Astronomy 101
Observational Astronomy
Ann Esin
14 Sept 2009
Solutions to Problem Set 1
1. Planning for Observations [10 points]
The answers to these questions will help you plan for you first lab.
(a) [4 points] Calculate a range of RA for the portions of the sky which cross the
meridian between 9pm and 12am on Wednesday, Sept 16, 2009.
On Sept 21, RA(Sun)/simeq12h . A correction for 5 days comes to about 24h ·5d/365d = 20m ,
so for Sept 16 we get RA(Sun) = 11h 40m .
On this day 9pm PST is actually closer to 8?
Our target RA can be found by setting RA(target) = HA(Sun) + RA(Sun), for 9pm this
gives us
RA(9pm) = 9h + 11h 40m = 20h 40m
(1)
and for 12am
RA(12pm) = 12h + 11h 40m = 23h 400 .
(2)
So the range we want is roughly between 21h and 24h of RA.
Keep in mind that we are off by about an hour here. This is because currently the time on
our clocks is actually pretty far off from the local mean solar time (LMST) which gives us the
HA of the Sun. In other words, we assumed that the Sun is on the meridian at 12pm, which
is reality is not the case. You can use the Daily Almanac on our course webpage (found
under Astrophysical Resources, Almanac Tools) to figure out the difference between
LMST and local clock time. I found that right now LMST is about 40 min behind local
time, i.e. when it is 9pm according to your clock, it is only about 8:20pm, according to the
Sun. The difference is mostly due to the Daylight savings time change which affects us in
the summer. so this implies that the good RA range is actually ∼ 20h to ∼ 23h of RA.
(b) [3 points] Calculate the plate scale, the field of view and the detector resolution
00
limit for the SBIG STL-11000M camera mounted in the focal plane of our 14
Celestron with an effective focal ratio f /10. The specifications for the camera
can be downloaded together with this homework.
The effective focal length of the telescope is equal to f = f # · D = 10 · 14 in = 3556 mm.
00
The plate scale is then PS = 1/f = 59 /mm. The size of one pixel is 9µm by 9µm, so the
00
00
angle seen by a pixel is θmin = 59 /mm · 0.009 mm = 0.53 . You need at least three pixels to
00
actually “resolve” two objects, so the resolution limit with our Brackett setup is about 1.5 .
This seems bad, but the seeing in Claremont is very rarely better than this.
0
0
The size of the chip is 36 mm by 24.7 mm, so the full field of view is FOV = 35.4 × 24.3 .
1
(c) [3 points] Now repeat the calculations from part (b) for the Apogee Alta U16
00
camera mounted on a 1-meter (actually, the mirror diameter is precisely 40.9 )
telescope up at TMO. The 1-meter has an effective focal ratio f /9.5.
The effective focal length of the 1-meter is f = f # · D = 9.5 · 40.9 in = 9869 mm.
00
The plate scale is then PS = 1/f = 21 /mm. With pixel size 9µm by 9µm (same as SBIG
00
00
camera), the plate scale is also 0.19 per pixel, corresponding to a resolution of roughly 0.6 ,
which is plenty good even on a very good night.
0
0
The total size of the array is 4096 squared, which gives us FOV = 12.8 × 12.8 , quite a bit
smaller than for the Celestron/SBIG configuration at Brackett.
2. Messier 2 [10 points]
(a) [4 points] Use SIMBAD (under Astrophysical Resources on our course webpage) to look up the equatorial coordinates of Messier 2, a globular cluster in
Aquarius. What is the best time of year to observe this object from Claremont?
Please explain your reasoning carefully.
Typing ’M2’ into SIMBAD yields the following epoch J2000 coordinates for the cluster:
21 33 27.0 − 00 49 24.
To figure out what is the best time to observe this object our reasoning is essentially the
reverse of that in problem 1. For simplicity, let’s ignore the difference between PST and
Mean Solar Time. Let’s optimize out observing conditions for midnight, since this will give
us the longest possible observing window. So we need to find at what time of the year M2
transits the meridian (i.e. has HA = 0) at midnight. Recall that local sidereal time (LST)
is equal to the sum of RA and HA for any object. Then at transit, LST is equal to RA of
M2, and we need
LST = RA(M2) = 21.6h .
(3)
Let’s now relate LST to the Sun’s motion. We have LST = HA(Sun) + RA(Sun), and since
at midnight HA(Sun) = 12h ,
RA(Sun) = LST − 12h = (21.6h − 12h ) = 9.6h ,
(4)
which is 2.4 hours of RA smaller that RA(Sun) at autumnal equinox. 2.4 hours is equivalent
to 1.6 months, so early August is the best time to observe this cluster. According to our
results in problem 1(a), currently it transits around 10pm.
(b) [2 points] It is not practical to observe targets at altitudes a < 30◦ . What is
the maximum latitude in the northern hemisphere from which M2 can still be
studied?
M2 is at its maximum height, h, above the horizon at meridian transit. It is located in the
Southern hemisphere, so at transit its distance from zenith is equal to
z = LAT + |DEC|,
2
(5)
where LAT is the observer’s latitude (or observers angular distance from the equator), and
DEC is the declination of M2. We need z = 90◦ − h < 60◦ , so
LAT = z + DEC < 60◦ − 0.8◦ = 59.2◦ .
(6)
Thus only observers South of latitude +59.2◦ can study M2. In Claremont our latitude is
about +34◦ , so we are OK.
(c) [4 points] Calculate the galactic coordinates of M2. Sketch the galactic coordinate system, indicating the locations of the galactic poles, Galactic Center and
M2. Based on your result, does M2 lie in the Galactic plane?
Let’s use the transformation equations given in lecture. We have α = 21.5575h = 323.329◦
and δ = −0.8233◦. Then the galactic coordinates are given by:
b = arcsin [sin δ cos 62.6◦ − cos δ sin (α − 282.5◦ ) sin 62.6◦ ] = −35.95◦ ,
l = 33◦ + arccos [cos δ cos (α − 282.25◦ )/ cos b] = 54.40◦ .
(7)
(8)
M2 lies quite far away from the Galactic plane(which has b = 0).
3. Eye as a Detector [10 points]
The human eye is the most fundamental astronomical detector. It is a fairly
complex system which can, to first order, be approximated as a converging lens
focusing an image onto the cornea which contains the actual light detector cells
(rods and cones). Rods are more sensitive in dim light and are therefore primarily used for night vision. A very detailed description of how your eye works
can be found in chapter 8.
(a) [5 points] Estimate the Rayleigh limit for the resolution of the normal and
dark adapted eye. (Hint: Look in the mirror to check out the size of your pupil
and iris.)
The normal pupil is about 3 mm in diameter, so the diffraction resolution limit for optical
light with a wavelength of 550 nm is about
αeye ' 1.22
λ
00
= 2.2 × 10−4 rad = 46
D
(9)
In the dark adapted eye, the pupil expands to about 1 cm in diameter, so the resolution limit
00
is a bit better at α ' 14 .
(b) [5 points] The actual resolution of your eye is about 1.5 arcminutes. Use
this number to estimate the distance between neighboring rods on your retina.
Explain your reasoning. (Hint: To estimate the focal length of your eye, keep in
mind that it has a roughly spherical shape.)
3
The actual resolution is a lot worse than the limit set by diffraction, and so must be set by the
structure of your eye. The simplest assumption is to blame the spacing of the light detecting
elements in your retina, called rods. Let’s assume that for two objects to be resolved, their
images must be separated by at least one rod, i.e the distance between the images on the
retina must be greater than ∼ 2a, where a is the average distance between the rods. Then
the rod separation is given by
α
,
(10)
a=
2 · PS
0
where α = 1.5 = 4.4 × 10−4 rad and PS is the ”plate scale” of the eye.
To calculate PS, we need to estimate the focal length of the eye. Since the focusing lens is
on the front of the eye and the retina is at the back, the focal length is roughly equal to eye
diameter, f ∼ 3.5 cm. Then we have
a=
αf
∼ 8 µm.
2
(11)
4. Telescopes as Light Buckets [8 points]
Recall that the magnitude of a star characterized by a flux F (flux = energy
detected per unit area per unit time) is given by
m = 2.5 log10 (F0 /F ),
(12)
where F0 is the flux from a reference star defined to have m0 = 0. Let’s assume
that the dimmest star your eye can detect in Claremont is about 4.5 (it is close to
7 in an area without light pollution). Estimate the limiting magnitude you can
00
see through the eye-piece of our 14 Celestron. What is the limiting magnitude
you can detect with the 1m telescope at the Table Mountain Observatory? You
can ignore the effects of magnification and assume that stars remain unresolved
point sources.
The main effect is the increase in collecting area from the pupil of the eye to the primary
mirror of the telescope. The limiting magnitude is set by the smallest number of photons
the eye can detect in a certain period of time (eye “readout” period), let’s call this quantity
Nlim . If Flim is the flux from a star with a limiting magnitude, we have Nlim = Flim A ∆t ∝
Flim D2 ∆t, where A is the area and D is a diameter of a light collecting aperture, and ∆t is
the readout time. This Nlim must remain constant, so Flim ∝ D−2 , when we change the size
of the aperture.
Using this relationship we can calculate the limiting magnitude for a star observed with
a telescope:
mtel − meye = 2.5 log10
F0
Flim,tel
− 2.5 log10
F0
Flim,eye
= 2.5 log10
so
mtel = meye + 5 log10
4
Dtel
Deye
Flim,eye
Dtel
= 5 log10
Flim,tel
Deye
(13)
(14)
For a Celestron at Brackett observatory meye = 4.5 mag and Dtel = 36 cm. Taking
Deye = 1 cm, we get
m14 in ' 4.5 + 8 = 12.5 mag.
(15)
At TMO, meye = 7 mag, and Dtel = 100 cm, so the limiting magnitude for eye-piece
observations is
m1 m ' 7 + 10 = 17 mag.
(16)
5. The Matter of Time [12 points]
(a) [7 points] Show that the TDB clock loses 49s per century as compared to
the TCB clock. Hints: Show that the Sun’s gravitational potential well is more
important than the Earth’s. Also, since both time dilation effects (moving clocks
and clocks in a gravitational potential) are quite small, the time changes they
produce can just be added to each other.
There are two effects we need to worry about: gravitational time dilation and time dilation
due to Earth’s orbital motion. Both slow down local clock (TDB) as compared to the clock
at infinity (TCB).
Let’s consider special relativity first. If tTCB is time interval according to a clock at infinity,
the corresponding time interval on Earth is tTDB = tTCB /γ, where γ = (1 − v 2 /c2 )−1/2 . Then
the difference between the two intervals is

∆tSR = tTCB − tTDB
v2
= tTCB 1 − 1 − 2
c
!1/2 

v2
1− 1− 2
2c
"
' tTCB
!#
= tTCB
v2
, (17)
2c2
where we used the binomial expansion since v c. Here v is the average orbital velocity of
the Earth around the Sun:
s
GM
v=
= 29.8 km/s.
(18)
aE
Substituting this in and setting tTCB = 100 years, we get ∆tSR = 15.6 s.
To calculate the effect due to general relativity, let’s follow the hint and compare the effects
due to the Earth and the Sun. The time period in the gravitational potential well a distance
r from mass M is given by t = tTCB (1 − 2GM/rc2 )1/2 (see e.g. Ch, 17 in Carroll and Ostlie).
Then the difference in time is
"
∆tGR = tTCB − tTDB = tTCB
= tTCB
2GM
1− 1−
rc2
GM
,
rc2
1/2 #
' tTCB
GM
1− 1− 2
rc
(19)
again using the fact that 2GM/rc2 1. For the Earth, we have GME /rE c2 = 6.64 × 10−10
and for the Sun,GM /aE c2 = 9.88 × 10−9 . Clearly, the effect due to the clock’s position in
the Sun’s potential well is about 15 times larger, so we can ignore the contribution from the
Earth. With this, ∆tGR ' 31.2 s when tTCB = 100 years.
5
Adding the two effects together we arrive at the total time difference
∆t = ∆tSR + ∆tGR ' 47 s.
(20)
This is quite close to the precise value of 49 s.
(b) [5 points] Using the fact that the period of precession for the Earth’s rotational
axis is about 25,770 years, estimate the rate at which vernal equinox moves along
the ecliptic. Show that this effect explains the difference between sidereal and
tropical years.
The period of precession is also the time for the vernal equinox to make one complete
revolution around the ecliptic. It’s rate of motion is then
ω=
360◦
00
= 0.01397◦ /yr = 50.29 /yr.
25, 770 yr
(21)
The Earth’s precesses in the clockwise direction looking down from the NCP, and the vernal
equinox moves in the same direction along the ecliptic. The Sun, on the other hand, moves
in the opposite direction along the ecliptic. Since the tropical year is measured between
successive passes through the vernal equinox, it would be a little shorter than the sidereal
year:
T Y = SY −
0.01397◦
= 365.25636 d − 0.01417 days = 365.24219 days.
360◦ /365.25 days
This agrees exactly with the length of the tropical year given in the textbook (p. 32).
6
(22)