Chapter 24: Normal and Active Galaxies
1
Chapter 23 Problems
1.
1 A.U. at 1 pc subtends an angle of 1" (by definition of the parsec).
100 A.U. at 100 pc will subtend
the same angle, 1".
This is 21600 times smaller than the angular size of the Andromeda galaxy.
2.
This would simply be a distance 21600 times closer than the 100 pc in the previous question, or
0.0046 pc or 950 A.U.
3.
20 - 0 = 5Log(
d
) - 5,
d
= 100,000 pc = 100 kpc.
4.
Using the inverse square law, if a Cepheid is 100 times brighter than an RR Lyrae variable, it should
be visible at 10 times the distance.
In other words, placing the Cepheid 10 times farther away will
make it 10
2
times fainter, or 100 times fainter.
Therefore it would equal the RR Lyrae star’s
brightness.
6.
25 - (-6.2) = 5Log(
d
) - 5 + 2.5
d
/1000,
7.24 = Log(
d
) +
d
/2000
Start an iterative solution with
d
=
10
4
, 7.24 = 9, then
d
= 5,000 gives 7.24 = 6.2.
Keep estimating and closing in on the solution until
you reach
d
= 6800 pc.
8.
At 20 kpc, the circumference of the orbit is 2
π
×
20,000 pc = 126,000 pc.
A parsec = 3.1
×
10
13
km,
so the circumference is 3.9
×
10
18
km.
At a rate of 240 km/s it will take 1.6
×
10
16
s to orbit once.
A
year has 3.2
×
10
7
s in it, so the period of the orbit is 509 million years.
Using Kepler’s third law, we have the period in years but we need the size of the orbit in A.U.
There
are 206,000 A.U. in 1 pc, so 20 kpc = 4.12
×
10
9
A.U.
Finally, we can apply Kepler’s third law; the
result will be in solar masses.
(5.09
×
10
8
)
2
= (4.12
×
10
9
)
3
/
M
,
M
= 270 billion solar masses.
10.
The results of this calculation will depend strongly on the assumptions made.
Here is one possible
solution.
For the Sun’s velocity around the Galaxy, the relationship derived in the previous question
becomes
P
= 27700
R
.
Using the often quoted period of 225 million years gives
R
= 8100 pc.
The
other star will be assumed to be at
R
= 8200 pc and with a period of 227 million years.
Assuming