Transcript numerical exercises

Astronomy 5500
Exercises
Problem. What is the distance to the star Spica (α
Virginis), which has a measured parallax according to
Hipparcos of πabs = 12.44 ±0.86 mas?
Solution.
The distance to Spica is given by the parallax equation,
i.e.
The uncertainty is:
The distance to Spica is 80.39 ±5.56 parsecs.
Problem. How distant is Spica (α Virginis), a B1 III-IV
star with apparent visual magnitude V = 0.91, given that
B1 III-IV stars typically have an absolute magnitude of
MV = –4.1 ±0.3 (Turner ZAMS)?
Solution.
The distance modulus for Spica is given by:
Thus:
The uncertainty is:
The spectroscopic parallax distance to Spica is 100.5
±13.9 parsecs. Note the small disagreement with the star’s
Hipparcos parallax distance of 80.39 ±5.56 parsecs.
Problem. Find the mass of the Galaxy given the local
circular velocity of 251 km/s at the Sun’s location roughly
8.5 kpc from the Galactic centre.
Solution.
Use Kepler’s 3rd Law in Newtonian fashion,
i.e. (MG + M)(in M) = a(in A.U.)3/P(in yrs)2.
For an orbital speed of 251 km/s and orbital radius of 8.5
kpc the orbital period is:
2R
2  8500 pc  206265 AU/pc 1.496 108 km/AU
P

251 km/s
251 km/s  3.1558 107 s/yr
 2.0805 108 yr
The semi-major axis is:
a  8500 pc  206265 AU/pc  1.7533109 AU
So the mass of the Galaxy is:
MG 
1.753310 
2.0805 10 
9 3
8 2
M Sun  1.24511011 M Sun
So ~1011 M is derived for the mass of the Galaxy
internal to the Sun. If the orbital velocity curve is flat to
~16 kpc from the Galactic centre, then one can redo the
calculations to find that ~21011 M is derived for the
mass of the Galaxy internal to ~16 kpc from the centre.
Where did the extra ~1011 M come from, or is it proper
to apply Kepler’s 3rd Law in situations like this? Recall
that it applies to the case of a two-body situation only,
not to a multi-body situation.
Mass/Light Ratios, M/L:
The mass-luminosity relation varies roughly as L ~ M4
(M3 for cool stars), so the mass-to-light ratio should vary
as M/L ~ 1/M3 or 1/M2. The typical star near the Sun is a
cool M-dwarf with a mass of only 0.25 M or less,
implying a typical mass-to-light ratio for our Galaxy of
~16. Since most stars are probably less massive than
that, the actual mass-to-light ratio for the Galaxy could
be in excess of ~25 or so.
Problem. What is the mass-to-light ratio for the Milky
Way, given the properties that have been derived for it,
i.e. estimated mass of 2  1011 M and LSR velocity of 251
km/s, and Mbol() = 4.79, from MV() = 4.82 and (B−V)
= 0.63?
Solution. Most supergiant Sb galaxies have MV  −20.5. If
the Milky Way is identical, then its luminosity in solar
units is:
So for the Milky Way, L/L = 10(25.32/2.5) = 1010.128 = 1.3 
1010, and M/L = 2  1011 M/M/1.3  1010 L/L  15,
larger than values typically adopted in research papers.
Problem. Use the rotation curve for stars within 1" of the
centre of M32 shown in Carroll & Ostlie to estimate the
mass lying within that region. Compare your answer with
the value obtained using the velocity dispersion data in
Example 25.2.1 of the textbook and with the estimated
range cited in that example.
Solution. We
did an example
using orbital
velocity to
estimate
galaxy mass.
Here, vr =
50 km/s and
r = 3.7 pc.
Use Kepler’s 3rd Law in Newtonian fashion,
i.e. (MG + Mstar) = a3/P2, for a in A.U. and P in years.
For an orbital speed of 50 km/s and orbital radius of 3.7
pc the orbital period is:
2R
2  3.7 pc  206265 AU/pc 1.496 108 km/AU
P

50 km/s
50 km/s  3.1558 107 s/yr
 4.5463 105 yr
The semi-major axis is:
a  3.7 pc  206265 AU/pc  7.6318 105 AU
So the mass of the Galaxy is:
MG 
7.6318 10 
4.546310 
5 3
5 2
M Sun  2.1506 106 M Sun
as compared to an estimate of ~107 M obtained from the
velocity dispersion and values ranging from 1.5−5106
M from the rotation curve, similar to the present value.
Problem. The star Delta Tauri is a member of the Hyades
moving cluster. It has a proper motion of 0".115/year, a
radial velocity of 38.6 km/s, and lies 29º.1 from the cluster
convergent point.
a. What is the star’s parallax?
b. What is the star’s distance in parsecs?
c. Another Hyades cluster member lies only 20º.0 from
the convergent point. What are its proper motion and
radial velocity?
Solution. The relationship applying to all stars in a
moving cluster is:
With the values given, the distance is found from:
The star’s parallax is:
And its space velocity is:
The other star must share the same space velocity as
Delta Tauri, so its proper motion and radial velocity are:
Problem. The open cluster Bica 6, at l = 167º, has a
radial velocity of 57 km/s, or 48 km/s relative to the LSR.
Its distance from main-sequence fitting is 1.6 kpc. Is the
cluster’s motion in the Galaxy consistent with Galactic
rotation?
Solution. See
predictions at right.
The cluster should
have a LSR velocity
of −9 km/s if it
coincided with
Galactic rotation.
The cluster’s actual
LSR velocity of 48 km/s is therefore completely
inconsistent with Galactic rotation. Can you think of a
reason why?
Problem. Interstellar neutral hydrogen gas at l = 45º has
a radial velocity of +30 km/s relative to the LSR. What
are its distance from Earth and the Galactic centre if R0 =
8.5 kpc and θ0 = 251 km/s?
  0 
Solution. Recall: vR  R0   0 sin l  R0  
 sin l
 R R0 
For a flat rotation curve, θ = θ0 = 251 km/s, so:
  0 
vR
30
   

 4.99 km/s/kpc

 R R0  R0 sin l 8.5 sin 45
1
1
 251

1
 

4
.
99
km/s/kpc

0
.
1375328
kpc


R 251 km/s  8.5

1 kpc
R 
 7.2709884 kpc
0.1375328
For plane triangles the cosine law is:
a  b  c  2bc cos A
2
2
2
Here: Angle A = 45º
Side b = 8.5 kpc = R0
Side a = 7.27 kpc = R
Side c = distance d to cloud
So:
7.27
Yielding:
2
 8.5  d 2  28.5d cos 45
2
d  12.020815d  19.3971  0
2
Which is a quadratic equation with solution:
2
12.020815  12.020815  419.3971
d
2
 6.0104075  12 8.1799507  kpc, giving
d  1.92 kpc and 10.10 kpc
Problem. The distance modulus of the Large Magellanic
Cloud (LMC) is 18.43, and its estimated mass is 2  1010
M. It orbits the Galaxy in the Magellanic Stream. What
is its time scale for dynamical capture by the Milky Way
for C = 23?
Solution. The distance to the LMC, its “orbital radius” r,
is determined from:
so:
The orbital speed of the LMC can be inferred from
Kepler’s 3rd Law:
for m1 and m2 in M, a in AU, P in yrs.
where:
Since 1 pc = 206265 AU, MMW  2  1011 M, then:
So:
and:
The time scale for dynamical capture can now be
evaluated:
2 vM r
tC 
CGM
2
2  1.3968  10  48.5  10  206265  1.496  10

23  6.6726  1011  2  1010  1.989  1030
1.9657  1048
3.2198  1016 s
9


 1.0203  10 yr
31
7
6.105  10
3.1557  10 s/yr
5
3

11 2
Which implies dynamical capture of the LMC in only
half of its orbital period. But that assumes that Milky
Way matter extends as far as the LMC so that there is
friction occurring constantly, whereas there is some
evidence that the LMC has orbited the Milky Way
several times.