Astronomy 160: Frontiers and Controversies in Astrophysics

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Astronomy 160: Frontiers and Controversies in Astrophysics
Homework Set # 6 Solutions
1) a) Faint “brown-dwarf ” stars have absolute magnitudes of around 17.5.
How many times fainter than the Sun are these stars?
MBD = 17.5
Msun = 5
17.5 − 5 = −5
log bb12
2
12.5×2
= log bb21
−5
−5 = log bb12
10−5 = bb12
So brown dwarfs are 105 fainter than the sun.
b) If one observes a nearby galaxy at a distance of 1 Mpc (= 106 parsecs)
what is the apparent magnitude of Sun-like stars in that galaxy?
distance = 106 parsecs
D
m − M = 5 log 10pc
m − 5 = 5 log 105
m − 5 = 25
So the apparent magnitude of Sun-like stars at a distance of 106 parsecs is 30.
c) The magnitude of the full Moon is around -14.5. How much brighter
does the Sun appear than the Moon?
The apparent magnitude of the full moon is -14.5. So we can use the brightness
equation along with the apparent magnitude of the sun (-27) to find the ratio in
brightnesses.
−27 + 14.5 = −5
log bb12
2
−12.5 × −2
= 5 = log bb12
5
b1
= 105
b2
The sun is 105 times brighter than the full moon.
d) The brightest stars are around 105 times brighter than the Sun. If the
apparent magnitude of these bright stars in some galaxy is 22.5, how far
away is the galaxy?
First we need to find the absolute magnitude of the bright stars.
log 105
M1 − 5 = −5
2
M1 = −5
× 5 + 5 = −12.5 + 5 = −7.5
2
Now use the distance equation and the given apparent magnitude to find the distance.
D
22.5 − (−7.5) = 5 log 10pc
1
D
22.5 + 7.5 = 30 = 5 log 10pc
D
= 106 so D = 107 parsecs
10pc
e) White dwarfs are 104 times fainter than “A-type stars”. the nearest
example of which is Sirius. Sirius has a distance of 3 parsecs, and an
absolute magnitude of M = 1. Sirius is orbited by a white dwarf known
as Sirius B. What is the apparent magnitude of Sirius B (note that
you can assume that the two stars in the Sirius system are at the same
distance from Earth).
First we ned to calculate the apparent magnitude of Sirius. Then we can use the
brightness equation to find the apparent magnitude of the companion white dwarf
(Sirius B).
3
m1 − 1 = 5 log 10
m1 − 1 = 5 × −1
2
+
1
=
−1.5
m1 = −5
2
−1.5 − m2 = −5
log 104
2
−5
−1.5 − m2 = 2 × 4
−10 = −1.5 − m2
m2 = 8.5
f) The faintest galaxies observed by the Hubble Space Telescope have apparent magnitudes around 30. Suppose these galaxies are ≈ 3 gigaparsecs away (3 × 109 parsecs). Assuming every star in these galaxies
emits about the same amount of light as the Sun (a false assumption,
but let’s make it just the same), how many stars would these galaxies
contain? (Hint, the number of stars in each galaxy will be equal to
amount by which the galaxy is intrinsically brighter than the Sun —
e.g. a galaxy with 10 Sun-like stars has a brightness 10 times that of
the Sun).
First we find the absolute magnitude of the faint galaxy, and then we can calculate
how much brighter it is than the sun.
D = 3 × 109 parsecs
D
30 − Mgal = 5 log 10pc
1
30 − Mgal = 5 log (10 2 × 108 )
30 − Mgal = 5 × 8.5 = 42.5
Mgal = −12.5
−12.5 − 5 = −5
log bb12
2
−17.5 × −2
= log bb12
5
7 = log bb12
2
The galaxy is 107 times brighter than the Sun, so it contains 107 Sun-like stars.
2) a) Suppose the entire difference between Hubble’s measurement of the
Hubble constant and the currently known value of 70 km/s/Mpc was
due to observing the wrong kind of Cepheid. What is the difference in
absolute magnitude between the two types of Cepheids? That is, compute ∆Cep = MC1 −MC2 , where MC1 and MC2 are the absolute magnitudes
of the two types of Cepheids.
If we assume that the entire difference between Hubble’s measurement of the Hubble constant and the currently known value is due to the difference in absolute
magnitude, we need to determine how the absolute magnitude can affect the Hubble constant. We can write down the distance modulus equation for both types of
Cepheids:
mC1 − MC1 = 5 log(d1 /10pc)
mC2 − MC2 = 5 log(d2 /10pc)
If we then subtract the two equations from one another, we end up with:
(mC1 − MC1 ) − (mC2 − MC2 ) = 5 log(dC1 /10pc) − 5 log(dC2 /10pc)
mC1 − mC2 − MC1 + MC2 = 5 log(dC1 /dC2 )
However, we know that the apparent magnitude of the two types of Cepheids are
the same, since this is just a measured quantity. This means that mC1 − mC2 = 0.
So, we then have:
−MC1 + MC2 = 5 log(dC1 /dC2 )
MC1 − MC2 = −5 log(dC1 /dC2 )
∆Cep = MC1 − MC2 = −5 log(dC1 /dC2 )
In order to find the difference in the absolute magnitude of the two types of
Cepheids, we need to relate the two distances to the two values of the Hubble
constant:
v = Hd
v
d=
H
There are two versions of this equation in this problem (one for dC1 and one for
dC2 ). Hubble assumed that he was looking at Type II Cepheids and got a value of
H = 500km/s/Mpc. Therefore, his version of this equation is:
dC2 =
vC2
500km/s/Mpc
3
Of course, we now know that he was looking at Type I Cepheids, and we get a
value of H = 70km/s/Mpc. So, our version of the equation is:
dC1 =
vC1
70km/s/Mpc
Plugging these two values into the equation with ∆Cep, we get:

∆Cep = MC1 − MC2 = −5 log 
vC1
70km/s/Mpc
vC2
500km/s/Mpc
500 vC1
= −5 log
70 vC2


However, once again, v is a quantity measurable from Earth and the intrinsic
properties of the Cepheid are not going to change our measurements. This means
that vC1 = vC2 and:
∆Cep = MC1 − MC2 =
=
=
=
500
−5 log
70
−5 log(7)
−5(0.8)
−4
So, the Type I Cepheids would have to be ∼ 4 magnitudes brighter than Type II
Cepheids to fully account for Hubble’s error.
b) Suppose there had been no problem with the standard candles Hubble
used, including the Cepheids, and that the error in the determination
of Hubble constant was due to some bizarre mistake in determine the
value of an Astronomical Unit (that is, the distance from the Earth to
the Sun). Explain why such an error could result in a mistaken value
for the Hubble constant. What would the mistaken value of an AU have
to be (in meters) to explain Hubble’s error?
This question is somewhat ambiguous and there are (at least) two possible answers
that I saw that got full credit.
The first answer (which accounted for the majority of the correct answers) had
to do with the use of parallax as the base of the distance ladder. The idea is
that the length of the AU is critical in determining the distance to nearby variable
stars, which can then be used to determine the distance to progressively further
and further away “standard candles.” Hubble’s value was too large by a factor
of 7, which means the distances he calculated were too small by a factor of 7.
These distance errors could be explained by an error in parallax (the base of the
distance ladder). We can consider two values for the AU: AUright (= 1.5 × 1011 m)
and AUwrong . Since, in each question, the parallax angle (as something that can be
measured), will be the same, we can write down two equations:
4
AUright
dright
AUwrong
α =
dwrong
α =
Since they both have the same value for α, we can set the right sides equal to one
another:
AUright
AUwrong
=
dright
dwrong
Since Hubble’s distances were too small by a factor of 7, this means that dright =
7dwrong . Therefore,
AUright
7dwrong
AUright
7
1.5 × 1011 m
7
AUwrong
=
AUwrong
dwrong
= AUwrong
= AUwrong
= 2 × 1010 m
The second answer to this problem (which was submitted by several people) pointed
out that, by redefining the AU, we redefine what the length of the parsec is. The
parallax equation is basically as:
α=
Earth-Sun Distance
Distance in parsecs
If the definition of the numerator on the right hand side of that equation changes,
the denominator must also change by the same amount (to keep the parallax angle
on the left hand side of the equation the same). A larger unit means that the
numerical value measured is smaller (for example, 24 inches is 2 feet, the smaller
unit going with the bigger number). Thus, if the measured AU is too big, the
number in the denominator (for a fixed length) is too small, making Hubble’s
constant too big. Since Hubble measured the constant to be too large by a factor
of 7, his AU (by this reasoning) must have been 7 times too big, or 1012 m.
Finally, I’ll point out that these two effects exactly cancel one another (i.e. 71 ∗ 7),
so the net effect is that changing the value of the AU does not change the value of
the Hubble constant, in the units commonly used (km/s/Mpc).
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