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Astronomy 120: Galaxies and the Universe
Fall 2016
Homework # 4 Solutions
1) (25 points)
Suppose the MESSENGER spacecraft, which recently made a close pass of Mercury,
decided to communicate with the Cassini probe, now exploring Saturn and its moons.
When Mercury is closest to Saturn in their orbits, it takes 76.3 minutes for the radio
signals from Mercury to reach Saturn. A little more than half a Mercurian year later,
when the 2 planets are furthest apart in their orbits, it takes 82.7 minutes.
a) What is the distance between Mercury and the Sun? Please give answers in both
light-minutes and astronomical units. Assume that the planets have circular orbits.
(It is easier to figure this one out with a figure!)
We are given,
D1 = 76.3 lm
and D2 = 82.7 lm
From the diagram, DMH (distance between Mercury and the Sun), is given by the equation
D2 = D1 + 2DMH
Therefore,
DMH
=
(D2 − D1 )/2
=
(82.7 lm − 76.3 lm)/2
= 3.2 lm
Distance travelled by light in 1 min
AU
1 lm
1.496 × 1011 m
8
−1
3 × 10 m s × 60 s
AU
= 3.2 lm ×
1 lm
1.496 × 1011 m
' 0.385 AU
=
3.2 lm ×
1
b) What is the distance between Saturn and the Sun?
From the figure, DSH (distance between Saturn and the Sun) is given by
DSH
=
D1 + DMH
=
76.3 lm + 3.2 lm
=
79.5 lm
=
79.5 lm ×
'
9.57 AU
3 × 108 m s−1 × 60 s
AU
1.496 × 1011 m
2) (20 points)
a) Stars Donald & Hillary have the same luminosity, but Hillary is 10 pc away while
Donald is 30 pc away. Which star appears brighter as seen from earth, and how many
times brighter is it?
We know that an object appears fainter the further away it is. We expect, therefore, that of two
stars with the same intrinsic luminosity, the closer one (in this case, star Hillary) should appear
the brighter of the two. The brightness of each star as a function of its luminosity and distance is
given by:
L
b=
.
(1)
4πd2
For each star individually, then, with LH = LD = L,
L
4πd2H
L
.
bD =
4πd2D
bH =
How many times brighter is star Hillary than star Donald? We take the ratio of the two brightnesses:
bH
4πd2D
d2
(30 pc)2
L
·
= D
=
= 9.
=
2
2
bD
4πdH
L
dH
(10 pc)2
We find that star Hillary appears brighter than Donald, as we expected, by a factor of 9.
b) Stars Spongebob and Patrick are equally bright as seen from the earth, but Spongebob
is 40 pc away while Patrick is 400 pc away. Which star has the greater luminosity,
and how many times greater is it?
As in the previous question, we examine the question qualitatively first: if an object appears fainter
as it is further away, the further away of the two stars with the same apparent brightness must
be the more luminous. We expect, therefore, to find that star Patrick is the more luminous of the
two. We solve equation 1 for luminosity, and find:
L = 4πd2 b.
To find how much more luminous star Patrick is, therefore, we take the ratio of the two luminosities:
LP
4πd2P b
d2P
(400 pc)2
=
=
=
= 100.
LS
4πd2S b
d2S
(40 pc)2
Star Patrick is 100 times more luminous than star Spongebob.
3) (25 points)
The angular resolution of a telescope (or other optical system) is a measure of the smallest details which can be seen. Because of the distorting effects of earth’s atmosphere, the
best angular resolution which can be achieved by optical telescopes from earth’s surface
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is normally about 1 arcsecond. This is why much clearer images can be obtained from
space. The angular resolution of the Hubble Space Telescope is about 0.03 arcseconds,
and the smallest angle that can be measured accurately with HST is actually a fraction
of one resolution element, or about 0.005 arcseconds.
a) What is the distance of a Cepheid variable star whose parallax angle is measured to
be 0.005 arcseconds? Express your answer in both light years and parsecs.
The easiest way to solve this problem is to solve it in parsecs, and then convert to light years. Use
1
the equation distance = parallax
, where distance is in parsecs and the parallax angle in arcseconds.
Plugging in the parallax angle gives
d =
1
pc = 200 pc
0.00500
To convert from parsecs to light years, note that 1 pc = 3.26 ly. So
200 pc × 3.26
ly
= 652 ly
pc
b) Suppose that the parallax angle is not measured with infinite precision, but has an
uncertainty of ± 0.001 arcseconds. What are the range of distances consistent with the
measured parallax angle of 0.005 ± 0.001 arcseconds [i.e., 0.004 to 0.006 arcseconds]?
Use the same method above for 0.004 and 0.006 arcseconds.
1
pc = 250 pc
0.004
1
pc = 167 pc
d =
0.006
The range of distances consistent with the measured parallax angle of 0.005 ± 0.001 arcseconds is
167-250 pc.
d =
4) (30 points)
The faintest stars that can be detected with the Hubble Space Telescope have apparent
brightnesses which are 4 × 1021 times fainter than the Sun!
a) How far away could a star like the Sun be, and still be detected with the Hubble
Space Telescope? Express your answer in light years.
The faintest stars that can be detected with the Hubble Space Telescope have apparent brightnesses
1
which are 4 × 1021 fainter than the Sun. Then limit apparent brightness blim = 4×10
21 b =
−22
2.5 × 10 b .
We also know that:
L = 4πd2 b
with L the luminosity and d the distance. Rearranging this equation, we see that
d=(
L 1/2
)
4πb
Using the ratio method, we can determine that
d
(L/4πb)1/2
L b 1/2
=
=(
)
d
L b
(L /4πb )1/2
For a star “like the Sun” (L = L ), we finally get,
d=(
b 1/2
) d = (6.3 × 1010 ) · (1 A.U.) = 6.3 × 1010 A.U. = 306 kpc = 1.00 × 106 light years
b
Since 206265 A.U. = 1 pc = 3.262 light years.
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b) Cepheid variable stars are very important distance indicators because they have large
and well-known luminosities. How far away could a Cepheid variable with 20,000 times
the luminosity of the Sun be, and still be detected with the Hubble Space Telescope?
Express your answer in light years.
For Cepheids, L = 2×104 L . Because we want to calculate the limiting distance for observing
this object. bcepheid = blim . Using the equation we derived in part a.),
d
L b 1/2
=(
)
d
L b
Plugging in the values we have for L and b,
d
= ((2 × 104 ) · (4 × 1021 ))1/2 = 8.94 × 1012
d
d = 8.94 × 1012 A.U. = 43.4 Mpc = 1.41 × 108 light years
c) Cepheid variables are luminous enough to detect in other galaxies. There are galaxies
in the Virgo cluster whose distances are estimated with Cepheid variables. Type Ia
supernovae have also been observed in some of the same galaxies. This is exciting
(for me, at least!) because it means that we can measure the distances to very remote
galaxies, using supernovae as standard candles. Type Ia supernovae have well-known
luminosities, and at their peak are 10 billion times more luminous than the Sun. How
far away could a Type Ia supernovae be, and still be detected with the Hubble Space
Telescope? Express your answer in light years.
For Supernova explosion, L = 10 × 109 L . Once again,
d
L b 1/2
=(
)
d
L b
Again, plugging in L and the limiting brightness for detection with HST,
d
= ((10 × 109 ) · (4 × 1021 ))1/2 = 6.32 × 1015
d
d = 6.32 × 1015 A.U. = 3.07 Gpc = 1.00 × 1011 light years
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