Download Ay 7A - Fall 2010 Section Worksheet 5 Telescopes

Ay 7A - Fall 2010
Section Worksheet 5
Telescopes
1. Are you going to approve?
You are on a funding committee that is going to approve or disapprove new telescope designs. Decide
whether you will approve or disapprove. Explain and justify the reason behind your decision.
(a) Build a 100 meter optical telescope(with no adaptive optics) on Earth.
The telescope’s resolution will be completely limited by atmospheric seeing. It will
be able to gather a lot of light quickly, but even with its huge size it will still
be limited to 1’’ -- 2’’ resolution. Shaving 10--20 meters off the size and investing
the money saved in an AO system would make a lot more sense.
(b) Build a 10 meter x-ray telescope on Earth.
The Earth’s atmosphere is opaque to x-rays. The telescope won’t be able to see anything,
because all the sources it could observe will eb blocked by the atmosphere.
(c) Build a 500 meter radio telescope in orbit.
Why build a radio telescope in orbit? The Earth’s atmosphere is very transparent
to radio waves, and seeing is not a problem in radio. It’s far cheaper to build something
on the ground than in space, and there’s no good reason to put it in space.
(d) Build a gamma-ray telescope in orbit using the same design as the Hubble Space Telescope but
with the CCD replaced by a gamma-ray detector.
The design for HST won’t work as a gamma-ray telescope because gamma-rays won’t reflect
off mirrors, especially not at the steep angles that HST’s mirrors use. The gamma-rays
will just scatter off the mirror and won’t make it to the detector.
2. Keck Telescopes vs. Hubble Space Telescope
The largest optical telescopes in the world are the twin 10 meter diameter Keck Telescopes on the
summit of Mauna Kea, in Hawaii. The focal length of Keck is 17.5m. In contrast, the Hubble Space
Telescope(HST) has a primary mirror 2.4m in diameter, and is in orbit around the Earth. The focal
length of HST is 57.6m.
(a) If it takes one of the Keck telescopes 1 hour to collect enough light to make a good image of a
particular object, how long would it take HST?
As you did in problem set 5, if we demand that a fixed number of photons be collected
by the same pixel, T collecting time is proportional to (f /D)2 . By taking the ratio,
(17.5/10)2
1 hour
TKeck
=
=
.
2
THST
(57.6/2.4)
x hours
Solving this, we get about 188 hours.
(b) For observations in red light(6500 Angstroms), what is Keck’s maximum angular resolution? What
is HST’s? (Keck has an adaptive optics system, but you may neglect its effects for this problem.)
1
Angular resolution is given by θres = 1.22λ/D.
Keck : θres = 1.22 ×
6500 × 10−10
≈ 7.9 × 10−8 rad ≈ 0.01600 .
10
HST : θres = 1.22 ×
6500 × 10−10
≈ 3.3 × 10−7 rad ≈ 0.06800 .
2.4
(c) Which telescope is better for observing faint objects? Which is better for observing objects of
small angular size? Why?
Keck is preferable for observing faint objects, because its larger area allows it
to gather light much faster than HST. HST is preferable for objects of small angular
size, because its angular resolution is far better than Keck’s.
3. Very Large Array
The Very Large Array(VLA), one of the world’s premier astronomical radio observatories, consists of
27 radio antennas on the Plains of San Agustin fifty miles west of Socorro, New Mexico. Each antenna
is 25 meters (82 feet) in diameter. There are 4 array configurations : A array, with a maximum
antenna separation of 36 km; B array – 10 km; C array – 3.6 km; and D array – 1 km. The VLA
is an interferometer; this means that it operates by multiplying the data from each pair of telescopes
together to form interference patterns. The structure of those interference patterns, and how they
change with time as the earth rotates, reflect the structure of radio sources on the sky: we can take
these patterns and use a mathematical technique called the Fourier transform to make maps1 .
(a) What diameter would a single-dish telescope have to be to have the same collecting area as VLA?
Since the VLA has 27 dishes, the collecting area of the VLA array is 27 times the
collecting area of a single dish. We want a single dish telescope that has the same
collecting area. The area of a radio dish is πr2 , where r is the radius, so:
Asingle
π
dish
2
dsingle dish
2
dsingle
dish
=
27AVLA dish
2
dVLA dish
= 27π
2
√
=
27dVLA dish ≈ 130 m .
(1)
(2)
(3)
Thus, a single dish telescope would have to be 130 m in diameter to have the same
collecting area as the VLA. This is much larger than VLA’s 25 m dishes.
(b) What diameter would a single-dish radio telescope have to be to have the same angular resolution
as VLA?
For an interferometer like the VLA, angular resolution is determined by the maximum
separation between the radio dishes. The VLA has the same angular resolution as a
single dish telescope whose diameter is equal to the maximum separation of the dishes.
Thus, a single dish telescope would have to be 36 km in diameter to have the same
angular resolution as the VLA’s A array configuration.
(c) The coverage of frequency of the VLA is from 74 to 43,000MHz. What is the best angular resolution of the VLA?
1 www.vla.nrao.edu
2
θres = λ/D. So, to get the best resolution, λ should be the shortest wavelength
which means the highest frequency because frequency f is c/λ and D should be the
longest baseline. Thus,
θres =
c
3 × 1010 cm
λ
=
=
≈ 2 × 10−7 rad ≈ 0.0400
D
fD
(43 × 109 ) × (36 × 105 )
(d) Consider only one pair of dishes in the VLA and a quasi-monochromatic interferometer, one that
responds only to radiation in a very narrow band centered on frequency ν = ω/(2π). Then the
output of antennas 1 and 2 can be written as V1 = V cos[ω(t − τ )] and V2 = V cos(ωt) where t
is time when we get the signal. Express τ , the geometric delay, in terms of the pointing angle of
two dishes, θ, and the baseline(the distance between two dishes), L. These two antennas give the
final output R = (V 2 /2) cos(ωτ ). (Hint: τ is related to the phase difference of the incident light.)
(e) Any real function can be described in Fourier series which is the a series of sines and cosines such
as
∞
∞
X
a0 X
+
an cos nx +
bn sin nx.
f (x) =
2
n=1
n=1
Knowing this and using the result you got in the previous question, explain the advantage of an
interferometer.
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