PROBLEM SET #8 SOLUTIONS
AST142
1. Free fall timescale and speed
Consider the accretion of material onto a gravitating object, a process that is important in star formation and in the emission of radiation by material surrounding compact
objects such as black holes.
(a) Suppose a particle starts from rest a distance r0 from a star with mass M and
radius R. It drops to the stars surface. How fast is it going when it gets there?
(b) How long does it take to get there? You can assume that r0 R.
It may be helpful to have the following integral handy
1/2
0
Z 0
Z x∗ x
2
dx =
2 cos udu = u + sin u cos u
1−x
u∗
1
u∗
with x∗ = R/r0 = cos2 u∗ . For x∗ → 0 we find u∗ → π/2 and The integral
approaches π/2.
(c) Consider the Sun as that star, and work out the arrival speed, and time of the trip
for particles originating at rest 1 AU away, and particles originating at rest 100 AU
away. R = 7 × 1010 cm.
———————————–
Solutions
(a) Energy
E=
r˙ 2 GM
GM
−
=−
2
r
r0
where I used initial condition to set the right hand side. At any time afterwards
the radial component of velocity r˙
r
1/2
2GM r0
−1
r˙ =
r0
r
When it reaches R
r
r˙ =
1/2
2GM r0
−1
r0
R
1
2
PROBLEM SET #8 SOLUTIONS
AST142
(b) Manipulating the previous equation and using r˙ = dr/dt
r
−1/2
r0 r0
−1
dt =
2GM r
Let x = r/r0 , and dx = dr/r0 .
r
dt =
r03
2GM
x
1−x
1/2
dx
x goes from 1 to a smaller value x∗ = R/r0 .
r
1/2
Z x∗ r03
x
∆t =
dx
2GM 1
1−x
Let x = cos2 u
Z x∗ 1
x
1−x
1/2
Z
dx =
u∗
u∗
2 cos udu = u + sin u cos u
2
0
0
cos2 u∗ .
(1)
with x∗ =
For x∗ → 0 we find u∗ → π/2 and the integral approaches π/2.
For x∗ small we can approximate
r
r
r03
π
r03 π 2
∆t ∼
× =
2GM
2
8GM
Notice that ∆t is independent of R. The central mass could be spread out anywhere.
As long as it remains inside the trajectory of our infalling particle the time is
unchanged and only depends on the mass interior to it.
Let us replace M = (4/3)πR3 ρ∗ .
r 3/2 r 3π
r 3/2
0
0
∆t ∼
=
tf f
R
32Gρ∗
R
where we have used the free fall time
r
3π
tf f =
32Gρ∗
.
(c) Plugging and chugging r˙ = 614.5, 615.9 km/s for starting at 1AU, 100AU respectively. And ∆t is 0.18 and 180 years starting at 1AU, 100AU respectively. You
don’t gain much in energy starting further out but it takes longer!
2. Virial theorem
Consider a spherical stellar cluster with N ( 1) members, core radius R, which on
average is at rest with respect to us. The stars in the cluster have typical mass m and
PROBLEM SET #8 SOLUTIONS
AST142
3
typical average value for the square of the velocity σ 2 ; the square root of this quantity
is called the velocity dispersion. Using each component of the velocity
σ 2 = σx2 + σy2 + σz2
Here
σx2 =
1 X 2
vi,x
N
i
is the variance of the x component of the velocity for a cluster that is at rest. We could
also take in account the mean velocity (if the center of mass of the cluster is moving)
with
1 X
(vi,x − v¯x )2
σx2 =
N
i
where v¯x is the mean of the distribution. The total mass of the cluster is M = N m.
(a) Taking into account all pairs of bodies, show that the total potential energy is
Gm2 N (N − 1)
2R
Here N (N − 1)/2 is the numbers of pairs of stars.
U =−
(b) Show that
U ≈−
GM 2
2R
(c) The virial theorem (for a gravitating system in equilibrium) gives
|U | ∼ 2K
with K the kinetic energy. Assuming that there is no rotation show that
GM 2
= M σ2
2R
(d) An isotropic velocity distribution has velocity dispersions in each direction that are
equal
σx2 = σy2 = σz2
Show that
GM
= 6σz2
R
This is useful as we can consider σz to be the dispersion of the line of sight velocity
component – and we can measure this spectroscopically using Doppler shifts.
(e) Use the virial theorem to find an expression for the total mass of the cluster, in
terms of the core radius and the velocity dispersion.
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PROBLEM SET #8 SOLUTIONS
AST142
(f) Consider the Orion Nebula Cluster which contains about 103 stars within a region
of 5pc. Its velocity dispersion measured along the line of sight is about 3km/s. Use
the virial theorem to estimate whether the cluster is likely to be bound (and in
equilibrium).
——————Solutions
Each pair has a potential energy −Gmm/R. The number of pairs is N (N − 1)/2.
Altogether we get U = −Gm2 N (N − 1)/R. Using M = N m we find U = −GM 2 /(2R).
Neglecting rotation the total kinetic energy is
K=
M σ2
2
The virial theorem gives
GM 2
= M σ2
2R
Using σ 2 = 3σz2 we find for an isotropic distribution
GM
∼ 6σz2
R
and
M=
6σz2 R
G
For the Orion nebula cluster, compute
6σz2 R
6 × (3 × 105 cm/s)2 × 5 × 3.1 × 1018 cm
M
=
×
= 63000M
−8
−1
3
−2
G
6.67 × 10 g cm s
2 × 1033 g
As this mass is above the 103 from the numbers of stars, the cluster is not bound.
3. A rotation curve model
A popular model for a spherical galactic mass distribution is one with circular velocity
r
vc = v0
r+a
Laplace’s equation in spherical coordinates gives
1 ∂
2
2 ∂Φ
∇ Φ= 2
r
= 4πGρ
r ∂r
∂r
where Φ(r) is the gravitational potential and ρ(r) the density.
PROBLEM SET #8 SOLUTIONS
AST142
5
(a) Show that this model exhibits solid body rotation in its core, r a, and has a flat
rotation curve at large radii, for r a.
(b) Show that for a central force law
vc2 = r
∂Φ
∂r
(c) Find M (r) the mass within radius r.
(d) Using Laplace’s equation show that the density distribution
ρ(r) =
v02 r + 3a
4πG (r + a)3
—————————
Solutions
For small r, vc = rv0 /a and this is proportional to r. Consequently Ω = v0 /a,
a constant angular rotation rate means solid body rotation. For large r the circular
velocity vc approaches v0 and the rotation curve is flat.
Force dΦ/dr equals acceleration is vc2 /r gives vc2 = rdΦ/dr.
The mass distribution
M (r) = rvc2 /G =
r3
v2
(r + a)2 0
To compute ρ we first compute
r2
∂Φ
r3 v02
= rvc2 =
∂r
(r + a)2
v02 3r2 (r + a)
2v02 r3
r + 3a
∂ r3 v02
=
−
= v02 r2
∂r (r + a)2
(r + a)3
(r + a)3
(r + a)3
Laplace’s equation gives
1 1 ∂
v02 r + 3a
2 ∂Φ
ρ(r) =
r
=
4πG r2 ∂r
∂r
4πG (r + a)3
4. Estimating the Number of Galaxies per cubic Mpc
Problem #2 of Workshop 8 http://astro.pas.rochester.edu/~aquillen/ast142/
Work/work8.pdf
———————————
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PROBLEM SET #8 SOLUTIONS
AST142
Solutions
At z = 0.2 300” gives 1Mpc or 3.325kpc/”. 0.1 degrees is 6’=360” = 1.2 Mpc. So our
angular area corresponds to 1.4 Mpc2 . z = 0.2 gives a comoving radial distance of 832
Mpc and z = 0.25 a radial distance of 1016 Mpc the difference is 184 Mpc. So our patch
corresponds to about 250 Mpc3 . I found about 8 galaxies with the query corresponding
to a galaxy density of
n = 0.03Mpc−3
Luminosity distance at z = 0.22 is DL = 1099.3Mpc. This corresponds to a
DM = 5 log10 (DL/10pc) = 40.2
One of the galaxies has mag of about 20 and correcting by the luminosity distance gives
an absolute magnitude of −20. These galaxies are about Milky Way sized.