Astronomy 11 - Fall 1998 (S.T. Myers)

1. The diffraction limit

   = ( / D ) rad ( x 206265 "/rad )

   gives for the D=2.4 m HST primary mirror:

   =	1150 Å	= 1.15 x 10-7 m	=>	= 0.0099 "
   =	5000 Å	= 5.00 x 10-7 m	=>	= 0.043 "
   =	13000 Å	= 1.30 x 10-6 m	=>	= 0.112 "

   
   

   To ensure that there are at least 3 pixels across one resolution element lambda/D, we set the pixel size to theta/3 at the shortest wavelength:

   _pix = /3 = (0.0099 ")/3 = 0.0033 " = 1.60 x 10^-8 rad

   and so we want one pixel in the focal plane to correspond to 3.3 milli-arcsec on the sky. The f-ratio f/24 means that the focal length of HST is

   f/D = 24 --> f = 24 D = 57.6 m

   so the plate scale tells us that the physical size of the pixel should be

   y = f _pix = 57.6 m x 1.60 x 10^-8 = 9.2 x 10^-7 m = 0.92 µm

   and thus our CCD pixels should be just under a micron square.

   At the 6565Å H-alpha line, HST has a resolution of

   = ( 6565 x 10^-10 m / 2.4 m ) = 2.74 x 10^-7 rad ( x 206265 = 0.0564 " )

   while at a wavelength of 21 cm you would need a diameter of

   D = / = 0.21 m / (2.74 x 10^-7) = 7.68 x 10⁵ m

   or 768 kilometers! This is why we use interferometers in radio astronomy.

2. Using the 2.4 meter HST primary mirror diameter again, we find

   =	0.8 µm	= 8.0 x 10-7 m	=>	= 0.069 "
   =	2.5 µm	= 2.5 x 10-6 m	=>	= 0.215 "

   In the focal plane, the 1 µm pixels correspond to an angle on the sky of

   _pix = y / f = 10^-6 m / 57.6 m x 206265 = 0.0036 "

   or 3.6 mas per pixel. Thus, a whole row of 256 pixels corresponds to a field of view of only

   FOV = 256 x _pix = 0.92 "

   which is ridiculously small! What's going on here? Well, I had assumed in the problem that the focal length of the NICMOS optics was the same as the WFPC, or 57.6 m. In fact, the NICMOS system has a set of re-imaging optics which makes the effective focal length shorter so that Camera 1 has 0.043" pixels and thus a 11" x 11" field-of-view! Oops :-(

3. The diameter of the Meade is 10" ( x 2.54 cm/in = 25.4 cm ) so

   f/D = 6.3 --> f = 6.3 x 0.254 m = 1.6 m

   (Note: the Meade tube is clearly not 1.6m long, so how do we accomodate this focal length in that short tube?) The diffraction limit of a 0.254 m telescope at 5000Å is

   = 5 x 10^-7 m / 0.254 m x 206265 = 0.41"

   in principle. However, turbulence in our atmosphere (especially in the vicinity of the telescope from the street, building vents, and nearby buildings which give off heat) will limit the seeing to 1" at best, as for most ground-based telescopes.

   The 9 µm pixels of our SBIG camera correspond to a focal plane scale of

   _pix = y / f = 9 x 10^-6 m / 1.6 m x 206265 = 1.16 "

   so the field of view is just

   FOV = ( 765 x 510 ) x 1.16" = 887.4" x 591.6" = 14.8' x 9.9'

   which is fairly large as far as CCD cameras go. (How would we make it larger without buying a new CCD?)

   The distance to Jupiter is

   r = 5.203 AU - 1 AU = 4.203 AU

   which is good enough for our estimate (you can also take into account the eccentricities of the Earth and Jupiter's orbits 0.017 and 0.048 respectively, which gives 3.934 AU). Thus, the angle corresponding to the diameter of Jupiter (142800 km from Table A3-3) at its distance of 4.2 AU (6.3 x 10^8 km) is

   = 142800 km / 6.3 x 10⁸ km = 2.27 x 10^-4 rad = 46.8"

   and thus if the pixel scale is 1.16"/pix, then the diameter of Jupiter in the focal plane is approximately 40 pixels!

   Using the data from Table A3-5 in the textbook, we have the following orbital radii in terms of Jupiters radius (71400 km, or 23.4" from above), and thus the apparent maximum offsets

   Io	a =	5.95	RJ =>	=	139.2"	= 120 pix
   Europa		9.47	RJ =>		221"	= 191 pix
   Ganymede		15.10	RJ =>		353"	= 304 pix
   Callisto		26.60	RJ =>		622"	= 536 pix

   where we used the 1.16"/pix scale.

4. The length of the umbra of the Moon is a maximum when the Earth (and thus the Moon) is at aphelion, when the distance from the Sun is

   d_sun = 1.017 AU = 1.521 x 10⁸ km

   for e=0.017 of Earth's orbit. Strictly, we should subtract off the Moon's distance from Earth to get

   d_sun = 1.521 x 10⁸ km - 384000 km = 1.517 x 10⁸ km

   but this is small in comparison anyway. The closest approach of the Moon to the Earth is at perigee

   d_moon = 384405 km ( 1 - 0.055 ) = 363263 km

   using the data from p.54 and A3-4 of the text. This is from the moon to the center of the Earth, so if we back this off by the Earth's radius we get

   d_obs = 363263 km - 6378 km = 356900 km

   (rounded) to the observer at the sub-lunar point on the surface of the Earth.

   We have a series of nested similar triangles (you should make the drawing) with ratio of sides corresponding to the apex of the umbra angle

   R_sun / ( d_sun + L ) = R_moon / L = R_umb / ( L - d_obs )

   for the three triangles, where R_umb is the radius of the umbra at the Earth's surface (what we want to find).

   We first solve for length L of the umbra, and uses the first two terms

   R_sun L = R_moon ( d_sun + L )

   so

   L = d_sun / ( R_sun/R_moon - 1 ) = 1.521 x 10⁸ km / ( 696500/1738 - 1 ) = = 3.798 x 10⁵ km

   or 379800 km which is (thankfully) longer than the distance from the Moon of 356900 km! The last two terms give us

   R_umb L = R_moon ( L - d_obs )

   so trivially

   R_umb = R_moon ( 1 - d_obs/L ) = 1738 km ( 1 - 356900/379800 ) = 105 km

   so the umbra should have a maximum diameter of 210 km on the surface of the Earth. This isn't very big, is it.

smyers@nrao.edu   Steven T. Myers