CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
1. Hubble Telescope
One of the early images taken by the repaired Hubble Telescope was of the "Butterfly Nebula"
(below), the remnants of a dying star that was once about five times the mass of the Sun
http://www.hubblesite.org/newscenter/archive/releases/2009/25/image/f/. The star is 38,000 ly
(light years) distant. The entire image spans 2.7 ly (2.4 arc minutes, or 6.98x10-5 radians). The
angular resolution (pixel size) of the Hubble is 0.085 arc seconds (4.12x10-8 rad).
a. [2] At this distance (D = 3,800 ly), what is the minimum
separation distance, Δx (in ly), of two stars that reside at the
center of a single pixel and are separated by a single pixel as
illustrated below? (This is the essence of the definition of
resolution.)
pixel
s
stars
b. [1] What is the solid angle subtended by a single Hubble telescope pixel? (Assume that the pixel
is square.)
c. If a polar orbiting satellite flying at an altitude of 800 km had an imaging system that was
capable of the same angular resolution:
i. [1] What would be the length of a single pixel on the earth immediately below the
satellite (viewing at nadir, or a viewing angle of 0)?
ii. [1] If the imaging system were pointed 60 off-nadir, what would be the area of a single
pixel along the optic axis? (Assume that the earth is flat to a good approximation?)
d. As a practical matter, the Hubble achieves this resolution by staring at a single spot for hours,
1.5 hours in the case of the Butterfly Nebula image. If a similar stare time were required for an
earth-viewing HubbIe, this would only be feasible from geostationary orbit (36,000 km).
i. [1] At that altitude, what would the resolution of the Hubble be?
ii. [1] What would be the GFOV of the instrument?
e. [1] The Hubble image was collected using several narrow-band filters centered at 373 nm, 469
nm, 502 nm, 656 nm, 658 nm, and 673 nm. Assuming that the earth-viewing Hubble was not
troubled by cloud cover during its 1.5 hour stare, what problems would have to be considered
for the earth-viewing system that would not be an issue for the space-viewing system?
2. [3] Radiometry
Given:
earth-sun distance
1.50E+11
solar constant
1.37E+03
Radius of the sun
6.95E+08
Boltzmann's constant 5.67E-08
m
W m-2
m2
W m-2 K-4
CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
Use the Stefan-Boltzman Law to estimate the temperature of the sun.
3. [2] Color
Spectra of 5 different minerals are shown below. Select three channels for display in RGB that
will clearly delineate among the minerals by presenting them as distinctly different colors, and
explain the rationale for your selection.
Note: These data are smoothed spectra from an AVIRIS image of Cuprite, Nevada. The image is
available on the course web site if you wish to try your selections. The original,
unsmoothed image is one of the default selections in ENVI. The spectral are also available
as a Spectral Library, which can be imported into the spectral viewer (the z-profile tool).
You may locate the pixels for which the spectra are shown using the pixel locator tool:
select Tools > pixel locator in the image window.
4. [Extra Credit] The International Space Station (ISS) is in low earth orbit at ~345 km. Some
pertinent characteristics are shown below (see: http://spaceflight.nasa.gov/realdata/tracking/ )
Length: 73 m Width: 108.5 m (356 ft)
Perigee: 341 km altitude (184 nmi)
Apogee: 353 km altitude (191 nmi)
Orbit inclination: 51.6419 degrees
Average speed: 27,743.8 km/h
Orbital period: c.91 minutes
That means that there is a chance that the ISS will pass through the field of view of any
virtually every other imaging satellite in orbit, whether in a polar or geostationary orbit. If that
should happen, discuss why you would (or would not) expect the ISS to be detected by:
i. [1] The earth-viewing Hubble in geosynchronous orbit.
ii. [1] The IKONOS of GeoEye-1 imaging systems (both pushbroom scanners, see:
http://www.geoeye.com/CorpSite/products/imagery-sources/Comparison.aspx
CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
SOLUTIONS
1. Hubble Telescope
a. [2] At this distance (D = 3,800 ly), what is the minimum separation distance, Δx (in ly), of two
stars that reside at the center of a single pixel and are separated by a single pixel
This is essentially the distance spanned by a single pixel.
Δx = D * tan(IFOV) ≈ D * IFOV (in radians)
Δx = 3.8x104 ly x 4.12x10-8 rad = 1.57x10-3 ly
b. [2] What is the solid angle subtended by a single Hubble telescope pixel? (Assume that the pixel
is square.)
The solid angle subtended by a single pixel is approximated (very accurately) by:
Solid angle  Area subtended by the pixel / square of the distance to the nebula
For a square pixel:  =
(1.84 103 ly) 2
x 2
=
= 1.70X10-15 sr
(38, 000 ly)2
D2
c. If a polar orbiting satellite flying at an altitude of 800 km had an imaging system that was
capable of the same angular resolution:
i. [1]What would be the length of a single pixel on the earth immediately below the
satellite (viewing at nadir, or a viewing angle of 0)?
Δx = altitude x IFOV (in radians)
= 8 x 105 m x 4.12x10-8 rad = 3.3*10-2 m
(The area of the pixel would then be 1.09 x 10-3 m2.)
ii. [1] If the imaging system were pointed 60 off-nadir, what would be the area of a single
pixel along the optic axis? (Assume that the earth is flat to a good approximation?)
The area of a surface on the earth and perpendicular to the viewing direction for the
imaging system viewing at angle  is:
A = distance2 * solid angle = (altitude * sec )2 * solid angle
= (8 x 105 m x sec )2 * 1.70x10-15 sr = 4.35x10-3 m2.
The area subtended by a section of the flat earth is then:
Area = A * sec 
= 4.35x10-3 m2 * 2 = 8.70x10-3 m2.
CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
d. As a practical matter, the Hubble achieves this resolution by staring at a single spot for hours,
1.5 hours in the case of the Butterfly Nebula image. If a similar stare time were required for an
earth-viewing HubbIe, this would only be feasible from geostationary orbit (36,000 km).
i. [1] At that altitude, what would the resolution of the Hubble be?
Δx = altitude x IFOV (in radians)
= 3.6 x 107 m x 4.12x10-8 rad = 1.48 m
ii. [1] What would be the GFOV of the instrument?
GFOV = altitude x FOV
(still a very small angle)
= 3.6 x 107 m x 6.98 x 10-5 rad = 2.51 km
e. [2] The Hubble image was collected using several narrow-band filters centered at 373 nm, 469
nm, 502 nm, 656 nm, 658 nm, and 673 nm, with all images acquired in a staring mode.
Assuming that the earth-viewing Hubble was not troubled by cloud cover during its 1.5 hour
stare, what issues might need to be considered for the earth-viewing Hubble that would not be
an issue for the space-viewing system?
i. Path radiance due to atmospheric scattering, especially for the UV (373 nm) and blue
(469 nm) channels.
ii. Variations in atmospheric composition that would affect the selected bands, e.g., ozone
absorption in the UV channel (373 nm).
iii. Changes in illumination (sun angle) during the long exposure.
CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
f. [Extra Credit] The International Space Station (ISS) is in low earth orbit at ~345 km. Some
pertinent characteristics are shown below (see: http://spaceflight.nasa.gov/realdata/tracking/ )
Length: 73 m Width: 108.5 m (356 ft)
Perigee: 341 km altitude (184 nmi)
Apogee: 353 km altitude (191 nmi)
Orbit inclination: 51.6419 degrees
Average speed: 27,743.8 km/h
Orbital period: c.91 minutes
That means that there is a chance that the ISS will pass through the field of view of any
virtually every other imaging satellite in orbit, whether in a polar or geostationary orbit. If that
should happen, discuss why you would (or would not) expect the ISS to be detected by:
i. The earth-viewing Hubble in geosynchronous orbit.
Although the ISS is certainly large enough to be resolved easily by the earth-viewing
Hubble, it would pass through the FOV at almost 28,000 km/h. Thus, it will only be
within the Hubble FOV for a few seconds at most. Since the Hubble system need to
stare for a much longer time, the passage of the ISS could do little more than add a
slight streak of noise.
ii. The IKONOS of GeoEye-1 imaging systems (both pushbroom scanners, see:
http://www.geoeye.com/CorpSite/products/imagery-sources/Comparison.aspx
Both the IKONOS and GeoEye are pushbroom scanners that are collecting individual
scan lines at millisecond rates. It is entirely conceivable that ISS would be detected, but
much less likely that it would appear as a recognizable image. The ISS orbit is nearly
perpendicular to that of the polar orbiters and would move a considerable distance
during a scan and even more between scans.
CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
2. Color
Spectra of 5 different minerals are shown below. Select three channels for display in RGB that
will clearly delineate among the minerals by presenting them as distinctly different colors, and
explain the rationale for your selection.
My selections, displayed above, are: Red = 2.05 ; Green = 2.17 ; Blue = 2.37 ..
CEE 6100 / CSS 6600
Exam 1
14-16 Oct 2009
5. [3] Radiometry
Given:
earth-sun distance
solar constant
Radius of the sun
Boltzmann's constant
Des
Esun
rsun
σ
1.50E+11
1.37E+03
6.95E+08
5.67E-08
m
W m-2
m2
W m-2 K-4
Use the Stefan-Boltzman Law to estimate the temperature of the sun.
a. The solar constant specifies the solar irradiance at the earth-sun distance. The same
energy impinges at every point of a sphere with that radius. The area of a sphere with
the radius of the earth-sun distance is:
b. 4  (rsun)2 =
c.
Area of sphere at Earth-Sun distance
Total energy on the surface of the sphere
Area of sphere at sun's surface
Emittance of the sun
T4
T
2.83E+23
3.87E+26
6.07E+18
6.38E+07
1.13E+15
5792.1973
m2
W
m2
W/m2
K4