Originally posted by Soothfast
Well, it's a good question, which I think can be roughly translated thus: how big must a telescope in space in the neighborhood of Earth be in order to capture the same amount of light from Pluto as the LORRI telescopic apparatus on the probe was getting at the time of Sonhouse's post?
At the time of Sonhouse's post, the New Horizons probe was roughl ...[text shortened]... he capacity to read billboards on a planet circling Alpha Centauri, but I haven't done the math.
There are two problems here.
One is light capture.
The other is diffraction of the light entering the telescope setting an upper limit on the angular resolution.
Light capture can be solved by both increasing telescope size, and in increasing exposure times and is
not the main limiting factor.
As for the other problem...
Pluto has a diameter of 2370km and is currently ~31.93 AU from the Earth. [~4,776,660,000 km]
A circle with a radius of 31.93 AU will have a circumference of 30,012,640,061 km
Divide that by the diameter of Pluto and you find the fraction of that circle occupied by Pluto...
1/12,663,561.2
The full circle has 360 degrees, 21,600 arc minutes, and 1,296,000 arc seconds. Pluto thus does indeed
occupy ~ 1/10th of an arc second.
1/10th of an arc second is 1/(60*60*10) of a degree, or 1/36,000 of a degree.
The maximum angular resolution T of any imaging device [in radians] is given by 1.220*(Y/D) where Y is the
wavelength and D is the diameter of the primary lens/mirror. [And is what we want to find].
The angular resolution for a given diameter decreases with wavelength so I will use red light 700nm as my baseline.
I also want this in degrees so I will add in a conversion factor of 180/Pi
So T = [1.220*(Y/D)]*[180/Pi]
Rearranging for D we get D = [1.220*180*Y]/[T*Pi]
So to have Pluto occupy 1 pixel we have this equation we have T=1/36000, and Y=700E-9.
Which gives a Diameter of ~1.762m ...
However we Want Pluto to be bigger than 1 Pixel. For sonhouse's challenge we want Pluto to look like this
http://pluto.jhuapl.edu/soc/Pluto-Encounter/data/pluto/level2/lor/jpeg/029861/lor_0298615084_0x630_sci_1.jpg
Which was how Pluto looked at about the time he set the challenge.
Pluto is about ~80 pixels across in that image.
So we multiply T by 1/80 [T=1/(36000*80)] to find the minimum Diameter needed...
Which gives a Diameter of ~141m
How about if we want an image like this
http://pluto.jhuapl.edu/soc/Pluto-Encounter/data/pluto/level2/lor/jpeg/029912/lor_0299123689_0x632_sci_3.jpg
Where Pluto is by my estimation ~633 pixels across...
We now need a Diameter of ~1,115 meters across.
But lets reverse things and ask, what would our angular resolution be if we had an optical telescope the size of the
Earth?
That makes D = ~12,742,000 m
Which gives us a T of ~3.84E-12
If we divide 1/36,000 by 3.84E-12 we can see how many pixels across Pluto would be...
The answer, 7,233,602 Pixels across.
Or 3052 Pixels per km, Or ~3 pixels per meter, ~9 pixels per square meter. [not accounting for spherical distortions as
you move away from the image centre].