together: the effective cross section of a star in the cluster is
σ? = πb2 = πR?2
2
7 2
vesc
11 2
2 (5 × 10 )
≈
π(10
)
cm
≈ 1026 cm2
v2
(106 )2
(36)
while the number density of stars is
n≈
N
106
=
= 10−51
d3
(1019 )3
(37)
thus the mean free path, l = (nσ)−1 ≈ 1025 cm
and finally, the collision rate
R=N
v
106
≈ 106 25 = 10−13 hz ≈ 1 per Myr
l
10
(38)