# starship-design: Into detail

```Kelly,

>Actually The largest scope I mentioned was Hundreds of thousands of miles
>across.

Ah, I must have read hundreds OR thousants.

>Thou the calculations are a good Idea.  To a degree you could scale
>up the scopes to any size nessisary.  How big would it need to be?

Well a factor 2 bigger 300,000 miles instead of 3,000 miles will resolve
objects of:

d=1.22 * 9.45E16 * 5.3E-7 / 5.556E8 = 110 metres

Besides Rayleigh's criteria, there is the point of brightness.

To create a visible image on each separate telescope I need to know the
minimal brightness (in Watt/m^2) that those CCD cameras can "see".
I don't know this "minimal brightness", but assume that one needs at least 1
photon per pixel.

Assuming that the planet we look at reflects 400 Watt per square meter (in
the visible range), than at Sol we have only 400/(2 pi r^2) = 400/(2 pi
9.45E16^2) = 7.1E-33 Watt/m^2
(I use 2*pi*r^2 which is the surface of half a sphere over which the light
is reflected)

Luckely we didn't want 1 meter detail, but110 meter detail, so we have 110^2
more Watts here at Sol: (110^2)*7.1E-33=8.6E-29 Watt/m^2

A single photon has an energy of
h*c/lambda=(1E-34)*(3E8)/(5.3E-7)=5.7E-20 Joule

So that means about 1 photon per second per 6.6E8 square meter.

Say we need 1E6 photons for a photograph (theoretical 1000x1000 pixels).

We make a 1 second photograph. (Can't do longer, otherwise the planet has
turned much more than 110 meters)

Therfore we need an aperture surface of 1E6*6.6E8=6.6E14 square meters.

Hmmm, that means a aperture radius of 14.5 kilometers. Most of my estimates
have been quite optimistic, so it could well be that one needs a few orders
bigger.

Timothy

P.S. If anyone has more hard data, I may be able to give a closer estimate.

```